Problems Plus In Iit Mathematics. Part 2 [2] 8177096575

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Problems Plus in

T Matiematics A Das Gupta

29881909

Part 2

A-150

Problems

a,

Let

and

B

be real and

i

a

i

Pls

=1and a be an arbitrary complex number

= 1.} B

f(a-106-1)- (B+1a-1)}a a(a

1

MB 1)-

(a

-

1MB

where the sets A

tAnB=o,

in lIT Mathematies 1

such that

(a-1KB-1)=0

-1)=0.

,p,

=

2,

then find the set of

possible values of a.

.

Find the quadratic equation whose roots are nth powers of the roots of r* - 12 x+1=0. t the point representing the complex number z, is a point on or inside the circle whose

l.

centre is the complex numberr+ i -0 and the radius is r then prove that the greatest value + of +3+ lisn(n +1).

n

Answers 1. (i) A,D

5.

(-b--2e

6.

1

9.

3

12.

(ü)C

(i) A.E

2.(i) 0 (i)

3z (ii) 0, 1, respectively

+20)= (C+ (6-b}, C0-

c-C

5 12

-1,1+ i, -i} 1

11. 4S

13.x-2r cos+1 =0

5. Permutation and Combination Recap of Facts and Formulae Counting of number of ways to do some work

3. Counting

consists of two parts W, IW, of which one aurt can be done in m ways and the other part in n ways

fa work

W

The number ofcombinations (selections)

"C,=n_ "C,!n-T) The total number of selections of one or more objects from n different objects

2-1="C

the work W can

be done in mn ways, if both the parts are to be done one after the other to do the work W. (Multiplication latw of counting)

number of ways to do the work under the restriction = (the number of ways to do the work without restriction) (the number of ways to do the work under opposite restriction).

formulae for permutation

Thenumber of permutations (arrangements) of ditferent things taking rat a time

"P,

n-r)!

wheren!=

..+"C,).

(when selection of 0 things is allowed) (when at least one thing is to be selected).

n+1

.The total number of selections from

the

. Counting

+ "C, +"C,+

.The total number of selections of any number of things fromn identical things

Similar is the law for works that have 3 or more parts.

fa work is to be done under some restriction then

of n diferent

things taking r at a time

ehen

the work W can be done in m+n ways, if by doing any of the parts the work W is done. (Addition lawofcounting)

formulae for combination

q

p like things, like things of another type and r distinct things

=(p+10q+1)2"-1

if at least one thing is

tobe selected) (P+1Mg+ 1)2'-2 (if none or all cannot be selected) The total number of selections of r things from n different things when each thing can be repeated unlimited number of times ="c,

n

For example If 8 balls are to be selected from balls of 3

1.2.3. ...

colours, balls of each colour being available unlimited number of times, then the number of selections

1.

ne number of permutations of n things taking all at a

38-1Cs

e

of which p things are identical, q things are Iaentical of another type and the rest are different

10!

10C%12! 4.

he

number of arrangements of n dijfërent thinigs round a closed curve -1)! if clockwise and anticlockwise arrangements are considered different,

herer

=8, n

=

3)

10 x9

245

Number of distributions The number of ways to distribute n dijiirent things between two persons, one receiving p things

other qthings,wherep+q=n

"Cx-C

2-1)! if clockwise and anticlockwise

n

(-p)! p!(n-p)! q!n-p-q)!

arrangements are considered identical. (Circular permutation) A-151

and the

A-152

Problems Plus in IIT Mathematics

= coefficient of

n=p+q

p!q!

r" in

Similarly for 3 persons, the number ways of

!p!q!r'

where p +q+r=n.

The number of ways to distribute

m

among n persons equally = mn)!

.

(m

r(r

xn

n

The number of ways to divide n different things into three bundles of p, q and r

things!air p

3!

different things

m * n

(m 1)

The total number of ways to divide among r persons

n identical things

=nr-'Cr-1 For example The number of different ways to divide 20 identical apples among 4 persons

=204-C4-, (* here n =20,r= 4)

23C23 *22 x21 = 23 x 11x7

n!

T(r+ 10r+ 2)... (r+n-1)

!"

The number of ways to divide into n equal bundles= n)

+1)r+2).. (r+n-1)

different things

(r+n-1)_n+-'C,n!(r-1) ! The number of integral solutions of

x+*2+X +

+I=n where x,21, x22 1,... ,x.>

is the same as the number of ways to distribute identical things amongr peršons each getting at least1, This is also equal to the coefficient of x" in the

expansion of

(r+x*+rt.)

=coeficient of r" =

in|

coefficient of x " in x'(1

-

x)

coefficient of r" in

r1++Tr+1),2,

6

1771.

rr+1)r+2).(+n-1),»,

.The toal number of ways to divide n identical things amongr persons so that each gets at least one

-Cr-

For example The number of different ways to distribute 20 identical balls in 4 different boxes so that no box remains empty =

20-1C-1

=

C

=

969.

l'

coefficient of "- in

1+

224.. rtr

here n =20, r=4)

-18 17

+ 1Xr+2). (r+n1

_r(r+ 1)(7+2)... (7+ n-r-1)

6

r(+

.

(n r)! 1(7 +2) (-1)D (-r)

5. Use of solution of linear

equations and coefficient of a power in expansions to find the number of ways of

distribution

The number of integral solutions of X +X2 +s+.. +X,=n where x 20, 20, ., x,20 is the same as the number of ways to distribute n identical things among r persons. This is also equal to the coefficient of x" in the

expansion of (x°+x'+x2+x*+.)

in 1-

=

coefficient of

=

coefficient of r" in

(1

-

x)

(n-1)! (n-r)!(r-1)!

n-1C-1

Note The number of solutions of x +X2 4+X3 + 4 wherex2 0, x21,x 23, x25 is equal to the coefficient of x in

2

(x+x+x2 r+x2+x3+ ..) x(r+x*+x5.. )(x+r+x+..) 6. Some useful results

C,="C-r .C,+"C-1="*'C

A-153 Permutation and Combination

1

"= 1-1)

-Co+"C,x +"*'C3x+"*C,x'+

+1 +2), n(+1)2, Y,2 (nn +10n

+ nx +

+"-Cxt

3!

2!

nn +10n+2). ( +r-),,

where

n is a

positive integer.

Selected Solved Exanples is divisible by ( )*. Showthat(kn)! 1. ... (kn) (kn)!= 1/2:3 Here u)l(n + 1)n + 2).. (n + = (1-2-3.

-'C+("* m -'C+" n)}

="*"C +""Ck-1

+ I(2n+ 1(2n + 2)... (2n n)..

(rn

Now,

+ 1)

[(k 1)n

x

+

1)(rn

1)n

|k-

N 2)... (rn + n), wherere (rH + n)

(rn +n)

(rn+1(rn +2)... by (1),

+)

Tm

and 2.

!

= (n !

M,

M,,

M)... (u! Mg-1 integer = (n M... M- !*x

divisible by )* (n*) ! is divisible by Putting k= u, we get

(n

j-0

'C+"Ck-1) +("*'Ck-1+""C-1t

Ck-1

+'Cg-1+"**Cg-t. n+m-C-1+""Ck-1

=("C+"*'C-1)

Ck1t+n-Ck-1+

n+Ck-1)

++-1C+"*"Ck-1)

="C+("*2Ck-1 +..+

r- 1) r)!

r-2)!

(r+2)! (-

1

or or

+(r*+3r+ 2) (n-r)-(2r+5)(n-r)

replacing r by Similarly, from (2)

(n-r-

1)*-(2r+7)(n

+I(r+1)+3(r+1)+21

or

(nr)- 2(n-r)+1

or

(n-r)-(2r

-

0

.

r+1 in (3), we get

r -1)

-

0

(2r+ 7)(m- r) =

+3 +2) 0 +(2r +7 +r+2r+1+3r = (4) 14) 0 +9(n )+(r* +7r+ -

identical (3) and (4) must be

+

+

(2)

(r+2)(r+ 1) 2 (n- r- 1)(n- r) (r+ 1)(-r- 1) 1)(0n- r) + 1)+(n-r2(r +2)(- r) = (r+2)(r . (3) =

'Ck-1+"**Ck-1t.

nnCk-1)

-

r!(n-

!.

2. Evaluate "C+ E"*lCk-1

"'C+("

.

"Cr+3

"C,2= "C,+

where M, is an integer

="C+ ("C+"u*

(1)

"C, + "C,+2

From(1),2r+1)!(1

!

(tn

is

in AP

"C,, = integer

is divisible byn

+n) =n

(rn

*

"Cr,2"C,,a are

2."C=

(kn)!=n! (n! M) (n!

Value

"Cr, l

"C

(rn)!

n!

( (rn +1)(ru +2)...

Note

found n, r cannot be

integers 3. Show that two positive "C,,2, "C,.s are in AP such that "C,,"C,,1,

+n)! +2).. (rn + )(rn (rm)!n! -

(kr)!

Ck

Ik- 1)n +n] (1)

+ 1)(rn+2).. 1-2-3. rn1-2-3 .. (rn)

(1 +1)(rn

+ 21

mCk. 1)

+

quadratic in (n- ),

127+5r'+3r+2

2r492+7r+14 2r 9 5

+9

2r +5

and

r+7r+ 14 =r+3r +2

= -3. or 4r=-12, i.e., r or are absurd. Both 9 5 and r =-3 hold at the same time. (1) and (2) cannot + 3 cannot be in AP.

"C"C

"C.2. "C

A-154 4.

Problems Plus in

IT Mathematics

How many numbers ofn digits can be made with the nonzero digits in which no two consecutive digits are

the same?

There are nine nonmzero digits, namely 1,2,3,... and 9.

Now, a number is divisible by 6 if it is even evor as well as divisible by 3. So, the number of 4-digit numbers divisiblot

by

6 that can be made with 1,2, 4, 5 =2 x 3! ( the numbers should have an even digit in the units place)

8

In order to make an n-digit number we have to fill n places by using the nine digits. As no two consecutive digits are to be the same, a digit used in a place cannot be used in the next place but it can be used again in the place

3P3

coming after the next place. So the first place can be filled in 9 ways; the second place can be filled in 8 ways

The number of numbers of 4 digits, divisible by can be made with 0, 3, 4,5

the required number of

=9 x8

5.

x

numbers Sx 8x... to n factors = 9 x 8"-.

How many numbers of 4 digits can be made with the digits 0, 1, 2, 3, 4, 5 which are being unrepeated in the same divisible by 3, digits number? How many of these wil1 be divisible by 6?

We know that a number is divisible by 3 if the sum of the digits in the number is divisible by 3.

Here 0+ 1 2 3+4 +5 omitted whose sum is 3 or

15; so two digits are to be 6 or 9.

Hence, the number of four digits can be made by either 1,2, 4,5 or 0,3,4,5 (omitting two digits whose sum is 3) 0,1,3,5 or 0,2, 3,4 (omitting two digits whose sum is 6) 0, 1,2,3 (omitting two digits whose sum is 9) The number of 4-digit numbers that can be made with

1,2, 4,5 =

*P, = 4 !.

The number of 4-digit numbers that can be made by the digits in any one of remaining four groups (each

containing =

the

0)

4!-3!

(':

the number of numbers beginning with 0=*P^ = 3 !)

required number of 4-digit numbers divisible by 3

= 4!+4(4! 24

+4(24

3) 6)

=

96.

that

= (3!-2!)+3! numbers should have 4 o 0 in units place and should not come in thousands place).

(rejecting the digit used in the first placc)

the third place can be filled in 7+1, i.e., 8 ways (rejecting the digit used in the second place but including the digit used in the first place) and so on.

6 .

(the

(

0-O|0)-Tn 2P2

3Pa

3P3

Similarly, the number of numbers of 4 digits, divisiblo by 6, that can be made with 0, 1,2,3

=(3!-2)+3!

The number of 4-digit numbers divisible by 6 that can be made with the digits 0, 1,3, 5 = 3 !. The number of numbers of 4 digits, divisible by 6, that canbe made with 0, 2,3, 4

= (3!-2!) +(3! -2!)

+3!

(the

numbers should have 4 or 2 or 0 in units place and 0 should not come in thousands place). the required 4-digit numbers divisible by 6 =2

x3! +(3!-2!)+3!+(3!-2!) +3!+3! +(3!-2

12+4

+6+4+6+6+4

6. How many words can be INTERMEDIATE if

+ (3!-2) +3! +4 +6 52. )

made with letters of the word

(i) the

words neither begin with I nor end with E (ii) the vowels and consonants alternate in the words (iii) the vowels are always consecutive

no vowel is between two consonants consonants does not change (vi) the order of vowels does not change? (i) The required number of words = (the number of words without restriction) the number of words beginning with 1) (the number of words ending with E) +(the number of words beginning with I and ending with E) (iv)

(v) the relative order of vowels and

A-155

Permutation and Combination

vords beginnin (because word

with I as well as words contain me words beginning with ding with Ewith E).

consonants if all the Consonants become consecutive the required number of words when all the the number of arrangements consonants are consecutive

uv) No vowel will be between two

ending number of words without restriction The there are 12 letters in which there 12-(

Iand

213:2!

number of words beginning with

The

312!

extreme lett place we are left to with I in the letters NTERMEDIATE in which there are ange 11 three Es). two Ts and number of words ending with E212121 The

the extreme right place we are left to MEDIAT in which there are arrange 11 letters Ts). Es and two two Is, two I number of words beginning with and ending

with

7!6! 2137

are two Is, three Es and two "Ts).

151200. and consonants will V The relative order of vowels of letters the not change if in the arrangement 1st, 4th, 7th, VOwels occupy places of vowels, i.e., occupy their 9th, 10th, 12th places and consonants 11th places. places, i.e., 2nd, 3rd, 5th, 6th, 8th, the required number of words

.

E in

The

6!6! 2!3!*2T

if no two order of vowels will not change arrangement vowels interchange places, i.e., in the all the vowels are treated as identical. have the same For example LATE, ATLE, TLAE, etc., ETLA, TLEA, order of vowels A, E. But LETA, E. So, etc., have changed order of vowels A, LATE is counted but LETA is not. V then LVTV If A, E, are taken as identical, say arrangement by does not give a new V.) V, interchanging The required number of words in the number of arrangements of 12 letters identical which 6 vowels are treated as

extreme in the extreme left and E in the 10 letters to arrange are left right places we Ts and two Es). NTERMEDIAT in which there are two

(with

I

of words the required number 11! 10! 12! 11! 2!2! 2!2!2! 2!3!2! 3!2! 10! 10 11 - 11 2-11.3+6)-85 24 213121(12 Gi)

There are 6 vowels and 6 consonants. So the number of words in which vowels and consonants

number of words in which vowels occupy odd places and consonants occupy even

(the

7 In how many ways can 5 identical black balls, identical red balls and 6 identical green balls be arranged in a row so that at least one ball is separated from balls of the same colour?

places) +(the number of words in which consonants occupy odd places and vowels occupy even places)

7.

6!

The required number of ways

2131 22

2131 43200.

219 (1)

are two Ts also)

there

alternate =

21600.

(vi) The

10 22!

with E

(as above)

Considering the 6 vowels IEEIAE as one thing, the consonants number of arrangements of this with 6 (

there are two Ts in the consonants).

=

without restriction) (the number of ways when balls of each colours

(the number of ways -

are consecutive)

18!-3!( 5!7!6!

6 consecutive in themselves VOwels can be arranged among 6! Ways.

For each of these arrangements, the

2!3 the

required number of words

2x

151200.

18!

5!6!76.

there are altogether 18 balls in which 5, 7 and 6 are identical; and considering balls of the same colour as one thing there are 3 things, there being no arrangement between balls of the same colour)

A1 Prews

12rs in

Ten

guests are to be seated adies 1he ladies insist on in a or of which tlhvee are sitting together while Iwo ot the gentlemen eiuse to take consecutive seats. in how many ways can the guests be seated?

Considering 3 ladius as one objort, the mumber ot arrangennents =*F

gentlemen

to be arrangad). and i But the 3 ladies can

goup of ladies, ie.,S things ar

be arrngei among themselves in

3! ways

the number of arrangenments in which 3 ladies ane together= *Psx3 Next, considering 3 ladies as

one object and 2 gentlemen efused to sit together) as another object, the number of arrangements (who

="P-

x3x2!.

the required number =

-

of arrangementss (the number of arrangements in which 3 ladies are together) {the

number of arrangements in which 3 ladies are together and 2 gentlemen (who refused to sit together) are together}

= $Ps

3!

-P,

3! -2!

=8!3!-7!.3! 2!

=7 !(Sx3!-3!2!)=5040(48 -12)

IT NMatlheratis After printing all these words, the wond cp RICKET

printed. the nJuirecd place ot the word CRICKET ( x4!+2!)+1 =531

5!+2

the wond will be printed in the 531th plae 10.

,

In a partieular progranming language. variable name can consist of a sequence of one lid alphanumerie characters A, B, C, tosix beginning with a letter. Find the Z, 0, 1,2 ,9 total valid variable names. ver

The number of valid variable names of one charaa

26

of

racter

(

there are 26 letters and the variable name must begin with a letter). The number of vali variablenames of two characters 26 x36 ( the first place shou have the second place can have a letter and any one of the26+10 alphanumeric characters,

repetition being allowed). The number of valid variable names of three characters ters 26 x 36 x 36 and so on. the total number of valid variable names of one to six characters

26+26 26(1+36

5040 x 36 181440.

Wilbe

x36+.. +36+.. +36 S)

x 36 +26

=26 36-1-2

+ 26 x 365

36-1).

9. A

dictionary is printed consisting of 7-lettered words that can be made with the letters of the word CRICKET. If the words are printed in the alphabetic order, as in an ordinary dictionary, find the position of the word CRICKET in that dictionary.

In alphabetic order the letters are C, C, E, I, K, R, T. Words beginning with C will be printed at begins with C. The number of wordsfirst. Our word also beginning with C and having C or E or I or K in the second place =

many numbers of five digits can be made with at least one repeated digit? Total number of numbers of five digits when digits can be repeated = 9x 10 ( there are ten digits 0, 1,2, ..., 9 and 0 cannot go in the ten thousands place) 11. How

4x5!.

E

x

x

K: x CRICx

CRIC

x

Exx

the number of numbers which may or may not have repeated digits = 9 x 10 But the number of numbers which do not have repeated

digits

1

T

The number of words beginning with CR and having C or E in the third place = 2 x 4!. The number of words beginning with CRIC and having E in the fifth place = 2!

Ps-P

( 10P

9PA

the required number of numbers having at least one repeated digit

A157 Permulation and Combination

4 in units place The number of numbers with

digits) of numbers of live digits having

nber of numbers

(total

of five

dhe nber (tle digit) no repeated P)

P-P the

9x10-"P 10-9.8.7.6(10- 1) 9x 27216 62784

=0x6+ 4x 4 =

.

The or

-

LL

10

10

10

used). 1,3,5, 7,9 will be number of numbers (20,000 being .the required excluded)

=4x 10 x 10 x 10 x 5-1= 19999. of four digits can be made How many even numbers numbers. 4? Find the sum of the with the digits 0,3, 5, 0 or 4 in the units place. The even numbers have

the

3

the 0 of four digits with in The number of even numbers .. (1)

units place = "Pa 4 in the of four digits with The number of even numbers (2)

units place="P,-?P2

4

= 2! in tens place

=

2.

= 32. =0x2+3x3 +4 x2 +5x3 hundreds place = 32.

Similarly, sum of the digits in place with 3 in thousands The number of numbers = 2x*P2 =4.

place with 5 in thousands The number of numbers

=2xP2 =4. The number of numbers 2! 2.

the

place with 4 in thousands

thousands place sum of the digits in

3x4+5 x4+4x2even40.numbers

.the

P=3!=6

=2.

with 5 in terns place The number of numbers caseof 3). 2!+1=3 (as in sum of the digits in tens place

13.

required sum of x 40x 1000 +32 x 100+32

10 + 16

43536.

of identical balls of four There are unlimited number arrangements of at most different colours. How many made by using them? 8 balls in a row can be arrangements of one ball = 4, The number of different balls. because there are only four balls of arrangements of two

14.

The number

2

numbers total number of even 10. = P3+ PP3 -2P2) = 6 +6-2 4 in units place. The even numbers have 0 or with 0 in units place 1he number of numbers

2

with 3 in tens place

with The number of numbers

55

=

=2!+1 =3.

filled in 4 ways. thousands place can be the ten thousands, ndreds and tens places can be Each of ways. filled in 10 10 x 10x 10 places can be filled in 4x four first the So ways. used is either these the sum of the digits After filling even or odd. if the sum can be filled in 5 ways (' 8 will be 4, 6, the last place 2, 0, digits even, one of the of the digits is digits the of one is odd, of the digits used and if the sum

P

16.

The number of numbers

Co

3

for all the in units place

0 in tens place The number of numbers' with

natural numbers are there lying between How many60,000, the sum of digits being eyen? 20.000 and five digits beginning with 2,3, 4 mbers will be of numt

12.

the digits

fron (2))

numbers

=

90000

sum of

=4

4, etc. arrangements the required number of = 4x 4 =

=4+43+4++4= 4(4"-) 41

from (1)1

-(4-1)=x

65535=4 x 21845 87380.

A-158 Problems

hen

rom

a

given

5

well-shuftled pack of

Pus

a

cards, player is cards. If the cands have consecutive values then it is said that the player has a run and if they are also trom the samne suit it is said that the player nasa runing tiash. In how player hold a (i) run ii) many different ways can a running flash?

are

4 suits ot having 13 cands, thecards ina pack

1( In a K.

A), 2, 3,

..

game of cards

values being

52

in

tT Mathematics The total nuniber of points of intersectinn on = the number of selections of 2 lines fro

om n

"C

lines

of 52 cards, cach suit

10, 11(= ). 12(= Q), 13(= K). A is also given

the next value above

our problem is to select 5 consecutive cards to have a unning flash when the cards of a suit are arranged as follows:

Clearly, each line will cut the

n-1 points.

out of "C points, n-

1

remaining n

points are on each

The number of lines obtained by joining any

ines

at

line. ithe

"C points n{n-1) fn(n -1)

"caC"

The nunmber of same suit from selections of 5 consecutive cards of the

the above arrangenment = 10, (1,2,3, 4, 5), (2,3,4, 5, 6), ..

mt-10n-n-2) 8

because they are like (10, 11, 12, 13, 14).

The number of selections of 5 cards of consecutive values from the pack 10 x 4, because in each place we have 4 choices of suits. the total number of running flashes 10

10

10

But from each of the n sets of (n

-1C lincs have been counted

=

10

x

4

=

the required number of new lines

n(n-10--2)1x"-'C

10240.

8

nn-10--2),1

16. There

are m bags which are consecutive positive integers numbered by m number k. Each bag contains starting with the flowers as the number levelled as many different against the bag. A boy has to pick up k flowers from any one of the bags. In how many different ways can he do the work? The total number of ways +***C C

=*C+'C

points

which are not new li nes. So nx"-C, lines are not new lines.

10 =40

and the total nunmber of runs

- 1) collinear.

x-D-2)

8

n(

+..

8

2-n -2-40n -2)}

=no-1%-

51+6)

=n-1)-2)01

-3).

