Maths Sample Paper 2020-21 CBSE

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MATHEMATICS

MTG Learning Media (P) Ltd. New Delhi | Gurugram

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Price : ` 200 Revised Edition : 2021 Published by

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PREFACE W

ith each passing year CBSE Board incorporates some changes in the pattern of Board question papers to prepare the students for the competitive world ahead. These changes aim to discourage students from

rote learning, develop their analytical skills and reasoning abilities, and to produce better results and improve the academic quality of institutions. This year’s exam pattern includes VSA, Case based MCQs, SA-I, SA-II and LA type questions. To keep you aligned with the latest pattern MTG brings you the completely revised edition of Score More 21 Sample Papers. These sample papers are prepared keeping in mind the various typology of questions like remembering, understanding, applying, analysing, evaluating and creating type. MTG Score More 21 Sample Papers are prepared, keeping all these aspects in mind. Solving more and more sample papers will increase your problem-solving speed and accuracy. By solving these sample papers, you can check your preparation level, and your strong and weak areas too. You can score more in the actual paper by getting to know your weak areas and working even harder on them. As you start attempting SQPs one by one your percentage scoring will improve. Your aim should be to reach the top position. After attempting all SQPs in exam-like environment the average score of 21 SQPs will give you the score which you are most likely to score in your actual exam. That is what MTG Guarantee. Practising these SQPs will definitely equip you to face your CBSE Board Class 12 exams with more confidence. Solutions of all SQPs are given as per the CBSE marking scheme. In MCQs section, students need to write only answer keys. Explanations for these MCQs are given for better understanding. The Self Evaluation Sheet provided after each Sample Question Paper (SQP) will help you to assess your performance. The performance analysis table will help you to check where you actually stand. Practice done with a proper planning promotes a person for a better performance. It is very necessary to practice in the right direction under a well guided guidance to reach to the goal. Practice means repeating an activity in the right direction which can sharpen the talent. Visit our website www.mtg.in to download the additional content associated with this book. Wishing our readers all the very best! MTG Editorial Board

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MTG’s VALUABLE TIPS BOARD EXAMS Exams are the part and parcel of our learning process. But most students suffer from exam phobia. They may be excellent at their work but when confronted by real exams, often they under-perform showing exam related stress. So, it’s time to strategize your learning and come out of the fear cropping up in your mind by combining self-confidence with disciplined study and by simply following these tips :

GET SET STARTED Now with the ending session you are close to completing your syllabus; it is time for effective revision of the relevant content for better results. Preparing a well-planned timetable is the need of the hour. Timetable should be judiciously planned to give enough time to different subjects. It should have scope of learning, practising as well as recapitulation.

DECK OUT WITH THE RIGHT TITLE Currently the market is flooded with many titles in different subjects. It calls for your wise decision to choose the right title. MTG gaining the trust of over 10 million readers across the country, now presents Score More 21 SAMPLE PAPERS covering all objective and subjective type of questions. The Book in itself is a complete package and provides an opportunity to all the readers to assess their progress as it includes questions based on the latest changed pattern of CBSE exams.

STRIVE FOR EXCELLENCE Start solving the sample papers provided in the book in an environment same as the real exam. Once you are done, check your answers to evaluate your learning. This will help you become aware of both your strong and weak areas. Work harder on your weak areas and repeat the process. You will master all concepts soon.

IT’S SHOW TIME Remember, there is no shortcut to success. However, these tips would surely help you a lot, if followed truly for at least a month prior to the exams. Organising your studies like this will help you deliver your best during exams and there is no doubt that you’ll reap richer benefits. So, gird up your lions and give your best shot!!!

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BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

1.

Relations and Functions

2.

Inverse Trigonometric Functions

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)#

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

–

2(2)

#

3.

Matrices

2(2)

–

–

4.

Determinants

1(1)

1(2)*

–

1(5)*

3(8)

#

–

1(2)

2(6)

–

3(8)

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)

1(2)

1(3)*

–

3(6)

#

1(2)*

–

–

4(5)

#

2(2)

1(2)

–

1(5)*

4(9)

–

–

–

1(5)*

1(5)

2(2) + 1(4)

1(2)

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

5.

Continuity and Differentiability

6.

10.

Vector Algebra

3(3)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

Probability Total

*It is a choice based question. # Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-1

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I  1. Find the magnitude of the vector a = 3ɵi − 2 ɵj + 6kɵ . OR 



Find the number of vectors of unit length perpendicular to both the vectors a = 2ɵi + ɵj + 2kɵ and b = ɵj + kɵ . 2. Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective fuses in the first two tests? 3. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} not to be transitive. OR Let R be the equivalence relation in the set A = {0, 1, 2, 3, 4, 5} given by R = {(a, b) : 2 divides (a – b)}. Write the equivalence class [0].

4. Find the direction cosines of the line

4 − x y 1− z = = . 2 6 3 Class 12

2

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0 a −3   5. If the matrix A = 2 0 −1 is skew symmetric, then find the value of ‘a’ and ‘b’. b 1 0  OR

Write the element a23 of a 3 × 3 matrix A = [aij] whose elements aij are given by aij =

|i − j | . 2

x sin θ cos θ 6. If − sin θ −x 1 = 8 , then find the value of x. cos θ 1 x 7. Find the value of 4

Evaluate :

∫

sin2 x − cos2 x sin2 x cos2 x

dx .

OR

x

∫ x 2 + 1 dx 2

8. Let A = {1, 2, 3}, B = {4, 5, 6, 7} and let f = {(1, 4), (2, 5), (3, 6)} be a function from A to B, state whether f is one-one or not. 9. If a line makes an angle q1, q2 and q3 with the x, y and z-axes respectively, then find the value of cos 2q1 + cos 2q2 + cos 2q3. OR x −3 z −5 = y+2 = , then find the numbers to which direction ratios of a 1 4 −2 line parallel to AB, are proportional. If the equation of a line AB is

10. Find the order and degree of y = px + x2

dy a 2 p 2 + b 2 , where p = dx . y2

= 1. b2    12. Find the sum of the vectors a = ɵi − 2 ɵj + kɵ , b = −2ɵi + 4 ɵj + 5kɵ and c = ɵi − 6 ɵj − 7kɵ .

11. Find the area enclosed by the ellipse

a2

+

13. Check whether the relation R in the set  of real numbers defined as R = {(a, b) : a < b} is (i) symmetric, (ii) transitive. 14. If A is a square matrix such that A2 = A, then show that (I – A)3 + A = I. 5 2 and P(A | B) = . 13 5    2  2 16. If (a × b ) + (a ⋅ b ) = 400 and a = 4, then find b .

15. Find the value of P(A ∪ B), if 2P(A) = P(B) =

Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Neelam and Ved appeared for first round of an interview for two vacancies. The probability of Neelam’s selection is 1/6 and that of Ved’s selection is 1/4. Based on the above information, answer the following questions : 3

Mathematics

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(i) The probability that both of them are selected, is 1 1 (a) (b) 12 24 (ii) The probability that none of them is selected, is 2 3 (a) (b) 7 8

(c)

1 6

(d)

1 2

(c)

5 8

(d)

1 3

(d)

1 3

(d)

2 7

(iii) The probability that only one of them is selected, is 2 5 2 (a) (b) (c) 3 8 7 (iv) The probability that atleast one of them is selected, is 3 1 3 (a) (b) (c) 8 8 7

(v) Suppose Neelam is selected by the manager and told her about two posts P and Q for which selection is 1 1 independent. If the probability of selection for post P is and for post Q is , then the probability that 6 7 Neelam is selected for at least one post, is 1 2 1 (b) (c) 3 (d) (a) 7 7 2 8 18. Sonam wants to prepare a sweet box for Diwali at home. For making lower part of box, she takes a square piece of cardboard of side 18 cm. Based on the above information, answer the following questions : (i) If x cm be the length of each side of the square cardboard which is to be cut off from corner of the square piece of side 18 cm, then x must lie in (a) [0, 18] (b) (0,9) (c) (0, 3) (d) None of these (ii) Volume of the open box formed by folding up the cutting corner can be expressed as x (a) V = x(18 – 2x)(18 – 2x) (b) V = (18 + x)(18 – x) 2 x (c) V = (18 – 2x)(18 + 2x) (d) V = x(18 – 2x)(18 – x) 3 dV = 0, are (iii) The values of x for which dx (a) 3, 2

(b) 0, 3

(c) 0, 9

(d) 3, 9

(iv) Sonam is interested in maximising the volume of the box. So, what should be the side of the square to be cut off so that the volume of the box is maximum? (a) 13 cm (b) 8 cm (c) 3 cm (d) 2 cm (v) The maximum value of the volume is (b) 232 cm3 (a) 144 cm3

(c) 256 cm3

(d) 432 cm3

PART - B Section - III 2

cos x

19. Differentiate sin x w.r.t. e 20. Find

∫

dx 5 − 4 x − 2x2

.

. Class 12

4

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OR sin 2 x Evaluate : ∫ dx π  π  sin  x −  sin  x +   3  3

21. Prove that the area enclosed between the x-axis and the curve y = x2 – 1 is 1 2

4 sq. units 3

1 2

22. Find the value of cos 1    2 sin 1   .       23. If a = 2ɵi + 3ɵj − kɵ and b = ɵi + 2ɵj + 3kɵ , find a × b and | a × b | . OR 2 Prove by vector method that the area of DABC is a sin B sin C . 2 sin A

24. If A and B are events such that P(A) =

1 1 1 , P(B) = and P(A ∩ B) = , then find P(not A and not B). 3 4 12

25. Show that the function f(x) = 4x3 – 18x2 + 27x – 7 is always increasing on R. 26. If the solution of the differential equation

dy ax + 3 represents a circle, then find the value of ‘a’. = dx 2 y + f

27. Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z. 28. If A is a non-singular 3 × 3 matrix such that |5 ⋅ adjA| = 5, then find |A|. OR 1 −2 5 If there are two values of a which makes determinant, ∆ = 2 a −1 = 86, then find the sum of these values. 0 4 2a

Section - IV 29. Find the value of k, for which  1 + kx − 1 − kx , if − 1 ≤ x < 0  x f (x ) =  2x + 1  , if 0 ≤ x < 1  x −1

is continuous at x = 0. π /2

30. Evaluate :

∫ sin 2x tan

−1

(sin x)dx

0

 a + b sin x  with respect to x. 31. Differentiate the function log   a − b sin x  OR Find

dy , when y = a + a + a + x 2 , where a is a constant. dx

32. Find area of the region in the first quadrant enclosed by the x-axis, the line y = x and the circle x2 + y2 = 32. 33. If the equation of tangent at (2, 3) on the curve y2 = ax3 + b is y = 4x –5, then find the values of a and b. 5

Mathematics

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34. Solve the differential equation : dy (1 + x 2 ) + 2 xy − 4 x 2 = 0, subject to the initial condition y(0) = 0. dx OR Find the particular solution of the differential equation

x(2 log x + 1) dy π , given that y = , when x = 1. = dx sin y + y cos y 2

35. Show that the function f : (–∞, 0) → (–1, 0) defined by f(x) =

x , x ∈ (–∞, 0) is one-one and onto. 1+ x

Section - V 8 4 2  36. Find the inverse of the matrix A = 2 9 4  .   1 2 8  OR  1 −2 2 3  –1 –1 –1 ,B=  If A =    , verify that (AB) = B A . 1 3 1 4 − −     37. Find the equation of the plane passing through the points (2, 2, –1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6. OR Consider the lines x +1 y + 2 z +1 x −2 y +2 z −3 L1 : = = , L2 : = = 3 1 2 1 2 3 Find the distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L1 and L2. 38. Solve the following LPP graphically. Maximize, Z = 150x + 250y Subject to the constraints x + y ≤ 35 1000x + 2000y ≤ 50000 x, y ≥ 0 OR Find the number of point(s) at which the objective function Z = 4x + 3y can be minimum subjected to the constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6; x, y ≥ 0.

Class 12

6

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SOLUTIONS  1. Here, a = 3ɵi − 2 ɵj + 6kɵ  \ Its magnitude = a

0 a −3  0 2 b      ⇒ 2 0 −1 +  a 0 1 = O b 1 0   −3 −1 0    

= 32 + (−2)2 + 62 = 9 + 4 + 36 = 49 = 7.

OR   Given, a = 2ɵi + ɵj + 2kɵ and b = ɵj + kɵ    a × b   Unit vectors perpendicular to a and b are ±     .  | a × b | So, there are two unit vectors perpendicular to the given vectors.

 0 2 + a b − 3  0 0 0      ⇒ a + 2 0 0  = 0 0 0  b − 3 0 0  0 0 0  \ a + 2 = 0 and b –3 = 0 ⇒ a = –2 and b = 3

2. Let D1, D2 be the events that we find a defective fuse in the first and second test respectively. \ Required probability = P(D1 ∩ D2)

Here, aij =

2 1 1 = P (D1 )P (D2 | D1 ) = ⋅ = 7 6 21 3. For transitivity of a relation, if (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R We have, R = {(1, 2), (2, 1)} ⇒ (1, 2) ∈ R and (2, 1) ∈ R but (1, 1) ∉ R \ R is not transitive. OR Here, R = {(a, b) ∈ A × A : 2 divides (a – b)}, which is an equivalence relation, where A = {0, 1, 2, 3, 4, 5}. Clearly, [0] = {a ∈ A : aR0} = {a ∈ A : 2 divides (a – 0)} = {a ∈ A : 2 divides a} = {0, 2, 4} \ Equivalence class [0] is {0, 2, 4}. 4. The given equation of line is 4 − x y 1− z = = 2 6 3 x−4 y z −1 x − 4 y z −1 ⇒ = = ⇒ = = −6 2 3 −2 6 −3

|i − j | 2

\ For i = 2, j = 3 we have, a23 =

| 2 − 3 | | −1 | 1 = = 2 2 2

6. Expanding the given determinant, we get x(–x2 – 1) –sin q(–xsin q – cos q) + cos q (–sin q + xcos q) = 8 ⇒ –x3 – x + x = 8 ⇒ x3 + 8 = 0 ⇒ (x + 2) (x2 – 2x + 4) = 0 ⇒ x+2=0 [ x2 – 2x + 4 > 0 ∀x] ⇒ x = –2 7. We have,

∫

sin2 x − cos 2 x

dx = ∫ (sec2 x − cosec 2 x )dx sin 2 x cos 2 x = tanx + cotx + C OR 4

...(i)

22 + (−6)2 + 32 = 7 2 −6 3 \ D.c’s. of (i) are , , . 7 7 7 5. Since, matrix A is skew symmetric matrix. \ A′ = –A ...(i)  0 2 b 0 a −3   ∴ A′ =  a 0 1  Now, as A = 2 0 −1    −3 −1 0 b 1 0      From (i), A + A′ = O Now, as

OR

x Let I = ∫ dx 2 2 x +1

1 ⇒ x dx = dt 2 Also x = 2 ⇒ t = 5 and x = 4 ⇒ t = 17 Put x2 + 1 = t

\ I=

1 2

17

∫

5

⇒

2xdx = dt

dt 1 = [log t ]17 5 t 2

1 1  17  = [log 17 − log 5] = log    5 2 2 8. We have, A = {1, 2, 3}, B = {4, 5, 6, 7} and f = {(1, 4), (2, 5), (3, 6)} \ f(1) = 4, f(2) = 5 and f(3) = 6. Since, distinct elements of A have distinct images in B, therefore, f is one-one function.

Mathematics

7

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9. Consider, cos 2q1 + cos 2q2 + cos 2q3 = 2(cos2q1+ cos2q2 + cos2q3) – 3 ( cos2x = 2cos2 x – 1) = 2(1) – 3 = – 1 OR Since direction ratios of line AB are 1, –2, 4, therefore the direction ratios of line parallel to AB will be proportional to 1, –2 and 4. 10. Given, y − px = a 2 p 2 + b 2 ⇒ (y – px)2 = a2p2 + b2 ⇒ (x2 – a2)p2 – 2xyp + (y2 – b2) = 0 2

 dy   dy  ⇒ (x2 – a2)   – 2xy  dx  + (y2 – b2) = 0    dx  Hence, order is 1 and degree is 2. 11. We know that, the area enclosed by ellipse 2 x 2  y  1 is pab sq. units. a 2 b2

12. The given vectors are    a = ɵi − 2 ɵj + kɵ , b = −2ɵi + 4 ɵj + 5kɵ , c = ɵi − 6 ɵj − 7kɵ  \ Required sum = a + b + c = (i − 2 j + k ) + ( −2i + 4 j + 5k ) + (i − 6 j − 7k ) = −4 j − k .

13. We have, R = {(a, b) : a < b}, where a, b ∈  (i) Symmetric : Let (x, y) ∈ R, i.e., x R y ⇒ x < y Then, y  x, so (x, y) ∈ R ⇒ (y, x) ∉ R Thus, R is not symmetric. (ii) Transitive : Let (x, y) ∈ R and (y, z) ∈ R ⇒ x < y and y < z ⇒ x < z ⇒ (x, z) ∈ R. Thus, R is transitive. 14. We have, A2 = A Now, L.H.S. = (I – A)3 + A = (I – A)(I – A)(I – A) + A = (I ⋅ I – I ⋅ A – A ⋅ I + A ⋅ A)(I – A) + A = (I – A – A + A)(I – A) + A [Q I ⋅ A = A ⋅ I = A and A2 = A] = (I – A)(I – A) + A = (I ⋅ I – I ⋅ A – A ⋅ I + A ⋅ A) + A = (I – A – A + A) + A = (I – A) + A = I = R.H.S. Hence, (I – A)3 + A =I 15. Given, P(A | B) =

P(A ∩ B) 2 2 ⇒ = P(B) 5 5

2 2 5 2 ⇒ P(A ∩ B) = P(B) = × = 5 5 13 13

Hence, P (A ∪ B) = P (A) + P (B) – P (A ∩ B) 1 5 5 2 5 1 5  = × + −  2P(A) = 13 , ∴ P(A) = 2 × 13  2 13 13 13 5 + 10 − 4 11 = = . 26 26

     16. We have, (a × b )2 + (a ⋅ b )2 = 400 and a = 4   2 2   We know that, (a × b )2 + (a ⋅ b )2 = a b 2 2 ⇒ 400 = (4)2 b ⇒ 16 b = 400 2  ⇒ b = 25 ⇒ b = 5 17. Let A be the event that Neelam is selected and B be the event that Ved is selected. Then, we have, 1 P(A) = 6 1 5 ⇒ P(A) = 1 − = = P (Neelam is not selected) 6 6 1 P(B) = 4 1 3 ⇒ P(B) = 1 − = = P (Ved is not selected) 4 4 (i) (b) : P (both are selected) = P(A ∩ B) = P(A)⋅P(B) 1 1 1 = × = 6 4 24 (ii) (c) : P (both are rejected) = P(A ∩ B) = P(A)⋅P(B) 5 3 5 = × = 6 4 8 (iii) (d) : P (only one of them is selected) = P(A ∩ B) + P(A ∩ B) = P(A)⋅P(B) + P(A)⋅P(B) 1 3 5 1 3 5 8 1 + = = = × + × = 6 4 6 4 24 24 24 3 (iv) (a) : P (at least one of them is selected) = 1 – P (Both are rejected) 5 3 = 1− = 8 8 (v) (b) : Let E1 be the event that Neelam is selected for post P and E2 be the event that Neelam is selected for post Q. \ P (Neelam is selected for atleast one post) = P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2) =

1 1 1 1 12 2 + − × = = 6 7 6 7 42 7

18. (i) (b) : Since, side of square is of length 18 cm, therefore x ∈ (0, 9). (ii) (a) : Clearly, height of open box = x cm Length of open box = 18 – 2x and width of open box = 18 – 2x \ Volume (V) of the open box = x × (18 – 2x) × (18 – 2x) (iii) (d) : We have, V = x(18 – 2x)2 dV = x ⋅ 2(18 – 2x)(–2) + (18 – 2x)2 \ dx = (18 – 2x)(–4x + 18 – 2x) = (18 – 2x)(18 – 6x) Class 12

8

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dV = 0 ⇒ 18 – 2x = 0 or 18 – 6x = 0 dx ⇒ x = 9 or 3 (iv) (c) : We have, V = x(18 – 2x)2 dV and = (18 – 2x)(18 – 6x) dx Now,

⇒

d 2V

= (18 – 2x)(–6) + (18 – 6x)(–2)

dx 2

= (–2)[54 – 6x + 18 – 6x] = (–2)[72 – 12x] = 24x – 144 For x = 3,

d 2V dx 2

and for x = 9,

0 dx 2 So, volume will be maximum when x = 3. (v) (d) : We have, V = x(18 – 2x)2, which will be maximum when x = 3. \ Maximum volume = 3(18 – 6)2 = 3 × 12 × 12 = 432 cm3 19. Let u(x) = sin2 x and v(x) = ecos x. du Now, = 2 sin x cos x dx dv cos x cos x =e (− sin x ) = − (sin x )e and dx du 2 cos x du dx 2 sin x cos x Thus, = = = − cos x dv dv − sin x e cos x e dx dx 1 dx 20. Let I = ∫ = ∫ 2 5 5 − 4 x − 2x 2 − 2x − x 2 2 dx 1 dx 1 = = ∫ ∫ 2 2 7 2  7 − 1 − 2x − x 2 2 2   − (x + 1) 2 =

 2  1 1  x + 1 sin −1 sin −1  (x + 1) + C +C =  7  2 2  7     2  OR

Let I =

∫

sin 2 x dx π  π  sin  x −  sin  x +   3  3

 π  π  sin  x −  +  x +    3  3 =∫  dx π  π  sin  x −  sin  x +   3  3

  π  π π  π   sin  x −  cos  x +  + cos  x −  sin  x +    3  3 3 3  dx =∫ π  π  sin  x −  sin  x +   3  3   π π   = ∫ cot  x +  + cot  x −   dx    3 3  π π   = log sin  x +  + log sin  x −  + C    3 3 21. The equation y = x2 – 1 represents a upward parabola with vertex at (0, –1). It cuts x-axis where y = 0 i.e., x2 – 1 = 0 ⇒ x = ± 1 \ Required area =

1

∫ (x

2

− 1) dx

Y

−1

1 X′

1

 x3  =   − [x]1−1  3  −1 =

–1

O

1

X

–1 Y′

1 2 4 (1 + 1) − (1 + 1) = − 2 = sq. units 3 3 3

22. We know that the range of principal value branch  π π of cos–1 and sin–1 are [0, p] and  − ,  respectively.  2 2 1 −1  1  Let cos   = x ⇒ = cos x  2 2 π 1  π Then, = cos   , where ∈[0, π]   2 3 3 1 −1  1  Let sin   = y ⇒ = sin y  2 2 1 π  π π  π Then, = sin   , where ∈  − ,    2 6 6  2 2 1 π π 2π 1     ∴ cos −1   + 2 sin −1   = + 2 ⋅ = .  2 3  2 6 3   23. Given, a = 2ɵi + 3ɵj − kɵ and b = ɵi + 2 ɵj + 3kɵ ɵi ɵj kɵ   Now, a × b = 2 3 −1 1 2 3 = (9 + 2) i − (6 + 1)j + (4 − 3)k = 11i − 7 j + k .   ∴ | a × b | = (11)2 + (−7)2 + (1)2 = 171 OR Area of the triangle ABC 1   1 = | BC × BA | = | ac sin B n | 2 2 9

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OR

A

1 1 c ac sin B = a sin B sin C 2 2 sin C 1 a = a sin B sin C [By sine rule] 2 sin A

=

c B

a 2 sin B sin C = 2 sin A

b a

C

1 1 1 × = = P ( A ∩ B) 3 4 12 ⇒ Events A and B are independent. ⇒ Events A and B are also independent. 24. Here, P(A) . P(B) =

Now, P(A ∩ B) = P(A) P(B) ( A and B are independent events) = (1 – P(A))(1 – P(B))  1  1  2 3 1 = 1 −  1 −  = × =  3  4  3 4 2 3

2

25. We have, f(x) = 4x – 18x + 27x – 7 ⇒ f ′(x) = 12x2 – 36x + 27 2

3 9   = 12  x 2 − 3x +  = 12  x −  ≥ 0 ∀ x ∈R  2 4 Hence, f(x) is always increasing on R.

dy ax + 3 = dx 2 y + f ⇒ (ax + 3)dx = (2y + f )dy

26. We have, 2

x + 3x = y 2 + fy + C (Integrating both sides) 2 a 2 ⇒ − x + y 2 − 3x + fy + C = 0 2 −a This will represent a circle, if = 1 ⇒ a = −2 2 [ coefficient of x2 should be equal to coefficient of y2] ⇒ a

27. Vector equation of the line passing through (1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z x −5 y −2 z x −5 y −2 z i.e., = = or = = 1 / 5 −1 / 7 1 / 35 7 −5 1  ɵ ɵ ɵ is r = (i + 2 j − k ) + λ(7ɵi − 5ɵj + kɵ ) 28. Given that |5 adjA| = 5 and A is non-singular matrix, i.e., |A| ≠ 0. Clearly, |5 adjA| = 5 1 ⇒ 53|adj A| = 5 ⇒ | adjA | = 2 5 1 3−1 ( |adj A| = |A|n–1, if |A| ≠ 0) ⇒ | A| = 2 5 1 ⇒ | A|= ± 5

1 −2 5 Given, ∆ = 2 a −1 = 86 0 4 2a Expanding along C1, we get 1(2a2 + 4) – 2(–4a – 20) = 86 ⇒ 2a2 + 4 + 8a + 40 = 86 ⇒ 2a2 + 8a – 42 = 0 ⇒ a2 + 4a – 21 = 0 ⇒ (a + 7)(a – 3) = 0 ⇒ a = 3, –7 Now, sum of values of a is 3 + (–7) = 3 – 7 = –4 29. Since, f(x) is continuous at x = 0 ∴ lim f (x ) = f (0) = lim f (x ) x → 0+

...(i)

x →0−

Here, f (0) =

2 × 0 +1 = −1 0 −1

...(ii) 2h + 1 = −1 h →0 h − 1

...(iii)

lim f (x ) = lim f (0 + h) = lim x →0

h →0

+

1 − kh − 1 + kh −h h →0

and lim f (x ) = lim f (0 − h) = lim x →0

h →0

−

1 − kh − 1 + kh 1 − kh + 1 + kh × h − h →0 1 − kh + 1 + kh (1 − kh) − (1 + kh) = lim h→0 −h[ 1 − kh + 1 + kh ] 2k 2k = =k = lim 2 h→0 1 − kh + 1 + kh = lim

...(iv)

Now, from (i), (ii), (iii) and (iv), we get k = –1 π /2

30. Let I = π /2

=

∫

∫

sin 2 x tan −1(sin x )dx

0

2 sin x cos x tan −1(sin x )dx

0

Let sinx = t ⇒ cosx dx = dt Also, x = 0 ⇒ t = 0 and x = 1

π ⇒t =1 2

1 −1

∴ I = ∫ 2t tan t dt = 2∫ t tan −1 t dt 0

0

Integrating by parts, we have 1

1

 t 2 t2 −1 − dt 2 I = 2  tan t   0 ∫0 2(1 + t 2) 2 1

t2  1 dt = 2  tan −11 − 0 − ∫  0 (1 + t 2) 2 1 1  π 1  1 π  t 2 +1 −1 1− dt = 2 ×  − ∫ = − dt ∫  2  2 4   1 + ( t )  2 4  0 (1 + t ) 0

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=

π − [t]10 + [tan −1 t]10 4

=

π π π −1+ = −1 4 4 2

Y

(4,4)

O

 a + b sin x  31. Let y = log   a − b sin x 

4

2

d  (a − b sin x) dx 

1 1 dy  (0 − b cos x) (0 + b cos x) −  = a b sin x − dx a + b sin x  

⇒

dy b cos x b cos x = + dx a + b sin x a − b sin x

dy  a − b sin x + a + b sin x  = b cos x   dx  (a + b sin x)(a − b sin x)  2ab cos x a − b 2 sin 2 x OR

We have, y = a + a + a + x 2 , where a is a constant. 1

2  ⇒ y = a + a + a + x 2 

 dy 1  ⇒ = a + a + a + x2  dx 2 

−

1 2

 d  a + a + a + x2  dx 

1 2

 1 a + a + x2 2 d × a + a + x2 dx

−

(

(

1 2

)

)

−

1 2

⇒

(

−

)(

π  1   = 8 + 16 ×  − 2 32 − 4 2 + 16 sin −1 2   2  = 8 + 8p – 8 – 4p = 4p sq. units



(

⇒

)

)

= 2a (2, 3)

So, equation of tangent at the point (2, 3) is y – 3 = 2a (x – 2) ⇒ y = 2ax – 4a + 3 ...(i) But we are given that equation of tangent at (2, 3) is y = 4x – 5 ...(ii) \ On comparing (i) and (ii), we get 2a = 4 ⇒ a = 2 Q Point (2, 3) lies on the curve y2 = ax3 + b, \ (3)2 = (2)3 a + b ⇒ 9 = 8a + b ⇒ 9 = 8 × 2 + b ⇒ b = –7 34. We have, (1 + x 2 )

1  −  1 a + a + x2 2   2  1  1 − ×  (a + x 2 ) 2 ⋅ 2x  2 

 dy 1  ⇒ = a + a + a + x2  dx 2 

4

4

 3ax 2   dy  ⇒   =  dx  (2, 3)  2 y 

2

 dy 1  ⇒ = a + a + a + x2  dx 2 

32 − x 2 dx

33. We have, y2 = ax3 + b Differentiating w.r.t. x, we get 2 dy 2y = 3ax 2 ⇒ dy = 3ax dx dx 2y

dy 1 1   = b cos x  +  dx a b x a b sin x + sin −  

=

∫

4 2 x x  32 x =   +  32 − x 2 + sin −1 2  2 0  2 4 2  4

⇒

⇒

X

4 2

0

dy 1 d 1  = × (a + b sin x) −  dx a + b sin x dx a b sin x −  ×

⇒

(4√2,0)

\ Required area = ∫ x dx +

Then, y = log(a + b sinx) – log (a – b sinx) ⇒

y=x

dy + 2 xy = 4 x 2 dx

dy 2x 4x 2 + y = dx 1 + x 2 1+ x2

This is a linear differential equation of the form dy 2x 4x 2 + Py = Q where P = Q = and dx 1+ x2 1+ x2 2x

1 −  2

dy 1  = x a + a + a + x 2 ⋅ a + a + x 2 ⋅ (a + x 2) dx 4 

32. We have, x2 + y2 = 32 ...(i), a circle with centre (0, 0) and radius 4 2 , and y = x ...(ii), a straight line. Solving (i) and (ii), we get point of intersection (4, 4) in the first quadrant.

∫ 2 dx Pdx 2 = e 1+ x ∴ I.F. = e ∫ = elog(1 + x ) = 1 + x2 So, the required solution is given by y(1 + x 2 ) = ∫

4x 2 1+ x2

(1 + x 2 ) dx + C

⇒ y(1 + x 2 ) = 4∫ x 2dx + C 11

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4x 3 +C 3 Given that y(0) = 0 \ 0(1 + 0) = 0 + C ⇒ C = 0 ⇒ y(1 + x 2 ) =

Thus, y =

4x

3

3(1 + x 2 )

Thus, A is a non-singular matrix and therefore it is invertible. Let Cij be cofactor of aij in A. Then, C11 =

is the required solution. OR

We have, dy = x(2 log x + 1) dx sin y + y cos y ⇒ (siny + y cosy)dy = x(2 logx + 1)dx On integrating both sides, we get – cosy + y siny – (– cosy)  1 x2  x2 x2 = 2  log x × − ∫ × dx  + +C 2 x 2   2 x2 x2 ⇒ y siny = x 2 log x − + +C 2 2 ⇒ y siny = x2 logx + C π Q When x = 1, y = 2 π π π ∴ sin = 1⋅ log(1) + C ⇒ = C 2 2 2 \ y siny = x2 logx + p/2 is the required solution. x 35. Given, f(x) = , x ∈(−∞, 0) 1+ x x = (Q x ∈ (–∞, 0), |x| = –x) 1− x For one-one : Let f(x1) = f(x2), x1x2 ∈ (–∞ , 0) x1 x ⇒ = 2 1 − x1 1 − x 2 ⇒ x1(1 – x2) = x2(1 – x1) ⇒ x1 – x1x2 = x2 – x1x2 ⇒ x1 = x2 Hence, if f(x1) = f(x2), then x1 = x2 \ f is one-one For onto : Let f(x) = y x ⇒ y= ⇒ y(1 – x) = x ⇒ y – xy = x 1− x y ⇒ x + xy = y ⇒ x (1 + y) = y ⇒ x = 1+ y Here, y ∈ (–1, 0) So, x is defined for all values of y. Also x ∈ (–∞, 0) for all y ∈ (–1, 0). \ f is onto. 36. We have, 8 4 2 | A | = 2 9 4 = 8(72 − 8) − 4(16 − 4) + 2(4 − 9) 1 2 8 = 512 – 48 – 10 = 454 ≠ 0

9 4 2 4 2 9 = 64, C12 = − = −12, C13 = = −5 2 8 1 8 1 2

C21 = −

4 2 8 2 = −28, C22 = = 62, 2 8 1 8

C23 = −

8 4 = −12, 1 2

C31 =

4 2 8 2 8 4 = −2, C32 = − = −28, C33 = = 64 9 4 2 4 2 9

 64 −12 −5  ′  64 −28 −2      ∴ adj A =  −28 62 −12 =  −12 62 −28  −2 −28 64   −5 −12 64  Hence, A

−1

 64 −28 −2  1 1  −12 62 −28 adj A = =  | A| 454  −5 −12 64  OR

 1 −2 2 3  and B =  Here, A =    1 −4   −1 3  ⇒ |A| = – 11 and |B| = 1 3 2  1 1  −4 −3 adj A = −  and B −1 =    |A| 11  −1 2  1 1  ⇒ R.H.S. = B–1 A–1 \ A −1 =

3 2  1   −4 −3  1   −14 −5 =  ⋅  −   −1 2  =  −   −5 −1 11  11  1 1   

...(i)

2 3   1 −2  −1 5  Now, A⋅B =  ⋅ =   1 −4   −1 3   5 −14  ⇒ |AB| = 14 – 25 = – 11  1   −14 −5 \ L.H.S. = ( AB) −1 =  −    11 −5 −1   –1 –1 –1 From (i) and (ii), (AB) = B A .

...(ii)

37. The equation of a plane passing through (2, 2, –1) is a(x – 2) + b(y – 2) + c(z + 1) = 0 ...(i) This plane also passes through (3, 4, 2). \ a(3 – 2) + b(4 – 2) + c(2 + 1) = 0 ⇒ a + 2b + 3c = 0 ...(ii) Now, plane (i) is parallel to the line whose direction ratios are 7, 0, 6 Therefore, 7a + 0(b) + 6c = 0 ...(iii) Class 12

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Solving (ii) and (iii) by cross-multiplication method, we get a b c = = (2) (6) − (0) (3) (7) (3) − (6) (1) (0)(1) − (2) (7) a b c ⇒ = = = λ (say) 12 15 −14 ⇒ a = 12l, b = 15l, c = –14l Substituting the values of a, b, c in (i), we get 12l(x – 2) + 15l(y – 2) – 14l(z + 1) = 0 ⇒ 12x – 24 + 15y – 30 – 14z – 14 = 0 [Q l ≠ 0] ⇒ 12x + 15y – 14z = 68 This is the required equation of plane.

The feasible region is the shaded region. We observe that the region is bounded. The corner points of the feasible region OBED are O(0, 0), B(35, 0), E(20, 15) and D(0, 25). The value of the objective function at corner points of the feasible region are :

OR Any plane through (–1, –2, –1) is a(x + 1) + b(y + 2) + c(z + 1) = 0 ...(i) D.R.’s of any normal to (i) are < a, b, c >. As this normal is perpendicular to both L1 and L2, therefore, 3a + 1b + 2c = 0 ...(ii) 1a + 2b + 3c = 0 ...(iii) Eliminating a, b, c between (i), (ii) and (iii), we obtain x +1 y + 2 z +1 3 1 2 =0 1 2 3 ⇒ (x + 1)(3 – 4) – (y + 2)(9 – 2) + (z + 1)(6 – 1) = 0 ⇒ – (x + 1) – 7 (y + 2) + 5 (z + 1) = 0 or x + 7y – 5z + 10 = 0 ...(iv) This is the required equation of plane. \ Distance of (1, 1, 1) from the plane (iv) =

| 1 + 7 − 5 + 10 |

=

13 units. 75

12 + 7 2 + (−5)2 38. The given problem is Maximize , Z = 150x + 250y Subject to the constraints x + y ≤ 35, x + 2y ≤ 50 and x, y ≥ 0 To solve graphically, we convert the inequations into equations to obtain the following lines : x + y = 35, x + 2y = 50, x = 0, y = 0 Let us draw the graph of these lines as shown below. Y 50 40

A (0, 35) 30 D(0, 25) 20

O(0, 0)

0

B(35, 0)

5250

E(20, 15)

6750 (Maximum)

D(0, 25)

6250

OR Let l1 : 3x + 4y = 24, l2 : 8x + 6y = 48, l3 : x = 5, l4 : y = 6, l5 : x = 0, l6 : y = 0 (l5) Y

X (l6)

For B : Solving l2 and l3, we get B(5, 4/3)  24 24  For C : Solving l1 and l2, we get C  ,   7 7 Shaded portion OABCD is the feasible region, where  24 24  O(0, 0), A(5, 0), B(5, 4/3), C  ,  and D(0, 6)  7 7 Now minimize Z = 4x + 3y Z at O(0, 0) = 0 Z at A(5, 0) = 4 × 5 + 3 × 0 = 20 4  4 Z at B  5,  = 4 × 5 + 3 × = 24  3 3 24 24  24 24  Z at C  ,  = 4 × + 3 × = 24  7 7 7 7

E(20, 15)

B(35, 0) C(50, 0) X 10 20 30 40 50 x + 2y = 50 x + y = 35 Y′

O

Value of Z = 150x + 250y

Clearly, Z is maximum at x = 20, y = 15.

10 X′

Corner points

Z at D(0, 6) = 4 × 0 + 3 × 6 = 18 Thus, Z is minimum at 1 point O(0, 0).

 13

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Self Evaluation Sheet Once you complete SQP-1, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Vector Algebra / Vector Algebra Probability Relations and Functions / Relations and Functions Three Dimensional Geometry Matrices / Matrices Determinants Integrals / Integrals Relations and Functions Three Dimensional Geometry / Three Dimensional Geometry Differential Equations Application of Integrals Vector Algebra Relations and Functions Matrices Probability Vector Algebra Probability Application of Derivatives

19

Continuity and Differentiability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Application of Integrals Inverse Trigonometric Functions Vector Algebra / Vector Algebra Probability Application of Derivatives Differential Equations Three Dimensional Geometry Determinants / Determinants Continuity and Differentiability Integrals Continuity and Differentiability / Continuity and Differentiability Application of Integrals Application of Derivatives Differential Equations / Differential Equations Relations and Functions Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score. Class 12

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SQP

2

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)#

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

5.

Continuity and Differentiability

–

1(2)

2(6)

–

3(8)

6.

Application of Derivatives

1(4)

1(2)

1(3)*

–

3(9)

7.

Integrals

1(1)*

1(2)

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)*

1(2)

1(3)*

–

3(6)

10.

Vector Algebra

3(3)

1(2)*

–

–

4(5)

11.

Three Dimensional Geometry

2(2)#

1(2)*

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

2(2) + 1(4)

1(2)*

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-2

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1 1. If the matrix A =  1   x

−2 2 2

3 1 is singular, then find x.  −3 OR

What positive value of x makes the following pair of determinants equal ? 2x 5

3 16 , x 5

8 2. If A = 4  3

3 2

0 2   −2 and B = 4   6   −5

−2 2  , then find the matrix X, such that 2A + X = 5B.  1 

3. Determine the order and degree of differential equation

d3x dt 3

+

d 2 x  dx  2 = et . + 2  dt  dt Class 12

16

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OR  d2 y  What is the degree of the differential equation  2   dx 

2 /3

+4−

3dy =0? dx

4. If R = {(x, y) : x + 2y = 8} is a relation on N, then find the range of R. 5. If a line makes angles 90°, 60° and 30° with the positive directions of x, y and z-axis respectively, then find its direction cosines. OR Find the direction cosines of the line passing through two points (2, 1, 0) and (1, –2, 3). 6. Find the area enclosed between the curve x2 + y2 = 16 and the coordinate axes in the first quadrant. 2

7. Evaluate: ∫ xe x dx OR Evaluate :

∫

(

cos x

)

x x 3 cos + sin 2 2

dx

8. If a, b, g are the angles made by a line with the co-ordinate axes. Then find the value of sin2a + sin2b + sin2g. 9. Check whether the relation R = {(1, 1), (2, 2) (3, 3), (1, 2), (2, 3), (1, 3)} on set A = {1, 2, 3} is an equivalence relation or not. OR Find the domain of the function f ( x ) =

1 where {·} denotes fractional part. {sin x } + {sin(π + x )}

   10. Find the value of (a ⋅ ɵi )2 + (a ⋅ ɵj)2 + (a ⋅ kɵ )2 . 11. A die, whose faces are marked 1, 2, 3 in red and 4, 5, 6 in green, is tossed. Let A be the event “number obtained is even” and B be the event “number obtained is red”. Find P(A ∩ B). 12. Find the value of p for which p(i + j + k ) is a unit vector. n + 1 , if n is odd  for all n ∈ N . Find whether the function f is bijective 13. Let f : N → N be defined by f (n) =  2  n , if n is even  2 or not. 14. If P ( A) =

12 7 9 and P ( A ∪ B) = , then evaluate P(A|B). , P ( B) = 13 13 13

             15. If a , b , c are unit vectors such that a + b + c = 0, then write the value of a ⋅b + b ⋅ c + c ⋅ a . 16. Write a 3 × 2 matrix whose elements in the ith row and jth column are given by aij =

(2i − j) . 2 17

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Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Arun and Richa decided to play with dice to keep themselves busy at home as their schools are closed due to coronavirus pandemic. Arun throw a dice repeatedly until a six is obtained. He denote the number of throws required by X. Based on this information, answer the following questions. (i) The probability that X = 3 equals 1 52 (b) 6 63 (ii) The probability that X = 5 equals 1 1 (b) (a) 4 6 66 (iii) The probability that X ≥ 3 equals 1 25 (b) (a) 36 216

(c)

(a)

5 6

(d)

3 (c)

54 5

(d)

6 (c)

5 36

(d)

1 63 5 64 25 36

(iv) The value of P(X > 3) + P(X ≥ 6) is (a)

53

(b) 1 –

53

(c)

53 × 61

65 65 65 (v) The conditional probability that X ≥ 6 given X > 3 equals (a)

36 25

(b)

52 2

(c)

5 6

(d)

(d)

53 64 1 6

6 18. Peter's father wants to construct a rectangular garden using a rock wall on one side of the garden and wire fencing for the other three sides as shown in figure. He has 100 ft of wire fencing. Based on the above information, answer the following questions.

(i) To construct a garden using 100 ft of fencing, we need to maximise its (a) volume (b) area (c) perimeter (d) length of the side (ii) If x denote the length of side of garden perpendicular to rock wall and y denote the length of side parallel to rock wall, then find the relation representing total amount of fencing wall. (a) x + 2y = 100 (b) x + 2y = 50 (c) y + 2x = 100 (d) y + 2x = 50 (iii) Area of the garden as a function of x i.e., A(x) can be represented as (a) 100 + 2x2 (b) x – 2x2 (c) 100x – 2x2 (d) 100 – x2 (iv) Maximum value of A(x) occurs at x equals (a) 25 ft (b) 30 ft (c) 26 ft (d) 31 ft Class 12

18

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(v) Maximum area of garden will be (a) 1200 sq. ft (b) 1000 sq. ft

(c) 1250 sq. ft

(d) 1500 sq. ft

PART - B Section - III 4

19. Evaluate :

(x 2 + x ) ∫ 2 x + 1 dx 2

20. The equation of a line is 5x – 3 = 15y + 7 = 3 –10z. Write the direction cosines of the line. OR x +1 y + 3 z + 5 x −2 y −4 z −6 Show that the lines and intersect. Also find their point of = = = = 3 5 7 1 3 5 intersection. 21. Find the area of the region bounded by the curve y = x2 and the line y = 4. 1 1 22. Prove that : 3sin–1x = sin–1(3x – 4x3), x ∈ − ,   2 2 23. An unbiased dice is thrown twice. Let the event A be ‘odd number on the first throw’ and B be the event ‘odd number on the second throw’. Check the independence of the events A and B. OR A bag contains 4 balls. Two balls are drawn at random (without replacement) and are found to be white. What is the probability that all balls in the bag are white ? 24. Determine the value of ‘k’ for which the following function is continuous at x = 3.  ( x + 3)2 − 36  f (x ) =  x −3  k

,

x≠3

,

x=3

25. Solve the differential equation : dy = 1 + x 2 + y 2 + x 2 y 2 , given that y = 1 when x = 0. dx 0 1 3 1 26. Find (AB)–1, if A =  and B −1 =   .  −4 2 5 2 27. Show that the function f(x) = x3 – 3x2 + 6x – 100 is increasing on R.    28. Prove that the points A, B and C with position vectors a , b and c respectively are collinear if and only if        a × b + b × c + c × a = 0. OR   ɵ  ɵ ɵ  ɵ ɵ ɵ If a = 7i + j − 4 k and b = 2i + 6 j + 3k , then find the projection of b on a . Section - IV x −1 , then show that f is one-one and 29. Let A = R – {2} and B = R – {1}. If f : A → B is a function defined by f ( x ) = x −2 onto. 30. Using integration, find the smaller area enclosed by the circle x2 + y2 = 4 and the line x + y = 2. π 31. Find the equation of tangent to the curve x = sin 3t, y = cos 2t at t = . 4 19

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OR Find the intervals in which the function f(x) = (a) strictly increasing

3x4

– 4x3 – 12x2 + 5 is (b) strictly decreasing

π /2

32. Evaluate :

dx 1 + tan x 0

∫

33. For what value of a is the function f defined by π  a sin 2 (x + 1), x ≤ 0 f (x ) =  tan x − sin x  , x>0  x3 is continuous at x = 0 ?    y 34. Solve the following differential equation :  x sin2   − y  dx + x dy = 0  x   OR dy π + y = x cos x + sin x, given that y = 1 when x = . Solve the differential equation x dx 2 35. Show that the function f(x) = |x – 1| + |x + 1|, for all x ∈ R, is not differentiable at the points x = –1 and x = 1. Section - V 1 36. If A = 3  2

2 2 −1

−3 −2 , then find A–1 and use it to solve the following system of the equations : 1 

x + 2y – 3z = 6, 3x + 2y – 2z = 3 2x – y + z = 2 4 4  1  −4  Determine the product −7 1 3  1    −3 −1 2  5 x – y + z = 4, x – 2y – 2z = 9, 2x + y + 3z = 1.

OR −1 1  −2 −2 and use it to solve the system of equations  1 3 

37. A variable plane which remains at a constant distance 3p from the origin cuts the coordinates axes at A, B, C. 1 1 1 1 Show that the locus of the centroid of triangle ABC is + + = . 2 2 2 x y z p2 OR Find the distance between the lines l1 and l2 given by  l : r = ɵi + 2 ɵj − 4 kɵ + λ (2ɵi + 3 ɵj + 6kɵ ) 1

 l2 : r = 3 ɵi + 3 ɵj − 5kɵ + µ (4ɵi + 6 ɵj + 12kɵ ). 38. Find graphically, the maximum value of z = 2x + 5y, subject to constraints : 2x + 4y ≤ 8, 3x + y ≤ 6, x + y ≤ 4; x ≥ 0, y ≥ 0 OR Maximise z = 8x + 9y subject to the constraints : 2x + 3y ≤ 6, 3x – 2y ≤ 6, y ≤ 1; x, y ≥ 0

Class 12

20

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SOLUTIONS 1. Given that matrix A is singular ⇒ |A| = 0 1 ⇒ 1 x

−2 2 2

3 1 =0 −3

So, PQ = (1 − 2)2 + (−2 − 1)2 + (3 − 0)2

⇒ 1(–6 – 2) + 2(–3 – x) + 3(2 – 2x) = 0 ⇒ –8 – 6 – 2x + 6 – 6x = 0 ⇒ –8x – 8 = 0 ⇒ x = –1 OR We have,

2 x 3 16 3 = 5 x 5 2

⇒ 2x2 – 15 = 32 – 15 ⇒ 2x2 = 32 ⇒ x2 = 16 ⇒ x = 4 [Q x > 0]

 10 − 16  =  20 − 8  −25 − 6

−2 8  2 − 2 4  3 1

0  −2  6 −10  14  −7 

3. Given differential equation is of order 3 and degree 1. OR 2 /3

=3

Thus, the direction cosines of the line joining two points −2 −1 3 − 0 −1 − 3 3 are < 1 − 2 , > i.e., < , , , > 19 19 19 19 19 19 6. Given curve is a circle with centre (0, 0) and radius 4. \ Required area 4

∫

16 x =  16 − x 2 + sin −1 2 2

2

(On cubing both sides)

\ Degree is 2. 4. Here, R = {(x, y) : x + 2y = 8}, where x, y ∈ N. For x = 1, 3, 5, ... ; x + 2y = 8 has no solution in N. For x = 2, we have 2 + 2y = 8 ⇒ y = 3 For x = 4, we have 4 + 2y = 8 ⇒ y = 2 For x = 6 , we have 6 + 2y = 8 ⇒ y = 1 For x = 8, 10, ... ; x + 2y = 8 has no solution in N. \ Range of R = { 1, 2, 3} 5. Let the direction cosines of the line be l, m, n. Then, 1 3 l = cos 90° = 0, m = cos 60° = and n = cos 30° = . 2 2 1 3 So, direction cosines are < 0, , >. 2 2

4

x = 4 π sq. units 4  0

2

7. Let I = ∫ xe x dx Put x2 = t ⇒ 2xdx = dt ⇒ x dx = 2

∴

et ex 1 I = ∫ et dt = + C = +C 2 2 2 OR

Let I = ∫

dy −4 dx

3  d2 y    dy ⇒  = 3 − 4    dx 2  dx 

16 − x 2 dx

0

−10 − 0  −6   10 + 4  =  12 5 − 12  −31

 d2 y  We have,  2   dx 

= 1 + 9 + 9 = 19

=

2. We have, 2A + X = 5B ⇒ X = 5B – 2A   2 ⇒ X = 5 4     −5

OR Here, P(2, 1, 0) and Q(1, –2, 3)

=∫

cos x x x   cos + sin  2 2

3

dx = ∫

dt 2

cos2 (x /2) − sin2 (x /2) {cos(x /2) + sin(x /2)}3

dx

cos( x /2) − sin(x /2)

dx 2 x x   cos + sin  2 2 1 x x x x Put t = cos + sin ⇒ dt =  − sin + cos  dx  2 2 2 2 2 dt −2 −2 +C ∴ I = 2∫ = +C = 2 cos( x /2) + sin(x /2) t t 8. a, b and g are the angles made by line with the co-ordinate axes. \ cos2 a + cos2 b + cos2 g = 1 ⇒ 1– sin2 a + 1 – sin2 b + 1 – sin2 g = 1 ⇒ sin2 a + sin2 b + sin2 g = 2 9. Reflexive : (1, 1), (2, 2), (3, 3) ∈ R, R is reflexive Symmetric : (1, 2) ∈R but (2, 1) ∉ R, R is not symmetric. Transitive : (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R, R is transitive. 21

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Since, R is not symmetric. So, R is not an equivalence relation. OR 1 1 f (x ) = = {sin x } + {sin(π + x )} {sin x } + {( − sin x )} 0, sin x is integer Now, {sinx} + {–sinx} =  1, sin x is not integer For f(x) to be defined, {sinx} + {–sinx} ≠ 0 ⇒ sinx ≠ integer ⇒ sinx ≠ ± 1, 0 nπ ⇒ x≠ 2  nπ  Hence, domain is R −  , n ∈ I  . 2    10. Let a = xiɵ + y ɵj + zkɵ ⇒ (a ⋅ ɵi )2 = x 2   Similarly, (a ⋅ ɵj )2 = y 2 and (a ⋅ kɵ )2 = z 2    2 ∴ (a ⋅ ɵi )2 + (a ⋅ ɵj )2 + (a ⋅ kɵ )2 = x 2 + y 2 + z 2 = a

11. We have, S = {1, 2, 3, 4, 5, 6} Let A be the event that number is even = {2, 4, 6} and B be the event that number is red = {1, 2, 3} Now, A ∩ B = {2} 1 ∴ P ( A ∩ B) = 6  12. Let a = (ɵi + ɵj + kɵ )  ɵi + ɵj + kɵ 1 ɵ ɵ ɵ So, unit vector of a = = (i + j + k ) 1+1+1 3 1 \ The value of p is . 3 1+1 2 13. Here, f (1) = = 1, f (2) = = 1 , 2 2 3+1 4 f (3) = = 2, f (4) = = 2 2 2 (2k − 1) + 1 = k and Thus, f (2k − 1) = 2 2k f (2k ) = =k 2 ⇒ f(2k – 1) = f(2k), where k ∈ N But, 2k – 1 ≠ 2k ⇒ f is not one-one. Hence, f is not bijective. 14. Given, 7 9 12 P ( A) = , P (B) = and P ( A ∪ B) = 13 13 13 \ P(A ∩ B) = P(A) + P(B) – P(A ∪ B) 7 9 12 4 = + − ⇒ P ( A ∩ B) = 13 13 13 13 P ( A ∩ B) 4 / 13 4 = = ∴ P (A | B) = P ( B) 9 / 13 9

   15. We have a , b , c are unit vectors.    Therefore, a = 1, b = 1 and c = 1     Also, a + b + c = 0 (given)  2 ⇒ a + b + c = 0 2         ⇒ a 2 + b + c 2 + 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0       ⇒ 1 + 1 + 1 + 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0       ⇒ 3 + 2(a ⋅ b + b ⋅ c + c ⋅ a ) = 0 3       ⇒ (a ⋅ b + b ⋅ c + c ⋅ a ) = − 2 2i − j 16. aij = , i = 1, 2, 3 and j = 1, 2 2 2−2 2(2) − 1 3 2 −1 1 = 0, a21 = = , a11 = = , a12 = 2 2 2 2 2 a22 =

2(2) − 2 2 2(3) − 1 5 = , = = 1, a31 = 2 2 2 2

a32 =

2(3) − 2 4 = =2 2 2

1 2 3 \ Required matrix =  2 5  2

0  1   2 

17. (i) (b) : P(X = 3) = (Probability of not getting six at first chance) × (Probability of not getting six at second chance) × (Probability of getting six at third chance) =

5 5 1 × × = 6 6 6

52 63

(ii) (c) : P(X = 5) =

5 5 5 5 1 × × × × = 6 6 6 6 6

54 65

(iii) (d) : P(X ≥ 3) = 1 – P(X < 3) = 1 – [P(X = 1) + P(X = 2)] 11 25 1 5 1 = 1−  + ×  = 1− = 36 36 6 6 6 5

6

5 1 5 1 (iv) (c) : P ( X ≥ 6) =   × +   × + ...∞ 6 6 6 6 = =

55 66

  5  5 2 1 + 6 +  6  + ... + ∞   

55  1   5  5 = 66  1 − 5   6   6  3

4

5 1 5 1 P ( X > 3) =   +   ⋅ + ...∞ 6 6 6 6 Class 12

22

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i.e., its direction ratios are proportional to 6, 2, –3.

5 3 1 5 =   × 1 + + ...   6  6  6 =

Now, 62 + 22 + (−3)2 = 7 \ Its direction cosines are

53  1   5  3 × = 64 1 − 5   6   6 

6 2 3 , ,− . 7 7 7

OR

5 3 5 5 5 3  52  \ P(X > 3) + P(X ≥ 6) =   +   =    2 + 1  6   6   6  6  =

53 × 61 65

5 (v) (b) : Required conditional probability = 6 5 6

5

( ) =5 () 6

2

3

2

18. (i) (b) : To create a garden using 100 ft fencing, we need to maximise its area. (ii) (c) : Required relation is given by 2x + y = 100. (iii) (c) : Area of garden as a function of x can be represented as A(x) = x ⋅ y = x(100 – 2x) = 100x – 2x2 (iv) (a) : A(x) = 100x – 2x2 ⇒ A′(x) = 100 – 4x For the area to be maximum A′(x) = 0 ⇒ 100 – 4x = 0 ⇒ x = 25 ft. (v) (c) : Maximum area of the garden = 100(25) – 2(25)2 = 2500 – 1250 = 1250 sq. ft

Any point on the line x +1 y + 3 z + 5 ...(i) = = = r (say) 3 5 7 is (3r – 1, 5r – 3, 7r – 5). Any point on the line x−2 y−4 z −6 = = = k (say) ...(ii) 1 3 5 is (k + 2, 3k + 4, 5k + 6) For lines (i) and (ii) to intersect, we must have 3r – 1 = k + 2, 5r – 3 = 3k + 4, 7r – 5 = 5k + 6 1 3 On solving these, we get r = , k = − 2 2 \ Lines (i) and (ii) intersect and their point of 1 3 1 intersection is  , − , −  2 2 2 21. We have, y = x2 and y = 4

19. Using integration by parts, we get 4 (x 2

∫

2

+ x)

2x + 1

4

4

dx = (x 2 + x ) ⋅ 2 x + 1  2 − ∫ (2 x + 1) ⋅ 2 x + 1 dx 2 4

= (60 − 6 5 ) − ∫ (2 x + 1)3/2 dx 2 4 1 = (60 − 6 5 ) − ⋅ (2 x + 1)5/2  2 5

  243 = (60 − 6 5 ) −  −5 5   5   57 − 5 5   57 =  − 5 =     5 5 20. The given line is 5x – 3 = 15y + 7 = 3 –10z 3 7 3 y+ z− 5 15 10 ⇒ = = 1 1 1 − 5 15 10 1 1 1 Its direction ratios are , , − 5 15 10 x−

Required area = area of shaded region 2

 32 x3  = 2 ∫ (4 − x )dx = 2  4 x −  = sq. units   3 0 3 0 2

2

22. Put sin–1x = q. Then x = sinq Now, sin3q = (3sinq – 4sin3q) = (3x – 4x3) ⇒ 3q = sin–1(3x – 4x3) ⇒ 3sin–1x = sin–1(3x – 4x3) [ q = sin–1x] Hence, 3sin–1x = sin–1(3x – 4x3) 23. If all the 36 elementary events of the experiment are considered to be equally likely, then we have 18 1 18 1 P ( A) = = and P (B) = = 36 2 36 2 Also, P(A ∩ B) = P(odd number on both throws) 9 1 = = 36 4 1 1 1 Now, P ( A) ⋅ P (B) = × = 2 2 4 Clearly, P(A ∩ B) = P(A) × P(B) Thus, A and B are independent events. 23

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OR Consider the following events. E : Two balls drawn are white A : There are 2 white balls in the bag B : There are 3 white balls in the bag C : There are 4 white balls in the bag P(A) = P(B) = P(C) = 2

P ( E /A ) = P ( E /C ) =

C2

4

C2

4

C2

4

C2

\ P (C /E ) =

=

1 3

3 1 , P ( E /B ) = C2 = 3 = 1 4 6 C2 6 2

=1

P (C ). P (E /C ) P ( A). P (E /A) + P (B). P ( E /B) + P (C ). P (E /C )

1 ×1 3 3 = = 1 1 1 1 1 5 × + × + ×1 3 6 3 2 3

24. Given, f(x) is continuous at x = 3. So, lim f (x ) = f (3) ⇒ x→3

⇒

⇒

(x + 3)2 − 36 =k x−3 x→3 lim

(x + 3)2 − 62 =k x−3 x→3 lim

lim

(x + 3 + 6) (x + 3 − 6) x−3

x→3

=k

⇒ 3+3+6=k ⇒ k = 12 25. We have,

dy = 1 + x2 + y2 + x2 y2 dx

dy = 1 + x 2 + y 2 (1 + x 2 ) = (1 + x2) . (1 + y2) dx dy ⇒ = (1 + x 2 ) dx 2 1+ y x3 Integrating both sides, we get tan −1 y = x + +C 3 When x = 0, y = 1 ∴

tan −1 1 = 0 + 0 + C ⇒ C =

π 4

1 π ∴ tan −1 y = x + x 3 + is the required solution. 3 4

 1 0 26. Given, A =   and B–1 =  −4 2 

3 1     5 2 

2 0 Now, |A| = = 2 and adj A =   −4 2  4 1  1

0

\ A–1 =

1 1 2 0  (adj A) =  A 2  4 1  

Now, (AB)–1 = B–1A–1

1  3 1   2 0  1 10 1   5 1/2     =  = 2  5 2   4 1  2 18 2   9 1  27. We have, f(x) = x3 – 3x2 + 6x – 100 ...(i) Differentiating (i) w.r.t. x, we get f ′(x) = 3x2 – 6x + 6 = 3(x2 – 2x + 1) + 3 = 3(x – 1)2 + 3 > 0 For all values of x, (x – 1)2 is always positive \ f ′ (x) > 0 So, f(x) is an increasing function on R. =

28. The points A, B and C are collinear   ⇔ AB and BC are parallel vectors.    ⇔ AB × BC = 0             ⇔ (b − a ) × (c − b ) = 0 ⇔ (b − a ) × c − (b − a ) × b = 0          ⇔ (b × c − a × c ) − (b × b − a × b ) = 0             ⇔ (b × c + c × a ) − (0 − a × b ) = 0 [∵ a × c = −(c × a )]    ⇔ a × b + b × c + c × a = 0 OR   ɵ ɵ ɵ Given, a = 7i + j − 4k and b = 2ɵi + 6 ɵj + 3kɵ     b ⋅a \ Projection of b on a =  |a | 14 + 6 − 12 8 = = 49 + 1 + 16 66 29. Here, f : A → B is given by f (x ) =

x −1 , x−2

where A = R – {2} and B = R – {1} Let f(x1) = f(x2), where x1, x2 ∈ A (i.e., x1 ≠ 2, x2 ≠ 2) ⇒

x1 − 1 x2 − 1 = x1 − 2 x2 − 2

⇒ (x1 – 1) (x2 – 2) = (x1 – 2) (x2 – 1) ⇒ x1x2 – 2x1 – x2 + 2 = x1x2 – x1 – 2x2 + 2 ⇒ – 2x1 – x2 = – x1 – 2x2 ⇒ x1 = x2 ⇒ f is one-one. Let y ∈ B = R – {1} i.e., y ∈ R and y ≠ 1 such that f(x) = y x −1 ⇔ = y ⇔ (x − 2) y = x − 1 x−2 ⇔ xy – 2y = x – 1 ⇔ x(y– 1) = 2y – 1 2y − 1 ⇔ x= y −1 Class 12

24

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\ f(x) = y when x =

2y − 1 ∈ A (as y ≠ 1) y −1

Hence, f is onto. Thus, f is one–one and onto.

Hence these points divide the whole real line into four disjoint open intervals namely (–∞, –1), (–1, 0), (0, 2) and (2, ∞)

30. The given curves are x2 + y2 = 4 and x + y = 2

Interval

Sign of f ′(x)

(– ∞, –1) (–1, 0) (0, 2) (2, ∞)

(–) (–) (–) < 0 (–) (+) (–) > 0 (+) (+) (–) < 0 (+) (+) (+) > 0

Nature of function Strictly decreasing Strictly increasing Strictly decreasing Strictly increasing

(a) f(x) is strictly increasing in (–1, 0) ∪ (2, ∞) (b) f(x) is strictly decreasing in (–∞, –1) ∪ (0, 2) π /2

\ Required area = area of shaded region

32. Let I =

0

2

= ∫  4 − x 2 − (2 − x ) dx 0

∫

1 + tan x

π /2

∫

⇒ I=

= 0 + 2 sin–1 (1) – 4 + 2 – 0

By the property,

0

π 2

31. The given curve is x = sin 3t; y = cos 2t dy dx ⇒ = 3 cos 3t ; = −2 sin 2t dt dt dy dy / dt 2 sin 2t ⇒ = =− dx dx / dt 3 cos 3t 3π π 1 π = sin = At t = , x = sin 4 4 4 2 π y = cos = 0 2 π 2 sin  dy  2 = − 2 ⋅1 = 2 2 =− and   3 3π  dx t = π 3 − 3 cos 4 4 2 \ Equation of the tangent to the given curve at π t = is 4 1    x −  ⇒ 3 y = 2 2x − 2 2 OR 4 3 We have, f(x) = 3x – 4x – 12x2 + 5 f ′(x) = 12x3 – 12x2 – 24x = 12x(x2 – x – 2) ⇒ f ′(x) = 12x(x + 1)(x – 2) Now, f ′(x) = 0 ⇒ 12x(x + 1)(x – 2) = 0 ⇒ x = –1, x = 0 or x = 2 y−0 =

2 2 3

∫

1+

dx

0

π  cos  − x  2 

0

π  π  cos  − x  + sin  − x  2  2 

∫

π /2

∫

=

sin x sin x + cos x

0

...(i)

a

π /2

I=

sin x cos x

f (x )dx = ∫ f (a − x )dx , we get

0

= 2 ⋅ − 2 = (π − 2) sq.units.

