IB Prepared IB Diploma Programme Mathematics Analysis and Approaches - Answers [1 ed.] 1382007221, 9781382007221

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M AT H E M ATI CS :

A N A LYSIS

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P R O G R A M M E

Paul Belcher

Ed Kemp

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IB PREPARED MAA WORKED SOLUTIONS TO THE END-OF-CHAPTER PRACTICE QUESTIONS Here are the worked solutions to the end-of-chapter practice questions from IB Prepared MAA. For direct access, click on the name of the chapter.

1 Number and algebra 2 Functions 3 Geometry and trigonometry 4 Statistics and probability 5 Calculus

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1

1 NUMBER AND ALGEBRA: END-OF-CHAPTER PRACTICE QUESTION WORKED SOLUTIONS SL Paper 1: Section A No technology allowed 1. a. u1 + 2d = 4, u1 + 5d = 19 ⇒ 3d = 15 ⇒ d = 5 b. u1 + 10 = 4 ⇒ u1 = −6 5 2

c. S5 = (2 × –6 + 4 × 5) = 20 2. a. u1r2 = 3, u1r5 = −24 ⇒ r3 = −8 ⇒ r = −2

3 4 5 3 ((–2) –1) 33 c. S5 = = 4 4 (–2 –1)

b. u1 (−2)2 = 3 ⇒ u1 =

3. Same 2nd term ⇒ 1 × r = 0 + d ⇒ r = d

1 × r3 = 0 + 9d ⇒ r3 = 9r  (because r = d)

r2 = 9 ⇒ r = ±3 r = 3, d = 3 or r = −3, d = −3 1 4 4. a. u1 = 5u1 ⇒ 1 – r = ⇒ r =

5 5 1 –r u1 2 b. = u1r ⇒ 1 = r(1 − r) ⇒ r − r + 1 = 0 1 –r This quadratic has negative discriminant and thus no real solutions, showing that this is not possible.

5. a. 24+3−6 = 21 = 2 b. 3−9+7+1 = 3−1 = c. 26 × 1 = 64

1 3

6. a. log 52 + log 4 = log 25 + log 4

= log (25 × 4)



= log 100



= log 102 = 2

b. ln (7 × 3) − ln 21 = ln 21 − ln 21 = 0 c. e(ln 7 + ln 3)–ln 21 = eln 21 – ln 21 = e0 = 1

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log 3 log 4 log 5 log 2 × × × =1 log 2 log 3 log 4 log 5

7. a.

b. log2 x + 2 logx 2 = 3 ⇒ log2 x +

2 =3 log 2 x

Letting log2 x = y 2 y + = 3 ⇒ y2 – 3y + 2 = 0 ⇒ (y – 2)(y – 1) = 0 ⇒ y = 1 or 2 y log2 x = 1 ⇒ x = 2, log2 x = 2 ⇒ x = 4

8. 4 C0 14 (–2x)0 + 4 C1 13 (–2x)1 + 4 C2 12 (–2x)2 + 4 C3 11 (–2x)3 + 4 C4 10 (–2x)4

= 1+ 4(–2x)+6(–2x)2 + 4(–2x)3 +1(–2x)4



= 1 − 8x + 24x 2 − 32x 3 + 16x 4

9. a. LHS ≡ (

x + 1)( x − 1) + x ( x − 1) + x ( x + 1) x ( x + 1)( x − 1) 2 2 x − 1 + x − x + x2 + x ≡ x(x 2 − 1) 2 3x − 1 ≡ 3 ≡ RHS x −x 3x 2 − 1 1 1 b. 3 = 0 ⇒ 3x 2 − 1 = 0 ⇒ x = or − x −x 3 3

SL Paper 2: Section A Technology required 10. a. V =

4π(8.5 × 10−16 )3 = 2.57 × 10−45 m 3 (3 sf) 3

b. r=3

3V 3 3 × 1.412 × 1027 = = 6.96 × 108  m m (3 sf) 4π 4π

11. a. Conjecture could be an arithmetic series with u1 = 3, d = 4

Re-check the 14 and 21 markings, expecting them to be 15 and 23. n b. Require (2 × 3 + (n − 1)4) = 253 ⇒ 2n2 + n − 253 = 0 2 Solving quadratic (and rejecting negative answer) n = 11

12. a. Require 500( 1.07 )n = 700

Solving (e.g. graph, table, logs) gives n = 4.97 ⇒ 5 years

b. The number of vouchers forms an arithmetic series with u1 = 1, d = 1 so S5 =

5 ( 1 + 5) = 15 2

500(1.07)5 − 700 = £1.28 (2 dp) c.

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2 13. a. 1 + 20 C1 ( 3x ) + 20 C2 ( 3x ) + ...

= 1 + 20 × 3x + 190 × 9x 2 + ... = 1 + 60x + 1710x 2 + ...

b. Term will be 2 × 60x + 1 × 1710x 2 = 1830x 2 r

2 14. a. General term is of the form Cr x ⎛⎜⎝ ⎞⎟⎠ x This simplifies to 6 Cr 2 r x 6−2r ⇒ power of x is 6 − 2r 6

6−r

6 − 2r = 4 ⇒ r = 1

⎛ 2⎞ The term in x4 is 6 C1 x 5 ⎜ ⎟ = 12x 4 ⎝ x⎠

6 − 2r = 0 ⇒ r = 3 b. 3



⎛ 2⎞ The constant term is 6 C3 x 3 ⎜ ⎟ = 20 × 8 = 160 ⎝ x⎠

HL Paper 1: Section A No technology allowed 15. Let P(n) be the statement that 2 n+1 + 32n−1 is exactly divisible by 7.

n = 1 ⇒ 2 2 + 3 = 7, so P(1) is true



Assume that P(k) is true so that 2 k+1 + 32 k−1 = 7s for s ∈!

Hence 2 k+1 = 7s − 32 k−1

Now consider P(k + 1):



2 k+1+1 + 32(k+1)−1 = 2 × 2 k+1 + 32 k+1 = 2 × ( 7s − 32 k−1 ) + 32 k+1



= 7 × 2s + 32 k−1 ( −2 + 9 )



= 7 ( 2s + 32 k−1 )

Since s ∈!, it follows that 2s + 32 k−1 ∈! and hence P(k + 1) is true. Since P(1) is true and P(k) true implies P(k + 1) true, by the principle of mathematical induction we have that P(n) is true for all n ∈!+ 4 ± 16 − 80 4 ± −64 4 ± 8i = = = 2 ± 4i 2 2 2 b. Let w + 1 = z 20 20 Then w − 3 + = 0⇒ z− 4+ = 0 ⇒ z 2 − 4z + 20 = 0 w+1 z So w = 1 ± 4i

16. a. z =

z z∗ z2 1+ ∗ 2 (z ) 2zz∗ = ∗ 2 2   (z ) + z 2w = 17. 1 + w2

2

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Letting z = a + bi 2w 2(a 2 + b 2 ) = 2 2 1+ w (a − 2abi − b 2 )(a 2 + 2abi − b 2 ) 2(a 2 + b 2 ) (a 2 + b 2 )2 2 = 2 2 (a + b ) =



This is purely real, giving the required result.

HL Paper 1: Section B No technology allowed 18. a. 2x 2 + 5x + 3 = ( 2x + 3 )( x + 1) 2x + 1 A B ≡ + 2x + 5x + 3 2x + 3 x + 1 2

⇒ 2x + 1 ≡ A(x + 1) + B(2x + 3) Equating coefficients: ⇒ 2 = A + 2B, 1 = A + 3B

Solving simultaneously: A = 4, B = −1 2x + 1 4 −1 ≡ + 2 2x + 5x + 3 2x + 3 x + 1

b. f ( x ) = 4( 2x + 3 )−1 − ( x + 1)−1

i. f ′(x) = −8(3x + 3)−2 + (x + 1)−2 4 1 ⎞ ii. ∫ f (x)dx = ∫ ⎛⎜ − dx = 2 ln 2x + 3 − ln x + 1 + c ⎝ 2x + 3 x + 1 ⎟⎠

19. p ( 2 − i ) + q ( 1 + i ) = 2 + 3i × −i ⇒ (2p + q) + (q − p)i = 3 − 2i 3+i i −i ( 1 + i )( 2 − i )

(2p + q) + (q − p)i = (3 + i)(3 − 2i) = 11 − 3i



Equating real and imaginary parts: 2p + q = 11, q − p = −3



Solving simultaneously gives: p =

14 5 ,q= 3 3

20. a. i. (z + w)∗ = ((a + c) + ( b + d ) i ) = (a + c) − (b + d)i = (a − bi) + (c − di) = z∗ + w∗ ∗

ii. (z − w)∗ = ((a − c) + (b − d)i ) = (a − c) − (b − d)i = (a − bi) − (c − di) = z∗ − w∗ ∗

iii. (zw)∗ = ((ac − bd) + (bc + ad)i ) = (ac − bd) − (bc + ad)i = (a − bi)(c − di) = z∗w∗ ∗



∗ ∗ z r r r cis(−θ ) z∗ b. i. ⎛⎜ ⎞⎟ = ⎛⎜ cis(θ − φ )⎞⎟ = cis(−θ + φ ) = = ∗

⎝ w⎠

ii. ( z

⎝s

⎠

) = ((r cisθ ) ) = ( r

n ∗

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n ∗

s

n

s cis(−φ )

w

cis nθ ) = r cis(−nθ ) = ( r cis(−θ )) = ( z∗ ) ∗

n

n

n

5

21. Let z = r cisθ 3



3π z 4 = 2 2 + 2 2 = 8 = 2 2 ;  z 4 is in the 2nd quadrant so its arg is 4 3 3π (r cisθ )4 = r 4 cis 4θ = 2 2 cis 4 Equating moduli and arguments: 3

3



r4 = 22 ⇒ r = 28 3π 3π nπ 4θ = + 2nπ, n ∈! ⇒ θ = + 4 16 2 Giving the four roots as



z = 2 8 cis



3

−5π 38 −13π 3π 38 11π 38 , 2 cis , 2 cis , 2 cis 16 16 16 16

22. a. Proof by contradiction.  Suppose  can be ordered.

Then either i = 0 or i < 0 or i > 0.

Case i. i = 0 ⇒ i 2 = 0 ⇒ −1 −1 ==00 (a contradiction) Case ii. i < 0 then using a property of ordering i × i > i × 0 ⇒ −1 > 0 (a contradiction) Case iii. i > 0 then using a property of ordering i × i > i × 0 ⇒ −1 > 0 (a contradiction) All three cases lead to a contradiction and so we conclude that  cannot be ordered.

b. i. no sense

ii. sense, 3 > 5 false

iii. sense, 5 > 13 false

4 12 4 12 iv. sense, arctan > arctan ⇒ > false 3 5 3 5 v. no sense ⎛ 1 −2 3 23. a. ⎜⎜ 2 1 2 ⎜⎝ 7 −4 13

9 11 λ



row 2 – 2 × row1



row 3 – 7 × row1

⎞ ⎟ ⎟ ⎟⎠

9 ⎞ ⎛ 1 −2 3 ⎜ 0 5 −4 − 7 ⎟ ⎜ ⎟ ⎜⎝ 0 10 −8 λ − 63 ⎟⎠

9 ⎞ ⎛ 1 −2 3 ⎜ 0 5 −4 − 7 ⎟ ⎜ ⎟ ⎜⎝ 0 0 0 λ − 49 ⎟⎠ row 2 – 2 × row1 So, to be consistent λ = 49

b. Let z = t

5y − 4t = −7 ⇒ y =

−7 4 + t 5 5

31 −7 ⎛ −7 4 ⎞ x − 2y + 3t = 9 ⇒ x = 2 ⎜ + t ⎟ − 3t + 9 ⇒ x = + t ⎝ 5 5 ⎠ 5 5 7 31 y+ 5 =z 5 = −7 4 5 5

x−

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HL Paper 2: Section A Technology required 24. a. 8 C4 = 70

b. 3 C2 × 5 C2 = 30

25. a. There are 4 vowels so 4 possibilities for the first letter.

This leaves 6 letters which can be arranged in 6! ways,



so the answer is 4 × 6! = 2880



b. The 4 vowels can be arranged in 4! ways.



The 3 constants can be arranged in 3! ways.



It can either be the vowels first or the constants first (2 arrangements)



so the answer is 2 × 4! × 3! = 288

26. a. ( 1 + 2x )

−

1 2

−1 −3 −1 −3 −5 × × × −1 2 2 2 2 2 (2x)3 + … = 1 + (2x) + (2x) + 2 2 1× 2 1× 2 × 3

3 5     = 1 − x + x2 − x3 + … 2 2 1 1 1 3 5 9 ! 1− + − = b. Putting x = 2 2 8 16 16 2 1 9 − 16 2 × 100% = 20.5% (3 sf) c. 1 2

27. a. p Ci =

p! (p − 1)! = p× i!(p − i)! i!(p − i)!

All of the prime factors of both i! and (p – i)! must be less than p since 1 ≤ i ≤ p – 1.