'C+**C.1- ** 'Ck.1)

Ck1**Ck-1)+. +(

C. 1-** m-'Ck.1)

18. There are two sets of parallel lines, being xcos a + ysin a = p; p=1,,

3,..., m and ycos a-Xsin a = ; q = 1, 2, 3,..., 1(n> m) where a is a given constant. Show that the lines form

because C-1= "Cn +"Cm- 1 i.e., "Cm="* "Cm-1"Cm 1 "*

k 17. There

m(m -1)(3n-m-1) squares.

Ck-1

are n coplanar straight lines, no two being parallel and no three are concurrent. How many different new straight lines will be formed by joining the intersection points of the given lines?

their equations

Clearly xcos a + ysin a = p; p = 1,2, 3,., m represent m parallel lines, the distance between consecutive lines being 1. = q; q = 1,2,3,.. ycos a xsina , n represent " parallel lines, the distance between consecutive lines being 1.

Also

A-159 Permutation and Combination 3 are ladies and 4 of them are relatives, them 19. A man has relatives, 3 of different ways 8entlemen; his wife has 7 how many 3 In gentlemen. 4 adies and 3 ladies and of party can they invite a dinner man's relatives 3 of the are there that so gentlemen relatives? and 3 of the wife's man's relatives by HL, lady man's relatives Let us denote the wife's gentleman gentleman relatives by HG, by WL. by WG and wife's lady relatives = 3. n(WG) = 4, n(WL) = 3, n(HG) = Here n(HL) 4,

m

7

n-1

each line of the first set

cot a;

the slope of Again, line of the second set = tan a. slopeofeach the (tan a) =- the two sets cut orthogonally (-cot a)

it, at

=-

3HL, 3WG

90o

consisti of two linessof the first set and two lines second set will form a square, if they are the distances. lines of equal taken at sides of length 1 unit number of squares of 2 The nunmber of selections of lines from m lines

theat

2 lines from unit distance and

distance

The

the

lines from

lines at

a

distance distance of 2 units ... (1,3), (2, 4), will be selections (°: ... = (M -2)(n-2) (m-2, m) and (1, 3), (2, 4), a

=

(1

number of squares 1Mn 1) + (m-2)(n -

-

2) + (m

1

-

21+ {mn -3(m +mn-(m

1)mn

-

1)mn-

(

=m(m 6

-

men if

-

1)(m +n)

The number and 4 men:

(m

1)(3n

=3x4x 4 x 3= 144 *C C C; xCx 3 6x 6x3 324

324

=

485.

Combinations

=7056

6w, 4m

+(m-1)9.

+n)

-

1)*)

(77-)m(21711)

2 + n)

m -1).

+21-1}

8:

CxC4!5!6!2! 9824 x*3528

5w, 5m

+ (m

x *C2

work with MrY insist to work together? least 4 women of committees of 10 with at

m-1)

(»1 +n)(1+2+3... +m-1)

6n-3(m

6

-

+ 1n) +31

+(1+2+3... (1

(n

16

(i) Ms X refuses to (ii) Ms X and Mr Y

+ n) +1)

(m

+{mm 2(n1 + n) +

(1

8

-3)n -3)

m-1)

=

of 10 be selected many ways can a committee women and 4 men from 9 with at least 4 women and

4w, 6m

+.. +(m = {mn

x

9!

and so on.

total

Cx'C,

x4

20. In how

Possibilities

n)

(-2,

4

=1 CC=1x1 'C,

1HG,2HL, 2WG, 1WL

16+1+144

sides of length 2 units number of selections of 2 lines from m lines at of 2 units and 2

2WL 2HG, 1HL, 1WG,

CC3

ways Therefore, required number of

umber of squares of =

3HG, 3WL

lines at unit

2), (2, 3), ... f: elections will be (1, (m 1, ni) from the first set and . n-1, n) from (1,2), (2, 3), the second set).

1(n-1)

(nt

n

Combinations

Possibilities

C C=5 24 CxCS,876-5-5880

at least 4 number of committees of 10 with women and 4 men 3528 7056 + 58S0 = 16464. in which Ms X and The number of such committees MrY are present: of committees of In this case we have to find the number 8 women and 8with at least 3 women and 3 men from committees in which 7 men. As above, the number of Ms X and Mr Y work together

total

="C^x

Cs+

"C,

x

"C+"Cs x 'C3

don uny ways can tvwO distinct subsets 3) elements be selectei so that the Aot k ney set etty owO nmon eleutents? n

of

hare

wv sutes

tt

e

allai P and Q Tha lements foe

ederettsut ot elenruts torboth th

r tte sudset tnielen

eNns

nents tr

nts

tnm theem and tien: can vary tum --eneas ir, te nunter ot slrtions s auss themunter of FRr a

ot any olrtins things trom r number

things

is

urber oi wlartions

the tal

--1 eadin

ut

te cas when both the subsets ant equal er having oy the two mon dlements, are every rair ot Qis apoaring twie like

.

Hene the myuirni number ot ways

end

osngC="C

C-2i.

*C.)-1}

D.2-1-1}4

hai of ie series of incmial coeficients

-2 Ifis

side, 4 on the row side and one to steer. There are 11 crew of which 2 can stroke only, 1 can row only while 3 can steer only. In how many ways the crew can be aranged for the boat?

2-

odd. ie.r

=

2

-

1

thenr m =

1.

the required number of subsets

m-C1

C

z-Co-1 2m-C

=

23. A boat is to be manned by 9 crew with 4 on the stroke

The number of ways to select the man to steer Of the remaining 8 persons, 2 (who

-=22=2-1

"C.

can stroke only) are already seletted for the stroke side and 1 is selected for the row side (who can row only). So, of the remaining 5 crew we have to select 2 for the stroke side and 3 for the row side.

ToW

half of the series of binomial coefficients

22

=

steer stroke

A-161

P'errutstion arid Cembinatin

t

identical), 13 are cards is (i.e.. two where each card ne same denomination irom 52 cards

nunber otselections- 'c, x 'C x'C uitd tho stroke side as well as nw tthe peisons on 4 ways tor every selection of the natranged in thenuirnd

wo in nunmber. selections of ne number of

'C4!s4!= 17280. «

elements.

A

aining contain isasetThe set A is nconstructed

subset

P of A is

many

In how mangoes and 4 apples. made if 27. There are 5 of fruits be selection a can different ways different samme kind are (i) fruits of the kind are identical? (ii) fruits of the same number of to select any ways of number () The mangoes

r

,

"Cs =25. =C+C+C+. number of ways to select any

The nunmber of

suisesand Prove from

apples

objects ol ways to select n that the number rest the which n are identical and 3n objects of

are difterent,is

=2x2-1

2**2r

n-r

0 to n. Herer varies from of selections the nequired number

2C-,

=2"C,+2"C,-+2"Cn-

=2-1. (i) The required

=(514

number 37800. Also number of factors of the proper divisors of the find the sum of the odd

2C%

number.

Here

3"C+"C,"C+... C

37SO0 = 378 x 100

coefficients+

C

23- 2 26.

shuffled together. Two different packs of cards are among 4 players, each getting

Cards are dealt equally a player get his cards 13 cards. In how many ways can samne same suit with the if no two cards are from the

denomination? each card being 2 in Here, there are 52 distinct cards, the same suit with number. As no two cards are to be of

2x5

2

5C,}+5.*C of the series of binomial

x 126 x

3

x5 =33x 42 x23x5 =3 x3x3x7 =233x5 x7factors

writing in reverse order

sum of the first

number of ways = x5-1 =29. 1)-1 6

28. Find the

r0

-C-3CChalf

C=2

fruits of ways to select number required the way in which 0 lencluding the selected} mangoes and 0 apples are

objects from u identical of selections of r number objects The objects from 2n different and objarts

=

2

C2"13!39

by eplacing the

A is chosen. Find the A subsc Q of lemen nts ot : of selecting and Q so that P and Q l' of ways nunnoninterseeting. ar elenents is ovailable for selection in making conmon t the But and Q must not have any vell as Q. element can be taken in 3 ways ch mt. EQ or e P, e Q) or Q which is can be taken in 3"-1 rays lementsumber ot ways to select nonintersarting are al to the case when P,Q ). Q (eacliuding the

o

distinct cards tron 52

13

cards C selected in 2 iways be can cards But each of the 13 packs). (belonging to either of the two ways the required nunmber of

number ot ways

C

A A

Caris ae to be seleted

the nequired number of from the total nunmber of selections =

(2, 2,2,), (G, 3, 3). (5, 5),. 7)

=3+13 1211

+

1)-2

(excluding 2°.30.5.70, i.e., 1 and 23 33 5.7, i.e., 37s00 as factors)

=4x4x 3 x2 The required sum

(333°

2 96-2

94.

-3%5°+ 5' 2

-5 x7° +7')-1

as a factor and 1 as a divisor are to be excluded)

A-162 Problems Plus in lIT Mathematics

3-

14141 12! =91

12!. However, the first r nethod metho is advised because it covers all situations.] When 3 girs seat together in the back row of an I: Possibilities Combinations Permutations Van I Van lI

8-1

(3-15-18 = (3

2x 4

-

1(5

1)-

-

Note

1

=

80

x

124

1

9919.

(i) If we have to find the number of odd proper

3g. 4b

5b

divisors then the required number = (3 +102+ 101+ 1)

3g, 3b

6b

we

3g. 2b

7b

-1 have toexclude 2s and the

number30.50.7)

(ii) If we have to find the number of even proper divisors

3(3

then the required number +102 + 101 +1)-1

weat

the

havetoexclude 23.33.52.7 and

+°Cx1 x2 9!

= 9!3!x7x6+9!x 14 =9!3!(42

5

Cs

CxCsx P;

x"Ps

12!

7!5! 77!

6

6

12CxC

C,xCx

1

x3!xx7!

x2x3!4!7!

x4!+9!x 3x24

56 +12) = 660

x9

the required number of ways

2x 660x9!= 1320x9!. 30.

1C7

girlsin

Similarly, the number of sitting arrangenments with3 girls in Van II = 660 x9!

Permutations

Van II

3

+1712x3x4!7!

persons = 3+9 I

9!

3!6!

29. In how many ways can 3 girls and 9 boys be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many

Van

number of sitting arangements with

=Cx1x2x314!xC,*1*2x3!x

least one 2 is to be selected). find the number of proper divisors divisible by 10 then the required number = 3 x (3+1)*2x(1 +1)-1 at least one 2 and one 5 are to be selected).

(ii) If we have to

There are 7 seats in each of the vans. The total number of = 12. Now: Possibilities Combinations

CCs CxCx (2x3!)4!xp, CxC,* 2x3)°P^ x7P, C CC CxCx (2 x 319 Pxp

because the 3 girls in the back row can take place i seatsremaining together in 2 x3! ways. in 4 Van I

sitting arrangements are possible if 3 girls should sit together in the back row on adjacent seats?

x

In the given figure you have the road plan of a city. A man standing at X wants to reach the cinema hall at Y by the shortest path. What is the number of different paths that he can take? UA

'P, x "P

L

12

6!67!.7!

5

total

12CxC

7

- 21 (12

!) +

12!.7.7.7.6 49 (12

!) +21 (12

2

L4

number of sitting arrangements

_121 7:6 [Note

CsxC x 'P; x"Pz 12 517727

!) = 91

(12

!)

We could have directly arranged 12 persons in 14 seats. The number of ways for this is "P12, i.e.,

L

As the man wants to travel by one of the many possible shortest paths, he will never turn to the right or turn downwards. So a travel by one of the shortest paths is to take 4 horizontal pieces and 4 vertical pieces of roads. As he cannot take a right turn, he will use only one or the five horizonal pieces in the same vertical column. Similarly, he wili use only one of the five vertical pieces

Permutation and Combination

row sanme horizonta L the in hortest path is an arrangement of eight things sho a so that the order of Ls and Us do U,, U,, U,U, L, L. clearly L cannot be taken without taking change (.'c not cannot be taken without taking U, etc.)

L Uy

Hence, the

=84x36 9 x (3360

Us do not change the order of of arrangements treating Ls as the number identical and Us as identical

1

1

five digits that can be The number of nunmbers of different digits where nmade with 1, 1,0 and two other

24

1,1

made numbers of five digits can be of How many identical digits?

How many two having exactly the repeated digits in consecutive these wil1 have

places?

digits be 0, 0 then three other digits "C ways. 1, 2,3, .,9 in can be made of numbers of five digits that The number digits other different with 0,0 and three can

5-4 the

without restriction number of arrangements 0 is 4 number of numbers beginning with

and the

is !)

that can be

number of numbers of five digits different digits made with 0, 0 and thrce other

the

-'C,

are consecutiv

aretreated as one object) two of numbers with Hence, the required number (4! -3!)

C2 x

1

(* 1,

identical digits consecutive

= "C,(4!-3!) +9/'C x 4!+ "C2 x (4!-3 ! =

84

x 18

+9(56

x 24

+

28

x

18)

18144. = 1512 + 16632 =

digits can be made with at most each of whih can be used

nunmbers of five 32. How many

the digits 1, 2, 3 thrice in a number?

make numbers digits 1, 1, 1, 2,2, 2, 3,3, 3 to will be as follows of five digits. The digits (constructionwise): (i) three identical, one pair We have the

(forexample: 1,

1,

1,2, 2, etc.)

different (i) three identical, two

4

The number of numbers with 1,1 and three other

made of five digits that can be 0 otlher than different digits

(forexample: 3,3, 3, 1,2, etc.)

(iii) two

5

pairs, one different

(for example: 2, 2, 1, 1, 3, etc.) possilbilities start with In order to cover all the the identical digits and go on reducing all

that can be made The number of numbers of five digits with 1,1,0 and two other different digits

5!

1

=*Cx4

8 7:6-5-2.7.5-70.

identical f the from selected be

45360.

one object). (.0,0 are treated ascan be made ="C44! -3) five digits that different The number of numbers of 9) and three other with 1, (or 2, 2 or ... or 9, 1, are consecutive digits other than 0, where as one object). C 1, are treated

U,

31.

=

1344)

of five number of numbers other Next, similarly as before, the with 0, 0 and three digits that can be made consecutive different digits when 0,0 are

of arrangements of L, L, Ly, L the number U U, where the order of Ls as well as

8!

42336

3024

number of shortest paths

4!4!

24)+9x {56 x 60 28x (60 12

84(60

4

digits that can be number of numbers of five made with 1,1 and three other different digits

the

ultimately number of identical digists, reaching all different digits. identical digits, The number of selections of three one pair = °C, xC. of Corresponding to each selection, the number

TIP G)

numbers that can be made

=

5!

3!2!

total number of numbers of three identical digits and one pair

the

etc. Similarly for numbers having 2, 2 or 3,3, having exactly two the required number of numbers

repeated digits

5! CC,3121 3x2x

60

ii) The number of selections of three identical

and two different digits - 'C,

x *C,

(1)

digits

A-i6

PobiemsPusim UT Mathemttis total numer of digits and two ditterent numlers ot thre digits the

-'c, (i)

idetntical

Pc-3

digit 'C 'C

'cC;

=

'c,

5

30=90.

the aquind nunmber of numbers 0 60

0=

x

3

identical,

1

CC

2 1

pairs, different

1

pair,

3 different

.the

CC CC

CC

CC2121 5!

CC

'C, +*C,

=

1

+2x

1+6

=

coefficient of

C

+ *C,

3+3x3+3 + 9+9+12

The required number

r

in

+r"x1-x3 -x* +x)

(1-x-

4 5.6 7445:6:7-8 ,5 4!

5

r

in

1+

4x

coefficient of

-xS)

+

10x+ 20x* +35+56x1

neglecting powers higher than -4 1

=41.

34. In how many different elements be partitioned

xC, +C,

« 3C

CC+C,°C3 =

(1-x1-xX1 -x*1 -x 1 -*)

= 56 10

required number of selections =

coetticient of r* in

x

3identical.

different

1- 1- x 1-x

(1--

C,C

+xr+r+x)

coefficient of xin

=

pair

2

=

'C

4 identical

+x"Mr+x'+x*+r+a

x(++x

Find the total number of selections of 5 letters from five As, four Bs, three Cs and two Ds. Also find the number of 5-letter words that can be made. There is one letter Ato take 5 identical letters, two ietters A, B to take 4 identical letters, three letters A. B, C to take 3identical ietters and four letters A, B. C, D to takepairs. Therc ane only four different letters. Possible structures, selections and arrangements are given in a tabuiar form as below: Possibilities Combinations Permutations 5 identical C

different

CCx

+30 + 90 + 180+ 360 + 240 =901. Note The number of selections is also equal to the coefficient of x* in 1

=

210.

33.

1

CC

ot two pairs, one dieent

the total nunmber of nunbers of two pairs and

onedifferent digit

CC

(2

Thenumber of soloctions

3!2

x

4

3 +6x2+4x

1

= 41.

of words (i.e., permutations)

5!

ways can a set A of 3n into 3 subsets of equal

number of elements? (The subsets partition if PuQUR =A, P nR =¢, QnR

P,

Q, R form a

=

¢, RnP=¢.) The number of ways to partition the set - the number of ways in which 3n different things can be divided in 3 equal groups = x (say). Then 3n different things can be distributed among

persons equally in xx (3

!)

ways.

But the number of ways to distribute 3n different things

equally among 3 persons

= 3"C

31

n!2n!

C,,x "C,

2! n!n!n

3n

!0!

(n )3

A-165 Permutation and Combination

a box giving 1 ball to equal after Similarly, identical. are two be divided in the remaining 4 are to groups. of ways the required number

3n

6(1!3

(

x3

boxes. Each can hold are to be placed in three 1h Fiveb slls balls. In ihow many different ways can we the five remains empty, if balls so that no box

the

place are all different balls and boxes i balls are identical but boxes are different (ii) but boxes are identical balls are fferent are identical boxes (ii) as as well ) balls identical but boxes are are boxes as as well (v)ballsin a row? kept can have balls in to remain empty, boxes box hox is numbers: no Asfollowing 1,2, 2. the or Possibilities 1, 1,3 bute the balls in number of ways to a) The

goups of1, 1,3 -

=CxCx C

2!

distribute Similarly, the total number of ways to 1,2,2 balls to the boxes

3!

required number of ways

. ..

36. Let

=

whole answer in tabular form, permutations possibilities combinations

Note Writing the

CxCxC

C x'C, x°C,x = 5

1,2,2

C x'Cx*C

,),

x4x3

= 60

C,x'C,x*C;x

x1

+2+*+...+Xk= =

y1+y2+ Y3t

or

.

k.

yk+ Let I = yi +1, x = y2 +2,.X Putting these in the equation, 1+1) + (2 + 2) + (y3 +3) +. +(

+k)

=

n

+

=n-(1+2+3...

+k)

+

,

=

+ n-kk21)

*2t1x23+3 (1) When balls are different and boxes are identical, atter giving 3 balls to a box, the remaining 2 are to

divided in two equal groups because the boxes

(1)

= m

equal to The number of solutions will be things identical m number of ways in which distributed the number 1) can be thing persons (here k variables) and this is

the total (here the

among k equal to

m+k-Ck-1 (1) number of non-negative integral solution of and hence that of the given equation

the

mi+K-"Ck-1

(m+k-1) (k-1) ! m where

= required number of ways 60 +90 150.

1) When balls are identical but boxes are different the number of combinations will be 1 in each case. the required number of ways

be

x21,

satisfying the condition

=5x6x3=90 the

-25

integers such that and k be positive of solutions Find the number integers *22,.. ,xk2 k, all

1+ 2+ V3t.

= 5x4x3+5 x6 x3 60 + 90 150.

3

n

n

x*C, x°C,xi+°Cx'CxC,x

-'C

10 + 15

0. m (say) where m is an integer2 non-negative Now, we have to find the number of integral solutions of the equation

-'C,x*C, x*C,xi the

-

G,c

the are identical, boxes as well as balls (V) When arrangements will be and combinations number of 1 each in both cases. ways the required number of =1x1+1x1 =2. will be treated in a row, they kept are boxes When ways (V the number of as different. So, in this case will be the same as in (ii).

CxC x'C. can interchange their content, no

the boxes a new way when boxes exchange giving numbers interchange. equal containing balls in ways to distribute 1, 1, 3 number the totalboxes of balls to the But

=Gx

m =

n- kk+1). 2

Note (1) The

number of solutions of

1+ 2+ 3 t.

+Uk= m where 1, y2 etc., and m are non-negative integers, can also be obtained by finding coefficient in a suitable expansion. Consider the expansion of (1 + x+**+...)*.

A-166 Problems Plus

(+x

+**+...) x{+x

n IT Mathematics

where n20, +a*t

.)X.

Then

(+a+a*+...)

there being k factors. The coefticient of a power of r, say 1 " in the RHS = 1number of ways in which m can be split ink parts (non-negative integers), sum of the parts being

+112 + 13 t

=

coefficient of x " in

=

coefficient of

x" in (1

coefficient of "

++r+...)a++*+...)

x(+r+x*+)(r+x'

.)

=

-

+k-1)!

! (k-1)! (2) The number of non-negative integral ofyi + y2 +.. +Y S m can be obtained bysolutions finding the number of non-negative integral solutions of V*y2+... +Y+Yx-1 =m (where yk. 1 is also non-negative integer). Their non-negative integral solutions will number of be equal. 37. Find the number of non-negative integral solutions of 2r+y + = 21.

Clearly a =0, 1,2, 3, ..., 10. Let = x k; then 0 sks When r = k, y +z = 21 2k. The number of non-negative integral solutions

10.

(103-) 45-6.. (100 k)! (103-)

the number of ways to distribute (21 -2k) identical things (each thing is the number 1)

among 2 persons

21-2k2-1C.-, = 22-*C

=22

required number of solutions

=

2(1 +2

=

=

-C,

k0

10

Cy+ 10

C.

+*C

coefficient of r* in I(1

)"

+ (1 + x)

coefficient of x in (1+x)3, =

coefficient of r* in

(1 +

x)*

= coefficient of x* in (1+x)

(1

..

-1

+x)

((1 + x)

+(1 +x)1

-1 11

104

C

39. How many triangles with sides of lengths in integral cm can we get if the length of the greatest side is given to be 2m cm? Also find how many of the

xsy

ni).

possible to make an unsuccessful attempt to open the lock.

A-170

Problems Plus in

IT Mathematics

is made of the words that can be ma 26 dictionary Atelegraph of the word PARKAR by arranging the letters N capable of taking 4 distinct positions, including the wvord "PARKAR" in that diaatis the position of the ionary position of rest. Ii 1023 diiferent signals can be sent in the:same order as that if words are printed in all then find the number of arms. of an ordinary dictionary? many 10-digit numbers can be made with odd How can 6 boy's and 4 girls sitin. 27, In how many ways digits so that no two consecutive digits are the row two girls? between is so that no boy same? each of wo books ana 28. There are three copies two many odd numbers of five digits can be How of three volumes. consisting works cach In now formed with the digits 1,2,3, 4,5 if the digits cannot many ways can a booksel arrange the 12 books in be repeated in the same number? a shelf of one row so that neither the copies of the same n E n N and 300 of Un where u, rove 7.

in

17. Find the coefficient of

-2)(n -3) n1-17(212

independent of 10

termn

o V

x in

the expansion of

A-198 Frubles Plus

(iv (1r)

-

(v23x 28. 29.

4

30. If the

P

4)

,,

coefficient

prove that

c-bd 3c consecutive terms in a binomial expansion with positive 280 and 560 integral index be respectively 14, 84, then find the expansion. Determine

o1o')

36.

38.

+

39.