0

cos x + sin x a

dx

∫

=

cos x

 2 2 =  x 4 − x + 4 sin −1 x − 2 x + x   2 2 2 2 0

2

π /2

dx

dx

dx

...(ii)

Adding (i) and (ii), we get π /2

2I =

∫ 0

π /2

=

∫

  cos x sin x +  dx  sin x + cos x   sin x + cos x

1 ⋅ dx = [x]0π / 2 =

0

π π ⇒ I= 2 4

33. For f(x) to be continuous at x = 0, we have lim f ( x ) = lim f ( x ) = f (0) x →0−

x →0+

...(i)

π =a 2  π lim f (x ) = lim a sin  (−h +1)   2 h→0 x →0− π π = lim a sin  − h  = lim a cos  π h  = a  2  h→0 2 2  h→0 tan h − sin h Now, lim f (x ) = lim + h→0 h3 x →0

Here, f (0) = a sin

25

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= lim

35. The given function is f(x) = |x – 1| + |x + 1|

 1  − 1 sin h   cos h 

sin h − sin h cos h

= lim h→0 h3 sin h 1 − cos h = lim × lim h→0 h h → 0 cos h ⋅ h2

 −2 x , x < − 1  −(x − 1) − (x + 1), x < − 1   −1 ≤ x < 1 =  −(x − 1) + x + 1 , − 1 ≤ x < 1 = 2, x − 1 + x + 1 , x ≥ 1 2 x , x ≥1  

h3

h→0

At x = 1, f (1 − h) − f (1) 2−2 = lim =0 − h h →0 h →0 −h f (1+ h) − f (1) 2(1+ h) − 2 2h f ′(1+ ) = lim = lim = lim =2 h h h →0 h →0 h →0 h f ′(1− ) = lim

h 2 sin2   2

1 1 = 1× = 2 2 h 4×  2 \ From (i), we get a = 1/2 Hence, f(x) is continuous at x = 0, if a = 1/2. 1 × lim h→0 cos h h → 0

= 1 × lim

2

\ f ′ (1–) ≠ f ′ (1+) ⇒ f is not differentiable at x = 1. At x = –1, f (−1 − h) − f (−1) −h h →0 −2(−1 − h) − (2) 2h = lim = lim = −2 −h h→0 h → 0 −h f ′(−1− ) = lim

y 34. We have, [ x sin2   − y] dx + x dy = 0 x  y  y dy ⇒ sin2   − + =0  x  x dx

... (i)

This is a linear homogeneous differential equation dy dv \ Put y = vx ⇒ =v+x dx dx \ (i) becomes dv sin2 v − v + v + x =0 dx dv dx ⇒ x + sin2 v = 0 ⇒ cosec2 vdv + =0 dx x Integrating both sides, we get y – cot v + log x = C ⇒ − cot   + log x = C x is the required solution. OR dy We have, x + y = x cos x + sin x dx dy y sin x ⇒ + = cos x + dx x x It is a linear differential equation. 1

∫ x dx

= e log x = x sin x   dx + c ∴ y ⋅ x = ∫ x  cos x +  x  = ∫ [ x cos x + sin x]dx + c I.F. = e

= x sin x − ∫ sin x dx + ∫ sin x dx + c = x sinx + c ⇒

y = sin x +

c x

π Given that, y = 1 when x = 2 c ∴ 1 = 1+ ⇒ c=0 π/2 \ y = sin x is the required solution.

f ′(−1+ ) = lim

h→0

f (−1 + h) − f (−1) 2−2 = lim =0 h h→0 h

\ f ′ (–1–) ≠ f ′ (–1+) ⇒ f is not differentiable at x = –1.  1 2 −3    36. Given, A =  3 2 −2   2 −1 1  1 2 −3 |A| = 3 2 −2 = 1(2 – 2) –2(3 + 4) – 3(–3 – 4) 2 −1 1 = –14 + 21 = 7 ≠ 0 \ A–1 exists Now, A11 = 0, A12 = –7, A13 = –7, A21 = 1, A22 = 7, A23 = 5, A31 = 2, A32 = –7, A33 = –4 0 1 2   \ adj A =  −7 7 −7   −7 5 −4  0 1 2 1 1  –1 ⇒ A = adj A =  −7 7 −7  7 A  −7 5 −4  The given system of equations is x + 2y – 3z = 6 3x + 2y – 2z = 3 2x – y + z = 2 The system of equations can be written as AX = B x 6  1 2 −3   3 2 −2  ; X =  y  , B =  3  where A =        z   2   2 −1 1  Q A–1 exists, so system of equations has a unique solution given by X = A–1B Class 12

26

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 0 1 2  6 x  7  1   1   1    ⇒  y  =  −7 7 −7   3  =  −35  =  −5  7 7  −7 5 −4   2   z   −35   −5  ⇒ x = 1, y = –5, z = –5 OR  −4 4 4   1 −1 1     We have,  −7 1 3   1 −2 −2   5 −3 −1  2 1 3 

where, a, b, c are variables. This meets X, Y and Z axes at A(a, 0, 0), B(0, b, 0) and C(0, 0, c). Let (a, b, g) be the coordinates of the centroid of triangle ABC. Then, a+0+0 a 0+b+0 b α= = ,β= = , 3 3 3 3 0+0+c c γ= = ...(ii) 3 3

 −4 + 4 + 8 4 − 8 + 4 −4 − 8 + 12  =  −7 + 1 + 6 7 − 2 + 3 −7 − 2 + 9   5 − 3 − 2 −5 + 6 − 1 5 + 6 − 3 

The plane (i) is at a distance 3p from the origin. \ 3p = Length of perpendicular from (0, 0, 0) to the plane (i)

8 0 0 =  0 8 0  = 8I  0 0 8 

⇒ 3p =

−1

2

2

2

2

2

1 1 1   +   +   a b c

 −4 4 4   1 −1 1  1   ⇒  −7 1 3   1 −2 −2  = I 8  5 −3 −1  2 1 3   1 −1 1    ⇒  1 −2 −2   2 1 3 

0 0 0 + + −1 a b c

 −4 4 4  1  =  −7 1 3  8  5 −3 −1

⇒

The given system of equations is x – y + z = 4, x –2y – 2z = 9, 2x + y + 3z = 1 and it can be written as AX = B where,  1 −1 1  x 4      A =  1 −2 −2  , X =  y  , B =  9   2 1 3   z   1 

1 1 1   +   +   a b c 1 1 1 1 = 2 + 2 + 2 2 9p a b c

1 p

2

=

1 α

2

+

1 β

2

+

γ2

...(i)

1 p

2

=

1 x

2

+

1 y

2

+

1 z2

OR Given lines are  l1 : r = ɵi + 2 ɵj − 4kɵ + λ(2ɵi + 3ɵj + 6kɵ );  l2 : r = 3ɵi + 3ɵj − 5kɵ + µ (4ɵi + 6 ɵj + 12kɵ )   \ We have a1 = ɵi + 2 ɵj − 4kɵ , b1 = 2ɵi + 3ɵj + 6kɵ   and a = 3ɵi + 3ɵj − 5kɵ , b = 4ɵi + 6 ɵj + 12kɵ 2

 24   3  1   =  −16  =  −2  ⇒ x = 3, y = –2, z = –1 8  −8   −1

...(iii)

1

So, the locus of (a, b, g) is

 −166 + 36 + 4  4  9  = 1  −28 + 9 + 3     8  20 − 27 − 1   1 

37. Let the equation of the plane be x y z + + =1 a b c

2

From (ii), we have a = 3a, b = 3b and c = 3g Substituting the values of a, b, c in (iii), we get 1 1 1 1 = 2 + 2 + 2 2 9p 9α 9β 9γ ⇒

Here, |A| = 1(–6 + 2) + 1(3 + 4) + 1(1 + 4) = –4 + 7 + 5 = 8 ≠ 0 So, the given system of equations has a unique solution given by X = A–1B. x  −4 4 4    1 ⇒  y  =  −7 1 3  8  z   5 −3 −1

1

⇒ 3p =

2

  So, a2 − a1 = 2ɵi + ɵj − kɵ     Also, b2 = 4 ɵi + 6 ɵj + 12kɵ = 2b1 ⇒ b1 || b2 Hence l1 and l2 are parallel lines. Shortest distance between two parallel lines is,    b × (a2 − a1 )  d= |b | 27

Mathematics

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⇒ d=

⇒ d=

\ The maximum value of z is 10, which is at A(0, 2). OR Let l1 : 2x + 3y = 6, l2 : 3x – 2y = 6, l3 : y = 1; x = 0, y = 0

(2 i + 3j + 6k ) × (2 i + j − k ) 22 + 32 + 62 −9 i + 14 j − 4k 7

(−9)2 + 142 + (−4)2 293 units. = 7 7 38. Let l1 : 2x + 4y = 8, l2 : 3x + y = 6, l3 : x + y = 4; x = 0, y = 0 Solving l1 and l2, we get B  8 , 6  ⇒ d=

5 5

Solving l1 and l3, we get D (1.5, 1)  30 6  Solving l1 and l2, we get C  ,   13 13  Shaded portion OADCB is the feasible region, where coordinates of the corner points are O(0, 0),  30 6  A(0, 1), D(1.5, 1), C  ,  , B(2, 0).  13 13  Shaded portion OABC is the feasible region, where coordinates of the corner points are O(0, 0), 8 6 A(0, 2), B  ,  , C(2, 0) 5 5 The value of objective function at these points are : Corner points O(0, 0) A(0, 2) 8 6 B ,  5 5 C(2, 0)

The value of the objective function at these points are :

Value of the objective function z = 2x + 5y 2×0+5×0=0 2 × 0 + 5 × 2 = 10 (Maximum) 2×

8 6 + 5 × = 9. 2 5 5

2×2+5×0=4

Corner points

Value of the objective function z = 8x + 9y

O (0, 0)

8×0+9×0=0

A (0, 1)

8×0+9×1=9

D (1.5, 1)

8 × 1.5 + 9 × 1 = 21

 30 6  C ,   13 13 

8×

30 6 + 9 × = 22.6 (Maximum) 13 13

B (2, 0)

8 × 2 + 9 × 0 = 16

 30 6  The maximum value of z is 22.6, which is at C  ,  13 13



Class 12

28

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Self Evaluation Sheet Once you complete SQP-2, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Chapter

Marks Per Question 1

Determinants / Determinants Matrices

Marks Obtained

1

Differential Equations / Differential Equations Relations and Functions Three Dimensional Geometry / Three Dimensional Geometry Application of Integrals Integrals / Integrals Three Dimensional Geometry Relations and Functions / Relations and Functions Vector Algebra Probability Vector Algebra Relations and Functions Probability Vector Algebra Matrices Probability Application of Derivatives Integrals Three Dimensional Geometry / Three Dimensional Geometry Application of Integrals Inverse Trigonometric Functions Probability / Probability Continuity and Differentiability Differential Equations Determinants Application of Derivatives Vector Algebra / Vector Algebra Relations and Functions Application of Integrals Application of Derivatives / Application of Derivatives Integrals Continuity and Differentiability Differential Equations / Differential Equations Continuity and Differentiability Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming Total

1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80 Percentage

.................. ..............%

Performance Analysis Table If your marks is

> 90% TREMENDOUS!

You are done! Keep on revising to maintain the position.

81-90% EXCELLENT!

You have to take only one more step to reach the top of the ladder. Practise more.

71-80% VERY GOOD!

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70% GOOD!

Revise thoroughly and strengthen your concepts.

51-60% FAIR PERFORMANCE!

Need to work hard to get through this stage.

40-50% AVERAGE!

Try hard to boost your average score.

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BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)*

–

1(5)*

3(8)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

2(2)#

1(2)*

1(3)*

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)*

1(2)

1(3)

–

3(6)

10.

Vector Algebra

3(3)

1(2)*

–

–

4(5)

11.

Three Dimensional Geometry

2(2)#

1(2)

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

1(2)

–

–

4(8)

10(20)

7(21)

3(15)

38(80)

13.

2(2)#

Probability Total

+ 1(4)

18(24)

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-3

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I π /2

1. Evaluate : ∫ x cos x dx 0

OR

Evaluate : ∫cos3x sin x dx 2. Check whether the function f : N → N defined by f(x) = 4 – 3x is one-one or not. 3. Solve the differential equation

dy y−x =2 . dx

dy  y  =  Solve the differential equation dx  x 

OR

1/3

.

sec θ tan θ   − tan θ − sec θ 4. Simplify : tan θ  + sec θ     tan θ − sec θ  − sec θ tan θ  31

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5. Find the direction cosines of the line that makes equal angles with the three axes in space. OR Find the vector equation of the symmetrical form of equation of straight line 6. Prove that the function f(x) =

x −5 y +4 z −6 = = 3 7 2

 π π 3 sin 2x – cos 2x + 4 is one-one in the interval  − ,  .  6 3

7. A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement, then find the probability of getting exactly one red ball. OR 1 2 3 and P ( A ∩ B ) = , then find the value of P(A′ | B′). If P ( A) = , P (B) = 5 5 10     ^ ^ ^ ^ ^ ^ 8. Find the angle between the vectors a and b if a = 2 i − j + 2 k and b = 4 i + 4 j − 2 k .

9. A matrix A of order 3 × 3 has determinant 5. What is the value of | 3A |? OR (1 + x )17

(1 + x )19

(1 + x )23

23 If f (x ) = (1 + x )

(1 + x )29

(1 + x )34 = A + Bx + Cx2 + ....., then prove that A = 0.

(1 + x )41

(1 + x )43

(1 + x )47

10. Find the vector in the direction of the vector i − 2 j + 2k that has magnitude 9. 11. If E and F are events such that 0 < P(F) < 1, then prove that P (E | F ) + P (E | F ) = 1 12. Find the direction cosines of the line joining A(0, 7, 10) and B(–1, 6, 6). 13. If g(x) = x2 – 4x – 5, then prove that g is not one-one on R.  ^ ^ ^  ^ ^ ^ 14. Find the projection of the vector a = 2 i + 3 j + 2 k on the vector b = i + 2 j + k . 15. If a matrix has 12 elements, then it has _______ possible orders. 2x

x

2 2 x 16. Evaluate : ∫ 2 2 2 dx

Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. The Government declare that farmers can get ` 200 per quintal for their potatoes on 1st February and after that, the price will be dropped by ` 2 per quintal per extra day. Ramu’s father has 80 quintal of potatoes in the field and he estimates that crop is increasing at the rate of 1 quintal per day. Based on the above information, answer the following question. (i) If x is the number of days after 1st February, then price and quantity of potatoes respectively can be expressed as (a) ` (200 – 2x), (80 + x) quintals (b) ` (200 – 2x), (80 – x) quintals (c) ` (200 + x), 80 quintals (d) None of these Class 12

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(ii) Revenue R as a function of x can be represented as (b) R(x) = –2x2 + 40x + 16000 (a) R(x) = 2x2 – 40x – 16000 (d) R(x) = 2x2 – 40x – 15000 (c) R(x) = 2x2 + 40x – 16000 (iii) Find the number of days after 1st February, when Ramu’s father attain maximum revenue. (a) 10 (b) 20 (c) 12 (d) 22 (iv) On which day should Ramu’s father harvest the potatoes to maximise his revenue? (b) 20th Febraury (c) 12th February (d) 22nd February (a) 11th February (v) Maximum revenue is equal to (a) `16000 (b) `18000

(c) `16200

(d) `16500

18. In an annual board examination, in a particular school, 30% of the students failed in Chemistry, 25% failed in Mathematics and 12% failed in both Chemistry and Mathematics. A student is selected at random. (i) The probability that the selected student has failed in Chemistry, if it is known that he has failed in Mathematics, is 12 3 (b) (a) 25 10 1 (c) (d) 3 4 25 (ii) The probability that the selected student has failed in Mathematics, if it is known that he has failed in Chemistry, is 3 12 2 (b) (c) (d) (a) 22 25 25 5 25 (iii) The probability that the selected student has passed in at least one of the two subjects, is 3 88 43 22 (b) (c) (d) (a) 75 125 100 25 (iv) The probability that the selected student has failed in at least one of the two subjects, is 43 3 2 22 (a) (b) (c) (d) 100 5 5 25 (v) The probability that the selected student has passed in Mathematics, if it is known that he has failed in Chemistry, is 2 3 4 1 (b) (c) (d) (a) 5 5 5 5

PART - B Section - III 19. Find

dy π at x = 1, y = , if sin2y + cos xy = K. 4 dx

20. Find the value of

∫

dx x +3 x

. OR

Evaluate :

1

∫ 1 + 3 sin2 x + 8 cos2 x dx 33

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21. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then find the probability that both drawn balls are black. 22. Find the number of solutions of the equation 2 cos −1 x + sin −1 x = 23. Evaluate the determinant ∆ =

π 11π , if sin–1 x + cos–1 x = . 2 6

log 3 512 log 4 3 . log 3 8 log 4 9 OR

If x is a complex root of the equation 1 1 x 1− x 1 = 0 , then find the value of x2007 + x– 2007. x + 1 1− x 1 1 1− x 1 2 2 dy x + y + 1 = 24. Find the solution of the differential equation satisfying y(1) = 1. 2 xy dx 1 x x

x 1 x

25. The x-coordinate of a point on the line joining the points P(2, 2, 1) and Q(5, 1, –2) is 4. Find its z-coordinate. 26. Find the point on the curve y = (x – 3)2 where the tangent is parallel to the chord joining (3, 0) and (4, 1). 27. Find the area bounded by the curve y2 = x, line y = 4 and y-axis. 28. Find a unit vector perpendicular to the plane ABC, where A, B and C are the points (3, –1, 2), (1, –1, –3), (4, –3, 1) respectively. OR 







Let a = iˆ + 2 jˆ − 3kˆ and b = 3iˆ − jˆ + 2kˆ be two vectors. Show that the vectors (a + b ) perpendicular to each other.





and (a − b ) are

Section - IV 29. Show that the height of the closed cylinder of given surface area and maximum volume, is equal to the diameter of base. 30. Solve : (x x 2 + y 2 − y 2 )dx + xy dy = 0 31. If y = ex sin x3 + (tan x)x, then find

dy . dx

OR

If x = 3 sin t – sin 3t, y = 3 cos t – cos 3t, then find

d2 y dx 2

at t =

π . 3

32. Find the area bounded by the lines y = 1 – ||x| – 1| and the x- axis. π   x + a 2 sin x , 0 ≤ x < 4   π π 33. Let f ( x ) =  2 x cot x + b, ≤ x ≤ 4 2  π  a cos 2 x − b sin x , 2 < x ≤ π be continuous in [0, p], then find the value of a + b. π /2

sin2 x dx 0 (1 + sin x cos x )

34. Evaluate : ∫

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OR 3 π/ 4

Find the value of

∫ π/ 4

x dx . 1 + sin x

35. Show that the relation R in the set of real numbers, defined as R = {(a, b) : a ≤ b2} is neither reflexive, nor symmetric, nor transitive. Section - V  −5 1 1 3    36. Find the product BA of matrices A =  7 1 −5 , B =  3  1 −1 2 1 equations : x + y + 2z = 1 ; 3x + 2y + z = 7 ; 2x + y + 3z = 2.

1 2 1

2  1 and use it in solving the 3

OR  − 1 −2 −2   Find the adjoint of the matrix A =  2 1 −2 and hence show that A⋅(adj A) = |A|I3.  2 −2 1  37. Solve the following linear programming problem graphically. Minimize Z = x – 7y + 227 subject to constraints : x+y≤9 x≤7 y≤6 x+y≥5 x, y ≥ 0 OR Solved the following linear programming problem graphically. Maximize Z = 11x + 9y subject to constraints : 180x + 120y ≤ 1500 x + y ≤ 10 x, y ≥ 0 38. If the lines

x −1 y − 2 z − 3 = = and x − 1 = y − 2 = z − 3 are perpendicular, then find the value of k and −3 −2k 2 k 1 5

hence find the equation of plane containing these lines. OR Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence find the distance of point P(–2, 5, 5) from the plane obtained above.

35

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SOLUTIONS 1.

π /2

π /2

0

0

π /2 ∫ x cos x dx = [x sin x]0 − ∫ 1 ⋅ sin x dx

[Integrating by parts] π π = + [cos x]0π /2 = − 1 2 2 OR We have, ∫ cos3 x sin x dx Put cos x = t ⇒ sin x dx = –dt 3 3 \ ∫ cos x sin x dx = − ∫ t dt = −

OR x − x1 y − y1 z − z1 is = = a b c  r = (x1 ɵi + y1 ɵj + z1 kɵ ) + λ(aiɵ + b ɵj + ckɵ )

The vector form of

\ Required equation in vector form is  r = (5ɵi − 4 ɵj + 6kɵ ) + µ(3ɵi + 7 ɵj + 2kɵ )   3 1 6. We have, f (x ) = 2  sin 2 x − cos 2 x  + 4   2 2

t4 +C 4

π π = 2  cos sin 2 x − sin cos 2 x  + 4   6 6

1 = − cos 4 x + C 4



2. We have, f : N → N, f (x) = 4 – 3x Let f(x1) = f(x2) ⇒ 4 – 3x1 = 4 – 3x2 ⇒ x1 = x2 \ f is one-one. dy dx dy y−x ⇒ y = x =2 dx 2 2 Integrating both sides, we get

Q sin x is one-one in  − π , π  .  2 2

∴

3. We have,

−2 − y −2 − x = +C log 2 log 2 –y –x –x –y ⇒ –2 + 2 = C log2 = k(say) ⇒ 2 – 2 = k

π

⇒ f(x) = 2 sin  2 x −  + 4  6

−

π π π  π π ≤ 2x − ≤ ⇒ x ∈ − ,  2 6 2  6 3

7. Required probability = P{(RBB), (BRB), (BBR)} = P(RBB) + P(BRB) + P(BBR) 5 3 2 3 5 2 3 2 5 5 15 = × × + × × + × × = 3× = 8 7 6 8 7 6 8 7 6 56 56

OR

OR

1/3

dy  y  =   ⇒ y–1/3 dy = x–1/3 dx dx  x  Integrating both sides, we get 3 2/3 3 2/3 2 y = x + k ⇒ y 2/3 − x 2/3 = k = c (say ) 2 2 3 We have,

Hence, required solution is y2/3 – x2/3 = c 4. We have,  sec θ tan θ  − tan θ − sec θ + sec θ  tan θ     tan θ − sec θ  − sec θ tan θ  − tan θ sec θ  tan θ sec θ tan2 θ +  =  − sec2 θ  tan2 θ − tan θ sec θ   0 −1 = .  −1 0 5. Since l = m = n and l2 + m2 + n2 = 1 1 ⇒ l=m=n=± . 3

− sec2 θ  tan θ sec θ

Given, P ( A) = 2 , P (B) = 3 , P ( A ∩ B) = 1 5 10 5 Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 2 3 1 1 = + − = 5 10 5 2

\ P ( A′ ∩ B′) = P (( A∪B)′) = 1 − P ( A∪B) = 1 − 1 = 1 2 2 3 7 Also, P (B ′) = 1 − P (B) = 1 − = 10 10 P ( A′ ∩ B ′ ) 1 / 2 5 = = P ( B ′) 7 / 10 7   ^ ^ ^ ^ ^ ^ 8. We have, a = 2 i − j + 2 k and b = 4 i + 4 j − 2 k   ^ ^ ^ ^ ^ ^ Now, a ⋅ b = (2 i − j + 2 k ) ⋅ (4 i + 4 j − 2 k ) ∴ P ( A′ | B ′) =

= 8 – 4 – 4 = 0.

 π  So, angle between a and b is . 2 9. Given, | A | = 5, order of A is 3 × 3. \ | 3A | = 33 | A | = 27 × 5 = 135. Class 12

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2x

OR Given that,

2x

(1 + x )

(1 + x )

Put 22

f (x ) = (1 + x )23

(1 + x )29

(1 + x )34

(1 + x )41

(1 + x )43

(1 + x )47

\ I=∫

17

(1 + x )

19

23

= A + Bx + Cx2 + ..... 1 On putting x = 0, we have f (0) = 1 1 ⇒ A=0  10. Let a = ɵi − 2 ɵj + 2kɵ  a = 1+ 4 + 4 = 9 = 3 \ Required vector =

1 1 1

1 1 =A 1

( R1 and R2 are identical)

9(i − 2 j + 2k ) = 3(i − 2 j + 2k ) 3

11. P (E | F ) + P (E | F ) =

x

16. Let I = ∫ 22 22 2 x dx

P (E ∩ F ) + P ( E ∩ F ) P ((E ∪ E ) ∩ F ) P (F ) = = =1 P (F ) P (F ) P (F )

12. Direction ratios of AB are (–1 – 0, 6 – 7, 6 – 10) or (–1, –1, – 4) (−1)2 + (−1)2 + (−4)2 = 3 2 −4  1  1 \ Direction cosines are  − ,− ,   3 2 3 2 3 2 1 4   1 or  , ,  3 2 3 2 3 2  Also,

13. Let g(x1) = g(x2) ⇒ x12 − 4 x1 − 5 = x 22 − 4 x 2 − 5 ⇒ x12 − x 22 = 4(x1 − x 2 ) ⇒ (x1 – x2)(x1 + x2 – 4) = 0 Either x1 = x2 or x1 + x2 = 4 Either x1 = x2 or x1 = 4 – x2 \ There are two values of x1, for which g(x1) = g(x2). \ g(x) is not one-one ∀ x ∈ R  ^  ^ ^ ^ ^ ^ 14. We have, a = 2 i + 3 j + 2 k and b = i + 2 j + k   ^ ^ ^ ^ ^ ^ ∴ a ⋅ b = (2 i + 3 j + 2 k ) ⋅ ( i + 2 j + k) = 2 + 6 + 2 = 10  and | b | = 12 + 22 + 12 = 6   a ⋅ b 10  So, projection of a on b =  = |b | 6 15. All possible orders of 12 elements are 1 × 12, 12 × 1, 2 × 6, 6 × 2, 3 × 4, 4 × 3 i.e., 6

2x

x

2 x 3 2 = t ⇒ 2 2 2 (log 2) dx = dt

1

x

2 1 1 22 + C dt = t +C = 3 3 3 (log 2) (log 2) (log 2)

17. (i) (a) : Let x be the number of extra days after 1st February \ Price = `(200 – 2 × x) = `(200 – 2x) Quantity = 800 quintals + x(1 quintal per day) = (80 + x) quintals (ii) (b) : R(x) = Quantity × Price = (80 + x) (200 – 2x) = 16000 – 160x + 200x – 2x2 = 16000 + 40x – 2x2 (iii) (a) : We have, R(x) = 16000 + 40x – 2x2 ⇒ R′(x) = 40 – 4x ⇒ R″(x) = –4 For R(x) to be maximum, R′(x) = 0 and R″(x) < 0 ⇒ 40 – 4x = 0 ⇒ x = 10 (iv) (a) : Ramu’s father will attain maximum revenue after 10 days. So, he should harvest the potatoes after 10 days of 1st February i.e., on 11th February. (v) (c) : Maximum revenue is collected by Ramu’s father when x = 10 \ Maximum revenue = R(10) = 16000 + 40(10) – 2(10)2 = 16000 + 400 – 200 = 16200. 18. Let C denote the event that student has failed in Chemistry and M denote the event that student has failed in Mathematics. 30 3 25 1 \ P (C ) = = , P (M ) = = 100 10 100 4 12 3 and P (C ∩ M ) = = 100 25 (i) (b) : Required probability = P(C | M) P (C ∩ M ) 3 / 25 12 = = 1 / 4 25 P (M ) (ii) (c) : Required probability P(M | C) =

P (M ∩ C ) 3 / 25 2 = = P (C ) 3 / 10 5 (iii) (a) : Revenue probability = P(C′ ∪ M′) = P[C ∩ M]′ 3 22 = 1 − P (C ∩ M ) = 1 − = 25 25 (iv) (d) : Required probability = P(C) + P(M) – P(C ∩ M) 3 1 3 43 = + − = 10 4 25 100 =

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(v) (b) : Required probability = P(M′ | C) P (M ′ ∩ C ) P (C ) − P (C ∩ M ) 9 / 50 3 = = = = P (C ) P (C ) 3 / 10 5 19. We have, sin2y + cos xy = K Differentiating w.r.t. x, we get dy   dy 2sin y cos y + (− sin xy )  x + y  = 0   dx dx dy y sin xy = ⇒ dx sin 2 y − x sin xy  dy  ⇒   π  dx   1,    4

20. Let I = ∫

π π sin π 4 = = 4 π π 4( 2 − 1) sin − sin 2 4

= 2t3 – 3t2 + 6t – 6log(t + 1) + C = 2 x − 3( 3 x ) + 6( 6 x ) − 6 log( 6 x + 1) + C OR 1

dx 1 + 3 sin x + 8 cos 2 x Dividing the numerator and denominator by cos2 x, we get I=∫ =∫

23. We have,

dx

x +3 x Putting x = t6 ⇒ dx = 6t5 dt, we get t3 6t 5 1   I = ∫ 3 2 dt = 6 ∫ dt = 6 ∫  t 2 − t + 1 −  dt  t 1 + t + 1 t +t

Let I = ∫

By multiplication rule of probability, we have 10 9 3 P(E ∩ F) = P(E) . P(F|E) = × = 15 14 7 11π 22. Given equation is 2 cos −1 x + sin −1 x = 6 11π −1 −1 −1 ⇒ cos x + (cos x + sin x ) = 6 π 11π  π −1 −1 −1 ⇒ cos x + =  Given cos x + sin x =  2 6  2 4π ⇒ cos −1 x = , 3 which is not possible as cos–1x ∈[0, p]. Thus, given equation has no solution.

2

sec 2 x sec 2 x + 3 tan 2 x + 8 sec 2 x

dx sec 2 xdx

dx = ∫ 1 + tan 2 x + 3 tan 2 x + 8 4 tan 2 x + 9 Putting tan x = t ⇒ sec2 x dx = dt, we get dt dt 1 1 1  t  +C = × tan −1  I =∫ 2 = ∫ 2 2  3/2  4t + 9 4 t + (3/2) 4 3/2 1 1  2 tan x   2t  +C = tan −1   + C = tan −1   3    6 6 3 21. Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P(EF). 10 Now, P(E) = P(black ball in first draw) = 15 When second ball is drawn without replacement, the probability that the second ball is black is the conditional probability of event F occurring when event E has already occurred. 9 \ P (F | E ) = 14

∆=

log 3 512 log 4 3 log 3 29 ⇒ ∆= log 3 8 log 4 9 log 3 23

1 log 2 3 2 ⇒ ∆= 3 log 3 2 2 log 3 2 2 9 log 3 2

log log

22

22

3

32

n   n  log a p m = p log a m  

 1 ⇒ ∆ = (9 log 3 2) × (log 2 3) −  log 2 3  (3 log 3 2)  2 3 ⇒ ∆ = 9(log 3 2 × log 2 3) − ( log 2 3 × log 3 2) 2 3 ⇒ ∆ = 9− [ logba × logab = 1] 2 15 ⇒ ∆= 2 OR

Expanding the two determinants, we get [1(1 – x2) – x (x – x2) + x (x2 – x)] + [(1 – x) [(1 – x)2 – 1) – 1 (1 – x – 1) + 1 (1 – 1 + x)] ⇒ (1 – 3x2 + 2x3) + (3x2 – x3) = 0 ⇒ x3 + 1 = 0 ⇒ x = – w, –w2, –1 \ x2007 + x– 2007 = – 1 – 1 = – 2. 2 2 dy x + y + 1 = dx 2 xy 2 2 ⇒ 2xydy = (x + y + 1)dx ⇒ 2xydy – y2dx = (x2 + 1)dx ⇒ xd(y2) – y2dx = (x2 + 1)dx  y2  xd( y 2 ) − y 2dx  1 1  = + 1 ⇒ dx d ⇒   = d  x −    2 2  x  x x x

24. Given,

Integrating both sides, we get 1 y2 = x − + C ⇒ y2 = x2 – 1 + Cx x x Now, given that y(1) = 1 \ 1=1–1+C⇒C=1 Thus, curve becomes y2 = x2 – 1 + x Class 12

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25. Given that P(2, 2, 1) and Q(5, 1, –2)

^

k

Let the point R on the line PQ, divides the line in the ratio k : 1. And x-coordinate of point R on the line is 4. So, by section formula 5k + 2 4= ⇒ k=2 k +1 Now, z-coordinate of point R, −2k + 1 −2 × 2 + 1 z= = = −1 k +1 2 +1 ⇒ z-coordinate of point R = –1 26. We have, y = (x – 3)2 dy = 2(x − 3) Slope of tangent = dx Slope of chord joining (3, 0) and (4, 1) = For parallel lines, slopes are equal

1− 0 =1 4−3 2

1 7  7 \ 2(x – 3) = 1 ⇒ x = and y =  − 3  = 2 4 2 7 Hence, required point is  , 2

1 . 4

27. We have, y2 = x, a parabola with vertex (0, 0) and line y = 4 y (0, 4)

x

Required area = area of shaded region 4

 3 2 ∫ y dy =  y3  = 643 sq. units 0 0   28. The vector AB × AC is perpendicular to the   vectors AB and AC .   AB × AC ... Required vector =   | AB × AC |  Now, AB = P.V. of B – P.V. of A ^ ^

^

^

^

^

= ( i − j − 3 k ) − (3 i − j + 2 k ) = −2 i + 0 j − 5 k  and AC = P.V. of C – P.V. of A ^

^

^

^

^

^

^

^

^

k −5 −1 ^

^

^

^

= (0 – 10) i − (2 + 5) j + (4 − 0) k = −10 i − 7 j + 4 k   ⇒ | AB × AC | = (−10)2 + (−7)2 + 42 = 165

Hence, required vector   1 AB × AC ^ ^ ^ (−10 i − 7 j + 4 k ) =   = | AB × AC | 165 OR  ^ ^ ^  ^ ^ ^ Given, a = i + 2 j − 3 k and b = 3 i − j + 2 k ^ ^ ^   Now, a + b = 4 i + j − k ^ ^ ^   Also, a − b = −2 i + 3 j − 5 k ^ ^ ^ ^ ^ ^     Now, (a + b ) ⋅ (a − b ) = (4 i + j − k ) ⋅ (−2 i + 3 j − 5 k ) = (4)(–2) + (1)(3) + (–1)(–5) = –8 + 3 + 5 = 0     Hence, (a + b ) and (a − b ) are perpendicular to each other. 29. Let r be the radius of the base and h be the height of a closed cylinder of given surface area S. Then,

^

...(i)

dV S = − 3πr 2 dr 2 For maximum or minimum value of V, we have ⇒

=

^

j 0 −2

S − 2 πr 2 2 πr Let V be the volume of the cylinder. Then, V = pr2h  S − 2 πr 2   rS − 2 πr 3  ⇒ V = πr 2    2 πr  =  2

y=4

O

^ ^

^

^

S = 2pr2 + 2prh ⇒ h =

y2 =x

4

i   \ AB × AC = −2 1

^

= (4 i − 3 j + k ) − (3 i − j + 2 k ) = i − 2 j − k

dV S = 0 ⇒ − 3πr 2 = 0 ⇒ S = 6 πr 2 2 dr 6 πr 2 − 2 πr 2 From(i), h = ⇒ h = 2r 2 πr d 2V = −6 πr < 0. Also, dr 2 Hence, V is maximum when h = 2r i.e., when the height of the cylinder is equal to the diameter of the base. 30. The given equation can be written as 2 2 2 dy y − x x + y , which is clearly homogeneous. = dx xy 39

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Putting y = vx ⇒

dy dv =v+x , we get dx dx

At t =

d2 y

dv v 2 x 2 − x x 2 + v 2 x 2 v+x = dx vx 2

dx

 dv  v 2 − 1 + v 2 dv − 1 + v 2 = − v ⇒ x =  dx  v dx v v dx ⇒ ∫ ⇒ 1 + v 2 = − log x + C dv = − ∫ x 1 + v2 x 2 + y 2 + x log x = Cx

31. Let u = ex sin x3 and v = (tan x)x Now, u = ex sin x3 Differentiating w.r.t. x, we get

{

}

d sin( x )3 du d x (ex) + sin x 3 ⋅ =e ⋅ dx dx dx = ex · cos x3 ⋅ 3x2 + sin x3 ⋅ ex du Hence, = 3x2 ⋅ ex cos x3 + ex sin x3 dx Again, v = (tan x)x ⇒ log v = x log (tan x) Differentiating w.r.t. x, we get 1 dv 1 = 1 ⋅ log(tan x ) + x ⋅ sec2 x v dx tan x dv = v [log (tan x) + x cot x ⋅ sec2 x] \ dx = (tan x)x [log (tan x) + x cot x sec2 x] dy du dv = + Now, y = u + v ⇒ dx dx dx dy ⇒ = 3x2excos (x3) + exsin (x3) dx + (tanx)x [log(tan x) + x cot x sec2 x] OR

dx

2

= −2 cosec2 2t ⋅ =–2

cosec22t

dt dx

1 ⋅ 3(cos t − cos 3t )

= −2 cosec2

2π 1 ⋅ 3π  π 3  3  cos − cos   3 3  2

32. We have, y = 1 – |x – 1| if x ≥ 0 1 − (x − 1), if x ≥ 1 2 − x , if x ≥ 1 = = if x < 1 1 + (x − 1), if x < 1  x , and y = 1 – |– x – 1| = 1 – | x + 1| , if x < 0 if x ≥ −1 1 − (x + 1), if x ≥ −1 − x , = = 1 + (x + 1), if x < −1  x + 2, if x < −1

2 1  Required area = 2  ∫ x dx + ∫ (2 − x ) dx   0  1 1

2

 x2   x2  = 2   + 2 2 x −  = 1 + 1 = 2 sq. units  2 0  2 1 33. Since, f(x) is continuous at x = p/4. π π  π   \ L.H.L.  at x =  = f   = R.H.L.  at x =    4 4 4 π π lim ( x + a 2 sin x ) = 2 × cot + b π 4 4 x→ 4

We have, x = 3 sin t – sin 3t dx = 3 cos t – 3cos 3t …(i) ⇒ dt dy y = 3 cos t – cos 3t ⇒ = – 3 sin t + 3 sin 3t …(ii) dt dy dy / dt sin 3t − sin t ∴ = = [Dividing (ii) by (i)] dx dx / dt cos t − cos 3t 2 cos 2t sin t = cot 2t = 2 sin 2t sin t Differentiating w.r.t. x, we get d2 y

2

 2  16 1 =–2 ⋅ =−  1  3 27   3  + 1 2 

⇒ x

⇒

π , 3

[From (i)]

π ...(i) 4 π Also, f(x) is continuous at x = . 2 π π  π   \ L.H.L.  at x =  = f   = R.H.L.  at x =   2 2  2 lim (2 × x cot x + b) = lim (a cos2x – bsinx) ⇒

x→

π π +a = +b ⇒ 4 2

a–b=

π 2

x→

π 2

π π π π ⇒ 2 × cot + b = a cos 2 × − b sin 2 2 2 2 ⇒ b = – a – b ⇒ 2b = – a π . From (i) and (ii), we get a + b = 12

...(ii)

Class 12

40

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π /2

sin2 x 34. Let I = ∫ dx ( sin cos ) + x x 1 0

...(i)

π /2

sin2[(π/2) − x] dx 0 1 + sin[( π/2) − x ]cos[( π/2) − x ] a   a  ∫ f (x )dx = ∫ f (a − x )dx   0  0

Then, I = ∫

π /2

cos2 x dx 0 (1 + sin x cos x ) Adding (i) and (ii), we get

⇒ I= ∫

π / 2 (sin2 x + cos2 x )

2I = ∫

0 π /2

= ∫

(1 + sin x cos x )

...(ii)

π /2

dx 0 (1 + sin x cos x )

dx = ∫

sec2 x

dx (sec2 x + tan x ) (By dividing numerator and denominator by cos2x) 0

π /2

sec2 x

0

(1 + tan2 x + tan x )

= ∫

dx

Putting tanx = t ⇒ sec2x dx = dt, π When x = 0 ⇒ t = 0 and x = ⇒ t → ∞ 2 ∞ ∞ dt dt ∴ I=∫ 2 =∫ 2 2  1 3 0 (t + t + 1) 0  + + t     2 2 ∞

 2  2t + 1  =  tan −1   3   0  3 =

2  −1 2  π π  2π −1  1   ⋅ −  =  tan (∞) − tan    = 3 3  3 2 6 3 3 OR

x ...(i) dx 1 + sin x  3π π  + − x    3 π/ 4 4 4 ⇒ I=∫ (By property) dx π/ 4  3π π  1 + sin  + − x   4 4  3 π/ 4 ( π − x ) dx ...(ii) ⇒ I=∫ π/ 4 1 + sin x Adding (i) and (ii), we get dx 3 π/ 4 (1 − sin x ) 3 π/ 4 2 I = π ∫π/4 = π ∫π/4 dx 2 1 + sin x cos x 3 π/ 4 2 = π ∫π/4 (sec x − sec x tan x ) dx 3 π/ 4

Let I = ∫ π/ 4

3 π/ 4

= π[tan x − sec x]π/4 = π{(−1 + 2 ) − (1 − 2 )} ⇒ 2 I = π(2 2 − 2) ⇒ I = π( 2 − 1)

35. Given relation is R = {(a, b): a ≤ b2} Reflexivity: Let a ∈ real numbers. aRa ⇒ a ≤ a2 but if a < 1, then a 0 ∀x ∈R ⇒ f is increasing on R.   23. Here, a + 3b = ɵi + ɵj + 2kɵ + 3(3ɵi + 2 ɵj − kɵ ) = 10i + 7 j − k   and 2a − b = 2(ɵi + ɵj + 2kɵ ) − (3ɵi + 2 ɵj − kɵ ) = −ɵi + 5kɵ     ∴ (a + 3b ) ⋅ (2a − b ) = (10ɵi + 7 ɵj − kɵ ) ⋅ (− ɵi + 5kɵ ) = 10 × (–1) + 7 × 0 + (–1) × 5 = –15

Adding (i) and (ii), we get

0

2

The given expression becomes

ɵi   We have, a × b = 1 1

x →0

dx

1+ x 1

OR

∴ f (0) = lim f (x ) ⇒ k = 1 20. Let I =

sin(cot −1 x ) =

1+ x 2

2

 sin 2 x  = lim   =1 x →0  2 x 

f is continuous at x = 0.

2π

x

21. cos(cot −1 x ) =

K   x|x| = C + C  where 2 2  

Required area = area of shaded region 1

1

3 = ∫ (3 y 2 − 9)dy = y − 9 y 0 0

= |1 – 9| = 8 sq. units 53

Mathematics

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25. Let the random variable X be defined as the number of spades in a draw of 2 cards successively without replacement, then X can take values 0, 1, 2. P(X = 0) = P(drawing no spade cards) 39 C 19 = 52 2 = C2 34 P(X = 1) = P(drawing one spade and one non-spade card) 13

39

C1 × C1

13 34 C2 P(X = 2) = P(drawing both spade cards) =

52

=

13 C 1 = 52 2 = C2 17 \ The probability distribution of number of spades is

0 19 34

X P(X) tan–1x1/3

1 13 34

2 1 17

tan–1a1/3

26. Let y = + Differentiating w.r.t. x, we get  1 −1  dy 1 1  x3  = / 1 3 2 dx 1 + (x )  3  ⇒

 1  dy 1 1 = = 2 /3   2 / 3 2 / 3 dx 1 + x  3x  3x (1 + x 2/3 )

 2 −3 27. We have, A =    −4 7  2 −3 \ | A |= = 14 – 12 = 2 ≠ 0 −4 7 So, A is a non-singular matrix and therefore it is invertible.  7 3 \ adj A =    4 2 1 1  7 3 Hence, A −1 = adj A =   2  4 2 |A|

OR Since, the line is equally inclined to the axes. \ l=m=n …(i) The required equation of line is x +3 y −2 z +4 = = [using (i)] l l l x +3 y −2 z +4 ⇒ = = 1 1 1 ⇒ x+3=y–2=z+4 29. We have, L.H.L. (at x = 0) 1 + kx − 1 − kx

= lim− f (x ) = lim

(

= lim

x

x →0

x →0

1 + kx − 1 − kx x

x →0

)( (

) 1 − kx )

1 + kx + 1 − kx 1 + kx +

1 + kx − 1 + kx

= lim

x →0 x

(

1 + kx + 1 − kx

)

2k 2k 2k = = =k 1+ 1 2 x →0 1 + kx + 1 − kx

= lim

R.H.L. (at x = 0) = lim f (x ) = lim +

x →0

x →0

2x + 1 x −1

= − 1 and f (0) = −1

Since f(x) is continuous at x = 0. \ k = –1. ∀x≥3  x − 3, 30. We have, y =  − x + 3, ∀ x < 3 4 3 O

3 3 3

4

\ Required area = ∫ −(x − 3) dx + ∫ (x − 3) dx 2 2

3

3 4

 x  1 1 x = 3x −  +  − 3x  = + = 1 sq. unit  3 2 2 2 2  2 2

28. Let l, m, n be the direction cosines of the line 31. Let x and y be two arbitrary elements in A. perpendicular to each of the given lines. Then, x −2 y −2 ...(i) ll1 + mm1 + nn1 = 0 = Then, f(x) = f(y) ⇒ x −3 y −3 and ll2 + mm2 + nn2 = 0 ...(ii) ⇒ xy – 3x – 2y + 6 = xy – 3y – 2x + 6 On solving (i) and (ii), we get ⇒ x = y, ∀ x, y ∈A l m n = = So, f is an injective mapping. m1n2 − m2n1 n1l2 − n2l1 l1m2 − l2m1 Again, let y be an arbitrary element in B, then f(x) = y Hence, the direction cosines of the line perpendicular 3y − 2 x −2 to the given lines are proportional to (m1n2 – m2n1), ⇒ =y ⇒ x= y −1 x −3 (n1l2 – n2l1), (l1m2 – l2m1). Class 12

54

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3y − 2 ∈ A, thus for all y ∈B, y −1 there exists x ∈A such that 3y − 2 −2 3 2 − y −1 y   =y = f (x ) = f   y − 1  3 y − 2 − 3 y −1 Thus, every element in the co-domain B has its pre-image in A, so f is a surjective. Hence, f : A → B is bijective. Clearly, ∀ y ∈B, x =

OR  x 2 , x ≥ 0 f (x ) = x | x | =  2 − x , x < 0 y The graph shows f(x) is one-one, as any straight line parallel to x-axis cuts only O at one point. y = –x2 Here, range of f(x) ∈ [–1, 1]. Thus range = co-domain. Hence, f(x) is onto. Therefore f(x) is one-one and onto, i.e, bijective.

33. We have, =∫

tan x

∫ 1 + tan x + tan2 x dx

1 + tan x − 1 1 + tan x + tan 2 x

dx = ∫

1 + tan x + tan 2 x − sec 2 x 1 + tan x + tan 2 x

  sec 2 x sec 2 x dx x = ∫ 1 − = −  ∫ 1 + tan x + tan2 x dx  1 + tan x + tan 2 x  = x−∫

1 1+ t + t

2

dt (Putting tan x = t ⇒ sec 2 xdx = dt )

1   t+  2  +C tan −1  =x− = x−∫  2 2 3 /2  1  3   3 /2    t + 2  +  2    dt

y = x2 x

32. Let dimensions of the rectangle be x and y (as shown). \ Perimeter of window, x πx ...(i) P = 2y + x + p x/2 = 10 ⇒ y = 5 − − 2 4

1

 2 tan x + 1  tan −1   +C  3 3 

2

=x−

\ K = 2, A = 3 2 2 34. We have, y dx + (xy + x )dy = 0 ⇒

Put y = vx ⇒ \ v+x

dy dv =v+x dx dx

 2v 2 + v  dv dx  v +1  = − ⇒ ∫ dv = − ∫  v + 1   v (2v + 1)  dx x

⇒ x

1

1 

⇒

2  x πx  1 x ⇒ A = x 5 − −  + π  2 4  2 4

1 ⇒ log v − log 2v + 1 + log x = log c 2

2

\

2

dA πx = 5− x − dx 4

Now,

dA 20 =0 ⇒ x= dx 4+π

d2A

 π x = − 1 +  < 0 A B   4 dx O Thus, A is maximum for y 20 x= 4+π C D 10 m From (i), y = 4+π 20 10 So, x = m, y = m will give maximum light. 4+π 4+π 2

− y2 dy = dx xy + x 2

−v 2 x 2 dv dv −v 2 = 2 −v ⇒ x = dx vx + x 2 dx v + 1

1 x2 Area of window, A = xy + π 2 4

x πx = 5x − − 2 8

dx

∫  v − 2v + 1  dv = − log x + log c

⇒ log

v2x2 v2x2 2 2 = log c ⇒ =c 2v + 1 2v + 1

 2 2  2y ⇒ y = c  + 1 ⇒ xy2 = c2(x + 2y)   x OR  dy  We have, ln   = 3x + 4 y  dx  dy = e 3 x ⋅ e 4 y ⇒ e −4 y dy = e 3 x dx ⇒ dx On integration, we get −

1 −4 y e 3 x = +C e 4 3

At x = 0, y = 0; we have 1 1 7 − = +C ⇒ C =− 4 3 12 55

Mathematics

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e −4 y e 3 x 7 + = ⇒ 3e −4 y + 4e 3 x = 7 4 3 12 −(x − 3), if x < 3 35. We have, f (x ) =   x − 3, if x ≥ 3 Test for continuity : L.H.L. = lim f (x ) = lim − (x − 3) = – (3 – 3) = 0 ∴ Solution is

x → 3−

x →3

R .H.L. = lim f (x ) = lim (x − 3) = 3 − 3 = 0 x →3

x → 3+

Also, f(3) = 3 – 3 = 0 \ L.H.L. = R.H.L. = f(3) Hence, f(x) is continuous at x = 3. Test for differentiability : f (3 − h) − f (3) Lf ′(3) = lim −h h→0 −(3 − h − 3) − 0 = lim = −1 −h h→0 f (3 + h) − f (3) lim (3 + h − 3) − 0 Rf ′(3) = lim = h→0 =1 h h h→0 Thus, Lf ′(3) ≠ Rf ′(3) Hence, f(x) is not differentiable at x = 3. 36. First we draw the lines whose equations are x + 2y = 50, 2x – y = 0 and 2x + y = 100 respectively. (0, 100)

\ z has minimum value 50 at two consecutive vertices B and C. \ z has minimum value 50 at every point of segment joining the points B(0, 25) and C(10, 20). Hence, there are infinite number of optimal solutions. OR Convert the inequations into equations and draw the corresponding lines. x + y = 6, x = 4, y = 6 As x, y ≥ 0, the solution lies in the first quadrant.

We have seen that O, A, B, C are the corner points. Hence maximum value of the objective function z will occur at one of the corner points. B is the point of intersection of the lines x + y = 6 and x = 4 i.e., B (4, 2) We have points A(4, 0), B(4, 2) and C(0, 6) Now, z = 11x + 8y \ z(A) = 11(4) + 8(0) = 44 z(B) = 11(4) + 8(2) = 60 z(C) = 11(0) + 8(6) = 48 z(O) = 11(0) + 8(0) = 0 \ z has maximum value 60 at B(4, 2).

(25, 50)

37. We have, | A | = (0, 25)

(10, 20)

1 tan x = 1 + tan 2 x ≠ 0 − tan x 1

So, A is invertible. (50, 0)

The feasible region is BCPDB which is shaded in the figure. The vertices of the feasible region are B(0, 25), C(10, 20), P(25, 50) and D(0, 100). The values of the objective function z = x + 2y at these vertices are given below. Corner points Value of z = x + 2y B(0, 25) 50 (minimum) C(10, 20) 50 (minimum) P(25, 50) 125 D(0, 100) 200

T

tan x  − tan x   1  1 = ∴ adjA =   1  1   − tan x  tan x 1 −1 adj A Now, A = |A| ⇒

1   2 A −1 =  1 + tan x  tan x  1 + tan2 x

− tan x   1 + tan2 x  1  2  1 + tan x 

1   − tan x  1 + tan 2 x  1 ∴ AT A −1 =    tan x tan 1 x     1 + tan 2 x

− tan x   1 + tan 2 x  1  2  1 + tan x  Class 12

56

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⇒

 1 − tan 2 x  2 T −1  1 + tan x A A =  2 tan x   1 + tan 2 x

⇒

cos 2 x − sin 2 x  AT A −1 =  .  sin 2 x cos 2 x 

i.e. parallel to the lines whose direction ratios are 1 –3, 4, 1 and 1, –2, i.e., –3, 4, 1 and 2, –4, 1 2 \ The equation of the required plane is     (r − a ) ⋅ (b × c ) = 0 .

−2 tan x   2 1 + tan x  2 1 − tan x  1 + tan 2 x 

x −2 ⇒ −3 2

OR 2 0 1   A = 0 −1 2 1 0 1 |A| = 2(–1 – 0) – 0(0 – 2) + 1(0 + 1) = –2 + 1 = –1 ≠ 0 So, A–1 exists.  −1 0 1  ∴ adj A =  2 1 −4   1 0 −2  A

−1

 −1 0 1   1 0 −1 1 = (adj A) = − 1  2 1 −4  =  −2 −1 4  |A|  1 0 −2   −1 0 2 

 −1 0 1   −1 0 1   Now, (adj A) = 2 1 −4   2 1 −4      1 0 −2   1 0 −2  2

 2 0 −3 =  −4 1 6   −3 0 5  38. The required plane passes through the point with ^ ^  position vector a = 2 i − k i.e., the point (2, 0, –1) y − 2 z +1 x and is parallel to the lines = = and −3 4 1 x − 4 y −1 z = = −2 1 / 2 1

y −0 4 −4

z − (−1) 1 =0 1

⇒ (x – 2) (4 + 4) – y(–3 – 2) + (z + 1) (12 – 8) = 0 ⇒ 8(x – 2) + 5y + 4(z + 1) = 0 ⇒ 8x + 5y + 4z – 12 = 0. ^ ^  ^ Its vector equation is r . (8 i + 5 j + 4 k ) − 12 = 0 OR Let M be the foot of the perpendicular drawn from the point P(2, 4, –1) to the given line. The coordinates of any point on the line x +5 y +3 z −6 = = are M(l – 5, 4l – 3, –9l + 6) 1 4 −9 Direction ratios of PM are l – 7, 4l – 7, –9l + 7 The direction ratios of the given line are 1, 4, –9 Since PM is perpendicular to the given line. \ 1(l – 7) + 4(4l – 7) – 9(–9l + 7) = 0 ⇒ 98l – 98 = 0 ⇒ l = 1 Putting l = 1, we have M ≡ (–4, 1, –3) Now, equation of PM = equation of the perpendicular from P to the given line x − 2 y − 4 z +1 i.e., = = −4 − 2 1 − 4 −3 + 1 i.e.,

x − 2 y − 4 z +1 = = 6 3 2



57

Mathematics

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Self Evaluation Sheet Once you complete SQP-4, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Matrices / Matrices Vector Algebra Integrals / Integrals Three Dimensional Geometry Relations and Functions / Relations and Functions Determinants Three Dimensional Geometry / Three Dimensional Geometry Matrices Probability / Probability Relations and Functions Vector Algebra Integrals Probability Differential Equations Three Dimensional Geometry Relations and Functions Application of Derivatives Probability

19

Continuity and Differentiability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Inverse Trigonometric Functions Application of Derivatives Vector Algebra / Vector Algebra Application of Integrals Probability Continuity and Differentiability Determinants Three Dimensional Geometry / Three Dimensional Geometry Continuity and Differentiability Application of Integrals Relations and Functions / Relations and Functions Application of Derivatives Integrals Differential Equations / Differential Equations Continuity and Differentiability Linear Programming / Linear Programming Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

5

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

1.

Relations and Functions

2.

Inverse Trigonometric Functions

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)#

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

–

–

–

2(2)

1(2)

–

1(5)*

3(8)

–

1(2)

2(6)#

–

3(8)

3.

Matrices

2(2)#

4.

Determinants

1(1)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)*

1(3)*

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)

1(2)

1(3)

–

3(6)

10.

Vector Algebra

3(3)#

1(2)

–

–

4(5)

11.

Three Dimensional Geometry

2(2)

1(2)

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

1(2)*

–

–

4(8)

10(20)

7(21)

3(15)

38(80)

13.

2(2)#

Probability Total

+ 1(4)

18(24)

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-5

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. A random variable X has the following distribution. X 1 2 3 4 5 P(X) 0.15 0.23 0.12 0.10 0.20

6 0.08

7 0.07

8 0.05

For the event E = {X is a prime number}, find P(E). OR 1 5 If A and B are two events such that P(A|B) = p, P(A) = p, P(B) = and P(A ∪ B) = , then find the value of p. 3 9 2. If A and B are the points (– 3, 4, – 8) and (5, – 6, 4) respectively, then find the ratio in which yz-plane divides AB. cos α − sin α  3. If A =   , then for what value of a, A is an identity matrix?  sin α cos α  OR

i  −i  i 0 −i      i  and Q =  0 0 , then find the matrix PQ. If P =  0 −i  i −i   −i i 0 Class 12

60

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 4. Find the distance of the point (2, 3, 4) from the plane r ⋅ (3ɵi − 6ɵj + 2kɵ ) + 11 = 0. 5. Evaluate :

Evaluate :

∫

dx x+3x

OR

cot x

∫ 3 sin x dx

56a 2 6. Prove that the area bounded by the parabola y2 = 4ax and the line x = a and x = 4a is sq. units. 3   7. If the position vector a of a point (12, n) is such that | a | = 13, then find the value of n . OR Find the projection of vector 7i + j − 4k on the vector 2i + 6 j + 3k .

8. Find the value of l for which the vectors 3i − 6 j + k and 2i − 4 j + λ k are parallel. 9. The number of bijective functions from the set A to itself, if A contains 108 elements is n!. Find the value of n. OR If the set A contains 5 elements and the set B contains 6 elements, then find the number of one-one and onto mappings from A to B. 10. Find the order of the differential equation whose general solution is given by y = (A + B) cos (x + C) + Dex. 11. Construct a 2 × 3 matrix whose elements aij are given by aij = 2i – 3j. 12. The random variable X has the following probability distribution : X 0 1 2 3 4 5 P(X) 0.1 k 0.1 3k 0.3 k Find the value of k. | x − 1| . (x − 1) 6 − x 4  14. For what value of x, matrix A =   is a singular matrix? 3 − x 1  13. Find the range of the function f (x ) =

15. Let R be a relation on the set N be defined by {(x, y) : x, y ∈N, 2x + y = 41}. Show that R is neither reflexive nor symmetric. 3 −x y+2 z+2 = = . 3 −2 6 Section - II

16. Write the direction cosines of a line parallel to the line

Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. In a school, a football game is to be organised between students of class 11th and 12th. For which, a team from each class is chosen, say T1 be the team of class 11th and T2 be the team class 12th. These teams have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probability of T1 winning, drawing and 1 1 1 losing a game against T2 are , and respectively. 2 6 3 61

Mathematics

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Each team gets 3 points for a win, 1 point for a draw and 0 point for a loss in a game. Let X and Y denote the total points scored by team T1 and T2, respectively, after two games. Based on the above information answer the following : (i) P(T2 winning a match against T1) is equal to (a) 1/2 (b) 1/6

(c) 1/3

(d) none of these

(ii) P(T2 drawing a match against T1) is equal to (a) 1/2 (b) 1/3

(c) 1/6

(d) 2/3

(iii) P(X > Y) is equal to (a) 1/4

(b) 5/12

(c) 1/2

(d) 7/12

(iv) P(X = Y) is equal to (a) 11/36

(b) 1/3

(c) 13/36

(d) 1/2

(c) 13/36

(d) 7/12

(c) N(P – 500)

(d) none of these

(v) P(X + Y = 12) is equal to (a) 0 (b) 5/12 18. Mr. Vinay is the owner of apartment complex with 50 units. When he set rent at ` 8000/month, all apartments are rented. If he increases rent by ` 250/month, one fewer apartment is rented. The maintenance cost for each occupied unit is ` 500/month. Based on the above information answer the following : (i) If P is the rent price per apartment and N is the number of rented apartment, then profit is given by (a) NP (b) (N – 500)P

(ii) If x represent the number of apartments which are not rented, then the profit expressed as a function of x is (a) (50 – x) (30 + x) (b) (50 + x) (30 – x) (c) 250(50 – x) (30 + x) (d) 250(50 + x) (30 – x) (iii) If P = 8500, then N = (a) 50

(b) 48

(iv) If P = 8250, then the profit is (a) ` 379750 (b) ` 4,00,000

(c) 49

(d) 47

(c) ` 4,05,000

(d) ` 4,50,000

(v) The rent that maximizes the total amount of profit is (a) ` 5000 (b) ` 10500 (c) ` 14800

(d) ` 14500

PART - B Section - III 2 x − 1, x < 2  x = 2 is continuous at x = 2.  x + 1, x > 2 

19. Find the value of ‘a’ if the function f(x) defined by f (x ) =  a, 20. Evaluate : ∫ x 2 (ax + b)−2dx OR Evaluate :

sin 2 x

∫ a2 sin2 x + b2 cos2 x dx Class 12

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21. Find the area bounded by the line y = x, x-axis and lines x = –1 to x = 2.  3 1 –1 22. If A =   , find A .  −1 2 23. A machine produces parts that are either good (90%), slightly defective (2%), or obviously defective (8%). Produced parts get passed through an automatic inspection machine, which is able to detect any part that is obviously defective and discard it. What is the probability of the parts that make it through the inspection machine and get shipped? OR If A and B are two independent events such that P ( A ∩ B ) =

2 1 and P ( A ∩ B ) = , then find P (B) – P(A). 15 6

4  3 24. Simplify : cos −1  cos x + sin x   5 5 25. Find the points on the line

x + 2 y +1 z − 3 = = at a distance of 5 units from the point P(1, 3, 3). 3 2 2

3 26. Find the solution of differential equation y −

dy dy = x2 . dx dx

  27. If a = −3i + nj + 4k and b = −2i + 4 j + pk are collinear, then find the value of n and p. 28. Find the equation of the tangent to the curve y = 3x2 – x + 1 at P (1, 2). OR Show, that the function f (x) = x9 + 4x7 + 11 is increasing on R. Section - IV 29. Show that f(x) = [x] is not differentiable at x = 1. 30. Find the area bounded by the X-axis, part of the curve y = 1 +

8 x2

and the ordinates at x = 2 and x = 4.

31. Find the equation of normal to the curve 16x2 + 9y2 = 144 at (2, y1) where y1 > 0. OR Find the interval on which the function f (x ) =

3 4 4 3 36 x − x − 3x 2 + x + 11 is increasing. 10 5 5

dy + y cot x = 2 cos x. dx 33. Show that the relation R in the set A = {1, 2, 3, 4, 5, 6} given by R = {(a, b) : |a – b| is divisible by 2} is an equivalence relation. 32. Solve the differential equation

34. Show that the exponential function ax is (where a > 0) continuous at every point. OR  (3sin x − 1)2 , for x ≠ 0  is continuous or not. Check whether the function f (x ) =  x ⋅ log(1 + x ) 2 log 3, for x = 0  35. Evaluate : ∫

dx 1 − cos x − sin x 63

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Section-V 36. An amount of ` 5000, is put into three investments at the rate of interest of 6%, 7% and 8% per annum respectively. The total annual income is ` 358. If the combined income from the first two investments is ` 70 more than the income from the third, find the amount of each investment by matrix method. OR Express the following matrix as the sum of a symmetric matrix and a skew-symmetric matrix and verify your  3 −2 −4   3 −2 −5 result :    −1 1 2 x −2 y +1 z +3 = = . 37. Find the image of the point (1, – 2, 1) in the line 3 −1 2 OR ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^   The lines r = (2 j − 3 k) + λ (m i + 2 j + 3 k) and r = (2 i + 6 j + 3 k) + µ (2 i + 3 j + 4 k) are coplanar. Find the value of m.

38. Find the maximum value of Z = 4x + 6y subject to constraints 3x + 2y ≤ 12, x + y ≥ 4, x ≥ 0, y ≥ 0. OR Find the number of points at which he objective function Z = 4x + 3y can be maximized subject to the constraints 3x + 4y ≤ 24, 8x + 6y ≤ 48, x ≤ 5, y ≤ 6, x ≥ 0, y ≥ 0.

Class 12

64

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SOLUTIONS 1. P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 OR 5 1 and P ( A ∪ B) = 9 3 P ( A ∩ B) p Now, P ( A |B) = = p ⇒ P ( A ∩ B) = P ( B) 3

We have, P ( A) = p, P (B) =

Since, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 5 − 3 2p 1 p 5 2 3 1 ⇒ = p+ − ⇒ = ⇒ p= × = 9 3 3 9 3 9 2 3   2. Let a = − 3ɵi + 4 ɵj − 8kɵ , b = 5ɵi − 6 ɵj + 4 kɵ  Let C (c ) be the point in yz-plane which divides AB in the ratio r : 1. 5r − 3 Then, 0 = ( In yz-plane, x = 0) r +1 3 ⇒ 5r − 3 = 0 ⇒ r = 5 \ Required ratio is 3 : 5. cos α − sin α  3. Given, A =   . Now, A is an identity  sin α cos α  cos α − sin α  1 0 matrix then,  =   sin α cos α  0 1 ∴ cos α = 1 and sin α = 0 ⇒ α = 0.

OR  −i i   i 0 −i    Since, P = 0 −i i and Q =  0 0       i −i   −i i 0   i 0 −i   −i i  ∴ PQ =  0 −i i   0 0   −i i 0   i −i 

9 + 36 + 4

=

6 − 18 + 8 + 11 7

= 1 unit

dx

x+3x Put x = t6 ⇒ dx = 6t5 dt 6t 5 t3 1  dt dt = 6∫  t 2 − t + 1 − ⇒ I = ∫ 3 2 dt = 6∫  t 1 + t 1 + t +t = 2t3 – 3t2 + 6t – 6log(t + 1) + C = 2 x − 3(3 x ) + 6(6 x ) − 6 log(6 x + 1) + C. OR cos x

cot x dx = ∫ 1/3 dx Let I = ∫ 3 sin x sin x ⋅ sin x cos x =∫ dx = ∫ sin −4/3 x ⋅ cos xdx sin 4/3 x Put sin x = t ⇒ cos x dx = dt −3 t −1/3 +C = +C 3 −1 / 3 sin x 6. Required area = 2 × area of region PSRQP ⇒ I = ∫ t −4/3 dt =

4a

=2∫ a

4a

 x 3/ 2  4ax dx = 4 a    3 / 2 a

8 ( 3/ 2 3/ 2 ) a 8a − a = 3 56a 2 = sq. units 3 7. The position vector of the point (12, n) is 12i + nj.   (Given) ∴ a = 12ɵi + njɵ ⇒ | a | = 122 + n2 = 13 2 2 2 ⇒ 12 + n = 169 ⇒ n = 25 ⇒ n = ± 5 OR    a ⋅b  Projection of a on b =  |b | =

 −i 2 − i 2 i 2 + i 2   1 + 1 −1 − 1  2 −2   =  i2 −i 2  =  −1 1  =  −1 1   2  1   −1 1  −i 2   −1  i  4. Here, a = 2ɵi + 3ɵj + 4kɵ The distance of the point (2i + 3j + 4k ) is (2i + 3j + 4k ) ⋅ (3i − 6 j + 2k ) + 11

5. Let I = ∫

(7i + j − 4k ) ⋅ (2i + 6 j + 3k ) (2)2 + (6)2 + (3)2

=

14 + 6 − 12 8 = 7 7

  8. a = 3ɵi − 6 ɵj + kɵ and b = 2ɵi − 4 ɵj + λ kɵ      Since, a and b are parallel ∴ a × b = 0 ɵi ɵj kɵ  ⇒ 3 −6 1 = 0 2 −4 λ  ⇒ (− 6 λ + 4)ɵi − (3λ − 2)ɵj + (− 12 + 12)kɵ = 0 ⇒ (− 6 λ + 4)i + (2 − 3λ)j = 0i + 0 j Comparing coefficients of i and j , we get

–6l + 4 = 0 and 2 – 3l = 0 ⇒ l = 2/3

65

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9. Since number of one-one onto functions from a set A having n elements to itself is n!. \ Required value of n is 108. OR As A contains 5 elements. \ For any one-one onto mapping f : A → B, f (A) also contains 5 elements but B contains 6 elements. \ f (A) ≠ B. So, no one-one mapping from A to B can be onto. 10. Given y = (A + B) cos (x + C) + Dex or y = k cos (x + C) + Dex Now order of a differential equation is same as the number of arbitrary unknowns present in the solution. Hence order of differential equation is 3. 11. Here aij = 2i – 3j, i = 1, 2 and j = 1, 2, 3 \ a11 = 2⋅1 – 3⋅1 = –1, a12 = 2⋅1 – 3⋅2 = – 4, a13 = 2⋅1 – 3⋅3 = – 7, a21 = 2⋅2 – 3⋅1 = 1, a22 = 2⋅2 – 3⋅2 = – 2, a23 = 2⋅2 – 3⋅3 = – 5  −1 −4 −7  Hence, the required matrix is  . − − 1 2 5   12. 