Hence, none of these prime factors can divide p. (p − 1)! p Ci is an integer so is also an integer showing that p exactly  i!(p − i)! divides p C i



b. Using part (a), the first row in Pascal’s triangle that is worth considering is n = 9

9 C3 = 84 which is not exactly divisible by 9.

28. a. x = 2, y = 3, z = 5

b. No solutions



c. x = 27 − 5t, y = −12 + 3t, z = t  

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x − 27 y + 12 = =z −5 3

7

HL Paper 3: Technology required 29. a. i. x + y = 3.551 which to 3 significant figures is 3.55. ii. x = 1.235 → 1.24 and y = 2.316 → 2.32 so adding gives 3.56 iii. The two answers are different. Adding and then rounding is not the same as rounding and then adding.

iv. Method (i) is better, as it is closer to the true answer.

b. i. ε a+b = (A + B) − (a + b) = (A − a) + (B − b) = ε a + ε b

ii. ε a−b = (A − B) − (a − b) = (A − a) − (B − b) = ε a − ε b iii. The absolute maximum error is ε a + ε b in both cases.

c. i. ε a×b = AB − ab

Now AB = (a + ε a )(b + ε b ) = ab + aε b + bε a + ε aε b and ε aε b is small so ε ab ! (ab + aε b + bε a ) − ab = aε b + bε a ii. This looks similar to the product rule. iii. The absolute maximum error is a ε b + b ε a



A a a + ε a a ab + bε a − ab − aε b bε a − aε b bε − aε − = − = = so ε a ! a 2 b b(b + ε b ) b(b + ε b ) B b b + εb b b b

d. i. ε a = b

ii. This looks similar to the quotient rule. a εb + b ε a iii. The absolute maximum error is b2

e. i. ra×b =

ε ab ε ε a εb + b ε a = = a + b = ra + rb ab ab a b

εa ii. ra =

b

b

a b

=

a εb + b ε a b

2

×

b ε a εb = + = ra + rb a a b

They are added in both cases.

f. i. An = (a + ε a )n = a n + na n−1ε a + (terms that can be ignored)



So, the error ε an is ε an = A n − a n = (a n + na n−1ε a ) − a n = na n−1ε a

ii. The absolute relative error is ran =

na n−1ε a a

n

=n

εa = nra a

(which agrees with earlier work since an can be considered as repeated multiplication)

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2 FUNCTIONS: END-OF-CHAPTER PRACTICE QUESTION WORKED SOLUTIONS SL Paper 1: Section A No technology allowed 1. a. f ′(x) =

a(x + c) − (ax + b) ac − b = (x + c)2 (x + c)2

b. i. ac − b > 0

ii. ac − b < 0 iii. ac − b = 0 b = ac ⇒ f (x) = c.

ax + ac a(x + c) = = a , x ≠ −c x+c x+c

x+2 y+2    inverse given by x = x−1 y−1 xy − x = y + 2 ⇒ y(x − 1) = x + 2

2. a. y =

y =



x+2 x+2 so f −1 (x) = x−1 x−1

b. i. Domain {x ∈! x ≠ 1}

ii. Domain {x ∈! x ≠ 1}

{ } Range { y ∈! y ≠ 1}

Range y ∈! y ≠ 1

( f ! f )(x) = x , x ≠ 1 since f is self-inverse c.

3. a. x + c = 0 ⇒ x = −c so c = −3 As x → ± ∞, y → a so a = 4 4x + b = 0 ⇒ x =

b. x=0⇒y =

−b −b = 2 ⇒ b = −8 so 4 4

−8 ⎛ intercept on y-axis is ⎜ 0, ⎝ −3

8⎞ ⎟ 3⎠

4. a.  x-axis being a tangent implies the function has one repeated root so 49 8 2 49 49 ⎞ 7⎞ ⎛ 2 7 ⎛ 2 b. 2x + 7x + = 0 ⇒ 2⎜ x + x + ⎟ = 0 ⇒ ⎜ x + ⎟ = 0 ⎝ ⎝ 8 2 16 ⎠ 4⎠ 7 x=− 4 b 2 − 4ac = 49 − 8c = 0 ⇒ c =

5. Let u = x  u2 − 7u + 10 = 0 ⇒ (u − 2)(u − 5) = 0

u = 2 or 5



x = u2 ⇒ x = 4 or 25

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6. Discriminant of (x2 + 6x + 4) = 0 is 36 – 16 which is positive and so the equation has 2 real roots. Discriminant of (x2 + 6x + 10) = 0 is 36 – 40 which is negative and so the equation has no real roots.

Hence, two real roots in total. −b −8 = = −1 2a 2 × 4 b.  The concave up quadratic 4x2 + 8x + (b – 10) cannot have any zeros so 64 – 16(b – 10) < 0 ⇒ b > 14

7. a. Discriminant = 64 – 16a = 0 ⇒ a = 4, the root is



(or by considering a vertical translation of the quadratic in part a.)

8. a. Considering the line of symmetry, h = –2 and then k = –16 x = 0 ⇒ a (–2)2 –16 = 4 ⇒ a = 5 16 ±4 b. 5(x − 2)2 − 16 = 0 ⇒ (x − 2)2 = ⇒ x − 2 = 5 5 4 5 x=2± 5

SL Paper 1: Section B No technology allowed 5−3 1 = 5−1 2 1 5 1 5 1 The line y = x + c through (1, 3) gives 3 = + c ⇒ c = ⇒ y = x + 2 2 2 2 2

9. a. Gradient is



b.  Gradient is – 2  The line is y = –2x + d through (5, 5) gives 5 = –10 + d ⇒ d = 15

⇒ y = –2x + 15

c. (2, p) lies on this line, giving p = – 4 + 15 = 11

1 2

d.  The line y = x + f through (2, 11) gives 11 = 1 + f ⇒ f = 10 1 x + 10 2 e.  The line y = –2x + g through (1, 3) gives 3 = –2 + g ⇒ g = 5

y =



y = –2x + 5

1 2

f. Solving y = x + 10 and y = –2x + 5  2.5x = –5 ⇒ x = –2, y = 9 D = (–2, 9)

⎛ g. The mid-point is ⎜ ⎝

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−2 + 5 9 + 5 ⎞ ⎛ 3 , ⎟ =⎜ , 2 2 ⎠ ⎝2

⎞ 7⎟ ⎠

10

10. a. Maximum domain {x ∈  | x ≠ 0}, range {y ∈  | y ≠ 0}

b. Maximum domain {x ∈  | x ≥ – 5}, range {y ∈  | y ≥ 0}

c. Maximum domain {x ∈  | x ≠ – 3}, range {y ∈  | y ≠ 0} d. Maximum domain {x ∈  | x > 3}, range {y ∈  | y > 0} e. Maximum domain {x ∈  | x > 2}, range  f. Maximum domain , range {y ∈  | y > 0} x−3 y−3 ⇒ y = 4x + 3    inverse given by x = 4 4 f −1 (x) = 4x + 3 ⎛ x−3 ⎞ − 3⎟ ⎜ ⎠ x − 15 x − 3 ⎛ ⎞ ⎝ 4 ( f  f )( x) = f ⎜ = = b. ⎟ ⎝ 4 ⎠ 4 16 y − 15 x − 15 ⇒ y = 16x + 15 y= c.    inverse given by x = 16 16 ( f ! f )−1 (x) = 16x + 15

11. a. y =

d. Given any functions f and g, you know that (g ! f )−1 = f −1 ! g −1

So, it follows that ( f −1 ! f −1 )(x) = ( f ! f )−1 (x) = 16x + 15 1 ⎞ 1 +3 = ⎝ x + 4 ⎟⎠ x + 4

12. a. (h ! g ! f )(x) = h(g( f (x))) = h(g(x + 4)) = h ⎛⎜

b. Domain {x ∈! x ≠ −4}. Range {y ∈! y ≠ 3}



⎞ −4 c. (h ! g ! f )−1 (x) = f −1 (g −1 (h−1 (x))) = f −1 (g −1 (x − 3)) = f −1 ⎛⎜⎝ ⎟= x − 3⎠ x − 3

1

{

1

}

Domain { x ∈! x ≠ 3}. Range y ∈! y ≠ −4 1 x+4

d. + 3 =

3x + 13   a = 3, b = 13, c = 4 x+4

13. a.

y 8 7 6 5 4 3 2 1

–4

–3

–2

–1

0

x 1

2

3

3

–1



Horizontal asymptote y = 1

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b. i. Vertical translation up by + 2

ii. Horizontal translation to the right by 1 1 iii. Horizontal stretch by a factor of 4 iv. 2 x+2 + 4 = 2 x 2 2 + 4 = 4(2 x + 1) so vertical stretch by a factor of 4

14. a.

C



B 

A



From the diagram, h(θ ) = AB + BC = 1cosθ + 1sin θ = cosθ + sin θ

b.

h

1

0

2

4

6

θ

–1



c.

C  

B A

From the diagram, h(θ ) = AC − BC = 1cosθ − 1sin θ = cosθ − sin θ

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d.

h

1

0

2

4

6

θ

–1

SL Paper 2: Section A Technology required 15. Using a calculator to graph the functions y = e x and y = 4sin x y 8 7 6 5 4 3 2 1

–1



0

x 1

2

3

4

–1

From calculator, they intersect at x = 0.371 and x = 1.37 (3sf) (3 sf)

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SL Paper 2: Section B Technology required 16. a. g (0) = 10, r (0) = 1000

b. 10 years = 1 decade ⇒ g (1) = 10



c. Solving g (t) = 20, t = 1.44, 14.4 (3 sf) years



d. Solving g (t) = r (t), t = 2.98, 29.8 (3 sf) years

e. Solving g (t) = 2r (t), t = 3.29, 32.9 (3 sf) years f. Solving r (t) < 0.5, t = 10.97, 110 (3 sf) years

HL Paper 1: Section A No technology allowed 17. a. Remainder = p(1) = 1 + 2 − 5 − 6 = −8 q(+1) = q ( −1) ⇒ 1 + 1 + c + 1 = −1 + 1 − c + 1 ⇒ c = −1 b.

18. a. x 2 + 2x + 5 = (x + 1)2 + 4, so the vertex is a minimum at (−1, 4) b. Maximum possible domain is { x ∈! x ≥ −1 } 2 c. f (x) is given by y = ( x + 1) + 4

The inverse is given by x = (y + 1)2 + 4 ⇒ x − 4 = (y + 1)2 ⇒ y = −1 + x − 4 So f −1 ( x ) = −1 + x − 4; the domain is { x ∈! x ≥ 4}, the range is

{y ∈! y ≥ −1}

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19. a.

y 10 9 8 7 6 5 4 3 2 1

–5

–4

–3

–2

–1

0

x 1

2

3

4

5

–1 –2 –3 –4 –5



b. Intersection when 2x + 4 = ±1 ⇒ x =

−5 −3 or 2 2

−5 −3 f (x) > g (x) in the intervals ⎤⎥ , − 2 ⎡⎢ and ⎤ , ∞ ⎡ ⎥ ⎢⎣ ⎣ ⎦2 ⎦2

20. a. y = 0 ⇒ x − 4 = 0 ⇒ x = 4 ⇒ x = ±4 So, x-axis intercepts are (− 4, 0) and (4, 0) x=0⇒y=4 So, y-axis intercept is (0, 4)

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b. For x > 0, x = x ⇒ y = x − 4 = x − 4 10

5

0

5

10

15

For x < 0, x = −x ⇒ y = x − 4 = −x − 4 = x + 4 10

5

–15

–10

0

–5

So, the required graph of y = x − 4 will be: y

6

4

2

–10

–5

© Oxford University Press 2021

0

x 5

10

2

16

HL Paper 1: Section B No technology allowed y

21. a. 5 4 3 2 1

–5

–4

–3

–2

x

0

–1

1

2

3

4

5

–1 –2 –3 –4 –5



Horizontal asymptote y = 0; vertical asymptote x = 0

b. i. horizontal translation to the left by + 1

ii. vertical stretch by a factor of 2 (or horizontal stretch by a factor of 2) 1 1 iii. horizontal stretch by a factor of (or vertical stretch by a factor of ) 3 3 iv. vertical translation up by + 2



c. i. no

ii. yes



d. i. neither ii. odd

e.

iii. yes

iv. no

iii. odd

iv. neither

y 5 4 3 2 1

–5

–4

–3

–2

–1

0

x 1

2

3

4

5

–1

Horizontal asymptote y = 0; vertical asymptote x = 0

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f. i. not self-inverse (the inverse does not exist without a domain restriction)

ii. even function 22. a. i. f (x) = f (−x) ax + b −ax + b ≡ ⇒ −ax 2 + (ca − b)x + bc ≡ −ax 2 − (ca − b)x + bc x+c −x + c Equating coefficients: ca − b = 0 ax + ac ii. f (x) =    f (x) = a , x ≠ −c a horizontal straight line x+c b. i. f (x) = − f (−x) ax + b −(−ax + b) ≡ ⇒ −ax 2 + (ca − b)x + bc ≡ ax 2 − (b − ac)x − bc −x + c x + c

Equating coefficients: x2 : a = 0 1 : bc = 0 ii. a = 0 and b = 0 gives f (x) = 0, x ≠ −c (a special case of a. ii.) b a = 0 and c = 0 gives: f (x) = , x ≠ 0 x

23. a. Possible factors are x ± 1, x ± 2, x ± 3 p(−1) = −1 + 2 + 5 − 6 = 0, so x + 1 is a factor p(2) = 8 + 8 − 10 − 6 = 0, so x − 2 is a factor b Sum of roots is − = −2, so 3rd root is − 3, so x + 3 is a factor a (or using product of roots)

Factorises as (x + 1) (x – 2) (x + 3)

b. 2x 3 + 3x 2 − x − 2 ≥ x 3 + x 2 + 4x + 4 ⇒ x 3 + 2x 2 − 5x − 6 ≥ 0

(x + 1) (x – 2) (x + 3) ≥ 0

Intercepts are at x = –3, −1, 2



Sign of p (x) is given by the table

x p(x)

x < 3 – ve

– 3 0

–3 < x < –1 +ve

– 1 0

–1 < x < 2 –ve

2 0

x > 2  ve +

The inequality is true in the intervals [ −3, − 1] and [ 2, ∞[

dy d2 y = 3ax 2 + 2bx + c ii. = 6ax + 2b dx d x2 −2b −b −2b = b. i. sum of roots is ii. sum of roots is 6a 3a 3a c. Let α and β be the x-values of the two turning points of y. dy Then α and β are the roots of the quadratic equation =0 dx −2b So α + β = 3a Let X be the x-value of the point of inflection of y. d2 y Then X is the root of the linear function 2 = 0 dx

24. a. i.