40.

expansion of

1)th and

the

(r +

(1

+x)"

the coefficient of the 2)th terms are equal, findr.

Prove that in the expansion of (1 x)", cocfficient of x" is double the coefficient of x" expansion

of

t

the in the

2x+ x*"

If in the expansion of (1 +a)", 5th,

the coefficients of the the 6th and the 7th terms are in AP, findn.

Prove that three consecutive binomial coefficients in the

.

expansion of (1 + x)", t e N cannot be in GP for Is there any n such that they are in AP or HP? 41. Givcn positive integers coefficients of 3rth term and r> 1,n > 2 and the (r binomial expansion of (1 x)" +2)th term in the are equal, find r. Ff any

in the

tha

are

expansion of (1+x)", the three consecutive Cuefficients are 165, 330 and 462 then find n.

.e

H

in

b, c, d

c)

xb-

2(a3-

46. Let (1 )(1 in AP, find

bd).

=

If a, az and

#.

are

n

47. If n be a

positive integer then prove that at the integral part P of (5-2v6)* is an od integer. If f be the fractional part of (5+2v6)". prove that = P

45. If (9

45)"

integers and

that(1 Bxp 49. If

+

(2r

)

in

4

(1

37. In the

a,

(

the value of r in the expression if the third term in the expansion is

Prove that the coefficient of the middle term of a)" is equal to the sum middle terms of (1 x)2'-1 of the coefficients of

the

be any tour onsecutive cocfticients ticients in a binomial expansion, show that

45. If

34. If four

10,00,000.

a is

2

+

binomial expansion then

tthat

Also show that

.

a

three consecu then prove

the expansion. 44. If a. b, c. d be four consecutive cocfficients binomial expansion of (1 - r)" then prove

4th term in the binomial expansion o

in

of (1 x)", a, b,c

baC. b-ac

1

is

(x

be

2ac

independent of x, find the value of Also calculate p if the fourth term is 31. Prove that there (independent of x) cannot be a constant term unless na is an integral multiple ain (xx" of b. 32. Find the term which does not contain irrational expression in the expansion of (3 33. If a, b, c 2. and d be the 6th, the 7th, the 8th and the terms respectively 9th

35.

expansion

coefficients

10

x

Find the term independent of x in the expansion or 4

tT Mathemuztics 43. If in the

1

Deternune the constant term in the expansion ot

(

in

50.

=p B

B

where r and

is a positive

B)

=

1

p

are positive proper fraction, prove

represents the least integer greater than then prove that ((fv3+ 1")), n e N, is divisible by 21 Find the greatest coefficient in the expansion ((x)

of

(1z) 51. If x

=find

the greatest term in the expansion of

(14x 52.

(a)

Find the nunerically greatest term in the expansiom of (3 2x) " when x 1.

b)

Find the value of the greatest term in the

expansion of v3 |1 53. In the

expansion of (r + a), if the eleventh term is the geometric mean of the eighth and the twelfth terms, which term in the expansion is the greatest?

54. In the expansion of

when

r it -

isknown

that the 6th term is the greatest term. Find the

possible positive integral values

of n.

At

Binomial 7heerem for Postex Integral Index

geatest cocfficient

the

in

expansion at the greatest the coetficient double in the 1)*"is

s

h

( of enpansion of(1 x) sum Find the . C,

C

Sum the

b6. Find the

series

65. Sum the series

1

57.

31(n-3)1 5 (-5)1

1!(n-)!

58.

Find the

"C; 2

."C"C,+2 P'rove that

59.

(1

.

Provethat "Co "C,-"C;"C, 61.

are

.!, If to

terms

the

s

in

C 1

the epansion of

x("C "C+

'C,+ "**C,++ C

'C:

Prove that

72.

+ If (1 x) " Co Cr+

"*

k

C

2

71.

Ca+.

+

+ (n

+

)"C

(1)a, +

++ (1 I*)=d

(iv)a4s4

(iii)a, +a,

4,X

+

then tind

76. P'rove

that "C,

1-(1+) 77. If

(it) 4d and

()4 b4.

ats-

2-

pove that

a

a,(3"+a,)

coefficients in the The sum of the binomial coefficient of is 1024. Find the expansion

ofa

the binomial expansion. exceeds that of another The eaponent of a binomial coefficients in

in 65.

binomial by 3. The sum of the taken together is 144. expansions of both binomials twoexponents. Find the smaller of the

..

"C-2"C,+3. "C s-

+

"C, +2.

,

then find (IV)a

+

(2n+

(1

+2x)

"C, + 3.

+

= (n+212"

= (u+

1) "C,

75. Prove that (i1)

+

+

n- 1d)C

+ C+{a +d)C +(a 2d)C,++(a+

"C+3.C, +5. "C,+

1f

C,"

series. find the sum of the following

74. Prove that

63.

""C

C+

+

C,

C,+

Co+2-"C, +3. "C,+

then find

"C

"C,

73. Prove that

prove that

(a)then

"C

C+

"C,

-t"C

"C)? *C, 12C,"C

"C,+ "C;+"C,

1001

Prove that

70.

sum:

"C,

3 1

'C

2

c,C,

(n- 1)!1 were

the exprst*

ain

Evaluate

69.

ever positive integer. is an

expansion o

the

+21(1 1)*

x)

(1

in

(1

Find the coefficient of

2nC

"

x

(1x

(1) 67.

C; 'c-

coetficient of

"C -(1 "C t

"C, =0

1

(-1) "(o 31)

12

"C,+= 0

+"C

Also prove that

("C,)"s "C, ("C) "C)'... 78. Find the sum:

1"Cot2. "C2 +3. "C+4 79. Show that

"C

M-

80. Sum the

"C, (m

-

"C,+

.

-2)+(-1) "C, {-

)"C2

(

)0

series

3.C-8.

"C +13. "C

18 "C,+

to(n1) term.

A-200 Problems Pius in irT Mathematices

81.

Evaluate x

82.

first r natural numbers. Prove by binomial expansion that 2

**. "C =

."C,

wherep, denotes the sum of the

214+1

nta

94.

1)(b-

-

1)+

S4. Prove that 1+3n+ 3

(a

2)-

95.

=

2 r=0

86.

96.

2 +8)

ne N.

97.

(r+ 1)2."C, r

then prove that

-*+

Ca+33

+.

+C"

..

10C10

90.

(b)

100.

"CT

Prove that 2n 1CR+21+1C?+2 =

101.

"Cak-

k=

C

(+ 1)2

r=0

91. Find

2

02

2

C,x'. prove that

3"+'-2"1

0

C

(b) (6)

2k

(-1)

2 T

Find the sum:

=

+2C .

c.. 21

21C?+2C3+

+2

C

ic3 +...

+ 27

1c? 1

Find the sum

C1015C +20C 5C

+1

+Cg 1C +20C%.

Ci0

k

(-3),

3nC2r-1 =0

r=1

(k+ 1)k + 2)

"

n!n!

102. Prove that 2

Ck

If (1 +x)

+2"CŽ= (-1)".2n)!

the sum of the products of binomial coefficients "Co, "C, "C "C taken two at a time.

11

Find the sum:

k1

n10.

99. Find

3"+-11 (a)

-

2)! (-)" 2) 2[I2n)2+(-1)". (n !)3

4-1 +2-

even.

... 2n*1c?

-

Prove that

98.

n

+3.

*..

2."C+23+2

(a)

C?+27 +1C

2 C 2c+ 2C3-..

(-1)"

+3

(ii) Prove that

89.

21

4n)+

87. (i) Prove that

k-0

according as n is odd od or

C+2C? + 2"C? +...

(Gi)Co-C

88.

"C

Prove that

Co

3 .

or (-1)2.

21CR

2" -s,

"C7 +2.

Prove that

1

If (1+x) "=Co + Cr+Cr+Czx+... )

0

2

)n +2)

(n+7n

=

-"Cf+"Cs-. +(-1)" ."c

5 n(n-1 M-2)

t Evaluate

Prove that "C =

3!

85.

n!n!

2

Prove that 2 "C "Ck,2

n)(b- 1) =0, n e N.

-

"C;+ "CZ+... + "C}=2n)!

k 0

a-2)(

+(-1)"

+

93. Prove that "C

83. Prove that

ab

+... +(-1) " Cu .

-.

n(n + 1)2

92. Evaluate

k +2

C Cr

where

k=and n is an even integer.

103. Prove that 1

2."C-2

"C-1+3.4."C,-2

+(-1)'(r+ 104. Show

that

C,(1-)-1-

(n+1)2 "1

"C,x ."c,x".

3.

1)(r+2) =2."-cC

-

"C3

+-1)

,1-,

-1. (1-x)"

A-201

Binomial Theorem for Positive Integral Index

Prove

that

i)

-npq+np 2."C,p'q"

105.

2."C?+3. "C} + ... +n.

106.

Prove that 107.

3

C2

+"C2-*C,-+. +"C2.2C,+ .). + 11) a,=2"("C, "C.2-C,

if p+q=1.

Provethat

"C=2n-1)! I(n-1)

Objective Questions

C

C

Fill in the blanks.

+T"C-1

The coefficient of a

118.

"Co

Prove that 10S. "C("CG "Co+

"CC

+1)."C"C2."C

x-when expanded in

..."C 120.

s.e1.1 n1C+'C2 S1 +"* 'C3

s2+ ... +H*'C

.1S,

2

Prove that

+(-1)".

S

"C (n

+1)

is

in the polynomial sum of the coefficients 2163 is_ expansion of (1 +r-3x2) in the coefficients (6) The sum of the numerical expansion of (2x + 3y) is of (a 2 +a)" does 122. If the 5th term in the expansion not contain a then the value ofn is. the expansion of 123. The sum of the rational terms in

(V2+35 10is

p1 2"*

m=P

C+2"*'C+2 1c?+.. 4 2n +Ic? C"C; where 0sis10,0sjs10.

112.

Evaluate

113.

Evaluate 2

114.

If (1+x+x*)"a +a1+ a2x+.

ij

aaa-ay43 +a2.. (ii) a= a2n -i* (ii)

126. If

x ocurs in the expansion

of

then

its

coefficient is

the sum of the binomial coefficients in greatest expansion of (a + b)" is 4096 then the binomial coefficient in the expansion is expansion of 128. The number of terms in the (1+x5) 5 which are free from radicals is . x the expansion 129. If n is even then the coefficient of in

+za +@21-2an 4+1

(1+x+x*+ x*) " = ao+a,x + azx+... +n " #3n

-r

= ag t

y +ay

t.

+2n

then prove that a, = a2n -r Also prove that + + 30 + 5a2 +..+ (4n +1)42 (2n +1)(2 p).

+2)"=ag + a,x + at nEN then prove that

117. If (1 +2x

of the positive integer. If the coefficients of (1 + x)" 2nd, 3rd and 4th terms in the expansion are in AP then the value of n is

-

a3-a+a2-a3 +

If(1 +py+ y"

x+is

127. If the

+421

then prove that

prove that a,=

middle term in the expansion of

125. Let n be a

"Cm"Cp|

111. Evaluate

116.

descending powers of

xis 31. Then n is equal to , 9950 100 S0 and 101 The larger among

124. The

115. If

expansion of

121. (a) The

Prove that

)

of (a + b) in the expansion

the The coefficient in the third term of

119.

snl+9+9**..+9",

110.

b

1s

+"C3)... ("C-1+"C,)

Given

109.

a, = "Co 2C,+"C,.2-*C-2

t

.

2n

of (1+)" |1-is 130. The sum of 2Co+2C +2C,+.. +2C0 is equal 131. The sum of 132. Take the

Co-2C

+2C

to

...+2C1o is equal to

expansions

(1+)"= "C, + "C,x+"Cx+ ... + "Cx" +"C (y+1)"="Coy"+"Cy"-l+"C2y-2

A-2202

P'nblems Plus

+)=

"C"+

"C"y+

The coetficient of

a1+y(r+ Choose

thhe

where

""

the expansion

in

(1+52a)+(1 (a) 5 135. In

(c) 2 in

coctficicnts of r" and enpansion of (l + a)"' " are (a) equal (b) equal bul opposite in sign (c) reciprvcal to each other (d) none of these The number of dissimilar terms

1

142.

(c)9

the expansion ofja*

(a + 2h+3c)

(d) 10

independent of r

(c)-C

(d) none of these The largest cocfficient in the expansion of

156.

is

(a) 137.

C

(Cu

(b) 6

(d)C

(c) 5

The sum of the series

"C C, +C+.

(a)2

(b) 2

(c)2+

"C

of

(c)

(-1) ",

m

(d) none of these

integral part of (8 +3V7)

is even.

there is

independent of x and a expansion

lernm

a term

independent of y but

of x and y both.

no term independent

of (1

+ 27+*°there

term whose coctficient coefficient of any other term. one

coefficients in the expansion

or false.

the expansion of

146. In the

the tern

is

is

is

exactly

not equal

to

1

(b) 2

(c) 2 1-.16 2 (8 =

4

147. In

the expansion

"C,, the sum of the series

148.

If fx)

of

every term

-

function of x.

(d) none of these

)

If C,

C

145. In

19

(d)2 19

139. The sum of the last eight (1 +) is equal to

(a) 2

+"C0

10

(b) 2C

(a)

144. The

(d) 2

2"-

State whether the statements are true

(d) 3

+

is

140.

)

35 when divided by 8 leaves the remainder 1

158.

(b)*C

(1 +

in th

in the expansi.. siono

(c) 45

expansion of (1+)

143. In the

(b) 0

p,9N)

is

(b) 24

(a) 9

the constant term is

-

C

(a)

(d) 2+ the expansion ot

-5v21)° is

(b) 7

(-1) " n

141. The

of x, is

number of terms

(cd)

2)

(e) none of these

number of terms in the exPansion ot (2a+*)" wrhen expanded in descending

(a)

()(-1)2 (n+1)

(c)-1)" (+

v))" is the series.

1

even integer, is

n is

(a) 0 ot

133. The

134. The

T Mathematus

"C"*v*t +"C"

correet option(s).

powes

in

x+

=

then

fa)

is a

is

polynomial function which is an even function.

IC6-2C

+3C

-

... +(-1) "(n + 1)C

149.

6Co-°C+bC2-... + "Cg =0.

Answers 2. 198

3.

84 9.1255 2

2.0020001 11. (i) -20

4. 2

(ii) (-1)

",

(21)

8.

*

10.

672, 5376

(iv)

(2

+ 1)

n! (n +1)!*

"

(iii) -560xs, 280x

21+1)! +1). n!

(

*

a

A-203

Binomal Theorem for P'ositiue Integrl Inulex

(v)(-1)

.

il

nis even,

3

63, (6i)

(i)

(iv) 3"-1 2

(ii)

1

(n+1) 2

(-1)

and

Do-1*

(-1)2

x

if n is odd

65. 4

1002! 67. 501952

68."'"Cm

12"n(1 (2"- 1 Ja (n 2"-

72.

12!8 2. 20!

13.

14. (i) 56 (ii) 0

77. (n-

45

80. 0

16.30! 155!

no

15.

(21)! 17 -1) ! (n +

18.714 1)

(2"- - 1) "C,

24.

4x+

25. 14

10+

5

27. (i)

16x+.. ; 19

(ii)

+)

n

34.

14

14 +84 + 280 + 560

i.e.,

(1

35. 10, 10 40.

+

2

"

39. 7 or 14

124.

possible to be in APbut not in HP

41.n 46.

2,3,4

52. (a)

53.

provided is even and n>2 n

54.

8th term

()(i)

57 (ii) (iv) 0

()

+2)x

1 (1+)

1+ x)"

(n +1)x

2)x"

(271)

99.

111.3"-2"

1132

(b) 5

2

42! !2

(21

120. 101 122. 10

0

10! (5 12

123. 41

125. 7

11

57344

243

127. 924

128. 12

20

132. "C

131.

20!

(10

27 x7! 13!

n= 49, 50, 51,

58.3

+

126. 42.

51. 6th term, 1.e.,

4th and 5th, i.e., 489888

56. 2 2

62.

50."C

(1

119. 32

121. (a) -1

37.14

-

(4 + 2) 1(21+1)!

118. 126

+2)

3

+3

35 10!25!

112.

672 + 448+128,

(b) n

2)

1(1 4 )"

+

2)x

(21+1)!

101.

32.22

32.

n= 6,p=

4

2 (u !2

(v)-336

28. 11851

+

(n + 1)(n

92,

29. 168

1)(1

(4)"*2

o1 1.

25215.3 10! 15!

81.(n +2)2"-7-5

22+ 1

78. (n + 4)2 "-

+1

2"''+1 +1(u +2) ( (a)+1 b)

89.

(211)(iv)(m 30.

88. (a)

20.0 23. Bn

1

al=

1)2 "

3)!4

2" +1)d

-

(n+21+ 1)x+ (3n

85.

19. 924

21.

4

-

I|

1)--

'C

"

69.

(

"

64. 120

..,

59

!

133. (d) 137. (d) 141. (a)

true 149. false

145.

129. 0 +

"C+ "C2+.

130. 2 2 + "CA

134. (a) 138. (c) 142. (c)

135. (c) 139. (c)

136. (c) 140. (e)

143. (c)

144. false

146. true

147. true

148. true

Encl

Ter

for Pesae integr i-dz

Chapter Test Time:75 mirztes

folowing, eech statement is incormplete. Fll in the blank so that ezcr sultin statemant becomes corect. () E the sum of binomial coeicients in the expansion of (a y"is 1024 then tne grea binomial coeficient occurs in the th term. () The coeficient of expansion in the of (2-z-3x) is. * (ii) If < (2-1)

1+a+ + a+... + a +

(1)

1X+b-1+ i.e.,

+

3 '+am

37amx

relations a, = a-1+bn-1 etc.)

bt +

Cmo

+

+1

(a +a24+...

Hence, the problem.

x-1

a+...

..

(3)

2

+a")+a

+2

+am- ') +a2+ ... +a ") + am-2

(a 1+x=x, given)

8M2|a+a*+ m1 +

Pm+1) is true.

.+a ,*2, a" +

+a *4+a

(1)

for alln e

(2)

Also from (3),

N.

or

for all ne

a1-a

-2 a+a+... +a ") Ta 1

N.

or

2>0

m(m + 1)

Prove using induction that

1+a+ a

a+..

a+a... (

.

+a?+... +a m-)21+a+a2+... +a"-2

P(m+1) is true; thus, P(1) is true by PMI, P7) is true

+a"1

If (2) is not true then

i.e.,

a

m

aa?+...

b, +

=x31+x x3"*, using =x3 (1x) =x 3" -x3

2.

+

bnX +Cm

+

=x3"xja,,x

a2

we

From (1),

using (1)

x=x+ 1

Let

(1)

true. For this

m+3

I

1+a +a*+ ... +a "**>

T+a

13.

+a"")

+b)x +ambn+Cm+ mCm

(

(from given

.

+a2

Now, we want to prove that P(m +1) is have to prove that

l

Now, a =

becausea22

P(1) is true.

n = 1,

When

+1)+3a

3>3a

x" =a,x b, + n

Let P):

n= 1,

1, co=F0.

=

1+2

+ a"

+a +...

a

1+a+a(a-24

b-1+ Ca -1 Cn-1

a

1+a+at... +a"

Let P():

-2(a+

+

1-a "2>

1

a+... +a") + m{m

+

3)a "*

m(m + 1)a " *2> 00

0

Principle of Mathematical

12+... +a")+ mna" +a ") +ma" or +a 2a possible for a 22 or

"l(m

+

1)a

"l(m

+

1a

1)-21 n

for all n e N.

58. Prove

(n

.

).

+.

that

46.

Prove thatfor all n

47.

49.

50.

E

1+V4a+1

,

b=perfect square and and

a"- b" Vr

ab is an integer

a"+b" =c" +d" where n e = 0, prove by 5x+1 xiof roots B the be If a, not divisible by induction that o"+B" is an integer N.

63.

for all

where ne N. '" Prove that 10" -(5+ V17) by 2"*for all n e N.

4 64.

N.

65. If

x+is

N.

y" Prove, by mathematical induction, that n. integer odd is divisible by x + y for any positive prove that then If n is any odd positive integer nn-1) is divisible by 24. Prove, by induction, that x(r"-- na"-1) + (n -1)a" is divisible by (x a) for all n e N, n>1.

"+

Prove that (v2 - 1)" can be expressed as Nt- Nt-1, positive where t is a positive integer for all odd integral values of n.

= 1,

66. If

-

(5

-

V17)" is divisible

induction, that an integer, prove by N. an integer for all n e

"+is

-

51.

+abi +bs +...+b)

where r is not a integer. Use induction. then prove that a,, is an a2+b?=c2+d2 for the real that 62. If a +b=c+d and prove by, induction, numbers a, b, c, d then

4 Prove, by induction, that "> "C

for all n 48.

E

+.. +a,b,) < (a +a +..

1.

61. If

are positive integers. where a 2, etc.,

s

+ a2b2

a

N.

2 31+

at least

1

+X2+ (a,b

Prove, by induction, that

=

1.

1

..+ 45.

all >

where each x>0 and x *z 32 If x, then prove, by induction, that one all n 2 2. X3 +. +Xn > n for ... x,

60. Prove

Va + Na + Va +... to n times 44. where a> 0.

13

-> vn for

that-

forn E

1+ nr

n e N.

n

+2)"23"

+x)">

-

induction or otherwise that

for all n

" > 1)! for n 2 2, n e N and that for n > 1, 57. Prove by induction

all "

Prove that 2"x (n

56.

inductiory that|

Prove, by

54. Prove that 55.

nduction, that n!Sn'" forall

Prove, by

>

3"

induction, that 3

.by by Prove,

for

-

induction, that

Prove, by

andb

53.

24-1+l,b,=a-120-1

and

integral values of all for Prove that 10"-2> 81n n25. ! n 2 7. Prove that 3"< n for ( !)2>n" for n 23.

U2 =

1

and u +2= lUn1

then

for n21

+ 1l

prove by induction,

-

=

l2 5 and -1=

67. Let

Show,by induction, that 68. If

for all 21.

,

+61l- , 122. =

3"

= 3un-7t-1 +5t-2 -3 0,

18,

112

forn20.

-

(-2)" if

23 and = 17 then prove that 2u,, 20n 7 n

n

21.

=

1,

+32-"

A-230

.

Problems Plus in IT Mathematics

a

If

# = 3 and a, = 34,,-1-2n-2 for 1 22 then prove by induction that a,, =2"+1 for all n 2 0. 70. Use induction to prove that =2,

s

forall

sin(ax 72. Prove, by

+ b)}

=a

1,

(a)

"sin|ax + b+n

induction, that for

m

=

J

mm,

prove that

m 0, 1, 2,..

Prove, by mathematical induction, that

75. Let

1)(2cos 20 1)

-1)... (2cos 2"- "0-1) +sin nx

u

=

J sin

4

Ccos x sin"-'r+

Sin. 77.

2

2

2 2.

Sin

4

sin n*| *

is

s not

=

!log

| sin x|,

an

+

(1

cos

X

forall ne

integral

(-1)"-1,

Sin 2110

2cos 0

n times

2cos

2

220..

cos

2-1e=Sin 29

2 "sin 6

that

2.cos+ 4

isin

4

cot11

+ n(u + 1)}

=

tan(

+

1)-neN. 4

92. Prove, by induction, that

multiple

to prove + cos 2x + cos 3x +.+cos

cos

20 cos

+i)"=2

=

where n is any integer. 91. Prove, by induction, that

N.

n e N.

mathematical induction

cos

1)0

prove that

90. Prove, by induction,

for all even positive integral values

n

lpx +(1 -p)yl

prove that

89. Using induction,

cos

of n.