5

∑ P( X = x) = 1 x=0

⇒ P(X = 0) + P(X = 1) + .... + P(X = 5) = 1 ⇒ 0.1 + k + 0.1 + 3k + 0.3 + k = 1 ⇒ 0.5 + 5k = 1 ⇒ 5k = 1 – 0.5 = 0.5 ⇒ k = 0.1  x − 1, x ≥ 1 1 − x , x < 1

13. We have, x − 1 =  \ f (x ) =

x − 1  1, x ≥ 1 = (x − 1) −1, x < 1

\ Range (f ) = {– 1, 1} 14. Matrix A is singular, when |A| = 0 6−x 4 =0 3− x 1

⇒ 6 – x – 12 + 4x = 0 ⇒ 3x = 6 ⇒ x = 2 15. R = {(x, y) : x, y ∈N, 2x + y = 41} Reflexive : (1, 1) ∉ R as 2 ⋅ 1 + 1 = 3 ≠ 41. So, R is not reflexive. Symmetric : (1, 39) ∈R but (39, 1) ∉ R. So R is not symmetric. x −3 y +2 z +2 = = −3 −2 6 ⇒ Direction ratios are – 3, – 2, 6. −3 −2 6 \ Direction cosines are , , . 7 7 7 These are direction cosines of a line parallel to given line. 16. We have,

17. (i) (c) : Clearly, P(T2 winning a match against T1) 1 = P(T1 losing) = 3 (ii) (c) : Clearly, P(T2 drawing a match against T1) 1 = P(T1 drawing) = 6 (iii) (b) : According to given information, we have the following possibilities for the value of X and Y. X Y

6 0

4 1

3 3

2 2

1 4

0 6

Now, P(X > Y) = P(X = 6, Y = 0) + P(X = 4, Y = 1) = P(T1 win) P(T1 win) + P(T1 win) P(match draw) + P(match draw) P(T1 win) 1 1 1 1 1 1 3 +1+1 5 = ⋅ + ⋅ + ⋅ = = 2 2 2 6 6 2 12 12 (iv) (c) : P(X = Y) = P(X = 3, Y = 3) + P(X = 2, Y = 2) = P(T1 win) P(T2 win) + P(T2 win) P(T1 win) + P(match draw) P(match draw) 1 1 1 1 1 1 1 1 1 13 = ⋅ + ⋅ + ⋅ = + + = 2 3 3 2 6 6 6 6 36 36 (v) (a) : From the given information, it is clear that maximum sum of X and Y can be 6, therefore P(X + Y = 12) = 0 18. (i) (c) : If P is the rent price per apartment and N is the number of rented apartment, the profit is given by P(N) = NP – 500 N = N(P – 500) [Q ` 500/month is the maintenance charges for each occupied unit] (ii) (c) : Now, if x be the number of non-rented apartments, then N = 50 – x and P = 8000 + 250 x Thus, P = N(P – 500) = (50 – x) (8000 + 250 x – 500) = (50 – x) (7500 + 250 x) = 250(50 – x) (30 + x) (iii) (b) : Clearly, if P = 8500, then 8500 = 8000 + 250 x ⇒ x = 2 ⇒ N = 48 (iv) (a) : Also, if P = 8250, then 8250 = 8000 + 250 x ⇒ x = 1 and so profit P(1) = 250(50 – 1) (30 + 1) = ` 379750 (v) (b) : We have, P(x) = 250(50 – x) (30 + x) Now, P′(x) = 250[50 – x – (30 + x)] = 250[20 – 2x] For maxima/minima, put P′(x) = 0 ⇒ 20 – 2x = 0 ⇒ x = 10 Thus, price per apartment is, P = 8000 + 2500 = 10500 Hence, the rent that maximizes the profit is ` 10500. 19. For f to be continuous at x = 2, we must have lim f ( x ) = f (2) = lim f ( x )

x → 2−

x → 2+

Now, f(2) = a

...(i) ...(ii)

lim f (x ) = lim f (2 − h) = lim[2(2 − h) − 1] = 3

x → 2−

h →0

h →0

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OR Since A and B are independent events, therefore, Ac and B are independent and also A and Bc are independent. \ P (A ∩ B) = P(Ac ∩ B) = P (Ac) P(B) = (1 – P(A)) P(B) and P(A ∩ B ) = P(A ∩ Bc) = P(A) P (Bc) = P (A) (1 – P(B)) ( A = Ac and B = Bc) 2 ...(i) ⇒ (1 – P (A)) P (B) = P ( A ∩ B) = 15 1 and P (A) (1 – P (B)) = P ( A ∩ B ) = ...(ii) 6 Subtracting (ii) from (i), we obtain 4−5 2 1 ⇒ P (B) – P (A) = P (B) – P (A) = − 30 15 6 1 ⇒ P (B) – P (A) = − 30

lim f (x ) = lim f (2 + h) = lim[(2 + h) + 1] = 3 h →0

x → 2+

h →0

\ From (i) and (ii), we get a = 3 20. Let I = ∫

x2

dx (ax + b)2 1 Put ax + b = t ⇒ dx = dt a 1 (t − b)2 1  b 2 2b  dt ∴ I= 3∫ = ∫ 1 + −  dt a t2 a3  t 2 t  2 1   = 3  t − b − 2b log t  + C  a  t =

1   b2 − 2b log(ax + b) + C + − ax b  3 a ax + b OR sin 2x

Let I = ∫

a 2 sin 2 x + b 2 cos 2 x Let a2 sin2x + b2cos2x = t

2

dx

⇒ (a 2 − b2 )sin 2 x dx = dt 1 1 1 \ I= dt = 2 2 log t + C 2 2 ∫t (a − b ) (a − b ) 1 ⇒ I = 2 2 log a 2 sin 2 x + b 2 cos 2 x + C (a − b ) 21. We have, y = x

= cos–1(cosq cosx + sinq sinx) 4 = cos −1 {cos( x − θ)} = x − θ = x − tan −1   3

y y=x x′

25. Any point on the line

x

O x=2 x = –1 y′

\ Required area = area of shaded region 0

2

x2 = ∫ x dx + ∫ x dx = 2 −1 0

0

x2 + 2 −1

2

0

1 5 1 + 2 = 2 + = sq. units 2 2 2 22. |A| = 6 + 1 = 7 ≠ 0, \ A–1 exists. 2 −1 1 2 −1 −1 (adj A) =   ⇒ A =   1 3 7 1 3    = −

23. Let G, SD, OD be the events that a randomly chosen part is good, slightly defective, obviously defective respectively. Then, P(G) = 0.90, P(SD) = 0.02, and P(OD) = 0.08 Required probability = P(G | ODc) =

P (G ∩ OD c ) P (OD c )

=

2

3 4 24. Here,   +   = 1 5 5 3 4 4 Put = cos θ and = sin θ ⇒ tan θ = 5 5 3 4 ⇒ θ = tan −1   3 4  3 ∴ cos −1  cos x + sin x   5 5

0.90 90 P (G) = = = 0.978 1 − P (OD) 1 − 0.08 92

x +2 3

=

y +1 z − 3 = = λ(say) is 2 2

of the form Q(3l – 2, 2l – 1, 2l + 3) ...(i) Now, distance PQ, where P is (1, 3, 3), is 5. So, (3l – 2 – 1)2 + (2l – 1– 3)2 + (2l + 3 – 3)2 = 52 ⇒ 9l2 + 9 – 18l + 4l2 + 16 – 16l + 4l2 = 25 ⇒ 17l2 – 34l = 0 ⇒ 17l(l – 2) = 0 ⇒ l = 0 or l = 2 Putting values of l in (i), the required points are (– 2, – 1, 3) and (4, 3, 7). dy dy dy = x2 ⇒ y 3 = (1 + x 2 ) 26. We have, y 3 − dx dx dx dy dx − 1 = ∫ 3 + c ⇒ tan −1 x = 2 + c ⇒ ∫ 1+ x2 y 2y   27. We have, a and b are collinear.   ∴ a = λb ⇒ − 3i + n j + 4k = λ(−2i + 4 j + pk ) 3 ⇒ λ= 2 3 Also, n = 4 λ ⇒ n = 4 × = 6 2 2 8 And, λp = 4 ⇒ p = 4 × = 3 3 67

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28. y = 3x2 – x + 1 is the given curve. Differentiating w.r.t. x, we have dy  dy  = 6(1) − 1 = 5 = 6x − 1 ∴    dx  x = 1 dx ⇒ The equation of tangent is (y – 2) = 5(x –1) ⇒ 5x – y – 3 = 0

 4 5 \ Equation of the normal at  2,  is  3  4 5 y− 3 = 3 5 ⇒ 3 5x − 6 5 = 8 y − 32 5 x−2 8 3 ⇒ 9 5x − 18 5 = 24 y − 32 5

OR Here, f(x) = x9 + 4x7 + 11 \ f ′(x) = 9x8 + 28x6 = x6(9x2 + 28) > 0 for all x ∈R Thus, f(x) is increasing on R. 29. We have, f(x) = [x] f (1 + h) − f (1) h h →0 [1 + h] − [1] (Q [1 + h] = 1 and [1] = 1) = lim =0 h h→0 f (1 − h) − f (1) and Lf′ (1) = lim −h h →0

Rf ′(1) = lim

= lim

h→0

[1 − h] − [1] =∞ −h

(Q [1 – h] = 0 and [1] = 1).

Thus Rf ′ (1) ≠ Lf ′ (1). Hence, f(x) = [x] is not differentiable at x = 1. 4



8 

∫ 1 + x 2  dx

30. Required area =

2

⇒ 9 5 x − 24 y + 14 5 = 0 OR 3 4 36 We have, f (x ) = x 4 − x 3 − 3x 2 + x + 11 10 5 5 6 3 12 2 36 ∴ f ′( x ) = x − x − 6 x + 5 5 5 f ′(x) > 0 as f(x) is increasing 6 12 36 ∴ x 3 − x 2 − 6x + > 0 5 5 5 ⇒ 6x3 – 12x2 – 30x + 36 > 0 ⇒ 6(x3 – 2x2 – 5x + 6) > 0 ⇒ 6(x – 1)(x – 3)(x + 2) > 0 The possible intervals are (–∞, –2), (–2, 1), (1, 3) and (3, ∞) For f(x) to be increasing, f ′(x) > 0. ⇒ (x – 1) (x – 3) (x + 2) > 0 ⇒ x ∈ (–2, 1) ∪ (3, ∞) –∞

Y

–

+ –2

– 1

+ 3

∞

So, f(x) is increasing on x ∈ (–2, 1) ∪ (3, ∞). y  1 X

8 X

x=4

O x=2

dy + y cot x = 2 cos x ...(i) dx This is a linear differential equation of the form dy + Py = Q , where P = cot x ; Q = 2 cos x dx P dx cot x dx = e log sin x = sin x = e∫ Now, I.F. = e ∫ 32. The given D.E. is

x2

Y

4

4  8 x −1   = x + 8 ×  = x−  x 2 −1  2   = (4 – 2) – (2 – 4) = 4 sq. units.

31. When x = 2 we have y =

\ The solution of equation (i) is given by y × I.F. = ∫ Q × I.F. dx + c

144 − 16(2)2 9

=

4 5 3

 4 5 So the point of contact is  2,   3  dy −32 x 144 − 16 x 2 ⇒ = dx 6 144 − 16 x 2 3 −32 (2) −32 −8  dy  = ⇒   4 5  = =  dx   2, 6 144 − 64 3 × 4 5 3 5   Now y =

3

⇒ Slope of the normal is

3 5 . 8

⇒ y sin x = ∫ 2 cos x ⋅ sin x dx + c 1 = ∫ sin 2 x dx + c = − cos 2 x + c 2 1 ⇒ y = − cos 2 x cosec x + c cosec x 2 This is the required solution of the given differential equation. 33. We have, A = {1, 2, 3, 4, 5, 6} and R = {(a, b) : |a – b| is divisible by 2} (i) Reflexive : For any a ∈ A |a – a| = 0, which is divisible by 2. Thus, (a, a) ∈ R. So, R is reflexive. Class 12

68

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(ii) Symmetric : For any a, b ∈ A Let (a, b) ∈ R ⇒ |a – b| is divisible by 2 ⇒ |b – a| is divisible by 2 ⇒ (b, a) ∈ R \ (a, b) ∈ R ⇒ (b, a) ∈ R \ R is symmetric. (iii) Transitive : For any a, b, c ∈ A Let (a, b) ∈ R and (b, c) ∈ R ⇒ |a – b| is divisible by 2 and |b – c| is divisible by 2. ⇒ a – b = ±2 k1 and b – c = ± 2k2; k1, k2 ∈ N ⇒ a – b + b – c = ±2(k1 + k2) ⇒ a – c = ±2 k3, k3 ∈ N ⇒ |a – c| is divisible by 2 ⇒ (a, c) ∈ R \ R is transitive. Hence, R is an equivalence relation. 34. Let f(x) = ax We have,

...(i)

  ax −1 × x + 1 lim a x = lim(a x − 1 + 1) = lim  x →0  x x →0 x →0  ax −1 = lim × lim x + lim(1) x →0 x x →0 x →0

= log a × 0 + 1 = 1 \ lim a = 1 x

x →0

Let c be an arbitrary real number. h→0

= lim a

c −h

h→0

c

= lim a ⋅ a h→0

−h

c

= a lim

1

h→0 a h

= ac ×1 = ac = f(c) [By (i)] Likewise lim f (x ) = a c x →c +

\

lim f (x ) = lim f (x ) = f (c)

x →c −

x →c +

\ f is continuous at x = c, where c is an arbitrary real number. \ f (x) = ax is continuous at every point. OR Given, f (0) = 2 log 3 sin x

lim f (x ) = lim

x →0

…(i) 2

− 1) (3 log(1 + x )

x →0 x

2

 3sin x − 1   sin x  2  sin x  .  x    = lim x →0 (1 / x ) log(1 + x )

(log 3)2 . (1)2

= (log 3)2 1 From (i) and (ii), lim f (x ) ≠ f (0) =

…(ii)

x →0

\ f is discontinuous at x = 0. dx 1 − cos x − sin x x x 1 − tan 2 2 tan 2 2 Since, cos x = and sin x = 2x 2x 1 + tan 1 + tan 2 2 1 2x sec dx dx 2 ∴ I =∫ =∫ 2 x x  2x 2x tan − tan 2 tan 1 − tan 2 2 2 2− 1−   2x 2x  1 + tan  1 + tan 2 2 x 1 2x Put tan = t ⇒ sec dx = dt 2 2 2 dt 1 1 = ∫  ∴ I=∫ 2 − dt  t − 1 t  t −t 35. Let I = ∫

= log(t − 1) − log t + C = log

Then lim f (x ) = lim f (c − h) x →c −

2

 3sin x − 1   sin x  2  sin x  .  x    = lim log(1 + x ) x →0 x

t −1 +C t

x tan − 1 x 2 = log + C = log 1 − cot + C x 2 tan 2 36. Let x, y and z be the investments at the rate of interest of 6%, 7% and 8% per annum respectively. Then, x + y + z = 5000. 6x , Income from investment of ` x, ` y and ` z is ` 100 7y 8z ` and ` respectively. 100 100  6x 7 y 8z  \ Total annual income = `  + + 100 100 100  6x 7 y 8z ⇒ + + = 358 100 100 100 ⇒ 6x + 7y + 8z = 35800 Also, by given condition 6x + 7 y = 70 + 8z 100 100 100 ⇒ 6x + 7y – 8z = 7000 So, we obtain the following system of linear equations : x + y + z = 5000 6x + 7y + 8z = 35800 69

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6x + 7y – 8z = 7000 The system of equations can be written in matrix form as: 1 1 1   x   5000      6 7 8   y  = 35800 or, AX = B 6 7 −8  z   7000      1 1 1  x   5000       where, A = 6 7 8  , X =  y  and B = 35800 6 7 −8 z         7000 1 1 1 8 = – 16 ≠ 0. Now, |A| = 6 7 6 7 −8 So, A–1 exist and the solution of the given system of equations is given by X = A–1 B.  −112 96 0 ′  −112 15 1   \ adj A =  15 −14 −1 =  96 −14 −2  1 −2 1  0 −1 1      −112 15 1 1 1   (adj A) = −  96 −14 −2 So, A–1 = 16 A  0 1 −1  Hence, the solution is given by  −112 15 1  5000 − 1    96 −14 −2 35800 X = A–1 B = 16   0 1  7000 −1  x   −560000 + 537000 + 7000 1000  1       ⇒  y  = −  480000 − 501200 − 14000  = 2200 16 z    1800  0 − 35800 + 7000       ⇒ x = 1000, y = 2200, z = 1800 Hence, three investments are of ` 1000, ` 2200 and ` 1800 respectively. OR  3 −2 −4    Let A =  3 −2 −5 ⇒ A′ =  −1 1 2

3 −1  3  −2 −2 1   −4 −5 2

3 −1  3 −2 −4   3  3 −2 −5 +  −2 −2 1 \ A + A′ =     −1 1 2  −4 −5 2 1 −5  6  1 −4 −4  =    −5 −4 4 

1 5   3 2 − 2  1  1  −2 −2 ⇒ (A + A′) = 2  2   5  2  − 2 −2  which is clearly a symmetric matrix and 3 −1  3 −2 −4   3  3 −2 −5 −  −2 −2 1 A – A′ =     −1 1 2  −4 −5 2 0 −5 −3   =  5 0 −6  3 6 0 5 3   0 − 2 − 2 5  1 ⇒ (A − A′) =  0 −3 2 2  3  3 0  2  which is clearly a skew symmetric matrix. 1 1 Since A = (A + A′) + (A − A′) 2 2 5 3 1 5    3 2 − 2  0 − 2 − 2  1  5  \ A= −2 −2 +  0 −3  2  2   5  3  3 0 2   − 2 −2  2  Thus A has been expressed as the sum of a symmetric matrix and a skew-symmetric matrix. 37. Let P(1, –2, 1) be the given point and let Q be the foot of the perpendicular drawn from P on of the given line x −2 y +1 z +3 = = = λ (say) 3 −1 2 ⇒ x = 3l + 2, y = – l – 1, z = 2l – 3 Let the coordinates of Q be (3l + 2, – l – 1, 2l – 3) ...(i) So, direction ratios of PQ be (3l + 2 – 1, – l – 1+ 2, 2l – 3 – 1) i.e., (3l + 1, – l + 1, 2l – 4) Since PQ is perpendicular to given line. \ 3(3l + 1) – 1(– l + 1) + 2(2l – 4) = 0 ⇒ 9l + 3 + l – 1 + 4l – 8 = 0 ⇒ l = 3/7 Class 12

70

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Putting l = 3/7 in (i), we get the coordinates of Q as  23 10 15   7 , − 7 , − 7  Let R (x1, y1, z1) be the image of P(1, –2, 1) and as Q is the mid point of PR. x + 1 23 y1 − 2 10 z + 1 15 \ 1 = , =− , 1 =− 2 7 2 7 2 7 39 6 37 ⇒ x1 = , y1 = − , z1 = − 7 7 7 Hence, image of P(1, –2,1) in given line is  39 6 37   7 , − 7 ,− 7  . OR  ^ ^  ^ ^ ^ Here, a1 = 2 j − 3 k , b1 = m i + 2 j + 3 k  ^ ^ ^ ^ ^  ^ a2 = 2 i + 6 j + 3 k, b2 = 2 i + 3 j + 4 k   ^ ^ ^ a2 − a1 = 2 i + 4 j + 6 k ^

Now, Z = 4x + 6y Z(A) = 4(4) + 6(0) = 16 Z(B) = 4(0) + 6(6) = 36 Z(C) = 4(0) + 6(4) = 24 \ Z has maximum value 36 at B(0, 6). OR Converting inequations into equations and draw the corresponding lines 3x + 4y = 24, 8x + 6y = 48, x = 5, y = 6 x y x y i.e., + = 1, + = 1, x = 5, y = 6 6 8 8 6 As x, y ≥ 0, the solution lies in the first quadrant.

^

^

i j k   ^ ^ ^ b1 × b2 = m 2 3 = i (8 − 9) − j (4m − 6) + k (3m − 4) 2 3 4 ^

^

^

= − i − (4m − 6) j + (3m − 4) k

Since, given lines  are  coplanar   ∴ (a2 − a1 ) ⋅ (b1 × b2 ) = 0 ^

^

^

^

^

^

⇒ (2 i + 4 j + 6 k ) ⋅ (− i − (4m − 6) j + (3m − 4) k ) = 0 ⇒ (2)(–1) – 4(4m – 6) + 6(3m – 4) = 0 ⇒ –2 – 16m + 24 + 18m – 24 = 0 ⇒ 2m = 2 ⇒ m = 1 38. Converting inequations into equations and drawing the corresponding lines. x y x y 3x + 2y = 12, x + y = 4 i.e., + = 1, + =1 4 4 4 6 As x ≥ 0, y ≥ 0 solution lies in first quadrant. We have points A(4, 0), B(0, 6) and C(0, 4). Y

B is the point of intersection of the lines  4 8x + 6y = 48 and x = 5 i.e., B =  5,   3 C is the point of intersection of the lines 3x + 4y = 24 24 24  and 8x + 6y = 48 i.e. C =  ,  7 7   4   24 24  We have points O(0, 0) A(5, 0), B  5,  , C  ,  3   7 7  and D(0, 6). Now, Z = 4x + 3y \ Z(O) = 4(0) + 3(0) = 0 Z(A) = 4(5) + 3(0) = 20 4 Z(B) = 4(5) + 3   = 24 3

8

x+ y=

6 B

4

 24   24  Z(C) = 4   + 3   = 24 7  7 

4 C 2 A X

O

6

4

+ 3x = 2y

Y

2

8

X

12

Z(D) = 4(0) + 3(6) = 18 Z has maximum value at points B and C. Since both the points lie on the same line 8x + 6y = 48. \ Each point of the line 8x + 6y = 48 will give maximum value of Z. Therefore, objective function can be maximized at infinite number of points.

 71

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Self Evaluation Sheet Once you complete SQP-5, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Probability / Probability Vector Algebra Matrices / Matrices Three Dimensional Geometry Integrals / Integrals Application of Integrals Vector Algebra / Vector Algebra Vector Algebra Relations and Functions / Relations and Functions Differential Equations Matrices Probability Relations and Functions Determinants Relations and Functions Three Dimensional Geometry Probability Application of Derivatives

19

Continuity and Differentiability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Application of Integrals Determinants Probability / Probability Inverse Trigonometric Functions Three Dimensional Geometry Differential Equations Vector Algebra Application of Derivatives / Application of Derivatives Continuity and Differentiability Application of Integrals Application of Derivatives / Application of Derivatives Differential Equations Relations and Functions Continuity and Differentiability / Continuity and Differentiability Integrals Determinants / Matrices Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)*

–

3(9)

7.

Integrals

2(2)#

1(2)

1(3)

–

4(7)

8.

Application of Integrals

–

1(2)*

1(3)

–

2(5)

9.

Differential Equations

1(1)*

1(2)

1(3)

–

3(6)

10.

Vector Algebra

1(1) + 1(4)

–

–

–

2(5)

2(2)#

1(2)*

–

1(5)*

4(9)

–

–

–

1(5)*

1(5)

4(4)#

2(4)#

–

–

6(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

Probability Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-6

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Solve the differential equation

dy = 1 − x + y − xy. dx OR 2

 dy  d 2 y What is the degree of the differential equation 5x   − 2 − 6 y = log x ?  dx  dx  x + 3 y y   4 −1 2. If   =   , then find the values of x and y.  7 − x 4  0 4 

3. The random variable X has a probability distribution P(X) of the following form, where ‘k’ is some number, if x = 0  k, 2k , if x = 1  P( X = x) =  3k , if x = 2  0, otherwise Determine the value of ‘k’. Class 12

74

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OR Out of 8 outstanding students of a school, in which there are 3 boys and 5 girls, a team of 4 students is to be selected for a quiz competition. Find the probability that 2 boys and 2 girls are selected. 4. Check whether the relation R on the set A = {1, 2, 3} defined as R = {(1, 1), (1, 2), (2, 1), (3, 3)} is reflexive, symmetric and transitive. 5. Evaluate :

Evaluate :

x

∫ ∫

x2 − x

dx OR

cos 2 x + 2 sin2 x

dx cos 2 x x −1 y − 2 z − 3 x −1 y − 5 z − 6 6. If lines are mutually perpendicular, then find the value of k. = = and = = −3 2k 2 3k 1 −5 ^

7. Find the vector equation of a plane which is at a distance of 6 units from the origin and which has k as the unit vector normal to it. OR Find the vector equation of the plane which is at a distance of vector from the origin is 2i − 3j + 4k .

6 units from the origin and its normal 29

8. Prove that for any square matrix A, AAT is a symmetric matrix. 9. If ∆(x ) =

f (x ) g (x ) , then prove that ∫ ∆(x )dx = a b

∫ f (x )dx ∫ g (x )dx . a

b

OR If A is invertible matrix of order 3 × 3, then prove that |A–1| = |A|–1.

10. Two dice are thrown together. What is the probability that the sum of the numbers on the two faces is neither 9 nor 11 ? 11. Let f : [2, ∞) → R be the function defined by f (x) = x2 – 4x + 5, then find the range of f. 12. If A and B are two events such that P(A) = 0.2 , P(B) = 0.4 and P(A ∪ B) = 0.5 , then find the value of P(A/B). π/ 4

13. Evaluate :

∫ (sec

2

x + cosec2 x ) dx

π /6

14. A bag contains 3 white and 6 black balls while another bag contains 6 white and 3 black balls. A bag is selected at random and a ball is drawn. Find the probability that the ball drawn is of white colour.        15. If a, b, c are the position vectors of points A, B, C respectively such that 5a − 3b − 2c = 0, then find the ratio in which C divides AB externally. 16. For real numbers x and y, we write xRy ⇔ x − y + 2 is an irrational number. Prove that the relation R is not transitive. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A building of a multinational company is to be constructed in the form of a triangular pyramid, ABCD as shown in the figure. 75

Mathematics

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A

C B

D

Let its angular points are A(3, 0, 1), B(–1, 4, 1), C(5, 2, 3) and D(0, –5, 4) and G be the point of intersection of the medians of DBCD. Based on the above answer the following. (i) The coordinates of points G are 1 1 1  1 1 (a)  , ,  (b)  0, ,  3 3 3  2 3  (ii) The length of vector AG is (a)

17 units

(b)

51 units 3

4 1 8 (c)  , ,  3 3 3

(c)

3 6

units

4 8 1 (d)  , ,  3 3 3

(d)

59 units 4

(iii) Area of triangle ABC (in sq. units) is (a) 24

(b) 8 6   (iv) The sum of lengths of AB and AC is (a) 4 units (b) 9.1 units

(c) 4 6

(d) 5 6

(c) 8.7 units

(d) 6 units

(v) The length of the perpendicular from the vertex D on the opposite face is (a)

14 6

units

(b)

2 6

units

(c)

3 6

units

(d) 8 6 units

18. A concert is organised every year in the stadium that can hold 42000 spectators. With ticket price of 10, the average attendance has been 27000. Some financial expert estimated that price of a x , ticket should be determined by the function p(x ) = 19 − 3000 where x is the number of tickets sold. Based on the above information, answer the following questions. (i) The revenue, R as a function of x can be represented as x2 (a) 19 x − 3000

x2 (b) 19 − 3000

(c) 19 x −

1 30000

(ii) The range of x is (a) [27000, 42000] (c) [0, 42000]

(b) [0, 27000] (d) none of these

(iii) The value of x for which revenue is maximum, is (a) 20000 (b) 27000

(c) 28500

(d) 19 x −

x 3000

(d) 28000

(iv) When the revenue is maximum, the price of the ticket is (a) 8 (b) 5 (c) 9

(d)

(v) How many spectators should be present to maximize the revenue? (a) 25000 (b) 27000 (c) 22000

(d) 28500

9.5

Class 12

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PART - B Section - III 19. Solve the differential equation

dy = e x − y + x 2e − y . dx

20. Find the area of the larger part bounded by y = cos x, y = x + 1 and y = 0. OR Find the area enclosed by the lines y = 0, y = x, x = 1, x = 2. 3x − 8, if x ≤ 5 21. Consider f (x ) =  if x > 5  2k , Find the value of k, if f(x) is continuous at x = 5. 22. A random variable X has the following distribution. X P(X)

1 0.15

2 0.23

3 0.12

4 0.10

5 0.20

6 0.08

7 0.07

8 0.05

For the event E = {X is prime number} and F = {X < 4}, find P(E ∪ F). 23. Find the direction cosines of the line passing through the two points (–2, 4, –5) and (1, 2, 3). OR If O be the origin and the coordinates of P be (1, 2, – 3), then find the equation of the plane passing through P and perpendicular to OP. 24. Find cofactors of a21 and a31 of the matrix  1 3 −2  A = [aij ] =  4 −5 6  .  3 5 2  25. Evaluate :

b

∫a

x x + a +b− x

dx

26. Let k and K be the minimum and the maximum values of the function, f (x ) = respectively. Find the ordered pair (k, K).

(1 + x )0.6 1 + x 0. 6

27. Find the number of triplets (x, y, z) satisfying the equation sin–1x + sin–1y + sin–1z =

defined on [0, 1],

3π . 2

1 3 28. Given that the events A and B are such that P ( A) = , P ( A ∪ B) = and P(B) = p. Find p if A and B are 2 5 (i) mutually exclusive (ii) independent. OR A bag contains 12 white pearls and 18 black pearls. Two pearls are drawn in succession without replacement. Find the probability that the first pearl is white and the second is black. Section - IV 29. Find all points of discontinuity of f, where f is defined as follows :  | x | + 3, x ≤ −3  f (x ) =  −2 x , −3 < x < 3 6 x + 2, x≥3  77

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 e −2 x y  dx 30. Solve the differential equation  = 1, x ≠ 0. −   x x  dy π  2x  dy , find 31. If y = (logcosxsinx)(logsinx cosx)–1 + sin–1  at x = .  1 + x 2  4 dx OR If x = 2 cos q – cos2q and y = 2 sin q – sin2q, find

d2 y dx 2

at θ =

π . 2

x y x2 y2 + = 1 and the line + = 1 . 3 2 9 4 33. Let A and B be non-empty sets. Show that f : A × B → B × A such that f (a, b) = (b, a) is a bijective function.

32. Find the area of the smaller region bounded by the ellipse

34. If f(x) = a log |x| + bx2 + x has extreme values at x = – 1 and at x = 2, then find a and b. OR Show that f(x) = cos (2x + p/4) is an increasing function on (3p/8, 7p/8). π/4

35. Evaluate :

∫ 0

j

dx 3

cos x 2 sin 2 x

Section - V

^  ^  36. Find the cartesian equations of the plane through the intersection of the planes r . (2 i + 6 j ) + 24 = 0 and r 3 − ^  ^ ^ = 0 and r . (3 i − j + 4 k) = 0 , which are at a distance of 2 units from the origin. OR

x −1 y z x +1 y z − 3 is unity, then find the If the shortest distance between the lines L1 : = = and L2 : = = − λ 1 1 2 2 2 value of l. 37. Solve the following LPP graphically. Maximize Z = 50x + 40y Subject to constraints : 1000x + 1200y ≤ 7600 12x + 8y ≤ 72 x, y ≥ 0 OR Solve the following LPP graphically. Minimize Z = 5x + 7y Subject to constraints : 2x + y ≤ 8 x + 2y ≥ 10 and x, y ≥ 0 1 −1 0   2 2 −4      38. If A = 2 3 4  and B =  −4 2 −4  , then find AB. Hence, solve the system of equations : 0 1 2   2 −1 5  x – y = 6, 2x + 3y + 4z = 34, y + 2z = 14 OR 2 3 10 4 6 5 6 9 20 Solve the system of the following equations : + + = 4, − + = 1, + − = 2 x y z x y z x y z Class 12

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SOLUTIONS dy dy = (1 − x )dx = (1 − x )(1 + y ) ⇒ 1+ y dx dy x2 ⇒ ∫ = ∫ (1 − x )dx ⇒ log | 1 + y | = x − + C 1+ y 2 OR

1. We have

Since greatest power of highest order derivative is 1, therefore degree of the given differential equation is 1.  x + 3 y y   4 −1 2. Given  =   7 − x 4  0 4  ⇒ y = –1 and 7 – x = 0 ⇒ x = 7, y = –1 if x = 0  k, 2k, if x = 1  3. We have, P ( X = x ) =  if , 3 k x =2   0, otherwise

Since, SP(xi) = 1, therefore k + 2k + 3k = 1 ⇒ 6k = 1 ⇒ k =

1 6

OR Total number of students = 8 The number of ways to select 4 students out of 8 8! students = C4 = = 70 4! 4! 8

The number of ways to select 2 boys and 2 girls 3! 5! × = 3 × 10 = 30 2!1! 2!3! 30 3 \ Required probability = = . 70 7 = 3C2 × 5C2 =

4. Given, R = {(1, 1), (1, 2), (2, 1), (3, 3)} Since, (2, 2) ∉ R Therefore, R is not reflexive. But R is both symmetric and transitive as (1, 2) ∈ R ⇒ (2, 1) ∈ R and (1, 1) ∈ R, (2, 1) ∈ R ⇒ (1, 1) ∈ R. 5. Let I = ∫ =∫

x x2 − x

dx

dx x = 2 x −1 + C dx = ∫ x −1 x(x − 1) OR

Let I = ∫

cos 2 x + 2 sin2 x

dx cos2 x cos2 x − sin2 x + 2 sin2 x =∫ dx cos2 x

=∫

cos2 x + sin2 x 2

cos x = tan x + C

dx = ∫

1 2

cos x

dx = ∫ sec2 x dx

x −1 y − 5 z − 6 y −2 z −3 = = and will 6. Lines x − 1 = = 3k 1 −5 2k 2 −3 be perpendicular if a1a2 + b1b2 + c1c2 = 0

⇒ –3(3k) + 2k + 2(–5) = 0 ⇒ –9k + 2k – 10 = 0 10 ⇒ k=− 7 7. Here n = k and d = 6   \ r ⋅ nɵ = d ⇒ r ⋅ kɵ = 6 , which is the required equation of plane. OR  Let n = 2ɵi − 3ɵj + 4kɵ. Then,  n 2ɵi − 3ɵj + 4kɵ 2ɵi − 3ɵj + 4kɵ nɵ =  = = |n | 4 + 9 +16 29 Hence, the required equation of the plane is   2 ɵ (−3) ɵ 4 ɵ  6 r ⋅ i+ j+ k=  29 29 29  29 or r ⋅ (2i − 3j + 4k ) = 6 8. Clearly, (AAT)T = (AT)T ⋅ AT ⇒ (AAT)T = AAT ⇒ AAT is a symmetric matrix. 9. Given, D(x) =

[ (AB)T = BTAT] [ (XT)T = X]

f (x ) g (x ) = bf(x) – ag(x) a b

\ ∫D(x)dx = ∫{bf(x) – ag(x)}dx = b∫f(x)dx – a∫g(x)dx = OR

∫ f (x)dx a

∫ g (x)dx b

Given, A is a matrix of order 3 × 3, then A −1 =

1 1 2 adj A = 3 ⋅ A A A

[ If A is invertible matrix of order n, then |adj A| = |A|n – 1] 1 −1 = = A A 10. If two dice are thrown, then total number of cases = 36 Cases for total of 9 or 11 are {(3, 6), (4, 5), (6, 3), (5, 4), (6, 5), (5, 6)}, i.e., 6 in number. 79

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6 1 = 36 6 P(sum is neither 9 nor 11) = 1 – P(sum is 9 or 11) 1 5 =1− = 6 6 11. Given, f(x) = x2 – 4x + 5 Let y = x2 – 4x + 5 ⇒ y = (x – 2)2 + 1

\ P(total 9 or 11) =

⇒ (x – 2)2 = y – 1 ⇒ x – 2 = ⇒ x=

y −1 [Q x ∈ [2, ∞)]

y −1 + 2

For range y – 1 ≥ 0 ⇒ y ≥ 1 \ Range is [1, ∞).

P ( A ∩ B ) 0. 1 1 = = = 0.25 \ P(A/B) = P ( B) 0. 4 4 π/4

∫

(sec2 x + cosec2 x )dx = [tan x − cot x]ππ//64

π /6

2  1  . = (1 – 1) –  − 3 =  3  3 14. Let E1 be the event that bag I is selected E2 be the event that bag II is selected E be the event that the ball drawn is of white colour. By rule of total probability, P(E) = P(E1)·P(E|E1) + P(E2)·P(E|E2) 1 3 1 6 9 1 = ⋅ + ⋅ = = 2 9 2 9 18 2        15. Given, 5a − 3b − 2c = 0 ⇒ 2c = 5a − 3b        5a − 3b 5a − 3b 3b − 5a = = ⇒ c= 2 5−3 3−5 So, C divides AB externally in the ratio 3 : 5 16. Clearly, 2R1 and 1R 2 2 therefore R is not transitive.

but

2R 2 2,

17. (i) (c) : Clearly, G be the centroid of DBCD, therefore coordinates of G are  −1 + 5 + 0 , 4 + 2 − 5 , 1 + 3 + 4  =  4 , 1 , 8   3 3 3  3 3 3 4 1 8 (ii) (b) : Since, A ≡ (3, 0, 1) and G ≡  , ,  3 3 3  4 1 8 \ AG =  − 3  ɵi +  − 0  ɵj +  − 1 kɵ 3  3  3  =

−5  1  5  i+ j+ k 3 3 3

 51 ⇒ | AG | = 3 (iii) (c) : Clearly, area of DABC =

1   | AB × AC | 2

ɵi   Here, AB × AC = −1 − 3

ɵj

kɵ

4−0

1−1

5−3

2−0

3 −1

j

k

i

= −4

4

2

2

0 = −8 1 1 2

i

12. We have, P(A) = 0.2, P(B) = 0.4 and P(A ∪ B) = 0.5 P(A ∩ B) = P(A) + P(B) – P(A ∪ B) = 0.2 + 0.4 – 0.5 = 0.1

13.

 5 2 1 2 5 2 25 1 25 51 ⇒ | AG |2 =   +   +   = + + = 3 3 3 9 9 9 9

j

k

−1

0

1

1

= −8[i(−1 − 0) − j(1 − 0) + k (1 + 1)] = −8[−i − j + 2k ] = 8(i + j − 2k )   \ | AB × AC | = 8 | ɵi + ɵj − 2kɵ | = 8 1 + 1 + 4 = 8 6 1 Hence, area of DABC = × 8 6 = 4 6 sq. units 2  (iv) (b) : Here, AB = −4ɵi + 4 ɵj + 0kɵ  ⇒ | AB | = 16 + 16 = 32 = 4 2   AC = 2ɵi + 2 ɵj + 2kɵ ⇒ | AC | = 4 + 4 + 4 = 12 = 2 3   Now, | AB | + | AC | = 4 2 + 2 3 = 9.1 units (v) (a) : The length of the perpendicular from the vertex D on the opposite face    = | Projection of AD on AB × AC | (−3i − 5j + 3k ) ⋅ (i + j − 2k ) = 12 + 12 + 22 −3 − 5 − 6 14 = = units 6 6 18. (i) (a) : Let p be the price per ticket and x be the number of tickets sold. x   Then, revenue function R(x ) = p × x =  19 − x  3000  x2 3000 (ii) (c) : Since, more than 42000 tickets cannot be sold. So, range of x is [0, 42000]. = 19 x −

x2 (iii) (c) : We have, R(x) = 19x – 3000 x ⇒ R′(x) = 19 – 1500 For maxima/minima, put R′(x) = 0 Class 12

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x = 0 ⇒ x = 28500 1500 1 < 0. Also, R′′(x) = – 1500 (iv) (d) : Maximum revenue will be at x = 28500 28500 \ Price of a ticket = 19 − = 19 – 9.5 = 9.5 3000 (v) (d) : Number of spectators will be equal to number of tickets sold when revenue is maximum. \ Required number of spectators = 28500 ⇒ 19 −

dy = e x − y + x 2e − y dx ⇒ dy = (ex–y + x2 e–y)dx ⇒ ey dy = (ex + x2)dx 19. We have,

⇒

∫e

y

dy = ∫ (e x + x 2 ) dx

[Integrating both sides]

x3 + C , which is the required solution. 3 20. It can be observed that y = cos x and y = x + 1 meet at the point (0, 1). Also, y = x + 1 passes through the points (–1, 0) and (0, 1), −π π y = cos x meets X-axis at  , 0  and  , 0  . 2   2  ⇒ e y = ex +

π 2 cos x dx 0

0

\ Required area = ∫−1(x + 1) dx + ∫ π

0

  x2 =  + x  + [sin x]02  −1  2 1  = 0 −  − 1 + 1 2  3 = sq. units. 2

Given lines are y = x, x = 1, x = 2 and y = 0. \ Required area = ∫ x dx  x2  4 1 =  = −   2 1  2 2  3 = sq . units 2

x →5 −

∴ k=

x=1

y=x x=2

O

21. We have, L.H.L. (at x = 5)

7 2

22. Clearly, P(E) = P(X = 2) + P(X = 3) + P(X = 5) + P(X = 7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62 P(F) = P(X = 1) + P(X = 2) + P(X = 3) = 0.15 + 0.23 + 0.12 = 0.50 P (E ∩ F) = P(X = 2) + P(X = 3) = 0.23 + 0.12 = 0.35 \ P(E ∪ F) = P(E) + P (F) – P (E ∩ F) = 0.62 + 0.50 – 0.35 = 0.77. 23. We know that the direction cosines of the line passing through two points P(x1, y1, z1) and Q(x2, y2, z2) are given by x 2 − x1 y 2 − y1 z 2 − z1 , , . PQ PQ PQ where, PQ = ( x 2 − x1 )2 + ( y 2 − y1 )2 + (z 2 − z1 )2 Here, P is (–2, 4, –5) and Q is (1, 2, 3). So, PQ = (1 − ( −2))2 + (2 − 4)2 + (3 − ( −5))2 = 77 Thus, the direction cosines of the line joining two 3 −2 8 points are , , . 77 77 77 OR

M 21 =

3 −2 = 6 + 10 = 16 ⇒ C21 = − M 21 = −16 5 2

M 31 =

3 −2 = 18 − 10 = 8 ⇒ C31 = M 31 = 8 −5 6

X

25. Let I =

= lim f (x ) = lim (3x − 8) = 7 x →5 −

x →5

and R.H.L. (at x = 5) \

lim f (x ) = lim (2k ) = 2k and f (5) = 7 x →5 +

x →5

Since f(x) is continuous at x = 5.

x →5 +

24. Let Mij and Cij respectively denote the minor and cofactor of element aij in A. Then,

Y

1

2

lim f (x ) = lim f (x ) = f (5) ⇒ 7 = 2k

The direction ratios of OP are < 1 – 0, 2 – 0, – 3 – 0 > i.e.

\ The equation of the plane passing through P and perpendicular to OP is (1) (x – 1) + (2) (y – 2) + (– 3) (z + 3) = 0 ⇒ x – 1 + 2y – 4 – 3z – 9 = 0 ⇒ x + 2y – 3z – 14 = 0

OR 2

\

\ I=

b

∫a

b

∫a

x dx x + a+b− x a+b− x

a+b− x + x

dx

...(i) ...(ii)

b b    ∫a f (x )dx = ∫a f (a + b − x )dx  81

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On adding (i) and (ii), we get b−a 2I = ∫ dx = [ x]ba = b − a ⇒ I = a 2 3/5 (1 + x ) 26. We have, f ( x ) = 1 + x 3/5 3 3 1 + x 3/5 (1 + x ) −2/5 − (1 + x )3/5 ⋅ x −2/5 5 5 f ′( x ) = (1 + x 3/5 )2 b

(

)

Clearly f ′(x) = 0 ⇒

x=1

Also, f (0) = 1, and f (1) = \ f(x) ∈ (2–0.4, 1) Thus, (k, K) = (2–0.4, 1)

2 0. 6 = 2 −0.4 2

27. We have, sin–1x + sin–1y + sin–1z =

By multiplication rule of probability, we have 2 18 36 P ( A ∩ B) = P ( A). P ( B | A) = × = 5 29 145 x ≤ −3  − x + 3, 29. We have, f (x ) =  −2 x , −3< x < 3 6 x + 2, x ≥ 3  Clearly, the possible points of discontinuity of f are 3 and –3. [ For all other points f(x) is a linear polynomial, which is continuous everywhere] Continuity at x = – 3 : lim f (x ) = lim (− x + 3) = 3 + 3 = 6 x → −3 −

3π 2

π π −π π ≤ sin −1 y ≤ − ≤ sin −1 x ≤ , 2 2 2 2 −π π ≤ sin −1 z ≤ and 2 2 \ The above condition will true if sin–1x = sin–1y = sin–1z = π ⇒ x = y = z = 1 2 Thus, there is only one triplet. 28. (i) When A and B are mutually exclusive, then A ∩ B = f ⇒ P(A ∩ B) = 0 ⇒ P(A ∪ B) = P(A) + P(B) 3 1 6−5 1 3 1 = = +p ⇒ p= − = ⇒ 5 2 10 10 5 2 (ii) When A and B are independent, then P(A ∩ B) = P(A) P(B) ⇒ P(A ∪ B) = P(A) + P(B) – P(A) P(B) 3 1 1 3 1 2p − p ⇒ = + p− ⋅p ⇒ − = 5 2 2 5 2 2 p 6−5 2 1 ⇒ = ⇒ p= = 2 10 10 5 OR Let A and B be the events of getting a white pearl in the first draw and a black pearl in the second draw respectively. Now, P(A) = P(getting a white pearl in the first draw) 12 2 = = 30 5 When second pearl is drawn without replacement, the probability that the second pearl is black is the conditional probability of the event B occurring when A has already occurred. 18 \ P (B | A) = 29

x → −3

lim f ( x ) = lim (−2 x ) = 6 x → − 3+

x → −3

f (–3) = –(–3) + 3 = 3 + 3 = 6 Thus, lim f (x ) = lim f (x ) = f (−3) x → −3−

x →− 3+

\ f is continuous at x = – 3. Continuity at x = 3 : lim f (x ) = lim (−2 x ) = − 6 x →3

x →3−

lim f (x ) = lim (6 x + 2) = 6(3) + 2 = 20 x →3

x →3+

Thus, lim f (x ) ≠ lim f (x ) x →3−

x →3+

\ f(x) is discontinuous at x = 3. So, the only point of discontinuity of f is x = 3.  e −2 x y  dx =1 − 30. We have,    x x  dy dy e −2 x y dy y e −2 x + = = − ⇒ dx dx x x x x This is a linear differential equation of the form

⇒

dy 1 e −2 x + Py = Q, with P = and Q = dx x x 1 dx ∫ P dx I.F. = e ∫ = e x = e2 x \ The solution is given by e ye 2 x = ∫ e 2 x ⋅

−2 x

x

dx + C

1 ⇒ ye 2 x = ∫ dx + C ⇒ ye 2 x = 2 x + C x ⇒ y = (2 x + C ) e −2 x , solution.

which is the required

31. We have,  2x  y = (log cos x sin x )(log sin x cos x ) −1 + sin −1   1 + x 2  Class 12

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log sin x  log cos x  = log cos x  log sin x 

−1

 2x  + sin −1   1 + x 2 

2

 log sin x   2x  + sin −1  =  log cos x  1 + x 2     log sin x  dy × = 2  log cos x  dx

∴

1 1   cos x − (log sin x ) (− sin x )   (log cos x ) sin x cos x   2   (log cos x )  2 (1 + x )2 − 2 x(0 + 2 x ) 1 + × 2 (1 + x 2 )2  2x  1−   1 + x 2  =

B (0, 2) X

4

32

4 32 32 8 + 2 = 2 − log(1 / 2 ) π + 16 π + 16 log 2

+ 2 π + 16

OR We have, x = 2 cosq – cos2q and y = 2sinq – sin2q dx = −2 sin θ + 2 sin 2θ dθ dy and = 2 cos θ − 2 cos 2θ dθ  dy  dy  dθ  2 cos θ − 2 cos 2θ ⇒ = = dx  dx  −2 sin θ + 2 sin 2θ   dθ 3θ θ cos θ − cos 2θ 2 sin 2 sin 2 3θ = = = tan θ θ 3 sin 2θ − sin θ 2 2 cos sin 2 2 Differentiating w.r.t. x, we get Here,

3θ 3 dθ 3 ⋅ = ⋅ 2 2 dx 2

1 1 ⋅  3θ  −2 sin θ + 2 sin 2θ dx cos 2    2 3 3 1 1 1 = ⋅ = ⋅ θ θ 8 4  3θ   3θ   3θ  cos 2   2 cos   sin cos 3   sin    2  2 2 2 2 2

= sec2

A (3, 0) Y

{cot x(log cos x ) + tan x(log sin x )} {log(cos x )}3 2(1 − x 2 ) + | 1 − x 2 |(1 + x 2 )

d2 y

X

O

2 log(sin x )

2 log(1 / 2 )  dy  {log(1 / 2 ) + log(1 / 2 )} ∴  = π  dx  x = {log(1 / 2 )}3

=

 d2 y  3 3 ∴  = =  3  dx 2  θ= π 8 cos 3  3π  sin π  1  1   − 8 2   4 4 2 2 3× 4 3 =− =− 8 2 2 2 x y + =1 32. Given ellipse is 9 4 x y and line is + = 1 3 2 Y

Required area = Area of shaded region 3

 2 = ∫ 2 1 − x − 2  1 − x   dx  3   9 0  3

3

0

0

2  x =∫ 9 − x 2 dx − ∫ 2  1 −  dx  3 3 3

3  2 x x2  2 9 −1 x  x =  9 − x + sin − − 2   32 2 3  0 6 0 

2 9 π   3 =  ×  − 2 3 −  3 2 2   2 3π − 6 3 3 = π −3 = = (π − 2) sq. units. 2 2 2 33. Injectivity : Let (a1, b1) and (a2, b2) ∈ A × B such that, f(a1, b1) = f(a2, b2) ⇒ (b1, a1) = (b2, a2) ⇒ b1 = b2 and a1 = a2 ⇒ (a1, b1) = (a2, b2) Thus, f(a1, b1) = f(a2, b2) ⇒ (a1, b1) = (a2, b2) for all (a1, b1), (a2, b2) ∈ A × B. So, f is an injective function. Surjectivity : Let (b, a) be an arbitrary element of B × A, where, b ∈ B and a ∈ A ⇒ (a, b) ∈ A × B Thus, for all (b, a) ∈ B × A, their exists (a, b) ∈ (A × B) such that, f (a, b) = (b, a) So, f : A × B → B × A is an onto function. Hence, f is a bijective function. 34. Clearly, domain (f ) = R – {0} We have, f(x) = a log |x| + bx2 + x a ⇒ f ′(x) = + 2bx + 1 x Since f(x) has extreme values at x = –1 and x = 2. Therefore, f ′(–1) = 0 and f ′(2) = 0 83

Mathematics

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a + 4b + 1 = 0 2

⇒ –a – 2b + 1 = 0 and

⇒ 12 = (2 + 3λ)2 + (6 − λ)2 + 16λ 2

⇒ a + 2b = 1 and a + 8b = –2 Solving (i) and (ii), we get 1 a = 2 and b = − . 2 OR

...(i) ...(ii)

We have, f(x) = cos(2x + p/4) ⇒ f ′(x) = –2 sin (2x + p/4) Now, x ∈ (3p/8, 7p/8) ⇒ 3p/8 < x < 7p/8 ⇒ 3p/4 < 2x < 7p/4 ⇒ p/4 + 3p/4 < 2x + p/4 < 7p/4 + p/4 ⇒ p < 2x + p/4 < 2p ⇒ sin (2x + p/4) < 0 [ sine function is negative in third and fourth quadrants] ⇒ –2 sin (2x + p/4) > 0 ⇒ f ′(x) > 0 Hence, f(x) is increasing on (3p/8, 7p/8). π/4

35. Let I =

0

π/4

=

∫ 0

=

1 2

∫

dx 3

cos x 2 sin 2 x dx

3

cos x 2 ⋅ 2 sin x cos x

π/4

=

1 2

π/4

dx

∫ 0

7

1

cos 2 x ⋅ sin 2 x

dx

∫

7 0 cos 2

π/4

1 x ⋅ tan 2

1 x ⋅ cos 2

x

π/4

sec 4 x ∫ cos 4 x tan x ∫ tan x dx 0 0 2 Put tan x = t ⇒ sec xdx = dt π Also x = 0 ⇒ t = 0 and x = ⇒ t = 1 4 =

1 2

dx

=

1

∴ I=

1 2

1

1

3

1 (1 + t 2 )dt 1 − 2 2 = ∫ (t + t ) dt 2 ∫0 20 t 1

1  t 1/2 t 5/2  6 =  +  = 2 1 / 2 5 / 2 0 5 36. Any plane through the line of intersection of the  ^ ^  ^ ^ ^ planes r ⋅ (2 i + 6 j) + 24 = 0 and r ⋅ (3 i − j + 4 k ) = 0 is given by  ^ ^  ^ ^ ^ r ⋅ (2 i + 6 j) + 24 + λ[r ⋅ (3 i − j + 4 k )] = 0  r ⋅ ((2ɵi + 6 ɵj) + λ(3ɵi − ɵj + 4kɵ )) + 24 = 0  ...(i) ⇒ r ⋅ ((2 + 3λ)ɵi + (6 − λ)ɵj + 4 λ kɵ ) + 24 = 0 Given, distance of (i) from origin is 2. 24 \ =2 (2 + 3λ )2 + (6 − λ )2 + 16λ 2

Squaring both sides, we get 144 = 4 + 9l2 + 12l + 36 + l2 – 12l + 16l2 ⇒ 26l2 = 104 ⇒ l2 = 4 ⇒ l = ±2 Now, from (i), we get  ^ ^ ^ r ⋅ (8 i + 4 j + 8 k ) + 24 = 0 and  ^ ^ ^ r ⋅ (−4 i + 8 j − 8 k ) + 24 = 0  ^ ^ ^ ⇒ r ⋅ (2 i + j + 2 k ) + 6 = 0 and  ^ ^ ^ r ⋅ (− i + 2 j − 2 k ) + 6 = 0 ⇒ 2x + y + 2z + 6 = 0 and x – 2y + 2z – 6 = 0 OR x −1 y z = = −1 2 1 x +1 y z − 3 and = = 2 2 λ Here, x1 = 1, y1 = 0, z1 = 0 a1 = 1, b1 = –1, c1 = 2 x2 = –1, y2 = 0, z2 = 3 a2 = 2, b2 = 2, c2 = l

…(i)

The lines are

x 2 − x1 a1 Now, a2

y 2 − y1 b1 b2

z 2 − z1 −2 0 c1 = 1 −1 c2 2 2

…(ii)

3 2 λ

= –2(–l – 4) + 0 + 3(2 + 2) = 2l + 8 + 12 = 2l + 20 and (b1c2 – b2c1)2 + (c1a2 – c2a1)2 + (a1b2 – a2b1)2 = (–l – 4)2 + (4 – l)2 + (2 + 2)2 = l2 + 16 + 8l + 16 + l2 – 8l + 16 = 2l2 + 48 Shortest distance between lines 2λ + 20 =1 [Given] = 2λ 2 + 48 ⇒ 2λ + 20 = 2λ 2 + 48 ⇒ 4l2 + 400 + 80l = 2l2 + 48 ⇒ 2l2 + 80l + 352 = 0 ⇒ l2 + 40l + 176 = 0 ⇒ l =

−40 ± 1600 − 4(1) (176)

2 −40 ± 1600 − 704 = = −40 ± 896 2 2

−40 ± 2 224 = −20 ± 224 2 37. The given problem can be written as Maximize Z = 50x + 40y subject to constraints : 5x + 6y ≤ 38 3x + 2y ≤ 18 =

Class 12

84

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x ≥ 0, y ≥ 0 Now, let us draw the lines l1 : 5x + 6y = 38 x y or + =1  38   19   5   3  l2 : 3x + 2y = 18

The coordinates of the corner points of the feasible region ABC are A(0, 5), B(0, 8) and C(2, 4). The values of the objective function Z = 5x + 7y at the corner points of the feasible region are given in the following table.

(l3)

Corner Points

Value of Z = 5x + 7y

A(0, 5)

5 × 0 + 7 × 5 = 35 (Minimum)

B(0, 8)

5 × 0 + 7 × 8 = 56

C(2, 4)

5 × 2 + 7 × 4 = 38

(l4)

x y + =1 6 9 l3 : x = 0 and l4 : y = 0 Lines l1 and l2 intersect at E(4, 3). The shaded region OCEB is the feasible region which is bounded. or

Corner points of the feasible region are O(0, 0), C(6, 0),  19  E(4, 3) and B  0,   3 The value of the objective function Z = 50x + 40y at corner points are given below: At O, Z = 50 × 0 + 40 × 0 = 0 At C, Z = 50 × 6 + 40 × 0 = 300 At E, Z = 50 × 4 + 40 × 3 = 320 (Maximum) 19 At B, Z = 50 × 0 + 40 × = 253.33 3 Clearly, the maximum value is 320 at E(4, 3). OR The given problem is Minimize Z = 5x + 7y subject to 2x + y ≤ 8 x + 2y ≥ 10 and x, y ≥ 0 To solve this LPP graphically, we first convert the inequations into equations to obtain the following line x y l1 : 2x + y = 8, or + = 1 4 8 x y l2 : x + 2y = 10, or + =1 10 5 l3 : x = 0 and l4 : y = 0 Y(l3) 8 B(0, 8) 7 6 5 A(0, 5) 4 3 2 1 O

Thus, Z is minimum when x = 0 and y = 5. 1 −1 0   2 2 −4     38. AB = 2 3 4   −4 2 −4  0 1 2   2 −1 5      2 + 4 + 0 2 − 2 + 0 −4 + 4 + 0  6 0 0     =  4 − 12 + 8 4 + 6 − 4 −8 − 12 + 20 = 0 6 0  0 − 4 + 4 0 + 2 − 2 0 − 4 + 10  0 0 6 = 6 I3 1 1  ⇒ A  B  = I 3 ⇒ A−1 = B 6  6 (By definition of inverse)

 2 2 −4  1  ⇒ A =  −4 2 −4  6  2 −1 5  The given system of equations is x – y + 0z = 6 2x + 3y + 4z = 34 0x + y + 2z = 14 This system of equations can be written as −1

1 −1 0   x   6       2 3 4   y  = 34  or AX = C 0 1 2   z  14 

x 6     where X =  y  and C = 34   z  14  As A–1 exists, therefore X = A–1 C

C (2, 4) (10, 0) (4, 0) 1 2 3 4 5 6 7 8 9 10 l2 l1

X (l4)

 2 2 −4   6  1   ⇒ X =  −4 2 −4  34  6  2 −1 5  14  85

Mathematics

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x   12 + 68 − 56   24   4  1 1    ⇒  y  =  −24 + 68 − 56 =  −122 =  −2 6 6 z   12 − 34 + 70   48   8          ⇒ x = 4, y = – 2, z = 8 Hence, the solution of the given system of equations is x = 4, y = – 2, z = 8. OR The equations can be written in the form AX = B, 2 1/ x  4 3 10       where, A = 4 −6 5 , X = 1/ y  and B =  1   6 9 −20  1/ z   2

 75 150 75 1 1   Hence, A = 110 −100 30 (adj A) =  | A| 1200  72 0 −24  −1

As, AX = B

⇒ X = A–1B

1 x  75 150 75  4    1 1    ⇒  = 110 −100 30  1   y  1200   72 0 −24   2    1    z  300 + 150 + 150 1   = 440 − 100 + 60 1200   288 + 0 − 48

2 3 10 Now, | A | = 4 −6 5 6 9 −20 = 2(120 – 45) – 3(–80 – 30) + 10(36 + 36) = 2(75) – 3(–110) + 10(72) = 150 + 330 + 720 = 1200 ≠ 0 \ A–1 exists. 72 ′  75 150 75  75 110     0 = 110 −100 30 ∴ adj A = 150 −100   30 −24   72 0 −24   75

⇒

1 1 x 2 600      1 1   1  = 400 =    y  1200  3 240      1 1   5  z 

1 1 1 1 1 1 = , = , = x 2 y 3 z 5 Hence, x = 2, y = 3, z = 5

Thus,



Class 12

86

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Self Evaluation Sheet Once you complete SQP-6, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Differential Equations / Differential Equations Matrices Probability / Probability Relations and Functions Integrals / Integrals Three Dimensional Geometry Three Dimensional Geometry / Three Dimensional Geometry Matrices Determinants / Determinants Probability Relations and Functions Probability Integrals Probability Vector Algebra Relations and Functions Vector Algebra Application of Derivatives

19

Differential Equations

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Application of Integrals / Application of Integrals Continuity and Differentiability Probability Three Dimensional Geometry / Three Dimensional Geometry Determinants Integrals Application of Derivatives Inverse Trigonometric Functions Probability / Probability Continuity and Differentiability Differential Equations Continuity and Differentiability / Continuity and Differentiability Application of Integrals Relations and Functions Application of Derivatives / Application of Derivatives Integrals Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming Determinants / Determinants

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50% Mathematics

Try hard to boost your average score. 87

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SQP

7

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

–

1(2)

2(6)

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

2(2)#

2(4)#

1(3)

–

5(9)

8.

Application of Integrals

–

–

1(3)*

–

1(3)

9.

Differential Equations

1(1)*

1(2)

1(3)*

–

3(6)

10.

Vector Algebra

1(1)* + 1(4)

–

–

–

2(5)

11.

Three Dimensional Geometry

2(2)

1(2)*

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

4(4)#

2(4)#

–

–

6(8)

18(24)

10(20)

7(21)

3(15)

38(80)

13.

Probability Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-7

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Evaluate :

0

dx

∫−1 2x + 3

OR

2

Evaluate :

sec x

∫ 2 + tan x dx

2. Let A = {1, 2, 3, 4} and R be a relation in A given by R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1)}. Then, show that R is reflexive.  1 −1 1   3. Show that the matrix A =  2 3 0  is not invertible. 18 2 10  OR Find the value of the determinant

x 1

−5x . x + 10

4. Find the equation of a plane with intercepts 2, 3 and 4 on the X, Y and Z-axes respectively. 89

Mathematics

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 d3 y  sin + 5. Find the degree of the differential equation  3  = 0 . dx 4 dx d4 y

OR 2

 dy   dy  Find the order of the differential equation x +   = 1 +   .  dx   dx  6. Let A = {1, 2, 3} and B = {1, 2, 4}. Check whether f = {(1,1), (1,2), (2,1), (3, 4)} from A to B is a function or not. 7. Let E and F be events associated with the sample space S of an experiment. Show that P(S|F) = P(F|F) = 1. OR Let A and B be two events associated with an experiment such that P(A ∩ B) = P(A)P(B). Show that P(A|B) = P(A) and P(B|A) = P(B). 0 1  2 8. If A =   , then find the value of A . 1 0       9. Find the angle between two vectors a and b with magnitude 1 and 2 respectively, such that a ⋅ b = 1 . OR   ^ ^ ^ Find the projection of a = 2 i − j + k on b = i^ − 2 ^j + k^ .