© Oxford University Press 2021

18

−b 3a

So X =

α +β , showing the desired result. 2 d. i. p′(x) = nan x n−1 + (n − 1)an−1x n−2 + ... + 2a2 x + a1

Hence α + β = 2X ⇒ X =



ii. p′′(x) = n(n − 1)an x n−2 + (n − 1)(n − 2)an−1x n−3 + ... + 2a2 −(n − 1)(n − 2)an−1 −(n − 2)an−1 −(n − 1)an−1 = ii. T = n(n − 1)an nan nan −(n − 1)(n − 2)an−1 (n − 2) T= = S, as required f. n(n − 1)an (n − 1)



e. i. S =

y

25. a. 5 4 3 2 1

–1

0

x 1

2

3

4

5

6

7

8

9

10

–1 –2 –3 –4 –5

x-intercept is (1, 0)

© Oxford University Press 2021

19



b.

y 5 4 3 2 1

–1

x

0

1

2

3

4

5

6

7

8

9

10

2

3

4

5

6

7

8

9

10

–1 –2 –3 –4 –5

x-intercept is (3, 0)

c.

y 5 4 3 2 1

–1

0

x 1

–1 –2 –3 –4 –5

x-intercept is (0.5, 0)

© Oxford University Press 2021

20



d.

y 5

–10

x

0

–5

5

10

5

x-intercepts are (1, 0) and (−1, 0)

e.

y 5 4 3 2 1

–1

0

x 1

2

3

4

5

6

7

8

9

10

–1

x-intercept is (1, 0) 1 x−2 Inverse given by x =

26. a. y =

f −1 (x) =

1 1 1 ⇒y−2= , y = +2 y−2 x x

1 + 2, domain: { x ∈! x ≠ 0}, range y ∈! y ≠ 2 x

© Oxford University Press 2021

{

}

21



b. Sketching f (x)and f −1 (x) y 5 4 3 2 1

–5

–4

–3

–2

–1

x

0

1

2

3

4

5

6

7

8

–1 –2 –3 –4 –5

Solving f (x) = x (Note: you can solve f (x)= f −1 (x) to find the intersection, but these curves are inverses, so they meet on the line y = x. Hence, it’s easier to solve f (x) = x) 1 = x ⇒ x 2 − 2x − 1 = 0 ⇒ x = 1 ± 2 x−2 f (x) > f −1 (x) in the intervals ]1 − 2 , 0[ and ]2, 1 + 2[

27. a. x 2 − 4x + 4 = (x − 2)2, so x = 2 is a vertical asymptote. As x → ±∞, the quadratic dominates the linear function so y = 0 is a horizontal asymptote. x → 2 + ⇒ y ⇒ +∞, x → 2 − ⇒ y ⇒ +∞ x → +∞ ⇒ y ⇒ 0+ , x → −∞ ⇒ y ⇒ 0− 3 3 b. x = 0 ⇒ y = so ⎛⎜⎝ 0, ⎞⎟⎠ is the intercept on the y-axis 4 4 y = 0 ⇒ x = −3 so (−3, 0) is an intercept on the x-axis dy 1(x 2 − 4x + 4) − (x + 3)(2x − 4) −(x − 2)(x + 8) −(x + 8) = = c. = dx (x 2 − 4x + 4)2 (x − 2)4 (x − 2)3 dy = 0 ⇒ x = −8 dx Creating a sign table:

x

x < –8

–8

–8 < x < 2

2

x > 2

dy dx

– ve

0

+ve

undefined

–ve

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−1 ⎞ ⎛ So ⎜ −8, ⎟ is a minimum ⎝ 20 ⎠ d. i. ] − 8, 2[ ii. ] − ∞, −8[ ∪ ]2, ∞[

e.

y 4

2

–10

0

–5 –8,

x 5

10

–1 20

28. a.  x2 + x + 4 = 0 has a discriminant of 1 − 16 = −15, so it has no roots and there are no vertical asymptotes As x → ± ∞ the quadratic dominates the linear function so y = 0 is a horizontal asymptote. x → + ∞ ⇒ y ⇒ 0+ , x → −∞ ⇒ y ⇒ 0− 3 ⎛ 3⎞ x = 0 ⇒ y = so ⎜ 0, ⎟ is the intercept on the y-axis b. 4 ⎝ 4⎠ y = 0 ⇒ x = −3 so (−3, 0) is an intercept on the x-axis dy 1(x 2 + x + 4) − (x + 3)(2x + 1) −x 2 − 6x + 1 c. = = 2 dx (x 2 + x + 4)2 (x + x + 4)2 dy −6 ± 36 + 4 = 0 ⇒ x 2 + 6x − 1 = 0, x = = −3 ± 10 dx 2 So, there are stationary points when x = −3 ± 10

Creating a sign table:

x dy dx

x < −3 − 10 −3 − 10 −3 − 10 < x < −3 + 10 −3 + 10 x > −3 + 10 – ve

0

+ve

0

–ve

So, x = −3 − 10 gives a minimum, x = −3 + 10 gives a maximum



d. i. ] − 3 − 10, − 3 + 10[

ii. ] − ∞, − 3 − 10[ ∪ ] − 3 + 10, ∞[

© Oxford University Press 2021

23

e.

y 2

1

–10

0

–5

x 5

10

–1

–2

29. a. x + 3 = 0 ⇒ x = −3, so x = −3 is a vertical asymptote The quadratic dominates the linear function as x becomes large, so f (x) → ± ∞, as x → ± ∞. Hence no horizontal asymptotes. x−7 2 x + 3 x − 4x + 4 x 2 + 3x − 7x + 4 −7x − 21 25 x 2 − 4x + 4 25 , so y = x − 7 is an oblique asymptote. = x−7+ x+3 x+3 4 4 b. x = 0 ⇒ y = , so intercept on y-axis at ⎛⎜ 0, ⎞⎟ ⎝ 3⎠ 3 x 2 − 4x + 4 = (x − 2)2 = 0 ⇒ x = −2 so intercept on x-axis at (−2, 0)

c. f (x) =

(x − 2)2 2(x − 2)(x + 3) − (x − 2)2 × 1 (x − 2)(x + 8) ⇒ f ′(x) = = x+3 (x + 3)2 (x + 3)2

f ′(x) = 0 ⇒ x = 2 or − 8

Using a sign table:

−3 x < −8 −  8 −8 < x < −3 3 B > C ⇒ sin A > sin B > sin C a b c a sin A b sin B = = ⇒ = and = sin A sin B sin C b sin B c sin C

Giving the desired result that a > b > c

b. The result is still true. A + B + C = π ⇒ π − A = B + C So π − A > B > C ⇒ sin(π − A) > sin B > sin C Also sin(π − A) = sin A so the proof now follows the last lines of that in part a.

7. a. Applying the cosine rule twice 2

a ⎛ a⎞ ˆ b 2 = AD2 + ⎜ ⎟ − 2AD cos ADC ⎝ 2⎠ 2 2 a ⎛ a⎞ ˆ c 2 = AD2 + ⎜ ⎟ − 2AD cos ADB ⎝ ⎠ 2 2

ˆ = 180° − ADC ˆ ⇒ cos ADB ˆ = − cos ADC ˆ ADB

Adding the first two equations: b 2 + c 2 = 2AD2 +

4AD2 = 2b 2 + 2c 2 − a 2, as required.

a2 giving 2

b. Smallest median of length l will have 4l 2 = 2 × 32 + 2 × 42 − 52 = 25 So l =

5 2

1 2 lθ ii. lθ 2 1 b. Curved area of cone will also be l 2θ 2 Arc length will become the circumference of the circular base. So lθ = 2πr 1 1 1 Area = l 2θ = l(lθ ) = l2πr = πrl, as required. 2 2 2

8. a. i.

SL Paper 1: Section B No technology allowed 9. a. P = (cos B, sin B), Q = (cos A, sin A) PQ = (sin 2 A − sin 2 B) + (cos 2 A − cos 2 B) b. 2 2 2 2    = sin A − 2 sin A sin B + sin B + cos A − 2 cos A cos B + cos B

   = 2 − 2 sin A sin B − 2 cos A cos B

c. PQ 2 = 12 + 12 − 2 × 1 × 1cos ( A − B ) = 2 − 2 cos ( A − B ) d. Equating PQ 2, we have 2 − 2 sin A sin B − 2 cos A cos B = 2 − 2 cos(A − B) Giving cos(A − B) = cos A cos B + sin A sin B, as required

© Oxford University Press 2021

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SL Paper: 2 Section A Technology required h = tan 55 ⇒ h = 15tan 55 = 21.4222... = 21.4 m (3 sf) 15 b.  The angle of depression from top of the tree to Pam’s feet (call it α) equals the angle of elevation from Pam’s feet to the top of the tree. h 21.4222... = tan α ⇒ α = arctan = 47.0° (3 sf) 20 20

10. a.

1+ 3 2 + 4 3 + 7⎞ , , ⎟ = (2, 3, 5) ⎝ 2 2 2 ⎠

⎛ M=⎜ 11. a.

CD = (3 − 1)2 + (4 − 1)2 + (7 − 1)2 = 4 + 9 + 36 = 49 = 7 b. AC = 32 + 42 + 7 2 = 9 + 16 + 49 = 74 c. AD = 12 + 12 + 12 = 3

28 7 2 = 74 + 3 − 2 × 74 3 cos A ⇒ cos A = 2 × 74 3 ˆ CAD = 20.0° (3 sf)

12. a.

N

H

180º

3

x

1 A

B

3 + 1 = 10 = 3.16 km (3 sf) b. 2

2

1 3

c. Bearing will be 180 − arctan = 162° (to nearest degree) 4 3

13. a. Volume of cuboid = 2 × 3 × 4 = 24m 3   Volume of sphere = π × 13 24 = 5.729...   So maximum number of spheres is 5. ⎛ 4π ⎞ ⎜⎝ ⎟⎠ 3 4π 24 − 5 × = 3.0560... = 3.06m 3 (3 sf) b. 3 x 3 = 3.0560... ⇒ x = 1.45 m (3 sf) c. 1 3

1 3

14. a. πr 2 h = πr 2 ( 2r ) = 100 ⇒ r 3 =

150   r = 3.62783... = 3.63 cm (3 sf) π

l = r 2 + (2r)2 = 5r b.

1 πrl + πr 2 3 1 Surface area = π 5(3.62783...)2 + π(3.62783...)2 = 72.2 cm 2 (3sf) (3 sf) 3 Surface area =

© Oxford University Press 2021

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1 2

1 2

l 2θ 2 × sin 69° = 46.679… 15. a. Area of triangle POQ = OP × OQ × sin POQ =  × 10 69 = 60.213… 360 Area between arc and chord = 60.213… – 46.679…= 13.5cm 2 (3 sf) 13.534... b. 2 × 100% = 4.31%(3 sf) π × 10

Area of sector = π × 102 × 

SL Paper 2: Section B Technology required 1 5 Length of string equals major arc AB plus 2 times distance AP

ˆ = α   cos α = 16. Let AOP

Angle at centre of major arc is 2π − 2α   AP = 52 − 12 = 24 1⎞ ⎛ Length is 1 × ⎜ 2π − 2arccos ⎟ + 2 24 = 13.3 m (3 sf) ⎝ 5⎠

HL Paper 1: Section A No technology allowed 17. a. tan 4θ =

sin 2(2θ ) 2 sin 2θ cos 2θ 4sin θ cosθ (1 − 2 sin 2 θ ) = = 1 − 2 sin 2 2θ cos 2(2θ ) 1 − 8sin 2 θ cos 2 θ

    =

4sin θ cosθ − 8sin 3 θ cosθ , as required 1 − 8sin 2 θ cos 2 θ

4sin θ cosθ − 8sin 3 θ cosθ = 1 − 8sin 2 θ cos 2 θ ⇒ tan 4θ = 1 b. π 5π 4θ = , 4 4 π 5π θ= , 16 16 π

π

π

3

1

18. cos ⎛⎜ x + ⎞⎟ = cos x cos − sin x sin = cos x − sin x ⎝ 2 2 6⎠ 6 6



3 1 cos x − sin x = 5cos x 2 2 sin x = ( 3 − 10)cos x



tan x = 3 − 10



1 2

19. a. Area of AOP = rl22θsin θ TP = r tan θ b.

1 2 1 l θ tan θ) = rl22θtan θ Area of POT = r(r 2 2 1 c. Area of sector OAP = rl22θ θ 2

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d. Area of triangle OAP < area of sector OAP < area of triangle POT 1 2 1 1 r sin θ < r 2θ < r 2 tan θ 2 2 2 ∴sin θ   2.60) = 17. a. i. 7

b. i. 25 − 1 = 24



c.

1

18. a.

5

P(2.60 < D < 2.70) = 0.789 (3 sf) 0.5

ii. 8

iii. 10.8

ii. 19 − 5 = 14

iii. 7.13 (3 sf)

8

Height, h m 1.05 < h ≤ 1.10 1.10 < h ≤ 1.15 1.15 < h ≤ 1.20 1.20 < h ≤ 1.25 1.25 < h ≤ 1.30 1.30 < h ≤ 1.35

19

25

Frequency 1 20 30 25 22 2

Height, h m Cumulative frequency 1 h ≤ 1.10 21 h ≤ 1.15 51 h ≤ 1.20 76 h ≤ 1.25 98 h ≤ 1.30 100 h ≤ 1.35

b. Modal class is 1.15 < h ≤ 1.20



c. Using mid-points of class widths:

i. h ≃ 1.2 m ii. σ ≃ 0.0854 m (3 sf)

© Oxford University Press 2021

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SL Paper 2: Section B Technology required 19. a. 50% b. P(–l < X < l) = 0.5 ⇒ l = 0.674489… = 0.674 (3 sf) c. Outlier if: y > Q3 + 1.5 × IQR = Q3 + 1.5 × (Q3 − Q1) = 2.5Q3 − 1.5Q1 or y < Q1 + 1.5 × IQR = Q1 + 1.5 × (Q3 − Q1) = 2.5Q1 − 1.5Q3

as required

d. By symmetry, P(X > 4l or X < −4l) = 2P(X > 4l) = 0.00698 (3 sf) e. Similarity is between Q1 and −l and also Q3 and l.