( "log x)

ys sin

N2+V2 +V2+.. to +(n + 1)

27|2 dx =0

p) sin

88. Use induction to

x

where n e N.

0)" = cos n0+ isin 0 for all positive as well as negative integral values of n. 87. Use induction to prove that sin sin 30+sin 50 ..

u.

e*sin

A

(cos 6+ isin

for n EN =

Use induction

1.

and 0 Sx, y sT.] 86. Prove, by induction, that

where 0 sps1

cos "x sin nx dx

79.

1

= 1,2, 3, ...,

+(-1)"- sin(21

dx =

e"(sin x + cos x)}

78.

r

i

sin A, Su sin

p sin x+ (1

"x dx. Prove, by induction, that

sin(21-1) 1)x

for

cos(1+1)x1 Sin a

[Given the fact

(-1)"-1.-1)!

for all positive integral values of n Prove, by inductio1, that

82. If

that

x + sin *2x + sin *3x+..

dx. -cos*

dxoa)=

|

-

(2cos

1

85. Let 0

X cos

23. Solve: cos

+ cos 2

.

0Sx sT. 1.

+ cOs 3x > 0, xe [0, n|

24. Solve: cos 3x +V3 sin 3x 25. Solve:

tan x +3cot

x>

+v2 0, cos I

1, cos x*1

x*nn

Sin 2x +lo81o cos x=

t

x+y*7+2**y**.y2**

sinx

x=10= 100.

so

orlo810 5in 2x+ log10

AM2 GM 3. Given

log

7, Solve for

Again, take the positive numbers x+*z,

43) = logn (43) = RHS.

are real if sin But here the logarithms

ie, 5-3 = 3-2 or ie, y=2 2=0, yThis is possibleif

log +log

z=0 log(r y+3.y3**.z*)=0,

log

43

Logarithm

x=log2a

4, y =

log,a 24 and z = lo843a

then prove that xyz + 1=2yz.

=loga

(ab), (bc), y = log, (ca) and z=loge

x+1 y+1 4.

Prove that loga(bc) logb(ca) log.(ab) =2+ log,(bc) +logs(ca) +log (ab).

-61 B-60

5. If

Problems Plus in

2+2-)_y2**-)ZT *Y2, log log y log

prove that

2

y-zV=z*.x*=xV.y .

If

O8= O8 V_Og E, y-2

-y

Z-X

8. Solve: logio sin x + logi00 (Bcos x)

-

logy =1.

Logarithm

1213 then Iflog 20.30103, log3=0.4771213 then th. thenumbey of digits in the number equal to 3 12 x 2*s 17. If log 3 0.4771213 then the number 16.

between the decimal point and the first digit in the value of 9-16 is_

prove that*+yy+zi23.

7. Prove thatx °%-logz.ylog-log .zlog

9. Solve: logz tan x +

IT Mathematics

Iflog2 sin

18.

-1 then the set of general aluesof vali

19. 1f 2loga+2log

Objective Questions

cos.log coslogsinlog(

+21og a"

- 55loga

(a) 10

(b)

(c) 20

(d)5

81 =

20. If

12. If logx + log y = log(x + y)

o-logy -C

then y as a function of x is

(a)

given by y=_

log z

C-a

The sum of the series

loge'

1

a?+b2=23ab then log

(a) b

is the AM of

22. If a, b, c

then

but

log(1+**)= log(.

log, *(log, a +log,

211 +

9 11.

3

2n7+cos 10.

(a) 1

(b) 2

(c) 3

(d) none of these

; neZ

272

2nm+12neE

16.9 S

18.

4nn,

1

17.15

+112+5,3T

4n

19.

13. log.9

14. log a, log b

20.(6

21.

()

22. (b)

+2,

a >0

c) is equal

Answers 8.

4.

is equal to

are positive numbers (#1) in GP and

1, the value of

nEZ

A + log

to

tan B+ log tan C.

triangle ABC?

log, sin+log sin +log 3. Prove that 4+

the value of xyz is

(b) c (d) none of these

()d

AABC, Prove that in an acute-angled log tan C)= tan log(tan A + tan B+

Can this hold for any

b)3

21.loga bx logbCX iO8

and 15. If0 0.

11. logs log2

Time: 30 minutes

gnifican

is

=logo3

Chapter Test

sin

3x > 0,0sxs 21. Solve: 1+log2 sinx+ log2 sin

+log2 sin =log25.

Problems Plus in IT

B-62

Mathematics

Answers

1cos0

)sin

2.

no

6.Properties of Triangle Recap of Facts and Formulae Angles ofa triangle triangle ABC, three angles are A,

1.

Ina

A+B+C=

sin(B +C)= sin(t-A)

and

C.

24 etc.

Vs(s-a)(s - b(s -c)

.A

= sin A

B cos(C+A) = cos(t- B) = -cos

in4sin

formula BC in the ratio m:n at D and In the AABC, AD divides = B. LBAC in two parts ZBAD = a, LCAD

4. Ratio

c

conco .COS 2 2.

=

.sin A B

in

Trigonometrical relations between sides and angles

For any AABC we have

m

2RR

where R = circumradius (Sine-rule)

b+c2-a2etc. .cos A =

(Cosineformulae)

2hc

If LADB

= 0

D

then

(m

+ n)cot 6

.(m

+ m)cot

=n cot p- m cot a = m cot C- n cot B

acos B+bcos A = c, etc.

in-

etc.,

bc

5. Some

where 2s =a+b+c

etc.

bC

tan-6-bMs-0_ s(s--a)

sin

-)4sts-a)

(s-bXs

tan 3.

2A + sin 2B + sin 2C

=

4sin A sin

1+cos2A+ cos 2B+ cos 2C= 4cos

etc,

sin A+ sin B + nC=4cos

where A= area of AABC

cot ASS-a)

important identities for angles of a triangle

In a AABC, we have

etc.

otetc.

B

sinC

A cos

B

cos C

A cos B cos C

+cos B + cos C=1+4sin

sin

sin

cos

A

cos

A + cos B + cos"C+2cos A cos B cos C = 1

A+ tan tan C= tan A tan B tan C, ie., cot B cot C+cot C cot A + cot A cot B 1

tan

B+

tantan-1,

Area of a triangle

the area of a AABC is denoted by A then sin A, etc.

Abe

B-63

B-64

Problems Plus in iIT

Mathematics

B-65 Properties of Triangle

an

6. Circumradius, inradius and exradii

stan

In the AABC, let the circumradius = R, inradius = r ana the three exradii corresponding to the vertices A, B and Cbe r and r respectively. Then

AAabc

R-2sin

-4Rsin

7.

A

6-tan-(6-tan=6-tan

=32R 28

coo

2. Prove

Regular polygon

i

the 20AA

ZA,0A2=2

that in a

= k(cos

b

-

LHS LHS

-+8-b)2

la

Also,tan Each

3. In the AABC,

2-2290.

cos

o

B + COS C)

BC.cot2

tan2

A)

os

COSAA

HSsin

cot

-2ksinsin

+b-c)

Dlcosin

+b+c-2c) =

ccot

in

2ksin

sin B).

C-cos

=k(cos

A

Similarly for others.

y+tancot

prove that

CotA +cot B+cot

interior angle of the regular polygon

cos B)+

+b-(acos B +bcos A))

+b+C-a-b)(a

exradius

Ckgiven)

A

2s-2)AD

cos Bcos

cosA

sin C.

sin

learly, in

coc

sin

B

AABC,

(2--b(a

tan-4R

3sin A sin

c

because AOAA, will *

one of the n equal triangles with a vertex at 0.

inradius

2-

+ z)tan

Aregular polygon of n sides will have s vertic a circle. If O be the centre and rbe the radius o circle, and a be the length of each side then

4Rsinsin ID

-4Rsincos

c-4+6°+c

=2 k(cos

4

cos B cos C sin B sin C

=

C-cos

B)

-cos

kl(cos C- cos B) + (cos A

C)

+(Cos

Each exterior angle of the regular polygon=2x

=kx0

=

B-cos A)

0.

-+

2R

Selected Solved Examples 1. In a AABC, prove that

sin sin A

A cos(B-)

3sin A sin B sin

C.

A sin

=

sin A

=

sin

sin

sin(B + C) cos(B-C)

sin

A (sin 2B + sin 2C).

R l

sin +

(bcos B+ccos

*A(sin 28+ sin 2C)

sin "B(sin 2C+ sin 24) +

c

cos A

22

A 2sin(B + C) cos(B C)

Similarly for others

LHS=

cos C+22

2422

A cos(B-C)

A sin

B

dbc2-a2+c?+a'-b+a?+b-c abc

A

2R sin C

acos A + bcos B)

8R3 1aDacos B+ bcos A) + ac(acos C+CcOs +bc(bcos C+cos

sin C(sin 2A + sin 28)

8R3 2bc

acos B+ bcos A =c, etc

atc Na-ac

2c0s Zcos-C

+b2+C2 a2bt2 be 42R be 45 sin A

ab+-RHS. 44 and z are proportional to cosines of the angles A, B and Cofa AABC respectively, prove that

y

8))

2+)tan

co0.

+c

= Here A, B, C are in AP; so B A -a,

4+C_180 Also B

abe la6+c3-a+b+ Rbe

+21

C)+b(ccos C+acos ) +c

In a AABC, the angles A, B, Care in AP. Show that

(a+b-c3 2R

cos(B C)

sin

5.

2a

B

90°

B2

2

90

B=60

-a

=

and C=A

60°, i.e.,

C=A -2a.

B

A = 60° +a

- 2a = 60° +a-2a

=

60°- a

C =2a.

A-

sin(60°+a)

sin 60°

sin(60°-a)

sin(60°+a)

a+CL 3 3 cosa 2cos

2

+C +

.

sin(60°

a

-

a)

a+C

v3 cos a

B-66

Problems Plus in IT

B67

Matkematics Properties of riangle

Now,

1-cosa-cos b- cos'c+2cos a c

a+

a +C

2-ac+C*

sin b sin c

VE)-4-E

1-cos a- cos "b

2tan -

cos

sin

= 2cos a (60° +a)sin(60° +a) 60° sin 60°

sin

sin

sin60

sin

60°

bcos

sina sin b sin'c

sin'a

sinB sin

(s-a)s-b%6=C)_ 5=1-4 1s6-aNs-bs-c)

sin

a -cos

Thus,

a-

sin

b+cisrational,becausea is

But

2cos

Vacos'asin'a-cos eos

sin A

sin

= sin C

B

-

=

cos b

4

6

sin

c-

-

-

rational rational

B + Sin

rational. But

a is

rational. So sinA is

Clearly

aa=

A;

Similarly (ii).

45in D= rational,

(i) a,

c

tantan

are rational

:Now, tan

-),

tan

and and

.

C(s-aXs-b

s

cos c)-cos 'a - cos b cos

+2cos a cos b cos c

Thus(Gi) 8.

=

s

a,b, c;A, are rational.

rational.

Thus ()

such that LOAC= 20CB= 2OBA = a. A Prove that cot a = cot + cot B+ cot C.

10. Let O be a point in the AABC

sin A = rational. ).

B)=

a

find the area of the triangle

Here, cos(A-B)-: By componendo and

oc

14AOC,-AC AAOC sin From AOCsin a

1- tan

A-B5

But LAOC=x-(2OCA + LOAC)

=T-(C-a +a)=r-C AC OC

dividendo,

sin(t-

).

Let(i)be true,

ie, a, tan

or

tanbe rational. Or

Now, sin B=

tan2

zC sin

tan

2Brational,

ceuse

tanis rational

But

tan

sina

C)

OC

(1)

sin

Similarly, from AOBC we get

2tan

tan

+cot

If in a triangle ABC, two sides are a = 6, b=3 and cos(A

nand tan C are rational because sum, differene So tanand

B

rational

2

Now, )

(cot

A + cot

festablish as in solved example 3).

4

sin A

Abe A

and

Sn

44

A

AsininC

AABC ) The sides a, b, c and the area A are rational

Let ) be true, ie., a, b, c and A be rational numbers.

bcos c

e

sin "b sin 'c

sin

rational.

a, sin A, sin B, sin C are rational. Prove that i) (i) (i)

+2cos a cos b cos c

b1

(cot

rational.

b

b+C

product and quotient of nonzero rational number

1 cos

A

from the altitudes of the AABC If a, B and y are the that show then respectively C vertices A, B and A+ cot B+ cot

sin sinC

isSln

()

cos a-cos "b cos

sinb-sin

sin

7. Consider the following statements concerning

Sinb Sin c sinA =1 - cos A _Cos a+cos b cos C-2cos a cos b cos c sin'b sinc

sin b

gin A

C63 sin 9

B, sin C be rational. Let (ii) be true, i.e., a, sin A, sin

cos c

sin b sinc =OS +Os cos C-2cos a cos

But

Thus(i)

thatsinasini

Cos

in AABC,

...\\

2cos a =2cos

cos a= cos b cos c + sin b sinc cos A cos b=cos c- cos a + sinc sin a cosB cos c= cos a cos b +sin a sin b cos C then prove

24A

sinb sin c

aina

6. If in a AABC,

Here, cos A

sin Bsin c

SinA

sina

9.

s is rational

e

area of the triangle

-zabsin

S

1-cos'a-cos b-cos C+ 2cos a cosb Cos

C=

cot1;

the

(s-cXs-a) (5-a\s-)

sin 'a sin "b sinc

cos a

tanis rational

Now, tan 2

C Similarlyinbsin'c

2cos

or

because

+2cos a cos

sinA

sin(60-a), (sin(60-a)

2 =rational,

1+tan

t:

OC

A> Bl.

sin

2)

Bsin(B-a)

(1)+(2)

bsin Bsin(B-a) asin C Sin a

B-69

Problems Plus in lIT Mathematics

B-68

Properties of Triangle

sinA sin C

sin

a

inA

sin B sinc cot a -cot B sin(A+C)= acot B sin

OinA sin

Cot

A sinC

cos C+ cos A sin A sin C

sin A or

cot C+ cot A cot a

=

=

sin Ccot

ar(AABC)yz n-(A +C)=B))

a

-

ABDM,

:In

L BC. So BD =DC.!

BDtan tan A

D

cot B

cot A + cot B+ cot C.

or

=

tan. A, ie,

-; LBMC=

2y

tan

A + tan B +

and tan

A

tan B, tan C

tanA + tan B+ tan C=

sin

sina cos C+ cos a sin

a,

cos

cos a sinC

or

tan A tan B. tan C

c-

or

cos C= sin "C

12. If

C= a"b sin C.

the distances of the sides of a AABC from its x,

circumcentre be

ay

c

y, z then prove that

abc 4xyz

a

AC

From the ACAD, cos C

CD

9BC-2a+c)-b3

=b

2b

2

BG=2a+2-b). 26-c Similarly CGor

... (2)

a

BG2CG2-BC2

13. 1f x, y, z are

the distances of the vertices of the AABC respectively from the orthocentre then prove that

BD From the AABDsin(A 90°)

cos A Let Hbe the orthocentre.

-oBG*+bi

.(1) 2bc

a+

Now, cos a =

2BG

CC

2a+22-b)+2a2b3-c)-a2

AB

sin ADD

2BG CG

-5a2

b

sin (90° + C)

2BG CG Then

ACos C

2cos

acosCOSC -2c

-,

a

-2c

c

LBHC=180° 2HBC-LHCB

180°

(90-C)-(90°-B) =B+C=t-A. ar(ABHC)= BH-CH sin LBHC

zyz sin(r-A) =;yzsinA.

b2+c2-a2-2b2

or

c-a2-3b2

Now,cos

A.cos

2ca2 3ca

ar(AABC)

2b_b2+c2-a2

cOs Sin

from (3)

= 3 ar(ABGC)}

a'+c-5a a

2bc

3b2+3(2-4)a-c+3¢82, 3ca

a

ar(AABC) .(3)

C+c-4

sin a

ar(ABGC)

c2-5jsin

(1)b+c-a-b 2bc

or

sin a

bc2-5a)sin a

from (2)

from

C1-0

ap+b*q-2alqcos

or

C=4C42

cos A

cos 3ca In a AABC prove that A and AD L AC. where AD is the median through

Cos

or

2 Now, BC2+ BA =2(BE 2+CE)

xyz

C

Cc

C+cos a sin C

BG= BE, CC-CF.

ie, 4 =

R

.abe4xyz

sin(a+)

AD

nd AG

From the AABC, cos A

But in a triangle ABC,

AC sin a,

geometry

AB2+AC2=2(AD?+ BD ), etc.,

or

anC =

b

We know from

a

abc

yz

tanC 14.

Let 2ACE =a. Clearly, from the figure, we get

+5zxsin B+xy sinC

(1 gives, 4R 4R

tan A.

tan B,

vertex C then prove that

sinC=a'p*+b'q*-2abpgcos C.

Also, we kmow that

C

Similarly,

sin A

... (1)

fromgeometry, ZBMD

cot a- cot B

11. If p, q are perpendiculars from the angular points A and B of the AABC drawn to any line through the

ab

,

ar(aAH)2ysinC Let Mbe the circumcenre. MD

sin A

(:

15.

ar(aCHA)=zxsin B Similarly

cot a sin B- cos B or

ABC make angles a The medians of a triangle that Prove other. each with coty+cot A + cot B+ cot C=0. cot a + cot ß+ etc. AABC and LBGC =a, Here, G is the centroid of the

BT

sin B cos a-cos B sin a

sin B

r

cota

124

bc2-52 12

3ca

Similarly, cot p-

5,

12A

coty2+b2-52 124

B-70

Problems Pius in

Properties of Triangle

.

cot a +cot ß + cot y=-

ab+c 44

Again gain cot A +cot + cot B+ B + cot C=OS 4

COS

(b+)sin B,

Cgin Asin B

R.

cos C sin C

Similarly

2R C

-(+c?-a2+c+a2-b?+a2+b?-c) h

+b2+c)

(a + 8) sin

Now,

R

sin(PIC

+

2QIC) 17.

44

0

(6

The internal bisectors of the angles of the AABC meet the sides BC, CA and AB at P, Q and R respectively. Show that the area of the APQR is equal to

+ c) sin

206 + eMe

2abc A (a+b(b +c)(C+a)

21BA +

2IAB-+

-BIP-

LBID

sin

+oysin

sin

o

bcr coDs 2c+aa

+b) sin

sin

2a + bX0 +) sin

sin2

Now, ar(APQR) = ar(APIQ)+ar(4QIR) + ar(ARi) abcr2 2(a + b)(b+c)c+a)

sin

sin

sin

cos

=

in 2(a+ b)(b+ cMc +0)

or

in

asinB-+in2vde SNa

r4Rsin

(7-cos a)r*=1-cos a

cOS

7-Cos

-cos

rTe x= where

.:

1, 1+x are in AP and from the question Clearly the sides a, b, c of the triangle are in AP. Hence

sinsin

a a

19. Prove that, fora scalene AABC,

cos a

a, c, b are in AP

cos A +cos B = 4sin

1-x,

abCk 1-r11+x

(suppose)

We have to prove that x

c

cos A +cos

B = 4sin

2cos A+B cos

=4sin2

=

2sinco-2sin0

N7-cos a

k{1+)_3k

cos

2

C is the is the greatest side. So +a. is the least angle. Hence C=A

cos

sin A+ sin

oninin

a+b

r: A+B=r-c

A--2sin 2

2sin

a=CA

n Cb n Ain

a 1-7x =

its greatest 18. If the sides of a triangle are in AP and the sides are angle exceeds the least by a, show that

aKing X Positive,

4R

cos

(1-x)

a+b+c-24 b+C-a

8earestangle and A

abcr

-1-

-2

+3a

S+b*K1-x)+k+ 2 2

COs 6

a -2cos-1

s(S-a)s-bS-c)S-a

Now,a=k(1-3), b=k, c=k(1 +X) Now, from ADIP,

1-X

V-2)1+2)1-x

A

inthe ratio (1-):1:(1+)

carcos2

ar(ARIP)

cot2

44 2. 34-a2a

sin

(3k-k) k(1-x) k(1 + X)

2.

9_s(6-b(s -c)

2a

Similarly, ar(AQIR) =-

-90P.

V

2S

+b+C

2(6+ cX+ a) sin

Let ZDIP =0, PI= x, Q! = y and RI =z. Then =

6-.6+s-b) b

are sides of a AABC and 3a = b+c,

that cot

LHS

abr cos

Let the bisectors of the angles meet at I. Then I is the incentre. Let ID L BC. Then ID =inradius = r. =

(c+a) sin

abr

Ifa, band c show

br

a+b+c

44

Clearly, 2BIP

abcr

a+bb +cXc +) 2abc A (a+bb+ ¢ +a)

ar(aPIQ)=PI1Q sinLPIQ

*y

-bX2.6-bX6-2

ab

abc r 4+bNb+Mc+a)tb+c)

and z=-

cot + cot y + cot A + cot B+ cot C

a+bC 16.

y=

+b+c)

bc

cot +

abcRr (a+b(b+C\C+a)

br

(c+a) sin

a+b 2ab

-

B-71

IT Mathematics

=

2c

B

=

4sin

4cos

sin=

2sin C

a,c,b are in AP.

8-7 Properties of Triangle

20. In a AABC, if atanA + btan B =(a

A+B

+bjtan2

sinA sinB sin?c

then prove that the triangle is isosceles.

Here, a tan A

"cos A

tan A

cos

=

X

B, cos Care also in AP. prove that cos A, cos

B

1

-coA1.1

sinA

nn

Sin

sin A

or

anun

i

tan

-

sin2B

1-cos "B=1-(2cos A A

1-4cos

A4+

a sin

bsin

A+B cos A cos

r

But cos

-B,

sin sin

cos

-14

-bcos A} =

{acos B

A cos

sin(A

-

B- sin

B)

=

B

x+6x+9

(1+ 9

cos A) =0

sin C

+x)

2+3)2

1+x 9 +x)

1

from (2)

1+cot C

from

2

tangents of the angles of a triangle are in AP prove that the squares of the sides are in the ratio

+9):(x+3):91+r) the greatest tangent.

Here tan A, tan B, tan 2tan

B

or

C are in AP;

tan A + tan

22.

A+C=T-B}

A

2cos A cos

sin B. cot B C=cos B =-cos(A +

or

2cos A cos C=-cos A cos C+ sin A sinC 3cos A cos

C= sin A sin

sin C

x9+

2+3)

... (1)

cos A+ Cos

9+x

c

+c2-a2,

2bccos A, 2cacos B, 2abcos C are in AF b

Also, s =

are in AP {dividing

2nbo)

COS COS eia are in AP Sin A 2R sin B2R tA, cot B, cot C are in AP (multiplying by

n+5

and

a+b+br b+br-a +br-b +b-br 2

a'h

x(2ab

But cos 20 2cos0-1;

2

(a+b2- b',}

4ah2 =4a2-(a2+b2-b?r32

or

=4ab- ta2+(1 r*)b2 (1-)5*-(4a-2(1-*ab+a'+ 4a h?=0

or

(1-O)*-2(1

+ra*b2+a*a2+ 4h3) =0o.