10. If P(A) =

3 5 3 , P(B) = and P(A ∪ B) = , then find P(A | B) . 8 8 4

11. Suppose you have two coins which appear identical in your pocket. You know that, one is fair and one is 2 headed. If you take one out, toss it and get a head, what is the probability that it was a fair coin? 12. Show that f(x) = x2 + 1 is not one-one? 13. Suppose that five good fuses and two defective ones have been mixed up. To find the defective fuses, we test them one-by-one, at random and without replacement. What is the probability that we are lucky and find both of the defective fuses in the first two tests? 2π

14. Evaluate :

dx

∫ e sin x + 1 0

15. What is the distance between the planes 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0 ?  −3 6 0 1   16. For matrix A =  4 −5 8 , find ( A − A ′) (where A ′ is the transpose of matrix A). 2  0 −7 −2 Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. If two vectors are represented by the two sides of a triangle taken in order, then their sum is represented by the third side of the triangle taken in opposite order.     (i) If p, q , r are the vectors represented by the sides of a triangle taken in orders, then p + q =    (a) r (b) 2r (c) −r (d) None of these   (ii) If ABCD is a parallelogram and AC and BD are its diagonals, then AC + BD =     (a) 2DA (b) 2AB (c) 2BC (d) 2BD Class 12

90

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      (iii) If ABCD is a parallelogram, where AB = x and BC = y , then AC − BD =   (a) x (b) 2 x (c) y (d) 2 y     (iv) If ABCD is a quadrilateral whose diagonals are AC and BD , then BA + CD =

  (a) AC + DB

D

C

A

B

  (b) AC + BD

  (c) BC + AD   (v) If S is the mid point of side QR of DPQR, then PQ + PR =

  (d) BD + CA

P

Q

 (a) 2QS

S

 (b) 2PS

R

 (c) 2SR

(d) None of these

18. A can manufacturer designs a cylindrical can for a company making sanitizer and disinfector. The can is made to hold 5 litres of sanitizer or disinfector. (i) If r cm be the radius and h cm be the height of the cylindrical can, then the surface area expressed as a function of r as (a) 2πr2 (b) 2πr2 + 5000 5000 10000 (c) 2πr2 + (d) 2πr2 + r r (ii) The radius that will minimize the cost of the material to manufacture the can is (a)

3

500 cm p

(b)

500 cm p

(c)

3

2500 cm p

(d)

2500 cm p

(iii) The height that will minimize the cost of the material to manufacture the can is (a)

3

2500 cm p

(b) 2 3

2500 cm p

(c)

2500 p

(d) 2

2500 p

2500 ≈ 9, then minimum cost is (iv) If the cost of material used to manufacture the can is ` 100/m2 and 3 π approximately (a) ` 16.7 (b) ` 18 (c) ` 19 (d) ` 20 (v) To minimize the cost of the material used to manufacture the can, we need to minimize the (a) volume (b) curved surface area (c) total surface area (d) surface area of the base

PART - B Section - III  π 19. Separate the interval 0,  into sub-intervals in which the function f(x) = sin4x + cos4x is increasing or  2 decreasing. 91

Mathematics

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20. If a discrete random variable X has the following probability distribution : X=x 1 2 3 4 5 6 7 P ( X = x ) k 2k 2k 3k k 2

2k 2

7k 2 + k

Find the value of k. OR An urn contains 10 black balls and 5 white balls. Two balls are drawn from the urn one after the other without replacement. Find the probability that both drawn balls are black. 21. Evaluate :

∞

∫0

dx 2

(x + 4)(x 2 + 9)

3  22. Solve for x : cos(tan −1 x ) = sin  cot −1  .  4 23. Evaluate : ∫ (e x log a + e a log x + e a log a ) dx OR Evaluate :

∫

dx 2 x3

2 x3

−4

24. Find the solution of the differential equation

dy = cos ( x + y ) . dx

1 2  –1 25. If A =   , then find A . 0 4   26. A box contains N coins, of which m are fair and the rest are biased. The probability of getting head when a 1 2 fair coin is tossed is , while it is when a biased coin is tossed. A coin is drawn from the box at random 2 3 and is tossed twice. The first time it shows head and the second time it shows tail. Find the probability that the coin drawn is fair.  x−4 + a, if   | x − 4 | 27. Find the values of a and b such that the function f defined by f (x ) =  a + b, if  x−4  + b, if | x − 4 | at x = 4.

x4

^

^

^

28. Find the cartesian equation of the plane passing through a point having position vector 2 i + 3 j + 4 k and ^

^

^

perpendicular to the vector 2 i + j − 2 k . OR Find the distance of the point (–1, –5, –10) from the point of intersection of the line ^ ^ ^ ^ ^ ^  ^ ^ ^  r = 2 i − j + 2 k + λ (3 i + 4 j + 12 k) and the plane r . (i − j + k) = 5 . Section - IV 29. Let N be the set of all natural numbers and let R be a relation in N × N defined by (a, b) R(c, d) ⇒ ad = bc for all (a, b), (c, d) ∈ N × N, show that R is an equivalence relation on N × N. Class 12

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30. Show that the surface area of a closed cuboid with square base and given volume is minimum, when it is a cube. 31. Solve : (x + y )2

dy = a2 dx

OR

Find one-parameter family of solution curve of the differential equation 32. If

dy cos 2 x = tan x − y . dx

d2 y −b 4 y2 = . , then prove that − = 1 dx 2 a 2 y 3 a 2 b2 x2

33. Evaluate :

1 + x cos x

∫ x(1 − x 2e 2 sin x ) dx

34. Find the area enclosed by y = 3x – 5, y = 0, x = 3 and x = 5. OR Using integration, find the area of the region bounded by the lines y – 1 = x, the X-axis and the ordinates x = –2 and x = 3. 35. If x =

1−t2 1+ t

2

and y =

2t

, then find dy . dx 1+ t 2

Section - V 36. The points A(4, 5, 10), B(2, 3, 4) and C(1, 2, –1) are three vertices of a parallelogram ABCD. Find the vector and cartesian equations of the sides AB and BC and find coordinates of D. OR If the planes x – cy – bz = 0, cx – y + az = 0 and bx + ay – z = 0 pass through a straight line, then find the value of a2 + b2 + c2 + 2abc. 37. Find the maximum value of Z = x + y subject to x + y ≤ 10, 3x – 2y ≤ 15, x ≤ 6, x ≥ 0, y ≥ 0. OR Find the maximum and minimum values of Z = 7x1 – 3x2 subject to x1 + 2x2 ≤ 2, 2x1 + 4x2 ≤ 8, x1 ≥ 0, x2 ≥ 0. 3 2   4 1 38. Find a matrix X such that, X  = . 1 −1  2 3 OR 1 0 1  Find the inverse of the matrix 0 2 3 , if exists. Also show that |adj A| = |A|2. 1 2 1

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SOLUTIONS 0

∫

1. Let I =

−1

dx 2x + 3

Also, P ( F | F ) =

0  log 3 log 1  log 3  log(2 x + 3)  = − = =   −1  2  2 2  2

OR 2

sec x dx 2 + tan x Put 2 + tanx = t ⇒ sec2x dx = dt dt ∴ I = ∫ = log | t | + C = log | 2 + tan x | + C t Let I = ∫

2. Here, A = {1, 2, 3, 4} and R = {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 1), (3, 1)} Now, (1, 1), (2, 2), (3, 3), (4, 4) ∈R; \ R is reflexive.  1 −1 1   3. A =  2 3 0  18 2 10 \ |A| = 1(30 – 0) + 1(20 – 0) + 1(4 – 54) = 30 + 20 – 50 = 0 So, A–1 does not exist. OR We have,

x 1

−5x = x(x + 10) + 5x x + 10 = x2 + 10x + 5x = x2 + 15x = x(x + 15)

4. As the plane has intercepts 2, 3 and 4 on X, Y and Z axes respectively. \ The required equation of the plane is x y z + + = 1 ⇒ 6x + 4y + 3z = 12 2 3 4 5. Since, the given differential equation cannot be expressed as a polynomial. So, its degree is not defined. OR  dy   dy  We have, x +   = 1 +    dx   dx 

2

dy . So, its order is 1. dx 6. Here, f is not a function from A to B as f (1) is not unique. Highest order derivative is

7. We know that, P (S ∩ F ) P (F ) P (S | F ) = = =1 P (F ) P (F )

P (F ∩ F )

=

P (F ) Thus, P(S | F) = P(F | F) = 1

P (F ) =1 P (F )

OR Since, P(A ∩ B) = P(A)P(B), therefore, A and B are independent events. P ( A ∩ B) P ( A)P (B) ∴ P ( A | B) = = = P ( A) P ( B) P ( B) Similarly, P(B|A) = P(B). 0 1  8. We have, A =   1 0  0 1  0 1   0 + 1 0 + 0  1 0  A2 =   = =  1 0  1 0  0 + 0 1 + 0  0 1      9. We are given | a | = 1, | b | = 2 and a ⋅ b = 1   a ⋅b 1 1 = Now, we have cos θ =   = | a | |b | 1× 2 2 π 1   or θ = cos −1   = . 2 3 OR   ^ ^ We have, a = 2 i^ − j^ + k and b = i^ − 2 j^ + k   a ⋅ b  Projection of a on b is  |b |   ^ ^ ^ ^ ^ a ⋅ b = (2 i − j + k ) ⋅ (i − 2 j^ + k ) = 2 + 2 + 1 = 5  | b | = 12 + (−2)2 + 12 = 1 + 4 + 1 = 6  5  . \ Projection of a on b is 6 10. P(A ∪ B) = P(A) + P(B) – P(A ∩ B) 3 5 3 3 1 \ P(A ∩ B) = + − = 1 − = 8 8 4 4 4 P ( A ∩ B) 1/4 2 = = . \ P ( A | B) = P ( B) 5/8 5 11. Let E1 be the event that fair coin is drawn, E2 be the event that 2 headed coin is drawn, and E be the event that tossed coin get a head. \ P(E1) = 1/2, P(E2) = 1/2, P(E|E1) = 1/2 and P(E|E2) = 1 Now, P(E1|E) =

P (E1 ) ⋅ P (E | E1 ) P (E1 ) ⋅ P (E | E1 ) + P (E2 ) ⋅ P (E | E2 )

1 1 1 1 ⋅ 1 2 2 4 = =4= = 1 1 1 1 1 3 3 ⋅ + ⋅1 + 2 2 2 4 2 4 Class 12

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12. Since f(x) = f(–x) = x2 + 1 for all x ∈R, therefore, f is not one-one. 13. Let D1, D2 be the events that we find a defective fuse in the first, second test respectively. Required probability = P(D1 ∩ D2) 2 1 1 = P (D1 )P (D2 | D1 ) = ⋅ = 7 6 21 2π

14. Let I =

dx

∫ esin x + 1

...(i)

0

⇒ I=

2π

dx

∫ esin(2 π− x ) + 1 0

2π

⇒ I=

dx

 Q 

⇒

∫ e − sin x + 1

I=

0

 f (x )dx = ∫ f (a − x )dx   0

a

a

∫

0 2π

e sin x

∫ e sin x + 1 dx

...(ii)

0

Adding (i) and (ii), we get 2π

2 I = ∫ 1 ⋅ dx = 2 π

\ I=π

0

15. Given planes are 2x + 2y – z + 2 = 0 and 4x + 4y – 2z + 5 = 0 Let point P(x1, y1, z1) lie on plane 2x + 2y – z + 2 = 0 ⇒ 2x1 + 2y1 – z1 = –2 4 x1 + 4 y1 − 2z1 + 5 2(2 x1 + 2 y1 − z1 ) + 5 = ∴ d= 16 + 16 + 4 42 + 42 + (−2)2 =

2(−2) + 5 36

1 = unit 6

 −3 6 0  −3 4 0     16. A =  4 −5 8 ⇒ A ′ =  6 −5 −7   0 −7 −2  0 8 −2  0 2 0  0 1 0 1 1    ∴ ( A − A′) =  −2 0 15 =  −1 0 15 / 2 2 2  0 −15 0  0 −15 / 2 0 17. (i) (c) : Let OAB be a triangle B such that     r OA = p , AB = q , OB = – r        Now, p + q = OA + AB    O p = OB = −r (ii) (c) : From triangle law of vector addition,       AC + BD = AB + BC + BC + CD D

C

A

B

 q

   = AB + 2 BC + CD     = AB + 2 BC − AB = 2 BC    (iii) (b) : In DABC, AC = x + y x D

…(i) C

 y

 y A

   and in DABD, y = x + BD

B

 x

…(ii) [By triangle law of addition] Adding (i) and (ii), we have      AC +y = 2 x+ y + BD ⇒ AC − BD = 2 x    (iv) (d) : In DABC, BA + AC = BC …(i) [By triangle law]    …(ii) In DBCD, BC + CD = BD     From (i) and (ii), BA + AC = BD − CD       ⇒ BA + CD = BD − AC = BD + CA (v) (b) : Since S is the mid point of QR   So, QS = SR       Now PQ + PR = (PS + SQ) + (PS + SR) (By triangle law)       [∴ SQ = −QS] = 2PS + SQ + SR = 2PS 18. (i) (d) : Given r cm is the radius and h cm is the height of required cylindrical can. Given that, volume = 5 l = 5000 cm3 (Q 1 L = 1000 cm3) 5000 ⇒ πr2h = 5000 ⇒ h = pr 2 Now, the surface area, as a function of r is given by  5000  S(r) = 2πr2 + 2πrh = 2πr2 + 2πr  2   pr  10000 = 2πr2 + r 10000 (ii) (c) : Now, S(r) = 2 πr 2 + r 10000 ⇒ S′(r) = 4 πr − 2 r To find critical points, put S′(r) = 0 4 πr 3 − 10000

A

  [∴ AB = −CD]

⇒

r2

=0 1/3

10000  2500  ⇒ = ⇒ r=  π  4π 20000 × 4 π Also, S ″(r ) r = 3 2500 = 4 π + 10000 π = 4π + 8π = 12π > 0 Thus, the critical point is the point of minima. r3

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(iii) (b) : The cost of material for the can is 2500 minimized when r = 3 cm and the height is π 5000 2

= 23

2500 cm. π

 2500  π3   π  2 πr 3 + 10000 (iv) (a) : We have, minimum surface area = r 2500 2π ⋅ + 10000 15000 π = = = 1666 .67 cm2 9 2500 3 π Cost of 1 m2 material = `100 1 \ Cost of 1 cm2 material = ` 100 1666.67 \ Minimum cost = ` = 16.7 100 (v) (c) : To minimize the cost we need to minimize the total surface area. 19. Here f(x) = sin4x + cos4x ...(i) On differentiating (i) w.r.t. x, we get f ′(x) = 4sin3x cos x – 4 cos3x sin x = 4 sinx cosx (sin2x – cos2x) = –2 sin2x cos2x = –sin4x π Put f ′(x) = 0 ⇒ x = 4 Intervals Conclusion Sign of f ′(x) –ve as 0 < 4x < π f is decreasing in  π  0,   π 4  0,  4 f is increasing in  π π  +ve as π < 4x < 2π ,   π π 4 2  ,  4 2

By multiplication rule of probability, we have 10 9 3 P(E ∩ F) = P(E) ⋅ P(F|E) = × = 15 14 7 ∞ dx 21. Let I = ∫0 2 (x + 4)(x 2 + 9) ∞ 1 ∞ 1 1  =  ∫0 2 dx − ∫0 2 dx  5 x +4 (x + 9)  ∞ ∞  1  1 1 −1 x  −1 x  =   tan tan −  3  0  5  2 2  0  3 1  1 π   1 π  1  π π  π =   ⋅ − 0  −  ⋅ − 0   =  −  = 5 2 2 3 2 60  5 4 6 3 3 22. Let cot −1 = q ⇒ cot q = 4 4 9 5 \ cosec q = 1 + cot 2 θ = 1 + = 16 4 4 −1 4 \ sin θ = ⇒ θ = sin 5 5  −1 4  4  −1 3  So, sin  cot  = sin  sin = 4 5 5 Let tan–1 x = f. Then, tan f = x

\ sec f = 1 + tan2 φ = 1 + x 2 1 \ cos φ = 1 + x2 1 So, cos(tan −1 x ) = cos φ = 1 + x2 1 4 1 16 Thus, = ⇒ = ⇒ 16 x 2 = 9 2 2 5 25 x + 1 1+ x 9 3 2 ⇒ x = ⇒x=± 16 4 23. Let I = ∫ (e x log a + e a log x + e a log a ) dx x

7

20. Since

∑ P( X = x) = 1 x =1

\ k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1 ⇒ 10k2 + 9k – 1 = 0 ⇒ (10k – 1)(k + 1) = 0 1 ⇒ k= or k = −1 10 Since probability cannot be negative. 1 ∴ k= 10 OR Let E and F respectively denote the events that first and second ball drawn is black. We have to find P(E ∩ F) or P(EF). 10 Now, P(E) = P(black ball in first draw) = 15 9 P ( F | E ) = and 14

a

a

= ∫ (e log a + e log x + e log a ) dx = ∫ (a x + x a + aa ) dx

[

= ∫ a x dx + ∫ x a dx + ∫ a a dx =

e log λ = λ ]

ax x a +1 + + aa x + c log a a + 1

OR −2

Let I = ∫

dx 2 x3

=∫

2 x3

−4

x3 2 x3

dx

− 22

−2

1

Put x 3 = t ⇒ x 3 dx = 3dt dt \ I = 3∫ = 3 log t + t 2 − 4 + c 2 2 t −2 = 3 log

1 x3

+

2 x3

−4 +c Class 12

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dy = cos ( x + y ) dx dy du Put x + y = u ⇒ = −1 dx dx du du = dx + c − 1 = cos u ⇒ ∫ 1 + cos u ∫ dx 24. We have

⇒

1

∫ 2 sec

u   du = x + c ⇒ tan   = x + c 2 2

2 u

x+ y = x +c ⇒ tan   2  25. | A | =

\ A–1 exists.

 4 −2 ∴ adj A =   0 1  1 1  4 −2 1 −1 / 2 ⇒ A −1 = (adj A) =  =  |A| 4  0 1  0 1 / 4  26. Let E be the event that the coin tossed twice shows first head and then tail and F be the event that the coin drawn is fair. P (F ) . P (E/F ) P(F/E) = P (F ) . P (E/F ) + P (F ) P (E/F ) m.1.1 m/4 N 2 2 = = m.1.1 N −m.2.1 m/4 + 2(N − m)/9 + N 2 2 N 3 3 9m = . m + 8N 27. We have, L.H.L. (at x = 4) x−4 + a = −1 + a = lim− f (x ) = lim− x→4 x→4 | x − 4| R.H.L. (at x = 4) x → 4+

x −4

x → 4+

|x − 4|

...(i)

Required distance = (2 + 1)2 + (−1 + 5)2 + (2 + 10)2 = 9 + 16 + 144 = 13 units

1 2 = 4−0 = 4 ≠ 0 0 4

= lim f (x ) = lim

OR Any point on line ^ ^ ^ ^ ^ ^  r = 2 i − j + 2 k + λ(3 i + 4 j + 12 k ) is (2 + 3l, –1 + 4l, 2 + 12l)  ^ ^ ^ If it lies on the plane r ⋅ (i − j + k ) = 5, then, 1(2 + 3l) – 1(–1 + 4l) + 1(2 + 12l) = 5 ⇒ 2 + 3l + 1 – 4l + 2 + 12l = 5 ⇒ 11l = 0 ⇒ l = 0 Putting l = 0 in (i), we get (2, –1, 2)

+b =1+b

and f(4) = a + b Since f(x) is continuous at x = 4. \ –1 + a = a + b ⇒ b = –1 and 1 + b = a + b ⇒ a = 1 Thus, f(x) is continuous at x = 4 if a = 1 and b = –1. 28. Here, (x1, y1, z1) = (2, 3, 4), a = 2, b = 1, c = –2 Cartesian equation is a(x – x1) + b(y – y1) + c(z – z1) = 0 ⇒ 2(x – 2) + 1(y – 3) –2(z – 4) = 0 ⇒ 2x – 4 + y – 3 – 2z + 8 = 0 ⇒ 2x + y – 2z = –1

29. Given (a, b) R(c, d) ⇒ ad = bc ∀ (a, b), (c, d) ∈ N × N Reflexive : Let (a, b) ∈ N × N when a ∈ N, b ∈ N (a, b) R (a, b) ⇒ ab = ba, which is true So, R is reflexive. Symmetric : Let (a, b), (c, d) ∈ N × N (a, b) R (c, d) ⇒ ad = bc ⇒ cb = da ⇒ (c, d) R(a, b) So R is symmetric. Transitive : Let (a, b), (c, d), (e, f ) ∈ N × N (a, b) R(c, d) ⇒ ad = bc (c, d) R (e, f) ⇒ cf = de ⇒ adcf = bcde ⇒ af = be ⇒ (a, b) R (e, f ) So R is transitive. Since R is reflexive, symmetric and transitive, so R is an equivalence relation. 30. Let V be the volume of a closed cuboid with length x, breadth x and height y. Let S be the surface area of cuboid. Then x2y = V and S = 2(x2 + xy + xy) = 2(x2 + 2xy) 2V  V   ∴ S = 2  x 2 + 2x ⋅ 2  = 2  x 2 +  x   x  2V  dS  ∴ = 2 2 x − 2  dx  x  dS \ = 0 ⇒ x3 = V = x2y ⇒ x = y dx d2S  4V  Now, 2 = 2 2 + 3  > 0 dx  x  \ x = y will give minimum surface area and x = y, means all the sides are equal. \ Cube will have minimum surface area. 31. Given equation is (x + y )2

dy = a2 dx

dy du = dx dx du  du  \ u 2  − 1 = a 2 ⇒ u 2 = u2 + a2  dx  dx Put x + y = u ⇒ 1+

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u2

du = ∫ dx ⇒ ∫ du − ∫ 2

⇒

∫ u2 + a

⇒

u 1 u − a2 × tan −1 = x + C a a x+y  = x +C x + y − a tan −1   a  x+ y =C y − a tan −1   a  x+ y +C y = a tan −1   a 

⇒ ⇒ ⇒

a2du 2

u +a

2

= x +C

...(i)

I.F. = e ∫ = e tan x \ The solution is given by sec2 x dx

tan x

32. Given

y2

x2

=1 a 2 b2 Let x = a sec q and y = b tan q dy dx ⇒ = a sec θ tan θ and = b sec2 θ dθ dθ dy dy / dθ ∴ = dx dx / dθ b sec2 θ dy dy b sec θ ⇒ = ⇒ = dx a sec θ tan θ dx a tan θ dy b cosec θ ⇒ = dx a Differentiating with respect to x, we get d 2 y −b cosec θ cot θ dθ × = a dx dx 2 2 d y −b cosec θ cot θ −b ⇒ = = 2 2 a (a sec θ tan θ) a tan3 θ dx −

y  From(i), tan θ =  b 

3

a ×y d2 y −b 4 ∴ = dx 2 a2 y 3 1 + x cos x

x(1 − x 2 e 2 sin x )

dx

Put xesinx = t ⇒ esinx(1 + x cosx)dx = dt dt dt =∫ ∴ I=∫ 2 t (1 + t )(1 − t ) t (1 − t ) 1 A B C Let = + + t (1 + t )(1 − t ) t 1 + t 1 − t

We have,

y ⋅(e tan x ) = ∫ tan x ⋅ sec2 x ⋅ e dx + c Let I = ∫ tan x ⋅ sec2 x ⋅ e tan x dx , cos x ≠ 0 Put tanx = t ⇒ sec2x dx \ I = ∫ tet dt = tet − e t ⇒ I = tan x . e tan x − e tan x Putting this value in (ii), we get ye tan x = (tan x − 1)e tan x + c ⇒ y = (tanx –1) + ce–tanx

2

33. We have, I = ∫

OR dy cos2 x = tan x − y dx dy  1  tan x ⇒ + 2  y = dx  cos x  cos2 x This is a linear differential equation of the form dy 1 tan x + Py = Q, where P = ,Q = 2 dx cos x cos2 x

=

−b × b3

...(ii)

⇒ 1 = A(1 – t2) + B(t – t2) + C(t + t2) Equating the coefficients of t2, t and constant terms, we get A + B – C = 0, B + C = 0 and A = 1 Solving these equations, we get −1 1 A = 1, B = and C = 2 2 dt dt 1 dt 1 dt ∴ I=∫ =∫ − ∫ + 2 t 2 1+ t 2 ∫ 1− t t (1 − t ) 1 1 = log t − log | 1 + t | − log | 1 − t | + c 2 2 1 sin x = log | xe | − log | 1 − x 2e 2 sin x | + c 2 34. We have, y = 3x –5, y = 0, x = 3, x = 5

...(i) 5

  3x 2 Required area = ∫ (3x − 5) dx =  − 5x  3  2 3 3 2 2 = (5 – 3 ) – 5(5 – 3) 2 3 = (25 – 9) – 5 × 2 = 24 – 10 = 14 sq. units 2 OR The line y – 1 = x meets the X-axis at the point where y = 0, i.e., where 0 – 1 = x, i.e., at the point (–1, 0). It meets Y-axis at the point where x = 0, i.e., where y – 1 = 0, i.e., at the point (0, 1). Required area (shown shaded) is given by 5

−1

=

∫ −2

3

(x + 1) dx + ∫ (x + 1) dx −1

Class 12

98

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−1

 r = 2i + 3j + 4k + λ(i + j + 5k )

3

 (x + 1)2   (x + 1)2  + =      2  −2  2  −1 1 1 (0 − 1) + (42 − 0) 2 2 1 17 = + 8 = sq. units 2 2 =

1− t2 2t and y = 2 1+ t 1+ t2 Putting t = tanq in both the equations, we get 1 − tan2 θ ...(i) x= = cos 2θ 1 + tan2 θ 2 tan θ = sin 2θ and y = ...(ii) 1 + tan2 θ Differentiating (i) and (ii), we get dy dx = −2 sin 2θ and = 2 cos 2θ dθ dθ dy dy dθ cos 2θ x Therefore, =− = =− dx sin 2θ y dx dθ 36. In a parallelogram, diagonals bisect each other. \ Mid point of BD = Mid point of AC  x + 2 y + 3 z + 4   4 + 1 5 + 2 10 − 1  , , , , ⇒   =  2 2 2   2 2 2  35. We have x =

(x, y, z) D

A (4, 5, 10)

(1, 2, –1) C

B (2, 3, 4)

⇒ x + 2 = 5, y + 3 = 7, z + 4 = 9 ⇒ x = 3, y = 4, z = 5 So coordinates of D are (3, 4, 5). Cartesian equation of side AB is x − 4 y − 5 z − 10 = = 2 − 4 3 − 5 4 − 10 x − 4 y − 5 z − 10 x − 4 y − 5 z − 10 ⇒ = = ⇒ = = −2 −6 1 1 3 −2 and vector equation of side AB is  r = 4i + 5j + 10k + λ(i + j + 3k ) Again, cartesian equation of side BC is x −2 y −3 z −4 = = 1 − 2 2 − 3 −1 − 4 x −2 y −3 z −4 x −2 y −3 z −4 ⇒ = = ⇒ = = 1 1 5 −1 −5 −1 and vector equation of side BC is

OR Given planes are x – cy – bz = 0 ...(i) cx – y + az = 0 ...(ii) bx + ay – z = 0 ...(iii) , The d.r. s of normal to plane (i), (ii) and (iii) are (1, –c, –b), (c, –1, a) and (b, a, –1) respectively. All planes pass through same line, then the line is perpendicular to each of the three normals. The d.r's. of line from planes (i) and (ii) are −c −b 1 −b 1 −c ,− , −1 a c a c −1 i.e., – ac – b, – a – bc, –1 + c2 ...(iv) , The d.r. s of line from planes (ii) and (iii) are −1 a c a c −1 ,− , a −1 b −1 b a i.e., 1 – a2, c + ab, ac + b ...(v) , The d.r. s in (iv) and (v) are in proportion, then −ac − b −a − bc −1 + c 2 = = c + ab ac + b 1 − a2 −ac − b −a − bc ⇒ = c + ab 1 − a2 ⇒ ac2 + bc + a2bc + ab2 = a + bc – a3 – a2bc ⇒ c2 + abc + b2 = 1 – a2 – abc ⇒ a2 + b2 + c2 + 2abc = 1

37. Converting inequations into equations and drawing the corresponding lines, we have x + y = 10, 3x – 2y = 15, x = 6 x y x  y  i.e., + = 1, +  = 1, x = 6 10 10 5  −7.5  Also, x ≥ 0, y ≥ 0 solution lies in first quadrant

B is the point of intersection of the lines x = 6 and  3 3x – 2y = 15 i.e., B ≡  6,  .  2 C is the point of intersection of the lines x = 6 and x + y = 10 i.e., C ≡ (6, 4). 99

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 3 We have points A(5, 0), B  6,  , C(6, 4) and D(0, 10).  2 Now Z = x + y \ Z(O) = 0 + 0 = 0 Z(A) = 5 + 0 = 5 3 Z(B) = 6 + = 7.5 2 Z(C) = 6 + 4 = 10 Z(D) = 0 + 10 = 10 \ Z has maximum value 10 at two points C(6, 4) and D(0, 10). OR Converting inequations into equations and drawing the corresponding lines, we have x1 + 2x2 = 2 ; 2x1 + 4x2 = 8 x x x x i.e 1 + 2 = 1 , 1 + 2 = 1 4 2 2 1 The constraints are shown by the graph.

Also, x1 ≥ 0, x2 ≥ 0 solution lies in first quadrant. We have points A(2, 0) and B(0, 1). Now, Z = 7x1 – 3x2 \ Z(O) = 7(0) – 3(0) = 0 Z(A) = 7(2) – 3(0) = 14 Z(B) = 7(0) – 3(1) = –3 \ Z has maximum value 14 at point (2, 0) and minimum value –3 at point (0, 1). 3 2   4 38. We have, X  = 1 −1  2 3 2  4 and B =  Let A =   1 −1 2

1 3 1 3

\ |A| = –3 – 2 = –5 ≠ 0, so A–1 exists. Now, the given matrix equation is XA = B \ X(AA–1) = BA–1 ⇒ X = BA–1

...(i)

Now, cofactors of matrix A are given by A11 = –1, A12 = –1, A21 = –2, A22 = 3 1

 −1 −1  −1 −2 \ adj A =   =   −2 3   −1 3  1 1  −1 −2 (adj A) = −  \ A–1 =  A 5  −1 3  Now, from (i), X = BA–1  1   4 1  −1 −2 ⇒ X = −    5   2 3  −1 3   1   −4 − 1 −8 + 3 ⇒ X = −    5   −2 − 3 −4 + 9 1 1  ⇒ X=   1 −1 OR 1 0 1  Let A = 0 2 3 1 2 1 \ |A| = 1(2 – 6) – 0 + 1(0 – 2) = –4 – 2 = –6 ≠ 0 So, A–1 exists. Let Aij be the co-factor of aij in |A|. Then A11 = –4, A12 = 3, A13 = –2 A21 = 2, A22 = 0, A23 = –2 A31 = –2, A32 = –3, A33 = 2  −4 2 −2  ∴ adj A =  3 0 −3  −2 −2 2   −4 2 −2 −1  1 −1 (adj A) = 3 0 −3 Hence, A = 6  |A|  −2 −2 2   4 −2 2  1  =  −3 0 3  6  2 2 −2 Now, |adj A| = –4(0 – 6) –2(6 – 6) –2(–6 – 0) = 24 – 0 + 12 = 36 = 62 = (–6)2 = |A|2



Class 12

100

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Self Evaluation Sheet Once you complete SQP-7, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Chapter

Marks Per Question 1

Integrals / Integrals Relations and Functions

Marks Obtained

1

Determinants / Determinants Three Dimensional Geometry Differential Equations / Differential Equations Relations and Functions Probability / Probability Matrices Vector Algebra / Vector Algebra Probability Probability Relations and Functions Probability Integrals Three Dimensional Geometry Matrices Vector Algebra Application of Derivatives Application of Derivatives Probability / Probability Integrals Inverse Trigonometric Functions Integrals / Integrals Differential Equations Determinants Probability Continuity and Differentiability Three Dimensional Geometry / Three Dimensional Geometry Relations and Functions Application of Derivatives Differential Equations / Differential Equations Continuity and Differentiability Integrals Application of Integrals / Application of Integrals Continuity and Differentiability Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming Determinants / Determinants Total

1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80 Percentage

.................. ..............%

Performance Analysis Table If your marks is

> 90% TREMENDOUS!

You are done! Keep on revising to maintain the position.

81-90% EXCELLENT!

You have to take only one more step to reach the top of the ladder. Practise more.

71-80% VERY GOOD!

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70% GOOD!

Revise thoroughly and strengthen your concepts.

51-60% FAIR PERFORMANCE!

Need to work hard to get through this stage.

40-50% AVERAGE! Mathematics

Try hard to boost your average score.

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101

SQP

8

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

–

1(3)

–

3(5)

2.

Inverse Trigonometric Functions

1(1)

1(2)

–

–

2(3)

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

–

4(9)

5.

Continuity and Differentiability

1(1)

1(2)

2(6)#

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

1(1)*

1(2)*

2(6)#

–

4(9)

–

1(2)

–

–

1(2)

7.

Integrals

8.

Application of Integrals

9.

Differential Equations

1(1)*

1(2)*

1(3)

–

3(6)

10.

Vector Algebra

1(1)

1(2)*

–

–

2(3)

+ 1(4)

–

–

1(5)*

4(11)

–

–

–

1(5)*

1(5)

4(4)#

2(4)

–

–

6(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

2(2)#

Probability Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-8

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Evaluate :

Evaluate :

∫ ∫

(a x + b x )2 axbx

dx OR

dx 1 − 2x − x 2

1 3  2 2. If A =   and A – kA – 5I = O, then find the value of k. 3 4   3. A black and a red die are rolled together. Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4. OR If P(A) = 0.4, P(B) = 0.8 and P(B | A) = 0.6, then find P(A ∪ B). 2

 2 tan x  4. Differentiate the function   w.r.t. x.  tan x + cos x  103

Mathematics

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y+z z+x . x+y

1 x 5. Find the cofactors of the element of third row and second column of the following determinant 1 y 1 z OR If A is a matrix of order 3 × 3 and |A| = 5, then find the value of |adj A|.

6. Set A has three elements and set B has four elements. Find the number of injections that can be defined from A to B. 7. Find the solution of the differential equation

dy = x 3 e −2 y . dx OR

Find the solution of y′ = y cot 2x. 8. Find the principal value of cot −1 (− 3 ) . 9. Find the direction cosines of a line, for which a = b and g = 45°. OR If P = (–2, 3, 6), then find the d.c.'s of OP. 10. How many equivalence relations on the set {1, 2, 3} containing (1, 2) and (2, 1) are there in all ? 11. If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(a) with x-axis, then find the value of a . 12. If A and B are two independent events such that P(A ∪ B) = 0.6 and P(A) = 0.2, then find P(B).  2x + y 3 y   6 0 ′ = , then x = _____. 13. If  4   6 4   0 14. If A and B are events such that P(A) > 0 and P(B) ≠ 1, then prove that P(A′ | B′) = 15. Find the value of k in the following probability distribution. X=x P(X = x)

0.5 k

1 k2

1.5 2k2

1 − P ( A ∪ B) . P ( B ′)

2 k

π 16. If the angle between ɵi + kɵ and ɵi + ɵj + akɵ is , then find the value of a. 3

Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A poster is to be formed for a company advertisement. The top and bottom margins of poster should be 4 cm and the side margins should be 6 cm. Also, the area for printing the advertisement should be 384 cm2. Based on the above answer the following : (i) If a be the width and b be the height of poster, then the area of poster, expressed in terms of a and b, is given by (a) 288 + 8a + 12b (b) 8a + 12b (c) 384 + 8a + 12b (d) none of these (ii) The relation between a and b is given by 12b 288 + 12b (b) a = (a) a = b−8 b−8

(c) a =

12b b+8

(d) none of these Class 12

104

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(iii) Area of poster in terms of b is 2 288b + 12b 288b + 12b2 (b) (c) (a) 12b b+8 b−8 b−8 (iv) The value of b, so that area of the poster is minimized, is (a) 24 (b) 36 (c) 18

2

(v) The value of a, so that area of the poster is minimized, is (a) 24 (b) 36 (c) 18

(d)

12b2 b+8

(d) 22 (d) 22

18. Consider the earth as a plane having points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6) on it. A mobile tower is tied with 3 cables from the point A, B and C such that it stand vertically on the ground. The peak of the tower is at the point (6, 5, 9), as shown in the figure. (6, 5, 9)

B(5, 2, 4) A(3, –1, 2)

C(–1, –1, 6)

Based on the above answer the following : (i) The equation of plane passing through the points A, B and C is (a) 3x – 4y + 3z = 0 (b) 3x – 4y + 3z = 19 (c) 4x – 3y + 3z = 0

(d) 4x – 3y + 3z = 19

(ii) The height of the tower from the ground is 5 6 units (d) units 34 34 (iii) The equation of line of perpendicular drawn from its peak to the ground is x −6 y −4 z −9 x −6 y −5 z −9 (a) (b) = = = = −5 3 3 −4 3 3 x −6 y −5 z −9 (c) x − 6 = y − 4 = z − 9 (d) = = 3 4 3 3 5 3 (iv) The coordinates of foot of perpendicular are 91 93 144  93 97 144  144 97 93  (b)  (a)  , , (c)  , , (d) none of these , ,  17 17 17   17 17 17   17 17 17  (v) The area of DABC is (b) 2 34 sq. units (c) 17 sq. units (d) 2 7 sq. units (a) 34 sq. units (a) 6 units

(b) 5 units

(c)

PART – B Section III 19. Find the derivative of the function 20. Evaluate : Evaluate :

∫

10 x 9 + 10 x log e 10 10 x + x10

∫ sin x +

1 3 cos x

a + a + x w.r.t. x.

dx OR

dx 105

Mathematics

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21. A random variable X has the following probability distribution: X

0

1

2

3

4

5

6

7

P(X)

0

K

2K

2K

3K

K2

2K2

7K2 + K

Determine: (i) K

(ii) P(X < 3)

22. If sin [cot–1 (x + 1)] = cos (tan–1x), then find x. 23. Solve the differential equation cos2 ( x − 2 y ) = 1 − 2

dy . dx

OR dy Find the solution of the differential equation x + y = sec( x 2 + y 2 ) . dx 2

2

24. Find the equation of normal to the curve x 3 + y 3 = 2 at (1, 1). 25. If P(not A) = 0.7, P(B) = 0.7 and P(B | A) = 0.5, then find P(A |B) and P(A ∪ B).  2 −3 26. Find the inverse of the matrix A =  .  −4 7  27. Compute the shaded area shown in the given figure. Y

x= 8 O

X

1

y  x3

    28. Find | a × b |, if a = ɵi + 3ɵj − 2kɵ and b = −ɵi + 3kɵ . OR   Find the angle between two vectors a and b having the same length

2 and their scalar product is –1.

Section - IV 29. Let a relation R on the set A of real numbers be defined as (a, b) ∈ R ⇒ 1 + ab > 0 for all a, b ∈ A. Show that R is reflexive and symmetric but not transitive. 2

30. Sketch the graph y = | x + 1|. Evaluate 2 31. Evaluate : ∫ x + 9 dx x 4 + 81

∫ | x + 1 | dx. −4

OR Evaluate : ∫ x 2 sin 2 x dx −1  dy  32. Solve : sin   = x + y  dx 

Class 12

106

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 2 1  x sin   , for x ≠ 0 33. If f (x ) =  , then show that the function is discontinuous at x = 0. x 1, for x = 0 34. If (ax + b)

ey/x

2

d2 y

 dy  = x, then show that x = x − y . 2  dx  dx 3

OR Find

dy t 1  , when x = a cos t + log tan2  and y = a sin t.  dx 2 2

35. Show that the condition that the curves ax2 + by2 = 1 and a′x2 + b′y2 = 1 should intersect orthogonally is 1 1 1 1 − = − . a b a ′ b′ Section-V 1 −1 0   36. If A = 2 5 3 , then find A–1. Hence find |adj A| and |A–1|. 0 2 1 OR  3 Find the inverse of A =  −2  2

−10 8 −4

−1  2  . Hence find (A–1)2. −2 

 37. Find the vector equation of the plane passing through the intersection of the planes r ⋅ (2ɵi − 7 ɵj + 4kɵ ) = 3  and r ⋅ (3ɵi − 5ɵj + 4kɵ ) + 11 = 0 and passing through the point (–2, 1, 3). OR y +3 z +5 = Find the co-ordinates of the points on the line x − 2 = , which are on either side of the point 2 −2 A(2, –3, –5) at a distance of 3 units from it. 38. Solve the following LPP graphically : Maximize Z = 600x + 400y subject to the constraints : x + 2y ≤ 12, 2x + y ≤ 12 5 x + y ≥ 5 and x, y ≥ 0. 4 OR Find the number of points at which the objective function z = 3x + 2y can be maximized subject to 3x + 5y ≤ 15, 5x + 2y ≤ 20, x ≥ 0, y ≥ 0.

107

Mathematics

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SOLUTIONS 1. We have,

∫

(a x + b x )2 axbx

dx = ∫

a2 x + b2 x + 2a x b x

x

axbx

dx

x

b  a      a   b   b a = ∫    +   + 2  dx = + 2x + C, a ≠ b +      b  a b a log logg b a OR dx dx Let I = ∫ =∫ 2 2 − (x 2 + 2 x + 1) 1 − (x + 2x ) x

=

∫

x

dx 2 − (1 + x )

2

=∫

 2 tanx  4. Let y =    tan x + cos x 

Differentiating w.r.t. x, we get (tan x + cos x ) ⋅ 2 sec2 x − 2 tan x ⋅  2 tan x  (sec2 x − sin x ) dy ⋅ = 2  dx (tan x + cos x )2  tan x + cos x  =

8 tan x (cos x sec2 x + tan x sin x ) (tan x + cos x )3

dx ( 2 ) − (1 + x )2 2

=

8 tan x (sec x + tan x sin x )

Let 1 + x = z ⇒ dx = dz ∴ I=∫

dz

= sin −1

( 2 )2 − z 2

−1  1 + x  +c + c = sin   2  2

z

2. Given, A2 – kA – 5I = O ⇒ kA = A2 – 5I 1 3  1 3   1 0  ⇒ kA =    − 5  3 4  3 4  0 1  1 3  10 15  5 0  5 15  = 5 − = =  = 5A    3 4  15 25 0 5 15 20 ⇒ kA = 5A \ k = 5 3. Let E : ‘a total of 8’ and F : ‘red die resulted in a number less than 4’ i.e., E = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} and F = {(x, y) : x ∈ {1, 2, 3, 4, 5, 6}, y ∈ {1, 2, 3}} i.e., F = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1),

(3, 2), (3, 3), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (6, 1), (6, 2), (6, 3)} Hence, E ∩ F = {(5, 3), (6, 2)} P(E) = 5/36, P(F) = 18/36, P(E ∩ F) = 2/36 \ Required probability = P (E | F) P (E ∩ F ) 2 / 36 2 1 = = = = . P (F ) 18 / 36 18 9 OR Given, P(A) = 0.4, P(B) = 0.8 and P(B|A) = 0.6 Clearly, P(A ∩ B) = P(B|A)P(A) = 0.6 × 0.4 = 0.24 Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.4 + 0.8 – 0.24 = 0.96

2

5. M32 =

(tan x + cos x )3 1 y+z =z+x–y–z=x–y 1 z+x

⇒ c32 = – M32 = y – x OR |adj A| = |A|n–1 = 5 (3–1) = 52 = 25 6. Since 3 < 4, injective functions from A to B are defined and the total number of such functions is 4P3 =

4! = 4 × 3 × 2 × 1 = 24. (4 − 3)!

dy = x 3 e −2 y ⇒ e2y dy = x3dx dx e2 y x 4 On integrating, we get = + C′ 2 4 ⇒ 2 e2y = x4 + C, where C = 4 C′ 7. We have,

OR dy = y cot 2x We have, y′ = y cot 2x ⇒ dx dy ⇒ = cot 2x dx y Integrating both sides, we get dy = ∫ cot 2x dx y 1 ⇒ log |y| = log |sin 2x| + log c 2

∫

⇒ log |y| = log sin2x + log c ⇒ log |y| = log c sin 2 x ⇒ y = c sin2 x Class 12

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π 5π  5π = cot  π −  = cot ⇒ θ= ∈ (0, π)   6 6 6 5π \ Principal value of cot–1 − 3 is . 6 9. Since, cos2 a + cos2 b + cos2 g = 1 ⇒ 2cos2 a + cos2 45° = 1 (∵ a = b) 1 1 1 ⇒ 2 cos2 a = 1 − = ⇒ cos2 a = 2 2 4 1 ⇒ cos a = ± 2 1 1 1 So, dc’s are  ± , ± ,   2 2 2

( )

OR Here, O ≡ (0, 0, 0) and P ≡ (–2, 3, 6) Direction ratios of OP are –2 –0, 3 – 0, 6 – 0 i.e., –2, 3, 6 \ Direction cosines of OP are −2 3 6 , , > < 2 2 2 2 2 2 2 (−2) + 3 + 6 (−2) + 3 + 6 (−2) + 32 + 62 −2 3 6 , , > 7 7 7 10. Possible equivalence relations are {(1, 2), (2, 1), (1, 1), (2, 2), (3, 3)} and {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (2, 3), (3, 2)} Hence, there are two possible equivalence relations. i.e.,
[ Perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (6, 5, 9), therefore its equation is x −6 y −5 z −9 = = −4 3 3 (iv) (a) : Let the coordinates of foot of perpendicular are (3l + 6, –4l + 5, 3l + 9) Since, this point lie on the plane 3x – 4y + 3z = 19, therefore we get 3(3l + 6) – 4(–4l + 5) + 3(3l + 9) – 19 = 0

1 ɵ ɵ ɵ (2i + 3 j + 2k ) × (−4ɵi + 4kɵ ) 2

ɵi 1 = 2 2 −4

18. (i) (b) : The equation of plane passing through three non-collinear points is given by x − x1 y − y1 z − z1 x2 − x1 y2 − y1 z 2 − z1 = 0 x3 − x1 y3 − y1 z 3 − z1

1   | AB × AC | 2

ɵj

kɵ

3

2

0

4

1 ɵ | 12i − 16 ɵj + 12kɵ | 2 1 = 122 + 162 + 122 2 1 = 544 = 2 34 sq. units 2 =

1/2

(

)

19. Let y = a + a + x = a + a + x Differentiating w.r.t. x, we get 1 −1 d dy 1 = (a + a + x ) 2 (a + a + x ) dx 2 dx 1   1 −1 d 1 = (a + x )  (a + x ) 2 dx  2 a + a + x  2 1

=

(0 + 1) =

4 a+x a+ a+x 20. Let I = ∫

10 x 9 + 10 x log e 10 10 x + x10

1 4 a+x a+ a+x dx

Put 10x + x10 = t ⇒ (10x loge10 + 10x9)dx = dt dt ∴ I=∫ t = loget + c = loge(10x + x10) + c OR Let I = ∫

1 sin x + 3 cos x

dx =

1 2∫ 1 2

dx sin x +

3 cos x 2 Class 12

110

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⇒ I=

⇒ I=

1 2∫

1 π  sin  x +   3

dx =

1 π  cosec  x +  dx ∫  2 3

24. Differentiating x 2 / 3 + y 2 / 3 = 2 with respect to x, we get 1/ 3

dy 2 −1/ 3 2 −1/ 3 dy y + y =0 ⇒ = −  x x dx dx 3 3

1 x π log tan  +  + C 2 6 2

21. (i) Since SP(X) = 1 \ 0 + K + 2K + 2K + 3K + K2 + 2K2 + 7K2 + K = 1 ⇒ 10K2 + 9K – 1 = 0 −9 ± 81 + 40 −9 ± 11 1 , ⇒ K= = = −1 20 20 10 Since the probability of the event lies between 0 and 1. 1 So, K = . 10 (ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) 1  3 ∵ K =  = 0 + K + 2K = 3K = 10 10 22. We have, sin[cot–1 (x + 1)] = cos (tan–1x) ... (i) –1 –1 Let cot (x + 1) = A and tan x = B 1 ⇒ x + 1 = cot A ⇒ sin A = (x + 1)2 + 1 1 Also, x = tan B ⇒ cos B = x2 +1 Now, sin A = cos B [From (i)] 1 1 ⇒ = ⇒ (x + 1)2 + 1 = x2 + 1 2 2 (x + 1) + 1 x +1 ⇒ 1 + 2x = 0 ⇒ x = −

1 2

dy dx 2dy du = x – 2y = u ⇒ 1 − dx dx

23. Given, cos2 (x – 2y) = 1 − 2 Let,

\ equation (i) becomes cos2 u = ⇒

2 ∫ dx = ∫ sec u du

...(i)

\ Slope of the tangent at (1, 1) = – 1 Also, the slope of the normal at (1, 1) is given by −1 =1 slope of the tangent at (1, 1) Therefore, the equation of the normal at (1, 1) is y – 1 = 1(x – 1) ⇒ y – x = 0 25. We have, P(not A) = 0.7 or P(A) = 0.7 ⇒ 1 – P(A) = 0.7 ⇒ P(A) = 0.3

∵P(A) + P(A) = 1  

P(A ∩ B) P(A) P(A ∩ B) ⇒ 0. 5 = ⇒ P(A ∩ B) = 0.15 0. 3 P(A ∩ B) 0.15 3 \ P ( A | B) = = = P(B) 0.7 14 and P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = 0.3 + 0.7 – 0.15 = 0.85 Now, P ( B | A) =

26. We have, | A | =

2 −3 = 14 − 12 = 2 ≠ 0 −4 7

So, A–1 exists 7 3 ∴ adj A =   4 2 1 Hence, A−1 = (adj A) |A| 1  7 3  7 / 2 3 / 2  = . = 1  2 4 2  2 27. Required area 0

0

du dx

⇒ x = tanu + c ⇒ x = tan(x – 2y) + c OR dy = sec( x 2 + y 2 ) dx dy 1 du Put x 2 + y 2 = u ⇒ x + y = dx 2 dx

=

∫ −8

x

1/3

 x 4 /3  3 dx =   = [0 − (−8)4/3 ] 4 / 3  −8  4

3 3 [−(−2)4 ] = × 16 = 12 sq . units 4 4   28. We are given, a = ɵi + 3ɵj − 2kɵ and b = −ɵi + 3kɵ =

We have x + y

ɵi ɵj kɵ   ∴ a × b = 1 3 −2 −1 0 3

\ 1 du = sec u ⇒ ∫ cos u du = 2 ∫ dx 2 dx

= (9 − 0)ɵi − (3 − 2)ɵj + (0 + 3)kɵ = 9ɵi − ɵj + 3kɵ   \ | a × b | = 92 + (−1)2 + 32 = 81 + 1 + 9 = 91

⇒ sin u = 2x + c ⇒ sin(x2 + y2) = 2x + c

111

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OR

  Let q be the angle between vectors a and b .

    We have, | a | = | b | = 2 and a ⋅ b = − 1   a⋅b −1 1 =− \ cos θ =   ⇒ cosθ = 2 | a || b | 2× 2 2π ⇒ θ = 2π 3 3  2π  Hence, the angle between a and b is . 3 ⇒ cos θ = cos

30. We have, y = | x + 1|  −(x + 1) x < − 1 \ y =  (x + 1) x ≥ − 1 The rough sketch of the curve y = |x + 1| is shown in figure. Y

–(

x = –4 X

x+

( y=

1)

(–4, 0)

x+

1) x= 2

.

(–1, 0) O

(2, 0)

X

Y

−1

2

\

2

∫ | x + 1|dx = ∫ −(x + 1)dx + ∫ (x + 1)dx −4

−1

−4 −1

 x2  = −  + x +  2  −4

2

 x2   2 + x   −1

 1   16  = −  − 1 −  − 4  + 2 2    

 4  1   2 + 2 −  2 − 1   

1 9 9    1 = −  − − 4 + 4 +  = + = 9 2 2 2 2   

⇒ I= ∫

x 4 + 81

dx ⇒ I = ∫

1 + 9 / x2 2

9 x 2 +   − 18 + 18 x

1 + 9 / x2 dx 81 x2 + x2 1 + 9 / x2

dx = ∫

2

9   x −  + 18 x

dx

9 9   = t ⇒ 1 +  dx = dt x  x2  dt dt ⇒ I=∫ \ I=∫ 2 2 t + 18 t 2 + (3 2 ) 1  t  tan −1  +c ⇒ I=  3 2  3 2 Let x −

29. Reflexive : Let a be any real number, then ( a2 > 0 for all a ∈ A) 1 + aa = 1 + a2 > 0 So, R is reflexive. Symmetric : Let (a, b) ∈ R, then 1 + ab > 0 = 1 + ba > 0 ( ab = ba for all a, b ∈ A) ⇒ (b, a) ∈ R Thus, (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈ A. Hence, R is symmetric. Transitive : We observe that   1 1  1, 2  ∈R and  2 , − 1 ∈R but (1, – 1) ∉ R because 1 + 1 × (–1) = 0 0 Hence, R is not transitive on A.

y=

31. Let I = ∫

x2 + 9

⇒ I=

 x2 − 9  tan −1  +c  3 2x  3 2 1

OR 2

Let I = ∫ x sin 2 x dx I

II

 − cos 2 x   − cos 2 x  = x2   dx  − ∫ 2 x ⋅   2 2  −1 = x 2 cos 2 x + ∫ x cos 2 x dx 2 I II  −1 sin 2 x   sin 2 x  −∫ = x 2 cos 2 x +  x  dx   2 2   2   −1 2 x sin 2 x 1 = x cos 2 x + + cos 2 x + c 2 2 4 2 −x x cos 2 x \ I= cos 2 x + sin 2 x + +c 2 2 4  dy  32. We are given that sin −1   = x + y  dx  dy ...(i) ⇒ = sin(x + y) dx dy dv dy dv ⇒ Let x + y = v. Then, 1+ = −1 = dx dx dx dx dv ∴ From (i), − 1 = sin v dx dv dv = 1 + sin v ⇒ ⇒ = dx dx 1+ sin v 1 [Integrating both sides] ⇒ ∫ dv = ∫ dx 1+ sinv 1 − sin v 1 − sin v dx = ∫ dv ⇒ ∫ dx = ∫ dv ⇒ ∫ cos 2 v 1 − sin 2 v ⇒ ∫ dx = ∫ (sec2 v − tan v sec v)dv

⇒ x = tanv – secv + C ⇒ x = tan(x + y) – sec(x + y) + C, which is the required solution. Class 12

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33. We have, f(0) = 1

...(1)

1 lim f (x ) = lim x 2 sin   x x →0 x →0 1 1 We have, −1 ≤ sin ≤ 1 ⇒ − x 2 ≤ x 2 sin ≤ x 2 x x 1   ⇒ lim (− x 2 ) ≤ lim x 2 sin   ≤ lim x 2  x  x →0 x →0 x →0 1 ⇒ 0 ≤ lim x 2 sin   ≤ 0 x x →0 1 2 ⇒ lim x sin   = 0 x x →0 ⇒ lim f (x ) = 0

dx a cos2 t ⇒ = dt sin t ...(2)

x →0

Differentiating w.r.t.t, we get dy dx 1 t 1  = a − sin t + = a cos t sec2 ⋅  and dt dt tan t / 2 2 2  1 dx   ⇒ = a − sin t +  2 sin(t / 2)cos(t / 2)  dt  1  dx   − sin2 t + 1 dx ⇒ = a ⇒ = a − sin t +    sin t  dt dt   sin t 

From (1) & (2), lim f (x ) ≠ f (0)

⇒

dy dy / dt a cos t = = = tan t dx dx / dt a cos2 t sin t

x →0

\ f is discontinuous at x = 0 34. Given, (ax + b)ey/x = x x ⇒ ey/x = ax + b Taking log on both sides, we get y x ⋅ log e = log x ax + b ⇒ y = log x − log (ax + b) ( x Differentiating w.r.t. x, we get dy − y ⋅1 1 x⋅ 1 dx = − ⋅a 2 x ax + b x ⇒ x dy − y = x 2 ⋅ ax + b − ax dx x(ax + b) ⇒ x dy − y = bx dx ax + b Differentiating again w.r.t. x, we get d 2 y dy dy (ax + b) ⋅ b − bx ⋅ a x 2+ ⋅1 − = dx dx dx (ax + b)2 ⇒ x

d2y dx 2

⇒ x3

=

d2y

b2

ax  dy  ⇒ m1 =   =− 1  dx ( x , y ) by1 1 1

  dy = x − y 2  dx  dx

...(i)

y

2

2

(Using (i)) OR

We have,

{ { {

}

1 t x = a cos t + log tan2 and y = a sin t 2 2 1 t ⇒ x = a cos t + ⋅ 2 log tan 2 2 t ⇒ x = a cos t + log tan 2

}

}

...(v)

Differentiating (ii) w.r.t. x, we get dy dy a ′x 2a ′ x + 2b ′y =0 ⇒ =− dx dx b ′y

 bx  ⇒ x = 2  ax + b  dx 3d

(ax + b)2

2

log e = 1)

35. We have, ax2 + by2 = 1 ...(i) ...(ii) and a′x2 + b′y2 = 1 Let (x1, y1) be the point of intersection of the given curves. Then, ...(iii) ax12 + by12 = 1 ...(iv) a′x12 + b′y12 = 1 Differentiating (i) w.r.t. x, we get dy dy ax = 0⇒ =− 2ax + 2by dx dx by

a ′x  dy  ⇒ m2 =   =− 1  dx ( x , y ) b ′y1 1 1

...(vi)

The two curves will intersect orthogonally, if m1m2 = –1  ax   a ′x  ⇒  − 1  ×  − 1  = − 1 ⇒ aa′x12 = –bb′y12 ...(vii)  by1   b ′y1  Subtracting (iv) from (iii), we get (a – a′)x12 = –(b – b′)y12 ...(viii) Dividing (viii) by (vii), we get a − a ′ b − b′ 1 1 1 1 = ⇒ − = − . aa ′ bb ′ a b a ′ b′ 1 36. We have, A = 2 0 

−1 5 2

0 3 1

|A| = 1(5 – 6) + 1(2 – 0) + 0(4 – 0) = –1 + 2 + 0 = 1 ≠ 0 \ A–1 exists 113

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 −1 Now, adjA =  1  −3 

′

−2 1 −3

4   −1 1 −2 =  −2 1 7   4 −2 1 3  −1 1 −1 So, A = adj A =  −2 1 −3 |A|  4 −2 7  

−3 −3 7 

−1 1 −3 Now, |adj A| = −2 1 −3 4 −2 7

^

(

 3 −10 −1  We have, A =  −2 8 2   2 −4 −2  ⇒ |A| = 3(–16 + 8) + 10(4 – 4) –1(8 – 16) = –24 + 8 = –16 ≠ 0. So, A–1 exists  −8 −16 −12 ∴ adj A =  0 −4 −4   −8 −8 4 

 −8 −1  . = 0 16   −8

1 2  Now, (A–1)2  0  1  2 5 8  1 = 8 1  8

1 . (adj A) | A|

9 8 3 16 1 2

1 1 4 1 2 7 16   0  9  16 

3  1 4  2  1  0 4  1  1 −  4 2

^

^

) (

) (

)

\ − 2 i^ + ^j + 3 k^ ⋅  2 i^ − 7 ^j + 4 k^ + λ 3 i^ − 5 ^j + 4 k^  = 3 − 11λ

OR

  −16 −12  −4 −4  =   −8 4    

So, equation of a plane passing through intersection of both planes is   ^ ^ ^ ^  ^ ^ r ⋅ (2 i − 7 j + 4 k) − 3 + λ r ⋅ (3 i − 5 j + 4 k) + 11 = 0 ^ ^ ^ ^    ^ ^ ⇒ r ⋅ (2 i − 7 j + 4 k) + λ (3 i − 5 j + 4 k) = 3 − 11λ ...(i) Since it passes through (– 2, 1, 3) i.e., − 2 i + j + 3 k

= –1(7–6)–1(–14+12)–3(4–4) = –1 + 2 = 1 −1 1 3 Also, |A–1| = −2 1 −3 = |adj A| = 1 4 −2 7

Hence, A−1 =

37. Vector equation of given planes are ^ ^  ^ ^  ^ ^ r ⋅ (2 i − 7 j + 4 k) = 3 and r ⋅ (3 i − 5 j + 4 k) + 11 = 0

1 2

1

0 1 2 1 1 4 1 2

1 4 1 2 3 4  1 4 −1   4

3  4   1  4  1 −  4

⇒ – 4 – 7 + 12 + l(– 6 – 5 + 12) = 3 – 11l ⇒ 1 + l = 3 – 11l ⇒ 12l = 2 ⇒ l = 1/6 Putting value of l in (i), we get  ^ ^ ^ ^ 3i − 5 j + 4k  11   ^ ^ r ⋅ 2 i − 7 j + 4 k + = 3− 6 6    ^ ^ ^ ⇒ r ⋅  (12 + 3) i − (42 + 5) j + (24 + 4) k  = 18 − 11  6 6   ^ ^ ^ i − j + k 15 47 28  =7 ⇒ r ⋅   6 6

(

)

^ ^ ^ ⇒ r ⋅ 15 i − 47 j + 28 k = 7

OR x −2 y +3 z +5 is the given line ...(i) = = 1 2 −2 Let A(2, –3, –5) lies on the line. Direction ratios of line (i) are 1, –2, 2 1 −2 2 \ Direction cosines of line are , , 3 3 3 \ (i) may be written as x −2 y +3 z +5 ...(ii) = = 1 2 2 − 3 3 3 Coordinates of any point on the line (ii), may be taken as −2 2  1  3 r + 2, 3 r − 3, 3 r − 5  2 −2  1 Let Q =  r + 2, r − 3, r − 5  3 3  3 Given |r| = 3, \ r=±3 Putting the values of r, we have Q ≡ (3, – 5, –3) or Q ≡ (1, – 1, –7) 38. Maximize, Z = 600x + 400y subject to the constraints : x + 2y ≤ 12

...(i) Class 12

114

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2x + y ≤ 12 5 x+ y≥5 4

...(ii) ...(iii)

x, y ≥ 0 ...(iv) Let us draw the graph of constraints (i) to (iv). ABCDEA is the feasible region (shaded) as shown in the figure. Observe that the feasible region is bounded, and coordinates of the corner points A, B, C, D and E are (5, 0), (6, 0), (4, 4), (0, 6) and (0, 4) respectively.

OR Converting inequations into equations and drawing the corresponding lines. 3x + 5y = 15, 5x + 2y = 20 x y x y i.e. + = 1, + = 1 4 10 5 3 As x ≥ 0, y ≥ 0 solution lies in first quadrant. Let us draw the graph of the above equations.

B is the point of intersection of the lines 3x + 5y = 15 and 5x + 2y = 20, i.e. B =  70 , 15   19 19  Let us evaluate Z = 600x + 400y at these corner points. Corner Points

We clearly see that the point (4, 4) is giving the maximum value of Z.

 70 15  We have points O(0, 0) A(4, 0), B  ,  and C(0, 3) 19 19  Now z = 3x + 2y \ z(O) = 3(0) + 2(0) = 0 z(A) = 3(4) + 2(0) = 12  70   15  z (B) = 3   + 2   = 12.63  19   19  z(C) = 3(0) + 2(3) = 6 \ z has maximum value 12.63 at only one point i.e.  70 15  B ,  19 19 



115

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Self Evaluation Sheet Once you complete SQP-8, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Integrals / Integrals Matrices Probability / Probability Continuity and Differentiability Determinants / Determinants Relations and Functions Differential Equations / Differential Equations Inverse Trigonometric Functions Three Dimensional Geometry / Three Dimensional Geometry Relations and Functions Three Dimensional Geometry Probability Matrices Probability Probability Vector Algebra Application of Derivatives Three Dimensional Geometry

19

Continuity and Differentiability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Probability Inverse Trigonometric Functions Differential Equations / Differential Equations Application of Derivatives Probability Determinants Application of Integrals Vector Algebra / Vector Algebra Relations and Functions Integrals Integrals / Integrals Differential Equations Continuity and Differentiability Continuity and Differentiability / Continuity and Differentiability Application of Derivatives Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score. Class 12

116

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SQP

9

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

1(2)

–

–

3(4)

4.

Determinants

1(1)

–

–

1(5)*

2(6)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)*

1(2)*

1(3)*

–

3(6)

10.

Vector Algebra

1(1)*

1(2)*

–

–

2(3)

+ 1(4)

–

–

1(5)*

4(11)

–

–

–

1(5)*

1(5)

4(4)#

2(4)

–

–

6(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

2(2)#

Probability Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-9

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I  dy  1. Solve the differential equation sin   = a .  dx  OR Solve the differential equation

1 − y2 dy + =0. dx 1 − x2

2. Check whether the function f : z → z, defined by f(x) = x2 + 5 ∀x ∈ z is one-one or not. 3. A line makes angles a, b and g with the coordinate axes. If a + b = 90°, then find the value of angle g. OR Find the distance of the plane 5x – y + 6z – 12 = 0 from the origin.  a + b 2   6 5 ′ =  , then find the value of a. b  2 2 

4. If   5

      5. If a − b = a = b = 1, then find the angle between a and b . Class 12

118

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OR 



Find the magnitude of each of the two vectors a and b , having the same magnitude such that the angle between them is 60° and their scalar product is

9 . 2

6. Let A = {1, 2, 3}. Check whether the relation {(1,1),(2,2),(3,3),(1,2),(2,1)} is an equivalence relation on A or not. 7. Evaluate :

1

Evaluate :

dx

∫ ∫ 0

x 2 − 3x + 2

{

e x + sin

OR

}

πx dx 4

8. Find the area of the shaded region, shown in the given figure. Y

X

O

A x+

y+ 1=

B

0

9. A bag contains 12 balls, of which 5 are red and 7 are blue. If 2 balls are drawn at random then find the probability of getting atleast 1 blue ball. OR If A and B are two independent events such that P(B) ≠ 0, then find P(A | B). 10. If A and B are symmetric matrices of the same order, then show that AB + BA is a symmetric matrix. 11. The probability distribution of a random variable X is given below : X

2

3

4

5

P(X)

5 k

7 k

9 k

11 k

Find the value of k. 12. If the equation of a line is line.

2x − 5 4

=

y+4

=

3

6−z 6

, then find the direction cosines of a line parallel to this

13. If P(not E) = 0.36 and P(F | E) = 0.5, then find P(E ∩ F). 14. Check whether the following arrow diagram represents a function or not. A a

f

B 1 4

b

2

c

3

5

15. If A and B are two independent events, then show that the probability of occurrence of atleast one of A and B is given by 1 – P(A′) P(B′). 16. The value of the determinant of a matrix A of order 3 × 3 is 4. Find the value of |5A|. 119

Mathematics

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Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Two cars A and B are running at the speed more than allowed speed on the road along the lines   r = ɵi + ɵj − kɵ + λ(3ɵi − ɵj) and r = 4ɵi − kɵ + µ(2ɵi + 3kɵ ), respectively.