So 2.5Q3 − 1.5Q1 is similar to 2.5l − 1.5(−l) = 4l W−µ ∼ N(0, 12 ) f. If W ~ N (μ, σ 2) then X = σ P(W > μ + 4lσ or X < μ − 4lσ) = P(X > 4l or X < −4l)

The answer is also 0.00698 (3 sf)

20. a. X = 1 comes from sequence 6, probability is

1 6

5 1 5 × = 6 6 62 5 5 1 25 X = 3 comes from sequence 666, probability is × × = 3 6 6 6 6 5 5 5 125 X = 4 comes from sequence 666 anything, probability is × × = 3 6 6 6 6 X = 2 comes from sequence 66, probability is

1 1 6

x P(X = x)

2 5 36

3 25 216

4 125 216

b. Sum of probabilities should be 1. Checking:

c. i. Mode is 4

36 + 30 + 25 + 125 216 = 3 =1 63 6

ii. By calculator, E(X) =

iii. By calculator, Var(X) = 1.37 (3 sf)

21. a. A

671 = 3.11 (3 sf) 216

10 9

frequency

8 7 6 5 4 3 2 1 0 0

5

© Oxford University Press 2021

10

15

20 25 score

30

35

40

45

B

10 9 8

frequency

7 6 5 4 3 2 1 0 0



5

10

15

b. A:

20 25 score

30

35

40

i. 20.5  ii. 20.5  iii.  40 − 1 = 39   iv.  30.5 – 10.5 = 20   v. 8.46 (3 sf) B: i. 20.5  ii. 20.5  iii.  40 − 1 = 39   iv.  30.5 – 10.5 = 20   v. 14.8 (3 sf)

c. Visually, the histogram of set B is more dispersed (spread-out) than that of set A.

The ranges and the interquartile ranges are the same. It is only the standard deviation that make a distinction, with that of set B being greater. So, the standard deviation was the best indicator of dispersion.

22. L ~ N (20, 0.5)2 a. P(19.6 < L < 20.8) = 0.733 (3 sf) b. P(L > l) = 0.95 ⇒ l = 19.2 cm (3 sf)

L−µ   Z ~ N (0, 12) 0.5 19 − µ ⎞ 19 − µ ⎛ P(L > 19) = 0.8 ⇒ P ⎜ Z < = −0.84162… ⎟⎠ = 2 ⎝ 0.5 0.5 New μ = 19.4208... = 19.4 cm (3 sf)

c. L ~ N (μ, 0.52)  Z =

d. L ~ N (19.4208..., 0.52)  P(L < 20) = 0.877 = 87.7% (3 sf) 23. a. r = − 0.840 (3 sf)

b. Fairly strong, negative, linear correlation



c. Require the regression line y on x

y = − 0.1440... x + 16.58… Substituting x = 105 gives y = 1.46, so the estimate is 1, to the nearest whole number

d. Require the regression line x on y

x = − 4.894… y + 110.03… Substituting y = 5 gives estimate as 85.6 (3 sf), which is 86 to the nearest integer.

© Oxford University Press 2021

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HL Paper 1: Section A No technology allowed 24. a. A and B being independent implies P(A ∩ B) = P(A) × P(B)  P(A′) × P(B) = (1 – P(A)) P(B) = P(B) − P(A) P(B) = P(B) – P(A ∩ B) = P(A′ ∩ B)



Showing that A′ and B are also two independent events.

b.  From part a., A and B independent implies A′ and B independent which implies B and A′ independent.

Part a. shows X and Y independent implies X′ and Y independent. Substituting in respectively, B and A′ independent implies B′ and A′ independent which in turn implies A′ and B′ independent. So A and B independent implies A′ and B′ independent

An alternative proof is:

P(A′) × P(B′) = (1 − P(A))(1 − P(B)) = 1 − P(A) − P(B) + P(A)P(B)

= 1 − P(A) − P(B) + P(A ∩ B) = 1 − P(A ∪ B) = P((A ∪ B)′)



Which upon considering a Venn diagram = P(A′ ∩ B′) as required.

25. a.  There are 3! = 6 ways of putting the three S’s next to each other. Think of this as being one entity. We have 5 objects to be placed in a line which can be done in 5! ways. The original seven letters could be placed in 7! ways. The answer is: 6 × 5! 1 = 7 7!

b.  There are 2 ways of putting the C’s at the ends. The remaining 5 letters

can be placed in the middle in 5! ways. The answer is: 2 × 5! 1 = 21 7!

26.

Choice

Revealed Goat2

1 3

Goat1

1 6

Goat2

1 6

Goat1

1 3

Goat1

£1 million

Goat2

Note that the game’s host would never reveal the million pounds, as that would ruin the game. © Oxford University Press 2021

47

2 In the top and bottom branches, which have a combined probability of , 3 it would be better to make the change. In the middle two branches which 1 have a combined probability of it would not be a good idea make 3 the change.

So, advise the contestant to change their choice.

27. a. r = –0.148 (3 sf)

b. Extremely weak, negative, linear correlation



c. As established in part a., there is essentially no correlation, so the line of best fit is meaningless.



This would be extrapolation it is well away from the data set.

It is the height that is known, so should be using the x on y regression line (if any).

HL Paper 1: Section B No technology allowed b

b

28. a. ∫ (x − µ )2 f (x)dx = ∫ (x 2 − 2xµ + µ 2 )2 f (x)dx a b

a

∫x

b

2

a

f (x)dx − 2 µ ∫ xf (x)dx + µ a

b

2

b

∫ f (x)dx = ∫ x a

2

f (x)dx − 2 µ( µ ) + µ 2

a

(since the area under the curve must equal 1) b

= ∫ x 2 f (x)dx − µ 2, as required. a

b. Var(X) = E(X2) – (E(X))2 Var(X) ≥ 0 ⇒ E(X2) ≥ (E(X))2, as required.



b

c.  Equality implies Var(X) = ∫ (x − µ )2 f (x)dx = 0. Now since (x − μ)2 ≥ 0 a

and f (x) ≥ 0 we must have (x − μ)2 f (x) = 0 for all x ∈ [a, b]. So when x ≠ μ b

then f (x) = 0 which makes it impossible to have ∫ f (x)dx = 1. a

By contradiction, we have shown that equality is not possible for a continuous random variable.



d. i. E(Y) = 7 × 1 = 7    ii.  E(Y 2) = 72 × 1 = 49    iii.  Var (Y) = 0

So E(Y 2) = (E(Y))2 is possible for a discrete random variable.

© Oxford University Press 2021

48

HL Paper 2: Section A Technology required 29. a.

1 4

Sends back

3 4

Eats

1 3

Sends back

2 3

Eats

1 6

Sends back

5 6

Eats

Burger 1 2 1 3

Chicken

1 6 Lasagna







1 1 1 4 3 3 1 P(C ∩ S) 9 8 = = c. P(C|S) = 19 19 P(S) 72 1 3 P(B ∩ S′) 2 × 4 = = d. P(B|S′) = 19 P(S′) 1− 72

30. i.

1 6

1 2

1 6

b. P(sends back) = × + × + × =

6

C3 4 = C3 91

3 8 = 27 53 53 72

5 ii. C152 × 4 = 8



15

C3 C + C + C3 34 iv. 3B, 3R or 3G 3 15 3 = C3 455 6

5

19 72

91



iii.

4 × 5 × 6 24 = 15 C3 91

4

HL Paper 2: Section B Technology required 31. a. Probabilities must add to 1, so

4+3+2 4 + k = 1 ⇒ k = , as required 13 13

32 = 2.46 (3 sf) 13 ii. By calculator, Var (X) = (0.1216…)2 = 1.48 (3 sf) c. 1 1 1 1 y 3 2 4 1 4 3 2 4 P(Y = y) 13 13 13 13 43 By calculator, E(Y) = = 0.551 (3 sf) 78 43 13 d. ≠ 78 22 e. E(W) = 0 since the game is fair 1 1 f. E ⎛⎜ ⎞⎟ will exist but is undefined. ⎝W⎠ 0



b. i. By calculator, E(X) =

© Oxford University Press 2021

49

32. a.

2π

∫ 0

π

2π

f (x)dx = 1 ⇒ ∫ k sin xdx − ∫ k sin xdx = 1 0

0

[ −k cos x ]0 + [ k cos x ]π = 1 ⇒ (k + k) + (k + k) = 1 ⇒ 4k = 1 1 ⇒ k = , as required 4 y b. π

2π

0.3 0.2 0.1

0 –0.1



π

π 2

3π 2

2π

x

c. i. E(X) = π, by symmetry

ii. Median = π, by symmetry iii. There are maximums of height

d. Var(X) =

2π

∫ 0

33. a.

1 π 3π at both x = and x = 4 2 2

x2 |sin x|dx − π 2 = 2.93 (3 sf) 4 chocolate

3 6

chocolate

3 5 3 6

plain

3 4 2 5

chocolate

plain 1 4

plain

1

chocolate



Let C = chocolate, and P = plain in the following. 1 b. X = 0 comes from sequence C, probability is 2 X = 1 comes from sequence PC, probability is 1 × 3 = 3 2 5 10 1 2 3 3 X = 2 comes from sequence PPC, probability is × × = 2 5 4 20 1 2 1 1 X = 3 comes from sequence PPPC, probability is × × × 1 = 2 5 4 20 x P(X = x)

0 1 2

1 3 10

2 3 20

3 1 20

10 6 3 1 20 + + + = =1 Sum of probabilities should be 1. Checking 20 20 20 20 20 3 c. By calculator E(X) = 4 63 d. By calculator Var(X) = 80

© Oxford University Press 2021

50

34. a.

Test says has virus

0.98 Actually has virus 0.2

0.02

Test says has virus

0.01

0.8 Actually does not have virus



Test says does not have virus

0.99

Test says does not have virus

b. Probability test says has virus = 0.2 × 0.98 + 0.8 × 0.01 = 0.204

P(V ∩ test says yes) 0.2 × 0.98 = = 0.961 (3 sf) P(test says yes) 0.204 P(V ∩ test says no) 0.2 × 0.02 P(V|test says no) = = = 0.00503 (3 sf) d. P(test says no) 1 − 0.204

c. P(V|test says yes) =

e. Let X be the number of pupils that the tests says have the virus. X ~ B(300, 0.01)

Expected value E(X) = np = 300 × 0.01 = 3 pupils

f. P(X ≥ 1) = 1 − P(X = 0) = 0.951 (3 sf) 35. a. i.

6 5 4 5 × × = 8 7 6 14

b. x

ii. 0 0

1 6

P(X = x)

0

C1 × C2 3 = 8 C3 28 2

2 6

C2 × C1 15 = 8 C3 28 2

3 6 8

C3 5 = C3 14

3 × 1 + 15 × 2 + 10 × 3 63 9 = = 28 28 4 3 3 c. i. ⎛⎜ 6 ⎞⎟ = 27 ii. ⎛⎜ 2 ⎞⎟ = 1 ⎝ 8⎠ ⎝ 8⎠ 64 64 6 9 d. This situation is binomial E(Y) = 3 × = 8 4 e. The expected value is the same regardless as to whether there is replacement or not.

E(X) =

HL Paper 3: Technology required 1+ 2 + 3 + 4 + 5 + 6 7 = 6 2 12 + 2 2 + 32 + 42 + 52 + 62 49 35 − = ii. Var(X) = E(X 2 ) − (E(X))2 = 6 4 12

36. a. i. E(X) =

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b. Y\X 1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

Counting P(T = 6) =

c.

2 1 36

t P(T = t)



4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12

6 1 = 36 6

3 2 36

4 3 36

5 4 36

6 5 36

7 6 36

8 5 36

9 4 36

10 3 36

11 2 36

12 1 36

Mode is 7

35 6 7 7 e. E(T ) = 7, E(X) + E(Y) = + = 7 confirming the result. 2 2 f. X and Y are independent 35 35 35 35 + = confirming the result. Var(T) = , Var(X) + Var(Y) = 12 12 6 6 ⎛ 35 ⎞ g. , P(6.8 ≤ T ≤ 7.2) = 0.592 (3 sf) T ≈ N ⎜ 7, ⎝ 600 ⎟⎠ h. Considering another lattice diagram

d. Using a calculator:   i.  E(T ) = 7,   ii.  Var(T) =

Y\X 1 2 3 4 5 6

1 1 1 1 1 1 1

2 1 2 2 2 2 2

3 1 2 3 3 3 3

4 1 2 3 4 4 4

5 1 2 3 4 5 5

6 1 2 3 4 5 6

9 1 P(M = 2) = = 36 4

i. 2M = T ⇒ 2X = X + Y or 2Y = X + Y ⇒ X = Y

So, the probability is

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6 1 = 36 6

52

5 CALCULUS: END-OF-CHAPTER PRACTICE QUESTION WORKED SOLUTIONS SL Paper 1: Section A No technology allowed 1. a.

dy = 6x + 2 dx

dy = 3x 2 − 3x −4 dx 1 dy 1 −23 y = x3 ⇒ = x c. dx 3 y = x 3 + x −3 ⇒ b.