4(1+a-4(1-)a

2

2n(n+1)

so 21

n+5-1

1

n+2)=nl(n + 5)-2(n +2)9

(2)

(2ab+ (a2+ b2-br*))

Now, the values of b* are real, so

2n(n +1)

(+5)

.

-

21(n+1)

220+

br-(a-b

or

(n+1)(n-3)-3

(-3

(1)

(a+b+br

2(n+2)

+2)2 cos 20-+(n+1)-(n

n-2n-3

or

..

2)

n+1(n+5) 2n+1(n+2)

c+a?62, a2+b2-care in A

cosA, cOsB, cos C

Cos A

B

)+(n +2)-n n+6n+5 2(n+1)(n +2)

Now, cos e-

laddinga2+ b*+c

. (2)

B

From (1) and (2),

Imultiplying by -2

Let the altitude be h.

Now, (area of the A) =|

.

and

b, br.

A, cos B, cos C are in AP.

2n+1M+

are in AP

of the two sides is In a triangle of base a, the ratio triangle is less r1). Show that the altitude of the

Let the other sides be

2cos

C=2cos

=4; hence sides are 4,5, 6.

(a +b)-b

nis

9(1+x)

cot A, cot B, cot C are in AR

-2a2,-26,-2

n

neN

Ifa,b,care in AP then show that

C

tan A tan C =3 Let tan A be the greatest or the least tangent. Then tan A =x.

b Now,sinA sin B

(1+r9+2)

or

4

i.e,-and-3 which are not natural numbers

than or equal to

Cos B

side From the question, the largest angle opposite to the side n+2 is 20 while the smallest angle opposite to the

a,b,carein AP

C=

or

9

Hence, the problem.

sin(A +C)2 cos A cos C

nB cos cos C 2cos A cos

1+

where x is the least or

+3)2

-7t V4924

4

Cos

2cos

= =4 or 2n+7n +3 0.

Roots of 2n+7n +3 =0 are

are three 24. The lengths of sides of a triangle consecutive natural numbers and its largest angle is of the sides the twice the smallest one. Determine triangle Let the lengths of the'sides be n, n +1,n +2 where

3), (4), (5) and (6) we get

21. If the

(n-4)(2n+7n + 3) = 0

25.

A-C

or

COs

0

Therefore, the triangle is isosceles.

nn B Sin Cos

or

or

n

*-90-

cossi

-3)n+4n

-

or

cos + coscosco6

in2n

or

from (2))

(n

:or

C

2Sn

5

COS

or

0

B

2cos

17) +4) = n(-n + 2n + 17n -n*+2n+ 12= n*+n*-8n 2n-n*-25n- 12 = 0

or

C

r or

=0, ie., A =B.

A-B

or

cos C

lfroe

4x2 1- (1+ 9+x) (1+X9 +x)-4 (1+x9+x

sin A- B

cos C)

B

0.

sin

or

2

2sin

11+tanA 1+1tan'C

A +B

.cos(acos B- bcos A) =0.

sin

or

cos

2 tan+tan

Here, 2tan

X

2

or

iftantantan9are in AP,

AABC, 23. In a

(a2 + 4h3)20

or

(1+ra*-(1-75-(a2+4h3)20

or

i(l+r-1--5la-

or or

ra2-1-r3h*20

(1-

4h220

8-74

Problems Plus in iIT Mathematics

B-75 Properties of Triangle

ra

1-h.

120°

y sin

ar(AADC)=

Hence, the altitude is less than or

we get (from the question),

3 =xy

1

equal to

ar(ABCD)=4V3 =* CL

... (1)

xy =6 26. If in a triangle ABC

C

cos A cos B+sin A sin B sin "C=1,

In

AAC'B, LABC'= ZB'OC+ LACB

n eN then prove that the sides are in the ratio 1:1:v2.

Weknow that in AABC, 12 0
0

S cos A cos

B

In

from the question, 1 S cos(A

But cos(A

=

B

and

Ab

sin(C-0)sin Bsin

B).

and thenC=

C-

2AR

8=

or

from similar

C

0,

y> 0)

sin2cossin= sinsin sin sin n sin

-t:

(3)

triangles

sin(0 + C) and BC' = 21R sin(C-0) BC'=2ARisin(C +0) +sin(C-8)

4

7T T;

or

*3

n=7

4R sin C

41R sin C

regular polygon of n sides has the circumradius R and inradius r then prove that each side of the

-yt1 using (3), we get x3 y=2 or x= 2, y=3.

15

29, Let ArAy Ay .,

30. If a

polygon is equal to 2(R +) tan

An be the vertices of an n-sided

regularpolygon such that AA

b

Let AAz be a side and 0 be the centre. Let OB LA,A2.

1

AA AA4

Find the value of n.

ninn

cos

fOis the centre of the n-sided regular polygon, then

A,O42A,OA,

r

28. The

theratio of thesides =1:1:v2.

two adjacent sides of a cyclic quadrilateral are and 5 and the angle between them is 60°. If the area or the quadrilateral is 4v3, ind the renaining sides. Let the cyclic quadrilateral be ABCD in whc artABCD) =43, AB= 2, BC =5 and ZB = 60° Then D = 120°.

Let

OAOA==

LBAC

BCA =C.

=

A, ZA'BC' = LABC= B,

2r sin

Also, ZA,OA2

An

Similarly

Let AD =x, CD =y.

Now, ar(aABC)=2-5

AA=2r sin

and A1A=2r sin sin 60°=

=and

and

AB

Otanie,

AA

AABC and AA'B'C' are similar where

B'A'C'=

Clearly, OA =R and OB =r. =

LA,OB

InAA,OB,cos LA,OB=cos

cos =5

ZBC'A=

... =* OAn =r.

learly we get AA =24,M

27. In the AABC, a similar A A'B'C' is inscribed so that B'C BC. If B'C is inclined at an angle 0 with BC,

prove that

Or

13

sin

Also (r-y)=(r+y)-4xy =5-4.6=1

=2AR 2sin C cos 8 COS

or

(r+y)=13+2xy = 13 +2.6=25

cAB=AC+

.

AB

(+-2xy

or

Thus, we get

AC 2R A-B =0

(2)

+y=5

A'C

CAC B) =1

2+y+y=19

C -0

sin

sinsin-sin sin

From(1)and (2), *+y2=13

BC

B) s 1. -

=

sinsin

120°

y-2iyay

BC A'C sin(C-0) sin B

B'C

Now, cos(A - B) =1 A

-

23R. 2R.

ABAC, ZBA'C"= LA'C'B'-LA'OC

+sin A sin B = cos(A -B),

equality holding when C=;:

So, we get cos(A

sin A

sin(0+C)sin A

sinsinsin T

AC2-AD+DC-2AD DC cos

ABC

or

cos 60°

-4+25-2-2.5.-19

B'C

AC

+sin A sin B sin "CC

0+C

sin(0+C) sin A

sin "Cs1

COSA

=

BC Also, AC=AB+BC-2AB

2rsin

2rsin

2rsin

(1)

AB =r tan n

24,3 =2r tan

Now, 2(R +) tan

2

.

anfrom

(1)

(2)

B-77

Problems Plus in IIT Mathea ics

B-76

2r

1+ cos

1-cos

cos

sin

ung1-cos

sin 0

and y= rcot

a

Ina AABC, we have the identity

a Prove that the harmonic mean of the exradii of inradius. triangleis three times the

Let the

exradii be r, 2 3 and inradius=r.

equilateral 37. Show that the AABCis Here R 2

2r COS 7

in

CoS

tan=A^Az

oraB-a

Thus,

Ts are the inradius and the respectively of a AABC then prove that

OC = R, BC =a.

3

2.4sin

(6-)+(6-(6-

A sin B sin C,

36.

=4s-25ta +b +c) +a2+b*+c}

28+ sin 20)

-2

2+a+b+c

p,y:

then prove thatr=

a+B4y

wnere

r= inradius,

Let the incircle touches the side AB at P where AP = a. Let I be the incentre. Then Al bisects the BAC.

Now, r+ 34. If the exradii r, 7,

of a AABC are in H,

its sides a, b, care in AP

r

We know that

23

A

a

h

are in HP

are

in HP

arein AP from the right-angled APA,

= tan

a=rcot

s-carein AP -c are in AP

S-,

s-b,

b, a, b,

care in AP.

s-c

Now,

show

-sineo

T2

-co

and inradius =,

circumradius R. Then we have to prove that r,

bc)-RHS. 32. If the distances of the vertices of a triangle from the points of contact of the incircle with the sides be a,

.

sinsinsin

Prove that in a triangle, the sum of exradii exceeds the inradius by twice the diameter of the circumcircle. (or, Prove that 7 +2+7=r+4R.)

Let the exradii be r,

(using identity)

sinini R-

r-Brsinsin sin

Hence, the problem.

2ks= sin 2C

2A + sin

Ca

3r.

or

26sin

Vs-a)Ns-b)

ab

R

(a+b+c

LHS

and

V-6 -a)

16-as-b

S-a+Sb+5-C 3s-(a +b+c)

exradi

We know thatr=

sin 28

aXs-b\s-)

sA

-bN9.e-

24 lusing sine rule in ABOC)

Similarly, 2R

s(s

A4

1,1 Now

33. If r,ry

R =

=

s-bs-c)

13r

1

atB+7

circumradius and R, Ry R, are the circumradii of the triangles OBC, OCA and OAB respectively then

Clearly, in the AOBC, 2BOC =24, OB

we have to prove that

ay

31. If in the AABC, O is the circumcentre and R is the

2in

(given).

6-X6-bX6-)

sin

from (2).

rove that

if its circumradius

is double of the inradius

We know thatr

Sin

2

2r

Similarly, P-rcot

:

1-cos

Properties of Triangle

+72+r=r+

sinin

4R.

equality holding when

But

+3r

co-

max

B = C.

gan-ni)-ng-sin]

than

*

stls6 ) f(s-b)(s-c)

for 25=a +b+c

+c)+bc +s-as sS-a)(-b(s - c)

A.SS{b

Aa2-s(a+b+)+be

(25-2 +a=r+4R.

be)=4R:R-A

equality holding when sin A

2-

sin

sin equality holding when

From (1), equality must hold.

and

sin

(1)

Problems Plus in IIT

B-78

B

C

and sin

ie.,=30°

A

So

=

B+C

=60°

Mathematics

B-79

or

ccos A=+a> 0

or

a-csin?A >0

or

a>csin A.

a, B and y be three angles given by cos 16. If

a > Csin A

From the question cos A)b

180° - A = 120°;

b2-(2

Properties of Triangle

Pc+a

+(c-a)=0

So the triangle is equilateral.

According to the question, the values of baare b,, ;

38. In a triangle ABC, the sides

17.

=2c cos A and b, 2b, = e?-a? cos A and 2b2=c2-a2

In a AABc, if +C

18.

a+b+c

C+a

or

=

orc-

2h0

b-(2c cos A)b+(c2-a)=0.

or

AABC,

cos B+ y cos C= 4R () a cos A+B R, the the altitudes of a triangle and are p Ps If Pz radius of its circumcircle, prove that

if c*-2(a+bJe +ad+ab2+ b=0

A=VRP

C-or

thenprovethat

32.

followingholds:

2c08

Prove that cosec

A

proportional to the (b) cosines of two angles are

33. In a APQR, QR

opposite sides.

1.

for each of the admissible values of

4'sin(B-C)

that Ina AABC, prove tnatsin =0. B+sin C

2 In a AABC, show that

11.

4.

ab

Ina AABC. prove that a sin

(a

+b) sin=ccos

cot A

en prove

Ifin the AABC,

=

sin

12.

=

7. In a AABC, prove that

(b+

7

13.

In a AABC, if

the

24.

C

25

tan= 5 and tan B2then 23

p

15.

=

+B) sin(C-A) = -sin(B+2) Band Determine the measures of the angles A, cosine the that tha Can a triangle be constructed such s

ofis

angles are

B

cot

C

=v3 then prove

f in a

tan,

AABC,

then show that the

In a AABC, prove that

and

cosec

7

reason Give

26. 1n

B

2

cosec

2C in are

cot

+cot

C = cot

o+ cot

C) = csin(A

-

B),

prove that

B.

angle A of AABC meets BC atD then prove that

38. The bisector of the = m

bc

39. If

(i) m-

2bc cos b

the median AD of AABC be perpendicular to AB

thenprove that tan A +2tan +2cos B+ cos C=2, prove that

abcare

length of the side BC. on In a 4ABC, 2C =60°, LA = 75°. If D is a point AC such that the area of ABAD is v3 times the area of ABCD, find LABD. 37. AD is a median of AABC and LBAD = 0, 2CAD= ¢ then prove that

If AD

unun 27 in a AABC, if cos A a,b, care in AP. 28. In a AABC, if asin(B -

and it

the median AD=

divides the angle A into angles 30° and 45°. Find the

AP.

aAABC, if sin A, sin B, sin C are in AP, show that

in AP

12,

36.

a",b,c*are in AP. cotA,cot B, cot C are in AP in HP, prove that c a b, are a, AABC, sides if the 25. In cosec

=

and point on PQ such that MP : MQ =^:1(a>1) Find the line RM divides the 0

In a

19.

a

tan

tan

a+b+c

31.

This is a quadratic equation in b. So, bwill have two values if D

tan

sides a, b, c are in AP are in HP hen prove that the the a AABC from of altitudes 30. If a, B,y are the prove that (where respectively, C B vertices A, and Ris the circumradius)

en prove that

a=+8sinA. We have, cos A

triangle ABC, the quantities

triangle are in AP and its area is th The sides of a triangle of the same perimeter. an equilateral of are in the ratio 3:5:7. Prove that the sides

2c

of which given. If the side b has two possible values one is double the other then prove that

tntn

tantantanž=tan

has two real values.

b+2b,

b+c

and cosy= +b where a, b, c are the then prove that AABC a sides of

sB CoS

because

but B = C; so B = C= 60°. Hence A= B=C= 60°.

are a, c and the angle A

29. If in a

B = 0.

40. In a AABC, the lengths of the bisectors of the anglees A, B and C are x, y andz respectively. Show that

B-80

Problems Plus in

IT Mathematics B-8T

41.

Properties of Triangle

In a AABC, the median AD and the altitude AM divides the angle A in three equal parts. Show that

cossin2

R

53.

where A = area of the triangle. 42. Acyclic quadrilateral ABCD of

3N3

areais inscribed

(4-V10+2V5)42.

quadrilateral ABCD, if AB = 10, 2DAB =75°; 2CBA = 90°, 2CAB=45° and ZDBA =30° then findCD. 46. Prove that the sum of the radii of the circles, inscribed in and circumscribed about a regular polygon of n sides, is a is the length

cotwhere

of a side of the polygon. 47. The ratio of the area of the regular polygon ofn sides circumscribed about a circle to the area of the

regular polygon of equal number of sides inscribed in the circle is 4:3. Find the value ofn.

L441=1,1,1 55.

1B-IC=abctan LA

sides are equal.

a of the circle Prove that area of the polygon tan49. Three points A', B' and C' are taken on the plane

nn

of a AABC such that AA'BC, AABC and AABC are equilateral

with their circumradii R,

r

T,r

Rg, R; inradii respectively.Prove that

R): (rrr)=1:8:27.

', are exradii and r Is the inradius of a triangle then prove that

50. If

tan

60. If in a

AABC1

two angles of a triangle are 30° and 45° and the included side is (V3 + 1) cm then the area of the

82.

triangle of side Acircle is inscribed in an equilateral the circle is The area of any square inscribed in

4.

83. In a AABC, a

:b:c=4:5:6. The ratio of the radius

incircle is of the circumcircle to that of the

then

A =

A, B

and

C

)3

B =

are in AP and

-B)then

85. Ina AABC,

(b-c)*+ 4bc sin

++)tan2A=4a2

m4 m 1PrOve

72.

If in a AABC; 2tan

73.

In any AABC,

that there are

values ofc, one of which is (m-1) times tneother

Ifin AABC,a:b:c=8:10:12 a ihen the greatest

and the lea angle in terms offa is

a=0 then the value of

c

= 87. In a AABC, B

AD =h. Then

and

(a) 1:2

and the altitude

(b) 2:1 (d) none of these

(c)1 4 a cos 8. If in &ABC,a

cos2=bthenthesides

(a)AP

(b) GP

(c)HP

(d) none of these

B

tan

89. If in a AABC, e(a +b) 76.

C=

h:a is equal to

trigonometrical ratios of the angles, is

If in the AABC, cos A

=and

cos C-

then

4--then e:a=,

In a AABC, if (a +b +c(b+ c-a)= 3bc then 90.

B c-b na AABC,if tan A-tan A + tan B c

tan

tnen

a

AABC,

4oS

COS B

ZCOS

a

A=

a AABC, AD is the altitude from A. Given b> c

cC-23

and AD

-then

2B-.

then the

In a AABC, if a* +b°+c*= 2e"(a+b) then the measure of C is

(b) ()3

A= in

cos=be+a) cos

triangle is (a) isosceles (b) right angled (c) isosceles or right angled (d) none of these

a:b:c

On angle s

abx2-cx + ab =0 (b) abx3-(a2+bx+ ab =0 (c)c*-abx+ c2=0 (d) ax bx+a=0

a, b, c are in

If in a AABC, c= 2a then tan

are the

(a)

sin A +sin B+ sin'C in terms of products of

then

Fill in the blanks. 64.

75.

a AABC, |b

B

roots of the equation

sin C

oL

Objective Questions

In a AABC, C=90°. Then tan A and tan

S

A= two

86.

cotthen b: c=_

a'sin(BC b sin(C-A) sin B sinA (A-B) PGc2sin +

78.

that

circumircum-

The smallest angle of the triangle, whose sides are 7,413and V13, is

7. If in the ABC,

le-o=2. Na-bsin A :63. In a AABC;a , prove that

=

B

or

centre. The angle of elevation of the aeroplane whet nearest to the man, is a and when farthest it isp"

(1)

(3)

, i.e,sin

from 0-

sin(a

o.

or

p

Again, from APOC, if LPOC

3. Aman standing at a distance d from a tower of height h finds that an aeroplane is describing a horizonlal circle of radius r with the top of the tower at 1s

The corner of the top of the tower nearest toPis E and F is one of the outer corners.

22BOR =2(180°20) 2T-40. 40 (formula) we get 2n-40 =

.(3)

0

cross-multiplication we get from (3) and (4),

sina

-212a + V404

sin e=aS

0

sin From -

x2a

- V2a +V2. V5a =av2(V5-1).

The base of a vertical tower 1s a square. A man on a diagonal (produced) of the square is at a distance 2a from the tower. He observes that the angle of elevation of each of the two outer corners of the top of the tower is 30° while that of the nearest corner is 45. Prove that the breadth of the tower

Let the breadth of the tower = BC =x.

(2)

24. A

Vhcota -d

av2(vs-1).

=

(d + r)sin ß=hcos

+22ax-8a2=0

-22a+

-2Vhcota-d-d tan B-h) =

2

(d-r)sina

(2)

or

135°+(20)-(2av3)

=-h

POB=20 radian. Then, clearly hv3

... (1)

=hcos a

-22at N8a-4:-84 )

B

-

180-45 135

-824ar

r

tan

Also, in AAEP,

==

B=

rsin a-dsin a +hcos a

or

BC=x, CP=24, BP =2av3

and BCP

"a-d2

BP =2av3.

3

BP

hcot a

have, tan a

2a= FB.

:From the right-angled AFBP,

the right-angled ABFP PE tan a

:in

Now, from the right-angled AECP

Let

cos=t;

then v3(2t- 1) =t

2v3-t-v3=0 1+43,4N3 2131+5

Problems Plus in IIT

8--100

Mathematics Heights and Distances

flagstaff is mounted in the middle a Squ. based vertical tower.A man standino on the midway in front of one side of the t ower, groun ound finds u elevation of the top of the tower to be hthe A p flagstaff to be from a distance d. hat of the ter getling a tower by the distance to a, he finds loser thatthe top of the flagstaft just disappears. Prove thathe B sin(B 4sin height of the flagstaff is a) cosa sin(0 B) where (d-a)tan 8=dtan a.

26. A

r

10

Also,

h3

radius

cos

3

h=r=

60

of the pond = height of the pole

'

&0

Here, QR is the height of the tower and the ao. PM =h (say). Clearly, from the figure, MN = QR.

25. A circular plate touches a vertical wall. The plate is fixed horizontally at a certain height above the ground. A lighted candle of a length equal to half of the height of the plate above the ground, stands at the centre of the plate. Prove that the breadth of the

A

Clearly from the figure, the required breadth of the shadow is the length of the chord PQ of the circle whose centre is O and radius = OD. Now, from similar triangles ACB and AOD, we get

OD =3a.

Clearly, the distance of the chord PQ from the centre O is OR = radius of the plate = a.

aA The man is at A initially and at disappearS.

B

when the

QR QR tan 6RB d-a

dl

QR

From (1) and (2), tan 0

-

tan

PQ=2. RQ=2. Hence, the problem.

are inclined to of equal heights a their noon the breadths of

sun, are

BC-b, QR =c. When the 1espectively along the breadths are BD and QS cross-section of the shadows.

P

At that time the second man is atB where 2BOA =B (from the question). Let the height of B BD =x. =

OB = OA

hcosec a

¢

DBC=

from

and 0

if>0,

c

Coordinates and Straight Lines

of the pentagon ABCDE bewhere

A=1,3), B=(-2,5), C=(3,-1), D =02)* E-(2, t).

Being equilateral triangle, the angle between (1) and (2) will be 60° tan 60°=

1+m(-1

or

S.ABCisa variable triangle with the fixed vertex C(1,2) (cos t, sin nd A B having the coordinatesparameter. Find oinf-cos t)respectively where tis a thelocus of the centroid of the

AABC.

LetGbethe centroid in any sition.

2

or

(by angle formula)

N3=

m+1

or

)I

=

V3(1 -m),

-V3(1 m)

(1+N3ym =v3-1 and (1-v3)m = - v3-1

3-1 33-1 1 N3

C-S

. 3-1)2, 3 2

2-v3, the

1)2

equation of BD is y 2x-4.

the

The distance of

As ABCD is a rectangle,

2

PB-PDAC-6-10-

equations of the other two sides are -2)

andy-3 =(2+ v3)x -2) Now, the slope of DB is tan 0 =2. Points on DB at distances v5 from P(3, 2) have h he coordinates

3-2N3 (3-

3

v3)x

v5cos 0, 2t v5sin 0),

the

y=2-x=2-3-1

ie,

3-1

C=[Y2 v3 C=|3-1 v3-1

coordinates of the other vertices are (3 + 1,2+2), (4, 4) and (2, 0).

equations of two sides of a square are 3x +4y-5=0 and 3r+4y-15 0. The line along the third side has a point (6, 5) on it. Find the equations of this and the remaining side of the square.

8. The

each side= AB or AC

Clearly 3x+4y

-5 =0

3X+4y-5=0

-64-23 3-1 6V3-17-

(2)

5T7a

and

4x-3y= 19.

or

whether P lies between = AD = PA + PD may

lead to wrong result lying in the first quadrant, are 9. Two sides of a thombus, length of 0 and 12r- 5y =0. If the given by 3x4 12, find the equations of the the longer diagonal is other two sides of rhombus. the first quadrant and their

As the sides are

vertex O(0, 0) is acute. So the longer diagonal passes through O0, 0.

(5, 1). The middle point

l ot AC

15

whose equation is y = 2x +C.

C(5,1)

y 2x+C

A.