Based on the above answer the following : (i) The cartesian equation of the line along which car A is running, is x +1 y +1 z −1 x −1 y −1 z +1 = = = = (b) (a) 3 0 3 0 −1 −1 x −1 y −1 z +1 = = (d) None of these 3 0 −1 (ii) The direction cosines of line along which car A is running, are 3 −1 , ,0 (a) 3, –1, 0 (b) – 3, –1, 0 (c) 10 10 (iii) The direction ratios of line along which car B is running, are (a) 3, 0, 2 (b) 2, 3, 0 (c) 2, 0, 3 (iv) The shortest distance between the gives lines is (c)

(a) 4 units (b) 2 3 units (v) The cars will meet with an accident at the point (a) (–1, 0, 4) (b) (4, 0, –1)

(iii) Find the range of x. (a) (2, ∞)

−3

,

10

−1

,0

10

(d) 0, 3, 2

(c) 3 2 units

(d) 0 units

(c) (4, –1, 0)

(d) does not exist

18. Neeta has a rectangular painting having a total area of 24 ft2 which includes a border of 1 ft on the left, right, bottom and a border of 2 ft on the top inside it. Based on the above information, answer the following questions : (i) If Neeta wants to paint in the maximum area, then she needs to maximize (a) Area of outer rectangle (b) Area of inner rectangle (c) Area of top border (d) None of these (ii) It x is the length of the outer rectangle, then area of inner rectangle in terms of x is  24  (a) (x + 3)  − 2  x 

(d)

2 ft

1 ft

1 ft

1 ft

 24  (b) (x − 2)  + 3  x 

 24  (c) (x − 2)  − 3  x 

 24  (d) (x − 2)   x 

(b) (2, 8)

(c) (– ∞, 2)

(d) (–2, 8) Class 12

120

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(iv) If area of inner rectangle is maximum, then x is equal to (a) 2 ft (b) 3 ft (c) 4 ft (d) 5 ft (v) If area of inner rectangle is maximum, then length and breadth of this rectangle are respectively (a) 2 ft, 3 ft (b) 3 ft, 4 ft (c) 1 ft, 2 ft (d) 2 ft, 4 ft

PART - B Section - III 3 1 1 19. If P ( A) = , P ( B) = and P ( A ∩ B) = , then find P ( A | B ) and P (B | A). 8 2 4 1

20. Evaluate :

x tan −1 x

∫ (1 + x 2 )3/2

dx

0

OR 1

Evaluate :

2x

∫ 1 + x 2 dx 0

21. Find the equation of the normal to the curve y = x2 + 4x + 1 at the point where x = 3. 2 0 −2  −2 2 0  and C =  . 22. Find a matrix A such that 2A – 3B + 5C = O, where B =   3 1 4 7 1 6              23. If a + b + c = 0, then prove that a × b = b × c = c × a . OR π If the angle between i + k and i + j + ak is , then find the values of a. 3  π π  1 − sin + 1 + sin  2 2 . 24. Find the principal value of cot −1   π π  1 − sin 2 − 1 + sin 2   

1 4 2 25. Let X and Y be two events such that P ( X ) = , P (Y ) = and P (Y | X ) = . Then find P ( X ′ | Y ) 3 15 5 26. Find the area bounded by the parabola y = 2x – x2 and x-axis.  ax2 + b, if  27. Determine the constants a and b such that the function f (x ) =  2, if 2ax − b, if  dy 28. Solve the differential equation ( x − 1) = 2 xy. dx OR

Solve the differential equation 5

x>2 x = 2 is continuous at x = 2. x. \ Direction cosines are 2 3 −6 , , 22 + 32 + (− 6)2 22 + 32 + ( − 6)2 22 + 32 + (− 6)2 2 3 −6 , , 7 7 7 So, direction cosines of a line parallel to given line are 2 3 −6 , , . 7 7 7 or

13. Clearly, P(E) = 1 – P( E ) = 1 – 0.36 = 0.64 Now, P(F | E) = 0.5 P (E ∩ F ) ⇒ = 0.5 ⇒ P ( E ∩ F ) = P(E) × 0.5 P (E) = 0.64 × 0.5 = 0.32 14. As f(a) is not unique, therefore f is not a function. 15. P(atleast one of A and B) = P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) [ A, B are independent] = P(A) + P(B) [1 – P(A)] = [1 – P(A′)] + P(B) P(A′) = 1 – P(A′) + P(B) P(A′) = 1 – P(A′) [1 – P(B)] = 1 – P(A′) P(B′) 16. Given, A is a 3 × 3 matrix and |A| = 4 ⇒ |5A| = 53 ⋅ |A| = 125 × 4 = 500. 17. (i) (b) : The line along which car A is running, is  r = ɵi + ɵj − kɵ + λ (3ɵi − ɵj), which can be rewritten as (xi + y j + zk ) = (1 + 3λ)i + (1 − λ)j − k ⇒ x = 1 + 3λ, y = 1 − λ, z = −1 y −1 x −1 = λ, = λ, z + 1 = 0 3 −1 Thus, the required cartesian equation is x −1 y −1 z +1 = = 3 0 −1 (ii) (c) : Clearly, D.R.’s of the required line are 3 −1 , ,0> \ D.C.’s are < 32 + (−1)2 + 02 32 + (−1)2 + 02 ⇒

i.e.,


(iii) (c) : The line along which car B is running, is  r = 4ɵi − kɵ + µ (2ɵi + 3kɵ ), which is parallel to the vector 2i + 3k \ D.R.’s of the required line are    (iv) (d) : Here, a1 = ɵi + ɵj − kɵ , a2 = 4ɵi − kɵ , b1 = 3ɵi − ɵj and  b = 2ɵi + 3kɵ 2

  ∴ a2 − a1 = 3ɵi − ɵj ɵi ɵj kɵ   and b1 × b2 = 3 −1 0 = −3ɵi − 9 ɵj + 2kɵ 2 0 3     Now, (a2 − a1 ) ⋅ (b1 × b2 ) = (3ɵi − ɵj) ⋅ (−3ɵi − 9 ɵj + 2kɵ ) = –9 + 9 = 0 Hence, shortest distance between the given lines is 0. (v) (b) : Since, the point (4, 0, –1) satisfy both the equations of lines, therefore point of intersection of given lines is (4, 0, –1). So, the cars will meet with an accident at the point (4, 0, –1). 18. (i) (b) : In order to paint in the maximum area, Neeta needs to maximize the area of inner rectangle. (ii) (c) : Let x be the length and y be the breadth of outer rectangle. \ Length of inner rectangle = x – 2 and breadth of inner rectangle = y – 3 \ A(x) = (x – 2) (y – 3) [ xy = 24 (given)]  24  = (x − 2)  − 3  x  (iii) (b) : Dimensions of rectangle (outer/inner) should be positive. 24 −3 > 0 x ⇒ x > 2 and x < 8 \ Range of x is (2, 8). \ x – 2 > 0 and

 24  (iv) (c) : We have, A(x) = (x – 2)  − 3   x  −24   24  48 + − 3 = −3 ⇒ A′(x) = (x – 2)   x2  x 2   x −96 and A″(x) = 3 x For A(x) to be maximum or minimum, A′(x) = 0 48 48 ⇒ −3+ 2 = 0 ⇒ x = ± =±4 3 x \ x=4

[Since, length can't be negative] Class 12

124

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−96

Also, A″(4) =

0 When x = , f ′′(x ) = e 1  e  \ By the second derivative test, f is minimum at 1 x= e 1 Minimum value of f at x = is e 1 1 1 1 1 log   = .log(e −1 ) = .(−1) log e = − e e e e e  −3 2  2 1  and C =  36. Let B =    3 2  5 −3 Since, |B| = (4 – 3) = 1 ≠ 0 and |C| = (9 – 10) = –1 ≠ 0 \ B–1 and C–1 exist. Now, the given matrix equation becomes BAC = I2 ⇒ B–1 (BAC) C–1 = B–1 I2 C–1 ⇒ (B–1B) A(CC–1) = B–1 (I2C–1) ⇒ I2 AI2 = B–1C–1 ⇒ A = B–1C–1 ...(i) 2 − 1   Now, adj B =    −3 2  ∴ B

−1

 2 −1 1 . adj B =  =  |B|  −3 2 

[Q |B| = 1]

 −3 −2 Again, adj C =    −5 −3 3 2 1 . adj C =  ∴ C −1 =  [Q |C| = –1] |C | 5 3  2 −1 3 2 Now from (i), A = B −1C −1 =      −3 2  5 3

Now, A11 = 0, A12 = 2, A13 = 1, A21 = –1, A22 = –9, A23 = –5, A31 = 2, A32 = 23, A33 = 13 0 −1 2    \ adj A = 2 −9 23 1 −5 13  0 −1 2  1   adj A = (−1) 2 −9 23 = Now, A 1 −5 13   0 1 −2    =  −2 9 −23  −1 5 −13 The given system of equations is 2x – 3y + 5z = 11, 3x + 2y – 4z = –5, x + y – 2z = –3 This system of equations can be written as AX = B, A–1

2 −3 5  x   11        where A = 3 2 −4  , X =  y  , B =  −5 1 1 −2   z   −3 Since A–1 exists therefore solution given by X=

A–1B

 0 1 −2   11     =  −2 9 −23  −5 =  −1 5 −13  −3

1    2  3

⇒ x = 1, y = 2 and z = 3.

37. Given L.P.P. is Maximize Z =

2x y + 25 10

... (i)

subject to constraints x ≥ 2000 ... (ii) y ≥ 4000 ... (iii) and x + y ≤ 12000 ... (iv) On plotting inequalities (ii), (iii) and (iv), we get the shaded region, which is bounded in the following figure.

1 1  =  1 0  OR 2 −3 5    We have, A = 3 2 −4  1 1 −2  2 −3 5 ⇒ |A| = 3 2 −4 = 2(–4 + 4) + 3(–6 + 4) + 5(3 – 2) 1 1 −2 = – 6 + 5 = –1 ≠ 0 \ A–1 exists. Class 12

128

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 r ⋅ (3ɵi − ɵj + 4kɵ ) = 0 is   [r ⋅ (2ɵi + 6 ɵj) + 12] + λ [r ⋅ (3ɵi − ɵj + 4kɵ )] = 0  ⇒ r ⋅[(2 + 3λ )ɵi + (6 − λ )ɵj + 4 λ kɵ ] + 12 = 0

Now, we evaluate Z at the corner points A(2000, 10000), B(8000, 4000) and C(2000, 4000). Corner Points

Value of Z =

2x y + 25 10

A(2000, 10000)

1160 (Maximum)

B(8000, 4000)

1040

C(2000, 4000)

560

...(i) These planes are at a unit distance from the origin. Therefore, length of the perpendicular from the origin on (i) is 1 unit. 12 ⇒ =1 (2 + 3λ )2 + (6 − λ )2 + 16λ 2

Clearly, maximum value of Z is 1160 at x = 2000 and y = 10000. OR The given problem is Minimize Z = x – 7y + 190 subject to x + y ≤ 8, x + y ≥ 4, x ≤ 5, y ≤ 5 and x ≥ 0, y ≥ 0 To solve this LPP graphically, let us first covert the inequations into equations and draw the corresponding lines.

The feasible regions ABPQCDA. P is the point of intersection of x = 5 and x + y = 8 i.e., P(5, 3) and Q is the point of intersection of y = 5 and x + y = 8 i.e., Q(3, 5). The corner points of the feasible region are A(4, 0), B(5, 0), P(5, 3), Q(3, 5), C(0, 5) and D(0, 4). The value of the objective function Z = x – 7y + 190 at these points are Z(A) = 4 – 7(0) + 190 = 194 Z(B) = 5 – 7(0) + 190 = 195 Z(P) = 5 – 7(3) + 190 = 174 Z(Q) = 3 – 7(5) + 190 = 158 Z(C) = 0 – 7(5) + 190 = 155 Z(D) = 0 – 7(4) + 190 = 162 Clearly, Z is minimum at x = 0, y = 5. The minimum value of Z is 155. 38. The equation of the planes through the intersection  of the planes r ⋅ (2ɵi + 6 ɵj) + 12 = 0 and

⇒ 144 = (2 + 3l)2 + (6 – l)2 + 16l2 ⇒ 144 = 40 + 26l2 ⇒ 26l2 = 104 ⇒ l2 = 4 ⇒ l = ± 2 Putting the values of l in (i), we obtain  r ⋅ (8ɵi + 4 ɵj + 8kɵ ) + 12 = 0  or r ⋅ (− 4ɵi + 8 ɵj − 8kɵ ) + 12 = 0 These equations can also be written as  r ⋅ (2ɵi + ɵj + 2kɵ ) + 3 = 0  or r ⋅ (− ɵi + 2 ɵj − 2kɵ ) + 3 = 0 , which are the equations of the required planes. OR Let M be the foot of the perpendicular drawn from the point P(2, 3, 4) to the given line. The coordinates of any point on the line 4 − x y 1− z = = 2 6 3 x − 4 y z −1 = = i.e. on the line are given by −2 6 −3 i.e. –2l + 4, 6l, –3l + 1 Let M = (–2l + 4, 6l, –3l +1) ...(i) \ Direction ratios of PM are –2l + 2, 6l – 3, –3l – 3 The direction ratios of the given line are –2, 6, –3 Since PM is perpendicular to the given line \ –2(–2l + 2) + 6(6l – 3) – 3(–3l – 3) = 0 13 ⇒ 49l – 13 = 0 ⇒ λ = 49 13 Putting λ = in (i), we get 49  170 78 10  M≡ , ,  49 49 49  \ Required length, 2

2

2

  170   78   10 PM =  − 2  +  − 3 +  − 4  =    49   49  49 =

44541 2401

3 101 units. 7

 129

Mathematics

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Self Evaluation Sheet Once you complete SQP-9, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Differential Equations / Differential Equations Relations and Functions Three Dimensional Geometry / Three Dimensional Geometry Matrices Vector Algebra / Vector Algebra Relations and Functions Integrals / Integrals Application of Integrals Probability / Probability Matrices Probability Three Dimensional Geometry Probability Relations and Functions Probability Determinants Three Dimensional Geometry Application of Derivatives

19

Probability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Application of Derivatives Matrices Vector Algebra / Vector Algebra Inverse Trigonometric Functions Probability Application of Integrals Continuity and Differentiability Differential Equations / Differential Equations Integrals Continuity and Differentiability Differential Equations / Differential Equations Application of Integrals Relations and Functions Continuity and Differentiability / Continuity and Differentiability Application of Derivatives Determinants / Determinants Linear Programming / Linear Programming Three Dimensional Geometry / Three Dimensional Geometry

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

10

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)#

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

1(5)*

3(7)

4.

Determinants

1(1)

1(2)

–

–

2(3)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)*

–

3(9)

7.

Integrals

2(2)*

1(2)*

1(3)

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)*

1(2)*

1(3)

–

3(6)

10.

Vector Algebra

1(1)

–

–

–

1(1)

1(2)

–

1(5)*

5(13)

–

–

–

1(5)*

1(5)

4(4)#

2(4)#

–

–

6(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

2(2)# +

Probability Total

1(4)

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-10

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Show that the relation R in the set A = {x ∈ Z : 0 ≤ x ≤ 12}, given by R = {(a, b) : a = b} is both symmetric and transitive. OR Give an example of a relation, which is transitive but neither reflexive nor symmetric. 2 −1 2. If A =   and B = 3 2  3. Evaluate :

x−4

∫ (x − 2)3 ⋅e

x

 0 4 2  −1 7  , then find the matrix A – B.  

dx OR

1

Evaluate : ∫ x e x dx 2

0

4. Find the values of x for which

3 x 3 2 = . x 1 4 1 Class 12

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5. Find the equation of a line passing through (1, 2, –3) and parallel to the line

x − 2 y +1 z −1 = = . 1 3 4

OR ^

^

^

Find the vector equation of the plane passing through a point having position vector 2 i + 3 j + 4 k and ^

^

^

perpendicular to the vector 2 i + j − 2 k . 6. Prove that the function f : R → R defined by f(x) = 3 – 4x is onto.  d3 y  7. Find the degree of the differential equation  3   dx 

2 /3

+5−2

d2 y dx 2

=0.

OR  e −2 x y  dx Find the integrating factor of the differential equation  −  = 1.  x x  dy 8. An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement, then find the probability that both drawn balls are black. 9. Prove that if E and F are independent events, then the events E′ and F′ are also independent. OR Given two independent events A and B, such that P(A) = 0.39 and P(B) = 0.6. Find P(A′ ∩ B′). 10. The cartesian equation of a line is

x +3 y −5 z +6 . Find its vector equation. = = 2 4 2

11. Let R be a relation on N defined by R = {(1 + x, 1 + x2) : x ≤ 5, x ∈N}. Then, verify the following : (a) R is reflexive (b) Domain of R = {2, 3, 4, 5, 6} 2

 1− x  x 12. Evaluate : ∫  e dx  1 + x 2  13. If A and B are two events such that P(A) = 0.53, P(B) = 0.24 and P(A ∩ B) = 0.42, then find P(B′ ∩ A).       14. If u = ɵi + 2 ɵj, v = − 2ɵi + ɵj and w = 4ɵi + 3ɵj . Find scalars x and y respectively such that w = xu + yv . 15. An urn contains 6 balls of which two are red and four are black. Two balls are drawn at random. Find the probability that they are of the different colours. 2 5   −1 2  16. Find the additive inverse of A + B, where A and B are given as A =  and B =   . 9 3   3 −9  Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A cricket match is organised between students of Class XI and Class XII for which a team from each class is chosen. Remaining students of Class XI and XII are respectively sitting on the plane’s represented by the equation   r ⋅ (ɵi + ɵj + 2kɵ ) = 5 and r ⋅ (2ɵi − ɵj + kɵ ) = 8, to cheers the team of their own class. Based on the above answer the following : (i) The cartesian equation of the plane on which student of class XI are seated is (a) 2x – y + z = 8 (b) 2x + y + z = 8 (c) x + y + 2z = 5

(d) x + y + z = 5 133

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(ii) The magnitude of the normal to the plane on which student of Class XII are seated, is (a)

5

(b)

(c)

6

3

(d)

2

(iii) The intercept form of the equation of the plane on which student of Class XII are seated, is y x z x y z x y z x y z + + =1 + + =1 (b) (c) (d) (a) + + =1 + + =1 4 (−8) 8 5 5 5 5 5 5/2 4 8 8 (iv) Which of the following is a student of Class XII? (a) A sitting at (1, 2, 1) (b) B sitting at (0, 1, 2)

(c) C sitting at (4, 1, 1)

(d) none of these

(v) The distance of the plane, on which student of Class XII are seated, from the origin is 8 5 (a) 8 units (b) (c) (d) none of these units units 6 6 18. A mobile company in a town has 500 subscribers on its list and collects fix charges of ` 300 per year from each subscriber. The company proposes to increase the annual charges and it is believed that for every increase of ` 1, one subscriber will discontinue service. Based on the above information answer the following questions: (i) If x denote the amount of increase in annual charges of each subscriber, then revenue, R, as a function of x can be represent as (a) R(x) = 300 × 500 × x (b) R(x) = (300 – 2x) (500 + 2x) (c) R(x) = (500 + x) (300 – x) (d) R(x) = (300 + x) (500 – x) (ii) If mobile company increases ` 50 as annual charges, then R is equal to (a) ` 157500 (b) ` 167500 (c) ` 17500

(d) ` 15000

(iii) If revenue collected by the mobile company is ` 156,400, then value of amount increased as annual charges for each subscriber, is (a) 40 (b) 160 (c) Both (a) and (b) (d) None of these (iv) What amount of increase in annual charges will bring maximum revenue? (a) 100 (b) 200 (c) 300

(d) 400

(v) Maximum revenue is equal to (a) ` 15000 (b) ` 160000

(d) ` 25000

(c) ` 20500

PART - B Section - III 1 a 2    19. For what value(s) of 'a' the matrix 1 2 5  will not be invertible. 2 1 1  20. Evaluate :

x

∫ 1 − sin 2x dx OR

Evaluate :

dx

∫ 1+ tan x

21. Find the equation of the tangent to the curve y = x4 – 6x3 + 13x2 – 10x + 5 at the point (1, 3). Class 12

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22. Using integration, find the area of the region bounded by the line 2y = 5x + 7, x-axis and the lines x = 2 and x = 8. 3 5 3 and P ( A ∪ B) = . Find the value of A and B be two events such that P ( A) = , P (B) = 4 8 8 P(A| B)⋅P(A′|B).

23. Let

OR Two events E and F are independent. If P(E) = 0.3, P(E ∪ F) = 0.5, then find P(E | F) – P(F | E).  3  1  − 3 sin −1  24. Find the value of sin −1  .   2  2  25. A speaks truth in 60% of the cases and B in 90% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact? 26. Find

dy for the equation x3 + y3 = sin(x + y). dx

  27. Find the distance between the lines given by r = ɵi + ɵj + λ(ɵi − 2 ɵj + 3kɵ ) and r = (2ɵi − 3kɵ ) + µ(ɵi − 2 ɵj + 3kɵ ).  dy  28. Solve log   = ax + by . dx OR 2

Solve the differential equation ( x + y )

dy = 1. dx Section - IV

π

29. Evaluate :

x tan x

∫ sec x + tan x dx 0

30. Consider f : R+ → [4, ∞) given by f(x) = x2 + 4. Show that f is bijective function. 31. Find the local maxima or local minima of f(x) = x3 – 6x2 + 9x + 15. Also, find the local maximum or local minimum values as the case may be. OR Find the values of x for which the function f(x) = xx, x > 0 is (a) increasing (b) decreasing. 32. Find a particular solution of the differential equation cos x

33. If y = xx , then find

 dy  = a(a ∈ R); y = 1 when x = 0 .  dx 

dy . dx

34. If a, b are the roots of ax2 + bx + c = 0 and f(x) is continuous at x = a, where f (x ) = b2 − 4ac for x ≠ a, then prove that f(a) = . 2 OR If f (x ) =

2 − (256 − 7 x )1/8 (5x + 32)1/5 − 2

1 − cos(ax 2 + bx + c) (x − α)2

,

, x ≠ 0 then for f to be continuous everywhere, what should be the value of f(0).

35. Find the area of the region bounded by the parabola y2 = 4ax, its axis and two ordinates x = a and x = 2a. 135

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Section - V 36. Solve the following LPP graphically : Maximize Z = x + y Subject to the constraints, 2x + 5y ≤ 100 y x + ≤1 25 40 x ≥ 0, y ≥ 0 OR Find the maximum value of Z = 5x + 2y subject to constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0. 37. Find the equation of the plane passing through the point A(1, 2, 1) and perpendicular to the line joining the x +3 y −5 z −7 points P(1, 4, 2) and Q(2, 3, 5). Also, find the distance of this plane from the line = = . 2 −1 −1 OR Find the coordinates of the points on the line

x + 2 y +1 z − 3 , which are at a distance of 2 units from = = 3 2 6

the point (–2, –1, 3).  3 −5 2 3 38. If A =   and f(x) = x – 5x – 14, find f(A). Hence obtain A . − 4 2   OR

Solve the following system of equations by matrix method : 2x + 5y = 1, 3x + 2y = 7

Class 12

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SOLUTIONS 1. The set A = {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2, ..., 12} R = {(a, b) : a = b} = {(0, 0), (1, 1), (2, 2),..., (12, 12)} (i) Let (a, b) ∈ R ⇒ a = b ⇒ b = a ⇒ (b, a) ∈ R. So, R is symmetric. (ii) Let (a, b) ∈ R and (b, c) ∈ R ⇒ a = b = c ⇒ a = c ⇒ (a, c) ∈ R. So, R is transitive. OR Let A = {1, 2, 3} and defined a relation R on A as R = {(1, 2), (2, 2)}. Then, R is transitive, as (1, 2), (2, 2) ∈ R ⇒ (1, 2) ∈ R But R is not reflexive, as 1 ∈ A but (1, 1) ∈ / R. and also R is not symmetric, as (1, 2) ∈ R but (2, 1) ∈ / R. 2 −1 2 −1  4 − 3 −2 − 2   1 −4   = =  3 2  3 2  6 + 6 −3 + 4  12 1 

2. A2 = 

 1 −4  \ A2 – B =  − 12 1   1 − 0 −4 − 4   1 = = 12 + 1 1 − 7  13 3. Let I =

x−4

∫ (x − 2)3 ⋅e

x

 0 4  −1 7    − 8 − 6

dx

 x −2 2  x = ∫ − e dx 3 3  (x − 2) (x − 2)   1 2  x ex = ∫ = − e dx +C  2 3 (x − 2)2  (x − 2) (x − 2)  ∵ [ f (x ) + f ′(x )]e x dx = e x f (x ) + C   ∫  OR 1

1

dt Let I = ∫ x e x dx = ∫ et 2 0 0 [Putting x2 = t ⇒ 2xdx = dt] 1 1 = [et ]10 = (e − 1) 2 2 2

4. We have, ⇒

3 x 3 2 = x 1 4 1

3 − x2 = 3 − 8 ⇒ x2 = 8. Hence, x = ±2 2 .

5. Since, the line is parallel to the line x − 2 y +1 z −1 = = 1 3 4 \ D.R.’s of the required line are

Hence, equation of the line passing through (1, 2, –3) x −1 y − 2 z + 3 with d.r.’s is = = 1 3 4 OR Vector equation of plane passing through a point   having position vector a and perpendicular to n is     given by r ⋅n = a ⋅n   Here a = 2 ɵi + 3ɵj + 4kɵ and n = 2 ɵi + ɵj − 2kɵ ^  ^ ^ \ Required equation is r . (2 i + j − 2 k ) = 4 + 3 − 8 = −1 6. Let y ∈R be any real number, such that f(x) = y. \ y = 3 – 4x 3− y ⇒ 4x = 3 − y ⇒ x = 4 3− y Since, for any real number y ∈R, there exists ∈R 4 3− y  3− y  = 3− 4 = 3−3+ y = y such that f    4   4  Hence, f is onto.  d3 y  7. We have,  3   dx 

2 /3

=2

2

d2 y dx 2

−5

3

  d3 y   d2 y ⇒  3  = 2 2 − 5 (On cubing both sides)   dx   dx Clearly, degree is 2. [∵ Power of highest order derivative is 2] OR dy 1 e −2 x + y= dx x x This is a linear differential equation of the form

We have,

dy 1 e −2 x ,Q = + Py = Q, where P = dx x x

\ I.F. = e ∫

Pdx

⇒ I.F. = e

∫

1 x

dx

= e2

x

8. Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F). 9 10 , P (F | E ) = Now, P(E) = 14 15 By multiplication rule of probability, we have 10 9 3 P(E ∩ F) = P(E).P(F|E) = × = 15 14 7 9. Since, E and F are independent events. \ P(E ∩ F) = P(E) P(F) ...(i) 137

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Now, P(E′ ∩ F′) = 1 – P(E ∪ F) [∵ P(E′ ∩ F′) = P((E ∪ F)′)] = 1 – [P(E) + P(F) – P(E ∩ F)] = 1 – P(E) – P(F) + P(E) P(F) [Using (i)] = (1 – P(E)) (1 – P(F)) = P(E′) P(F′) Hence, E′ and F′ are also independent events. OR Since A and B are independent events, therefore A′ and B′ will also be independent. So, P(A′ ∩ B′) = P(A′).P(B′) = (1 – P(A)) (1 – P(B)) = (1 – 0.39) (1 – 0.6) = 0.61 × 0.4 = 0.244 10. The given cartesian equation is x+3 y−5 z+6 = = 2 4 2 The line passes through the point (–3, 5, –6) and is parallel to vector 2ɵi + 4 ɵj + 2kɵ . Hence, the vector equation of the line is  r = − 3ɵi + 5ɵj − 6kɵ + λ (2ɵi + 4 ɵj + 2kɵ ). 11. Clearly, R = {(2, 2), (3, 5), (4, 10), (5, 17), (6, 26)} Domain of R = {x : (x, y) ∈R} = {2, 3, 4, 5, 6} and R is not reflexive, as 1 ∈ N but (1, 1) ∈/ R.  1 + x 2 − 2x  x 12. Let I = ∫   e dx  (1 + x 2 )2  2x  x 1  1 ex +C = ∫ e dx = − 2 2 2  1+ x 1 + x2 (1 + x )  ∵ e x [ f (x ) + f ′(x )]dx = e x f (x ) + C   ∫  13. P(B′ ∩ A) = P(A – B) = P(A) – P(A ∩ B) = 0.53 – 0.42 = 0.11    14. We have, w = xu + yv ⇒ 4ɵi + 3ɵj = x(ɵi + 2 ɵj) + y(− 2ɵi + ɵj)  ⇒ (x − 2 y − 4)ɵi + (2 x + y − 3)ɵj = 0 ⇒ x – 2y – 4 = 0 and 2x + y – 3 = 0 ⇒ x = 2 and y = – 1 15. Total number of possible outcomes = 6C2 = 15 Number of favourable outcomes = 2C1 4C1 = 2 × 4 = 8 \ Required probability =

8 15

1 7  16. Let C = A + B =   12 −6  −1 −7  Now, (–C) =   , which is the additive inverse  −12 6  of A + B.

17. (i) (c) : Clearly, the plane for Class XI students is  r ⋅ (ɵi + ɵj + 2kɵ ) = 5, which can be rewritten as (xiɵ + y ɵj + zkɵ ) ⋅ (ɵi + ɵj + 2kɵ ) = 5 ⇒ x + y + 2z = 5, which is the required cartesian equation. (ii) (b) : Clearly, the plane for Class XII students is  r ⋅ (2ɵi − ɵj + kɵ ) = 8, which is of the form   r ⋅n = d  \ Normal vector to the plane is n = 2ɵi − ɵj + kɵ and its  magnitude is | n | = 22 + 12 + 12 = 6 (iii) (b) : The cartesian form is 2x – y + z = 8, which can be rewritten as 2x y z − + =1 8 8 8 x y z + =1 + ⇒ 4 −8 8 (iv) (c) : Since, only the point (4, 1, 1) satisfy the equation of plane representing Class XII, therefore C is the student of XII. (v) (b) : Equation of plane representing Class XII is  r ⋅ (2ɵi − ɵj + kɵ ) = 8, which is not in normal form, as  |r |≠1 On dividing both sides by 22 + 12 + 12 = 6 , we get  2 ɵ 1 ɵ 1 ɵ 8 , r ⋅  i− j+ k =  6 6 6  6  which is of the form r ⋅ nɵ = d 8 Thus, the required distance is units. 6 18. (i) (d) : If x be the amount of increase in annual charges of each subscriber, then number of subscriber reduces to 500 – x \ Revenue, R(x) = (300 + x) (500 – x) = 150000 + 200x – x2, 0 < x < 500 (ii) (a) : Clearly, at x = 50 R(50) = 150000 + 200(50) – (50)2 = 150000 + 10000 – 2500 = ` 157500 (iii) (c) : Since, 150000 + 200x – x2 = 156400 (Given) ⇒ x2 – 200x + 6400 = 0 ⇒ x2 – 160x – 40x + 6400 = 0 ⇒ x(x – 160) – 40(x – 160) = 0 ⇒ x = 40, 160 dR d2R = 200 – 2x and = –2 < 0 dx dx 2 dR For maximum revenue, =0 dx ⇒ x = 100 \ Required amount = 100 (iv) (a) :

(v) (b) : Maximum revenue = R(100) = (300 + 100) (500 – 100) = 400 × 400 = ` 160000 Class 12

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1 a 2 19. The matrix will not be invertible if 1 2 5 = 0 2 1 1 ⇒ 1(2 – 5) – a (1 –10) + 2(1 – 4) = 0 ⇒ –3 + 9a – 6 = 0 ⇒ a = 1 x x(1 + sin 2 x ) 20. Let I = ∫ dx = ∫ dx 1 − sin 2 x cos2 2 x   = ∫ x  sec2 2 x + sec 2 x tan 2 x  dx   I II  tan 2 x sec 2 x  = x +  2 2   log sec 2 x log sec 2 x + tan 2 x  − +  + C  4 4 x 1 \ I = (tan 2x + sec 2x) – log |sec2 2x + sec 2x tan 2x| + C 2 4 OR cos x dx Let I = ∫ = dx 1 + tan x ∫ cos x + sin x 1 2cos x = ∫ dx 2 cos x + sin x =

1 (cos x + sin x ) + (cos x − sin x ) dx 2∫ (cos x + sin x )

1 1  cos x − sin x   =  ∫ dx + ∫   dx 2  cos x + sin x   2 =

x 1 + log | cos x + sin x | + C 2 2

21. Here, y = x4 – 6x3 + 13x2 – 10x + 5 Differentiating (i) w.r.t. x, we get

...(i)

dy = 4 x 3 − 18 x 2 + 26 x − 10 dx

 dy  ∴   = 4 − 18 + 26 − 10 = 2  dx (1, 3) Hence the equation of the tangent to (i) at (1, 3) is y – 3 = 2(x – 1) ⇒ y = 2x + 1 22. Let us draw the graph of given lines, as shown below:

\ Required area (shown in shaded region) 8 8  5x + 7  = ∫ y dx = ∫   dx 2 2 2  8

 1  5(64) 1  5x 2    5(4) =  + 7 x  =  + 14 + 56 −  2  2    2  2 2  2 1 1 192 = [(160 + 56) − (10 + 14)] = (216 − 24) = 2 2 2 = 96 sq. units. 3 5 3 23. Given, P ( A) = , P (B) = and P ( A ∪ B )= . 8 8 4 Clearly, P(A ∩ B) = P(A) + P(B) – P(A ∪ B) 3 5 3 3+5−6 1 = + − = = 8 8 4 8 4 Also, we know that P(A′ ∩ B) + P(A ∩ B) = P(B) [As A′ ∩ B and A ∩ B are mutually exclusive events] 5 1 3 \ P(A′ ∩ B) = P(B) – P(A ∩ B) = − = 8 4 8 P ( A ∩ B ) P ( A′ ∩ B ) Now, P ( A | B) ⋅ P ( A′ | B ) = ⋅ P ( B) P ( B) 1 / 4 3 / 8 3 64 6 = ⋅ = × = 5 / 8 5 / 8 32 25 25 OR Since, E and F are independent events. \ P(E ∩ F) = P(E) P(F) ⇒ P(E|F) = P(E) and P(F|E) = P(F) Now, P(E ∪ F) = P(E) + P(F) – P(E ∩ F) ⇒ 0.5 = 0.3 + P(F) – 0.3 P(F) ⇒ P(F)(1 – 0.3) = 0.5 – 0.3 ⇒ P (F ) = 0.2 = 2 0. 7 7 \ P(E | F) – P(F | E) = P(E) – P(F) 3 2 1 2 − = = 0. 3 − = 7 10 7 70  3 π  1  π 24. sin −1  − 3 sin −1   = 4 − 3  3    2 2   π −3π = −π= 4 4   −1  1  π ∵ Principal value of sin   = and   2 4      3 π  that of sin −1   =   2  3   25. Let E = the event of A speaking the truth and F = the event of B speaking the truth 60 3 90 9 Then, P (E ) = = and P (F ) = = 100 5 100 10 Required probability = P (A and B contradicting each other) = P (EF or EF ) = P (EF ) + P (EF ) 139

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= P ( E ) ⋅ P (F ) + P ( E ) ⋅ P (F ) [∵ E and F are independent events] = P(E) · [1 – P(F)] + [1 – P(E)] · P(F) 3 9 3 9 21 42 = = 1 −  + 1 −  ⋅ = 5  10   5  10 50 100 Thus, A and B are likely to contradict each other in 42% cases. 26. We have, x3 + y3 = sin(x + y) On differentiating both sides w.r.t. x, we get dy d 3x 2 + 3 y 2 = cos( x + y ) (x + y ) dx dx  dy  = cos( x + y ) 1 +   dx  dy dy ⇒ 3x + 3 y ⋅ = cos( x + y ) + cos(x + y ) dx dx dy dy 2 − cos( x + y ) = cos(x + y ) − 3x 2 ⇒ 3y dx dx dy  2 2 ⇒ 3 y − cos( x + y ) = cos(x + y ) − 3x dx dy cos( x + y ) − 3x 2 ⇒ = dx 3 y 2 − cos( x + y ) 27. The given lines are parallel.    Here, a1 = ɵi + ɵj, a2 = 2ɵi − 3kɵ and b = ɵi − 2 ɵj + 3kɵ   Now, a2 − a1 = (2ɵi − 3kɵ ) − (ɵi + ɵj) = ɵi − ɵj − 3kɵ    b × (a2 − a1 ) = ɵi(6 + 3) − ɵj(−3 − 3) + kɵ (−1 + 2) = 9ɵi + 6 ɵj + kɵ  | b | = 12 + (−2)2 + 32 = 1 + 4 + 9 = 14    | b × (a2 − a1 )| | 9ɵi + 6 ɵj + kɵ |  Shortest distance = = |b | 14 2

=

2

1 14

(9)2 + (6)2 + (1)2 =

59 118 = units 14 7

 dy  28. We have log   = ax + by dx dy = e ax +by ⇒ dy = eax eby dx ⇒ e–by dy = eax dx ⇒ dx e −by e ax [Integrating both sides] = +C , ⇒ −b a which is the required solution. OR dy 2 =1 ...(i) We have (x + y) dx dy du = Let x + y = u ⇒ 1 + dx dx 1 + u2 du 1 du 1 = + 1 = −1 = 2 ⇒ (i) becomes, dx u2 dx u u2

⇒ ⇒ ⇒ ⇒

u2 + 1 − 1 u2 ∫ u2 + 1 du = ∫ dx + C ⇒ ∫ u2 + 1 du = x + C u – tan–1(u) = x + C (x + y) – tan–1(x + y) = x + C y – tan–1(x + y) = C is the required solution π

29. Let I =

x tan x

∫ sec x + tan x dx 0

π

x sin x ⇒ I= ∫ dx 1 + sin x 0

...(i)

π

⇒ I=

(π − x )sin(π − x ) dx 1 + π − x sin( ) 0

∫ π

(π − x )sin x dx 1 x + sin 0 Adding (i) and (ii), we get ⇒I=

∫

π

2I =

...(ii) π

π sin x π sin x(1 − sin x ) ∫ 1 + sin x dx ⇒ I = 2 ∫ (1 + sin x )(1 − sin x ) dx 0 0

⇒I=

π π π sin x − sin2 x π  sin x sin2 x  ⇒ = − dx I  dx  2 ∫ cos2 x 2 ∫  cos2 x cos2 x  0 0

π

π = ∫ (tan x sec x − tan2 x ) dx 20 π

=

π π [tan x sec x − (sec2 x − 1)]dx = [sec x − tan x + x]0π ∫ 20 2

π π = [sec π − tan π + π] − [sec 0 − tan 0 + 0] 2 2 π π π2 π π = [ −1 − 0 + π] − [1 − 0 + 0] = − + − 2 2 2 2 2 =

π2 π − π = (π − 2) 2 2

30. We have, f : R+ → [4, ∞) defined by f(x) = x2 + 4. (i) Let x1, x2 ∈ R+ s.t. f(x1) = f(x2) ⇒ x12 + 4 = x22 + 4 ⇒ x12 = x22 ⇒ x1 = x2

(∵ x1, x2 ∈ R+)

⇒ f is one-one (ii) y = f (x), where y ∈ [4, ∞), i.e., y ≥ 4 ⇒ x2 + 4 = y ⇒ x =

y−4

Now, x is defined if, y − 4 ≥ 0 and y − 4 ∈ R+ Thus, for each y ∈ [4, ∞), we have x = y − 4 ∈ R+ 2

such that f ( y − 4 ) = ( y − 4 ) + 4 = ( y − 4) + 4 = y ⇒ f is onto. \ f is one-one and onto. ⇒ f is bijective function Class 12

140

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31. Given that, f(x) = x3 – 6x2 + 9x + 15 ⇒ f ′(x) = 3x2 – 12x + 9. For local maxima or minima, we must have f ′(x) = 0. Now, f ′(x) = 0 ⇒ 3(x2 – 4x + 3) = 0 ⇒ 3(x – 3)(x – 1) = 0 ⇒ x = 3 or x = 1

 dy  32. We have, cos   = a  dx  dy ⇒ = cos −1 a ⇒ dy = cos −1 a dx dx Integrating (i) both sides, we get −1 –1 ∫ dy = cos a ∫ dx ⇒ y = x cos a + C

Case I : When x = 3 In this case, when x is slightly less than 3 then f ′(x) is negative and when x is slightly more than 3 then f ′(x) is positive. Thus, f ′(x) changes sign from negative to positive as x increases through 3. So, x = 3 is a point of local minima. \ Local minimum value = f(3) = 15

When x = 0, y = 1 ⇒ 1 = C Thus, particular solution is y = x cos–1 a + 1 y −1 ⇒ (y – 1) = x cos–1 a ⇒ cos −1 a =   x  y −1 ⇒ a = cos   x 

Case II : When x = 1 In this case, when x is slightly less than 1 then f ′(x) is positive and when x is slightly more than 1 then f ′(x) is negative. Thus, f ′(x) changes sign from positive to negative as x increases through 1. So, x = 1 is a point of local maxima. \ Local maximum value = f(1) = 19. OR Given, f(x) = xx ⇒ f(x) = ex logx d ...(i) (x log x ) = xx (1 + loge x) dx (a) f(x) is increasing ⇒ f ′(x) ≥ 0 ⇒ xx(1 + logex) ≥ 0 [From (i)] ⇒ (1 + logex) ≥ 0 [ xx > 0 when x > 0]

x

33. Given, y = xx Taking log on both sides, we get log y = xx log x Again, taking log on both sides, we get log (log y) = log(xx log x) ⇒ log (log y) = log xx + log (log x) ⇒ log (log y) = x (log x) + log (log x) 1 1 dy  1   1 1 ⋅ ⋅ =  x ⋅ + log x ⋅1 +  ⋅ log y y dx  x   log x x  ⇒

dy 1 1 ⋅ = 1 + log x + y log y dx x log x

⇒

dy 1   = y log y 1 + log x + dx x log x  

[ xx > 0]

⇒ logex ≤ –1 ⇒ x ≤ e–1 ⇒ 0 < x ≤  1 \ f(x) is decreasing on  0,  .  e

x x  1  = x ( x ) log x ( x ) 1 + log x + x log x  

34. Given, a, b are the roots of ax2 + bx + c = 0, therefore ax2 + bx + c = a (x – a) (x – b) Since f (a) is continuous at x = a, therefore f (α) = lim f (x ) x →α

⇒ f ′(x) ≤ 0 ⇒ xx(1 + logex) ≤ 0 ⇒ (1 + logex) ≤ 0

...(i)

On differentiating (i) both sides w.r.t. x, we get

⇒ f ′(x) = ex logx ⋅

1  ⇒ logex ≥ – 1 ⇒ x ≥ e–1 ⇒ x ∈  , ∞  e  1  \ f(x) is increasing on  , ∞  . e  (b) f(x) is decreasing

...(i)

 1 1 ⇒ x ∈ 0,  e e

1  Hence, f(x) is increasing on  , ∞  and decreasing e   1 on  0,  .  e

 1 − cos(a(x − α)(x − β))  ⇒ f (α) = lim   x →α   (x − α)2   1 − cos(a(x − α)(x − β)) 2 ⋅ a (x − β)2  = lim  2 2 2 x → α  a ( x − α ) ( x − β)  =

a2 (α − β)2 2

Now (a –

b)2

…(i) = (a + =

b)2

b2 − 4ac

2

c  −b  – 4ab =   − 4   a  a 

a2

From (i), we get f(a) =

b2 − 4ac . 2 141

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OR Consider, lim f (x ) = f (0) x →0

(5x + 32)1/5 − 321/5

(256 − 7 x )1/8 − 2561/8 × (−7 x ) (256 − 7 x ) − (256)

x →0

0 20

= − lim

x →0

= 5y

x →0

(256 − 7 x )1/8 − 2561/8

+ 8x

⇒ lim f (x ) = − lim

2x + 5y = 1

(5x + 32)1/5 − 3221/5 × 5x (5x + 32) − (32)

00

1

−1 7 1 ⋅ ⋅ 256 8 7 = =5 8 1 64 −1 1 ⋅ 32 5 5

35. Equation of parabola is y2 = 4ax Its axis is y = 0 and vertex is (0, 0) \ Required area ABCDA = =2 a

∫ y dx

a

2a

∫

2a

x dx

[

y > 0]

a 2a 2 = 2 a .  x 3/2  a 3 2 = 2 a . (2a)3/2 − (a)3/2  3 4 a a3/2 (23/2 − 1) = 3 4 = a2 [2 2 − 1] sq . units 3 36. Given problem is Maximize Z = x + y Subject to the constraints, x ≥ 0, y ≥ 0, 2x + 5y ≤ 100, y x + ≤ 1 ⇒ 8x + 5y ≤ 200 25 40 Let us convert the system of the inequations into equations. l1 : 2x + 5y = 100 and l2 : 8x + 5y = 200

 50 40  Both the lines intersect at B  ,  .  3 3 The solution set of the given system is the shaded region OABC. The coordinates of corner points O, A, B, C are (0, 0),  50 40  (25, 0),  ,  and (0, 20) respectively. 3 3

Corner Points O(0, 0) A(25, 0)

Value of Z = x + y 0 25

 50 40  B ,  3 3 

30 (Maximum)

C(0, 20)

20

So, Z = x + y is maximum when x =

50 40 and y = . 3 3

OR Let us convert the given inequations into equations and draw the corresponding lines. We have, 3x + 5y = 15 and 5x + 2y = 10 x y x y i.e., + = 1 and + = 1 5 3 2 5 As x ≥ 0, y ≥ 0, solution lies in first quadrant.

Here, B is the point of intersection of the lines  20 45  3x + 5y = 15 and 5x + 2y = 10 i.e., B =  ,  19 19   20 45  We have points A(2, 0), B  , and C(0, 3).  19 19  Now, value of Z = 5x + 2y at these points are given below: Z(O) = 5(0) + 2(0) = 0 Z(A) = 5(2) + 2(0) = 10 20 45 Z (B) = 5   + 2   = 10  19   19  Z(C) = 5(0) + 2(3) = 6 Class 12

142

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Thus, Z has maximum value 10 at two points A(2, 0)  20 45  and B  , .  19 19  37. The line joining the given points P(1, 4, 2) and Q(2, 3, 5) has direction ratios i.e., The plane through (1, 2, 1) and perpendicular to the line PQ is –1(x – 1) + 1(y – 2) – 3(z – 1) = 0 ⇒ x – y + 3z – 2 = 0 x +3 y −5 z −7 Now, direction ratios of line are = = −1 2 −1 2, –1, –1. Since 2(1) + (–1) (–1) + (3) (–1) = 2 + 1 – 3 = 0 \ Line is parallel to the plane. Since, (–3, 5, 7) lies on the given line. \ Distance of the point (–3, 5, 7) from plane is −3 − 5 + 3(7) − 2 d= 1+1+ 9 11 ⇒ d= = 11 units. 11 OR x + 2 y +1 z − 3 = = is the given line ...(i) 3 2 6 Let P(–2, –1, 3) lies on the line. The direction ratios of line (i) are 3, 2, 6 3 2 6 \ The direction cosines of line are , , 7 7 7 Equation (i) may be written as x + 2 y +1 z − 3 = = ..(ii) 3 2 6 7 7 7 Coordinates of any point on the line (ii) may be taken as 6 2  3  7 r − 2, 7 r − 1, 7 r + 3  2 6  3 Let Q ≡  r − 2, r − 1, r + 3   7 7 7

3 ⋅ 3 + (−5) ⋅ (−4) 3 ⋅ (−5) + (−5) ⋅ 2  29 −25 =  =  −4 ⋅ 3 + 2 ⋅ (−4) −4 ⋅ (−5) + 2 ⋅ 2   −20 24  Given f(x) = x2 – 5x – 14

⇒ f(A) = A2 – 5A – 14I2 1 0   29 −25  3 −5 −5 − 14  =     −20 24   −4 2  0 1   29 −25  15 −25 14 0  =  − −   −20 24   −20 10   0 14  −25 − (−25) − 0 0 0  29 − 15 − 14 =O = = 24 − 10 − 14  0 0  −20 − (−20) − 0 ⇒ A2 = 5A + 14I2 ...(i) \ A3 = AA2 = A(5A + 14I2)

(Using (i))

= A(5A) + A(14I2) = 5AA + 14(AI2)

= 5A2 + 14A = 5 (5A + 14I2) + 14A

 3 −5 1 + 70  = 39A + 70I2 = 39    −4 2  0

(Using (i)) 0 1

 117 −195 70 0   187 −195 = + =   −156 78   0 70  −156 148  OR The given equations can be written as 2 5  x  1   3 2   y  = 7       2 5 x  1  or AX = B, where A =  , X =   and B =    3 2  y 7  Now, on premultiplying the above matrix equation by A–1, we get (A–1A)X = A–1B ...(i) ⇒ X = A–1B

Given |r| = 2, \ r = ± 2 Putting the value r, we have  −8 −3 33  Q≡ , ,   7 7 7

2 5  2 −5 , |A| = –11 and adj A =  Now as A =    3 2  −3 2  1  2 −5 1 (adj A) = ∴ A−1 =   −11  −3 2  |A|

 −20 −11 9  or Q ≡  , ,  7 7 7 

Now, X =

 3 −5 38. We have, A =    −4 2   3 −5  3 −5 \ A2 = AA =     −4 2   −4 2 

⇒

1  2 −5 1      −11  −3 2  7 

[Using (i)]

 x  1  33  x   3   y  =  −11 ⇒  y  =  −1   11      

Hence x = 3 and y = –1.

 143

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Self Evaluation Sheet Once you complete SQP-10, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Relations and Functions / Relations and Functions Matrices Integrals / Integrals Determinants Three Dimensional Geometry / Three Dimensional Geometry Relations and Functions Differential Equations / Differential Equations Probability Probability / Probability Three Dimensional Geometry Relations and Functions Integrals Probability Vector Algebra Probability Matrices Three Dimensional Geometry Application of Derivatives

19

Determinants

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Integrals / Integrals Application of Derivatives Application of Integrals Probability / Probability Inverse Trigonometric Functions Probability Continuity and Differentiability Three Dimensional Geometry Differential Equations / Differential Equations Integrals Relations and Functions Application of Derivatives / Application of Derivatives Differential Equations Continuity and Differentiability Continuity and Differentiability / Continuity and Differentiability Application of Integrals Linear Programming / Linear Programming Three Dimensional Geometry / Three Dimensional Geometry Matrices / Determinants

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

11

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)*

–

–

1(2)

1.

Relations and Functions

2.

Inverse Trigonometric Functions

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

2(2)#

1(2)*

1(3)*

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)*

1(2)

1(3)

–

3(6)

Vector Algebra

3(3)#

1(2)*

–

–

4(5)

11.

Three Dimensional Geometry

4(4)#

–

–

1(5)*

5(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

1(4)

2(4)

–

–

3(8)

18(24)

10(20)

7(21)

3(15)

38(80)

10.

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-11

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 2 λ −3 1. If A = 0 2 5  , then for what value of l, A–1 will exist?   1 1 3  OR

Find the values of x for which

3 x

x 3 = 1 4

2 . 1

2. Show that the function f : R → R, given by f(x) = |x| is neither one-one nor onto. ^ ^  ^ 3. Find the direction cosines of the perpendicular from the origin to the plane r . (2 i − 3 j − 6 k ) = 5 .

OR Find the equation of plane passing through the point (1, 2, 3) and the direction cosines of the normal as l, m, n. 0 1  4. Check whether the matrix A =   is a symmetric matrix. 1 0  Class 12

146

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1 5  3 −5 +  dx 5. Evaluate : ∫  5x + 2 x − 7 x +  x x Evaluate : ∫

OR

sin( x − a) dx sin( x + a)

6. Find the number of bijective functions from set A to itself when A contains 106 elements.     7. If a = 2ɵi + ɵj + 3kɵ and b = 3ɵi + 5ɵj − 2kɵ , then find | a × b | . OR Find the value of l so that the vectors 2ɵi − 4 ɵj + kɵ and 4ɵi − 8 ɵj + λ kɵ are perpendicular. π /2

8. Evaluate :

∫

cos 2 x dx

π/ 4

9. Write the direction cosines of a line parallel to z-axis. 3 2 d y   dy   2 10. Determine the order and degree of 5 2 = 1 +    .  dx   dx   2

OR dy + (sec x ) y = tan x . Find the integrating factor of the differential equation dx         11. Let a and b are non-collinear. If c = (x − 2)a + b and d = (2 x + 1)a − b are collinear, then find x. 12. Find the equation of the plane with intercept 2, 3 and 4 on the x,y and z-axis, respectively. 2 x : x > 3  2 13. Let f : R → R be defined by f (x) =  x : 1 < x ≤ 3 . Find f (– 1) + f (2) + f (4). 3x : x ≤ 1  14. The equation of a line is 5x – 3 = 15y + 7 = 3 – 10z. Write the direction cosines of the line. 15. Find the projection of the vector ɵi + 3ɵj + 7kɵ on the vector 2ɵi − 3ɵj + 6kɵ . 3 4  16. If A =   , then find A + A ′ , where A ′ is the transpose of matrix A. 2 3  Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Ajay wants to construct a rectangular fish tank for his new house that can hold 72 ft3 of water. The top of the tank is open. The width of tank will be 5 ft but the length and heights are variables. Building the tank cost 10 per sq. foot for the base and 5 per sq. foot for the side. Based on the above information, answer the following question : (i) In order to make a least expensive fish tank, Ajay need to minimize its (a) Volume (b) Base (c) Curved surface area (d) Cost 147

Mathematics

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(ii) Total cost of tank as a function of h can be represented as (a) c(h) = 50 h – 144 – 720/h (b) c(h) = 50 h – 144 h – 720 h2 720 (c) c(h) = 50 + 144 h + 720 h2 (d) c(h) = 50 h + 144 + h (iii) Range of h is (a) (3, 5) (b) (0, ∞) (c) (0, 8) (d) (0, 3) (iv) Value of h at which c(h) is minimum, is (b) 12.2 (a) 14.4

(c) 14.5

(d) 12.5

(v) The cost of least expensive tank is (b) 502.04 (a) 500

(c)

(d)

523.47

600.05

18. A night before sleep, grandfather gave a puzzle to Rohan and Payal. The probability of solving this specific puzzle 1 1 independently by Rohan and Payal are and respectively. 2 3 Based on the above information answer the following : (i) Probability that both solved the puzzle, is 1 (a) 2 1 (c) 6

1 3 5 (d) 6 (b)

(ii) Probability that puzzle is solved by Rohan but not by Payal, is (a)

1 2

(b)

1 6

(c)

3 5

(d)

1 3

(d)

5 6 5 6

(iii) Find the probability that puzzle is solved. 1 1 2 (b) (c) 2 3 3 (iv) Probability that exactly one of them solved the puzzle. (a)

1 1 (b) 3 2 (v) Probability that none of them solved the puzzle. (a)

(a)

1 2

(b)

1 3

(c)

1 6

(d)

(c)

2 3

(d) None of these

PART - B Section - III 19. Find the area enclosed by the line y = 3x, the x-axis, and the ordinates x = 1 and x = 4.  −1   −1  20. Find the principal value of cos −1   + 2 sin −1   . 2  2 

OR Find the two branches other than the principal value branch of tan–1x. 21. A coin is tossed and then a die is thrown. Find the probability of obtaining a ‘6’ given that head came up. Class 12

148

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3 , then find A–1. 10 

1 22. If A =  3 3

23. Evaluate :

1

∫ x(1 + log x)2 dx 1

Evaluate :

OR

x π e  +  dx 2 4

2π x

∫0

24. If y = cotx, then show that

d2 y dx

2

+ 2y

dy = 0. dx

25. Find the order and degree of the differential equation given by x3

y2 dy 3y dx

2x 2 5x

3 0 = 0.

 d 2 y  dy  2  2y 2 +    0  dx    dx

2 . 3 27. A couple has 2 children. Find the probability that both are boys, if it is known that (i) one of them is a boy, (ii) the older child is a boy.

26. Find the point at which the tangent to the curve y = 4 x − 3 − 1 has its slope

28. If A, B, C have position vectors (2, 0, 0), (0, 1, 0), (0, 0, 2), show that DABC is isosceles. OR               If a × b = c × d and a × c = b × d , then show that (a − d ) is parallel to (b − c ) , it is being given that a ≠ d   and b ≠ c . Section - IV 29. Find the equations of tangent and normal to the curve 2x2 + 3y2 – 5 = 0 at (1, 1). 30. Show that the family of curves for which the slope of the tangent at any point (x, y) on it is by x2 – y2 = Cx.

x2 + y2 , is given 2xy

 sin(a + 1)x + sin x , x0  OR  b x3 If 1 − x 2 + 1 − y 2 = a(x − y ) , then prove that

1 − y2 dy . = dx 1 − x2

32. Find the area bounded by the curve y = sinx between x = 0 and x = 2π. 33. Let P be the set of all the points in a plane and the relation R in set P be defined as R = {(A, B) ∈ P × P | distance between points A and B is less than 3 units}. Show that the relation R is not an equivalence relation. 149

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34. Evaluate :

π

x sin x

∫0 1 + cos2 x dx OR 3

Evaluate : ∫ | x2 − 2 x | dx 1

35. If x = a(cos 2q + 2q sin 2q) and y = a(sin 2q − 2q cos 2q), then find

d2 y dx

2

at θ =

π . 8

Section - V 36. Solve the following problem graphically. 4x 6y Minimize Z = + 1000 1000 subject to constraints : 0.1 x + 0.05 y ≤ 50 0.25 x + 0.5 y ≥ 200 x, y ≥ 0 OR Solve the following problem graphically. Minimize Z = 150x + 200y subject to constraints : 6x + 10y ≥ 60 4x + 4y ≤ 32 x, y ≥ 0 37. The management committee of a residential colony decided to award some of its members (say x) for honesty, some (say y) for helping others and some others (say z) for supervising the workers to keep the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for helping others and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. OR 2 3 4    If A = 1 −1 0  , then find A–1. Hence, solve the system of equations 2x + 3y + 4z = 17; x − y = 3;y + 2z = 7. 0 1 2  38. Find the shortest distance between the lines x −2 y −4 z −5 x −1 y − 2 z − 3 and = = . = = 3 4 5 2 3 4 OR  If the points (1, 1, p) and (–3, 0, 1) be equidistant from the plane r ⋅(3ɵi + 4 ɵj − 12kɵ ) + 13 = 0, then find the value of p.

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SOLUTIONS 1. We know that A–1 exists if |A| ≠ 0. 2 λ −3 \ 0 2 5 ≠ 0 ⇒ 2(6 – 5) + 1(5l + 6) ≠ 0 1 1

3

⇒ 2 + 5l + 6 ≠ 0 −8 ⇒ l≠ 5

⇒

5l ≠ –8

OR We have, 3 – x2 = 3 – 8 ⇒ x2 = 8 ⇒ x = ±2 2 2. Since, f(1) = f(–1) = 1, therefore f is not one-one. Also, Range (f) = [0, ∞) ≠ R, therefore f is not onto. 3. The perpendicular from the origin to plane  r ⋅ (2ɵi − 3ɵj − 6kɵ ) = 5 is along the vector 2ɵi − 3ɵj − 6kɵ . 2 −3 −6 \ Its d.c’s are , , 4 + 9 + 36 4 + 9 + 36 4 + 9 + 36  2 −3 −6  = , ,  7 7 7 

sin t cos 2a cos t sin 2a dt − ∫ dt sin t sin t = cos 2a ∫ dt − sin 2a ∫ cot t dt = tcos2a – sin2a log|sint| + C = (x + a)cos2a – sin2a log|sin(x + a)| + C =

∫

6. The total number of bijections from a set containing n elements to itself is n! Hence, required number = (106)! ^

i

0 1  4. Since, A′ =  = A 1 0  \ A is a symmetric matrix.

1 5  5. We have ∫  5x 3 + 2x −5 − 7x + +  dx  x x 1 = 5∫ x 3dx + 2∫ x −5dx − 7∫ xdx + ∫ x −1/2dx + 5∫ dx x 4 2 1/2 −4 x x x x = 5⋅ + 2⋅ − 7⋅ + + 5 log | x | + C 4 2 (1 / 2) (−4) 5x 4 1 7x 2 − 4− + 2 x + 5 log | x | + C 4 2 2x OR sin(x − a) Let I = ∫ dx sin(x + a) Put x + a = t ⇒ x = t – a ⇒ dx = dt sin(t − 2a) ∴ I=∫ dt sin t sin t cos 2a − cos t sin 2a =∫ dt sin t =

^

j

k

  7. We have, a × b = 2 1

3

3 5 −2 ^

^

^

^

^

= i (−2 − 15) − (−4 − 9) j + (10 − 3) k = −17 i + 13 ^j + 7 k   Hence, | a × b | = (−17)2 + (13)2 + (7)2 = 507

OR The given vectors will be at right angles if their dot product vanishes, i.e., (2ɵi − 4 ɵj + kɵ ) ⋅ (4ɵi − 8 ɵj + λ kɵ ) = 0 ⇒ 8 + 32 + l = 0 ⇒ l = –40

OR Equation of plane passing through (1, 2, 3) having direction cosines of normal as l, m, n is l(x – 1) + m(y – 2) + n(z – 3) = 0 ⇒ lx + my + nz = l + 2m + 3n

^

π /2

8. We have, I =

π /2

 sin 2 x  ∫ cos 2x dx =  2  π/4 π/4

π   sin π sin 2  1 1 = −  = 0− = − 2  2 2  2 9. We know that, two parallel lines have same set of direction cosines. Therefore, required direction cosines are the direction cosines of Z-axis, i.e., 0, 0, 1 10. The given differential equation can be written as 3

  dy  2   d 2 y  2 1 +    =  5 2    dx    dx  Clearly, it can be observed that order of differential equation is 2 and degree is 2. OR dy Given, + (sec x ) y = tan x , dx which is a linear differential equation of the form dy + Py = Q dx Here, P = sec x and Q = tan x secx dx Pdx log |sec x + tan x| \ I.F. = e ∫ = e ∫ =e = secx + tanx 151

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      11. Given, c = (x − 2) a + b, d = (2 x + 1) a − b are  c = md collinear, therefore     ⇒ (x − 2)a + b = m ((2 x + 1)a − b ) ⇒ m = –1 1 and m (2x + 1) = x – 2 ⇒ – 2x – 1 = x – 2 ⇒ x = 3

12. Here, a = 2, b = 3 and c = 4 x y z \ Required equation of plane is + + = 1 . 2 3 4 ⇒ 6x + 4y + 3z = 12, which is the required equation of plane. 13. Clearly f(–1) = 3(–1) = –3 ; f(2) = (2)2 = 4 and f(4) = 2(4) = 8 \ f(–1) + f(2) + f(4) = –3 + 4 + 8 = 9 14. The given equation of line can be written in standard form as 3 3 7    5  x −  = 15  y +  = − 10  z −   10    5 15  3 7 3 y+ z− 5= 15 = 10 ⇒ 1 1 1 − 5 15 10 3 7 3 x− y+ z− 5= 15 = 10 ⇒ 6 2 −3 Thus, d.r.'s of given line are 6, 2, –3. x−

Hence, d.c.'s of given line are

6 2 −3 , , . 7 7 7

  15. Let a = ɵi + 3ɵj + 7kɵ and b = 2ɵi − 3ɵj + 6kɵ .   a ⋅ b  Now, projection of a on b =  |b | ɵ ɵ ɵ ɵ ɵ ɵ (i + 3 j + 7k ) ⋅ (2i − 3 j + 6k ) 2 − 9 + 42 35 = = = =5 2 2 2 4 + 9 + 36 7 2 + (−3) + 6  3 2 3 4  16. We have, A =  therefore A′ =     4 3 2 3   3 4   3 2  6 6  ⇒ A + A′ =   = + 2 3   4 3  6 6 

17. (i) (d) : In order to make least expensive tank, Ajay need to minimize its cost. (ii) (d) : Let l be the length and h be the height of the tank. Since breadth is equal to 5 ft. (Given) \ Two sides will be 5h sq. feet and two sides will be l h sq. feet. So, the total area of the sides is (10 h + 2 l h)ft2 Cost of the sides is ` 5 per sq. foot. So, the cost to build the sides is (10h + 2lh) × 5 = ` (50h + 10lh)) Also, cost of base = (5 l) × 10 = ` 50 l

Total cost of the tank in ` is given by c = 50 h + 10 l h + 50 l Since, volume of tank = 72 ft3 72 \ 5 l h = 72 ft3 \ l = 5h 72 72 \ c (h) = 50 h + 10   h + 50    5h   5h  = 50 h + 144 +

720 h

(iii) (b) : Since, all side lengths must be positive. 72 \ h > 0 and >0 5h 72 > 0, whenever h > 0 Since, 5h \ Range of h is (0, ∞) dc =0 (iv) (a) : To minimize cost, dh ⇒ 50 − 720 =0 h2 ⇒ 50 h2 = 720 ⇒ h2 = 14.4 ⇒ h = ± 14.4 ⇒ h = 14.4 [ height can not be negative] (v) (c) : \ Cost of least expensive tank is given by 720 c ( 14.4 ) = 50 14.4 + 144 + 14.4 = 100 × 14.4 + 144 ≈ 523.47 18. Let E1 be the event that Rohan solved the puzzle and E2 be the event that Payal solved the puzzle. Then, P(E1) = 1/2 and P(E2) = 1/3 (i) (c) : Since, E1 and E2 are independent events \ P(both solved the puzzle) = P(E1 ∩ E2) 1 1 1 = P(E1) ⋅ P(E2) = × = 2 3 6 (ii) (d) : P(puzzle is solved by Rahul but not by Payal)  1 1 2 1 1 = P( E 2)P(E1) =  1 −  ⋅ = ⋅ = 3 2 3 2 3 (iii) (c) : P(puzzle is solved) = P(E1 or E2) = P(E1 ∪ E2) = P(E1) + P(E2) – P(E1 ∩ E2) 1 1 1 4 2 = + − = = 2 3 6 6 3 (iv) (b) : P(Exactly one of them solved the puzzle) = P(E1 and E 2) or (E2 and E 1) = P (E1 ∩ E 2) + P(E2 ∩ E 1) = P(E1) × P( E 2) + P(E2) × P( E 1) 1 2 1 1 = × + × [ P( E 1)= 1 – P(E1)] 2 3 3 2 1 1 3 1 = + = = 3 6 6 2 Class 12

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(v) (b) : P(none of them solved the puzzle) 1 2 1 = P (E1 ∩ E2 ) = P (E1 ) ⋅ P (E2 ) = × = 2 3 3 19. Area enclosed by line y = 3x, x-axis, x = 1 and x = 4 is shown in figure. 4

\ Required area =

∫ 3x dx 1

4

 3x  3 =  = [16 − 1]  2 1 2 2

3 45 × 15 = = 22.5 sq . units 2 2 −1 2π 20. Principal value of cos −1   is  2  3 =

−1 −π Principal value of sin−1   is   2  6  −1 −1 \ Principal value of cos −1   + 2 sin −1    2   2  − π  2π π π 2π  = + 2 × − = =  3 6  3 3 3 OR We known that, principal value branch of tan–1x is  π π  − 2 , 2  .

\ Other two branches are  − π − π, π − π  and  − π + π, π + π   2   2  2 2 3π π π 3π i.e.,  − , −  and  ,  .  2 2 2  2 21. The sample space S associated to the given random experiment is given by S = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)} and Let the event B = {(H, 6), (T, 6)} and A = {(H, 1), (H, 2), (H, 3), (H, 4), (H, 5), (H, 6)} \ Required probability = P(B | A) 1 P(A ∩ B) 12 1 = = = P(A) 6 6 12 1 3  22. Let A =   3 10

\ |A| = 10 – 9 = 1 ≠ 0 So, A–1 exists.