2. a.

dy = 4x 3 dx

dy = 4   The equation of tangent is y = 4x + c dx Through (1, 2) ⇒ 2 = 4 + c ⇒ c = −2, equation of tangent is y = 4x – 2 At x = 1,

An equally good alternative is to give the answer in the form y − 2 = 4(x − 1) dy 1 b. At x = –1, = −4, gradient of normal is , equation of normal is dx 4 1 y = x+d 4 1 9 1 9 Through (−1, 2) ⇒ 2 = − + d ⇒ d = , equation of normal is y = x + 4 4 4 4 An equally good alternative is to give the answer in the form 1 y − 2 = (x + 1) 4

3. a.

dy dy = x 2 − 3x − 10 = (x + 2)(x − 5)   = 0 ⇒ x = −2or 5 dx dx x dy dx

x < –2

–2

–2 < x < 5

5

x>5

+ve

0

–ve

0

+ve

Maximum when x = –2   Minimum when x = 5



b. i. increasing for ]−∞, − 2[ ∪ ]5, + ∞[  ii.  decreasing for ] − 2, 5[

dy d2 y = 6(x − 1) = 0 ⇒ x = 1, so the point of inflexion is (1, 5) dx dx 2 b. Using a sign table

4. a.  = 3(x − 1)2 ,

x

x1

–ve

0

+ve

2

dy dx 2

There is a sign change at x = 1, hence there is a point at x = 1. dy = 0 at x = 1 so the tangent is a horizontal one. dx © Oxford University Press 2021

53



c. i. concave up for {x ∈! x > 1}

ii. concave down for {x ∈! x < 1} 5.

y

x



f ′(x) is in blue and passes through the origin.



f ′′(x) is in green.



These graphs were obtained using the following information.

f ′(x) will have zeros at turning points of f (x), it will be positive when f (x)  is increasing and negative when f (x) is decreasing. f ′′(x) will have zeros at the inflection points of f (x) (which will correspond  to the turning points of f ′(x)), it will be positive when f (x) is concave up ( f ′(x) is increasing) and negative when f (x) is concave down ( f ′(x) is decreasing).

6. a. Integrating f ′(x) gives f (x) = x 3 + e x−1 + x + c f (1) = 2 ⇒ 1 + 1 + 1 + c = 2 ⇒ c = −1 f (x) = x 3 + e x−1 + x − 1

b. Differentiating f ′(x) gives f ′′(x) = 6x + e x−1 7. a. ∫ x 2 + x −2 dx = 2 3

x3 − x −1 + c 3

b. ∫ x dx = x + c 1 2

3 2

1 3

c. ∫ sin x + cos 3x dx = − cos x + sin 3x + c

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8. a.

y 25 20 15 10 5

–3

–2

–1

x

0

1

2

3

–5 2



2

b. Require ∫ 20 − 3x 2dx = ⎡⎣ 20x − x 3 ⎤⎦ −2 = (32) − (−32) = 64 −2

9. a. By inspection, ∫ sin x (cos x)5 dx = 1 2

−1 (cos x)6 + c 6

b. By inspection, ∫ xe x +3 dx = e x +3 + c 2

2

x2 −1 dx = ∫ x 2 (x 3 + 2)−2 dx = (x 3 + 2)−1 + c by inspection. 2 (x + 2) 3 Note all the above could also be done by integration by substitution.

c. ∫ 3

SL Paper 1: Section B No technology allowed dy = 3x 2 cos x − x 3 sin x dx dy 2 sin x − (2x + 1)cos x b. = dx sin 2 x 2 dy c. = 2xe(x +1) dx dy 1 d. = ln(3x) + x × × 3 = ln(3x) + 1 dx 3x

10. a.

11. a.  V = x2 h

2x + h ≤ 1 so, for a maximum h = 1 –2x

V = x2 (1 – 2x) = x2 – 2x3 dV 1 = 2x − 6x 2 = 2x(1 − 3x) = 0 ⇒ x = 0 or . dx 3 Can reject x = 0 as x > 0 in the context of the problem.

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Checking with a sign table (An alternative would be to look at the sign of the second derivative) x

0

0

0

–ve

1 1 x = does give a maximum and the corresponding value of h =  3 3 3 1 3 ⎛ 1⎞ b. i. Maximum volume is ⎜⎝ ⎟⎠ = m 3 27 2 1 2 ii. surface area is 6 × ⎛⎜⎝ ⎞⎟⎠ = m 2 3 3 y

12. a. 15 10 5

–6

–4

–2

0

x 2

4

6

–5 −2

⎤ ⎡ x3 7 x3 ⎛ −8 ⎞ − 4x ⎥ = ⎜ + 8⎟ − ( −9 + 12 ) = x − 4dx = − 4x   b. ⎢ ∫ ⎝ ⎠ 3 3 3 ⎣3 ⎦ −3 2

2

⎡ x3 ⎤ ⎛ 8 ⎞ ⎛ −8 ⎞ −32 ⎢ 3 − 4x ⎥ = ⎜⎝ 3 − 8⎟⎠ − ⎜⎝ 3 + 8⎟⎠ = 3 ⎦ −2 ⎣ 4

⎡ x3 ⎤ ⎛ 64 ⎞ ⎛ 8 ⎞ 32 ⎢ 3 − 4x ⎥ = ⎜⎝ 3 − 16⎟⎠ − ⎜⎝ 3 − 8⎟⎠ = 3 ⎦2 ⎣ 7 32 32 71 + = So total area is + 3 3 3 3

SL Paper 2: Section A Technology required dy at x = 2 which is −0.628 (3 sf) dx −1 = 1.59 (3 sf) b. Gradient of normal is −0.628279...

13. a.  Require

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14. a.

y 10

5

x

0

1

2

1

b. Definite integral required is ∫ e2 x dx = 2.36 (3sf) 2

0

y

15. a. 5 4 3 2 1

–3

–2

–1

0

x 1

2

3

–1 –2



b. Intersections (−1.96, 0.140) and (1.06, 2.88) (3 sf)



c. Require

1.0580...

∫

(4 − x 2 ) − e x dx = 6.43 (3 sf)

−1.9646...

dv = 12 cos 4t dt 3 −3 −3 3 b. s(t) = ∫ vdt = cos 4t + c  t = 0, s = 0 ⇒ c =   s(t) = cos 4t + 4 4 4 4 c. s(5) = 0.444 m (3 sf)

16. a. a(t) =

5

Require  ∫ 3sin 4t dt = 9.44m (3 sf) 0

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SL Paper 2: Section B Technology required 17. a. y(0) = 5 ⇒ c = 5 dy = 3x 2 + 2ax + b = 0 at a turning point so dx −15 3 + 2a + b = 0, 48 + 8a + b = 0  solving a = , b = 12 2 b. Maximum is (1, 10.5), minimum is (4, –3) d2 y 5 c. 2 = 6x − 15 = 0 ⇒ x =    point of inflexion is (2.5, 3.75) dx 2 The curve must have a point of inflexion between a maximum and a minimum.

HL Paper 1: Section A No technology allowed 18. lim h→0

f (x + h) − f (x) ((x + h)2 − 4) − (x 2 − 4) x 2 + 2xh + h2 − x 2 = lim = lim h→0 h→0 h h h



2xh + h2 = lim 2x + h = 2x  (as expected) h→0 h

lim h→0

19. Applying the substitution u = 3x + 4:  x = Limits: x = −1 ⇒ u = 1, x = 0 ⇒ u = 4

u− 4 dx 1    = 3 du 3



so integral becomes



4 ⎡ 2u 2 8u 2 ⎤ u− 4 1 1 u 2 4u− 2 × × du = ∫1 3 u 21 3 ∫1 9 − 9 du = ⎢⎣ 27 − 9 ⎥⎦ 1



⎛ 16 16 ⎞ ⎛ 2 8 ⎞ −10 − = =⎜ − − ⎝ 27 9 ⎟⎠ ⎜⎝ 27 9 ⎟⎠ 27

4

1

1

3

1

4

0 ln(1 + x) − x is of the form 2 0 x 0 (1 + x)−1 − 1 It equals lim , which is still of the form and so equals x→0 0 2x −(1 + x)−2 −1 lim = x→0 2 2 1 ∞ ln x 1 b. lim is of the form . It equals lim x = lim 2 = 0 x→∞ x 2 x→∞ x→∞ ∞ 2x 2x

20. a.  lim x→0

tan 4 x +c 4 tan 4 x +1 y = 0 when x = 0 gives c = 1   e y = 4 4 ⎛ tan x ⎞ y = ln ⎜ + 1⎟ ⎝ 4 ⎠

21. ∫ e y dy = ∫ sec 2 x(tan x)3 dx  e y =

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HL Paper 1: Section B No technology allowed dy dy y2 tan x + y sec 2 x + 2y arctan x + =0 dx dx 1 + x2 dy y2 (tan x + 2y arctan x) = −y sec 2 x − 1 + x2 dx ⎛ y2 ⎞ 2 y sec x + ⎜⎝ 1 + x 2 ⎟⎠ dy =− ( tan x + 2y arctan x ) dx

22. a.

y = arccosec x ⇒ cosec y = x b. − cosec y cot y

dy dy −1 = 1 ⇒   = dx dx cosec y cot y

Using 1 + cot 2 y = cosec 2 y ⇒ cot y = cosec 2 y − 1 For x > 1, y is in the first quadrant so cot y will be positive dy −1 = dx x x2 − 1 1 dx = arctan x + c from the formula book 1 + x2 x 1 2 b. ∫ 1 + x 2 dx = 2 ln(1 + x ) + c by inspection

23. a. ∫

c. Using partial fractions:

1 1 A B = ≡ + ⇒ A(1 − x) + B(1 + x) ≡ 1 2 (1 + x)(1 − x) 1 + x 1 − x 1 − x 1 1 1 1 1 1 dv 2 ln 1 − x + c Solving A = , B =    ∫ 2 + 2 dx = ln 1 +u x= − = e2 x 2x 1+ x 1− x 2 2 2 2 dx dv e2 x du 2 u = =2x4x v = = e2 x dx 2 dx 24. Using integration by parts and ILATE: dv e2 x22xx du 2 2x 2 2x 2x 2 dv 2x e dx = x e − 2xe dx u = 2x = 4x v = == ee u = 2x ∫ ∫ dx 2 dx dx 2x du dv= ee2 x 2 x du = 4x v u = =2x2 v==e dx dx 22 dx Using integration by parts again on 2xe2 xdx

∫



and ILATE:



2x 2x 2x 2x ∫ 2xe dx = xe − ∫ e dx = xe −

So ∫ 2x 2e2 x dx = x 2e2 x − xe2 x +

25.

2

dy y ⎛ y⎞ = + 1− ⎜ ⎟ ⎝ x⎠ dx x dv v+x = v + 1 − v2 dx



arcsin v = ln x + c



y = x sin(ln x + c)

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e2 x +c 2

e2 x +c 2

homogeneous, so let v = 1 dv = ∫ dx x 1− v ⎛ y⎞ arcsin ⎜ ⎟ = ln x + c ⎝ x⎠

∫

du =2 u = 2x dx du =2 dx

1

dv e2 x v = = e2 x dx 2 e2 x v= 2

y x

2

59

26. a. 1st order linear   Integrating factor = e ∫ e x y = ∫ 4xe x dx = 2e x + c 2

x2

2

2

2

= ex   ex

y = 5 when x = 0 ⇒ c = 3

x2

e y = 2e + 3

2 xd x

y = 2 + 3e− x

2

2 2 dy + 2xe x y = 4xe x dx

2

2

b. As x → ±∞ , e− x → 0, y(x) → 2

x3 x5 + − ... 3 5 dv u = arctan x =1 b. dx xdx   using integration by parts with ILATE ∫ arctan xdx = ∫ 1 × arctan du 1 dv v = x= 1 = u = arctan x dx 1 + x 2 dx dv du 1 u = arctan x2 = v = =x 1 dx dx 1 + x x 1 duarctan1xdx = xarctan x − dx = xarctan x − ln(1 + x 2 ) + c = v=x ∫ 2 ∫ 2 1 + x 2 dx 1 + x

27. a. arctan x = x −

c. Integrating the Maclaurin expansion for y = arctan x

1 x 2 x 4 x6 + − ...  putting x = 0 ⇒ c = 0 xarctan x − ln(1 + x 2 ) = c + − 2 12 30 2 1 x 2 x 4 x6 xarctan x − ln(1 + x 2 ) = − + − ... 2 12 30 2 x 4 x6 xarctan x = x 2 − + − ... d. 3 5 e. Differentiating the answer to part d. x 4x 3 6x 5 arctan x + = 2x − + − ... 3 5 1 + x2 x2 x3 f. ln(1 + x) = x − + − ... 2 3 1 x 2 x 4 x6 2 ln(1 + x ) = − + − ... 2 4 6 2

⎛ ⎞ ⎛ x 2 x 4 x6 ⎞ x 2 x 4 x6 1 x 4 x6 − + − ... + − ...⎟ − ⎜ − + − ...⎟ = ⎝ ⎠ ⎝ 2 ⎠ 2 12 30 3 5 4 6 2 ⎛ x 2 x 4 x6 ⎞ x 2 x 4 x6 − ...⎟ = − + − ...,  as required. − ⎜ − + ⎝ 2 ⎠ 2 12 30 4 6

g. xarctan x − ln(1 + x 2 ) = ⎜ x 2 −

28. a. f (x) = a x

f ′(x) = (ln a)a x

f ′′(x) = (ln a)2 a x

f ′′′(x) = (ln a)3 a x

f (0) = 1

f ′(0) = (ln a)

f ′′(0) = (ln a)2

f ′′′(0) = (ln a)3

f (x) = a x = 1 + ( ln a ) x +

(ln a)2 2 (ln a)3 3 x + x + ... 2! 3!