It passes through (6, 5). So 4 x 6-3x5

;:

equation of AD is 4x 3y =9. As BC | AD, the equation of BC is 4x-3y=

the

Let PQ1 BC.

ThenPO= 4.6-3.5-2

distance between the parallel lines

ABandDC

Now, any point on AB 0. is

=7

>»

6130

a

11

theequation of CB, from

(2), is

18 130 5

3r-4y-

condition for the quadrilateral to be concyclic whose consecutive sides are

10. Find the

ax+by +c=0;r=1,2,3, 4. equationss Let L,a,x + b,y + c, =0 be the sides and the L =0, L2 = 0, L = 0, of AB, BC, CD, DA be respectively 3x-4y

= 0

L4

0

=0.

Let 2ABC

=

0; then LADC = T-6.

the slope of OC is

and

La-0

0

side AD is on a line through Pl6, the perpendicular to 3x + 4y 15 0. Any line perpendicular to 3x+ 4y 15 0 has u :

12

equation of AB, from (1), is 12x -5y

the

the angle at the

and tan

The slope of OA is tan

equation 4x- 3y =

a130 65

in

inclinations are tan

00)0

3x+ 4y-15

or5+9)-12

B

v6.

3,2. It lies on the other diagonal BD

A(1,3)

But OB =12;

Letthe length of each side be a.

points (1, 3) and (5, 1) are two opposite vertices of a rectangle. The other two vertices lie on the line Find c and the remaining vertices. y 2r+c.

ie., c--4.

BC is equationi of the remaining side

the 4x-3y-1

5

B-

19

910-i

P(6,5)

7. The

2-2x3+c,

9--+10

2

V3-V24-12v3 124-123

=

are wo

Let

a)-213-3

A = (1, 3) and C

and 3+4y 15 0

these sides parallel lines, their 'm' being equal to be AB and DC in the square ABCD and the point P(6, 5) be on the side AD.

-N

ie,

.

3r-4y14

99a Solving (1) and (2), y=65

19-Al=10

Note As ithas not been verified PQ A and D or not, use of

ie3525

Let

equation of CB is

the

-

2-x=(2-v3)x -1+213

B or

2

2or

erce

1 +2v3 and y = (2 +v3)x or -1 -23. Solving one of these, say, y= (2- 3)x-1 +213 with x+ y =2 we get

or

this point from DC

3x0-4

V3

2+

y-3-2-9x y=2-V3)x

or

Coordinates and Straight Lines

Problems Plus in lIT Malhematics

tan

A(0+acos Similaly, C

L20

0,0+asin 6)=

0+ acos , 0+ asin )=1

0

12a

13

NOw, ABis a line parallel to OC and passing through

tan 0= But tan

2

|1+mm2

theequation of AB is 12r-

125 the

5y =k where

and

ork

when 35

(1)

where

.Mef

equation of AB is 12r -5y=

e equation of CB is 3x-4y

n(t-0)-1mm 1+mm2|1+

mm

0 is acute,

T

is obtuse)

C-10 Problems Plus in IIT Mathematics

-11

Coordinates and Straight Lines

55425 cos(-a)=14

or or

A(0-3)

where cos d or

by4-a,ba

or

a,ba2b

or

Gba)an,

b-abe +

a142 + bba

ab4ab

a,42 + b,ba

+ cos

ag4,+ bsb t

bbi)

+

(agh-abaa,

*

y-2

11. Find the

equation of the line passing through the point P(1, 2) cutting the lines x+y-5=0 and 2x-y=7 at A and B respectively such that the harmonic mean of PA and PB is 10. Let the equation of the line passing through P(1, 2) be y-2

Let PA

m(x-1) PB

=r

.(1)

tan cos1

+cosiA

V146

-1).

and orthocentre of the triangle formed by the lines 3x-2y=6,3x+4y+12=0 and 3 -8y + 12 =0.

B= (1+r2cos 6,2+2sin 0), where tan 6 =m.

3x-8y +12 0

a

(2)

e

3r+4y+12=0a |2x-y-7

1+cos +2+7sin

cos 0+ sin 0)

=

.. )

6-5 0

2(1+r2cos 0)-(2+r,sin

3x-8y+

Tz(2cos -sin

12cos 2

-

6)

7 =0 (3)

r

2cos

-

0

C

C=(4,0) B

=4;

Solving (1), (3) we get y =-3,x = 0; To find

Now, HM of PA, PB is 10

or

12

Solving (1), (2) we gety =3, x

)=7

102PAPA +PB PB

A

- sin

0)+ 35(cos

10(2cos

or

55cos 0+25sin

)

+

8a +

the centroid

'

(1)

a-4

mof AC

we get, BH I AC or

2

4--4

a-4

or

3(B -3);

8a +30B +3)

4a- 3ß 7

t2,

sn21t *h*1s 3

ie, ie,

=

0,i.e.,

8a +3ß =-9

or

6y -9

=

-16x

16x+6y-9=0 Again, E midpoint of AC

.

and 3x + 4y +12 0. 3x-2y-6=0 Any line through A is a line passing through the intersection of the above lines. So the equation of the altitude through A is

(iii)

3x-2y-6+A(3x

+4y+ 12) = 0

Being altitude, it is perpendicular to BC whose slope 8

[from (2), (3)1

sin )=14

So,theslope of the equ side Ac

ACis

To find the circumcentre

ie

Then MA

=

= MB MC.

g

The slope of a)

is-

by condition of perpendicularity,

ME is

on of the perpendicular bisector ME of the

0 = 14

Let M(a, p) be the circumcentre.

..(

Alternatioely The equations of AB, AC are respectively

or

y1), etc:

...()

orthocentre is

the

equation of the perpendicular bisector MD of the the side BCis

iheslopeof ACis

-1

-1

Again AH 1 BC or

-

Solving () and (I) we get a=

is

0

4(a4)

=

0--4)

ofBC

of AH

(ii)

D-midpoint of BC

2

'm of BH=

16

and (i) we get a=

recumcentre

(4,3) =(0,3)

where vertices are (x1

(cos 0+ sin 8)1,

or

8a+12p-16 =0, i.e., 2a + 3B-4 0

8a-6B+7=0

=

We know that centroid G

102

sin

6 +9

or

So,the slope of MD

Solving (2), (3) we get y=0, x= -4;

sin 6

Now,

and

Solving

C-4,0)

B(4,3)

+16 and

6+9=-8a-68+25

Theslope of BCis

.(2)

As B is on the line 2x-y-7=0,

or

6B +9

68 +9=-8a 6B+25 8a

or

Now,

2

-

A(0-3)

Allernatively Take the perpendicular bisectors of the sides BC, CAwhich are respectively MD and ME.

0

cos +sin

AH L BC.

=a+p2-8a -6+25 =a+p+8a +16

respectively.

As Ais on the line x +y-5 =0,

or

a

12. Find the coordinates of the centroid, circumcentre

3x-2y-6=0e

P(12) x*y-5 0/

(a-0)+(+3)2 (a-4)+(-3) or

A=(1 +71cos 6,2+r^sin 6)

Tofind the orthocentre Let Ha, B) be the orthocentre. Then BHL AC

= (a +4)+(B-0)2

Let the sides AB, BC and CA have the equations

Then

=r

5V146

8

circumcentre-12

the (-4,0)

B(4,3)

5146cos 14+cos

iv

0

Solving ii) and (iv) we get x

55-23

fom (1), the equation of the line is

bê)=0.

+7

8x-6y

cos(-a)5146

or

|a44 bab|

=

or + 2), ie, 6y +9 -8(x + 2)

or

3(3+3A) =8(4-2) = 9+9N 32A- 16

(a)

C12

or

Problems Plus in IIT Mathematics

23

the

.

25;

L

equation of the altitude through A is

3x-2y-6+3x4y+12) =0 or

69x-46y-

or

144xr +

138+75x + 100y +300 0

Itis to

or 8x + 3y +9 = 0 (6) Again, any line through the intersection of BC and BA is

k Sin

BC whose

,.

e, + Acos

cos 0(cos

or

cos(-0)+Acos(-0)=0

Puing in (1), the equation of the altitude through becomes or

9+9

the

equation of the altitude through B is 3x-8y+12 +23(3x 2y-6) =0

or

=

8

=

4(21 +8)

=

72r-54y-126

cos(-8

s,-,0

23

+32;

m' ofGO=

Cos

or 4x-3y -7 0 Solving (6) and (c) we get the coordinates of

orthoentre. We get x

=y

and

ortboemtre-(

L,=Xcos 6,+ ysin 6,-P=0;r=1,2,3. Show that the orthocentre of the triangle is given by = Lcos(® -9,)= Lgcos(@,-0 Lcos-0,) =

0, L=0,

L

B

= 0.

L1-0 Any line passing through A, ie, the intersection of the lines L2 = 0, Ly = 0 is

COS

Cos

d2t

COS

altitudes,

and (4). So the orthocentre is given by

L,(0, 0)0

dg

(0,0) and (-1,-4) are on the opposite sides of

a, + sin

L 3x+2y +6=

l23x+2y+6 0

a+

Here, the circumcentre O =(0, 0). So OA= OB

EX-2y-2

NOw

L0,3)-1-6 2

tan y;

slope of the refracted ray =

the

is N5.

53 6 3

21. Find the

y+

8+5a

11

3(5-2N3)x313y +13v3 - 50

:or

23.

0.

ExercisesS

1. Ifa vertex of a

triangle be (1, 1) and the middle points of two sides through it be (-2, 3) and (5, 2) then find the centroid and the incentre of the triangle. 2. If two vertices of a parallelogram are (3, 2), (-1,1) and the diagonals cut at (2, -5), find the other vertices of the parallelogram. triangle has its orthocentre at (1, 1) and

3. If a

circumcentre at

11.

,

Prove that the area of the triangle whose vertices bsin 8,);r=1,2,3 is equal to (acos

2ab

sin--

triangle. does not cut the line segmentjoining the points (-2,-3) and (1,-2). intermal bisector of the the and a triangle 5. ABC is angle A cuts the side BC at D. If A= (3,-5) AAD. B=(-3,3), C= (-1,-2) then find the length of

4. Prove that the line 2x-3y +6

6. Find the point P if Q

The coordinates of the points .A, B and C respectively (6, 3), (-3, 5) and (4, -2) and coordinates of P are (x, y), prove that ar(APBC): ar(AABC) = | x+y-2|:7.

=

0

(-2, 4) is the point on the line

segment OP such that OQ

=

OP,

triple its length, then find the new position of B. are A(1,2 8. The coordinates of the extremities of a rod and B(3, 4). S(0, 0) is a point source of light. The rod AB is parallel to the wall and is midway between the point source and the wall. CD is the shadow of AB on the wall. S, AB and CD are in the same horizontal plane. Find the ends C, D of the shadow. 9. Find the circumcentre of the triangle whose vertices 4) are (2, 1), (5, 2) and (3, 10. Find the centroid and the incentre of the triangle whose vertices are (2, 4), (6, 4) and (2,0).

the

sin

-)sin

are

@-

y

ae-1

15. Find the locus of the point at which the line segment joining the given points (o, B) and (7, 8) subtends a

right angle 16.

LinesL

make an angle 8 with intersect at the point P and a line different from each other. Find the equation of P and makes the same L2 which passes through

A rod of lengthI cm slides on two mutualy perpendicular lines. Find the locus of the middle point

position of a moving point in the r-y plane at time t is given by (ucos a t,1usin oa t-pt) where 1, a, p are constants. Find the equation of the locu of the moving point. 18. The straight line lis perpendicular to the Ine 5x -y =1. The area of the A formed by the line and the coordirnate axes is 5. Find the equation of tne line. 17. The

l

= 5 and 4x-3y 15 intersect at 24. Straight lines 3r+ 4y lines such that A. Points B and Care chosen on these possible equations of the AB AC. Determine the line BCpassing through the point (1, 2). bisectors of the 25. The equations of perpendicular y + 5 =0 and sidesAB and AC of a AABC are (1, -2), find +2y 0 respectively. If the vertex A is the equation of the line BC. BC and CA of a 26. The equations of the sides AB, AABC are respectively x +y =1, x-2y =3 and the foot of the perpendicular from A 2x-y 0.1 to the side BC, and DE is drawn parallel to BA cutting AC at E. Find the point E. 27. The coordinates of the foot of the perpendicular drawn from the point (x1, y1) on the line + my + n =0 are (h, k). Prove that

x-

Hence or otherwise prove that no three distint points of the above type can be collinear 14. Show that the equation of the locus ofa point, whih moves so that the difference of its distances tirom two given points (ne, 0) and (-ae, 0) is equal to 24. is

O being the origin

7. One end of a thin straight elastic string is fixed at A(4, -1) and the other end B is at (1,2) in the unstretched condition. If the string is stretched to

are

triangle have integral coordinates prove that the triangle cannot be equilateral.

13.

10.

angle 8 with L.

12. If the vertices of a

then find the centroid of the

=

between the lines is such that its segment 22 A line bisected at the 5r-y+4= 0 and 3x + 4y-4 =0 is equation. point (1, 5). Obtain its = =ax + by+c=0 and L = lx + my +n 0

the equation of the refracted rayis

7+5V3)(19 +53)208+1303 286 192-(5V3)

lie a unit points on the line r+y=4 that at

2distance from the line 4r +3y

lr

zh-k

m

Iatmyt+ 12+ m

28. Equations of two straight lines are Xcos a+ ysin a = p and xcos ß+ ysin ß = p'. Show that the area of the quadrilateral formed by the two lines and the perpendiculars drawn from the origin to the lines is

2sin(B-2Pp-(p*+p)cos(a- B)]. 29. The sides AB, BC, CD and DA of a quadrilateral = have the equations x+2y 3, x=1, x-3y =4 and respectively. Find the angle between thediagonals AC and BD. 30. Let P, Q and R be three collinear points on BC, CA and AB of AABC. Show that BP CQ AR+ PC QA RB =0, Proper sign for extermal section being taken into account.

5x+y+12 =0

AC: If D is the triangle with AB = perpendicular of the midpoint of BC, E the foot the midpoint of DE, F drawn from D to AC and prove that AF L BE. a and b on the Let the line 1, have intercepts passes through the coordinate axes. Another line 2 on, between the axes. middle point of the intercept the axes respectively then IFh cuts intercepts p, g on

31. Let ABC be a

19.

80+5030

be y then tan

1536-1 13

553+88-8-53

11-

and the line through

3-

A(10/3,-7/3) with the slope

tan y=

8+53 5V3+8-

As

6

Coordinates and Straight Lines

32.

prove that ag + bp = 2p. = 0, 5x -y =6 and Show that the lines 4x+y-9 intercepts on any line of x-2y +3 =0 make equal gradient 2. lines are 7x +y= 16, 34. The equations of three = 0. Any line through 5x-y-8=0 and x -5y +8 the lines at P, Q, R the point A= (c+ 1, 2c) meets harmonic mean respectively. Prove that AP is the

33.

between AQ and AR. 3 off equal that the lines y = m,x;r= 1,2, cut if 1+ m intercepts on the transversal x+y=1 l+mi2 1 + ma are in HP. cuts the lines 36. A straight line through A(-2,-3) B C respectively X+3y 9 and x +y +1 = 0 at and Find the equation of the line if AB. AC=20. meets the lines 37. A line through the point A(-5,-4) = 0 at the x+3y +2 0, 2x +y +4 = 0 and x-y -5 IF points B,C and D respectively.

35. Prove

find the equation of the line.

In the AABC, the equations of AB and AC are 2x+3y 29 and x +2y 16.If the median through A cuts BC at (5, 6), find the equation of the side BC. 39. Find points equidistant from the lines 4x +3y 0 and 3x-4y =3, the points being at a distance 2 from the point of intersection of the given lines.

38.

equations of the lines which pass through (4,5) and make equal angles with the lines 5y 12x+6 and 3r = 4y +7. 41. Find the equations of the bisectors of the angles between the lines x+ y-3 =0 and 7x -y+5 = 0, and indicate which of them bisects the acute angle between the lines. 42. The bisectors of the angle between the lines y= v3x +3 and v3y =r+313 meet the x-axis at P and Q. Find the length PQ. 43. Prove that the bisector of the angle containing the origin is also the bisector of the acute angle between the lines. a^x+ biy + C =0 and ar + b,y +c2=0 provided C, C2 are of the same sign and a142 +bba 0

The equation of a circle whose centre is the origin and radius is a, is

4. Location of a circle

The general equation of a circleis =

be two circles. Let D be the discriminant for the quadratic equation in x (or y) obtained by eliminating y (or x) from the two equations of the circle. Then they are two intersecting circles if D >0 they are nonintersecting (no common points) if D 0

a point circle (i.e., a circle of zero radius) if

no real circle if g+f0 where D is the discriminant of the quadraticif C43

44

Problems Plus in lIT Mathematics

x (or y) obtained by eliminating y (or x)

tonin from L 0, S

0;

=

equivalently, the length of the perpendicular from the centre to the line is less than the radius. The line touches the circle at a point if D=0;equivalently, the length of the perpendicular from the centre to the line is equal to the radius. The line does not cut or touch the circle if D0;

+2a>0

or

-8

or

a4

(a+2)(a-2)>0 (2)

by

-*-

.Utv2a1-2

aE

sign-scheme for quadratic expression, we get

,

by 2x(x-a) + 2y-b)=0, a #0, b#0. Find the condition on a and bif two chords, each bisected by the x-axis, can be drawn to the circle from

where

a

2(01+ 1)

the point

2

centre=

2tr+y)-2ax-

Mis the middle point of PA, CM L PA

or

2a 4

Or

+

Qm

+

1+2a=0

by =0

x*+yax-5y=0

Its centre=

V20t

ine

a2

Here, the equation of the circle is

1137g

and. the Foint

a>2.

9. Let a circle be given

2(m+ 1)

M=(a,-a)

AS

0r

(-,-2) u (2, +«).

2a (+m)r

2

Now, C

must be on the same side of t

*-y+1=0. .(9)

-

point of interscction of (1) and the line

0

or

u,

For two chords like PA, the equation must have two real and uncqual roots. So D>0

ac-2

(5a-9)(a+1) 0.

ly 0

Here the circles

Any line through ,has

the equation

y-m-) y =0

Asatisfes(1) Thecentre

willbe

perpendicular to the line (2).

the

through the

m or

bm+2am -(4am-2b) bm2-2am +2b 0

radius = v1+2-(-20) =5. So, the required circle of radius 5 must touch the given circle externally

( A* )

y=tv-e

centre-0of the circle 4)

will cut the

Y5.5) (a,p)

(1.2)

Let the centre of the required circle be (a, B).

circle (2) at real points.

5

Now, any chord through

required circle is

and

5

a-9,B=8

ym

( +3)3

0

For the given circle, centre = (1,2) and

of intersection will be imaginary and hence the circles will be nonintersecting be within the other or fully will circle Hence, one outside the other. If (1) is within (2) then any chord

- -

theline

is 5 and the equation of the circle whose radius 0 at which touches the circle x+y*-2x-4y-20 the point (5, 5).

13. Find

ifc>0, the points

a-r= distance between the centres

20

circle are obtained by

u0.

c0,

(2)

ux+c=0 of the

x=0

y+c=0

If the required circle has the radius a then

is

and the centre

(1)

Puttingx =0 in (1),

lies on the ircle.

at(m:0

+y x=0;

(1)-(2)

at(a-2m0

As the chord along (2) is bisected

joining

are:+ y*+ Ax +c=0

The points of intersection solving (1) and (2).

.2

(2) cuts the I-ais, ie,

Hence, one circle is within the other if

+y+x

for the circle, centre = (9, 8) and radius

=

5.

.

the equation of the required circle is

or or

or

=

0

.

roots,i.e., 4a-4 b-2b >0 or a-2b>0

equation of the circle having the pair of lines its normals and having the size just sufficient to contain the circle

x+27y+3x+6y = 0 as

two normals of the circle are x+3 =0, x+2y=0.

their

lis

or

A-B

4-

H

+MAx2Hxy +By2+2Gx +2Fy+C)=0 :

the centre of the circle.

The given circle is r*+y*-4x-3y0

(1tm)*+amtHx+

...

(1)

or

m

+2m

a-ma(1

or

xy=0.

a+A =b+AB and 2h+2 2H=0

a-b=(B-A) and"hleRH 1 -AH

4-8 H

2x3-5-1)+1829 radius va2+(29)

+m

=

29

V38.

4c(1+m)>0 This is truefor all real m.

and

The length of the perpendicular from the to the chord

am)-4+m)

coefficient of r= coefficient of y

.

+C=0.

Both roots must be real and unequal. So D >0

As the points of intersection of the two curves concyclic, (1) must be a circle

coefficient of

+2y) =0

intersection|-3,

+ 2gx+ 2fy +c=0 and Ax+2Hxy+By+2Gx +2Fy+c=0 intersect at four

for some

xr-4)+yy-3)=0. Here the pair is x(r + 3)+2y(x +3)=0

the equation of the circle whose centre is (3,-1 and which cuts off a chord of length 6 on the line 2-5y +18 =0.

*uT+C=0

If the curves ar+2hxy+by

ax+2hay+by?+2gx +2fy+c

10. Find the

(r+3)r

Solving (2) and (3),

Any second degree curve passing throug intersections of the given curves is

or a 2b2

V2b

++6x-3y

14. Find 11.

(3)

Two chords are possible if (3) has two real and unequal

or

=5* -9)-8)* 120 =0. 16y+ or+y-18x

225 +6x+9+-3y+44

m

(2u-12-4c)+2-4c>0

Also, the circle (2)

Tadius

2x-5y+18 -0

is real if

-/2-c

Hence, (4) is true for

is real, i.e., 4c20.

allrealm if2A4 1-4c>0 or 23>3+4c; butc>0 44cs0 Au0.

.. (4)

the

required equation of the circle is

(-)1i)= pr

Or

(133)

2-6x+9+y?+2y+1= 38

+y-6x+2y-28 0.

ntre (3, -1)

c53 C-52

Circles

Problems Plus in UT Mathematics

Circle touches the line y =xat a point P such that OP=4V2 where O is the origin. The irdle contains the point (-10, 2) in its interior and the length of its chord on the line x + y=0 is 6v2. Find the equation of the circle.

A

Clearly y =x and x +y=0 are two perpendicular lines passing through the origin O. Let P (k, k) as it is on the line y =x. =4V2

OP=VR+R t2k=4v2 k=t4

a+B+8=0 and a-ß=t

or

2a8t a=1,-9 and 9,1 B)=(1,-9)

If points

or

or or

(-9,1).

the distance

betwee. (-10,2)and (1,-9) is vi1+11> 5v2

So(-10,2)

So (-10,2) will be an intenor

the distance

vi+1
52. So (-10,2) will not be an interior point.

1+(-1)7

V(4v2)+ (3v2)

=

anu

X-

N2

2,Y

=V16+7-2.16-4.7-20

15xV1+2-(-20)

the new coordinates the point (-4v2, in be an interior point e,

-6V)

required circle willbe in the new third

uadrant

the centre=(42s2) and radius= 5v2. tne equation of the circle, in the new coordinates

(X+412)+ (Y+5V2)=(52)

CQ2={{acos

8 -asin COs 0

(asin -a)

-a2+a2-2a(a -asin 6)- 2a sin

+

+a)

0

s |*a+a2-20(a

+ asin 6)

+2asin +a2+sin01

BC

=a22-2(1- sin 0)-2sin 75.

and BQ are fixed parallel tangents to a circle, and tangent at any. point C cuts them at P and Q respectively, Show that CP CQis independent of the position of Con the circle and POQ is a right angle. Let the parallel tangents touch the circle at M andN ESpectively. Then MN is a diameter through the centre O. 17, AP

61 +(asin

AB

15x radius

2--62

and

=225 15. So, ar(ABCD) =2 ar(AACB) =2

5/2.