10 −3 Now, adj A =    −3 1 

⇒ A−1 =

10 −3 1 (adj A) =  . | A|  −3 1 

3

1

∫ x(1 + log x)2 dx

23. I =

1

dx = dt x When x = 1, t = 1 and when x = 3, t = 1 + log 3

Put 1 + log x = t ⇒ 1+ log 3

∫

\ I=

1

dt t2

1+ log 3

 t −1  =    −1 1

log 3  1  = − − 1 =  1 + log 3  1 + log 3 OR

x π e  +  dx = 2 4

2π x

∫0

π x  e + e  dx 2 4

2π  x x

∫0

2π

  =   x  (e x ) −  1  (e x ) + π e x  2 4 0  2  1 π 1 π = πe 2 π − e 2 π + e 2 π + − 2 4 2 4  5π 1  1 π = e2 π  − + −  4 2  2 4 dy 24. We have, y = cot x ⇒ = − cosec2 x dx dy dy ⇒ = − (1 + cot2 x ) ⇒ = − (1 + y2 ) dx dx Differentiating w.r.t. x, we get d2 y dy d2 y  dy  + 2y =0 = − 2 y  ⇒ 2 2   dx dx dx dx 25. We have, x3 2x 2 5x

y2 3y

3

dy dx

0 =0

 d 2 y  dy  2  2 y 2 + 0  dx    dx

Expanding along C3, we get   d 2 y  dy  2  dy  2 3  4 x  y 2 +    − 15xy  = 0  dx   dx    dx ⇒ 4x2 y

d2 y

2

dy  dy  + 4 x 2   − 15xy =0 2  dx  dx dx

d2 y Now, highest order derivative is 2 . So, its order is 2 dx and degree is 1. 153

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26. Slope of tangent to the given curve at (x, y) is dy 1 = (4 x − dx 2

−1 3) 2 4

2 =

4x − 3

2 The given slope is . 3 2 2 = So, 3 4x − 3 Now, y =

dy dy −2 x = = −4 x ⇒ dx dx 3y −2(1) 2  dy  Now,   = = − = slope of the tangent at  dx (1,1) 3(1) 3 (1, 1). \ The equation of the tangent at (1, 1) is 2 y − 1 = − (x − 1) ⇒ 3y – 3 = – 2x + 2 3 \ 2x + 3y – 5 = 0 3 Now, the slope of the normal at (1, 1) is . 2 \ The equation of the normal at (1, 1) is 3 y − 1 = (x − 1) 2 \ 2y – 2 = 3x – 3 ⇒ 3x – 2y – 1 = 0 Hence, the equations of tangent and normal are 2x + 3y – 5 = 0 and 3x – 2y – 1 = 0 respectively. ∴ 6y

4x − 3 − 1

So, when x = 3, y = 4(3) − 3 − 1 = 2 Thus, the required point is (3, 2). 27. Let Bi(i = 1, 2) denote the ith child is a boy and Gi(i = 1, 2)denote the ith child is a girl respectively. Then sample space is, S = {B1B2, B1G2, G1B2, G1G2} Let A be the event that both are boys, B be the event that one of them is a boy and C be the event that the older child is a boy. A = {B1B2}, B = {G1B2, B1G2, B1B2} C = {B1B2, B1G2} ⇒ A∩B = {B1B2} and A∩C = {B1B2} (i) Required probability = P(A|B) P ( A ∩ B) 1 / 4 1 = = = P ( B) 3/4 3 (ii) Required probability = P(A|C) P(A ∩ C) 1 / 4 1 = = = P (C ) 2/4 2  28. We have, AB = P.V. of B – P.V. of A  ⇒ AB = (0ɵi + ɵj + 0kɵ ) − (2ɵi + 0 ɵj + 0kɵ ) = −2ɵi + ɵj + 0kɵ  \ AB = | AB | = (−2)2 + 12 + 02 = 5  BC = P.V. of C – P.V. of B  ⇒ BC = (0ɵi + 0ɵj + 2kɵ ) − (0ɵi + ɵj + 0kɵ ) = 0ɵi − ɵj + 2kɵ  2 2 2 \ BC = |BC| = 0 + ( − 1) + 2 = 5

2 2 dy x + y = dx 2 xy It is homogeneous equation. dy dv Putting y = vx and = v + x , we get dx dx

30. It is given that

2 2 2 1 + v2 dv x + v x = = dx 2v 2vx 2 2 dv 1 + v dv 1 − v 2 x = − v ⇒ x = ⇒ dx 2v dx 2v 2vdv dx 2vdv dx = = −∫ ⇒ ∫ 2 ⇒ 2 x x 1−v v −1

v+ x

⇒ log | v2 – 1 | = –log|x| + logC1 ⇒ log |(v2– 1)x| = log C1 ⇒ (y2 – x2) = C1x ⇒ x2 – y2 = Cx, where C = –C1 31. Here, f(0) = c L.H.L. at x = 0 sin(a + 1)x + sin x lim− f (x) = lim x x→ x →0

Since, AB = BC, therefore, DABC is isosceles. OR         Given, a × b = c × d and a × c = b × d     Now, (a − d ) × (b − c )         = (a × b ) − (d × b ) − (a × c ) + (d × c ) [By distributive law]         = (a × b ) + (b × d ) − (a × c ) − (c × d ) = 0          ⇒ (a − d ) × (b − c ) = 0 ⇒ (a − d )||(b − c )         [Here | a − d | ≠ 0 and | b − c | ≠ 0 as a ≠ d and b ≠ c ].

 sin(a + 1)x sin x  = lim−  + x  x  x →0  sin(a + 1)x sin x = lim− ⋅ (a + 1) + lim− x ( a + 1 ) x x →0 x →0 = (a + 1) + 1 = a + 2 R.H.L. at x = 0

29. We have, 2x2 + 3y2 – 5 = 0 Differentiating both sides w.r.t. x, we get dy −0 = 0 4 x + 6x dx

 1 + bx − 1  1 + bx + 1 = lim+   ×  1 + bx + 1 bx   x →0 1 + bx − 1 = lim+ x →0 bx( 1 + bx + 1)

...(i)

{

}

lim+ f (x) = lim−

x →0

x →0

x + bx 2 − x b x

3

= lim− x →0

x { 1 + bx − 1} bx x

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1 1 1 = =2 1 +1 1 + bx + 1 x →0 Now, f is continuous at x = 0 if lim f (x) = f (0) = lim f (x) = lim+

x →0 −

x →0+

1 1 3 ⇒ a = − and c = 2 2 2 Hence, for f (x) to be continuous at x = 0, we must have 3 1 a = − 2 , c = ; b may have any real value. 2 i.e., if a + 2 = c =

OR We have, 1 − x 2 + 1 − y 2 = a(x − y) Let x = sin A, y = sin B ∴ 1 − sin2 A + 1 − sin2 B = a(sin A − sin B) ⇒ cos A + cos B = a(sin A – sinB) A+B A−B ⇒ 2 cos  cos   2   2  A+B  A−B = 2a cos  sin  2   2  A−B A−B ⇒ cos  = a sin   2   2  A−B A−B ⇒ cot  = cot −1 a =a⇒  2  2 ⇒ A – B = 2 cot–1 a ⇒ sin–1 x – sin–1 y = 2 cot–1 a Differentiating w.r.t. x, we get 1 − y2 dy dy = ⇒ = 0 dx 1 − y 2 dx 1 − x2

1 ⇒

1

−

1 − x2

32. The given curve is y = sinx x y = sinx

0 0

π/2 1

3π 0

3π/2 –1

2π 0

B

–1

 2

3 2



D 2

x

C y

\ Graph between x = 0 and x = π, x = π and x = 2π has equal area above the x-axis and below the x-axis. \

2π

π

∫ sin x dx = 2(area OAB) = 2 ∫ sin x dx 0

x sin x

dx 1 + cos 2 x π (π − x)sin(π − x) 0

Then, I = ∫

...(i)

dx 1 + cos 2(π − x) π ⇒ I = ∫ (π − x)sin x dx ...(ii) 0 1 + cos 2 x Adding (i) and (ii), we get π (x + π − x)sin x π sin x 2I = ∫ dx = π∫ dx 2 0 0 1 + cos 2 x 1 + cos x π sin x ⇒ I = π∫ ...(iii) dx 2 0 1 + cos 2 x Put z = cos x ⇒ dz = – sin x dx Also, when x = 0, z = 1 and when x = p, z = – 1 \ From (iii), π −1 −dz π −1 1 π I= ∫ =− ∫ dz = − [tan −1 z]1−1 2 1 1+ z2 2 1 1+ z2 2 0

π π  π π  π2 = − [tan −1(−1) − tan −1(1)] = −  − −  = 2 2  4 4 4 Let I = ∫

A

O

π

3 2 x − 2x 1

y

x

34. Let I = ∫

OR

and –1 ≤ y = sinx ≤ 1 1

For reflexivity : (A, A) ∈ R is true as distance between points A and A is 0, which is less than 3 units for all A ∈ P. Hence, R is reflexive. For symmetry : Let A, B ∈ P and (A, B) ∈ R ⇒ distance between points A and B is less than 3 units. ⇒ Distance between B and A is less than 3 units. So, (B, A) ∈ R Hence, R is symmetric. For transitivity : Let points A, B and C are collinear. B is mid-point of AC such that distance between A and B is 2 units and between B and C is also 2 units, i.e., (A, B) ∈ R and (B, C) ∈ R, we notice distance between A and C is 4 units ⇒ (A, C) ∉ R. Hence, R is not transitive. Hence, R is not an equivalence relation.

0

= 2[− cos x]0π = 2[− cos π + cos 0] = 2 [1 + 1] = 4 sq. units. 33. Given, R = {(A, B) ∈ P × P | distance between points A and B is less than 3 units}

dx

−(x 2 − 2x) , when 1 ≤ x < 2 x 2 − 2x =  2  (x − 2x) , when 2 ≤ x ≤ 3

∵

2

3

⇒ I = ∫ |x2 – 2x|dx + ∫ | x 2 – 2x|dx 2

1

3

2

= ∫ − (x2 – 2x)dx + ∫ (x 2 − 2 x ) dx 2

1

2

3

 x3   x3  = −  − x2  +  − x2  3 1  3 2  4 2 4 6 = − − +  +   = = 2  3 3 3 3 155

Mathematics

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35. Given, x = a(cos 2q + 2q sin 2q) dx = a(–2 sin 2q + 2 sin 2q + 4q cos 2q) ⇒ dθ dx = a(4q cos 2q) ⇒ dθ Again, y = a(sin 2q – 2q cos 2q) dy = a(2cos 2q + 4q sin 2q – 2 cos 2q) ⇒ dθ dy = a(4q sin 2q) ⇒ dθ From (i) and (ii), we get dy a(4θ sin 2θ) = = tan 2θ dx a(4θ cos 2θ) Differentiating again w.r.t. x, we get

find the value of Z at each corner point.

...(i)

...(ii)

Corner Points A (0, 400) B (400, 200) C (0, 1000)

Value of Z 2.4 2.8 6

Minimum

Thus, Z has minimum value 2.4, when x = 0 and y = 400 OR The given problem is Minimize Z = 150x + 200y subject to constraints 6x + 10y ≥ 60 ; 4x + 4y ≤ 32 ; x ≥ 0, y ≥ 0 As x ≥ 0, y ≥ 0 \ Solution lies in first quadrant Y

2

d y

dθ 2 = 2 sec 2 θ . dx dx 2 2 1 d y ⇒ = 2 sec2 2θ ⋅ 2 a(4θ cos 2θ) dx  d2 y  8 2 1 π \  = = 2 sec2 ⋅ 2 π π πa 4 π  dx  θ = a cos 8 2 4 36. The given problem is 4x 6y + Minimize Z = 1000 1000 Subject to constraints: 0.1 x + 0.05 y < 50 0.25 x + 0.5 y ≥ 200 x, y ≥ 0 Convert the inequations into equations and draw the graph of lines: 0.1x + 0.05y = 50 ; 0.25x + 0.5y = 200

1000 900 800 700 600 500 400 A(0, 400) 300 200 100 100 200 300 400 500 600 700 800

As x ≥ 0, y ≥ 0 \ Solution lies in first quadrant Here, the shaded region is the feasible region. Now, we

C(0,8) A (0,6) X′

B(5,3) O(0,0) Y′

X 6x + 10y ≥ 60 4x + 4y ≤ 32

Convert the inequations into equations and draw the graph of lines: 6x + 10y = 60 ; 4x + 4y = 32 Here, shaded region is the feasible region. Corner points of feasible region are A(0, 6), B(5, 3) and C(0, 8). Value of Z at these corner points are: Corner Points Value of Z A(9, 6) 1200 (minimum) B(5, 3) 1350 C(0, 8) 1600 Thus, Z has minimum value 1200 when x = 0 and y = 6. 37. According to given conditions, we have x + y + z = 12, 2x + 3(y + z) = 33, x + z = 2y i.e., x + y + z = 12, 2x + 3y + 3z = 33, x – 2y + z = 0. The given system of equations can be written as AX = B 12  x  1 1 1 2 3 3 , X =  y  and B = 33 where A =       0 z  1 −2 1     1 1 1 Now, | A | = 2 3 3 1 −2 1 = 1 (3 + 6) – 1 (2 – 3) + 1 (–4 –3) =9+1–7=3≠0 \ A–1 exists and the solution is given by X = A–1 B. Class 12

156

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 9 −3 0    Now, adj A =  1 0 −1  −7 3 1     9 −3 0  1 1  −1 \ A = adj A =  1 0 −1 | A| 3  −7 3 1     108 − 99 + 0   9 −3 0  12  1 1    −1 \ X = A B =  1 0 −1 33 =  122 + 0 + 0  3 3  −7 3 1   0   −84 + 99 + 0      x   9  3   1    ⇒  y  = 12 =  4  ⇒ x = 3, y = 4 and z = 5. 3 z  15  5        \ The number of awardees for honesty is 3, for helping others is 4 and supervising the workers is 5. OR 2 3 4  We have, A =  1 −1 0  0 1 2    \ |A| = 2(–2 – 0) –3(2 – 0) + 4(1 – 0) = –6 ≠ 0 \ A–1 exists. Cofactors are A11 = –2, A12 = –2, A13 = 1, A21 = –2, A22 = 4, A23 = –2, A31 = 4, A32 = 4, A33 = –5  −2 −2 1  ′  −2 −2 4  ∴ adj A =  −2 4 −2  =  −2 4 4   1 −2 −5   4 4 −5       −2 −2 4  adj A 1  −2 4 4  ∴ A−1 = = | A | −6  1 −2 −5    System of equations can be written as AX = B, 2 3 4  x 17       Where A = 1 −1 0 , X = y , B = 3  0 1 2  z  7       Now, AX = B ⇒ X = A–1B  −2 −2 4  17  1  ⇒ X= −2 4 4   3  −6  1 −2 −5   7    

38. The equation of two given lines are x −1 y − 2 z − 3 ...(i) = = 2 3 4 x −2 y −4 z −5 and ...(ii) = = 3 4 5 Line (i) passes through (1, 2, 3) and has direction ratios proportional  to 2, 3, 4. So, its vector equation is   ...(iii) r = a1 + λb1 

where, a1 = ɵi + 2 ɵj + 3kɵ and b1 = 2ɵi + 3 ɵj + 4kɵ Line (ii) passes through (2, 4, 5) and has direction ratios proportional to 3, 4, 5. So, its vector equation is    ...(iv) r = a2 + µb2   ɵ ɵ ɵ where, a = 2i + 4 j + 5k and b = 3ɵi + 4ɵj + 5kɵ 2

2

The shortest distance between the lines (iii) and (iv) is given by     (a2 − a1) ⋅ (b1 × b2)   ...(v) S.D. = | b1 × b2 | We have,   a2 − a1 = (2ɵi + 4 ɵj + 5kɵ ) − (ɵi + 2 ɵj + 3kɵ ) = ɵi + 2 ɵj + 2kɵ ɵi ɵj kɵ   and b × b = 2 3 4 = − ɵi + 2 ɵj − kɵ 1

2

3 4 5   \ | b1 × b2 | = 1 + 4 + 1 = 6

...(i)

[From (i)]

 −12   2  1  −34 − 6 + 28  −1  ⇒ X = −  −34 + 12 + 28  = 6  =  −1 6 17 − 6 − 35  6  −24   4       

x  2  ⇒ X =  y  =  −1 z   4      ⇒ x = 2, y = –1 and z = 4

    and (a2 − a1 ) ⋅ (b1 × b2 ) = (ɵi + 2 ɵj + 2kɵ ) ⋅ (−ɵi + 2 ɵj − kɵ ) =–1+4–2=1     Substituting the values of (a2 − a1 ) ⋅ (b1 × b2 ) and   b1 × b2 in (v), we get S.D. = 1 / 6 unit. OR  ɵ ɵ The given plane is r ⋅ (3i + 4 j − 12kɵ ) + 13 = 0 ⇒ (xiɵ + y ɵj + zkɵ ) ⋅ (3ɵi + 4 ɵj − 12kɵ ) + 13 = 0 ⇒ 3x + 4y – 12z + 13 = 0 3(1) + 4(1) − 12( p) + 13 3(−3) + 4(0) − 12(1) + 13 = Now, 9 + 16 + 144 9 + 16 + 144 (Given points are equidistant from the given plane) ⇒ |20 – 12p | = | –8 | ⇒ 20 − 12 p = ± 8 ⇒ 5 − 3p = ±2 ⇒ – 3p = 2 – 5 or – 3p = – 2 – 5 ⇒ – 3p = –3 or – 3p = – 7 7 Hence, p = 1 or p = 3

 157

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Self Evaluation Sheet Once you complete SQP-11, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Determinants / Determinants Relations and Functions Three Dimensional Geometry / Three Dimensional Geometry Matrices Integrals / Integrals Relations and Functions Vector Algebra / Vector Algebra Integrals Three Dimensional Geometry Differential Equations / Differential Equations Vector Algebra Three Dimensional Geometry Relations and Functions Three Dimensional Geometry Vector Algebra Matrices Application of Derivatives Probability

19

Application of Integrals

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Inverse Trigonometric Functions / Inverse Trigonometric Functions Probability Determinants Integrals / Integrals Continuity and Differentiability Differential Equations Application of Derivatives Probability Vector Algebra / Vector Algebra Application of Derivatives Differential Equations Continuity and Differentiability / Continuity and Differentiability Application of Integrals Relations and Functions Integrals / Integrals Continuity and Differentiability Linear Programming / Linear Programming Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

–

1(3)

–

3(5)

2.

Inverse Trigonometric Functions

1(1)

1(2)

–

–

2(3)

3.

Matrices

2(2)

–

–

1(5)*

3(7)

4.

Determinants

1(1)

1(2)*

–

–

2(3)

5.

Continuity and Differentiability

1(1)*

1(2)

2(6)

–

4(9)

6.

Application of Derivatives

1(4)

1(2)

1(3)*

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)*

1(2)

1(3)*

–

3(6)

Vector Algebra

2(2)#

1(2)*

–

–

3(4)

11.

Three Dimensional Geometry

5(5)#

–

–

1(5)*

6(10)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

1(4)

2(4)

–

–

3(8)

18(24)

10(20)

7(21)

3(15)

38(80)

10.

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-12

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I kx 2 , if 1. If the function f (x ) =   3, if

x≤2 is continuous at x = 2, then find the value of k. x>2 OR

If y = log7 (log x), then find

dy . dx

2. If tan–1(cotq) = 2q, then find the value of q. 3. Find the value of (i + j) × (j + k ) ⋅ (k + i) . OR If lines

x −1 y − 5 z − 6 x −1 y − 2 z − 3 are mutually perpendicular, then find the value of k. and = = = = 3k 1 −5 2k 2 −3

4. If a line makes angles 90°, 135°, 45° with the X, Y, Z axes respectively, then find its direction cosines. Class 12

160

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5. Evaluate :

dx

∫ 5 − 8x − x 2 OR π/ 4

Evaluate :

∫

sin x dx

− π/ 4

 3 4 −2 1   6. For matrix A =  −4 5 −3 , find ( A − A ′). (where A′ is the transpose of the matrix A) 2  2 7 9 7. Find the direction cosines of the side AC of a ∆ABC whose vertices are given by A (3, 5, 4), B (–2, –2, –2) and C (3, –5, 4). OR Show that three points A(–2, 3, 5), B(1, 2, 3) and C(7, 0, –1) are collinear. 8. If A = {1, 5, 6}, B = {7, 9} and R = {(a, b) ∈A × B : |a – b| is even}. Then write the relation R. 2

d2 y  dy  9. Find the degree and order of the differential equation : 5x   − 2 − 6 y = log x.  dx  dx OR dy = ey. dx 10. If A and B are the points (– 3, 4, – 8) and (5, – 6, 4) respectively, then find the ratio in which yz-plane divides the line joining the points A and B. Solve the differential equation (1 + x 2 )

11. If A is a square matrix such that A2 = A, then find (I + A)3 – 7A. 12. A line makes an angle of p/4 with each of X-axis and Y-axis. What angle does it make with Z-axis? 10 −2 –1 13. If P =   , then check whether P exists or not. 5 1 −       ^ ^ ^  ^ ^  ^ ^ ^ ^ 14. Write the projection of b + c on a , where a = 2 i − 2 j + k , b = i + 2 j − 2 k and c = 2 i − j + 4 k . 15. Let n(A) = 4 and n(B) = 6, then find the number of one-one functions from A to B. 16. A line makes 45° with OX, and equal angles with OY and OZ. Find the sum of these three angles. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A card is lost from a pack of 52 cards. From the remaining cards of pack two cards are drawn and are found to be both spades.

161

Mathematics

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Based on the above information, answer the following questions : (i) The probability of drawing two spades, given that a card of spade is missing, is 21 1 22 23 (a) (b) (c) (d) 425 425 425 425 (ii) The probability of drawing two spades, given that a card of club is missing, is 22 19 23 26 (b) (c) (d) (a) 425 425 425 425 (iii) Let A be the event of drawing two spades from remaining 51 cards and E1, E2, E3 and E4 be the events 4

that lost card is of spade, club, diamond and heart respectively, then the value of ∑ P ( A / Ei ) is i =1

(a) 0.17

(b) 0.24

(c) 0.25

(d) 0.18

(iv) All of a sudden, missing card is found and, then two cards are drawn simultaneously without replacement. Probability that both drawn cards are aces is 1 1 2 1 (b) (c) (d) (a) 121 221 221 52 (v) If two card are drawn from a well shuffled pack of 52 cards, with replacement, then probability of getting not a king in 1st and 2nd draw is 144 64 12 (b) (c) (d) none of these (a) 169 169 169 18. Arun got a rectangular parallelopiped shaped box and spherical ball inside it as his birthday present. Sides of the box are x, 2x, and x/3, while radius of the ball is r cm. Based on the above information, answer the following questions : (i) If S represents the sum of volume of parallelopiped and sphere, then S can be written as 4 x3 2 2 2 x2 4 2 + πr + πr (b) 3 2 3 3 3 2x 4 3 2 4 + πr (c) (d) x + πr 3 3 3 3 (ii) If sum of the surface areas of box and ball are given to be constant, then x is equal to (a)

(a)

k2 − 4 πr 2 6

(b)

k2 − 4 πr 6

(c)

k2 − 4 π 6

(d) none of these

(iii) The radius of the ball, when S is minimum, is k2 k2 k2 k2 (b) (c) (d) 4π + 3 54 + 4 π 64 + 3π 54 + π (iv) Relation between length of the box and radius of the ball can be represented as r r (c) x = (d) x = 3r (a) x = 2r (b) x = 2 2 (v) Minimum volume of the ball and box together is (a)

(a)

k2 2(3π + 54)2/3

(b)

k (3π + 54)3/2

(c)

k3 3(4 π + 54)1/2

(d) none of these

PART - B Section - III 19. Find the intervals on which the function f(x) = 2x3 + 9x2 + 12x + 20 is increasing. Class 12

162

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   20. A vector r is inclined at equal angles to OX, OY and OZ. If the magnitude of r is 6 units, then find r . OR   Find the value of l such that the vectors a = 2ɵi + λ ɵj + kɵ and b = ɵi − 2 ɵj + 3kɵ are perpendicular to each other. 1 1 21. If A and B are two independent events, such that P ( A) = and P ( B) = , then find the value of 2 5 P(A|A ∪ B). 22. If x ∈ [0, 1], then find the value of 23. Evaluate :

∫ π /2

16 + (log x )2 x

1 1− x  cos −1  .  1 + x  2

dx OR

sin x

∫ 1 + cos2 x dx

Evaluate :

0

dy 3e 2 x + 3e 4 x = x −x dx e +e 25. A and B are two events such that P(A) ≠ 0. Find P(B/A) if (i) A is a subset of B (ii) A ∩ B = f 24. Solve the differential equation :

2 −1 26. Find the derivative of [ 1 − x sin x − x] w.r.t. x.

27. Find the area bounded by the curve x2 + y2 = 1 in the first quadrant. 2  28. Compute the adjoint of the matrix 5   1

0 1 1

−1 0  . 3  OR

1 a 2  If the matrix 1 2 5 is not invertible, then find the value of a. 2 1 1 Section - IV 29. Let A = R – {2} and B = R – {1}. If f : A → B is a mapping defined by f(x) =

x −1 , then show that f is bijective. x −2

 cos2 x − sin2 x − 1 , for x ≠ 0  . If f (x) is continuous at x = 0, then find the value of k. 30. Consider f (x ) =  x2 + 1 − 1 k , for x = 0 

31. Find the values of x for which f (x) = (x (x – 2))2 is an increasing function. Also, find the points on the curve, where the tangent is parallel to x-axis. OR An open box with a square base is to be made out of a given quantity of cardboard of area c2 square units. c3 Show that the maximum volume of the box is cubic units. 6 3 1

32. Evaluate : ∫ {tan −1 x + tan −1 (1 − x )} dx 0

163

Mathematics

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2

2  x   dy  3 d y y = x , log 33. If = x − y .   then prove that x 2   a + bx  dx dx x 2 dy e (sin x + sin 2 x ) = 34. Solve the differential equation . dx y(2 log y + 1) OR

Find the solution of the equation

y2 − y − 2 dy = 2 . dx x + 2 x − 3

35. Find the area enclosed between the curve y = loge (x + e) and the coordinates axes. Section - V  36. Find the image of the point having position vector i + 3j + 4k in the plane r ⋅ (2ɵi − ɵj + kɵ ) + 3 = 0. OR Find the points on the line

x + 2 y +1 z − 3 = = at a distance of 2 units from the point (–2, –1, 3). 1 2 2

37. Solve the following linear programming problem (LPP) graphically. Maximize Z = 4x + 6y Subject to constraints: x + 2y ≤ 80, 3x + y ≤ 75 ; x, y ≥ 0 OR Solve the following linear programming problem (LPP) graphically. Minimize Z = 30x + 20y Subject to constraints : x + y ≤ 8, x + 4y ≥ 12, 5x + 8y ≥ 20 ; x, y ≥ 0  0 6 7 0 1 1  2       38. If A =  −6 0 8  , B = 1 0 2 and C =  −2 , then calculate AC, BC and (A + B)C. Also verify that  7 −8 0  1 2 0  3  (A + B)C = AC + BC. OR 1 2   4 7  1 0  Find the matrix A satisfying the matrix equation  A = . 2 3  3 5  0 1 

Class 12

164

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SOLUTIONS 1. Since, f (x) is continuous at x = 2. ∴ lim− f (x ) = lim+ f (x ) x →2

⇒

x →2

3 k (2) = 3 ⇒ k = 4 2

OR y = log7 (log x) = \

log(log x ) log 7

dy dy 1 1 1 1 = ⋅ ⋅ ⇒ = dx log 7 log x x dx x log 7 log x

2. We have, tan–1(cotq) = 2q ⇒ cot q = tan 2q  π ⇒ cot q = cot  − 2θ   2 π π π – 2q ⇒ 3q = ⇒q= ⇒ q= 2 2 6 3. We have, (i + j) × (j + k ) ⋅(k + i) = (i × j + i × k + j × k ) ⋅(k + i) = (k − j + i) ⋅(k + i) = k ⋅ k + i ⋅i

(∵ ɵi ⋅ ɵj = ɵj ⋅ kɵ = kɵ ⋅ ɵi = 0)

2 2 = k + i = 1 + 1 = 2

OR x −1 y − 2 z − 3 = = and 2k 2 −3 x −1 y − 5 z − 6 = = are perpendicular if 3k 1 −5

Lines

 3  6. We have, A =  −4  2  0 1 1 ( A − A ′) =  −8 ∴ 2 2  4

 3 4 −2   5 −3 ⇒ A ′ =  4  −2 7 9 8 −4   0 4  0 −10 =  −4 0 10 0  2 5

7. The direction cosines of the line AC are 3−3 −5 − 5 , , 2 2 2 2 0 + (−10) + (0) 0 + (−10)2 + 02 4−4 02 + (−10)2 − 02 = 0, –1, 0 OR Direction ratios of the line AB = 3, –1, –2, Direction ratios of the line BC = 6, –2, –4 3 −1 −2 Now, = = 6 −2 −4 Since the direction cosines of the line AB and BC are proportional and B is the common point. Hence, the points are collinear. 8. We have, A × B = {(1, 7), (1, 9), (5, 7), (5, 9), (6, 7), (6, 9)} \ R = {(1, 7), (1, 9), (5, 7), (5, 9)}

a1a2 + b1b2+ c1c2 = 0. 10 ⇒ –3(3k) + 2k + 2(–5) = 0 ⇒ k = − 7

9. Here, highest order derivative is

4. Here a = 90°, b = 135°, g = 45° Direction cosines are l = cos a = cos 90° = 0, −1 1 m = cos β = cos 135° = , n = cos γ = cos 45° = 2 2 dx dx =∫ 5. Let I = ∫ 21 − (x + 4)2 5 − 8x − x 2

is 2 and power of

=∫

dx ( 21) − (x + 4) 2

2

=

1 2 21

log

21 + x + 4 21 − x − 4

OR π/ 4

Let I =

∫

π/ 4

| sin x | dx = 2

− π/ 4

∫ sin x dx 0

= 2 [ − cos x ]0

π/ 4

 1  = −2  − 1 = 2 − 2  2 

−4 2  5 7 −3 9 −2 −5 . 0

+C

d2 y dx 2

d2 y dx 2

, so its order

is one, so its degree is 1. OR

dy = ey dx dy dy dx dx ⇒ y = ⇒ ∫ y =∫ 2 e e 1+ x 1 + x2

We have, (1 + x 2 )

⇒ –e–y = tan–1x + C ⇒ e–y + tan–1x + C1 = 0. 10. Let l be the ratio in which yz-plane divides the line joining the points (–3, 4, –8) and (5, –6, 4). The co-ordinates of any point on the line joining the two  5λ − 3 −6 λ + 4 4 λ − 8  points are  , , . If the point is  λ +1 λ +1 λ + 1  in yz-plane, then its x-coordinate should be zero. 165

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5λ − 3 3 = 0 ⇒ 5λ − 3 = 0 ⇒ λ = λ +1 5 So, the required ratio is 3 : 5.

(iii) (b) : We have , P(E1) = P(E2) = P(E3) = P(E4) 13 1 = = 52 4

11. We have, A2 = A …(i) Now, (I + A)3 – 7A = I3 + A3 + 3A2I + 3AI2 – 7A = I + A2A + 3A2I + 3AI – 7A = I + AA + 3A + 3A – 7A [Using (i)] 2 =I+A –A=I+A–A [Using (i)] =I

P ( A / E1 ) =

∴

12. Let g be the required angle. Then cos2a + cos2b + cos2g = 1 1 1 ⇒ cos2 γ = 1 − − = 0 ⇒ cos γ = 0 2 2 π ⇒ γ= 2 10 −2 13. Since | P | = = 10 − 10 = 0 −5 1

22 + (−2)2 + 12

6 = =2 3

15. Number of one-one functions from A to B = 6P4 = 6 · 5 · 4 · 3 = 360 16. Here a = 45° and b = g 1 \ cos a = and cos b = cos g 2 Since, cos2a + cos2b + cos2g = 1 ⇒ 1/2 + cos2b + cos2 b = 1 1 ⇒ 2 cos2b = 1/2 ⇒ cosβ = ⇒ b = g = 60° 2 \ a + b + g = 45° + 60° + 60° = 165° 12

17. (i) (b) : Required probability =

51

=

C2 C2

22 425

=

26 425

13

P ( A / E2 ) =

51

C2

26 425 4 22 26 26 26 100 ∴ ∑ P ( A / Ei ) = + + + = = 0.24 425 425 425 425 425 i =1 P ( A / E3 ) = P ( A / E4 ) =

4

 ^ ^ ^ 14. Here, a = 2 i − 2 j + k  ^ ^  ^ ^ ^ ^ b = i + 2 j − 2 k and c = 2 i − j + 4 k   ^ ^ ^ ⇒ b + c = 3i + j + 2k    ∴ Projectionof b + c on a  (b + c ) ⋅ a (3 i^ + ^j + 2 k^) ⋅ (2 i^ − 2 ^j + k^) = =  a ^ ^ ^ 2i − 2 j + k 3 × 2 + 1 × (−2) + 2 × 1

51

C2

(iv) (b) : P(getting both aces) =

\ P–1 does not exist.

=

12

C2

C2 12 × 11 22 = = 51 × 50 425 13 C2 13 × 12 26 (ii) (a) : Required probability = = = 51 C2 51 × 50 425

C2

52

=

4×3 1 = 52 × 51 221

C2 4 1 (v) (a) : P(drawing a king) = = 52 13 1 12 \ P(not drawing a king) = 1 − = 13 13 12 12 144 × = \ Required probability = 13 13 169

18. (i) (c) : Let S be the sum of volume of parallelopiped and sphere, then 2x 3 4 3  x 4 ... (1) + πr S = x(2 x )   + πr 3 =  3 3 3 3 (ii) (a) : Since, sum of surface area of box and sphere is given to be constant. x x   ∴ 2  x × 2 x + 2 x × + × x  + 4 πr 2 = k 2 (say )   3 3 ⇒ 6x2 + 4pr2 = k2 k 2 − 4 πr 2 k 2 − 4 πr 2 ⇒ x= 6 6 (iii) (b) : From (1) and (2), we get ⇒ x2 =

2  k 2 − 4 πr 2  S=   3 6 

3/ 2

4 + πr 3 3

2

4 (k 2 − 4 πr 2 )3/2 + πr 3 3 3×6 6 dS 1 3 2 = (k − 4 πr 2 )1/2 (−8 πr ) + 4 πr 2 dr 9 6 2

= ⇒

... (2)

  1 = 4 πr r − k 2 − 4 πr 2    3 6 dS For maximum/minimum, =0 dr −4 πr 2 ⇒ k − 4 πr 2 = −4 πr 2 3 6 Class 12

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⇒ k2 – 4pr2 = 54r2 2

⇒ r2 =

2

k k ⇒ r= 54 + 4 π 54 + 4 π

... (3)

 k2   k 2 − 4 πr 2 1  2 = k − 4 π   6 6   54 + 4 π   [From (2) and (3)]  k2  9k 2 = 9r 2 = (3r )2 = = 9  54 + 4 π  54 + 4 π 

(iv) (d) : Since, x 2 =

⇒ x = 3r (v) (c) : Minimum volume is given by 2 4 2 4 V = x 3 + πr 3 = (3r )3 + πr 3 3 3 3 3 4 4   = 18r 3 + πr 3 =  18 + π r 3  3 3  4   k2   = 18 + π    3   54 + 4 π 

Now, =

3/ 2

[Using (3)]

k 1 3 (54 + 4 π)1/2

19. Given, f(x) = 2x3 + 9x2 + 12x + 20 ⇒ f ′(x) = 6x2 + 18x + 12 = 6(x2 + 3x + 2) = 6(x + 1)(x + 2) For f (x) to be increasing, f ′(x) > 0 ⇒ 6(x + 1)(x + 2) > 0 ⇒ (x + 1) (x + 2) > 0 ⇒ x + 1 > 0, x + 2 > 0 or x + 1 < 0, x + 2 < 0 ⇒ x > –1 or x < –2 ⇒ x ∈( –1, ∞) or x ∈ (– ∞, – 2) \ f is increasing in (–∞, –2) ∪ (–1, ∞).  20. Suppose r makes an angle a with each of the axes OX, OY and OZ. Then, its direction cosines are l = cos a, m = cos a, n = cos a ⇒ l = m = n 1 Now, l 2 + m2 + n2 = 1 ⇒ 3l2 = 1 ⇒ l = ± 3   ^ ^ ^ \ r = | r | ( l i + m j + nk )   1 ^ 1 ^ 1 ^ ^ ^ ^ i± j ± k  = 2 3 (± i ± j ± k ). ⇒ r =6 ±   3 3 3 OR   If the vectors a and b are perpendicular to each other,   then a ⋅ b = 0 . ⇒ (2i + λ j + k ) ⋅ (i − 2 j + 3k ) = 0

⇒ (2) (1) + l(–2) + (1) (3) = 0 5 ⇒ –2l + 5 = 0 ⇒ λ = 2

1 1 1 1 = + −   ⋅   (A and B are independent events) 2 5 2 5 3 = 5 P ( A ∩ ( A ∪ B)) ∴ P ( A / A ∪ B) = P ( A ∪ B) P ( A) 1/ 2 5 = = = P ( A ∪ B) 3 / 5 6 22. Let x = tan2 θ ⇒ x = tan θ ⇒ θ = tan −1 x

3

=

1 1 21. We have, P ( A) = , P (B) = 2 5 Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

 1 − tan2 θ  1 1− x  1 cos −1  = cos −1    1+ x  2 2  1 + tan2 θ 

1 1 cos −1 (cos 2θ) = (2θ) = θ = tan −1 x 2 2

16 + (log x )2 dx x 1 Put log x = t ⇒ dx = dt x 23. Let I = ∫

∴

I = ∫ 16 + t 2 dt 16 t 16 + t 2 + log | t + 16 + t 2 | + c 2 2 1 I = log x 16 + (log x )2 2 =

∴

+8 log | log x + 16 + (log x )2 + c OR π /2

∫

sin x

dx 2 1 cos x + 0 Put cos x = t ⇒ – sin x dx = dt

Let I =

π When x = 0, t = 1 and when x = ,t = 0 2 0 0 dt = −  tan −1 t 1 \ I = −∫ 2 1 1+ t π = −[tan −1 0 − tan −1 1] = 4 dy 3e 2 x + 3e 4 x = x −x dx e +e 2x 3e (1 + e 2 x ) dx [Integrating both sides] ⇒ ∫ dy = ∫ − x 2 x e (e + 1) 24. We have,

⇒ y = ∫ 3e 3 x dx =

3e 3 x + c ⇒ y = e3x + c 3 167

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25. (i) Since, A is a subset of B. \ A ⊂ B ⇒ A∩B=A \ P(A ∩ B) = P(A) ... (i) P (B ∩ A) P ( A) Now, P (B / A) = [Using (i)] = P ( A) P ( A) =1 (ii) If A ∩ B = f ⇒ P(A ∩ B) = 0 P (B ∩ A) 0 ∴ P (B / A) = = =0 P ( A) P ( A) d [( 1 − x 2 ) sin −1 x − x] 26. We have, dx d d = 1 − x 2 ⋅ (sin−1 x ) + (sin −1 x ) ⋅ 1 − x2 − 1 dx dx =

(

)

(

1 − x2 ⋅

)

= 1−

x sin

(

1

(

−1

x

1 − x2

1− x

2

−1 =

)

)

1 + (sin−1 x ) ⋅ (1 − x 2 )−1/2 ⋅ (−2 x ) − 1 2

− x sin −1 x 1 − x2

27. We have, x2 + y2 = 1, a circle with centre (0, 0) and radius = 1. Required area = area of shaded region 1

A = ∫ 1 − x 2 dx 0 1

x 1 x =  1 − x 2 + sin −1  2 1 0 2  1 π π 1 =  sin −1 1 =  ×  = sq. unit 2  2 2  4 2  28. Let A = 5  1 1 A11 = (−1)1+1 1 A13 = (−1)1+3

0 −1 1 0  1 3  0 5 0 = 3 ; A12 = (−1)1+2 = −15 3 1 3

5 1 0 −1 = 4 ; A21 = (−1)2+1 = −1 1 1 1 3

OR 1 a 2 The matrix is not invertible if 1 2 5 = 0 2 1 1 ⇒ 1(2 – 5) – a (1 –10) + 2(1 – 4) = 0 ⇒ –3 + 9a – 6 = 0 ⇒ a = 1 x −1 x −2 For one-one : Let x, y ∈ A and consider f(x) = f(y) x −1 y −1 = ⇒ x −2 y −2

29. We have, f (x ) =

⇒ (x – 1)(y – 2) = (x – 2)(y – 1) ⇒ xy – y – 2x + 2 = xy – x – 2y + 2 ⇒ x = y Thus, f(x) = f(y) ⇒ x = y for all x, y ∈ A So, f is one-one. For onto : Let y be an arbitrary element of B. Then, 1− 2y x −1 = y ⇒ (x – 1) = y(x – 2) ⇒ x = f(x) = y ⇒ 1− y x −2 1− 2y is a real number for all y ≠ 1. Clearly, x = 1− y 1− 2y 1− 2y = 2, ≠ 2 for any y, for, if we take Also, 1− y 1− y then we get 1 = 2, which is wrong. So, f is onto. Hence, f is a bijective. 30. f(0) = k (Given) Since, f(x) is continuous at x = 0. ∴ f (0) = lim f (x ) = lim f (x ) x →0 −

= lim

2 −1 2 0 = −2; = 7 ; A23 = (−1)2+3 1 3 1 1

A31 = (−1)

0 −1 2 −1 = 1 ; A32 = (−1)3+2 = −5; 1 0 5 0

A33 = (−1)3+3

2 0 =2 5 1

3+1

′

 3 −15 4   3 −1 1   ∴ adj A =  −1 7 −2 =  −15 7 −5  1 −5 2   4 −2 2 

x →0

x2 + 1 − 1 ×

x2 + 1 − 1

x →0

= lim

cos2 x − sin2 x − 1

1 − sin2 x − sin2 x − 1

x2 + 1 + 1 x2 + 1 + 1

−2 sin2 x( x 2 + 1 + 1)

x →0

A22 = (−1)2+2

x →0 +

Now, lim f (x ) = lim x →0

= lim − 2

x2 +1 −1 sin2 x 2

.( x 2 + 1 + 1)

x  sin2 x  = −2  lim 2  × lim( x 2 + 1 + 1)  x →0 x  x →0 = –2(1)2 (1 + 1) = –4 From (i) and (ii), we get k = –4. x →0

…(i)

…(ii)

31. Given, f (x) = (x(x – 2))2 = x2(x – 2)2, Df = R. Differentiating w.r.t. x, we get f ′(x) = x2 · 2(x – 2) + (x – 2)2 · 2x = 2x(x – 2)(x + x – 2) = 2x(x – 2)(2x – 2) = 4x(x – 1)(x – 2) Class 12

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Now, the given function f is (strictly) increasing iff f ′ (x) > 0 ⇒ x ∈ (0, 1) ∪ (2, ∞) –

–

+

+

0 1 2 Further, the tangents will be parallel to x-axis iff f ′(x) = 0 ⇒ x = 0, 1, 2 The given curve is y = x2(x – 2)2 When x = 0, y = 0; When x = 1, y = 12(1 – 2)2 = 1 × (–1)2 = 1 × 1 = 1; When x = 2, y = 22 (2 – 2)2 = 4 × 0 = 0. \ The points on the given curve, where the tangents are parallel to x-axis are (0, 0), (1, 1) and (2, 0).

OR Let h be height and x be the side of the square base of the open box. (given) Then its area = x × x + 4 h × x = c2 2 2 c −x ⇒ h= 4x h Now V = volume of the box x x 2 2 c − x 1 = x 2h = x 2 ⋅ = (c 2 x − x 3 ) 4 4x 2 dV 1 2 d V 1 −3 x ⇒ = (c − 3x 2 ) and 2 = (−6 x ) = dx 4 4 2 dx c2 dV = 0 ⇒ x2 = For maxima or minima dx 3 c ⇒ x= ( x 1  bx + 2, Section-V 2 1  −3 2  1 2  36. Find the matrix P satisfying the matrix equation  . = P 3 2  5 −3 2 −1 OR Two factories decided to award their employees for three values of (a) adaptable to new techniques, (b) careful and alert in difficult situations and (c) keeping calm in tense situations, at the rate of ` x, ` y and ` z per person respectively. The first factory decided to honour respectively 2, 4 and 3 employees with a total prize money of ` 29,000. The second factory decided to honour respectively 5, 2 and 3 employees with the prize money of ` 30,500. If the three prizes per person together cost ` 9,500; then (i) Represent the above situation by a matrix equation and form linear equations using matrix multiplication. (ii) Solve these equations using matrices. 177

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37. Find the distance of the point (–2, 3, –4) from the line 4x + 12y – 3z + 1 = 0.

x + 2 2 y + 3 3z + 4 measured parallel to the plane = = 3 4 5

OR Find the equation of the plane passing through three given points −2ɵi + 6 ɵj − 6kɵ , − 3ɵi + 10 ɵj − 9kɵ , −5ɵi − 6kɵ . 38. Solve the following linear programming problem (LPP) graphically. 35 x + 7y Maximize Z = 2 Subject to constraints : x + 3y ≤ 12; 3x + y ≤ 12; x, y ≥ 0 OR Solve the following linear programming problem (LPP) graphically. Maximize Z = 500 x + 150 y Subject to constraints : 2500 x + 500 y ≤ 50000 x + y ≤ 60; x, y ≥ 0

Class 12

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SOLUTIONS 1. We have,  sin θ − cos θ  cos θ sin θ cos θ  + sin θ   sin θ  − sin θ cos θ cos θ

, 5. Principal value branch of cosec–1x is  – {0}.  2 2  OR

 cos2 θ sin θ cos θ  sin2 θ − sin θ cos θ +  =  − sin θ cos θ cos2 θ  sin θ cos θ sin2 θ 

Let cos–1  2  = q  

 cos2 θ + sin2 θ sin θ cos θ − sin θ cos θ   =   − sin θ cos θ + cos θ sin θ cos2 θ + sin2 θ 1 0  =  0 1  OR 3

4

 0

1 

 −1 2 1   2 3 

Given, AT =  −1 2  and B =  1  −1 1   2  1 3 

⇒ BT =  2

 3 4   −1 1   4 3   \ A − B =  −1 2  −  2 2  =  −3 0   0 1   1 3   −1 −2  2. The projection of the vector (ɵi + ɵj + kɵ ) along the T

T

ɵj  vector ɵj is (ɵi + ɵj + kɵ ) ⋅   2 2 2  0 +1 + 0

  =1  

3. Let I = ∫ [sin(log x ) + cos(log x )] dx

− 3

t

= e sin t + C = xsin(logx) + C

1

1 + x2 Integrand is in the form ex[f(x) + f ′(x)] 1   \ ∫ e x  tan −1 x + dx = e x tan −1 x + C 2   1+ x 4. Given, A = {0, 1} Since, f : N → A such that f (2n − 1) = 0, f (2n) = 1 ∀n ∈N So, A = Range f, which is equal to Co-domain of A Hence, the mapping f : N → A is onto.

⇒ cos q =

− 3 2

5π π  π = cos  π −  = cos 6 6 6 5π ∈[0, π ] ⇒ q= 6 5π −1  − 3  . is \ Principal value of cos   6  2  = –cos

  6. Here a = ɵi + 3ɵj − 7kɵ , b = 5ɵi − 2 ɵj + 4kɵ    AB = b − a = 4i˘ − 5 ˘j + 11k˘ −5 −5 = Direction cosine along Y-axis = 16 + 25 + 121 162  π  7. Clearly, angle between a and b =   2 ⇒ a ⋅b = 0  2 2 2   ∴ a + b = a + b + 2a ⋅ b = 1+ 1 + 0 = 2   ⇒ a +b = 2

OR We have, a = 2iˆ + jˆ + 2kˆ  a = (2)2 + (1)2 + (2)2 = 4 + 1 + 4 = 9 = 3  2 1 2 a Required unit vector is aˆ =  = iˆ + jˆ + kˆ 3 3 3 a

\

8. M11 =

Put logx = t ⇒ x = et ⇒ dx = etdt \ I = ∫ (sin t + cos t )et dt OR 1   Let I = ∫ e x  tan −1 x +  dx  1 + x2  Consider, f(x) = tan–1x and f ′(x ) =

 −π π 

0 1 = 0 – 1 = –1 ⇒ C11 = M11 = –1 1 −1

M21 =

5 −1 = –5 + 1 = –4 ⇒ C21 = – M21 = 4 1 −1

M31 =

5 −1 = 5 – 0 = 5 ⇒ C31 = M31 = 5 0 1

2π π π 9. Here, α = , β = , γ = 3 3 4 Direction cosines of the line are cosa, cosb, cosg π 2π π 1 −1 1 , = cos , cos , cos = , 3 3 4 2 2 2 OR Direction cosines of the line joining P(4, 3, –5) and Q (–2, 1, –8) is x −x y −y z −z l = 2 1 ,m= 2 1 ,n= 2 1 PQ PQ PQ 179

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x2 – x1 = –6, y2 – y1 = –2, z2 – z1 = –3 PQ = 36 + 4 + 9 = 7 −6 −2 −3 \ l = ,m= ,n= 7 7 7 10. The equation of the plane passing through (2, 3, – 1) and perpendicular to the vector 3ɵi − 4 ɵj + 7kɵ is 3(x – 2) + (– 4) (y – 3) + 7(z – (– 1)) = 0 ⇒ 3x – 4y + 7z + 13 = 0 3 3 π 11. Let sin −1 = θ ⇒ sin θ = and θ ∈  0,  5 5  2 sin θ

3



sin θ

3/5

3

\ tan  sin −1  = tan θ = = = =  cos θ 5 1 − sin2 θ 4 / 5 4 2 3

12. Here,

∫ 1

x −1 x

2

2

dx = ∫ (x − x −2 ) dx 1

2

2

 x2 1   x 2 x −1   4 1 1  5 3 = −  =  +  =  +  −  + 1 = − = 1 2 −1 1  2 x 1  2 2   2  2 2

13. Direction ratios of the line AB = –3, –1, –4 Direction ratios of the line BC = 9, 3, l + 2 Since the points A, B and C are collinear, hence its direction ratios are proportional. −3 −1 −4 \ = = 9 3 λ+2 ⇒ l + 2 = 12 ⇒ l = 10 14. The cofactors of the elements of |A| are given by A11 = 3, A12 = –1, A21 = –5, A22 = 2 T

 3 −5  3 −1 = ∴ adj A =     −5 2   −1 2  15. The given equation of the plane is 5x – 7y + 2z = 3 or (xiɵ + y ɵj + zkɵ ) ⋅ (5ɵi − 7 ɵj + 2kɵ ) = 3  ⇒ r ⋅ (5ɵi − 7 ɵj + 2kɵ ) = 3 ,which is the required vector equation of plane. 16. The highest order derivative present in the differential equation is y′′′, so its order is three. The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined. 17. (i) (b) : Since, it rained only 5 days each year, therefore, probability that it rains on wedding day is 5 1 = 365 73 (ii) (c) : The probability that it does not rain on 1 72 360 = wedding day = 1 − = 73 73 365 (iii) (c) : It is given that, when it actually rains, the weatherman correctly forecasts rain 90% of the time. 90 9 = \ Required probability = 100 10

(iv) (a) : Let A1 be the event that it rains on wedding day, A2 be the event that it does not rain an wedding day and E be the event the weatherman predict rain. 5 360 Then we have, P(A1) = , P(A2) = , 365 365 9 1 P(E/A1) = and P(E/A2)= 10 10 Required probability P ( A1 ) ⋅ P (E / A1 ) = P(A1/E) = P ( A1 ) ⋅ P (E / A1 ) + P ( A2 ) ⋅ P (E / A2 ) 5 9 × 45 365 10 = ≈ 0.111 9 360 1 405 5 × + × 365 10 365 10 (v) (a) : Required probability = 1 – P(A1/E) = 1 – 0.111 = 0.889 18. (i) (b) : Let C(x) be the maintenance cost function, then C(x) = 4000000 + 140x – 0.04x2 We have, C(x) = 4000000 + 140x – 0.04x2 (ii) (b) : Now, C′(x) = 140 – 0.08 x For maxima/minima, put C′(x) = 0 ⇒ 140 = 0.08x ⇒ x = 1750 (iii) (b) : Clearly, from the problem statement we can see that we only want critical points that are in the interval [0, 5000] Now, we have C(0) = 4000000 C(1750) = 4122500 and C(5000) = 3700000 \ Maximum value of C(x)would be ` 4122500 (iv) (a) : The complex must have 5000 apartments to minimise the maintenance cost. (v) (a) : The minimum maintenance cost for each apartment woud be ` 740. 19. The given function is f(x) = x4 – 8x3 + 22x2 – 24x + 21 ⇒ f ′(x) = 4x3 – 24x2 + 44x – 24 = 4(x3 – 6x2 + 11x – 6) = 4(x – 1)(x2 – 5x + 6) = 4(x – 1)(x – 2)(x – 3) Thus, f ′(x) = 0 ⇒ x = 1, 2, 3. Hence, possible intervals are (–∞, 1), (1, 2), (2, 3) and (3, ∞). In the interval (–∞, 1), f ′ (x) < 0 In the interval (1, 2), f′(x) > 0 In the interval (2, 3), f′(x) < 0 In the interval (3, ∞), f′(x) > 0 \ f is increasing in (1, 2) ∪ (3, ∞).     20. Given, a = 2, b = 7 and a × b = 3ɵi + 2 ɵj + 6kɵ   \ a × b = 32 + 22 + 62 = 49 = 7 Class 12

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  Let q be the angle between a and b .   1 π 7 a ×b = ⇒ θ= Now, sin θ =   = 2×7 2 6 a b OR  Let a be the resultant of given vectors. Then,  a = (ɵi + 2 ɵj + 3kɵ ) − ɵi + 2 ɵj + kɵ + (3ɵi + ɵj) = 3ɵi + 5ɵj + 4kɵ  \ a = 32 + 52 + 42 = 5 2  3 ɵ 5 ɵ 4 ɵ  a Now, unit vector along a =  = i+ j+ k. a 5 2 5 2 5 2 2 −3 21. A =   5 −7  \ |A| = –14 + 15 = 1 ≠ 0 So, A–1 exists.  −7 3 ∴ adj A =    −5 2

=∫

sec 2 x dx = dx 3 sin 2 x + 4 ∫ 3 tan 2 x + 4 sec 2 x sec 2 x

 7 tan x  1 dt tan −1   + c 2 =  2 7 2 4 + 7t OR

π/ 4

Let I =

∫ (tan x + cot x)

2

dx

π /3 π/ 4

=

∫

π/ 4

(tan 2 x + 2 + cot 2 x ) dx =

π /3

∫ (sec π /3

2

2

x + cosec x )dx

3

=

e −e

−2 3

−x

e x + e−x

0, x≤0   x −x f (x ) =  e − e  x −x , x > 0 e + e ⇒ f(x) is many one. For x > 0, f (x ) =

e2x − 1 2x

= 1−

2 2x

e +1 e +1 Range of f(x) ∈[0, 1) ≠ codomain of f(x) ⇒ f(x) is into. So, f(x) is many one into.

y

dx 4 + 7 tan 2 x Put tan x = t ⇒ sec2x dx = dt ∴ I=∫

24. f : R → R such that f (x ) =

dy

2 −3  −7 3   −5 0  Thus, A + A −1 =  + =   5 −7   −5 2   0 −5 22. Total number of ways in which Mother (M), Father (F) and Son (S) can be lined up at random in one of the following ways: MFS, MSF, FMS, FSM, SFM, SMF is 6. We have, A = {MFS, FMS, SMF, SFM} and B = {MFS, SFM} \ A ∩ B = {MFS, SFM} Clearly, n(A ∩ B) = 2 and n(B) = 2 n( A ∩ B) 2 = =1 \ Required probability = P(A/B) = n(B) 2

1

|x|

25. We have,

 −7 3 1 (adj A) =  ⇒ A −1 =  |A|  −5 2

23. Let I = ∫

= [ tan x − cot x ]ππ//34 = 1 − 1 − 3 +

dy = ex + y + eyx3 = ey(ex + x3) dx

= (ex + x3)dx

e Integrating both sides, we get dy 3 x ⇒ ∫ y = ∫ (e + x )dx e x4 x4 ⇒ –e–y = ex + + c1 ⇒ ex + e–y + =c 4 4 26. Given curves are, y2 = x, y = 0, x = 1 and x = 3 Required area = 2 (Area of shaded region) 3

= 2 ∫ x dx 1 3

 x 3/ 2  = 2   3 / 2 1 4 = (3)3/2 − (1)3/2  3 4 4  = (3 3 − 1) =  4 3 −  sq . units  3 3 27. Let E1 be the event that a student fails in physics and E2 be the event that a student fails in mathematics. 30 3 25 1 Then, P (E1 ) = = , P ( E2 ) = = 100 10 100 4 10 1 and P (E1 ∩ E2 ) = = 100 10 P (E1 ∩ E2 ) \ Required probability = P (E1 | E2 ) = P ( E2 ) 1 / 10 4 2 = = = 1 / 4 10 5 28. Clearly, f(0) = 0.  e1/h − 1  Now, lim f ( x ) = lim f (0 + h) = lim f (h) = lim   h→0 h→0 h→0  e1/h + 1  x →0+ 181

Mathematics

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1   1− 1 e1/h 1 − 1/h  1/h  e   = lim  e 1 = lim 1  h→0 1 + h→0 1/h  e 1 + 1/h   e1/h  e 

   =1 

 −1   e h −1 lim f ( x ) = lim f (0 − h) = lim f (−h) = lim   h→0 h→0 h→0  −1 x →0−   e h +1  1 −1  1/h  = lim  e 1  = −1. h→0  1/h + 1  e

Thus, lim f ( x ) ≠ lim f (x ) and therefore, lim f (x ) x →0+

x →0−

x →0

does not exist. Hence, f (x) is discontinuous at x = 0. OR y We have, (cosx) = (cosy)x ⇒ y log cosx = x log cos y  − sin y  dy dy  − sin x  ⋅ + (log cos x ) = x ⋅  ⇒ y⋅   cos x  dx  cos y  dx +(log cos y ) ⋅1 dy ⇒ (log cos x + x tan y ) = (log cos y + y tan x ) dx  + log cos tan x  y y dy = ⇒ dx  log cos x + x tan y  29. Let side of square = x and radius of circle = r. Perimeter of square = 4x and perimeter of circle = 2πr Let 4x + 2pr = k k − 2 πr ⇒ 4x = k – 2pr ⇒ x = ...(i) 4 Let A = Area of square + Area of circle ⇒ A = x2 + πr2 2

 k − 2 πr  + πr 2 ⇒ A=  4  Differentiating w.r.t. r, we get dA  k − 2 πr   2 π  = 2 + 2 πr −  4   4  dr dA − k π + 2 π2r + 8 πr ⇒ = 4 dr

2 πr + 8r − 2 πr = 2r 4 Hence, side of square is equal to diameter of circle when their combined area is least. ...(i) 30. Given parabola is y = x2 and the given line is y = x + 2 ...(ii) The line and the parabola meet where Y x2 = x + 2 (Eliminating y) 2 ⇒ x –x–2=0 2 x+ ⇒ (x – 2) (x + 1) = 0 y= ⇒ x = 2 or –1 (2,4) When x = 2, y = 2 + 2 = 4 y = x2 ( 1,1) and when x = –1, X O y = 2 + (–1) = 1 \ The line and the parabola meet at the points (–1, 1) and (2, 4). \ Required area

From (i), x =

2

=

2

2  1  22 + 2 × 2 −  (−1) + 2(−1) − (23 − (−1)3 ) 2  2  3 1 9 = 6 − + 2 − 3 = sq. units. 2 2 31. Given, xpyq = (x + y)p + q Taking log on both sides, we get log (xp × yq) = log (x + y)p + q ⇒ p log x + q log y = (p + q) log (x + y) Differentiating w.r.t. x, we get  1  dy   p q dy + = ( p + q)  1 +  x y dx  (x + y )  dx   p + q dy p q dy p + q ⇒ + = + x y dx x + y x + y dx

=

⇒

[From (i)]

dA For finding maxima or minima, =0 dr ⇒ –kπ + 2π2r + 8πr = 0 ⇒ –k + 2πr + 8r = 0 ⇒ k = 2πr + 8r ⇒ k = r(2π + 8) k ⇒ r= 2π + 8 2 2 Now, d A = 2 π + 8 π > 0, so, local minima 4 dr 2

2

2

 x3    x2 2 ∫ (x + 2) dx − ∫ x dx =  2 + 2x  −  3  −1 −1 −1 −1

dy  q ( p + q)   p + q p  − = − dx  y x + y   x + y x 

dy  qx + qy − py − qy   px + qx − px − py   =   dx  y(x + y ) x(x + y ) dy y ⇒ = dx x OR

⇒

(

)

2

Given, y =  log x + x 2 + 1  Differentiating w.r.t. x, we get dy 1 2x   ⋅ 1+ = 2 log ( x + x 2 + 1 ) ⋅   2 dx 2 x2 + 1  x + x +1  ⇒

(

2 dy 2 log x + x + 1 = dx x2 + 1

) Class 12

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dy = 2 log x + x 2 + 1 dx Squaring both sides, we get ⇒

(

x2 + 1

2

(

−2 x

)

∫ x 2 +1dx

)

 dy  (x +1)   = 4  log x + x 2 + 1   dx  2

2

2

 dy  2 ⇒ (x + 1)   = 4 y  dx  Differentiating w.r.t. x, we get 2 dy  dy   dy  d 2 y (x 2 + 1) 2   2 + 2 x   = 4  dx   dx  dx dx

⇒ (x 2 + 1)

d2 y dx

2

+ x⋅

dy =2 dx

3 2

32. Let I = ∫ | x cos πx | dx 0

 x cos π x ; 0 < x < 1  2 ∵ | x cos πx | =  1 3 − x cos π x ; 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

14

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

1(1)

–

1(3)

–

2(4)

2.

Inverse Trigonometric Functions

2(2)

1(2)

–

–

3(4)

3.

Matrices

2(2)#

1(2)

–

–

3(4)

4.

Determinants

1(1)

–

–

1(5)*

2(6)

5.

Continuity and Differentiability

–

1(2)*

2(6)

–

3(8)

6.

Application of Derivatives

1(4)

1(2)*

1(3)

–

3(9)

7.

Integrals

2(2)#

1(2)

1(3)*

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)

1(2)

1(3)*

–

3(6)

Vector Algebra

3(3)#

1(2)*

–

–

4(5)

11.

Three Dimensional Geometry

4(4)#

–

–

1(5)*

5(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

1(4)

2(4)

–

–

3(8)

18(24)

10(20)

7(21)

3(15)

38(80)

10.

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-14

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. If a matrix A is both symmetric and skew-symmetric, then show that A is a zero matrix. OR 1 2  3 1  7 11 If   =  , then find the value of k. 3 4  2 5 k 23 2 2. Evaluate : ∫ cos 2 x + 22 sin x dx cos x

3. If the direction ratios of a line are 1, –3, 2, then find its direction cosines. OR The coordinates of a point P are (3, 12, 4) w.r.t. origin O, then find the direction cosines of OP. 4. Let R be a relation defined on the set of natural numbers N as follow : R = {(x, y) | x ∈ N, y ∈ N and 2x + y = 24} Find the domain and range of the relation R. 5. Find the distance from the origin to the plane x + 3y – 2z + 1 = 0. Class 12

188

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OR Find the foot of the perpendicular from (0, 0, 0) to 3x + 4y – 6z = 0. 6. Construct a matrix A = [aij]2×2, where aij = i + j.  ^ ^ ^  ^ ^  ^ 7. If p = i − 2 j + k , q = i + 4 j − 2 k are the position vectors of points P, Q respectively and point R (r ) divides the line PQ internally in the ratio 2 : 1, then find the coordinates of R. OR   ^ ^    ^ ^ ^ ^  If a = i + 3 j, b = 2 i + 5 j, c = 4 i + 2 j and c = t1 a + t 2 b, then find the value of t1 and t2. 8. Find the principal value of tan −1 (− 3 ). π/ 4

9. Evaluate :

∫

tan x dx

0

OR

Evaluate : ∫ x cot

−1

x dx

^ ^

^

^ ^

^

^ ^

^

10. Find the value of i ⋅ ( j × k ) + j⋅ (i × k ) + k⋅ (i × j). 11. If (2, 4, –3) is the foot of the perpendicular drawn from the origin to a plane, then find the equation of the plane. 12. Find the domain of f(x) = sin–1 x + tan–1 x + sec–1x. 13. Find the equation of the line in symmetric form which passes through the points A(–2, –1, 5) and B(1, 3, –1).  3− x 2 2  4−x 1  is singular. 14. Find the value of x for which the matrix A =  2   −4 −1 − x   −2 15. The cartesian equations of a line are 6x – 2 = 3y + 3 = 2z – 4. Find the direction ratios of the line. dy y 16. Find the integrating factor of the differential equation + = x. 2 dx 1− x

Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. In a family there are four children. All of them have to work in fields to earn their livelihood at the age of 15. Based on the above information, answer the following questions : (i) Probability that all children working in fields are boys if it is given that elder child working in fields is a boy, is (a) 3/8 (b) 1/8 (c) 5/8 (d) none of these (ii) Probability that all children working in fields are grass, if first two children working in fields are girls, is (a) 1/4 (b) 3/4 (c) 1/2 (d) none of these (iii) Find the probability that two middle child working in fields are boys if it is given that first child working in fields is a girl. (a) 0 (b) 3/4 (c) 1/4 (d) none of these (iv) Find the probability that all children working in fields are girls if it is given that at least one of the children working in fields is a girl. (a) 0 (b) 1/15 (c) 2/15 (d) 4/15 (v) Find the probability that all children working in fields are boys if it is given that at least three of the children working in fields are boys. (a) 1/5 (b) 2/5 (c) 3/5 (d) 4/5 189

Mathematics

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18. In a street two lamp posts are 300 feet apart. The light intensity at a distance d from the first (stronger) lamp post is 1000 , the light intensity at distance d from the

d2 second (weaker) lamp post is 125 (in both cases the light d2 intensity is inversely proportional to the square of the distance to the light source). The combined light intensity is the sum of the two light intensities coming from both lamp posts. Based on the above information, answer the following. (i) If you are in between the lamp posts, at distance x feet from the stronger light, then the formula for the combined light intensity coming from both lamp posts as function of x, is 1000 125 1000 125 1000 125 + 2 + + (a) (c) (d) None of these (b) 2 2 x (300 − x ) x (300 − x )2 x2 x2 (ii) The maximum value of x can not be (a) 100 (b) 200 (c) 300 (d) None of these (iii) The minimum value of x can not be (a) 0 (b) 100 (c) 200 (d) None of these (iv) If I(x) denote the combined light intensity, then I(x) will be minimum when x = (a) 100 (b) 200 (c) 300 (d) 150 (v) The darkest spot between the two lights is (a) at a distance of 100 feet from the weaker lamp post. (b) at distance of 100 feet from the stronger lamp post. (c) at a distance of 200 feet from the weaker lamp post. (d) None of these

PART – B Section - III 4

19. Evaluate

x

∫ x 2 + 1 dx by using substitution method. 2

20. If

21. 22. 23.