a = eln a ⇒ a x = (eln a )x = e(ln a)x b. 1 2 1 3 x + x + ...  replacing x with (ln a) x gives 2! 3! (ln a)2 2 (ln a)3 3 x + x + ...,  as required. a x = 1 + (ln a)x + 2! 3! ex = 1 + x +

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HL Paper 2: Section A Technology required 1

29. a. Require ∫ 2 x dx = 1.29 (3sf) 2

0 1

1

b. Require ∫ π(2 x )2 dx = ∫ π2 2 x dx = 5.47 (3sf) 2

0

2

0

30. Using xn+1 = xn + 0.25 yn+1 = yn + 0.25( xn yn + 1)

y(1) ! 2.68 (3sf)

HL Paper 2: Section B Technology required 31. Let the origin be where the wall meets the ground, the distance from the origin to the foot of the ladder be x and the distance from the origin to the top of the ladder be y. dx dy −0.5x dx dy = 0.5 = 2x + 2y = 0 x 2 + y 2 = 132 a.  dt dt y dt dt dy −0.5 × 5 When x = 5, y = 132 − 52 = 12 = = −0.208m s −1 (3 sf) dt 12 The negative indicates (as expected) that she is descending dy −0.5 × 168 = = −6.48m s −1 (3 sf) dt 1 dx c.  As x → 13, y → 0 ⇒ → −∞ which is impossible (not to mention very dt painful for Almu)

b. When y = 1, x = 132 − 12 = 168

x3 x2 y ′′(0) + y ′′′(0) + ... 2! 3! y(0) = 1   y ′ = x 2 y 2 + 1    y ′(0) = 1

32. a. y(x) = y(0) + xy ′(0) +

y = 2xy 2 + x 2 2yy ′    y ′′(0) = 0 ′′ ′′′ y = 2y 2 + 8xyy ′ + x 2 (2( y ′)2 + 2yy ′′)   y ′′′(0) = 2 x2 x3 x3 × 0 + × 2 + ...    y(x) = 1 + x + + ... 2! 3! 3 1 7 b. y(1) ! 1 + 1 + = 3 3

y(x) = 1 + x +

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HL Paper 3: Technology required 33. a. i.  x = 2 is the only solution to y = 0 So, only x-axis intercept is (2, 0) dy = 2(x − 2)(x 2 + 3) + (x − 2)2 2x = (x − 2)(4x 2 − 4x + 6) ii. dx 4x 2 − 4x + 6 = 0 has discriminant of 16 − 96 = −80 so this quadratic has  no real roots and the only stationary point is (2, 0).

Looking at the sign table x dy dx





x2

–ve

0

+ve

So (2, 0) is a minimum.

b. i.  x = 2 is the only solution to y = 0

So, only x-axis intercept is (2, 0) dy = 3(x − 2)2 (x 2 + 3) + (x − 2)3 2x = (x − 2)2 (5x 2 − 4x + 9) ii. dx 5x 2 − 4x + 9 = 0 has discriminant of 16 − 180 = −164 so this quadratic  has no real roots and the only stationary point is (2, 0).

Looking at the sign table x dy dx



x2

+ve

0

+ve

So (2, 0) is a horizontal point of inflexion.

c. Let p(x) = (x − c)2 q(x) then p′(x) = 2(x − c)q(x) + (x − c)2 q′(x) p(c) = 0, p′(c) = 0+ 0 = 0, so (c, 0) is a stationary point.

d. Let p(x) = (x − c)3 q(x) then p′(x) = 3(x − c)2 q(x) + (x − c)3 q′(x) p (x) = 6(x − c)q(x) + 6(x − c)2 q′(x) + (x − c)3 q′′(x) ′′ p (c) = 0 + 0 + 0 = 0 ′′



e. i. From part c. we know that (c, 0) is a stationary point.

For n = 2 p′′(x) = 2q(x) + 4(x − c)q′(x) + (x − c)2 q′′(x) So p′′(c) = 2q(c) ≠ 0 and so (c, 0) cannot be a point of inflection. p′(x) = (x − c)(2q(x) + (x − c)q′(x)) = (x − c)2q(x) + (x − c)2 q′(x)

Case 1: If q(c) is positive then the sign table will be x p′(x)



xc +ve

so there is a minimum at (c, 0)

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Case 2: If q(c) is negative then the sign table will be: x p′(x)



xc –ve

so there is a maximum at (c, 0)

Note in both cases the (x − c)2q(x) will be the dominant term as x → c so the sign of q′(c) does not matter.

ii. From parts c. and d. we know that p′(c) = 0 and p′′(c) = 0. So, at the point (c, 0) the graph is momentarily horizontal and straight. p′(x) = n(x − c)n−1 q(x) + (x − c)n q′(x) = (x − c)n−1 (nq(x) + (x − c)q′(x))  If n is odd and q(c) is positive, then the sign table will be: x p′(x)

xc +ve

so there is a horizontal point of inflexion at (c, 0).

 If n is odd and q(c) is negative, then the sign table will be: x p′(x)

xc –ve

so there is a horizontal point of inflexion at (c, 0).  If n is even and q(c) is positive, then the sign table will be: x p′(x)

xc +ve

so there is a flat bottom minimum at (c, 0).

 If n is even and q(c) is negative, then the sign table will be: x p′(x)

xc –ve

so there is a flat top maximum at (c, 0).

Note n(x − c)n−1 q(x) will be the dominant term as x → c. Also note that in this question we are dealing with polynomials and so we know that all derivatives exist.

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IB PREPARED MAA WORKED SOLUTIONS AND MARKSCHEMES FOR THE PRACTICE EXAM PAPERS Here are the worked solutions and markschemes for the practice exam papers from IB Prepared MAA. These worked solutions are in a format that is similar to an IB markscheme. It shows how marks are to be gained in the different parts of a question. Examiners use annotations when marking and some of their abbreviations are shown here: M1 indicates a Method Mark, A1 an Achievement Mark and R1 a Reasoning Mark. If a mark is shown in brackets, e.g., (M1), it implies that the mark can be given even if the method has not been shown but is implied by subsequent student working. The abbreviation AG indicates “As Given” and is used at the end of a “show that …” problem. This means no marks would be given for this line and examiners know that they have to check the student’s work carefully here, to see that the student has not just written down the required answer. For direct access, click on the name of each practice exam paper.

SL Practice paper 1 SL Practice paper 2 HL Practice paper 1 HL Practice paper 2 HL Practice paper 3

© Oxford University Press 2021

1

SL PAPER 1 MARKSCHEME Section A: Short answers 1. a. u1 + 4d = 18, u1 + 9d = 38 ⇒ 5d = 20 ⇒ d = 4 [(M1)A1] 

[2 marks]

d = 4 ⇒ u1 + 16 = 18 ⇒ u1 = 2 [(M1)A1] b. 

[2 marks]

u101 = 2 + 100 × 4 = 402 [(M1)A1] c. 

[2 marks]



[TOTAL 6 marks]

dy 1 [M1A1] = dx cos 2 x π dy = 2[A1] at x = , 4 dx Equation of tangent is y = 2x + c π π ⎛π ⎞ Through ⎜ , 1⎟ ⇒ 1 = 2 × + c ⇒ c = 1 − [M1] ⎝4 ⎠ 4 2

2. a.



Equation of tangent is

π⎞ 1⎛ π oryy−–11==− 2 ⎜ x − ⎟ [A1] 4⎠ 2⎝ 2  [5 marks] 1 b. Gradient of normal is − [A1] 2 1 Equation of normal is y = − x + d. 2 π π ⎛π ⎞ Through ⎜ , 1⎟ ⇒ 1 = − + d ⇒ d = 1 + [M1] ⎝4 ⎠ 8 8 y = 2x + 1 −



Equation of normal is π⎞ 1⎛ 1 π y = − x + 1 +  or y − 1 = − ⎜ x − ⎟ [A1] 4⎠ 2⎝ 2 8  [3 marks] 

3. a. LHS =

[TOTAL 8 marks] (x + 2) + x(x + 1) x 2 + 2x + 2 1 x = RHS ≡ ≡ + (x + 1)(x + 2) (x + 1)(x + 2) x+1 x+2

 x 2 + 2x + 2 10 = ⇒ x 2 + 2x − 8 = 0  b. (x + 1)(x + 2) (x + 1)(x + 2)

[M1A1] [2 marks] [M1A1A1]

⇒ (x + 4)(x − 2) = 0 ⇒ x = −4 or 2 [M1A1A1] 

[6 marks]



[TOTAL 8 marks]

© Oxford University Press 2021

2

1 4

4. a. By inspection or substitution, ∫ cos x sin 3 x dx = (sin 4 x) + c[(M1)A2]  [3 marks] π 2 1 1 ⎤ ⎡ 4 b. ⎢⎣ 4 (sin x) ⎥⎦ 0 = 4 [M1A1]  [2 marks] 

[TOTAL 5 marks]

5. a. i. {x ∈!} [A1] ii. {y ∈! y > 4} [A1] 

[2 marks] − 4 − 4 x x ⎛ ⎞ = e y ⇒ y = ln ⎜ b. Inverse given by x = 3e y + 4 ⇒ [M1A1] ⎝ 3 ⎟⎠ 3 ⎛ x − 4⎞ [A1] f −1 (x) = ln ⎜ ⎝ 3 ⎟⎠  [3 marks]

c. i. {x ∈! x > 4} [A1] ii. {y ∈!} [A1]



[2 marks]



[TOTAL 7 marks]

6.

D 1 2 3 4 5 6

1 0 1 2 3 4 5

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 5 4 3 2 1 0

[(M1)A1]

From the lattice diagram above (or otherwise) the probability distribution table for D is d P(D = d)

0 6 36



1 10 36

2 8 36

3 6 36

4 4 36

5 2 36 [M1A1]

6 10 8 6 4 2 + 1× +2× +3× + 4× +5× 36 36 36 36 36 36 10 + 16 + 18 + 16 + 10 70 ⎛ 35 ⎞ =   = ⎜ = ⎟ [M1A1] 36 36 ⎝ 18 ⎠  [TOTAL 6 marks] E ( D) = 0 ×

© Oxford University Press 2021

3

Section B: Long answers 7. a. f ′(x) = 2x cos x − x 2 sin x [M1A1A1] 

[3 marks]

f ′′(x) = 2 cos x − 2x sin x − 2x sin x − x 2 cos x = 2 cos x − 4x sin x − x 2 cos x b. 

[M1A1A1A1]



[4 marks]



c. i. 0

ii. 0 iii. 2

[A1A1A1]



[3 marks]



d.  Since f ′(0) = 0 and f ′′(0) is positive, the point (0, 0) must be a minimum.

[R1A1A1]



[3 marks]



[TOTAL 13 marks]

8. a.

1 2

Sleep

S

1 6

1 2

Not sleep

C

1 2

1

Sleep

R

1 3

3 4

Sleep



1 4

Not sleep

[M1A1A1]

 [3 marks] 1 1 1 1 3 1 + 6 + 3 10 5 b. × + × 1 + × = = = [M1A1] 6 2 2 3 4 12 12 6  [2 marks] 1 P(C ∩ Sleep) 2 3 = = [M1A1] c. P(C Sleep) = 5 5 P(Sleep) [2 marks] 6 

1 P(S ∩ Not Sleep) 12 1 = = [M1A1] d. P(S Not Sleep) = 1 2 P(Not Sleep)  [2 marks] 6

e. P(S ∩ not S) + P(C ∩ not C) + P(R ∩ not R)

1 5 1 1 1 2 5 + 9 + 8 22 11 × + × + × = = = [M1A1A1] 36 36 18 6 6 2 2 3 3  [3 marks]

= 

© Oxford University Press 2021

[TOTAL 12 marks]

4

3 2 1 ii. y = [A1A1] 2  [2 marks]

9. a. i. x =



b. i. (–1, 0)





−1 ii. ⎛⎜⎝ 0, ⎞⎟⎠ [A1A1]



3

[2 marks] 1(2x − 3) − (x + 1)2 −5 c. [M1A1] f ′(x) = = (2x − 3)2 (2x − 3)2  [2 marks]

d. f ′(x) < 0 for all x, so graph is decreasing

[R1AG]



[1 mark] 20 f ′′(x) = e. [M1A1] (2x − 3)3  [2 marks] 3 f. If x > , f ′′(x) > 0, so graph is concave up [A1] 2 3 [A1] If x < , f ′′(x) < 0, so graph is concave down 2  [2 marks] 5 1 5 (2x − 3) + 1 2 ≡ x + 1 [M1A1AG] g. + 2 ≡ 2 2 2x − 3 2x − 3 2x − 3 

[2 marks] 5 1 x 5 h. ∫ 2 + 2x2− 3 dx = 2 + 4 ln ( 2x − 3) + c [M1A1]  [2 marks] 

© Oxford University Press 2021

[TOTAL 15 marks]

5

SL PAPER 2 MARKSCHEME Section A: Short answers 1. a. i. 7

[A1]

ii. 8 

[A1]



[2 marks]

b. Q1 = 3, Q3 = 8, IQR = 5[M1A1]



8 + 1.5 × 5 = 15.5 so 16 is an outlier

[R1A1]



[4 marks]



[TOTAL 6 marks]

4 5 ˆ = 126.5...°) [M1A1A1] ˆ = 53.46...°,(ACB ⇒ CDB = ˆ ˆ sin 40 sin(ACBorCDB) Triangle CBD isosceles ⇒ CD = 2 × 4cos 53.46...° = 4.76(3 s.f.)[R1M1A1A1]

2.