Also, when x=-10, y=2 we get

the =

=VS(16,7)

Then we have to find a circle touching X-axis cutting off 6v2 from Y-axis.

B-8-a=-1,9. B) =

and

=/sin

asin 0)

*acos -4tasin cos

to the circle

So, take the following equations of transformation:

,

4x7+2016 3

Now, BC length of the tangent from C(16, 7)

radiuoQP or 18(ap-zta-p* 188-^a-p), using ()

asin 66 x t Cos 0

and

C=(acos

2

3)

d,

a-asin 6 y=a and x=- cos

cos

to the chord

:ovy

100;

acos 6 +y asin 0=a2

y=-a

C (16,7)

(a-P)=

I

p-(a-asin 8,.

a-4-1=1

(2)

asin 0).

or rcos 6 + ysin 6 =a Solving (2) and (3),

(2)

3x-4y-20 = 0

,

The tangent to the circle at Cis

are tvo

y=*

y-0

Hence, (a,

20 =0,

0

a +B 8 r The length of perpendicular from Q (a,

From (1)

-

y = ta and the parallel tangernts as (acos is circle the Any point Con

A

Taking P=(4, 4),

or

. (1)

y-7=0

ie, 5y-35=0

as the y-axis

+y=a2

circle whose equation is be the centre of the

Let the centre Q of the circle = (, P).

QP LOP

and MN Taking the centre O as the origin the circle as we can take the equation of

=0.

tangents at the points B(1, 7) andd Suppose that the the point C Find the D4,-2) on the circle meet at ABCD. area of the quadrilateral tangents at B, D are respectively The equation of the x1+y:7-(r+1)-2y +7)-20 0

and

and

10)* = 100

+y+18x-2y+ 32

16. Let A

(r+9)+-1)2= (5v2)2 Second Method Here x+y=0 perpendicular lines.

(+y+8)+ (x-y+

2x+2y+36x-4y+ 64 = 0

r+y-2x-4y-20=0.

pont

the required circle has centre (-9,1), the equation of the required circle is

o(ap)

th

will not be an interior point.

If the centre = (-9,1) then points(-10,2) and (-9, 1) is

(-10,2)

100

10

the centre =(1,-9) then

e.ra

the equation

Taking P=(4,4) as before, we get

(a,

or (4,4).

P=(4,4)

tance bet between If the centre = (l,)tnen the distance th 5V2. points (-10, 2) and (1,9) is Y9+7> not interior an be point. will So-10,2)

0+e

xa2-2(1+sin@)+2sin

&1+sin 02

asin cos

cos

"0)2

a.sin 0 cos

CP2=a*=constant.

*cos

Froviems Plus in lit MathematicS

C55

Circles

Also'm' of PO =m

a-0

-asin

1-sin

cos

'm

of QO= m2*

-0

or As (

=--OS

a+asin

1+sin 9

6 cos sin

m21-sin 01+

hence

0)

65-4

a point on the line 4x-3y=6 tangents are drawn to the circle x*+y-6x-4y +4 = 0 which

or

9x+4x-6)54x

or

25x-150x=0

or

z*-6)=0

between them. Find the

Let the angle between the tangents be 0.

4x

c

radius=CA =v3+2-4=3

+y-6x-4/1+4

tan 2+yi

or or

x(7x-24y-48)=0

25

is

2 +y-6-41

+4-9

+y-6x-4y+

X+yí-6x,-4y +4+9

+y-6K-4y-5 x+yf-6x1-4y1+13

= +-64y +y-6x-4y+13

7x +7y-42x,

(

=

an or

-28y1+91

25x7+25y?-150x,100y

or 125

the

=3.

V3+62 36

=3.

02)

2

=u.

pair of tangents from

4) 36 +36-66-4-6+4

(2)

L-3-6)

a+p*-1+22

5+20-P(-1)=

=0

(3)

LetP(a, B) be the centre of the circle passing through the centres of the three given circles. Then

8a+6B +14 0

+7-0

c32a 9a+ (4a

..(4)

7

-9

+7)+18a

25a+170a+ 280

a+(-22=(a +3)2+(B+6)2

+ 24(4a +7)

= (a

4-4ß 6a +9 +12ß+36 = 6a+16+41 =0

and

6a-8ß-5 =0

+63 = 0

0

5a+34a +56 0 or Sa+20d +14a +56 =0 (5c14)a+4)=0

(1)-(2)

(2)

3p-(4a +7)

--16+7),

-4

-9, D

6a-8

or 6a+-50; 46

the

+36+ 4B +4

(1)

0

-

50

a= 18

centre of the circle passing through centres of given circles is

3,7/5.

TOm (1), the equations of required circles are

+y2 8x-6y +24 0

12a

(2)

24B+46

-12

a-4-14/5

(4) gives

+6)+(B +2), =r)

=

or

or

(3x+4y-26)2

tangents are x-6=0,7x-24y+102 =0.

88+7=0

Puting in (2) from (4),

or

916y2+676+24xy-156x-208 7x-24xy + 60x+144y -612 = 0 (7x-24y +102)(-6)=0 ds

centre (3,-6) and radius =

(4)=(a2+p2-1)+8

a+p+10a-2+21

4a+3p

or

6+:6-3(x +6)-2y+6)+4) 16(+y2-6x-4y +4)

or

2-)

is

+24xy 48x 64y

tangents from (0,-2) arer=0,7x-24y-48

-6x-4y+4

.(1)

the second circle orthogonally if

(3)-(2)

=(-3x-4y +8)

9x+16y+64 7x-24xy 48x =0

a

(1) cuts

or

+4) (0+4-0+8+4)

Similarly, the equation of the

1+tan 2

(0, -2)

16+y-6x-4y+4)

length of tangent = vxf+yi-6x-41+4

tanPA

co

y-2ar-2py+a+p2-1=0 1+2-P)

V6+2-31

=

For the third circle,

cuts the first circle orthogonally if

or

-0+-2)-3(r+0)-2y-2)+43 or

centre= (-6,-2) and radius

be

fusing 28,8, +2ff=ci+ C}

xx+yy-3(x+X) -2(y +yi)4

(+y-67-4y

3y =6

1

(1)

2:-a)

the equation of the pair of tangents from

Now, centre C= (3, 2) and

or

required circle of radius

or

T

N(-2)+5 = 3.

centre (0, 2) and radius =

(-a)+-)-12 y1) is

6x+ 12y +36=0.

For the second circle,

circles touch externally, not internally.

Let the

x+y-6x-4y +4 S-+yi6x 4y+4

(3,2)

x+y+

For the first circle,

d--1+5)+(4-1)

(6, 6).

Ti

x+y-4y-5-0, r+y+12r+4y +310 and

The distance between the centres = V16 +9 =5

where S

P(11

1-tan

circle,

(-5,1) and radius r= V5+1-22 =2.

the

S51

circle of minimum radius which contains the three circles

dnt

)=(0,-2),

:ie, 5(+y)+28xr - 14y +44 0.

-

The equation of the pair of tangents from (r,

Then tan 6=

or

-12(4x1-6) 108=0

-26

Let P(x, y) be a point on the line 4x -3y =6 having the given property.

=

For the second

centre

0,6

coordinates of all such points and the equations of tangents.

and PA

120

and

20. Find the equation of the

(1, 4) and radius r = v12+4-8=3.

centre

18.From

o

Show that the circles +y* +2x-8y +8 = 0 and +y+10x-2y+22=0 touch each other. Do they touch intemaly? Also obtain the equations of circles of radius 1 which cut both the circles orthogonally

circle, For the first

Solving (1) and (2), -1;

P0Q=;

make an angle of tan

)

19.

is on the line 4 -3y = 6, we get

4x-3y16

CoS 6

Clearly, m, m,= Clearly, m1

18x+ 18y-108x1-72y1-216=0 x+yí-61 -4/1-12=0

or

coS 6

radius r= va+(B-2)

C-56 Problems Plus in IIT Mathenatics

C57

Circles

V

+y6r=0

7 36

23725-949

A

radius of the required circle

Here the circle x+y-6x

=

centre (3, 0) and radius

3.

The circle +y+2=

0

5

=3+949 and

has the

0

form

-

x-0

respect to the circle

and sum of radii = 3 +1=4 circles touch externally.

Clearly the inclination of y

(3,0)

and that of y= 3 Also, their point

41- 511+6l+1 0 lx +1ny +1=0 touches a and centre of the circle.

21. If

then show that the line fixed circle. Find the radius

m

51+61+1

It is possible to find a, The condition is

=0 B, a

nd

(1) or

C2mc 2)

or

the line touches the fixed circle, (x-3)+(y- 0) = (V52

+ y-

6x +4 =0

whose centre = (, P)

=

(3,0) and radius =n= Vs.

1

.

6mc+c=9 8mc=8;

3)gives

c

When

.

.

axby

always lies on the line ar = by which is a fixed

line.

equation of the circle described on the and common chord of the circles x*+y-4x-5=0 x+y*+8y +7=0 as a diameter.

25. Find the

Here the circles are 5, = +y2-4x-5=0

S2x*+y

and

+8y +7=0

. **

(1) (2)

(2)

N3,

mc

tV3.

m= 1N3.

from (3) we get

13

and

y

y-a?+Mxx It will pass through

2+y2-4x -5 +r2+y2+8y +7)=0

or

(1+)r* +y)-4x

or

-v3

(

+

yy-a)=0

y1) if

(iyi-ax1+3)=

But+ya?,

0.

for (u yi) is not on the circle.

-1 from (3), the circumcircle of APQR is

4 *y13**1

+8Ay +7

y

(3)

-5 =0

-0. 1+0.

nte-1

3)

(4)

The equation of the common chord of (1) and (2) is

S-S-0

+y-a+f+y?-a)=0 Or

3+2nt3=1; m=1N3 two commontangents äre y

or

Any circle through the intersection of (1) and (2), is

from (3) we get

3-2iV3 1; When c-3,

circumcircle of APQR is a circle passing through the intersection of the circle (1) and the line (2), and the pointP y

he

(4)

1.

c3

Any circle through the intersection of the circles (1) and (2) is given by S, +15a=0 a-1)

P(x1)

(6mc +e)- (c2nic)=9-1-8 or

i.e., a = V5. a-3-4, Also a = 3, B =0,a = v5 satisfies (3)

coefficients,

V1).

equation of QR is xx + yy1-a'=0

(31+ c)=9(1 +m).(2)

(1)

which implies

must be the same line.

(1)

from the point PX,

the

and+=1 b

SM +C

(2)

a =3

--1

+y-a2-0

the length of the perpendicular from (3, y= mx+cis equal to the radius of the first circle

if (1) and (2) are identical.

6 0 -20B2-2-1 01

Tangents PO, PR are drawn to the circle x+y=a2 from the point P(x y1) touching the circle at Q, R respectively. Find the equation of the circumcircle of the APOR.

=c2is

.(1)

ax=by=c;

(y)

of intersection is (-3,0).

QR is the chord of contact of the tangents to the circle

and

+(n2-B2)m2

-2Bim-2ia-2mB-1 =0

B=0,

t

+c)=1+m

a+m)= la +mß+1)

But 4

0)

23.

comparing or

as shown in the figure.

is

we get from the figure that the triangle is equilateral.

equal to the radius of the second circl

m or i-1

+mß+ 1|

(a-a y

Then, the length of the perpendicular from (-1, y=mx+cis

The line Ix +my+1 =0 touches thecircle if

or

Let y= +cbe acommon tangent to the circlesev at the point of contact max

Let the circle be (r-a+(y-B)=a?

or

with the x-axis is

6

Obviously, the point of contact = (0, 0).

-

=-V3

x+y

yc

Now, the line (1)

centres.

n

Xx +

X

In similar situations if the radii of the circles are different then add the largest radius to the radius of the circle passing through, the

a

1 that, for all e eR, the pole of the line on a fixed with respect to the circle x+y=clies line. y1) with Let the pole be (x1. y). Then the polar of (x1,

24. Prove

an equilateral triangle.

the distance between (3, 0) and (-1,0)

the

- yyi =0.

orx+y2-xx,

y=+13

and

3

=0,y

1.

-V3+1+07-4

equation of the required circle is

a)=0

+y-a3-1(x+y

tangent at the point of contact (0, 0) is = 0. 0+y:0+ (x +0) =0;

prove that the lines Now, we have to

has the

centre = (-1,0) and radius =

re-2

the

The

form an equilateral triangle.

-V the

common tangents to the Circles and x+y+2r=0

22. Show that the

ie,

or

+y-4x-5)-(x+

y2+8y +7)=0

4x +8y+12 0

or x+2y +3=0 .(5) The centre will lie o the common chord, ie., (3) will be a circle described on the common chord as a diameter if

C-5S

Problems Plus in

or

2-8

putting circle is

y+

or

++28)x

2xy)-4x+8y +2=0 y-2x

of the circles, passes through a fixed point.

27.

B=(

Va)

A(X11)

i+f

th.

A variable circle passes through the

point

ot

intersection O of two given straight lines and cuts off from them portions OP and OQ such that prove that the circle always m OP+n 00-1 passes through a fixed point other than 0.

Let O be selected as the origin and the lines be y =0 and y=ar where Pis on y=0 and Qis on y = ar.

Q=(0Qcos e, 0Qsin

0)

which is of the form S+ AL =0. circle passes through the intersection of the

and the line 2r +ky =0,both being

fixed.

r

+y+py = 0 cut But the line 2x +ky = 0 and the circle at two fixed points, one of them being the origin O. Hence, the circles pass through a fixed point other

(0,0)

S0

than O. 28.

tan

The equation of the circle with AB as a diameter, is

(-XMr

- x)+(y-y1y-y)=0

0=a)

If two circles cut a third circle orthogonally, prove that their common chord will pass through the centre of the third circde.

Let us take the equations of the

2) =aX

and .(3

Let the equation of the circle be

r+y+2gx+ 2fy+c=0

(x-xr-x)+0-y-y»

The chord of intersection any one of the circles (4) is given by

2gx + 2+c-

S =0

(+X2)r +(y1 +y2)y-xi*2-y1Y2

OP=

.

(.

+y+2g

"

og+00

V1

Thecircles

(1) and (3)

cut orthogonally .(4)

a+c from (4) and (5), 2g(-)

.

=0. But A

2

80

ience, thecentre of the third circle=(0,-f).

-0

he

+a

common chord of (1) and (2) has the equation

S-S=0 y+2x +a)-(r+y+ Or 201-)r=0 x=0

OP+n-00=1

1

0f Hence,

satisfies the equation x = the problem.

S-S

=0,

2gx +2fy +c+4 =0

It has to pass through the centre (0, 0) of the circle (1.

(2)

0

. ()

0+a

1+a

V1+a

+m

=

The circles (2) and (3) cut orthogonaly

0Q=2tfu) m-38)

.(1)

+a

Is-S10 The equation of the common chord of (1) and (4) is

ie,

For a general statement in geometry concerning two circles we can select axes suitably (the line of centres as x-axis and the point midway between the centres as origin) to get the above form of equations. Letthe third circle be x2 + y2 +2gx + 2fy +c=0. 3

2g +39 00

i

-28

and

But m

2gx+2y+-7

****

OP+2g OP =0;

t-1)

-yy-y)a or

C=0

(4)4) 1=0. 1

3 of the fixed circle

+y+

... (

As it passes through 0, P and Q we get

+

two circles as

+y+2x+a=0

The equation of the line AB is

2 circle passing through A and B (the points of any intersection of the circle 2) and the line (3) is given by

.(4)

of a The. circle (4) will cut the circle (1) at extremities diameter if the common chord of (4) and (1) passes (1). of through the centre

Now, P= (OP, 0) and

B(22

(3)

S=x+y2+2gx +2fy+c=0

+y+Py+8(2x +ky) = 0

this

(1

. (2)

Let the required circle be

xty+28x+ (kg+Ply =0

circle y+Py=0

at

Sx+y-4=0 Sax+y-6x -8y +10 = 0 S=x+y* +2x-4y-2=0

kg+p where k and p are constants.

or

0

diameter Here the circles are

from (1) and (2), the circle has the equation fixed pont

x+y-6x-8y+ 10 0

+y=4,

x*+y*+2r -4y 2

l+2 m

+G +92* 2)9-X¥2-yiyat c-o

Hence, their point of intersection is also a Thisproves the result

Let Sx+y+2gr+25+c0 be the fixed circle, and A = (*, 1),

i

11-0

and

+4y +1=0.

26. A fixed circle is cut by circles passing through wo fixed points A(r, y1) and B(r y). Show that the chord of intersection of the fixed circle with any one

circles

This represents a famuly of ines passing through intersection of the two fixed straight lines

8y +7)=0

which cuts each of the and the extremities of a

29. Find the equation of the circle 1

-12-1a +0-

=

+3(1 +) A=1 0; this value in (3), the equation of the required

y4-5 or

Circles

(r+2+28)* +U +2 +2f)y

or

(from (4) and (5))

+3=0

C-59

IT MathematicsS

(5)

(5) C+4=0 The equation of the common chord of (2) and (4) is

S-S 0, (28+ 6)x +(2f+8)y +C- 10 =0. It has to pass through the centre (3,4) of the circle (2) (2g+6)3 +(2f+ 8)4 +c- 10 = 0 i.e,

or 6g+ 8f+ C+ 40 =0 (6) The equation of the conmmon chord of (3) and (4) is

S-S0, (28-2)x +02f+4)y+c+2=0. Ithas to pass through the centre (-1,2) of the circle (3) (2f+ 4)2 +c+2 =0 28-2X-1)+ ie,

or-2g +4f+c+ 12 =0 Now, 5)

6g+8+

22+a) =0 0.

c-4; 36

=

(7)

so, (6) and (7) give

0, i.e., 3g +4f+ 18

=

0

and -28+4f+8 0 (8)-99*

58+10=0 and so f=-3. 8-2 therequired circleis x+y*-4x-6y

4=0.

.

(8)

.

(9)

C-60

C61

Problems Plus in lIT Mathematics

Circles

30. For

what values of Tand m the circle 5(x+y") + ly - m =0 belongs to the coaxal systemn determined by the circles x+y+2r+4y-6=0 and

24+y)-x=

P(a.B)B

Y-B

(*1

. (2) tp)X=a(a +p)+ B(B+g) or (+9)Y+(a origin and the intersection The pair of lines through the degree of (, (2) is found by making8 (), second the help of (2).

0? o(0,0)

If

the radical axis for each pair of the three given circles is the same then the result is established.

S

The centre of the circle is O0, 0). Clearly po perpendicular to AB.

y-x=0 Now,

=0 is

S-S2=0, ie,

+y+2x+4y-

the

5x +8y -12 =0

0B a-0 a

e,

y'--*yz-)5r+21y-2m

= 0

8

Y1)

2p(a+p)

-a-p-4

equation of the locus of P(a, ) is

Find the locus of the middle points of chords of a given circle r+y=a*which subtend a right angle at the fixed point (p, q)

ie, and

or

+p+9-a (a

the

-Bk-B) 2), (a-a)x

(*1-a)h-a)=0 -aa =0.

equation of the locus in X,

aa(k-p)-Bla(h-a)+

+p)* +(p+4))=0

Y coordinates

is

-aa(h a)

21X(X+p)+YY+4)

or

2+y2 =4, m =6.

S+S

in

a2

new (X, Y) coordinates, P=(0, 0)

and equations of transformation are

n-a2

.

of the middle points of the chords of the circle r*+y'=a" which pass through a given point ( y

Let Pa, B)be the middle point of any chord AB thrpugh

the given point (x1, y1).

..

the middle point M of the chord AB be (a, P) Let new coordinates.

(

or

+

(a-a)la(h-a)+PkBI12 ++az(h-a)(a-a)tk-B)Ph-a)

y)=0

2(+y)-(px+9y)

=2a k-6)-plath

or

a+P

AB is perpendicular to MC, the equationofor MC

+(-4a+a-aa)(h -a)+ P(a-a)k- p))2 p 0*-9)-aB(h a)

qyXr*+ y)

a+yi)=0 +(p*+9*-a)=0

+(p+92-a

the

Or

will

2+y-2(px+

2+y)-2(px+ qy)

+y 7+y-pX -qy+g2-a)=0.

-a)+p(k-p))

(a-a)k-B)-B(h a) (aa -p°%k-B)-aßh-a))2

+(p+92-a?xx +y)=0or

(1)

Then 'm' of CM=P1,

be

(a- a)Xk-B)-ph a)

+2(px +4y)lx* + y')-(px + qy))

FP-9)

31. Find the locus

+ B(k- P)}

+

+(p+9-ax+

=0, (

-1). =p+X,y=q+Y. If we can find a e R such that S +152 =0 and the equation of the circle becomes S0 are identical equations then we can say (p +0)+(q+ S =0, is a circle belonging to the coaxal system determined by S, = 0, S = 0. whose centreis C=

B)}

p)*»(Y +4))=0But x7 +y?= 2ax Using equations of transformation (0), we get the alk-B)-Blah-a)+pCk-B)12 equation of the locus of M in old coordinates X, y as (a a(k-B)-B(h-a) 2(T-px+y-9)y+24px +qyi{(r -p)x+y-9)¥}

+p+9-MX

Note The system of circles coaxal with the circldes S=0 and S2 =0 has the equation of the form

B(k-

+(a -a)lath-a)

+2p(X+P)+40Y+ 9MX(X +P)+Y(Y+9))

l

(3)

+Bys

From (3), (h-a)x +(k-By1-a(h-a)-p(-B)=0

+24p(a +p) + q(ß + q)}la(a +p) + B(B + )}

(1) and (2) must be iderntical. So, comparing them,

(using perpendicularity) =-1using

1-ah-a +

From

2(a(a+p) + B(B+ 4))

P.q)

. (2)

(h, k).

-a-p-qla(a +p)+(B + +)| or

B) be the foot of the perpendicular to (1) fromn Then (a, B) satisfies (1), (2) (-o)a + y1B-ax =0

Let P(a,

(a+p) (a(a+p)+p(B+9J

2g( +9) a(a +p)+B(B+4)

Change the origin to the point P(p, 4).

-0

or or

32.

Thk)

orala+P)+P(B+9)

The equation of the radical axis of circles S, = 0, S3 =0 is

S-S 0

. (1)

+ y1y-ax =0

between the lines if

-)+v-)=0.

(1)

+ yy1= a(r + X1)

(1-a)x

a-p?a(a+p)+(B +4

aa-z)+B(By)=0

or

XX1

the

The pair has a right angle coefficient of X* + coefficient of Y? = 0

-.-1

6- r

orx+4y-6-0 or

of OP

equation of y1) be any point on the circle. The tangent to the circle at (x, yi) is ie.,

(a+p)x

a-X

m S2

:

x+Y+2pX+q).t9Y+ a(a + p) + P(B +9)

is

m'of ABPi

SyygyThe equation of the radical axis of circles S, =0,

+y=2ax. Let(x

homogeneous with the equation of the pair PA, PB is

Let the circles be

S=x+y+2x +4y-6=0

perpendiculars from of the feet of the to the circle the point (h, k) to the tangents

:33. Find the locus

-a)

2al(aa

x

0)

the

k(a-a)-

B(h-a)}

equation of the locus of the feet is

(lax-y-y)-ryh-x)}

+lx(r-2a

h-x)+y«-«)k-y)N2

C62

Problems Plus in IIT Mathematics

-y xk- y)-xy(h -x)Wk(x-)-yh-a)).