24. 25.

y2

=

ax2

+ bx + c, then find the value of

d 3 . dx y y 2 OR

(

)

Differentiate ex log(sin 2x) w.r.t. x. A bag contains 6 red, 5 blue and 7 white balls. If three balls are drawn one by one (without replacement), then what is the probability that all three balls are blue? π Determine the area enclosed between the curve y = cos2 x, 0 ≤ x ≤ and the axes. 2   Find a unit vector perpendicular to the plane of a = 2ɵi − 6ɵj − 3kɵ and b = 4ɵi + 3ɵj − kɵ . OR       Find the angle between the vectors a + b and a − b , where a = ɵi + ɵj + 4kɵ and b = ɵi − ɵj + 4kɵ . dy Solve the differential equation +1 = e x+ y . dx An unbiased dice is thrown twice. Let the event A be ‘odd number on the first throw’ and B be the event ‘odd number on the second throw’. Check the independence of the events A and B.

  −3   26. Evaluate : sin 2 cos −1     5   Class 12

190

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0 −1 2 2 27. If A =   , then show that A + I ≠ A(A – I). 1 0   28. Find the values of x if f (x) = 6(x2 – 5x – 24) is an increasing function. OR A rod 108 metres long is bent to form a rectangle. Find its dimensions, if its area is maximum. Section - IV 29. If the tangent at P(1, 1) on

y2

= x(2 –

x)2

meets the curve again at Q, then find the point Q.

30. Show that the function f : R → R given by f(x) = x3 + x is bijective. xdy 31. If + 2 y = ln x, then find the value of e2 y(e) – y(1). dx OR dy Solve the initial value problem 2 xy + y 2 − 2 x 2 = 0, y (1) = 2 . dx  1 + kx − 1 − kx , if − 1 ≤ x < 0  x 32. Find the value of k, for which f (x ) =  is continuous at x = 0. 2x + 1  , if 0 ≤ x < 1  x −1

33. Find the area bounded by the circle x2 + y2 = 8x and the line x = 2. 34. Evaluate :

x2 + 9

∫ x 4 − 2x 2 + 81 dx

OR

π /2

Evaluate :

∫

x 2 sin 2 x dx

0

35. Find the second order derivative of a sin3t with respect to a cos3t at t = π / 4 . Section - V 36. Solve the following problem graphically : Maximize Z = 22x + 18y Subject to constraints : x + y ≤ 20 36x + 24y ≤ 576 x, y ≥ 0 OR Find the maximum value of z = 3x + 5y subject to x + 4y ≤ 24, 3x + y ≤ 21, x + y ≤ 9, x ≥ 0, y ≥ 0.   37. Find the equation of the plane containing the lines r = ɵi + ɵj + λ (ɵi + 2ɵj − kɵ ) and r = ɵi + ɵj + µ (−ɵi + ɵj − 2 kɵ ). Find the distance of this plane from origin and also from the point (1, 1, 1). OR  Find the length of the perpendicular drawn from the point (2, 4, –1) to the line r = ɵi + λ(2ɵi + ɵj + 2kɵ ) .  1 −1 1   38. If A = 2 1 −3 , then find A–1 and hence solve the system of linear equations x + 2y + z = 4, –x + y +  1 1 1   z = 0, x – 3y + z = 2. OR Solve the following system of equations : 3x – y + z = 5 2x – 2y + 3z = 7 x + y – z = –1 191

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SOLUTIONS 1. Given, A is a symmetric matrix. \ AT = A A is also a skew-symmetric matrix. \ AT = –A From (i) and (ii), A = –A ⇒ A = O Hence, A is a zero matrix.

…(i)

D.c.’s of normal are

…(ii)

Here, d =

1⋅ 3 + 2 ⋅ 2 1⋅1 + 2 ⋅ 5   7 11 ⇒  =  3 ⋅ 3 + 4 ⋅ 2 3 ⋅1 + 4 ⋅ 5 k 23

=∫ =∫

cos2 x 2 2 cos x − 1 + 2 sin2 x cos2 x 2(cos2 x + sin2 x ) − 1 cos2 x

dx

dx dx = ∫ sec2 x dx = tan x + c

(1)2 + (−3)2 + (2)2 = 14 1 −3 2 \ Direction cosines are , , . 14 14 14

 3 12 4   −3 −12 −4  , . \ Direction cosines are  , ,  or  , 13 13 13 13 13 13  4. Here, R = {(x, y) | x ∈ N, y ∈ N and 2x + y = 24} \ R = {(1, 22), (2, 20), (3, 18), (4, 16), (5, 14), (6, 12), (7, 10), (8, 8), (9, 6), (10, 4), (11, 2)} Domain of R = {1, 2, 3, 4, ..., 11} Range of R = {2, 4, 6, 8, 10, 12 ,..., 22} 5. Required distance, 2

2

=0 61 \ Foot of perpendicular is (ld, md, nd) i.e., (0, 0, 0)

OR    Here, c = t1 a + t 2 b ⇒ 4ɵi + 2 ɵj = t1 ɵi + 3ɵj + t 2 2ɵi + 5ɵj ⇒ t1 + 2t2 = 4 and 3t1 + 5t2 = 2 Solving (i) and (ii), we get t1 = – 16, t2 = 10

2

(1) + (3) + (−2)

=

1 14

OR Given plane is 3x + 4y – 6z = 0 The d.r.’s of normal to plane (i) are 3, 4, –6.

)

(

)

( )

32 + 122 + 4 2 = 13

0 + 0 − 0 +1

61

...(i) ...(ii)

−1 8. Let tan − 3 = α ⇒ tan α = − 3 = − tan

OR Direction ratios of OP are (3 – 0, 12 – 0, 4 – 0) or (3, 12, 4).

d=

61

−6

0

(

3. We have,

Also,

61

,

   2q + p 2ɵi + 8 ɵj − 4kɵ + ɵi − 2 ɵj + kɵ 7. Here r = = 2 +1 3 ɵ ɵ ɵ 3i + 6 j − 3k ɵ ɵ ɵ = = i +2j −k 3 ⇒ R ≡ (1, 2, – 1)

 7 11  7 11 ⇒  =  ⇒ k = 17 17 23 k 23 cos 2 x + 2 sin2 x

4

,

6. Here, a11 = 1 + 1 = 2, a12 = 1 + 2 = 3, a21 = 2 + 1 = 3 and a22 = 2 + 2 = 4 2 3  Hence, A =  . 3 4 

OR 1 2  3 1  7 11 Given,   =  3 4  2 5 k 23

2. Let I = ∫

3

...(i)

π 3

 π −π  −π π  = tan  −  ⇒ α = , ∈  3 3  2 2   −π  \ Principal value of tan −1 − 3 is   .  3 

( )

π/ 4

9. We have

∫

tan x dx = [log | sec x |]0π/ 4

0

= log sec

π 1 − log | sec 0 | = log | 2 | − log | 1 | = log 2 4 2 OR

Let I = ∫ x cot

−1

x dx

=

x2 x2 / 2 cot −1 x + ∫ dx 2 1+ x2

=

1 1 x2 1 x2 +1 cot −1 x + ∫ 2 dx − ∫ 2 dx 2 x +1 2 2 x +1

=

x2 x cot −1 x cot −1 x + + +C 2 2 2 Class 12

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10. ɵi ⋅ ɵi + ɵj − ɵj + kɵ ⋅ kɵ = 1 − 1 + 1 = 1

( )

11. Foot of perpendicular from (0, 0, 0) to the plane is ^ ^ ^  (2, 4 –3), then a = 2 i + 4 j − 3 k ^ ^ ^  Normal to plane is, n = 2 i + 4 j − 3 k     \ Equation of plane is, r ⋅ n = a ⋅ n ^

^

^

^

^

^

⇒ (x i + y j + z k ) ⋅ (2 i + 4 j − 3 k ) = 4 + 16 + 9 = 29 ⇒ 2x + 4y – 3z – 29 = 0 12. Given f(x) = sin–1x + tan–1x + sec–1x Domain of sin–1x = [–1, 1] Domain of tan–1x = (–∞, ∞) Domain of sec–1x = (–∞, ∞) – (–1, 1) Domain of f(x) = [–1, 1] ∩ (–∞, ∞) ∩ [(–∞, ∞) – (–1, 1)] = {–1, 1} 13. The symmetric form of the equation of line passing through A(–2, –1, 5) and B(1, 3, –1) is x − (−2) y − (−1) z − 5 x + 2 y +1 z − 5 = = ⇒ = = 1 − (−2) 3 − (−1) −1 − 5 3 4 −6 14. A is singular \ |A| = 0 3− x 2 2 ⇒ 2 4−x 1 =0 −2 −4 −1 − x ⇒ (3 – x) (–4 – 4x + x + x2 + 4) – 2(–2 – 2x + 2) + 2(–8 + 8 – 2x) = 0 2 ⇒ (3 – x) (x – 3x) + 4x – 4x = 0 ⇒ (x – 3)2 x = 0 ⇒ x = 0, 3 15. The equations of the given line are 6x – 2 = 3y + 3 = 2z – 4 1 x− 3 = y +1 = z − 2 ⇒ 1 2 3 Clearly, the direction ratios of the given line are 1, 2, 3. dy y 16. The given equation is + = x, dx 1 − x2 1 where P = and Q = x 1 − x2 ∴ I.F. = e

∫ Pdx

=e

∫

dx 1− x2

= esin

−1

x

17. Let B and G denote the boy and girl respectively. If a family has 4 children each of four children can either boy or girl. Sample space is given by S = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG, GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG}

(i) (b) : Let A = All children are boys. \ A = {BBBB} i.e., n(A) = 1 B = Elder child is a boy \ B = {BBBB, BBBG, BBGB, BGBB, BBGG, BGBG, BGGB, BGGG} i.e., n(B) = 8 Now, n(A∩ B) = 1 ( A ∩ B) 1 ∴ P ( A / B) = = n(B) 8 (ii) (a) : Let A = All are girls. \ A = {GGGG} i.e., n(A) = 1 B = First two children are girl \ B = {GGBB, GGBG, GGGB, GGGG} i.e., n(B) = 4 Now, n(A∩ B) = 1 n( A ∩ B) 1 ∴ P ( A / B) = = n(B) 4 (iii) (c) : Let A = Two middle child are boys. \ A = {BBBB, BBBG, GBBB, GBBG} i.e., n(A) = 4 B = First child is a girl = 8 \ B = {GBBB, GBBG, GBGB, GBGG, GGBB, GGBG, GGGB, GGGG} i.e., n(B) = 8 Now, n(A∩ B) = 2 2 1 ∴ P ( A / B) = = 8 4 (iv) (b) : Let A = All are girls. \ A = {GGGG} i.e. n(A) = 1 B = At least one child is girl. \ B = {BBBG, BBGB, BGBB, GBBB, BBGG, GGBB, GBGB, BGBG, BGGB, GBBG, GGGB, GGBG, GBGG, BGGG, GGGG} i.e., n(B) = 15 Now, n(A ∩ B) = 1 1 ∴ P ( A / B) = 15 (v) (a) : Let A = All are boys. \ A = {B B B B} i.e. n(A) = 1 B = At least three of the children are boys. \ B = {BBBB, BBBG, BBGB, BGBB, GBBB} i.e., n(B) = 5 Now, n(A ∩ B) = 1 1 ∴ P ( A / B) = 5 18. (i) (c) : Since, the distance is x feet from the stronger light, therefore the distance from the weaker light will be 300 – x. So, the combined light intensity from both lamp posts 1000 125 . is given by 2 + (300 − x )2 x 193

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(ii) (c) : Since, the person is in between the lamp posts, therefore x will lie in the interval (0, 300). So, maximum value of x can't be 300. (iii) (a) : Since, 0 < x < 300, therefore minimum value of x can't be 0. 1000 125 (iv) (b) : We have, I (x ) = 2 + x (300 − x )2 −2000 250 and ⇒ I ′( x ) = + 3 x (300 − x )3 2000 750 ⇒ I ′′(x ) = 4 + x (300 − x )4 For maxima/minima, I′(x) = 0 2000 250 ⇒ = ⇒ 8(300 − x )3 = x 3 3 3 x (300 − x ) Taking cube root on both sides, we get 2(300 – x) = x ⇒ 600 = 3x ⇒ x = 200 Thus, I(x) is minimum when you are at 200 feet from the strong intensity lamp post. (v) (a) : Since, I(x) is minimum when x = 200 feet, therefore the darkest spot between the two light is at a distance of 200 feet from stronger lamp post, i.e., at a distance of 300 – 200 = 100 feet from the weaker lamp post. 4

x

∫ x 2 + 1 dx

19. Let I =

2

Put x2 + 1 = t ⇒ 2x dx = dt When x = 2, t = 5 and when x = 4, t = 17 \ I=

17

1  17  1 1 1 17 dt = [ log t ]5 = log   ∫ 2 5 25t 2

20. Given, y2 = ax2 + bx + c Differentiating both sides, we get 2yy1 = 2ax + b …(i) Again differentiating, we get 2yy2 + y1(2y1) = 2a 2

 2ax + b  2 (Using (i)) ⇒ yy 2 = a − y1 ⇒ yy 2 = a −   2 y  =

4 y 2a − (4a 2x 2 + b 2 + 4abx)

3 ⇒ y y2 =

⇒

4y2 4a(ax 2 + bx + c) − (4a 2 x 2 + b2 + 4 abx) 4ac − b2 = 4 4

d 3 ( y y 2) = 0 dx

OR Let y = ex log (sin2x) dy d x ∴ = e log(sin 2 x ) dx dx d d = e x ⋅ {log(sin 2 x )} + log(sin 2 x ) ⋅ (e x ) dx dx

{

}

{

}

1 x ⋅ cos 2 x ⋅ 2 + log(sin 2 x ) ⋅ e sin 2 x x = 2e cot2x + ex log (sin 2x)

= ex ⋅

= ex{2 cot2x + log(sin 2x)} 21. Let A, B and C be the events of drawing a blue ball in first, second and third draw respectively. Then probability of getting blue ball in all three draws = P(A ∩ B ∩ C) = P(A)⋅P(B|A)⋅P(C|A ∩ B) 4 5 3 and P (C | A ∩ B) = Now, P ( A) = , P ( B | A) = 17 18 16 5 4 3 5 ∴ P(A ∩ B ∩ C) = × × = 18 17 16 408 π 22. Given curve is y = cos2x, 0 ≤ x ≤ 2 π /2

Required area =

∫

cos 2 x dx

Y

0 π /2

=

∫ 0

 1 + cos 2 x    dx 2

y = cos2x X′

O

π /2

 x sin 2 x  = + 2 4  0

π 2

X

Y′

 π  π  =   − 0  − (0) = sq. units    4  4 ɵi ɵj kɵ   23. We have, a × b = 2 −6 −3 4 3 −1   ⇒ a × b = ɵi (6 + 9) − ɵj (−2 + 12) + kɵ (6 + 24)   ⇒ a × b = 15ɵi − 10ɵj + 30kɵ   and a × b = 152 + (−10)2 + (30)2 = 35   a ×b 3ɵi − 2ɵj + 6kɵ \ Required vector =   = 7 |a ×b | OR   ɵ ɵ ɵ Given that, a = i + j + 4k and b = ɵi − ɵj + 4kɵ     ∴ a + b = 2ɵi + 8kɵ and a − b = 2ɵj     Let q be the angle between a + b and a − b , then     (a + b ) ⋅ (a − b ) cosθ =     a +b a −b (2ɵi + 0ɵj + 8kɵ ) ⋅ (0ɵi + 2ɵj + 0kɵ ) 0+0+0 = =0 2 2 2 2 2 2 = 2 +0 +8 0 +2 +0 4 + 64 4 π ⇒ cosq = 0 ⇒ θ = 2 Class 12

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dy + 1 = ex + y dx dy du = Let x + y = u ⇒ 1 + dx dx du = eu ⇒ ∫ e − udu = ∫ dx ⇒ –e–u = x + c1 ⇒ dx ⇒ e–(x + y) + x = c, where –c1 = c 25. If all the 36 elementary events of the experiment are considered to be equally likely, we have 18 1 18 1 P ( A) = = and P (B) = = 36 2 36 2 Also, P(A ∩ B) = P(odd number on both throws) 9 1 = = 36 4 1 1 1 Now, P ( A) P (B) = × = 2 2 4 Clearly, P(A ∩ B) = P(A) × P(B) Thus, A and B are independent events. 24. We have

 −3  26. Let cos −1   = θ , where q ∈ [0, π]  5  −3 Then, cos θ = 5 Since q ∈ [0, π], we have sin q > 0 ∴ sin θ = 1 − cos 2 θ = 1 −

OR Let x be the length and y be the breadth of the rectangle \ 2x + 2y = 108 ⇒ y = 54 – x Now, area of rectangle = xy =x(54 – x) Let f (x) = 54x – x2 \ f ′(x) = 54 – 2x and f ′′(x) = – 2 For extreme values, f ′(x) = 0 ⇒ 54 – 2x = 0 ⇒ x = 27 f ′′(27) = –2 < 0 ...(i)

dy dy 3x 2 − 8x + 4 = 3x 2 − 8x + 4 ⇒ = 2y dx dx

3−8+ 4 1  dy  ⇒   = =− 2 2  dx  (1,1) \ Equation of tangent at P is 1 y − 1 = − (x − 1) ⇒ x + 2y – 3 = 0 2 3−x in (i), we get Using y = 2

...(ii)

2

 3− x 3 2  2  = x − 4x + 4x

0 −1 0 −1 1 0 L.H.S. = A2 + I =   +  1 0  1 0  0 1  ...(i)

R.H.S. = A( A2 − I ) 0 −1  0 −1 0 −1 1 0  =  −    1 0   1 0  1 0  0 1  0 −1   −1 0  1 0  0 −1  −2 0  =    =  −   1 0    0 −1 0 1  1 0   0 −2 ...(ii)

From (i) and (ii), we get R.H.S. ≠ L.H.S.

5 11 5 11 ⇒ x− > or x − < − 2 2 2 2 ⇒ x > 8 or x < – 3 \ f is increasing, if x < – 3 or x > 8.

⇒ 2y

  −3   ∴ sin 2 cos −1    = sin 2θ = 2 sin θ cos θ  5    4  −3   −24 = 2 × ×    =  5  5   25 0 −1 27. Here, A =   1 0 

 0 2 =   −2 0

2

5  121  25 25 ⇒ x −  > ⇒ x − 5x + > 24 +   2 4 4 4 2

\ Area is maximum when x = 27, y = 27 29. Here, y2 = x(2 – x)2 ⇒ y2 = x3 – 4x2 + 4x

9 16 4 = = 25 25 5

 −1 0  1 0 0 0 = + =   0 −1 0 1 0 0

28. f (x) is increasing if f ' (x) > 0 ⇒ 6 (x2 – 5x – 24) > 0 ⇒ x2 – 5x – 24 > 0 ⇒ x2 – 5x > 24

⇒ 9 + x2 – 6x = 4x3 – 16x2 + 16x ⇒ 4x3 – 17x2 + 22x – 9 = 0 which has two roots 1, 1 (Because of (ii) being tangent at (1, 1)) 17 17 9 Sum of 3 roots = ⇒ 3rd root = − 2 = 4 4 4 9 3− 4=3 Then, y = 2 8  9 3 So, Q is  ,  . 4 8 30. Here, f : R → R, f(x) = x3 + x One-One : x1, x2 ∈ R such that, f(x1) = f(x2) ⇒ x13 + x1 = x23 + x2 ⇒ x13 − x23 + x1 − x2 = 0

⇒ (x1 − x2 )(x12 + x1 x2 + x22 + 1) = 0 195

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⇒ x1 – x2 = 0 {∵ x12 + x1x2 + x22 ≥ 0 ∀ x1 , x2 ∈R} ⇒ x1 = x2 Thus, f(x1) = f(x2) ⇒ x1 = x2 So, f is one-one. Onto : Let y be any arbitrary element of R. Then, f(x) = y ⇒ x3 + x = y ⇒ y = x3 + x ⇒ x3 + x – y = 0 We know that an odd degree equation has at least one real root. Therefore, for every real value of y, the equation x3 + x – y = 0 has a real root a such that a3 + a – y = 0 ⇒ a3 + a = y ⇒ f(a) = y. Thus, for every y ∈ R, there exists a ∈ R such that f(a) = y. So, f is onto. Hence, f : R → R is a bijective function. dy 2 ln x 31. We have + y = . dx x x It is a linear differential equation. 2

I.F. = e

∫ x dx

= x2 ln x dx = ∫ x log x dx x

2

2h + 1 = −1 h→0 h − 1

lim f (x ) = lim f (0 + h) = lim x →0

h→0

+

lim f ( x ) = lim f (0 − h) = lim h→0

x → 0−

= lim h→0

h→0

1 + kh − 1 − kh h

1 + kh − 1 − kh 1 + kh + 1 − kh × h 1 + kh + 1 − kh (1 + kh) − (1 − kh) 1 + kh + 1 − kh]

2k

=

1 + kh + 1 − kh

2k =k 2

\ From (i), we get k = –1 33. Given curve is x2 + y2 = 8x

dy =0 dx

2

Required area = 2 ∫ 8 x − x 2 dx

2

Putting y = vx ⇒

2 × 0 +1 = −1 0 −1

e2 + 1 . 4 OR

We have 2 xy + y 2 − 2 x 2 dy 2 xy + y = dx 2x 2

x → 0−

Now, f (0) =

h→0 h [

2

So, e2y(e) – y(1) =

⇒

x → 0+

= lim

x2 x2 1 dx ln x − ∫ 2 2 x

...(i)

lim f ( x ) = f (0) = lim f ( x )

∴

h→0

x x ln x − + c 2 4 2 e e2 e2 2 \ x = e ⇒ e y(e) = − +c = +c 2 4 4 1 and x = 1 ⇒ y(1) = − + c 4 ⇒ yx2 =

32. Since, f(x) is continuous at x = 0.

= lim

\ Solution is, yx2 = ∫ x 2 . ⇒ yx2 =

Putting c = –1 in (i), we get 2x 2x − = log x − 1 ⇒ y = y 1 − log x Clearly, y is defined if x ≠ 0, and 1 – log|x| ≠ 0. Now, 1 – log |x| = 0 ⇒ log|x| = 1 ⇒ |x| = e ⇒ x = ± e. 2x Hence y = , where x ≠ 0, ± e gives the 1 − log x solution of the given differential equation.

2

= 2 ∫ (4)2 − (x − 4)2 dx

dy dv =v+x dx dx 2

0

0 2

 x − 4  2 16  x − 4  4 − (x − 4)2 + sin −1  = 2    4   0 2  2 

2

dv 2v + v dv 2v + v \ v+x = ⇒ x = −v dx 2 dx 2 1 1 ⇒ 2 ∫ 2 dv = ∫ dx x v 2x 2 = log x + c ⇒ − = log x + c ⇒ − y v It is given that y(1) = 2 i.e., y = 2 when x = 1 \ (i) becomes –1 = 0 + c ⇒ c = –1

...(i)

   −2   −1  = 2    16 − 4 + 8 sin −1   − 8 sin −1 (−1)     2   2 π 8π  16 π    = 2  − 12 − 8 × +  = 2  − 12 +  6 2   6    16 π = − 2 12  sq . units   3 Class 12

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34. Let I = ∫

x2 + 9 4

x − 2x + 81

1+ ⇒ I=∫

⇒ I=∫

Put x −

2

dx

9 x

1+

2

x2 − 2 +

1+ 2

81

dx = ∫

x2 9

x

9

x2 dx 2 81 x + 2 −2 x

d2y

2

9  x − x  + 18 − 2

dx 2

dx

9 9  = t ⇒  1 + 2  dx = dt x  x 

\ I= ∫

1 t = tan −1   + C 4 4 t + 16 dt

2

9  2 x− 1 −1   + C = 1 tan −1  x − 9  + C x = tan   4x  4 4  4 OR π 2

= − sec 2 t

dt − sec 2 t 1  sec 4 t  = = dx 3a cos 2 t(− sin t ) 3a  sin t   

 2  1 4 4 2 \  d y = ⋅ = 2 3 1 a 3a  dx  t = π 4 2 36. To solve this LPP graphically, we first convert the inequations into equations l1 : x + y = 20, l2 : 36x + 24y = 576, l3 : x = 0, l4 : y = 0. Draw the corresponding lines. The feasible region of the LPP is shaded in figure. The corner points of the feasible region OA2PB1 are O(0, 0), A2(16, 0), P(8, 12) and B1(0, 20).

π 2

∫x

 1 − cos 2 x  sin 2 x dx = ∫ x 2 ⋅   dx  2

2

0

π 2

=

dx dy = 3a cos 2 t(− sin t ) = 3a sin 2 t cos t and dt dt  dy  dy  dt  3a sin 2 t cos t \ = = = − tan t dx  dx  3a cos 2 t(− sin t )  dt  Again differentiating w.r.t. x, we get

0 π 2

1 1 x 2dx − ∫ x 2 cos 2 x dx ∫ 20 20

π π   π 2 32    sin 2 x  2 1 x 1  sin 2 x  dx =   −  x 2 ⋅ − ∫ 2x ⋅    2  3 0 2  2 0 0 2  π

 12 1  π3  1  π2 =   −  × 0 − 0 + ∫ x sin 2 x dx  20 2  24  2  4 π 2

  π  − cos 2 x   π 3 1   (− cos 2 x )  2 = + x⋅  dx  − ∫ 1 ⋅   48 2   2 2 0  0 π 2

π3 1  − π  1 = +  cos π − 0 + ∫ cos 2 x dx  40 48 2  4 π /2

3

=

π π 1  sin 2 x  − (−1) +  48 8 4  2  0 3

=

3

π π π π 1 + + [0] = + 48 8 48 8 8

35. Let y = asin3t and x = a cos3 t On differentiating w.r.t. t, we get

The values of the objective function Z at corner-points of the feasible region are given in the following table. Corner Points

Value of Z = 22x + 18y

O(0, 0)

0

A2(16, 0)

352

P(8, 12)

392 (maximum)

B1(0, 20)

360

\ Z has maximum value 392 at P(8, 12). OR Converting inequations into equations and drawing the corresponding lines. x + 4y = 24, 3x + y = 21, x + y = 9 x y x y x y i.e., + = 1, + = 1, + =1 24 6 7 21 9 9 197

Mathematics

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As x ≥ 0, y ≥ 0 solution lies in first quadrant

|(ɵi + ɵj + kɵ ) ⋅ (−ɵi + ɵj + kɵ )| 12 + 12 + 12

=

| −1 + 1 + 1 | 3

=

1 units 3

OR Let M be the foot of the perpendicular drawn from  P (2ɵi + 4 ɵj − kɵ ) on the line r = ɵi + λ(2ɵi + ɵj + 2kɵ ) Let the position vector of M be ɵi + λ(2ɵi + ɵj + 2kɵ )

The feasible region of the LPP is shaded in figure. The corner points of the feasible region OABCD are O(0, 0), A(7, 0), B(6, 3), C(4, 5) and D(0, 6). Corner Points O(0, 0)

Value of Z = 3x + 5y 0

A(7, 0)

21

B(6, 3)

33

C(4, 5)

37 (Maximum)

D(0, 6)

30

\ Z has maximum value 37 at C(4, 5). 37. We are given the equation of lines as  r = ɵi + ɵj + λ (ɵi + 2ɵj − kɵ ) and  r = ɵi + ɵj + µ (−ɵi + ɵj − 2kɵ ) We know that the equation of plane containing           r = a + λb1 and r = a + λb2 is (r − a) ⋅ (b1 × b2) = 0    Here, a = ɵi + ɵj, b = ɵi + 2ɵj − kɵ , b = − ɵi + ɵj − 2kɵ 1

2

ɵi ɵj kɵ   b1 × b2 = 1 2 −1 = (− 4 + 1)ɵi − (−2 − 1)ɵj + (1 + 2)kɵ −1 1 −2   ⇒ b1 × b2 = − 3ɵi + 3ɵj + 3kɵ Therefore, the equation of plane is,  (r − (ɵi + ɵj )) ⋅(−3ɵi + 3ɵj + 3kɵ ) = 0   ⇒ r ⋅ (−3ɵi + 3ɵj + 3kɵ ) + 3 − 3 = 0 ⇒ r ⋅ (−ɵi + ɵj + kɵ ) = 0 This is the equation of the required plane. We know that the distance of a point P with position   | a ⋅n − d |     vector a from the plane r ⋅ n = d is given by |n | Now, distance from origin is 0 and distance from the point (1, 1, 1) i.e., ɵi + ɵj + kɵ is

= (1 + 2 λ)ɵi + (λ)ɵj + (2 λ)kɵ ( M lies on the line)  Then PM = [(1 + 2 λ)ɵi + (λ)ɵj + (2 λ)kɵ ] − (2ɵi + 4 ɵj − kɵ )  ∴ PM = (−1 + 2 λ)ɵi + (−4 + λ)ɵj + (1 + 2 λ)kɵ  Since PM is perpendicular to the given line which is  parallel to b = (2ɵi + ɵj + 2kɵ )   ∴ PM ⋅ b = 0 i.e., [(−1 + 2 λ)ɵi + (−4 + λ)ɵj + (1 + 2 λ)kɵ ] ⋅ (2ɵi + ɵj + 2kɵ ) = 0 ⇒ 2(–1 + 2l) + 1 (–4 + l) + 2(1 + 2l) = 0 4 ⇒ 9l – 4 = 0 ⇒ λ = 9 Putting the value of l, we obtain the position vector of 8   17 4 M as  ɵi + ɵj + kɵ  9 9 9   Now, PM = −1 ɵi − 32 ɵj + 17 kɵ 9 9 9  1 1029 289 ∴ Required length PM = + + 81 81 81 =

1319 units 9

 1 −1 1   38. We have, A = 2 1 −3  1 1 1   1 −1

\ |A| = 2 1

1

1 −3 1 1

= 1(1 + 3) + 1 (2 + 3) + 1 (2 – 1) = 10 ≠ 0 So, A is invertible.  4 −5 1 ′  4 2 2     \ adj A =  2 0 −2 =  −5 0 5 2 5 3  1 −2 3  ⇒ A

−1

 4 2 2 1  1  (adj A) =  −5 0 5 = |A| 10  1 −2 3  

... (i)

Class 12

198

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Now, the given system of equations is expressible as  1 2 1  x   4        −1 1 1  y  =  0  1 −3 1  z   2      x 4     or A′X = B, where X =  y  and B =  0  z  2     Now, | A′ | = | A | = 10 ≠ 0. So, the given system of equations has a solution given by [ (A′)–1 = (A–1)′] X = (A′)–1 B = (A–1)′B x   4 2 2 ′  4  1       [Using (i)] ⇒ y =   10  −5 0 5  0  z   1 −2 3  2       

5 x  3 −1 1      where A =  2 −2 3  , X =  y  and B =  7   −1  z   1 1 −1 Now, |A| = 3(2 –3) – (–1) (–2 – 3) + 1 (2 + 2) = –4 ≠ 0 ⇒ A–1 exists and so the given system has a solution given by X = A–1B.  −1 0 −1 Now, adj A =  5 −4 −7     4 −4 −4  \ A

−1

 −1 0 −1 1 1 = adj A = −  5 −4 −7  4 |A|  4 −4 −4 

 −1 0 −1  5   1  1 \ X=A B=− 5 −4 −7   7  =  −1 4  4 −4 −4   −1  1  x  1   ⇒  y  =  −1 ⇒ x = 1, y = –1, z = 1.  z   1  −1

x   4 −5 6 10  42   1     ⇒  y  = 10  2 0 −2  04 = z  2 5 8 30  26   

 9 / 5    2 / 5  7 / 5   ⇒ x = 9 / 5, y = 2 / 5 and z = 7 / 5. OR The given system of equations can be written as AX = B

Hence, the solution of the given system of equations is x = 1, y = –1 and z = 1.



199

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Self Evaluation Sheet Once you complete SQP-14, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Matrices / Matrices Integrals Vector Algebra / Vector Algebra Relations and Functions Three Dimensional Geometry / Three Dimensional Geometry Matrices Vector Algebra / Vector Algebra Inverse Trigonometric Functions Integrals / Integrals Vector Algebra Three Dimensional Geometry Inverse Trigonometric Functions Three Dimensional Geometry Matrices Three Dimensional Geometry Differential Equations Probability Application of Derivatives

19

Integrals

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Continuity and Differentiability / Continuity and Differentiability Probability Application of Integrals Vector Algebra / Vector Algebra Differential Equations Probability Inverse Trigonometric Functions Matrices Application of Derivatives / Application of Derivatives Application of Derivatives Relations and Functions Differential Equations / Differential Equations Continuity and Differentiability Application of Integrals Integrals / Integrals Continuity and Differentiability Linear Programming / Linear Programming Three Dimensional Geometry / Three Dimensional Geometry Determinants / Determinants

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

15

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

–

1(3)

–

3(5)

2.

Inverse Trigonometric Functions

1(1)

1(2)

–

–

2(3)

3.

Matrices

2(2)

1(2)

–

–

3(4)

4.

Determinants

1(1)*

–

–

1(5)*

2(6)

–

1(2)

2(6)#

–

3(8)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)*

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)*

1(2)*

1(3)

–

3(6)

10.

Vector Algebra

3(3)#

1(2)*

–

–

4(5)

11.

Three Dimensional Geometry

4(4)

–

–

1(5)*

5(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

1(4)

2(4)

–

–

3(8)

18(24)

10(20)

7(21)

3(15)

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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38(80)

Subject Code : 041

SQP-15

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part -A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I  3 2 6 1. Write the cofactor of the element a31 in A =  5 0 7 .    3 8 5 OR If A is a square matrix of order 3 and |2A| = k|A|, then find the value of k.  1   2. Evaluate : tan−1 2 cos  2 sin−1    2    −2 x y  dx = 1(x ≠ 0). 3. Find the integrating factor of the differential equation  e  +  x x  dy OR 4

3

 3   2  Find order and degree of the equation  d y  +  d y  + dy + 4 y = sin x .  dx 3   dx 2  dx Class 12

202

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 2 3 4. If f (x) = x2 – 4x + 1, find f (A), where A =  .  1 2 ^ → ^ ^ 5. Find the unit vector in the direction of vector a = 2 i + 3 j + 4 k .

OR ^

Find the projection of the vector 7^i + ^j − 4k^ on 2^i + 6 ^j + 3k . 6. Check whether the function f(x) = x3 – 3x2 – x is one-one or not? 3

7. Evaluate : ∫ (x − 1)(x − 2)(x − 3)dx 1

OR π

Evaluate :

∫x

10

sin7 x dx

−π

8. Check whether the lines having direction ratios ( 3 − 1, − 3 − 1, 4) and (− 3 + 1, 3 + 1, − 4) are perpendicular to each other. 9. If the vectors 3ɵi + 2 ɵj − kɵ and 6ɵi − 4 x ɵj + ykɵ are parallel, then the values of x and y. OR Find the point which divides the line segment joining the points (–2, 3, 5) and (1, 2, 3) in the ratio 2 : 3 externally. 10. Find vector equation of the plane which is at a distance of the origin is 2ɵi − 3ɵj + 4kɵ .

6 from the origin and its normal vector from 29

11. Let A = {1, 2, 3, 4}. Show that f = {(1, 2), (2, 3), (3, 4), (4, 1)} is a bijection from A to A?     12. If a and b are unit vectors enclosing an angle q and a+b < 1, find the value of q. 13. Find the area of the region bounded between the line x = 2 and the parabola y2 = 8x. 14. Find equation of a line which passes through the point (1, 2, 3) and is parallel to the vector 3ɵi + 2 ɵj − 2kɵ . 6 2   2 2   15. If A =  −3 1 , B =  1 3 , such that A + B + C is a zero matrix, then find the matrix C. 0 4   4 0

16. If the line joining (2, 3, –1) and (3, 5, –3) is perpendicular to the line joining (1, 2, 3) and (3, 5, l), then find the value of l. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A teacher arranged a surprise game for students of a classroom having 5 students, namely Amit, Aruna, Eklavya, Yash and Samina. He took a bag containing tickets numbered 1 to 11 and told each student to draw two tickets without replacement. 203

Mathematics

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(i) Probability that both ticket drawn by Amit shows even number, is (a) 1/11 (b) 2/11 (c) 3/11

(d) 4/11

(ii) Probability that both tickets drawn by Aruna shows odd number, is (a) 1/11 (b) 2/11 (c) 3/11

(d) 4/11

(iii) When tickets are drawn by Eklavya, find the probability that number on one ticket is a multiple of 4 and on other ticket is a multiple 5. (a) 4/55 (b) 6/55 (c) 7/55 (d) None of these (iv) When tickets are drawn by Yash, find the probability that number on one ticket is a prime number and on other ticket is a multiple of 4 . (a) 3/11 (b) 5/11 (c) 6/11 (d) 2/11 (v) When tickets are drawn by Samina, find the probability that first ticket drawn shows an even number and second ticket drawn shows an odd number. (a) 2/11 (b) 3/11 (c) 5/11 (d) 8/11 18. An open water tank of aluminium sheet of negligible thickness, with a square base and vertical sides, is to be constructed in a farm for irrigation. It should hold 4000 l of water, that comes out from a tube well.

Based on above information, answer the following questions. (i) If the length, width and height of the open tank be x, x and y m respectively, then surface area of tank is given by (a) S = x2 + 2xy

(b) S = 2x2 + 4xy

(c) S = 2x2 + 2xy

(d) S = 2x2 + 8xy

(c) x2y2 = 4

(d) xy = 4

(ii) The relation between x and y is (a) x2y = 4

(b) xy2 = 4

(iii) The outer surface area of tank will be minimum when depth of tank is equal to (a) half of its width

(b) its width

1 (c)    4

th

of its width

(d)  1   3

rd

of its width

(iv) The cost of material will be least when width of tank is equal to (a) half of its depth

(b) twice of its depth

1 (c)    4

th

of its depth

(d) thrice of its depth Class 12

204

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(v) If cost of aluminium sheet is ` 360/m2, then the minimum cost for the construction of tank will be (a) ` 2320 (b) ` 3320 (c) ` 4320 (d) ` 5320

PART - B Section - III 19. Find the equations of the tangent and the normal to the curve y = x3 at the point P(1, 1).  cos x   − π 3π  , x ∈ , 20. Express tan −1  in the simplest form.  2   2  1 + sin x 

21. Find the area of region bounded by the curve y2 = x(1 – x)2, shown in following figure. Y X′

X O

Y

22. Suppose 5 men out of 100 and 25 women out of 1000 are good orator. If an orator is chosen at random, find the probability that a male person is selected. Assume that there are equal number of men and women. 23. Evaluate :

∫

1 1− sin x

dx OR

dx Evaluate : ∫ 1 − 2 sin x cos x 5a −b  T 24. If A =   and A adj A = AA , then find the value of 5a + b . 3 2   25. Find the projection of the vector 2ɵi − 3ɵj − 6kɵ on vector joining the points (5, 6, – 3) and (3, 4, – 2). OR     ^ ^ ^ ^ ^ b a | × 2 | . If a = 4 i + 3 j + 2 k and b = 3 i + 2 k , find 26. Suppose that two cards are drawn at random from a deck of 52 cards. Let X be the number of aces obtained. Then, find the probability distribution of X. d2 y

dy = 0. dx dx dy y (1 + x ) = 28. Find the solution of the differential equation . dx x ( y − 1) 27. If y = sin–1x, then show that (1 − x 2 )

2

−x

OR  dy  Find the particular solution of the differential equation log   = 3x + 4 y ; y = 0, x = 0 .  dx  Section - IV 29. Show that the curve for which the normal at every point passes through a fixed point is a circle. 30. Find the point on the parabola y2 = 2x which is closed to the point (1, 4). ax 2 bx c = + + + 1, then find dy . y 31. If (x − a)(x − b)(x − c) (x − b)(x − c) x − c dx 205

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OR Differentiate

2 2 x + x +1 ⋅ 4 x − 1

(x 2 − 1)3/2

w.r.t. x.

32. Let f : A → B be a function defined as f (x ) = f one-one and onto?

2x + 3 , where A = R − {3} and B = R − {2}. Is the function x −3

33. Find the area of the region bounded by the curve y = x2 + x, the x-axis and the lines x = 2, x = 5. 1

−1 34. Evaluate : ∫ tan x dx 0

OR

2 dx Evaluate : ∫1 x(1 + x 2 )

35. If f(x) is continuous at x = 0, where  sin x  x + cos x , for x > 0 f(x) =  , then find f(0).  4(1 − 1 − x ) , for x < 0  x Section-V 36. Find the vector and cartesian equation of the line through the point ɵi + ɵj − 3kɵ and perpendicular to the lines   r = 2ɵi − 3ɵj + λ(2ɵi + ɵj − 3kɵ ) and r = 3ɵi − 5ɵj + µ(ɵi + ɵj + kɵ ) . OR The four points A(3, 2, –5), B(–1, 4, –3), C(–3, 8, –5) and D(–3, 2, 1) are coplanar. Find the equation of the plane containing them. 37. Find the minimum value of Z = 3x + 4y + 270 subject to the constraints x + y ≤ 60 x + y ≥ 30 x ≤ 40, y ≤ 40 x ≥ 0, y ≥ 0 OR Find the point for which the maximum value of Z = x + y subject to the constraints 2x + 5y ≤ 100, y x + ≤ 1, x ≥ 0, y ≥ 0 is obtained. 25 50 2 3 7  38. If A =  3 −2 −1 , find A–1. Using A–1 solve the following system of equations :    1 1 2  2x + 3y + 7z = 12 3x – 2y – z = 0 x + y + 2z = 4  1 −2  If A =  −2 3  1 1

1 1 5

  –1  , find (adj A ). 

OR

Class 12

206

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SOLUTIONS  3 2 6 1. We have, A =  5 0 7    3 8 5

^

Therefore, a^ =

Cofactor of a31 = C31 = (−1)3+1

2 6 = 14 0 7

OR Given, A is a square matrix of order 3 \ |2A| = 23|A| = 8|A| = k|A| (Given) ⇒ k=8    1   2. We have, tan −1 2 cos  2 sin −1      2    π   = tan −1 2 cos  2 ×    6  

{

= tan −1 2 cos

π  −1 1 ∵ sin 2 = 6 

}

π π 1  = tan −1  2 ×  = tan −1 1 =  3 2 4

dy 1 e −2 x 3. We have, − ⋅y = dx x x −1 dy + Py = Q, where P = This is of the form dx x e −2 x and Q = . x ∫ Pdx I.F. = e ∫ =e

−1 dx x

= e −2

x

4. We have, f(x) = x2 – 4x + 1 ⇒ f(A) = A2 – 4A + I

∴

29

2 3 2 3  7 12 2 3  ∴ A2 =  A=  =   1 2  1 2   4 7  1 2  2 3 1 0   7 12 f ( A) =  − 4  +  1 2  0 1  4 7 

 7 12  8 12 1 0 0 0 = − + = .  4 7   4 8  0 1  0 0   5. The unit vector in the direction of a vector a is given  a by a^ =  . Now, | a | = (2)2 + (3)2 + (4)2 = 29 |a |

=

2 ^ 3 ^ 4 ^ i+ j+ k 29 29 29

OR  ^ ^  ^ ^ ^ ^ Let a = 7 i + j − 4 k and b = 2 i + 6 j + 3k .   ^ ^ Then, a ⋅ b = (7 i^ + ^j − 4 k ) ⋅ ( 2 i^ + 6 ^j + 3k ) = 14 + 6 – 12 = 8  Also, | b | = 22 + 62 + 32 = 7   a ⋅ b 8  \ Projection of a on b =  = 7 |b | 6. We have f(x) = x3 – 3x2 – x Clearly, f(1) = 1 – 3 – 1 = –3 and f(–1) = –1 – 3 + 1 = –3 ⇒ Distinct elements have same image, therefore f is not one-one. 3

7. Let I = ∫ (x − 1)(x − 2)(x − 3)dx 1

3

3

 x 4 6 x 3 11x 2  = ∫ (x − 6 x + 11x − 6)dx =  − + − 6x  3 2  4 1 1 3

2

 81 162 99  1 6 11   = − + − 18 −  − + − 6   = 0 4 3 2  3 2 4

OR

.

OR  d3 y  Highest order derivative is   . So, its order is 3  dx 3  and degree is 4.

Q

2 ^i + 3 ^j + 4 k

π

∫x

10

Let I =

sin7 x dx

−π

Also, let f(x) = x10sin7x Then, f(–x) = (–x)10 [sin(–x)]7 = –x10sin7x = –f(x) ⇒ f(x) is an odd function. π

∴

I=

∫x

10

sin7 x dx = 0

−π

8. Here, a1 = 3 + 1, b1 = − 3 + 1, c1 = 4 and a2 = − 3 + 1, b2 = 3 + 1, c2 = −4

a1 b c = −1, 1 = −1 and 1 = −1 a2 b2 c2 a b c ∴ 1= 1= 1 a2 b2 c2 ⇒ Direction ratios of lines are proportional. Hence, the lines are parallel to each other.

Since,

207

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  9. Let a = 3ɵi + 2 ɵj − kɵ and b = 6ɵi − 4 x ɵj + ykɵ   Since, a and b are parallel   ∴ a = mb , for some m ∈ R ⇒ 3ɵi + 2 ɵj − kɵ = m (6ɵi − 4 x ɵj + ykɵ ) 1 2 −4 x Also, − 4 xm = 2 ⇒ = 2 ⇒ x = −1 2 y and ym = −1 ⇒ = −1 ⇒ y = −2 2 ⇒ 3 = 6m ⇒ m =

OR Let C(x, y, z) divides the line segment joining the points A(–2, 3, 5) and B(1, 2, 3) in the ratio 2 : 3 externally.       2b − 3a , where a , b and c are position Now, c = 2−3 vectors of A, B and C respectively. = −1  2 ( ɵi + 2 ɵj + 3kɵ ) − 3 (−2 ɵi + 3ɵj + 5kɵ ) = − 1(8 ɵi − 5 ɵj − 9kɵ ) = − 8 ɵi + 5 ɵj + 9kɵ So, co-ordiantes of C ≡ (–8, 5, 9)  10. Let n = 2ɵi − 3ɵj + 4kɵ . Then,  ɵ ɵ ɵ ɵ ɵ ɵ ɵn = n = 2i − 3 j + 4k = 2i − 3 j + 4k |n | 4 + 9 + 16 29 Hence, the required equation of the plane is 2 ɵ 3 ɵ 4 ɵ 6  r ⋅ i− j+ k =  29  29 29 29 11. Here f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 1 Since no two elements have the same image. So f is one-one. Also, every elements has atleast one preimage. So, f is onto. Thus f is bijective.    2 12. a + b < 1 ⇒ a + b < 1 2 2     ⇒ a + b + 2a ⋅ b < 1 ⇒ 1 + 1 + 2a ⋅ b < 1 1 1     ⇒ a ⋅ b < − ⇒ a b cos θ < − 2 2 1 1 ⇒ 1 × 1 × cos θ < − ⇒ cos θ < − 2 2 1 2π ⇒ − 1 ≤ cos θ < − ⇒ π ≥ θ > 2 3 13. We have, y2 = 8x and x = 2 \ Required area = Area of shaded region 2 0

X′

X

O Y′

x=2

  14. Let a = ɵi + 2 ɵj + 3kɵ and b = 3ɵi + 2 ɵj − 2kɵ .  We know that the line which passes through point a     and parallel to b is given by r = a + λb , where l is a constant.  ∴ r = ɵi + 2 ɵj + 3kɵ + λ(3ɵi + 2 ɵj − 2kɵ ) is the required equation of the line. 15. We have, A + B + C = O ⇒ C = –[A + B]   2 2  6 2    8 4   −8 −4       ⇒ C = (−1)   −3 1 + 1 3   = (−1)  −2 4  =  2 −4    4 0  0 4    4 4   −4 −4      16. D.R.’s of the two lines are 1, 2, –2 and 2, 3, l – 3. Since, lines are perpendicular \ a1a2 + b1b2 + c1c2 = 0 ⇒ 1 × 2 + 2 × 3 – 2 (l – 3) = 0 ⇒ l = 7 17. (i) (b) : Total number of tickets = 11 Let the event A = First ticket shows even number and B = Second ticket shows even number Now, P(Both ticket shows even number) = P(A).P(B|A) 5 4 2 = ⋅ = 11 10 11 (ii) (c) : Let the event A = First ticket shows odd number and B = Second ticket shows odd number P(Both ticket shows odd number) 6 5 3 = × = 11 10 11 (iii) (a) : Required probability = P(one number is a multiple 4 and other is a multiple 5) = P(multiple of 5 on first ticket and multiple of 4 on second ticket) + P(multiple of 4 on first ticket and multiple of 5 on second ticket) 2 2 2 2 = ⋅ + × 11 10 11 10 4 4 8 4 = + = = 110 110 110 55 (iv) (d) : Required probability = P(one ticket with prime number and other ticket with a multiple of 4)

2

32 2  = 2 . ∫ 8 x dx = 4 2  x 3/2  = sq. units. 3 0 3

y2 = 8x

Y

=

5 2 2 5 10 10 2 × + × = + = 11 10 11 10 110 110 11 Class 12

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(v) (b) : Let the event A = First ticket shows even number and B = Second ticket shows odd number Now, P(First ticket shows an even number and second ticket shows an odd number) = P(A) . P(B|A) 5 6 30 3 = × = = 11 10 110 11 18. (i) (d) : Since the tank is open from the top, therefore the total surface area is = 2(x × x + 2 (xy + yx)) = 2(x2 + 2(2xy)) = 2x2 + 8xy ym

xm

xm

(ii) (a) : Since, volume of tank should be 4000 l. \ x2y m3 = 4000 l = 4 m3 [ 1 litre = 0.001 m3] So, x2y = 4 (iii) (a) : Let S be the outer surface area of tank. Then, S = x2 + 4xy 16 4 [ x2y = 4] ⇒ S(x) = x2 + 4x ⋅ 2 = x 2 + x x 32 d2S dS 16 = 2 x − 2 and 2 = 2 + 3 ⇒ dx x dx x For maximum or minimum values of S, consider dS =0 dx 16 ⇒ 2x = 2 ⇒ x3 = 8 ⇒ x = 2 m x 32 d2S At x = 2, 2 = 2 + 3 = 2 + 4 = 6 > 0 dx 2 \ S is minimum when x = 2 Now as x2y = 4, therefore y =1 Thus, x = 2y (iv) (b) : Since, surface area is minimum when x = 2y, therefore cost of material will be least when x = 2y. Thus, cost of material will be least when width is equal to twice of its depth. (v) (c) : Since, minimum surface area = x2 + 4xy = 22 + 4 × 2 × 1 = 12 m2 and cost per m2 = ` 360 \ Minimum cost is = ` (12 × 360) = = ` 4320 19. The given curve is y = x3. dy = 3x 2 ⇒ dx  dy  = 3(1)2 = 3 Slope of tangent at (1, 1) is    dx  (1,1) Equation of tangent at (1, 1) is y – 1 = 3(x – 1) ⇒ y – 1 = 3x – 3 ⇒ 3x – y = 2

Equation of normal at (1, 1) is −1 y −1= (x − 1) ⇒ x + 3y = 4 3  cos x  20. We write, tan −1    1 + sin x    x x cos2   − sin2     2 2 = tan −1    cos2  x  + sin2  x  + 2 sin  x  cos  x           2   2 2 2   x   cos  2 = tan −1    

x x    cos − sin   2 2   2 x x   + cos sin    2 2

+ sin

x  2

  cos = tan −1   cos 

x x x  − sin  1 − tan   2 2 2 = tan −1   x x  x  1 + tan  + sin  2 2 2   π x  π x = tan −1  tan  −   = − 4 2  4 2 

21. Given curve is y2 = x(1 – x)2 If y = 0, then x(1 – x)2 = 0 ⇒ x = 0, x = 1 1

\ Required area = 2 ∫ x (1 − x ) dx 0

Y

1

 x 3/2 x 5/2  = 2 −   3 / 2 5 / 2 0

X

2  2  8 = 2  (1) −  (1)   = sq . unit  5   15 3

A(1, 0) X O

Y

22. Let E1, E2 and A denote the events defined as follows : E1 = person selected is man E2 = person selected is woman A = person selected is good orator 1 1 We have, P (E1 ) = , P (E2 ) = 2 2 5 25 Now, P ( A | E1 ) = and P ( A | E2 ) = 100 1000 Required probability is P (E1 ) × P ( A | E1 ) P (E1 | A) = P (E1 )P ( A | E1 ) + P (E2 )P ( A | E2 ) 1 5 × 50 2 2 100 = = = 1 5 1 25 75 3 × + × 2 100 2 1000 209

Mathematics

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23. Let I =

∫

1 1− sin x

1

=∫

dx =

dx =

1

∫

1  π 1 + cos  + x   2 π

dx

x

∫ sec  4 + 2  dx 2

π x 2 cos2  +  4 2 2 π π x log tan  + +  + C = 4 8 4 2 =

4

Put tan x = t ⇒ dx =

dt 1+ t2 dt

and sin 2x =

−1 +C =∫ = \ I=∫ 2 2 t −1 (t − 1) 1 + t − 2t −1 +C ⇒ I= tan x − 1 dt

^

C1 × 48C1

4 C 1 32 P ( X = 1) = 52 and P ( X = 2) = 52 2 = = 221 C2 C2 221 The probability distribution of X is as follows:

OR

dx 1 − sin 2 x

^

26. Total no. of aces = 4 Also, X can take the values 0, 1, 2 48 C 188 , ∴ P ( X = 0) = 52 2 = C2 221

 3π x  2 log tan  +  + C  8 4

Let I = ∫

^

i j k  ^ ^ ^  Now, b × 2a = 3 0 2 = − 12 i + 4 j + 18 k 8 6 4  \ | b × 2a | = (−12)2 + 42 + (18)2 = 22

2t 1+ t2

T 5a −b   5a 3 24. We have, AA =     3 2   −b 2

25a2 + b2 15a − 2b  =  13   15a − 2b 0  5a −b   2 b  10a + 3b and A ⋅ (adjA) =  =    10a + 3b   3 2   −3 5a   0 Q A . (adj A) = AAT is known, so equating the two expressions, we get 25a2 + b2 15a − 2b  10a + 3b 0  =  10a + 3b  13   0  15a − 2b We have, 10a + 3b = 13 and 15a – 2b = 0 On solving, we get a = 2/5 and b = 3 Thus, 5a + b = 2 + 3 = 5  25. Let a = 2iɵ − 3 ɵj − 6kɵ , P ≡ (5, 6, – 3) and Q ≡ (3, 4, – 2)  \ PQ = (3 − 5)iɵ + (4 − 6)jɵ + (−2 + 3)kɵ = −2iɵ− 2 jɵ + kɵ   Now the projection of a on PQ   a ⋅ PQ − 4 + 6 − 6 4 =− =  = 3 4 + 4 +1 PQ OR  ^ ^ ^ ^ ^  We have, a = 4 i + 3 j + 2 k and b = 3 i + 2 k ^ ^ ^  \ 2a = 8 i + 6 j + 4 k

X

0

1

2

P(X)

188 221

32 221

1 221

27. We have, y = sin–1 x. dy 1 2 dy ∴ = =1 ⇒ 1− x 2 dx dx 1− x dy  d  1 − x2 ⋅  = 0  dx  dx  2 d y dy d ⇒ 1 − x2 ⋅ + ⋅ 1 − x2 = 0 2 dx dx dx 2 d y dy 2x ⇒ 1 − x2 ⋅ 2 − ⋅ =0 dx 2 1 − x 2 dx d2 y dy ⇒ (1 − x 2 ) 2 − x =0 dx dx

⇒

(

)

dy y(1 + x ) = dx x( y − 1)  y − 1  1+ x  dx dy =  ⇒   x   y 

28. We have,

⇒



1

1



∫ 1 − y  dy = ∫  x + 1 dx + C1

⇒ y – log |y|= log |x| + x + C1 ⇒ x – y + log |xy| = C, where C = –C1 OR dy  dy  = e3x + 4 y Given, log   = 3x + 4 y ⇒ dx dx dy = e 3 x e 4 y ⇒ ∫ e −4 y dy = ∫ e 3 x dx ⇒ dx e −4 y e 3 x = +C ⇒ −4 3 −1 1 −7 At x = 0, y = 0, then = +C ⇒ C = 4 3 12 Class 12

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e −4 y e 3 x 7 = − ⇒ 4e 3 x + 3e −4 y − 7 = 0 , which is −4 3 12 the required solution. \

29. Let P(x, y) be an arbitrary point on the given curve. The equation of the normal to the given curve at (x, y) is Y − y = − 1 ( X − x ) dy dx It is given that the normal at every point passes through a fixed point (a, b) (say). dx Therefore, β − y = − (α − x ) dy ⇒ (x – a)dx + (y – b)dy = 0 Integrating both sides, we get

∫ (x − α)dx + ∫ ( y − β)dy = C (x − α)2 ( y − β)2 + =C 2 2 ⇒ (x – a)2 + (y – b)2 = r2, where r2 = 2C Clearly, this equation represents a circle, having centre at (a, b) and radius r.

x3 ⇒ y= (x − a)(x − b)(x − c)   x3 ⇒ log y = log    (x − a)(x − b)(x − c)  ⇒ logy = 3logx – {log(x – a) + log(x – b) + log(x – c)} On differentiating w.r.t. x, we get 1 dy 3 1 1 1 = − + + y dx x x −a x −b x −c dy  1 1  1 1  1 1  ⇒ = y  −   +  −  +  −  dx x x −b x x − c   x x −a (−b) (−c )  dy  −a + + ⇒ = y  dx  x(x − a) x(x − b) x(x − c) 

{

⇒

2

 y2  Now, AB = (x − 1) + ( y − 4) =  − 1 + ( y − 4)2   2 4 ( y − 32 y + 68) = 4 y 4 − 32 y + 68 Let f ( y ) = 4 2

2

2

4 y 3 − 32 = y 3 − 8 and f ′′( y ) = 3 y 2 4 Now, f ′( y ) = 0 ⇒ y 3 − 8 = 0 ⇒ y = 2

Then, f ′( y ) =

Also, f ′′(2) = 3 × 4 = 12 > 0 So, y = 2 is a point of minima. y2 4 Now, y = 2 ⇒ x = = =2. 2 2 So, the required point is (2, 2). 31. We have, y=

⇒ y=

ax 2 bx c + + +1 (x − a)(x − b)(x − c) (x − b)(x − c) x − c ax 2 + bx(x − a) + c(x − a)(x − b) + (x − a)(x − b)(x − c) (x − a)(x − b)(x − c)

{

dy y a b c = + + dx x a − x b − x c − x

}

OR

⇒

30. Let A (x, y) be the required point which is closest to the point B(1, 4). Then, the distance AB should be minimum and therefore AB2 should be minimum.

}

2

Let y =

2 x + x +1 ⋅ 4 x − 1 (x 2 − 1)3/2

 x 2 + x +1  ⋅ 4x − 1  ⇒ log y = log  2  (x 2 − 1)3/2  1 3 = (x 2 + x + 1)log 2 + log(4 x − 1) − log( x 2 − 1) 2 2 Differentiating both sides w.r.t. x, we get 1 dy d 1 1 d ⋅ = log 2 ⋅ (x 2 + x + 1) + ⋅ × (4 x − 1) dx 2 4 x − 1 dx y dx 3 1 d ⋅ (x 2 − 1) − ⋅ 2 2 x − 1 dx 1 3 = (2 x + 1)log 2 + ⋅4 − ⋅ 2x 2(4 x − 1) 2(x 2 − 1) 2 3x  dy  − = y (2 x + 1)log 2 + 2 4 x − 1 x − 1  dx  2x + 3 32. Let y = f(x) = ...(i) x −3 Let x1, x2 ∈ A = R – {3} such that f(x1) = f(x2) 2x1 + 3 2x2 + 3 = ⇒ x1 − 3 x2 − 3 ⇒ (2x1 + 3)(x2 – 3) = (2x2 + 3)(x1 – 3) ⇒ 2x1x2 – 6x1 + 3x2 – 9 = 2x1x2 – 6x2 + 3x1 – 9 ⇒ –6x1 + 3x2 = –6x2 + 3x1 ⇒ 9x1 = 9x2 ⇒ x1 = x2 Now, f(x1) = f(x2) ⇒ x1 = x2 So f(x) is one-one. 2x + 3 For onto, let y = ⇒ xy – 3y = 2x + 3 x −3 ⇒

211

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⇒ xy – 2x = 3y + 3 ⇒ x(y – 2) = 3(y + 1) 3( y + 1) ⇒ x= ...(ii) ( y − 2) Equation (ii) is defined for all real values of y except 2 which is same as given set B = R – {2}. 3( y + 1) Thus, for every y ∈ B, there exist x = ∈ A such y −2 that f(x) = y Hence, function f is onto.

33. Given curve is y = x2 + x. Clearly, required area = Area of shaded region 5

2

= [ log x ]1 − ∫ 2

1

2

x 1+x

2

dx = log 2 − ∫ 1

x dx 1 + x2

Put 1 + x2 = t ⇒ 2x dx = dt When x = 1, t = 2 and when x = 2, t = 5 5 1 5 1 1 \ I = log 2 − ∫ dt = log 2 −  log t  2 2 2 2t 1 = log 2 − [log 5 − log 2] 2 8 1 5 1 = log 2 − log   = log     2 5 2 2

 x3 x2  = ∫ (x + x ) dx =  +  2 2 3 2  125 25  8   = + −  + 2 2  3   3

35. Since, f(x) is continuous at x = 0, therefore f(0) = lim f (x ) = lim f (x )

 250 + 75 14  = −  3  6

1 − (1 − x )   1−1+ x  = 4 lim   = 4 lim   x →0  x (1 + 1 − x )  x →0  x (1 + 1 − x ) 

5

2

=

x →0−

= 4  1  = 4 = 2  (1 + 1)  2 From (i), we get f (0) = 2.

1

34. Let I = ∫ tan

−1

−1 x dx = ∫ tan x ⋅1 dx 0

0

= [tan −1 x ⋅ x]10

1

1

∫ 1 + x 2 ⋅ x dx 0

1

π 1  π  1 2x dx = − I1 =  − 0  − ∫ 4 2 4 2 1 + x2 0 1

…(i)

2x

Consider I1 = ∫

dx 2 x 1 + 0 Put 1 + x2 = t ⇒ 2x dx = dt When x = 0, t = 1 and when x = 1, t = 2 2

1 2 ∴ I1 = ∫ dt = [ log t ]1 = log 2 − log 1 = log 2 t

...(ii)

1

π ⇒ I = − log 2 4

[From (i) and (ii)]

OR dx Let I = ∫ 2 1 x (1 + x ) A Bx + C 1 Consider, = + x (1 + x 2 ) x 1 + x 2 2

⇒ 1 = A (1 + x 2 ) + (Bx + C ) ⋅ x ⇒ 1 = x2(A + B) + Cx + A On equating the coefficient of x2, x and the constant term from both sides, we get A = 1, B = –1 and C = 0 \ I=

2

∫ 1

2

−x 1 dx dx + ∫ 2 x 1 1+ x

…(i)

 4(1 − 1 − x ) lim f (x ) = lim   − x x →0 x →0 

325 14 325 − 28 297 − = = sq . units 6 3 6 6 1

x →0+

36. Here we need to find, the equation of the line through the point (1, 1, –3) and perpendicular to the lines x −2 y +3 z −0 …(i) = = 2 1 −3 x −3 y +5 z −0 and …(ii) = = 1 1 1 Let the direction ratios of required line are a, b, c. Then, equations of this line is given by x −1 y −1 z + 3 …(iii) = = a b c Direction ratios of line (i) are 2, 1, –3 and line is perpendicular to line (iii) having direction ratios a, b, c \ 2a + b – 3c = 0 …(iv) Similarly a + b + c = 0 …(v) Solving equation (iv) and (v), we get a −b c = = 4 5 1 \ From equation (iii), required equation of the line x −1 y −1 z + 3 is = . = 4 −5 1  Its vector equation is r = (ɵi + ɵj − 3kɵ ) + λ(4ɵi − 5ɵj + kɵ ) OR The equation of the plane passing through the point A(3, 2, –5) is given by a(x – 3) + b(y – 2) + c(z + 5) = 0 ...(i) If it passes through B(–1, 4, –3) and C(–3, 8, –5), Class 12

212

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we get a(–1 –3) + b(4 – 2) + c(–3 + 5) = 0 ⇒ – 4a + 2b + 2c = 0 ⇒ 2a – b – c = 0 ...(ii) Also a(–3 – 3) + b(8 –2) + c(–5 + 5) = 0 ⇒ –6a + 6b + 0c = 0 ⇒ a – b – 0c = 0 ..(iii) Solving (ii) and (iii) by cross multiplication method, we get a b c = = (0 − 1) (−1 − 0) (−2 + 1) a b c = = = k (say) ⇒ a = k, b = k, c = k. 1 1 1 Putting a = k, b = k and c = k in (i), we get (x – 3) + (y – 2) + (z + 5) = 0 ⇒ x + y + z = 0. Thus, the equation of the plane passing through the points A(3, 2, –5), B(–1, 4, –3) and C(–3, 8, –5) is x + y + z = 0. Clearly, the fourth point D(–3, 2, 1) also satisfies x + y + z = 0. Hence, equation of the plane containing the given points is x + y + z = 0. ⇒

37. Converting inequations into equations, we get x + y = 60 ...(i) x = 40 ...(ii) y = 40 ...(iii) x + y = 30 ...(iv) x=0 ...(v) and y = 0 ...(vi) Let us draw the graph of equations (i) to (vi). The feasible region is shown in figure. Y (0, 60) (0, 40) E (0, 30) F X

O Y

x = 40 (20, 40) D

y = 40 (40, 20) Cx+ y= 60 A B X (40, 0) ( (30, 0) 60 , 0) x+ y= 30

The coordinates of the corner points of the feasible region are A(30, 0), B(40, 0), C(40, 20), D(20, 40), E(0, 40) and F(0, 30). Let us evaluate Z at these points. Corner points

Value of Z = 3x + 4y + 270

A(30, 0)

360 ← Minimum

B(40, 0)

390

C(40, 20)

470

D(20, 40)

490

E(0, 40)

430

F(0, 30)

390

From the table, the minimum value of Z is 360, which is attained at the point A(30, 0). OR Converting inequations into equations, we get 2x + 5y = 100, 2x + y = 50, x = 0 and y = 0 x y x y + = 1, i.e., + = 1 , x = 0 and y = 0 50 20 25 50 Let us draw the graph of above equations.