[7 marks]



[TOTAL 7 marks]

3. General term is 7 Cr (x 2 )7−r (x −1 )r[M1A1]

require 2(7 − r) − r = 2 ⇒ r = 4



Term is 7 C4 x 2, so coefficient is 35

[A1A1] [(A1)A1 ]



[6 marks]



[TOTAL 6 marks] y

4. 110

(8, 98)

(2, 8)

 –5

x –1

1

15

[(M1)A1]

8



A = ∫ [(x 2 + 5x − 6) − (2x 2 − 5x + 10)]dx [M1A1A1] 2

© Oxford University Press 2021

6

15

A = 36 0

x 1



10

–5

[(M1)]

A = 36

[A1] OR (without GDC)



2x 2 − 5x + 10 = x 2 + 5x − 6 ⇒ x 2 − 10x + 16 = 0 [M1]



(x − 2)(x − 8) = 0 ⇒ x = 2 or 8



[M1A1A1]

8

8

2

2

A = ∫ [(x 2 + 5x − 6) − (2x 2 − 5x + 10)]dx = ∫ [−x 2 + 10x − 16] dx = 36[M1A1A1]

 

[7 marks] [TOTAL 7 marks]

5. a. 500(1.04) = $740.12 10

[M1A1]



[2 marks]

b. Solving (X(1.03) − 200)(1.03) − 300 = 0 gives X = 395.75 5

5

[M1A1A2]



[4 marks]



[TOTAL 6 marks]

6. log 4 x − 3log x 4 − 2 = 0 ⇒ log 4 x −

3 − 2 = 0 [M1A1] log 4 x

⇒ (log 4 x)2 − 2 log 4 x − 3 = 0 ⇒ (log 4 x − 3)(log 4 x + 1) = 0 [M1A1M1] 1 [M1A1A1] log 4 x = 3 or − 1 ⇒ x = 64 or 4  [8 marks]





[TOTAL 8 marks]

Section B: Long answers 7. a. r = −0.856 (3 sf) 

b. Quite strong negative linear correlation. 

[A2] [2 marks] [R1] [1 mark]

y = −0.733x + 25.9 (3 sf)[M1A1A1] c.  © Oxford University Press 2021

[3 marks]

7

−0.733 × 31 + 25.9 = 3[M1A1] d. 

[2 marks]

e. Using the line x on y,  −0.998 × 11 + 30.6 = 20 [M1M1A1] 

[3 marks]

f. (x, y) = (17.4,13.1) (3 sf) [R1A1A1] 

[3 marks]



[TOTAL 14 marks]

8. a. i. P(24 < T < 27) = 0.533 (3 sf) [(M1)A1] ii. P(26 < T) = 0.309 (3 sf) [(M1)A1] iii. P(T < t) = 0.8 ⇒ t = 26.683... [(M1)A1]

so cut off time is 26 minutes, 41 seconds



[A1] [7 marks]

b. i. B(10, 0.8), P(X = 7) = 0.201 (3 sf)

[(M1)A1]

ii. P(X ≥ 7) = 1 − P(X ≤ 6) = 0.879 (3 sf)

[(M1)A1]





[4 marks] S− µ 2 c. S ∼ N( µ , 4 ), P(S < 60) = 0.25, Z = ∼ N(0, 1 ) [(M1)] 4 60 − µ ⎞ 60 − µ ⎛ P⎜ Z < = −0.67448... [M1A1] ⎟⎠ = 0.25⇒ ⎝ 4 4 2

µ = 62.697... 62 minutes, 42 seconds

[A1A1]



[5 marks]



[TOTAL 16 marks]

9. a. i. 1 ms [A1] −1

ii. −1 ms [A1] −1





b. i.

[2 marks]

10

∫ sin

t + 1 dt = 5.58 m (3 sf) [M1A1]

0

ii. 

10

∫ sin

t + 1 dt = 5.78 m (3 sf) [M1A2]

0

[5 marks] 1 c. Differentiating, acceleration is cos t + 1 [M1A2] 2 t+1  [3 marks] 

© Oxford University Press 2021

[TOTAL 10 marks]

8

HL PAPER 1 MARKSCHEME Section A: Short answers 1. a. u1 + 4d = 18, u1 + 9d = 38 ⇒ 5d = 20 ⇒ d = 4 [(M1)A1] 

[2 marks]

d = 4 ⇒ u1 + 16 = 18 ⇒ u1 = 2 [(M1)A1] b. 

[2 marks]

c. u101 = 2 + 100 × 4 = 402 [(M1)A1] 

[2 marks]



[TOTAL 6 marks]

dy 1 [M1A1] = dx cos 2 x π dy at x = , = 2 [A1] 4 dx

2. a.

Equation of tangent is y = 2x + c π π ⎛π ⎞ Through ⎜ ,1⎟ ⇒ 1 = 2 × + c ⇒ c = 1 − [M1] ⎝4 ⎠ 2 4

Equation of tangent is

π 1 π oryy−–1 1= =− 2 ⎛⎜ x − ⎞⎟ [A1] 4⎠ 2⎝ 2 [5 marks] 1 b. Gradient of normal is − [A1] 2 1 Equation of normal is y = − x + d. 2 π π ⎛π ⎞ Through ⎜ ,1⎟ ⇒ 1 = − + d ⇒ d = 1 + [M1] ⎝4 ⎠ 8 8 Equation of normal is 1 π π 1 y = − x + 1 +  or y − 1 = − ⎛⎜ x − ⎞⎟ [A1] 2 8 2⎝ 4⎠  [3 marks] y = 2x + 1 −



[TOTAL 8 marks] 1 4

3. a. By inspection or substitution, ∫ cos x sin 3 x dx = (sin 4 x) + c [(M1)A2]  [3 marks] π 2 1 1 ⎡ (sin 4 x) ⎤ = [M1A1] b. ⎢⎣ 4 ⎥⎦ 0 4  [2 marks]



[TOTAL 5 marks]

4. a. i. {x ∈!}[A1] ii. { y ∈! y > 4} [A1]  © Oxford University Press 2021

[2 marks]

9

x−4 ⎛ x − 4⎞ = e y ⇒ y = ln ⎜ [M1A1] ⎝ 3 ⎟⎠ 3 x − 4⎞ [A1] f −1 ( x ) = ln ⎛⎜ ⎝ 3 ⎟⎠  [3 marks]



b.  Inverse given by x = 3e y + 4 ⇒

c. i. {x ∈! x > 4} [A1] ii. {y ∈!} [A1]





[2 marks]



[TOTAL 7 marks]

5.

1 0 1 2 3 4 5

D 1 2 3 4 5 6

2 1 0 1 2 3 4

3 2 1 0 1 2 3

4 3 2 1 0 1 2

5 4 3 2 1 0 1

6 [(M1)A1] 5 4 3 2 1 0

From the lattice diagram above (or otherwise) the probability distribution table for D is d P(D = d)

0 6 36

1 10 36

2 8 36

3 6 36

4 4 36

5 2 36

[M1A1] 6 10 8 6 4 2 10 + 16 + 18 + 16 + 10 E(D) = 0 × + 1× +2× +3× + 4× +5× = 36 36 36 36 36 36 36 70 ⎛ 35 ⎞    = ⎜ = ⎟ [M1A1] 36 ⎝ 18 ⎠  [TOTAL 6 marks]

6. w∗ = 1 + 4i ⇒ w = 1 − 4i [(A1)] z 2 + 3i 1 + 4i 2 + 8i + 3i − 12 −10 11 = × = = + i [M1A1A1A1] w 1 − 4i 1 + 4i 17 17 17  [5 marks]





[TOTAL 5 marks]

f (x + h) − f (x) ((x + h)2 + (x + h) + 2) − (x 2 + x + 2) = lim [M1A1] h→0 h h 2xh + h2 + h h(2x + h + 1) = lim = lim = lim(2x + 1 + h) = 2x + 1 [A1(A1)A1] h→0 h→0 h→0 h h  [5 marks]

7. lim h→0



[TOTAL 5 marks]

8. p(1) = 1 + 1 + a + b = 7 ⇒ a + b = 5 [M1A1]

p(−2) = −8 + 4 − 2a + b = −8 ⇒ −2a + b = −4 [A1]



Solving simultaneously ⇒ a = 3, b = 2

[M1A1A1]



[6 marks]



[TOTAL 6 marks]

© Oxford University Press 2021

10

9. a. a i b = 0 ⇒ 2 + 2 − k = 0 ⇒ k = 4[M1A1A1] 

[3 marks]

c = λ b ⇒ λ = 2 ⇒ l = −2 [M1A1] b. 

[2 marks]

b × c = 0 because c and b are parallel c.

[R1]

⇒ a i (b × c) = 0 [A1] 

[2 marks]



[TOTAL 6 marks]

Section B: Long answers 10. a.

1 2

Sleep

S

1 6

1 2

Not sleep

C

1 2

1

Sleep

R

1 3

3 4

Sleep

1

Not sleep

 4

[M1A1A1]

 [3 marks] 1 1 1 1 3 1 + 6 + 3 10 5 = = [M1A1] b. × + × 1 + × = 6 2 2 3 4 12 12 6  [2 marks] 1 P(C ∩ Sleep) 2 3 = = [M1A1] c. P(C Sleep) = 5 5 P(Sleep) [2 marks] 6 

1 P(S ∩ Not Sleep) 12 1 = = [M1A1] d. P(S Not Sleep) = 1 2 P(Not Sleep)  [2 marks] 6

e. P(S ∩ not S) + P(C ∩ not C) + P(R ∩ not R)



1 5 1 1 1 2 5 + 9 + 8 22 11 × + × + × = = = [M1A1A1] 36 36 18 6 6 2 2 3 3 [3 marks]



[TOTAL 12 marks]

=

© Oxford University Press 2021

11

11. a. x 2 + x − 6 = (x + 3)(x − 2) [A1] 3x + 4 A B 2 ≡ + ⇒ 3x + 4 ≡ A(x − 2) + B(x + 3) [M1A1] x +x−6 x+3 x−2 x = −3 ⇒ A = 1, x = 2 ⇒ B = 2 [(M1)A1] 1 2 + [A1] f (x) = x+3 x−2 2 ⎞ ⎛ 1 ∫ ⎜⎝ x + 3 + x − 2 ⎟⎠ dx = ln(x + 3) + 2 ln(x − 2) + c [M1A1]  [8 marks] 1 2 b. [M1A1] f ′(x) = − − (x + 3)2 (x − 2)2 As squares are always non negative, this expression is always negative and hence the function is always decreasing. [R1] 

[3 marks]



[TOTAL 11 marks] n

1 4

12. a. Let P(n) be the statement that ∑ i 3 = n2 (n + 1)2. i=1

12 × 2 2 = 1, so P(1) is true. [M1A1] 4 Assume P(k) is true and attempt to prove for P(k + 1)[M1] k+1 k 1  LHS of P(k + 1) =∑ i 3 = ∑ i 3 + (k + 1)3 = k 2 (k + 1)2 + (k + 1)3[M1A1A1] 4 i=1 i=1

For n = 1, LHS = 13 = 1, RHS =

1 1 = (k + 1)2 (k 2 + 4k + 4) = (k + 1)2 (k + 2)2 = RHS of P(k + 1)[M1A1] 4 4 So since P(1) is true and P(k) true implies P(k + 1) true, by the principle of mathematical induction the statement has been proved for all n ∈!+.[R1] 

[9 marks] n b. ∑ i is an arithmetic sequence with u1 = d = 1 so Sn = 2 (1 + n)[M1A1] i=1 2 ⎛ n ⎞ n2 (n + 1)2 Hence ⎜ ∑ i ⎟ = as required. [2 marks] 4 ⎝ i=1 ⎠ n

c. Since “odds” equals “all” minus “evens” n

2n

n

2n

n

i=1

i=1

i=1

i=1

i=1

∑ (2i − 1)3 = ∑ i 3 − ∑ ( 2i )3 = ∑ i 3 − 8∑ i 3 [M1A1] ⎛ n2 (n + 1)2 ⎞ (2n)2 (2n + 1)2    = − 8⎜ ⎟⎠ [A1] ⎝ 4 4 2n

[Note: the step above uses ∑ i 3 = i=1

by the formula proved in part a.]

n (2n)2 (2n + 1)2 1 and ∑ i 3 = n2 (n + 1)2 4 4 i=1

Simplifying gives n

∑ (2i − 1)3 = n2 (2n + 1)2 − 2n2 (n + 1)2 = n2 (2n2 − 1) [A1A1] i=1

 

© Oxford University Press 2021

[5 marks] [TOTAL 16 marks]

12

13. a. ln y = ln x x = x ln x [A1] 1 dy = ln x + 1[M1A1A1] y dx dy = y(ln x + 1) = x x (ln x + 1) [A1] dx  [5 marks]



b. i. x x (ln x + 1) = 0 ⇒ ln x + 1 = 0 ⇒ ln x = −1 ⇒ x = e−1 [M1A1] −1

So point is (e−1 , e−e )[A1] ii.

x

x < e−1

x = e−1

x > e−1

dy dx

–ve

0

+ve

Sign diagram shows that the point is a minimum.

[M1A1]



[5 marks] −∞ ln x c. which is of the form [M1] lim(x ln x) = lim x→0 x→0 1 ∞ x 1 lim x = lim(−x) = 0 M1A1A1 x→0 −1 x→0 2  x [4 marks] ln x x → 0 ⇒ x x → 1 as x → 0 [M1A1] d. 

[2 marks]



[TOTAL 16 marks]

© Oxford University Press 2021

13

HL PAPER 2 MARKSCHEME Section A: Short answers 4

5

ˆ ˆ 1. sin 40 = ˆ ˆ ⇒ CDB = 53.46...°,(ACB = 126.5...°)[M1A1A1] sin(ACBorC DB) Triangle CBD isosceles ⇒ CD = 2 × 4cos 53.46...° = 4.76(3 s.f.) [R1M1A1A1] 

[7 marks]



[TOTAL 7 marks]

2. General term is 7 Cr (x 2 )7−r (x −1 )r 

[M1A1]

require 2(7 − r) − r = 2 ⇒ r = 4 [A1A1]

Term is 7 C4 x 2, so coefficient is 35

[(A1)A1]



[6 marks]



[TOTAL 6 marks]

3. a. 500(1.04)10 = $740.12 [M1A1] 



[2 marks]

b.  Solving (X(1.03) − 200)(1.03) − 300 = 0 gives X = 395.75 [M1A1A2] 5

5



[4 marks]



[TOTAL 6 marks]

4. log 4 x − 3log x 4 − 2 = 0 ⇒ log 4 x −

3 − 2 = 0 [M1A1] log 4 x



⇒ (log 4 x)2 − 2 log 4 x − 3 = 0 ⇒ (log 4 x − 3)(log 4 x + 1) = 0 [M1A1M1]



log 4 x = 3 or − 1 ⇒ x = 64 or x =



1 4

[M1A1A1]



[8 marks] [TOTAL 8 marks]

π2

5. a. i. π ∫ (sin x)2 dx [M1A1] 0

ii. 15.5 (3 sf) [A2] 

[4 marks]

b. i. Maximum of sin is 1, so maximum radius is 1

[R1] π π2 ii. Occurs when x = ⇒ x = [M1A1] 4 2  [3 marks] 

© Oxford University Press 2021

[TOTAL 7 marks]

14

6. a. A = πr 2 − l 2 [M1A1]

dA dr dl = 2πr − 2l = 0.4πr − 0.2l [M1A1A1] dt dt dt when r = 2 and l = 1, dA = 0.8π − 0.2 = 2.31m 2 s −1 (3 sf) [M1A1] dt  [7 marks] 10 = 0.4πr − 0.2l ⇒ l = 2πr − 50 [M1A1] b. 

[2 marks]



[TOTAL 9 marks]

7.

f ′(0) = 1 × 2 + 3 = 5 [A1]



f ′′(x) = f (x) + (x + 1) f ′(x) [M1A1]



f ′′(0) = 2 + 5 = 7 [A1]



f ′′′(x) = 2 f ′(x) + (x + 1) f ′′(x) [M1A1]

f ′′′(0) = 2 × 5 + 7 = 17 [A1] 7 17 f (x) = 2 + 5x + x 2 + x 3 + ...[A1] 2 6  [8 marks]



[TOTAL 8 marks]

Section B: Long answers 8. a. r = −0.856 (3 sf) [A2] 

b. Quite strong negative linear correlation. 

[2 marks] [R1] [1 mark]

y = −0.733x + 25.9 (3 sf) [M1A1A1] c. 

[3 marks]

−0.733 × 31 + 25.9 = 3 [M1A1] d. 

[2 marks]

e. Using the line x on y, −0.998 × 11 + 30.5 = 20 [M1M1A1] 

[3 marks]

f. (x, y) = (17.4, 13.1) (3 sf) [R1A1A1]  

© Oxford University Press 2021

[3 marks] [TOTAL 14 marks]

15

y dy 1 + x = 9. so it is homogeneous. [A1R1] dx 1 − y x y Letting v = x y = xv[M1] dv 1+ v x +v= [A1] dx 1− v dv 1 + v 2 x = [A1] dx 1 − v 1− v 1 ∫ 1 + v2 dv = ∫ x dx[M1] v ⎞ ⎛ 1 ∫ ⎜⎝ 1 + v2 − 1 + v2 ⎟⎠ dv = ln x + c [M1A1]

1 arctan v − ln(1 + v 2 ) = ln x + c [A1A1] 2



2 ⎛ y⎞ 1 ⎛ ⎛ y⎞ ⎞ arctan ⎜ ⎟ = ln ⎜ 1 + ⎜ ⎟ ⎟ + ln x + c [M1] ⎝ x⎠ 2 ⎝ ⎝ x⎠ ⎠

⎛ y⎞ 1 arctan ⎜ ⎟ = ln(x 2 + y 2 ) + c [A1] ⎝ x⎠ 2 π 1 x = 1, y = 1 ⇒ c = − ln 2 [M1A1] 4 2 π 1 ⎛ y⎞ 1 arctan ⎜ ⎟ = ln(x 2 + y 2 ) + − ln 2 [A1] ⎝ x⎠ 2 4 2  [15 marks]





[TOTAL 15 marks] 1

1

10. a. ∫ ae x dx = 1 ⇒ ⎡⎣ ae x ⎤⎦ 0 = a(e − 1) = 1 ⇒ a = 0





1

1 [M1A1AG] e−1 [2 marks]

xe x 1 ⎞ ⎛ dx = 0.582 (3 sf) ⎜ = ⎟ [(M1)A2] ⎝ e−1 e − 1⎠ 0

b. i. µ = ∫

1

x 2e x dx − (0.582…)2 = 0.0793 (3 sf) [M1A2] e − 1 0  [6 marks] M M ex 1 ⎡ ex ⎤ eM − 1 1 c. ∫0 e − 1 dx = 2 ⇒ ⎢⎣ e − 1 ⎥⎦ = e − 1 = 2 [M1A1] 0

ii. σ 2 = ∫

2e M − 2 = e − 1 ⇒ e M =  

© Oxford University Press 2021

e+1 ⎛ e + 1⎞ ⇒ M = ln ⎜ (= 0.620 (3 sf)) ⎝ 2 ⎟⎠ 2

[M1A1] [4 marks]

[TOTAL 12 marks]

16

⎛ 3⎞ ⎛ 2⎞ 11. a. Plane is perpendicular to ⎜⎜ 2 ⎟⎟ , line is parallel to ⎜⎜ 2 ⎟⎟ [R1R1] ⎜⎝ 1 ⎟⎠ ⎜⎝ 1 ⎟⎠

Using the dot product to find the angle between these two vectors

[M1]

6 + 4 + 1 = 4 + 4 + 1 9 + 4 + 1 cosθ [A1] 11 cosθ = ⇒ θ = 11.490... [(A1)A1] 3 14 So angle between line and plane is 78.5° (3 sf) [A1] 

[7 marks]

b. Since line and plane are not parallel they must cross.

[R1]

Points on the line have the form x = 1 + 3t, y = 2 + 2t, z = 3 + t[A1]

Substituting into the plane: 2(1 + 3t) + 2(2 + 2t) + (3 + t) = 9 ⇒ 11t = 0 ⇒ t = 0 [M1A1]

Intersection point is (1, 2, 3)

[A1]



[5 marks]

c. Let F be the foot of the perpendicular from P to the plane. ⎛ 2⎞ ⎛ 2⎞ ⎛ 2⎞ !!!" !!!" ⎜ ⎟ ⎜ ⎟ PF = λ 2 , OF = 1 + λ ⎜ 2 ⎟ [R1A1] ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜⎝ 1 ⎟⎠ ⎜⎝ 1 ⎟⎠ ⎜⎝ 1 ⎟⎠ 2 F lies in the plane so 2(2 + 2 λ ) + 2(1 + 2 λ ) + (1 + λ ) = 9 ⇒ λ = [M1A1] 9 !!!" 2 2 4 + 4 + 1 = [M1A1] Distance is PF = 9 3 (Note there are several other methods.) [6 marks] 

© Oxford University Press 2021

[TOTAL 18 marks]

17

HL PAPER 3 MARKSCHEME Section A: Short answers 1. a. i. f ′(x) = e x sin x + e x cos x [M1A1] ii. f ′′(x) = e x sin x + e x cos x + e x cos x − e x sin x = 2e x cos x[M1A1] iii. f ′′′(x) = 2e x cos x − 2e x sin x = 2e x (cos x − sin x)[M1A1] iv. f (4) (x) = 2e x cos x − 2e x sin x − 2e x sin x − 2e x cos x = −4e x sin x [M1A1] v. f (4n) (x) = (−4)n e x sin x vi. f

(4n+1)

[A2]

(x) = (−4) (e sin x + e x cos x) [A1] n

x

f (4n+2) (x) = (−4)n (2e x cos x) [A1] vii. f ( 4n+3) (x) = (−4)n (2e x cos x − 2e x sin x) [A1] viii. 



[13 marks]

b. i. f ′(x) = g ′(x) × h(x) + g(x) × h′(x) [A1]

ii. f ′′(x) = g ′′(x) × h(x) + 2 g ′(x) × h′(x) + g(x) × h′′(x) [M1A1] iii. f ′′′(x) = g ′′′(x) × h(x) + 3 g ′′(x) × h′(x) + 3 g ′(x) × h′′(x) + g(x) × h′′′(x) [M1A1] 

[5 marks]

f (x) = e e = e c. λx x

(1+ λ )x

[A1]

n (1+ λ )x

f (x) = (1 + λ ) e (n)

[M1A1]

g (n−i) (x) = λ n−ieλ x , h(i) (x) = e x

[A1A1]

So equation becomes n

n

i=0

i=0

(1 + λ )n e(1+λ )x = ∑ ai λ n−ieλ x × e x = ∑ ai λ n−ie(1+λ )x[M1A1] n

So (1 + λ )n = ∑ ai λ n−i[A1] i=0

Giving ai = Ci from the binomial expansion of (1 + λ )n[A1R1] n



[10 marks]



[TOTAL 28 marks]

2. a. | z | = 1 



b. i.

[A1]

[1 mark] 1 z n + n = (cisθ )n + (cisθ )− n = cis(nθ ) + cis(−nθ ) [M1A1] z = cos(nθ ) + i sin(nθ ) + cos(−nθ ) + i sin(−nθ )

= cos(nθ ) + i sin(nθ ) + cos(nθ ) − i sin(nθ )      = 2 cos nθ [M1AG]     

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ii. (cisθ )n − (cisθ )− n = cis(nθ ) − cis(−nθ ) [A1] = cos(nθ ) + i sin(nθ ) − cos(−nθ ) − i sin(−nθ ) = cos(nθ ) + i sin(nθ ) − cos(nθ ) + i sin(nθ )                       = 2i sin nθ [M1AG] 



[5 marks] 2 1⎞ 1 ⎛ c. i. ⎜ z + ⎟ = z 2 + 2 + 2 = 2 cos 2θ + 2 = 2(1 + cos 2θ ) [A1] ⎝ z⎠ z

Hence, (2 cosθ )2 = 4cos 2 θ = 2(cos 2θ + 1) [M1A1] 1 + cos 2θ [AG] ⇒ cos 2 θ = 2 2 1⎞ 1 ⎛ ii. ⎜ z − ⎟ = z 2 − 2 + 2 = 2(cos 2θ − 1) [A1] ⎝ ⎠ z z 2 2 (2i sin θ ) = −4sin θ = 2(cos 2θ − 1) [A1] 1 − cos 2θ [AG] ⇒ sin 2 θ = 2  [5 marks] 3 1 3 1 d. i. ⎛⎜ z + ⎞⎟ = z 3 + 3z + + 3 ⎝ z⎠ z z 1 1⎞ ⎛ = z 3 + 3 + 3 ⎜ z + ⎟ [A1]     ⎝ z z⎠ 3 (2 cosθ ) = 2 cos 3θ + 3 × 2 cosθ ⇒ 8cos 3 θ = 2 cos 3θ + 6cosθ [M1A1] 1 3 cos 3 θ = cos 3θ + cosθ [A1] 4 4 4 1 4 1 ii. ⎛⎜ z + ⎞⎟ = z 4 + 4z 2 + 6 + 2 + 4 ⎝ z⎠ z z 1 1⎞ ⎛    = z 4 + 4 + 4 ⎜ z 2 + 2 ⎟ + 6 [A1] ⎝ z z ⎠ (2 cosθ )4 = 2 cos 4θ + 4 × 2 cos 2θ + 6 ⇒ 16cos 4 θ = 2 cos 4θ + 8cos 2θ + 6 [A1] 1 1 3 cos 4 θ = cos 4θ + cos 2θ + [A1] 8 2 8  [7 marks] 3 1⎞ 3 1 ⎛ 3 e. i. ⎜ z − ⎟ = z − 3z + − 3 ⎝ z⎠ z z = z3 −    

1 1⎞ ⎛ − 3 ⎜ z − ⎟ [A1] 3 ⎝ z z⎠

(2i sin θ )3 = 2i sin 3θ − 3 × 2i sin θ ⇒ −8i sin 3 θ = 2i sin 3θ − 6i sin θ [A1] −1 3 sin 3 θ = sin 3θ + sin θ [A1] 4 4 4 1 4 1 ii. ⎛⎜⎝ z − ⎞⎟⎠ = z 4 − 4z 2 + 6 − 2 + 4 z z z 1 ⎛ 2 1⎞ 4 = z + 4 − 4 ⎜ z + 2 ⎟ + 6 [A1]       ⎝ z z ⎠ (2i sin θ )4 = 2 cos 4θ − 4 × 2 cos 2θ + 6 ⇒ 16sin 4 θ = 2 cos 4θ − 8cos 2θ + 6 [A1] 1 1 3 sin 4 θ = cos 4θ − cos 2θ + [A1] 8 2 8  [6 marks]

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cos 4 θ − sin 4 θ = (cos 2 θ + sin 2 θ )(cos 2 θ − sin 2 θ ) [M1A1] f. 2

2

      = cos θ – sin θ = cos 2θ[AG] From parts d. ii. and e. ii. 1 1 3 ⎛1 1 3⎞ cos 4 θ − sin 4 θ = cos 4θ + cos 2θ + − ⎜ cos 4θ − cos 2θ + ⎟ [M1] ⎝ 8 2 8 8 2 8⎠ [AG]       = cos 2θ, confirming the result



[3 marks]



[TOTAL 27 marks]

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