Clearly, the feasible region is OABCO, which is shaded in the figure. Here, B is the point of intersection of lines 2x + 5y = 100  75 25  and 2x + y = 50 i.e., B =  ,  4 2   75 25  We have corner points A(25, 0), B  ,  and 4 2 C(0, 20). The values of the objective function Z = x + y at these points are Z(A) = 25 + 0 = 25 75 25 Z(B) = + = 31.25 4 2 Z(C) = 0 + 20 = 20 Z(O) = 0 + 0 = 0 The maximum value of Z is 31.25, which is attained at  75 25  .  , 4 2 2 3 7  38. We have, A =  3 −2 −1    1 1 2  \

2 3 7 A = 3 −2 −1 1

1

2 213

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= 2(–4 + 1) – 3(6 + 1) + 7(3 + 2) = –6 – 21 + 35 = 8 ≠ 0. So A–1 exist. The cofactors of elements of A are C11 = –3, C21 = 1, C31 = 11 C12 = –7, C22 = –3, C32 = 23 C13 = 5, C23 = 1, C33 = –13  −3 1 11  \ adjA =  −7 −3 23     5 1 −13  and A

−1

 −3 1 11  adj A 1  = = −7 −3 23   A 8   5 1 −13 

Given system of equations is 2x + 3y + 7z = 12 3x – 2y – z = 0 x + y + 2z = 4 which can be written in matrix form as  2 3 7   x  12   3 −2 −1  y  =  0        1 1 2   z   4  ⇒ AX = B ⇒ X = A–1B  −3 1 11  12  1 ⇒ X = −7 −3 23   0    8   5 1 −13   4 

OR Here,  1 −2 1  A =  −2 3 1  1 1 5 We know, if A is non-singular matrix, then (adj A–1) = (adj A)–1, so we will find (adj A)–1. The cofactors of elements of A are A11 = 14 A12 = 11 A13 = –5 A21 = 11 A22 = 4 A23 = –3 A31 = –5 A32 = –3 A33 = –1 ′

14 11 −5 14 11 −5 ∴ B = adj A =  11 4 −3 =  11 4 −3  −5 −3 −1  −5 −3 −1 \ | B| = |adj A| = 14(–4 – 9) –11(–11 – 15) –5(–33 + 20) = –182 + 286 + 65 = 169 ≠ 0 Cofactors of B are B12 = 26 B13 = –13 B11 = –13 B21 = 26 B22 = –39 B23 = –13 B31 = –13 B32 = –13 B33 = –65 ′

 −13 26 −13  −13 26 −13 ∴ adj B =  26 −39 −13 =  26 −39 −13  −13 −13 −65  −13 −13 −65 and B

x  −36 + 0 + 44  1 1   Now, y = X = −84 − 0 + 92  = 1      8   z   60 + 0 − 52  1 Comparing we get x = 1, y = 1, z = 1

−1

 −13 1  = (adj A) = 26 169   −13  −1 2 1 =  2 −3 13  −1 −1 −1

26 −13 −39 −13 −13 −65 −1 −1 −5



Class 12

214

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Self Evaluation Sheet Once you complete SQP-15, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Chapter

Marks Per Question 1

Determinants / Determinants Inverse Trigonometric Functions

Marks Obtained

1

Differential Equations / Differential Equations Matrices Vector Algebra / Vector Algebra Relations and Functions Integrals / Integrals Three Dimensional Geometry Vector Algebra / Vector Algebra Three Dimensional Geometry Relations and Functions Vector Algebra Application of Integrals Three Dimensional Geometry Matrices Three Dimensional Geometry Probability Application of Derivatives Application of Derivatives Inverse Trigonometric Functions Application of Integrals Probability Integrals / Integrals Matrices Vector Algebra / Vector Algebra Probability Continuity and Differentiability Differential Equations / Differential Equations Differential Equations Application of Derivatives Continuity and Differentiability / Continuity and Differentiability Relations and Functions Application of Integrals Integrals / Integrals Continuity and Differentiability Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming Determinants / Determinants Total

1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80 Percentage

.................. ..............%

Performance Analysis Table If your marks is

> 90% TREMENDOUS!

You are done! Keep on revising to maintain the position.

81-90% EXCELLENT!

You have to take only one more step to reach the top of the ladder. Practise more.

71-80% VERY GOOD!

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70% GOOD!

Revise thoroughly and strengthen your concepts.

51-60% FAIR PERFORMANCE!

Need to work hard to get through this stage.

40-50% AVERAGE! Mathematics

Try hard to boost your average score.

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215

SQP

16

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

–

1(3)

–

3(5)

2.

Inverse Trigonometric Functions

1(1)

1(2)

–

–

2(3)

3.

Matrices

2(2)

1(2)

–

–

4(9)

4.

Determinants

1(1)

–

–

1(5)*

1(1)

5.

Continuity and Differentiability

2(2)#

1(2)

2(6)#

–

5(10)

6.

Application of Derivatives

–

2(4)

1(3)

–

3(7)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)

–

3(6)

9.

Differential Equations

1(1)*

1(2)*

1(3)*

–

3(6)

10.

Vector Algebra

1(1)* + 1(4)

–

–

–

2(5)

2(2)#

1(2)*

–

1(5)*

4(9)

–

–

–

1(5)*

1(5)

2(2) + 1(4)

1(2)

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

Probability Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-16

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. If y = xx, then find

dy . dx

2

OR

x d y If y = log   , then find . x dx 2 e  2

cos x − sin x  T 2. If A =   , then find AA . x x sin cos   3. Solve the differential equation secx dy + cosecy dx = 0. OR Solve the differential equation 4. Check whether f (x ) =

dy + y = e −2 x . dx

(4 x + 3) is one-one or not. (3x + 4)

     5. Vectors drawn from the origin O to the points A, B and C are respectively a , b and 4a − 3b . Find AC. 217

Mathematics

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OR       If a = 10, b = 2 and a ⋅ b = 12, then find the value of a × b . 6. If A and B are events such that P(A) > 0 and P(B) ≠ 1, then find the value of P(A′ | B′). 7. Evaluate :

∫

2 x + 3x 5x

dx OR

π/ 4

Evaluate :

∫

sin 2 x dx

0

5 3 8 8. If ∆ = 2 0 1 , then write the minor of the element a23. 1 2 3

9. Find the distance between the planes 2x – y + 2z = 5 and 5x – 2.5y + 5z = 20. OR If a line makes angles a, b, g with the positive direction of coordinate axes, then find the value of sin2a + sin2b + sin2g. −1  1  −1  1  10. Using principal values, write the value of cos   + 2 sin   . 2 2

 3π  11. If f(x) = –cos x, then find f ′   . 4 2 and the probability that he will not get 12. The probability that a person will get an electric contract is 5 2 4 plumbing contract is . If the probability of getting at least one contract is , what is the probability that 3 7 he will get both? 2a + b a − 2b   4 − 3 13. If  =  , then find the value of 3a – b + 9c + 2d.  5c − d 4c + 3d  11 24  14. Find the area of the region bounded by the curve x = 2y + 3 and the lines y = 1 and y = –1. 15. Find the distance of the plane 2x – 3y + 6z + 14 = 0 from the origin. 16. Let R = {(a, b) : |a – b| is divisible by 2} be any relation in the set A = {0, 1, 2, 3, 4, 5}. Write the equivalence class [0]. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Three hoardings are to be placed on a beach at the points A, B and C displaying A (Do not litter)), B (Keep you place clean) and C (Go green). The coordinates of these points should be (4, 2, 1), (4, 8, 2) and (8, 4, 3). Take centre of beach as origin. Based on the above information, answer the following questions :    (i) Let a , b and c be the position vectors of points A, B and C    respectively, then a + b + c is equal to (a) 8iˆ + 7 jˆ + 3kˆ

Keep Your Place Clean Do not litter

(b) 2(8iˆ + 7 jˆ + 3kˆ ) (c) 7iˆ + 8 jˆ + 3kˆ

Go green

(d) 2(7iˆ + 8 jˆ + 3kˆ ) Class 12

218

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(ii) Which of the following is not true?     (a) AB + BC + CA = 0     (c) AB + BC − CA = 0

    (b) AB + BC − AC = 0     (d) AB − CB + CA = 0

(iii) Area of triangle ABC is 1 692 (d) 2 (iv) Suppose, if the given hoardings are to be placed on a same line, then the value of will be equal to (a) –1 (b) 2 (c) 0 (d) (v) Equation of plane containing all the given point is (a) 5x + 2y – 12z = 0 (b) 5x + 2y – 12x = 12 (c) 5x + 2y = 0 (d) (a) 0

(b)

(c)

692

592       a ×b +b +c +c ×a –2 none of these

18. An advertising executive is studying television-viewing habits of married couples during prime time hours. Based on past viewing records he has determined that during prime time husbands are watching television 60% of the time. It has also been determined that when the husband is watching television, 40% of the time the wife is also watching. When the husband is not watching television, 30% of the time the wife is watching television. Based on the above information, answer the following questions : (i) The probability that the husband is not watching television during prime time, is (a) 0.6 (b) 0.3 (c) 0.4 (d) 0.5 (ii) If the wife is watching television, the probability that husband is also watching television, is 1 2 1 5 (b) (c) (d) (a) 3 3 6 6 (iii) The probability that both husband and wife are watching television during prime time, is (a) 0.24 (b) 0.2 (c) 0.3 (d) 0.4 (iv) The probability that the wife is watching television during prime time, is (a) 0.24 (b) 0.36 (c) 0.3 (d) 0.4 (v) If the wife is watching television, then the probability that husband is not watching television, is 1 2 1 5 (b) (c) (d) (a) 3 3 6 6

PART - B Section - III 19. If xy = ex–y, then prove that dy = dx

log x (1 + log x )2

.

x +2 y +1 z −3 = = at a distance of 5 units from the point P(1, 3, 3). 3 2 2 OR Find the direction cosines of the line passing through the two points (–2, 4, –5) and (1, 2, 3).

20. Find the points on the line

21. Find the maximum and minimum values (if any) of x + 1 in [–1, 1].  − cos x 22. If A =   − sin x

− sin x  , then find A–1. cos x  219

Mathematics

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23. Evaluate : ∫

x 3 − 3x 2 + 5x − 7 + x 2a x 2x 2

dx OR

Evaluate

2 cos x − 3 sin x

∫ 6 cos x + 4 sin x dx by using substitution method.

24. A four digit number is formed using the digits 1, 2, 3, 5 with no repetitions. Write the probability that the number is divisible by 5.  1 + cos x + 1 − cos x  π x π −1  25. Prove that tan  = + ,0 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

More ebooks at https://telegram.me/unacademyplusdiscounts

SQP

17

BLUE PRINT

Time Allowed : 3 hours

Maximum Marks : 80 VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

Relations and Functions

1(1)

1(2)

1(3)

–

3(6)

Inverse Trigonometric Functions

2(2)#

–

–

–

2(2)

3.

Matrices

2(2)#

–

–

1(5)*

3(7)

4.

Determinants

1(1)

1(2)

–

–

2(3)

–

4(9)

S. No. 1. 2.

Chapter

5.

Continuity and Differentiability

1(1)

1(2)

2(6)#

6.

Application of Derivatives

1(1)*

2(4)

1(3)

–

4(8)

1(3)*

–

4(8)

7.

Integrals

1(1)*

2(4)#

8.

Application of Integrals

1(1)

–

1(3)

–

2(4)

9.

Differential Equations

1(1)

1(2)

1(3)

–

3(6)

10.

Vector Algebra

1(1)* + 1(4)

–

–

–

2(5)

11.

Three Dimensional Geometry

2(2)

1(2)*

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

2(2) + 1(4)

1(2)*

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-17

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Find the area of a parallelogram whose adjacent sides are represented by the vectors 2i − 3k and 4 j + 2k . OR 







Write the direction ratios of the vector 3a + 2b , where a = ɵi + ɵj − 2kɵ and b = 2ɵi − 4 ɵj + 5kɵ . 2. If A is a square matrix satisfying A2 = I, then what is the inverse of A? 3. If the curve ay + x2 = 7 and x3 = y cut orthogonally at (1, 1), then find the value of a. OR  t   Find the slope of the tangent to the curves x = a sin t, y = a cos t + log  tan   at the point ‘t’. 2  

4. A flashlight has 8 batteries out of which 3 are dead. If two batteries are selected without replacement and tested, then find the probability that both are dead. 2 3   1

5. If   5 7   −2

−3  −4 6   , then find x. = 4   −9 x  Class 12

224

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OR  0 −5 8   0 12 is a skew symmetric matrix. Show that the matrix  5  −8 −12 0

6. Solve :

dy = e x − y + x 2e − y dx

7. Evaluate :

dx

∫ x(x 5 + 3)

OR

2

Evaluate : ∫ (x − [ x])dx 0

8. Find the area of the triangle formed by the straight lines y = 2x, x = 0 and y = 2 by integration.   π  9. Evaluate : tan −1 sin  −     2 

OR Write the domain of the function cos–1(2x – 1). 6   2 3 10. Find the distance of the plane r ⋅  ɵi + ɵj − kɵ  = 1 from the origin. 7 7 7 

11. Find the principal value of tan–1(–1). 12. If y = log10x + logey, then find

dy . dx

 1 3 2 13. If A =   , then find the value of |A – 2A|. 2 1  4 3 1 14. If A and B are two events such that P (B) = , P ( A | B) = and P ( A ∪ B) = , then find the value of P(A). 5 5 2

15. If (1, 3), (2, 5) and (3, 3) are three elements of A × B and the total number of elements in A × B is 6, then write the remaining elements of A × B. 16. Find the direction cosines of the normal to the plane 3x – 6y + 2z = 7. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. In a bilateral cricket series between India and England, the probability that India wins the first match is 0.5. If India wins any match, then the probability that it wins the next match is 0.4, otherwise the probability is 0.3. Also, it is given that there is no tie in any match. Based on the above answer the following : (i) The probability that India losing the second match, if India has already won the first match is (a) 0.5 (b) 0.4

(c) 0.3

(d) 0.6 225

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(ii) The probability that India winning the third match, if India has already loosed the first two matches is (a) 0.2 (b) 0.3 (c) 0.4 (d) 0.6 (iii) The probability that India winning the first two matches is (a) 0.1 (b) 0.2 (c) 0.4 (d) 0.01 (iv) The probability that India winning the first three matches is (a) 0.8 (b) 0.6 (c) 0.04 (d) 0.08 (v) The probability that India winning exactly one of the first three matches is (a) 0.205

(b) 0.21

(c) 0.405

(d) 0.312

18. Consider the following diagram, where the forces in the cable are given. z

E FEC FED

FEA FEB

D 12 m

A

C 6m

O 4m 4m

B

6m y

x

Based on the above answer the following : (i) The cartesian equation of line EA is x −6 y z x y z = = (a) (b) = = 2 6 −3 −3 2 6  (ii) The vector ED is (a) 6 i − 4 j + 12k (b) − 6 i − 4 j + 12k (iii) The length of the cable EB is (a) 14 units (b) 16 units (iv) The length of cable EC is equal to the length of (a) EA (b) EB (v) The sum of all vectors along the cables is (b) 48 j (a) 48 i

(c)

x y z − 12 = = −3 2 6

(d)

x y z − 12 = = 3 6 6

(c) −6 i − 4 j − 12k

(d) 6i + 4 j + 12k

(c) 17 units

(d) 15 units

(c) ED

(d) All of these

(c) −48 k

(d) 48 k

PART – B Section - III 1 −  19. If f (x ) =   

cos 4 x x2 4,

, when x ≠ 0

, then check whether function f(x) is continuous at x = 0 or not.

when x = 0 Class 12

226

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20. Show that the lines of these lines.

x −1 y − 2 z − 3 x − 4 y −1 and = = = = z intersect. Also, find the point of intersection 2 3 4 5 2 OR

 Find the equation of the plane which contains the line of intersection of the planes r . (ɵi + 2 ɵj + 3kɵ ) = 4,   r . (2ɵi + ɵj − kɵ ) + 5 = 0 and which is perpendicular to the plane r . (5ɵi + 3ɵj − 6kɵ ) + 8 = 0. 21. Find domain and range of the function

x . 1+ x

22. Find the maximum profit that a company can make, if the profit function is given by P(x) = 41 + 24x – 18x2. 4 23. The probability that a student selected at random from a class will pass in Mathematics is and the 5 1 probability that he/she passes in Mathematics and Computer Science is . What is the probability that 2 he/she will pass in Computer Science, if it is known that he/she passed in Mathematics ?

OR A speaks truth in 80% cases and B speaks truth in 90% cases. In what percentage of cases are they likely to agree with each other in stating the same fact? π /2

24. Evaluate :

∫ 0

1 + sin x dx 2 + sin x + cos x

1 0      and B = 0 1 and M = AB, then find M–1.  1 1  dy  26. Find the solution of the differential equation ( x − y ) 1 −  = e x .  dx  −1 −2

1 25. If A =  2

3 0

1 27. Find the interval in which the function x 4 − x 3 is increasing. 3 dx . 28. Find the value of integral ∫ 3 cos2 x + 5 OR cos( x + a) Evaluate : ∫ dx sin( x + b) Section - IV 29. If f (x ) =

π π tan 2 x π , x ≠ , is continuous at x = , then find the value of f   . 4 4 4  π cot  − x   4

30. Find the equation of the curve passing through origin if the slope of the tangant to the curve at any point (x, y) is equal to the square of the difference of the abscissa and ordinate of the point. π /3

31. Evaluate :

∫

1

1 + tan x π /6

dx OR 2

Evaluate : ∫ (3 sin x + 4cosec x ) dx 227

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32. Let f : R+ → [4, ∞) given by f(x) = x2 + 4. Is this function one one and onto? 33. Find the area of the region bounded by the parabola y2 = 4ax, its axis and two ordinates x = 5 and x = 8.   dy  2  1 +    dx  34. If (x – a)2 + (y – b)2 = c2, for some c > 0, then prove that  d2 y OR

3/2

is independent of a and b.

dx 2

3π dy −1  1 + sin x − 1 − sin x  If y = tan   , where π < x < , then find . 2 dx  1 + sin x + 1 − sin x  35. Find the values of x, for which the function f (x) = x3 + 12x2 + 36x + 6 is increasing. Section - V  36. Find the shortest distance between the lines r = (ɵi + 2 ɵj + 3kɵ ) + λ(2ɵi + 3ɵj + 4kɵ ) and  r = (2ɵi + 4 ɵj + 5kɵ ) + µ(4ɵi + 6 ɵj + 8kɵ ). OR Find the equation of the plane parallel to the line (5, 2, –1) and passes through the origin.

x − 2 y −1 z − 3 , which contains the point = = 1 3 2

37. Solve the following linear programming Problem (LPP) graphically. Maximize Z = 20x + 30y Subject to constraints : 2x + 3y ≥ 100 ; x + 2y ≤ 80 ; x ≥ 14 ; y ≥ 16, x, y ≥ 0 OR Solved the following linear programming problem (LPP) graphically. Maximize Z = 0.6x + 0.4 y Subject to constraints : x + y ≤ 500 ; 2x + 2y ≤ 800 ; x, y ≥ 0  1 2 3   38. If A =  3 −2 1 , then show that A3 – 23A – 40I = O.  4 2 1 OR  1 0 0  1 1 and 6A–1 = A2 + cA + dI, then find the values of c and d. If A = 0 0 −2 4  INSTANT

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Class 12

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Self Evaluation Sheet Once you complete SQP-17, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Chapter

Marks Per Question 1

Vector Algebra / Vector Algebra Matrices

Marks Obtained

1

Application of Derivatives / Application of Derivatives Probability Matrices / Matrices Differential Equations Integrals / Integrals Application of Integrals Inverse Trigonometric Functions / Inverse Trigonometric Functions Three Dimensional Geometry Inverse Trigonometric Functions Continuity and Differentiability Determinants Probability Relations and Functions Three Dimensional Geometry Probability Vector Algebra Continuity and Differentiability Three Dimensional Geometry / Three Dimensional Geometry Relations and Functions Application of Derivatives Probability / Probability Integrals Determinants Differential Equations Application of Derivatives Integrals / Integrals Continuity and Differentiability Differential Equations Integrals / Integrals Relations and Functions Application of Integrals Continuity and Differentiability / Continuity and Differentiability Application of Derivatives Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming Matrices / Matrices Total

1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80 Percentage

.................. ..............%

Performance Analysis Table If your marks is

> 90% TREMENDOUS!

You are done! Keep on revising to maintain the position.

81-90% EXCELLENT!

You have to take only one more step to reach the top of the ladder. Practise more.

71-80% VERY GOOD!

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70% GOOD!

Revise thoroughly and strengthen your concepts.

51-60% FAIR PERFORMANCE!

Need to work hard to get through this stage.

40-50% AVERAGE!

Try hard to boost your average score.

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SQP

18

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

–

1(3)

–

3(5)

2.

Inverse Trigonometric Functions

1(1)*

1(2)

–

–

2(3)

3.

Matrices

2(2)

–

–

1(5)*

3(7)

4.

Determinants

1(1)

1(2)

–

–

2(3)

5.

Continuity and Differentiability

1(1)*

1(2)

2(6)

–

4(9)

6.

Application of Derivatives

1(1)

2(4)

1(3)

–

4(8)

2(2)

1(2)*

1(3)*

–

4(7)

–

1(2)

1(3)

–

2(5)

#

7.

Integrals

8.

Application of Integrals

9.

Differential Equations

1(1)

1(2)

1(3)*

–

3(6)

10.

Vector Algebra

1(4)

1(2)*

–

–

2(6)

3(3)

–

–

1(5)*

4(8)

–

–

–

1(5)*

1(5)

2(2) + 1(4)

1(2)*

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

#

#

Probability Total

*It is a choice based question. # Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-18

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part -A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Evaluate :

Evaluate : 0 2. If A =  x

∫ ∫

dx x +1 + x + 2

OR

dx 2 + 4x − x2 0 , then find A16. 0

3. If an equation of the plane passing through the points (3, 2, –1), (3, 4, 2) and (7, 0, 6) is 5x + 3y – 2z = l, then find l. OR Find the distance of the plane 2x – 3y + 4z – 6 = 0 from the origin. 4. If A = {1, 2, 3}, B = {1, 4, 6, 9} and R is a relation from A to B defined by ‘x is greater than y’. Then find the 231

Mathematics

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range of R. 5. A box contains 3 orange balls, 3 green balls and 2 blue balls. Three balls are drawn at random from the box without replacement. Find the probability of drawing 2 green balls and one blue ball. OR If A and B are two events such that P(A) = 6. Find the order and degree of

d5 y dx

5

4 7 and P(A ∩ B) = , then find P(B|A). 5 10

+ e dy /dx + y 2 = 0 .

dy −1 −1  1  , then find 7. If y = tan ( 3 ) + tan  .  3  dx OR Show that f(x) = x3 is continuous at x = 2. 8. If a line makes angle a, b and g with the coordinate axes, then find the value of cos 2a + cos 2b + cos 2g. 9. Evaluate : cosec −1 (2/ 3 ) OR 2

–1

Evaluate : sec (tan 2) 1

10. Evaluate :

∫

tan −1 x

2 0 1+ x

dx

1 11. If matrix A = [aij]2 × 2, where aij =  0

if

i≠ j

if

i= j

then find A3.

12. If A and B are events such that P(A) = 0.4, P(B) = 0.3 and P(A ∪ B) = 0.5, then find P(B′ ∩ A). 13. How many one-one functions from set A = {1, 2, 3} to itself are possible? 14. Write the direction cosines of the line segment joining the points A(7, –5, 9) and B(5, –3, 8). 15. If the area of a triangle with vertices (–3, 0), (3, 0) and (0, k) is 9 sq. units, then find the value of k. 16. Find the interval on which f(x) = 2x3 – 6x + 5 is a strictly increasing function. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. A graduate student is preparing for competitive examinations. The probabilities that the student is selected in competitive examination of B.S.F., C.D.S. and Bank P.O. are a, b and c respectively. Of these examinations, students has 70% chance of selection in at least one, 50% chance of selection in at least two and 30% chance of selection in exactly two examinations. Based on the above answer the following : (i) The value of a + b + c – ab – bc – ca + abc is (a) 0.3 (b) 0.5

(c) 0.7

(d) 0.6

(ii) The value of ab + bc + ac – 2abc is (a) 0.5 (b) 0.3

(c) 0.4

(d) 0.6 Class 12

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(iii) The value of abc is (a) 0.2

(b) 0.5

(c) 0.7

(d) 0.3

(iv) The value of ab + bc + ac is (a) 0.1 (b) 0.9

(c) 0.5

(d) 0.3

(v) The value of a + b + c is (a) 1.9 (b) 1.5

(c) 1.6

(d) 1.4

18. Consider the following diagram, where the forces in the cable are given. Z D FDC FDB 24 m

FDA

Y′

X′ 16 m C 18 m

O

8m

X

B

A

5m

4m 6m

Y

Based on the above answer the following : (i) The equation of line along the cable AD is x y z − 24 x y z − 24 = = (b) = = (a) 5 8 24 8 5 24

(c)

(ii) The length of cable DC is (a) 43 m (b) 34 m

(c) 54 m

(d) 45 m

(c) 6i + 4 j + 24 k

(d) none of these

(c) 15i + 6 j − 72k

(d) none of these

(c)

(d) none of these

x y 24 − z = = 5 8 24

(d)

x y 24 − z = = 8 5 24

(iii) The vector DB is (a) −6i + 4 j − 24 k

(b) 6i − 4 j + 24 k

(iv) Find the sum of vectors along the cables. (a) 15i + 6 j + 72k

(b) 15i − 6 j − 72k

(v) The sum of lengths, i.e., OA + OB + OC, is 89 + 52 + 580 (b)

(a)

52 + 580 + 48

89 + 560 + 49

PART – B Section - III 19. Solve for x : cos(2 sin −1 x ) =

1 , x > 0. 9

20. A man speaks truth in 75% cases. He throws a die and reports that it is a six. Find the probability that it is actually a six. OR 1 Amit and Nisha appear for an interview in a company. The probability of Amit’s selection is and that of 5 1 Nisha’s selection is . What is the probability that only one of them is selected? 6 dy x + y y 1 21. If tan −1 . = = log ( x 2 + y 2 ), then prove that dx x − y x 2

()

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22. Find the point on the curve y = x3 – 11x + 5 at which the equation of tangent is y = x – 11. 23. Let ABCD be the parallelogram whose sides AB and AD are represented by the vector 2i + 4 j − 5k and   i + 2 j + 3k respectively. If a is a unit vector parallel to AC , then find a. OR    The vector i + x j + 3k is rotated through an angle q and doubled in magnitude, then it becomes 4i + (4 x − 2)j + 2k . Find the value of x. π  2 + sin x  dy = − cos x , y(0) = 1, then find y   . 24. If   2  1 + y  dx a  25. If A = 0  0

0 a 0

0 0 , then find |A| |adj A|.  a

26. Find the area bounded by the curve y = x4, x-axis and lines x = –2, x = 2. 1 27. Show that the function f ( x ) = 3 − 4 x + 2 x 2 − x 3 is decreasing on R. 3 1 + sin x 28. Evaluate : ∫ dx 1 + cos x OR If I1 =

2 ex dx I and = 2 ∫e log x ∫1 x dx , then show that I1 = I2. e2

Section - IV  1 + (ax )2 − 1   2x 1 − x 2  a −1  with respect to tan −1  at x = 0 is . 29. Prove that the derivative of tan    ax 4  1 − 2x 2  30. Find the intervals in which the function f(x) = (x – 1)3 (x + 2)2 is strictly increasing or strictly decreasing. Also, find the points of local maximum and local minimum if any. 31. Evaluate : ∫

5x + 3 x 2 + 4 x + 10

dx OR

π

Evaluate :

∫ x cos

2

x dx

0

32. Using integration, find the area bounded by the ellipse

2 x2 y + = 1. 4 25

33. Show that f : R → R, given by f(x) = x – [x], is neither one-one nor onto. 34. Find the particular solution of (x + y)dy + (x – y)dx = 0, given that y = 1 when x = 1. OR Solve the differential equation : xdy − ydx = x 2 + y 2 dx , 4  2 35. If f ( x ) = ax + b, cos x , 

if x ≤ −1 if − 1 < x < 0 is continuous. Find the value of a and b. if x ≥ 0 Class 12

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Section - V 1 36. If A =  2 1 If A = 1  2

−1 a and B =   −1 b 2 −1 3

1 and (A + B)2 = A2 + B2, then find the values of a and b. −1 OR

5 −1 , then find A–1. Hence solve the following system of equations :  −1

x + 2y + 5z = 10, x – y – z = – 2, 2x + 3y – z = – 11 37. Solve the following Linear Programming Problem (LPP) graphically. Maximize Z = 20x + 10y Subject to constraints : x + 2y ≤ 28; 3x + y ≤ 24; x, y ≥ 0 OR Solve the following Linear Programming Problem (LPP) graphically. Maximize Z = 4500x + 5000y Subject to constraints : x + y ≤ 250; 25000x + 40000y ≤ 7000000; x, y ≥ 0  38. Find the image of the point having position vector i + 3 j + 4 k in the plane r ⋅ (2ɵi − ɵj + kɵ ) + 3 = 0. OR Find the coordinates of the points on the line the point (1, 2, 3). INSTANT

x −1 y − 2 z − 3 = = , which are at a distance of 1 unit from 1 2 3

Download answers of this SQP from the given link https://bit.ly/2UHqmTA

235

Mathematics

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Self Evaluation Sheet Once you complete SQP-18, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Chapter

Marks Per Question

Integrals / Integrals Matrices Three Dimensional Geometry / Three Dimensional Geometry Relations and Functions Probability / Probability Differential Equations Continuity and Differentiability / Continuity and Differentiability Three Dimensional Geometry Inverse Trigonometric Functions / Inverse Trigonometric Functions Integrals Matrices Probability Relations and Functions Three Dimensional Geometry Determinants Application of Derivatives Probability Three Dimensional Geometry

19

Inverse Trigonometric Functions

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Probability / Probability Continuity and Differentiability Application of Derivatives Vector Algebra / Vector Algebra Differential Equations Determinants Application of Integrals Application of Derivatives Integrals / Integrals Continuity and Differentiability Application of Derivatives Integrals / Integrals Application of Integrals Relations and Functions Differential Equations / Differential Equations Continuity and Differentiability Matrices / Determinants Linear Programming / Linear Programming Three Dimensional Geometry / Three Dimensional Geometry

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score. Class 12

236

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BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

1.

Relations and Functions

2.

Inverse Trigonometric Functions

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

1(2)*

–

–

3(4)

–

–

1(5)*

2(6)

–

4(9)

3.

Matrices

2(2)#

4.

Determinants

1(1)

5.

Continuity and Differentiability

1(1)

1(2)*

2(6)#

6.

Application of Derivatives

1(1)*

2(4)

1(3)*

–

4(8)

7.

Integrals

2(2)#

1(2)*

1(3)

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)

1(2)

1(3)

–

3(6)

10.

Vector Algebra

3(3)#

–

–

–

3(3)

11.

Three Dimensional Geometry

1(4)

1(2)

–

1(5)*

3(11)

12.

Linear Programming

–

–

–

1(5)*

1(5)

1(2)

–

–

4(8)

10(20)

7(21)

3(15)

38(80)

13.

2(2)#

Probability Total

+ 1(4)

18(24)

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-19

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Suppose P and Q are two different matrices of order 3 × n and n × p, then find the order of the matrix P × Q. OR 3 4 7 1 2    Simplify :  + 4 2 2       9 2 1   5   

2. Write the general solution of differential equation 3. Prove that the function f (x ) = log(1 + x ) −

dy = ex+y. dx

2x is increasing on (–1, ∞). 2+x OR

Find the equation of the tangent to the curve y2 = 4ax at the point (at2, 2at). 4. Let A = {a, b, c} and R be the relation defined on A as follows : R = {(a, a), (b, c), (a, b)}. Write minimum number of ordered pairs to be added to R to make R reflexive and transitive. Class 12

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5. Evaluate :

x3 − x2 + x − 1 ∫ x − 1 dx

OR Evaluate : ∫ (4 x + 3x + 2 x + 4)dx 3

2

1 2   x   5  6. If     =   , then find the value of y. 2 1   y   4  14 1 , cos b = , then find cos g. 15 3   ɵ ɵ ɵ 8. Find the area of the parallelogram whose adjacent sides are a = i + 2 j + 3k and b = 3ɵi − 2 ɵj + kɵ .

7. If a, b, g are the direction angles of a vector and cos a =

OR         If a and b are two vectors such that a = 10, b = 15 and a ⋅ b = 75 2, then find the angle between a and b . 9. Two cards are drawn at random and one-by-one without replacement from a well-shuffled pack of 52 playing cards. Find the probability that one card is red and the other is black. OR When will be two events A and B independent? 10. Find the derivative of (4x3 – 5x2 + 1)4 w.r.t. to x. 11. Show that the relation R on the set {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3)} is reflexive but neither transitive nor symmetric. 2π

∫0

12. Find the value of

sin x dx.

13. Using determinants, find the area of triangle with vertices (2, –7), (1, 3), (10, 8). 





   14. If a × b + a ⋅ b = 400 and a = 5, then write the value of | b | . 2

2

15. Four cards are drawn successively without replacement from a deck of 52 cards. Find the probability that all the four cards are king. 16. If f(x) = [|x|], where [·] is the greatest integer function, then find f(–5/4). Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Suppose you visit to a hotel with your family and you observe that floor of a hotel is made up of mirror polished Kota stone. Also, there is a large crystal chandelier attached at the ceiling of the hotel. Consider the floor of the hotel as a plane having equation 3x – y + 4z = 2 and crystal chandelier as the point (3, –2, 1).

239

Mathematics

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Based on the above information, answer the following questions: (i) The d.r.’s of the perpendicular from the point (3, –2, 1) to the plane 3x – y + 4z = 2, is (a) (b) (c) (d) (ii) The length of the perpendicular from the point (3, –2, 1) to the plane 3x – y + 4z = 2, is 13 1 13 13 units (b) units (d) units (c) (a) 13 units 2 2 2 (iii) The equation of the perpendicular from the point (3, –2, 1) to the plane 3x – y + 4z = 2, is x − 3 y + 2 z −1 x − 3 y − 2 z −1 x + 3 y + 2 z −1 = = = = = = (b) (c) (d) None of these (a) 3 −1 4 3 −1 4 3 −1 4 (iv) The foot of the perpendicular drawn from the point (3, –2, 1) to the plane 3x – y + 4 z = 2, is 3 −3 −3 3 3 3 1 3 (a)  , , −1 (b)  , , −1 (c)  , ,−1 (d)  , , − 1 2 2   2 2  2 2  2 2  (v) The image of the point (3, –2, 1) in the given plane is (a) (0, 1, 3) (b) (0, –1, 3) (c) (0, 1, –3) (d) (0, –1, –3) 18. A factory has three machines A, B and C to manufacture bulbs. Machine A manufacture 25%, machine B manufacture 35% and machine C manufacture 40% of the bulbs respectively. Out of their respective outputs 5%, 4% and 2% are defective. A bulb is drawn at random from total production and it is found to be defective.

Based on the above information, answer the following questions : (i) Probability that defective bulb drawn is manufactured by machine A, is 41 25 16 (b) (c) (a) 69 69 69 (ii) Probability that defective bulb drawn is manufactured by machine B, is (a) 0.3

(b) 0.1

(c) 0.2

(d)

69 2000

(d) 0.4

(iii) Probability that defective bulb drawn is manufactured by machine C, is 16 17 25 42 (a) (b) (c) (d) 69 69 69 49 (iv) Probability that defective bulb is not manufactured by machine B, is 2 61 1 41 (a) (b) (c) (d) 69 69 7 69 (v) If a bulb is drawn at random, then what is the probability that bulb drawn is defective ? (a) 0.03 (b) 0.09 (c) 0.3 (d) 0.9

PART – B Section - III 19. If sin–1x + sin–1y + sin–1 z =

3π , then find the value of x2 + y2 + z2 + 2xyz. 2 Class 12

240

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20. The equation of the line in vector form passing through the point (−1, 3, 5) and parallel to line x −3 y −4 = ,z =2. 2 3 1 2

21. Discuss the continuity of the function f(x) at x = , when f(x) is defined as follows: 1 1  2 + x, 0 ≤ x < 2   1 f (x ) =  1, x= 2  3 1   2 + x , 2 < x ≤ 1

OR If y = ae2x + be–x, then show that

d2 y dx

22. Solve the differential equation

xe–y

2

−

dy − 2y = 0 . dx

dx + y dy = 0.

23. Evaluate : ∫ (1 − x )(2 + 3x )(5 − 4 x ) dx OR Evaluate : ∫

1 dx x 2 + 2 x + 10

24. Find the maximum value of slope of the curve y = – x3 + 3x2 + 12 x – 5. 25. Find the points on the curve y = x3 – 3x2 – 4x at which the tangent lines are parallel to the line 4x + y – 3 = 0. 26. Find the area enclosed between the curve y = x −1, the x-axis and the line x = 5. 27. Urn 1 contains 5 white balls and 7 black balls. Urn 2 contains 3 white and 12 black balls. A fair coin is flipped, if it is head, a ball is drawn from Urn 1, and if it is tail, a ball is drawn from Urn 2. Suppose that this experiment is done and a white ball was selected. What is the probability that this ball was in fact taken from Urn 2? 1

2  x 

28. If [2 x 3]  −3 0   3  = O, then find x.   

OR

1 3 3  1 0 0      Show that AB ≠ BA, where A = 1 4 3  , B = 0 2 0 1 3 4  0 0 3 Section - IV 29. Solve the differential equation : 30. Evaluate :

π /2

∫0

dy + y sec2 x = tan x sec2 x; y(0) = 1 dx

x dx sin x + cos x

 x 2 , x ≤ c 31. For what choices of a and b, the function f (x ) =  is differentiable at x = c ? ax + b, x > c OR x 3 2 Differentiate (e cos x sin x) w.r.t. x. 32. Let f : R → R be defined by f (x) = x + |x|. Show that f is neither one-one nor onto. 241

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33. Find the area enclosed between the circle x2 + y2 = 1 and the line x + y = 1 lying in the first quadrant. π 34. Find the equation of the normal to the curve y = 2 sin23x at x = . 6 OR Find the values of x for which the function f (x) = x3 + 12x2 + 36x + 6 is decreasing.  if x ≤ mx + 1, 35. If f (x ) =  sin x + n, if x > 

π π 2 is continuous at x = , then find the relation between m and n. π 2 2 Section - V

1 −1 2   −2 0 1     36. Use product 0 2 −3  9 2 −3 to solve the system of equations x + 3z = 9, –x + 2y – 2z = 4, 3 −2 4   6 1 −2

2x – 3y + 4z = –3. OR 2 0 1    If A = 2 1 3 , then find A2 – 5A + 4I and hence find a matrix X such that X = 5A – 4I – A2. 1 −1 0 37. Solve the following problem graphically. 1 Minimize Z = (1800000 + 30 x − 30 y ) 1000 subject to constraints: 0 ≤ x ≤ 15000 0 ≤ y ≤ 20000 15000 ≤ x + y ≤ 30000 OR Solve the following problem graphically. Maximize Z = x + y subject to constraints: 2x + 5y ≤ 100 x y + ≤1 25 40 x, y > 100  38. Find the vector equation of the plane passing through the intersection of the planes r ⋅ (ɵi + ɵj + kɵ ) = 6 and  r ⋅ (2ɵi + 3ɵj + 4kɵ ) = − 5 and the point (1, 1, 1). OR Find the vector and cartesian forms of the equation of the plane passing through the point (1, 2, –4) and   parallel to the lines r = ɵi + 2 ɵj − 4kɵ + λ(2 ɵi + 3ɵj + 6kɵ ) and r = ɵi − 3ɵj + 5kɵ + µ( ɵi + ɵj − kɵ ). INSTANT

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Class 12

242

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Self Evaluation Sheet Once you complete SQP-19, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Matrices / Matrices Differential Equations Application of Derivatives / Application of Derivatives Relations and Functions Integrals / Integrals Matrices Vector Algebra Vector Algebra / Vector Algebra Probability / Probability Continuity and Differentiability Relations and Functions Integrals Determinants Vector Algebra Probability Relations and Functions Three Dimensional Geometry Probability

19

Inverse Trigonometric Functions

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Three Dimensional Geometry Continuity and Differentiability / Continuity and Differentiability Differential Equations Integrals / Integrals Application of Derivatives Application of Derivatives Application of Integrals Probability Matrices / Matrices Differential Equations Integrals Continuity and Differentiability / Continuity and Differentiability Relations and Functions Application of Integrals Application of Derivatives / Application of Derivatives Continuity and Differentiability Determinants / Matrices Linear Programming / Linear Programming Three Dimensional Geometry / Three Dimensional Geometry

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

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SQP

20

BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

1.

Relations and Functions

2(2)

1(2)*

1(3)

–

4(7)

2.

Inverse Trigonometric Functions

1(1)

–

–

–

1(1)

3.

Matrices

2(2)

–

–

–

2(2)

4.

Determinants

1(1)*

1(2)

–

1(5)*

3(8)

5.

Continuity and Differentiability

1(1)*

1(2)

2(6)

–

4(9)

6.

Application of Derivatives

1(1)

2(4)

1(3)

–

4(8)

7.

Integrals

2(2)#

1(2)*

1(3)*

–

4(7)

8.

Application of Integrals

–

1(2)

1(3)

–

2(5)

9.

Differential Equations

1(1)

1(2)*

1(3)*

–

3(6)

10.

Vector Algebra

1(1) + 1(4)

–

–

–

2(5)

2(2)#

1(2)

–

1(5)*

4(9)

–

–

–

1(5)*

1(5)

1(2)

–

–

4(8)

10(20)

7(21)

3(15)

38(80)

11.

Three Dimensional Geometry

12.

Linear Programming

13.

2(2)#

Probability Total

+ 1(4)

18(24)

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-20

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks.

2.

Part-A has Objective Type Questions and Part-B has Descriptive Type Questions.

3.

Both Part-A and Part-B have internal choices.

Part - A : 1.

It consists of two Sections-I and II.

2.

Section-I comprises of 16 very short answer type questions.

3.

Section-II contains 2 case study-based questions.

Part - B : 1.

It consists of three Sections-III, IV and V.

2.

Section-III comprises of 10 questions of 2 marks each.

3.

Section-IV comprises of 7 questions of 3 marks each.

4.

Section-V comprises of 3 questions of 5 marks each.

5.

Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Find the number of discontinuous functions y(x) on [– 2, 2] satisfying x2 + y2 = 4. OR If y =

sin x + y , then find

dy . dx

 2 −2 2 2. If matrix A =   and A = pA, then find the value of p. − 2 2   3. Evaluate :

cot x

∫ 3 sin x dx OR 1

Evaluate : ∫ (x 2 + 2 x + 5) dx 0

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                4. a , b , c are three vectors, such that a + b + c = 0, a = 1, b = 2, c = 3, then find the value of a ⋅ b + b ⋅ c + c ⋅ a. 1 3 5. Given that, the events A and B are such that P(A) = , P ( A ∪ B) = and P(B) = p. Then find the value of 2 5 p, if A and B are mutually exclusive events. OR If E and F are event such that 0 < P(F) < 1, then show that P(E|F) + P(E|F) = 1. 6. Determine the degree and order of the differential equation 7. If

2x 8

d5 y dx 5

+ e dy /dx + y 2 = 0

5 6 −2 = , then find the value of x. x 7 3 OR

 −2 4  2 If A =   , then show that A is a null matrix. − 1 2   8. If x is real, then find the minimum value of f(x) = x2 – 8x + 17. 9. Find the shortest distance between the lines

x y z −a x y z +a . = = , = = m1 1 0 m2 1 0 OR

If the lines

x − 2 y − 9 z − 13 x − a y −1 z + 2 = = and = = are coplanar, then find the value of a. 1 2 3 1 3 −2

2 x + y 4 x   7 7 y − 13 10. If   , then find the values of x and y. =  5x − 7 4 x   y x + 6  π

11. Using properties of definite integrals, prove that

x tan x π2 . dx = ∫ sec x cosec x 4 0

12. If a line has the direction ratios 4, –12, 18, then find its direction cosines. 13. Find the value of tan–1(1) + tan–1(0) – tan–1 ( 3 ). 14. Let R be a relation on N defined by x + 2y = 8. Find the domain of R. 15. If = P ( A)

3 2 3 = , P ( B) and P ( A ∪ B) = , then find the value of P(B|A). 10 5 5

16. Find the number of reflexive relations on a set having 6 elements. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. Box A contain 1 white, 2 black and 3 red balls. Box B contain 2 white, 1 black and 1 red ball. Box C contain 4 white, 5 black and 3 red balls. One box is chosen at random and two balls are drawn with replacement. These happen to be one white and one red.

Class 12

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If E1, E2, E3 be the events that the balls drawn from box A, box B and box C respectively and E be the event that ball drawn are one white and one red. Based on the above information answer the following questions. (i) Probability that the ball drawn are one white and one red, given that the balls are from box A, is 1 2 3 1 (b) (c) (d) (a) 6 6 5 7 (ii) Probability that the ball drawn are one white and one red, given that the balls are from box B, is 1 1 3 3 (a) (b) (c) (d) 2 4 4 4 (iii) Probability that the ball drawn are one white and one red, given that balls are from box C, is 1 3 1 4 (a) (b) (c) (d) 12 11 6 11 3

(iv) The value of ∑ P (E | Ei ) is equal to i =1 5 2 1 7 (a) (b) (c) (d) 7 7 12 12 (v) The probability that the ball drawn are from box B, it is being given that the balls drawn are one white and one red, is 3 4 5 1 (a) (b) (c) (d) 7 7 7 7 18. A ship is pulled into harbour by two tug boats as shown in the figure. y 12 11 10 9 8 7 6 5 4 3 2 1 x'

O

A

B

C 1 2 3 4 5 6 7 8 9 10 11 12 y'

x

Based on the above information, answer the following questions : (i) Position vector of A is (a) 11i + 2 j

(b) 2i + 11j

(c) 2i − 11j

(ii) Position vector of B is (a) 4i + 4 j (b) 6i + 6 j (c) 7i + 7 j  (iii) Find the vector AC in terms of i , j . (b) −9 j (c) 9i (a) 9 j  (iv) If A = ɵi + 2 ɵj + 3kɵ , then its unit vector is 3i 2 j k 1i 2 j 3k 2i 3j k (a) (b) (c) + + + + + + 14 14 14 14 14 14 14 14 14   ɵ ɵ  ɵ ɵ  (v) If A = 2i + 3 j and B = 3i + 2 j , then | A | = | B | = ______ . (a)

5

(b)

7

(c)

11

(d) 11i − 2 j (d) 3i + 3j (d) None of these

(d) None of these

(d)

13 247

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PART - B Section - III 19. If y = 3cos(logx) + 4sin(logx), then show that x 2 dy = x log x . 20. Solve the differential equation dx

d2 y dx

2

+x

dy + y = 0. dx

OR

y  y  dy Find the solution of the differential equation x sin   = y sin   − x . x  x  dx 21. Find the maximum and the minimum values (if any) of f (x) = x2. 22. Find the area bounded by the curve y = x2 and the line y = 3x. 23. Prove that the inverse of an equivalence relation is also an equivalence relation. OR 13π   Using principal value, find the value of cos −1  cos . 6   24. The cartesian equation of a line is 3x + 1= 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios and also its vector equation. 25. Two balls are drawn one after another (without replacement) from a bag containing 2 white, 3 red and 5 blue balls. What is the probability that atleast one ball is red? 26. Find the interval in which f (x) = x3 – 3x2 – 9x + 2 is increasing. 2 3  27. If A =   , then find A–1. 1 5  28. Evaluate :

sin( x − α)

∫ sin(x + α) dx OR 2

Evaluate :

x − 3− x dx x(3 − x )

∫ 1+ 1

Section - IV

λ(x 2 + 2), if x ≤ 0 29. For what value of l, the function defined by f (x ) =  is continuous at x = 0? Hence check 4 x + 6 if x > 0 ,   the differentiability of f(x) at x = 0. 30. Draw the graph of y = |x + 1|and using integration, find the area below y = |x + 1|, above x - axis and between x = – 4 to x = 2. 31. Evaluate :

cos x

∫ (1 − sin x)(2 − sin x ) dx OR

Evaluate :

π/ 4

∫0

2 tan3 x dx

32. The perimeter of a triangle is 10 cm. If one of the side is 4 cm, then what are the two sides of the triangle for its maximum area? Class 12

248

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

33. If y = cot

−1 

 dy  , then find . dx + log x x 

x − log x x

 log e x 2 

2

34. Find the solution of the differential equation y − x

dy dy   = 2  1 + x 2  given that y = 1 when x = 1. dx dx  

OR Solve the following initial value problem : (xey/x + y)dx = xdy, y(1) = 1 35. Check whether the function f from the set of natural numbers to integers is defined by  n −1 , when n is odd  f (n) =  2 is one-one and onto or not.  − n , when n is even  2 Section - V  2 −1 1   3 1 −1     36. If A =  −1 2 −1 and B =  1 3 1  , then find the product AB and use this result to solve the following  −1 1 3   1 −1 2      system of linear equations 2x – y + z = –1, –x + 2y – z = 4 and x – y + 2z = –3. OR 1 1 1    If A = 1 0 2  , then find A–1. Hence, solve the system of equations x + y + z = 6, x + 2z = 7, 3x + y + z = 12. 3 1 1    37. Find the equation of plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line x + 3 y −3 z − 2 = = . 2 7 5 OR Find the image of the point (2, –1, 5) in the line

x − 11 y + 2 z + 8 = = . Also, find the equation of the line 10 −4 −11

joining the given point and its image. Find the length of that line segment also. 38. Determine graphically the minimum value of the objective function Z = 4x1 + 5x2 Subject to the constraints : 2x1 + x2 ≥ 7; 2x1 + 3x2 ≤ 15; x2 ≤ 3; x1, x2 ≥ 0. OR Solve the following LPP graphically. Maximize Z = x + 2y Subject to the constraints : x + 2y ≥ 100 2x – y < 0 2x + y ≤ 200 x, y ≥ 0. INSTANT

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Mathematics

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Self Evaluation Sheet Once you complete SQP-20, check your answers with the given solutions and fill your marks in the marks obtained column according to the marking scheme. Performance Analysis Table given at the bottom will help you to check your readiness. Q.No.

Chapter

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

Continuity and Differentiability / Continuity and Differentiability Matrices Integrals / Integrals Vector Algebra Probability / Probability Differential Equations Determinants / Matrices Application of Derivatives Three Dimensional Geometry / Three Dimensional Geometry Matrices Integrals Three Dimensional Geometry Inverse Trigonometric Functions Relations and Functions Probability Relations and Functions Probability Vector Algebra

19

Continuity and Differentiability

20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38

Differential Equations / Differential Equations Application of Derivatives Application of Integrals Relations and Functions / Inverse Trigonometric Functions Three Dimensional Geometry Probability Application of Derivatives Determinants Integrals / Integrals Continuity and Differentiability Application of Integrals Integrals / Integrals Application of Derivatives Continuity and Differentiability Differential Equations / Differential Equations Relations and Functions Determinants / Determinants Three Dimensional Geometry / Three Dimensional Geometry Linear Programming / Linear Programming

Marks Per Question

Marks Obtained

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 4 × 1 4 × 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 5 5 5 80

Total

Percentage

.............. ..............%

Performance Analysis Table > 90%

You are done! Keep on revising to maintain the position.

81-90%

You have to take only one more step to reach the top of the ladder. Practise more.

71-80%

A little bit of more effort is required to reach the ‘Excellent’ bench mark.

61-70%

Revise thoroughly and strengthen your concepts.

51-60%

Need to work hard to get through this stage.

40-50%

Try hard to boost your average score.

More ebooks at https://telegram.me/unacademyplusdiscounts

CBSE S Q P

2020-21 BLUE PRINT

Time Allowed : 3 hours S. No.

Maximum Marks : 80

Chapter

1.

Relations and Functions

2.

Inverse Trigonometric Functions

VSA / Case based (1 mark)

SA-I (2 marks)

SA-II (3 marks)

LA (5 marks)

Total

3(3)#

–

1(3)

–

4(6)

–

1(2)

–

–

1(2)

–

–

–

2(2)

1(2)*

–

1(5)*

3(8)

–

1(2)

2(6)#

–

3(8)

3.

Matrices

2(2)#

4.

Determinants

1(1)

5.

Continuity and Differentiability

6.

Application of Derivatives

1(4)

1(2)

1(3)

–

3(9)

7.

Integrals

1(1)*

1(2)*

1(3)

–

3(6)

8.

Application of Integrals

1(1)

1(2)

1(3)*

–

3(6)

9.

Differential Equations

1(1)*

1(2)

1(3)

–

3(6)

10.

Vector Algebra

3(3)

1(2)

–

–

4(5)

11.

Three Dimensional Geometry

2(2)

1(2)

–

1(5)*

4(9)

12.

Linear Programming

–

–

–

1(5)*

1(5)

13.

Probability

2(2) + 1(4)

1(2)*

–

–

4(8)

18(24)

10(20)

7(21)

3(15)

38(80)

Total

*It is a choice based question. #Out of the two or more questions, one/two question(s) is/are choice based.

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Subject Code : 041

SQP-21

MATHEMATICS Time allowed : 3 hours

Maximum marks : 80

General Instructions : 1. 2. 3.

This question paper contains two parts A and B. Each part is compulsory. Part-A carries 24 marks and Part-B carries 56 marks. Part-A has Objective Type Questions and Part-B has Descriptive Type Questions. Both Part-A and Part-B have internal choices.

Part - A : 1. It consists of two Sections-I and II. 2. Section-I comprises of 16 very short answer type questions. 3. Section-II contains 2 case study-based questions. Part - B : 1. It consists of three Sections-III, IV and V. 2. Section-III comprises of 10 questions of 2 marks each. 3. Section-IV comprises of 7 questions of 3 marks each. 4. Section-V comprises of 3 questions of 5 marks each. 5. Internal choice is provided in 3 questions of Section-III, 2 questions of Section-IV and 3 questions of Section-V. You have to attempt only one of the alternatives in all such questions.

PART - A Section - I 1. Check whether the function f : R → R defined as f(x) = x3 is one-one or not. OR How many reflexive relations are possible in a set A whose n(A) = 3 ? 2. A relation R in S = {1, 2, 3} is defined as R = {(1, 1), (1, 2), (2, 2), (3, 3)}. Which element(s) of relation R be removed to make R an equivalence relation? 3. A relation R in the set of real numbers R defined as R = {(a, b) : a = b} is a function or not. Justify OR An equivalence relation R in A divides it into equivalence classes A1, A2, A3. What is the value of A1 ∪ A2 ∪ A3 and A1 ∩ A2 ∩ A3. 4. If A and B are matrices of order 3 × n and m × 5 respectively, then find the order of matrix 5A – 3B, given that it is defined. 1 if i ≠ j 5. Find the value of A2, where A is a 2 × 2 matrix whose elements are given by aij =  . 0 if i = j OR Given that A is a square matrix of order 3 × 3 and |A| = –4. Find |adj A|. Class 12

252

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6. Let A = [aij] be a square matrix of order 3 × 3 and |A|= –7. Find the value of a11 A21+ a12A22+ a13 A23 where Aij is the cofactor of element aij. x 2 7. Find ∫ e (1 − cot x + cos ec x )dx.

OR π 2

Evaluate :

∫ −

x 2 sin xdx

π 2

8. Find the area bounded by y = x2, the x- axis and the lines x = –1 and x =1. 9. How many arbitrary constants are there in the particular solution of the differential equation dy = −4 xy 2 ; y(0) = 1? dx OR For what value of n is the following a homogeneous differential equation: dy x 3 − yn = dx x 2 y + xy 2 10. Find a unit vector in the direction opposite to −

3^ j. 4 ^

^

11. Find the area of the triangle whose two sides are represented by the vectors 2 i and − 3 j . ^ ^ ^ 12. Find the angle between the unit vectors a^ and b , given that | a + b | = 1.

13. Find the direction cosines of the normal to YZ plane. 14. Find the coordinates of the point where the line

x + 3 y −1 z − 5 cuts the XY plane. = = −1 3 −5

1 1 and respectively. If both of them try 3 4 to solve the problem independently, what is the probability that the problem is solved?

15. The probabilities of A and B solving a problem independently are

16. The probability that it will rain on any particular day is 50%. Find the probability that it rains only on first 4 days of the week. Section - II Case study-based questions are compulsory. Attempt any 4 sub parts from each question. Each sub-part carries 1 mark. 17. An architect designs a building for a multi-national company. The floor consists of a rectangular region with semicircular ends having a perimeter of 200 m as shown below: Based on the above information answer the following: (i) If x and y represents the length and breadth of the rectangular region, then the relation between the variables is (a) x + π y = 100 (b) 2x + π y = 200

y x

(c) π x + y = 50

(d) x + y = 100 253

Mathematics

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(ii) The area of the rectangular region A expressed as a function of x is 1 2 x (a) (b) (c) (100 x − x 2 ) (100 x − x 2 ) (100 − x ) π π π (iii) The maximum value of area A is 3200 2 5000 2 p (a) (b) (c) m2 p m p m 3200

2 (d) πy 2 + (100 x − x 2 ) π

(d)

1000 2 p m

(iv) The CEO of the multi-national company is interested in maximizing the area of the whole floor including the semi-circular ends. For this to happen the value of x should be (a) 0 m (b) 30 m (c) 50 m (d) 80 m (v) The extra area generated if the area of the whole floor is maximized is 5000 2 3000 2 (b) (a) p m p m 7000 2 (d) No change, Both areas are equal p m 18. In an office three employees Vinay, Sonia and Iqbal process incoming copies of a certain form. Vinay process 50% of the forms, Sonia processes 20% and Iqbal the remaining 30% of the forms. Vinay has an error rate of 0.06, Sonia has an error rate of 0.04 and Iqbal has an error rate of 0.03. (c)

Based on the above information answer the following: (i) The conditional probability that an error is committed in processing given that Sonia processed the form is (a) 0.0210 (b) 0.04 (c) 0.47 (d) 0.06 (ii) The probability that Sonia processed the form and committed an error is (a) 0.005 (b) 0.006 (c) 0.008 (d) 0.68 (iii) The total probability of committing an error in processing the form is (a) 0 (b) 0.047 (c) 0.234 (d) 1 (iv) The manager of the company wants to do a quality check. During inspection he selects a form at random from the days output of processed forms. If the form selected at random has an error, the probability that the form is NOT processed by Vinay is 30 17 20 (c) (d) (a) 1 (b) 47 47 47 (v) Let A be the event of committing an error in processing the form and let E1, E2 and E3 be the events that 3

Vinay, Sonia and Iqbal processed the form. The value of

∑ P(Ei | A) is i =1

(a) 0

(b) 0.03

(c) 0.06

(d) 1

PART - B Section - III π  cos x  −3π 19. Express tan −1  , < x < in the simplest form.   1 − sin x  2 2 20. If A is a square matrix of order 3 such that A2 = 2A, then find the value of |A|. OR  3 1 2 –1 If A =   , show that A – 5A + 7I = O. Hence find A . 1 2 −   Class 12

254

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21. Find the value(s) of k so that the following function is continuous at x = 0. 1 − cos kx if x ≠ 0  f (x ) =  x sin x 1 if x = 0  2 1 22. Find the equation of the normal to the curve y = x + , x > 0 perpendicular to the line 3x − 4y = 7. x 1 dx. 23. Find ∫ 2 2 cos x(1 − tan x )

OR 1

Evaluate : ∫ x(1 − x )n dx 0

24. Find the area of the region bounded by the parabola y2 = 8x and the line x = 2. 25. Solve the following differential equation: dy = x 3 cosec y, given that y(0) = 0 dx 26. Find the area of the parallelogram whose one side and a diagonal are represented by coinitial vectors ^

^ ^

^

^

i − j + k and 4 i + 5 k respectively.

^  27. Find the vector equation of the plane that passes through the point (1, 0, 0) and contains the line r = λ j .

28. A refrigerator box contains 2 milk chocolates and 4 dark chocolates. Two chocolates are drawn at random. Find the probability distribution of the number of milk chocolates. What is the most likely outcome? OR Given that E and F are events such that P(E) = 0.8, P(F) = 0.7, P(E ∩ F) = 0.6. Find P (E | F ). Section - IV 29. Check whether the relation R in the set Z of integers defined as R = {(a, b) : a + b is “divisible by 2”} is reflexive, symmetric or transitive. Write the equivalence class containing 0, i.e., [0]. 2 dy 30. If y = e x sin x + (sin x )x , find . dx

31. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 2 is not differentiable at x = 1. OR If x = a secq, y = b tanq, find

d2 y dx

2

at θ =

π . 6

 π 32. Find the intervals in which the function f given by f(x) = tanx – 4x, x ∈  0,  is  2 (a) strictly increasing (b) strictly decreasing 33. Find ∫

x2 + 1 (x 2 + 2)(x 2 + 3)

dx.

34. Find the area of the region bounded by the curves x2 + y2 = 4, y = 3x and x-axis in the first quadrant. OR Find the area of the ellipse x2 + 9y2 = 36 using integration. 255

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35. Find the general solution of the following differential equation: xdy – (y + 2x2)dx = 0 Section - V 1 2 0   36. If A =  −2 −1 −2 , find A–1. Hence solve the system of equations;  0 −1 1  x – 2y = 10 2x – y – z = 8 –2y + z = 7

OR 1 −1 0   2 2 −4      Evaluate the product AB, where A = 2 3 4  and B =  −4 2 −4  0 1 2   2 −1 5  Hence solve the system of linear equations x–y=3 2x + 3y + 4z = 17 y + 2z = 7   ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ 37. Find the shortest distance between the lines r = 3 i + 2 j − 4 k + λ( i + 2 j + 2 k ) and r = 5 i − 2 j + µ(3 i + 2 j + 6 k ). If the lines intersect, then find their point of intersection.

OR Find the foot of the perpendicular drawn from the point (–1, 3, –6) to the plane 2x + y – 2z +5=0. Also find the equation and length of the perpendicular. 38. Solve the following linear programming problem (L.P.P) graphically. Maximize Z = 3x + y subject to constraints ; x + 2y ≥ 100 2x – y ≤ 0 2x + y ≤ 200 x, y ≥ 0 OR The corner points of the feasible region determined by the system of linear constraints are as shown below: Answer each of the following: (i) Let Z =3x – 4y be the objective function. Find the maximum and minimum value of Z and also the corresponding points at which the maximum and minimum value occurs. (ii) Let Z = px + qy ,where p, q > 0 be the objective function. Find the condition on p and q so that the maximum value of Z occurs at B(4,10) and C(6, 8). Also mention the number of optimal solutions in this case.

Y 11 10 9 A(0, 8) 8 7 6 5 4 3 2 1 O

B(4, 10) C(6, 8)

D(6, 5)

E(4, 0)

X

1 2 3 4 5 6 7

Class 12

256

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SOLUTIONS 1. Let f(x1) = f(x2) for some x1, x2 ∈ R. ⇒ (x1)3 = (x2)3 ⇒ x13 – x23 = 0 ⇒ (x1 – x2)(x12 + x22 + x1x2) = 0 2 2   x  x  x  ⇒ (x1 − x2 )  x12 + 2 x1 ⋅  2  +  2  −  2  + x22  = 0  2 2 2  2  x  3x 2  ⇒ (x1 − x2 )   x1 + 2  + 2  = 0   2  4 

⇒ x1 = x2, hence f(x) is one-one. OR Number of reflexive relations on a set having n elements = 2n(n – 1) So, required number of reflexive relations = 23(3 – 1) = 26 2. We have, R = {(1, 1), (2, 2), (3, 3), (1, 2)} which is reflexive and transitive. For R to be symmetric (1, 2) should be removed from R. 3. Since a is not defined for a ∈ (–∞, 0) \ a = b is not a function. OR As we know that, union of all equivalence classes of a set is the set itself. \ A1 ∪ A 2 ∪ A 3 = A Also, A1 ∩ A2 ∩ A3 = f [Q Equivalence classes are either equal or disjoint] 4. For addition or subtraction of two matrices to be defined, the two matrices should be of same order. \ 3 × n = m × 5 ⇒ m = 3 and n = 5 So, order of matrix (5A – 3B) is 3 × 5. 1, if i ≠ j 5. Given, aij =  0, if i = j 0 1  0 1   1 0  0 1  ⇒ A2 =  ∴ A=  =   1 0  1 0   0 1  1 0  OR We know, |adjA| = |A|n – 1, where n × n is the order of non-singular matrix A. \ |adj A| = (–4)3 – 1 = 16 6. We know that, if elements of a row are multiplied with cofactors of any other row, then their sum is 0. \ a11A21 + a12A22 + a13A23 = 0. 7. Let I = ∫ e x (1 − cot x + cosec2 x ) dx = ∫ e x dx + ∫ e x ((− cot x ) + cosec2 x ) dx

= e x + e x (− cot x ) + C  ∫ e x ( f (x ) + f ′(x ))dx = e x f (x ) + C    = ex(1 – cotx) + C OR Q f(x) = x2sinx is an odd function. π /2

∴

2 ∫ x sin xdx = 0

− π /2 1

8. Required area, A = ∫ x 2 dx −1 1

1 2 2 ⇒ A = 2 ∫ x 2dx =  x 3  = sq. units   0 3 3 0

9. There is no arbitrary constant in a particular solution of differential equation. OR For n = 3, the given differential equation becomes homogeneous.  10. Let a be the unit vector in the direction opposite  3  to the given vector  − ^j  .  4   Then, a =

 3 ^ ^  − 4 j  = j

−1 3  4 

2

11. Area of the triangle 1 ^ 1 ^ ^ = | 2 i × (−3) j | = | −6 k | = 3 sq. units 2 2 ^

^

12. We have, | a + b |2 = 1 ^ ^ ⇒ a^2 + b 2 + 2 a^ ⋅ b = 1 ^

⇒ 2 a^ ⋅ b = 1 − 1 − 1

^

(| a^ | = | b | = 1)

−1  1 ^ −1 ^ ⇒ a^ ⋅ b = ⇒ | a^ || b | cos θ = ⇒ θ = cos −1  −   2 2 2 π 2π ⇒ θ = π− ⇒ θ= 3 3 13. Since, normal to the YZ plane is a line parallel to X-axis. \ Direction cosines of normal to YZ plane are 〈1, 0, 0〉. 14. The given line is x + 3 y −1 z − 5 = = −1 3 −5 257

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Now, as given line cuts XY-plane. So, z-coordinate of point of intersection = 0. y −1 x +3 ⇒ = 1 and = 1 ⇒ x = 0, y = 0 −1 3 \ Required point is (0, 0, 0).

dP 2x 40000 − 4 x 2 10000 − x 2 ∴ =− = dx π 4π π dP For maximum or minimum, =0 ⇒ x=0 dx

15. Required probability = 1 – P(problem is not solved) 2 3 1 = 1 − P ( A) ⋅ P (B ) = 1 − × = 3 4 2

−2