2al(ax

34. Show that the locus of the point, the tangents from which to the circle «+y*a^ include a constant angle a is

y-2a*

Then

Tet AB

y=a

(1)

be inclined at an angle a.

Let

+y=9.

to

+y-9%*+yí-9)= -9-9+yh

-9

(*x, +yy

S

S,

=

-h

(2)

-3m

k-mh

Dr

T

-9

2x1y1xy-18xx, 18p

a(1+m)=(k-mh)

or

a2-h ym2+2hkm+a2-k2=0

.

-m

mm)

or

(1+mm)* tan'u = (m, +m)*-4mm

(1+mm)

+yi-18

m

x2+y4=18

V+m ..(2)

+y+2Ax +41y 18(+1)=0

(3) is the family of circles

(2a-h?-k)tan

'a=4hk*-4(a-hKa?-k3) 4a h+k2-a)

equation of the locus is tan

a=4ar+y2-a

Second Method Any tahgent

to +y=a*is

y = mx +avi + m2

It passes through P(h, ) if k=mh +avi+

or (k-mh)*=a (1

+ m)

which is the same as (3).

the

C PN

touching (1) at (12, 4).

+y-2x-4y-200

the

angentially at leaves it point (-2,-2). After getting reflected the line it passes through the centre of the fromnd ciredicula equation of the straight line if its perpe distancefrom Pis the

ball

8+63

-

8v3 +6 + c(4 -

or

0

2cos

CoS

0

cos 20

20= sin

Now

'

14-213+c(4 +3v3) = t.10-+25 4+33

=ttan 20

As the ball is projected from a point in the third quadrant while moving in the anticlockwise sense, clearly the slope of the AB will be positive and it should cut negative intercept on the y-axis.

2sin6

cos 0

c23-39 4+33

8

the

cos(90°-0)

20

90°-; ofPC-2

+33)

16+ 27+2413+48+9-2413

c2V3-14 +25

1+2+2+2 92cos Sins 20 20

in anticlockwise direction

is P = (-2,-2)

or

tan 20

sec

the circle

Here, centre C=(1,2) and the point wne leaves the circle

5 . sec

Fromthe right-angled ACPN,

required equation of coaxal circles is (5)

36. A ball moving around

4N3-3

From the right-angled APLN,

2x +4y-18=0 from(3))

C+*5

-24+33

Now PN1 PC and PLLAB. So ZNPL=ZPNM = 0.

PN PLsec .. (3

4+313

4+3V3

2-2m1 +

.. ()

As (2) touches (1),

Dr

=-43+4V3 43 3

Let the equation of AB be y = mx +C where

P-2-21R

Also any two circles of (3) have the same radical axs

+y-24

4-313 4+313

=m3v3 -4

thetangent at (l2, 4) to the circlex*+y^= 181s

or

=

4v3m

3 43

AB "m' 'm' of of AB

+y-18 +(2x+4y18) =0

tan *a

=

3+413 3-413

The equation of circles passing through the intersection of (1) and (2) is

a1+ma

or

the

4733 = 3m t 47313 m=

Clearly/ (2, 4) is a point on x?+y2= 18.

This is quadratic equation in m whose roots m, m, are the slopes of the two tangents. As the angle between the two tangents is a,

or

4-3m

or

C( 1,2)

y-9+-9=0

y-18

or

t V3(3 + 4m)

or

As the pair is at right angle, a + b=0,

or

t 13

3+4m

equation of the locus of point of intersection of the perpendicular tangents is

V1+m

tan

4-3

ie., sum of coefficients ofx and y is zero

radius of (1) = length of the perpendicular from (0,0) si to the line (2)

=

60°= tan 60 3+4m

c1.2)

-

Any line through P(. ) is y-k=mx (2) will touch (1) if

tan(180°-90°+30)

2)

the equation of the pair of tangents is

= m

So ZPNM= 2CNM=0 (say).

(7 ) be the point of intersection of erpendeular

or

Let 'm' of MN

reflection.

tangents.

.

line and N be the point of

Let NM be the normal to the line AB at N.

At first we find the locus of point of intersecti ton f mutually perpendicular tangents.

ie,

(hay

of

h

intersection of mutually perpendicular tangents the circle

be the required

dence. PN is the line of incidence and NC is the line

35. Find the equation of the system of coaxal cirel are tangent at (2, 4) to the locus of

tan'a=4a'ta+ y-a').

Let P(h, k) be a point from which the tangents drawn to the circle

C63

Circles

proceed as above.

(-2)

30°

equation of the line

isy4+3134+313

Similarly, we can obtain the equation of the line when 'm of AB is 413 which is also

313-

positive.

C4 Problems Plus in iIT Mathematics

C65

Circles

Exercises Find the equation of the family of circles which touch the pair of linesx-y+2y-1=0. = 0 and 5x -12y- 40= 0 touch 29. Lines 5x + 12y-10 C lies in the circle C, of diameter 6. If the centre of C2 the first quadrant, find the equation of the circle which is concentric with C and cuts intercepts of length 8 on these lines. 28.

the equation of a circle whose diameter has the ind length 20 and the are 2x+ y=6 and

2.

14. Find the range of values of r

equations of two of its diameterss

the line

3x + 2y=4.

Find the equations of circles touching the y-axis at (0,3) and making an intercept of 8 units on thex-axis.

circles

5.

6.

1

9. Of the two

concentric circles the smaller one has the equation x+y*=4. If each of the two intercepts on the line x+y=2 made between the two circles is 1, find the equation of the larger circle. 2,3, 4 are

1,

3x + 4y +15

=

+y+2gr +2fy+c=0. 25. Find the

26.

0.

the four lines 47 -3y=5, 7x+y= 40 and x+3y+10=0 torm

13. Can

x-2y=10, the sides of a

cyclic quadrilateral? Justify your answer

b

contact. 21. Two circles, each of radius 5, touch each other t (1, 2). If the equation of the common tangent 4x +3y 10, find the equation of the circles. 22. Prove that x+y2=a2 and (x-20)2+y=a' ar wo equal circles touching each other. Find the equation of the circle(s) of equal radius touching both the circles. 23. Find the equation of the circle which touches5 circle r+y-6x+6y +17 0 externally ana which the lines x-3xy -3x+9y=0 are norm Find the area of an equilateral triangle insct ibed it 24. he circle

is

concyclic then prove that the product of their ordinates is equal to 1. 11. Find the angle that the chord of the circle x*+y-4y = 0 along the line r+y=1 subtends at the circumference of the larger segment. 12. Prove that the circlesx +y-9=4r*;r=1,2,3 cut off equal intercepts between the circles on the line

are

0.

line 31. A circle touches both the x-axis and the 4x 3y+4=0. Its centre is in the third quadrant and lies onthe line x - y -1 =0. Find the equation ot the circle w 32. Obtain the equations of the straight lines passing through the point A(2, 0) and making an angle 45° with the tangent at A: to the circle Find the equations of the 3)=25. x+2)+ circles, each of radius 3 whose centres are on these straight lines at a distance 52 fromA. 33.

The extremities of a diagonal of a, rectangle are 4 and (6, -1). A circle circumscribes the rectangle and cuts an intercept AB on the y-axis. Find the area of the triangle formed by AB and the tangents to the circle at A and B.

area of a square circumscribing the c rcle

30x+y6x+8y =0.

by Show that the area of the angle formed cinche tangents from the point (4,3) to the xty9and the line segmentjoiningtne points of contact is 7square

units.

:

27. Find the

3)

equations of the circles passing through 2 and touching the lines x + y=2 and x

x+y*=8

:44.

4

o'

1

(. 10. If the four distinct points m

th.

whether the circles x +y-21 - 4u( andr+y-8y-40 touch each other externall or intermally. Also find the point of tangency. 19. A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. Find the equation of the cirde with centre at (6, 5) and touching the above circle. 20. Show that the circles x+y-10x+ 4y-20 0 and x*+y+14x-6y +22 =0 touch each other. Find the equation of the conmmon tangent at the point

0.

2

the 32 but not

18. Examine

the points of intersection of C with the r-axis.

3

=

those values of pe Rfor which t chords can be drawm from the point (p, p) to the circle(r-p)+y=pboth ofwhich bisected

thelinex-2y+2

the equation of the circle passing through the points A(4,3) and B(2, 5), and touching the y-axis. Also find the point P on the y-axis such that LAPB has the greatest magnitude.

30. Find

values of m such that

17. Determine all

be two fixed points

1,2,3

un

h

largest circle

circle+y=4.

7. Find the equation of the circumcircle of the triangle formed by the lines y =x, y= 2x and y = 3r +2. 8. Prove that the equation of the circle passing through

Y;r=

ween

with thecentr e and lying between the Im

y=mx+4 cuts the circle x* +y*-4x

in a plane. Let C denote a cirele with centre B and passing through A. Prove that the real roots of the equation x+px +q=0 are given by the abscissae of

thenoncollinear points (z

2x+y3

16. Find the possible

The abscissa of two points A and B are the roots of the equation r2+2ax-b* =0 and their ordinates are the roots of the equation r+2px-q2=0. Find the equation and radius of the circle with AB as diameter. 2

points

lie

and x*+y?=2.

-y=1andr-y+3=0.

0.

Let A = (0, 1) and B =5

+=1

on the line

points (1,-2) and (4,-3), and whose centre lies on the line 3x +4y =7. 4. Find the equation of the circle passing through the intersection of the lines 3x + y=4 and x-3y +2=0 and concentric with the circle

20x+y)-3x +8y

3

15. Find the equation of the

3. Find the equation of the circle passing through the

for which then.

-2)=20-1)=r

the circle tangent drawn from the point (4, 0) to the first touches it at a point A in B on quadrant. Find the coordinates of another point the circle such that AB =4. (x -a)+y=b? 41. A tangent is drawn to the circle circle and a perpendicular tangent tothe their point of (x+a) +y= c2; find the locus of.bisectors of the intersection, and prove that the of angles between them always touch one or other two other fixed circles. circles 42. Find the coordinates of the point at which the 8y + 43 = 0 r+y-4x- 2y=4 and x+y-12x of common touch each other. Also find equations tangents touching the circles in distinct points. 43. A circle Cj of radius 5 has its centre at the origin A. Circles C2 and Cg with centres at B and C, and of radii 3 and 4 metres respectively touch the circle C and also touch the x-axis to the right of A, C2 being in the first quadrant and C3 in the fourth quadrant. Find the equations of the common tangents to the circles C2 and C3 except the x-axis. 40. A

at the 34. Tangents are drawn to the circle x+y=12 points where it is met by the circle whose equation x+y-5x+3y-2= 0; find the point of inter section of these tangents. 35,

circle 45. Find the equations of common tangents to circles +y=1 and (r- 1)2 (-3)2-4.

the

is

46. From any point on the circle + 2fy + c=0 r+y+28r

Find the equation of the tangent to the circle 60y +2100 = 0 at the point nearest to the origin.

tangents are drawn to the circle x*+y+2gx +2fy + csin'a+ (8+/cos 'a = 0. Prove that the angle between the tangents is 2a. 47. The chord of contact of tangents from a point on the circle 2+y=a2 to the circle x2+y^=b2 touches the circlex* +y*=ciShow that a, b, c are in GP 48. Show that the polar of the origin with respect to the circle +y?+2gx+2fy+c=0 touches the

r+y80x36.

A line segment AB is divided at C so that AC= 3CB. Circles are described on AC and CB as diameters and a common tangent meets AB produced at D. Show that BD is equal to the radius of the smaller

Prove that the line x+y =22 touches the circle +y=4. Also, if the point of contact be P then find the equations of the tangents to this circle which are at a distance 2 - 2 from the point P.

37. Prove that the tangent to the circle r+y*=5 point (1,-2) also touches the circle

-8x

+6y+ 20

at the

0

and find its point of contact. 38, AB is a diameter of a circle. CD is a chord parallel to AB and 2CD =AB. The tangent at B meets AC Produced at E. Prove that AE = 2AB. the point P(1, 0), a tangent PA is drawn to the circle + y2-8x + 12= 0, A being the point of contact. Find the equations of the tangents to the Circle from the middle point of PA if A is in the first

rom

quadrant

circle

r+y=a2

ifc2 =a?+s)

49. If the pole of a straight line with

+y=c

respect to the circle

lies onthecirclex+ y=9c* then prove that the line is a tangent to the circle 9x+9y2=c2

50.

Tangents are drawn to the circle x*+ y2=a2 from the point (h, k). Prove that the area of the triangle formed by them and their chord of contact is

ah2+k2-a ih2+k2

-66 Problems Plus in

51. Let C be the centre of a circle. Lines L and L are the polars of the points A and B with respect to the circle respectively. Perpendiculars AM and BN are dropped from A to the line L and from B to the line L, respectively. Prove that CA : CB =AM: BN. 52. Find the pole of the chord of the circle

x2+y2=81,

53.

the chord being bisected at the point (-2, 3). Find the equation of the circle which has for its diameter the chord cut off on the line pr + 4y-1=0 by the circle x*+y2-a2 =0.

IT Mathematics

circles. 64.

y+ax+

1-a bbc-c'

ABC

66. Prove

through two fixed points. Find the fixed points. x*+ y*-6x-4y +9 =0bisects the circumference of the circle x+y-8x -6y +23 = 0. 57. The polar lines ofapoint P with respect to two given intersecting circles meet in Q. Show that the common chord of the circles bisects PQ. 58. Find the equation to the circle whose diameter is the common chord of the circles (x-a)*+y2=a^ and x+(y- b)=bi Also find the. length of the common chord. 59. Prove that the length of the comnon chord of the

56. Prove that the circle

circles

+y+2hx

+a=0

and+y-2ky-a=0is

that x+y*+A(y-x-212) = 4 represen circles touching each other at a common point tor all real Also find the common point.

.

that the equation x*+y-2x-2y represents a circle passing through two fixed -S=0 points A, B for all real If the tangents at A, B to the circle meet on the line x + 2y +5= 0 then find the equation of that particular circle. 67. Find the equations of circles passing through the intersection of the circles x*+ y = 4 and

.

i

x+y-2x-2y+1

0

whose centre is at a distance 2v2 from the origin. 68. Prove that the common chord of the circle +y+6x+8y+7=0 and the circle which alwavs touches the line y = x and passes through (0, 0), will always pass through a fixed point. 69. Consider a family of circles passing through wo fixed points A(3,7) and B(6, 5). Show that the chords in which the circle x +y-4x-6y-3 = 0 cuts the members of the family are concurrent. Find the coordinates of the point of concurrency. 70. If the circle x+y2=16

60. Find the

equation of the circle passing through the points of intersection of the circles x 2+y2=4a and r*+y-2x-4y +4 0 and touching the line

61.

0.

intersects another circle C 0

radius 5 in such a manner that the common chord

maximum length'and has a slope show that the coordinates of the centre of the circle are

x+2y=0. Find the equation of the circle which passes through the points of intersection of the circles = +2y +4 -6 0 and +y*+2x-4y

x+y-6x

and whose centre lies on the liney=x. 62 Find the equation of the circle whose diameter is the common chord of the circles x+y+2x + 3y + 1 =0 and x+y*+4r+3y +2=0. 63. Do the circles

x+y2-8x

+y-6x-4y+9=0

and

-6y +23 =0 intersect each other? If they intersect, find the length of the common chord. touch, find the common tangent at the point If they

87.

+y 1

2 to two circles is 77. Prove that a common tangent bisected by the radical axis of the circles. the equation of the circle coaxal with the circles 78. Find

2x+2y-2x

and+y+4x

+

6y-3=0

+2y+1=0,

and whose centre is on the radical axis of the circles. 79. Three circles are given:

r+y=1,x*+y*-8x+ 15 and +y+10y +24 = 0.

0

Determine the radical centre (i.e., the point such that the tangents drawn from it to three circles are equal in length). From a point P, tangents drawn to the circles

= x+ytx-3 0, 3x+3y-5x+3y =0 and 4x2+4y2+8x+7y +9 =0 are equal. Find the

equation of the circle through P which touches the liner+y5 at the point (6, -1). 81.

82.

Prove that a common tangent to two circles of a coaxal system subtends a right angle at either imiting point of the system.

LetA=(-2,0) and B =(1,0) and Pis a variable point Such that LAPB = 60°. Prove analytically that the locusof Pis a circle. Find its radius and centre.

the

71 Find the equation

0

orthogonally

circles cutting the 76. Prove that the general equation of two circles x+y*+28X +2fy + C, = 0; r=1,2 orthogonally is

5

ofthe

86. The

the family of circles whose centres lie on 75. Prove that the line 2r-2y +9 =0 and each of which cuts the circlex+y=4 orthogonally, passes through two fixed points. Find the fixed points.

80.

of the circle passing througn u origin and cutting the circles + +y-4x+ 6y 10 0 and 2+y2+ 12y +6 at right angles. kRU 2OY 72. Find the equation of the circle which passes rougn theorigin, has its centre on the line x + y +4 cuts the circlex+y24r + 2y+4 0 orthogona 73 Prove that the two circles which pass throu the

onts (0,

a)

ca

(2+m ).

touch the n each other ortho8o lly

and (0,-a) and

mx t+c,will cut

83. A circle passes through the origin O and cuts two

lines r+y=0 and x

-

y

=

0in PandQrespectively.If

thestraightline PQ always passesthrough a fixed Point, find the locus of the centre of the circle.

8

Apoint moves such that the sum of the squares of its Cistances from the sides of a square of side unity is Equal to 9. Show that the locus of the point is a circle. its centre and radius. wo rods of lengths a and b slide along the coordinate axes in a manner that their ends are always concyclic. Find the locus of the centre of the Circle passing through these ends.

ind

5

any point of the circle point A(1, 5) is joined to middle point of AP Find the locus of the as P moves on the circle. the point P(1,2) and A variable circle passes through locus of the other touches the r-axis. Show that the end of the diameter through P is (r-1)8y. From the point A(0, 3) on the circle

S=0,5=0 have radii r and r'respectively. S S' circles =0 ill cut each other Prove that the Circles

74.

c=0 I

Prove that the equation 2(3 +p)x + 2(3-p)y +4=0 represents a circle for all values of p, passing

by +

Rand S are concyclic then show that

65. Prove

x+y+

e

in Pand Q.Another lineA *+B y+C'=0cu ircle 4y2+a'x +b'y+c=0 in R and S.. Ifts he P.O

equation of a chord of the circle +y-2ax = 0, prove that the circle of which this chord is a diameter has the equation

(1+m(x+y)-2a(r+ my) =0.

55.

the

The line Ax + By +C=0 cuts the circle

54. If y= mx be the

x

C-67 Circles

of contact. F they do not intersect or touch radius of the smallest circle touchined

+y=4.

88.

*+4x +(y 3)

= 0,

M such

to point a chord AB is drawn and extended the locus of M. that AM = 2AB. Find the equation of

a

circle origin, chords are drawn to the of the middle points locus the 1. Find = (r-1)+y of the chords. a fixed point 90. The base of a triangle passes through a, b) and its sides are respectively bisected at right y -8xy-9x=0. Prove that the angles by the lines locus of the vertex is a circle. Firnd its equation. O and 91. A circle of radiusr passes through the origin Cuts the axes at A and B. Find the locus of the

89. From the

centroid of AOAB. circle of radius r passes through the origin O and cuts the axes at A and B. Let P be the foot of the perpendicular from the origin to the line AB. Find the locus of P. 93. Find the locus of the middle points of the chords of thecirclex+y=a*subtendinga right angle at the point (a, 0). 94. Through a fixed point P a line is drawn perpendicular to any diameter I of a given circle to meet the diameter m at the point Q where m makes 45° with l in anticlockwise sense. Prove, that the locus of Q is a circle passing through the centre of the given circle. 95. The point A is one of the points of intersection of two given intersecting circles. Any line is drawn through A to cut the circles again at P and Q. Prove that the locus of the middle point of PQ is a circle. 96. Prove that the locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle is constant, is a circle whose centre is at the centroid of the triangle. 97. The tangent at any point P on the circle x+y?=2 cuts the axes in L and M. Find the locus of the middle point of LM. 98. A triangle has two of its sides along the axes; its third side touches the circle x*+ y*-2ax 2ay +a2=0. Find the equation of the locus of the circumcentre of the triangle.

92. A

na tne locus ot the foot of the perpendicular drawn from a fixed point on the r-axis to any tangent to the circle x +y=a2.A

C-68 Problems Plus in IIT Mathematics

C-69 Circles

00.

Let

S

=x*+y4+2gx + 2fy+c=0 be

a

The centre of a circle 1s (,1 and its rad units. If the centre is shifted along the lin through a distance V2 units, find the equati circle in the new position.

114.

given

circle Find the locus of the foot of the perpendicular drawn from the origin upon any chord which subtends a right angle at the origin. 101. Show that the locus of points from which the tangents drawn to a circle are orthogonal, is a concentric circle. 102. Find the locus of the point of intersection of tangents to the circlex=acos 0, y= asin 0 at two

Objective Questions

129.

cincles 130. The

115.

The circle x+ y- 4r 4y+ 4 =0 is inscribed in a triangle which has two of its sides along the axes of coordinates. If the locus of the circumcentre of the =0,

117.

118. The equation of

the circle which is concentric wth the circle 2+y2 5x-4y-1 0 and touches the 0 is . straight line 4r -3y-6

119. A circle

of:

(x- a)+y-B)=b2.

touches the z-axis at (2, 0) and has an intercept of 4 units on y-axis. Then the equation of 2 the circle is 120. The line L passes through the points of intersection of the circles +y=25 and +y2- 8t+7=0 The length of perpendicular from the centre ot the second circle on the line L is-

121.

109. Two equal circles touch one another. Find the locus of a point which moves so that the sum of the

tangents from it to the two circles is the constant k. 110. Find the locus of the centres of circles which cut the

112.

113,

- 6y +9 =0 +y+4x x+y*-4x +6y +4 =0 orthogonally

Find the locus of the point of contact of parallel tangents which are drawn to each of a series of coaxal circles.

Prove that the cirde (r - a)'+(y-a) =a touches the x-axis. If the circle is rolled on the x-axis in the positive direction through a complete revolution, find the equation of the circle in the new position.

The cirele z+y-4r-6y+ 16-0 rolls up the tangent to it at (2 +V3,3) by 2 units in the direction of increasingx. Find the equation of the circle in the new position.

The equation of the locus of midpoints of the chords of the circle 4x?+ 4y?-12x+ 4y +1 =0 that subtend an angle of

122.

at

its centre is

The sides of a' square ae x = 4, 1 = 7, y=l a The equation of the circumcircle of the squan

4

twocircless and

The point of intersection of the lines ax + by a =0 and bx-ay +b=0 is P. A circle with the centre et (1,0) passes throughP The tangent to the circle at P meets the x-axis at (k, 0). Then k = -

externally.

the origin O cuts the circle **+y*=2(x + y) at A. Prove that the locus of the centre of the circle drawn on OA as diameter is also a circle which touches the given circle internally 107. Find the locus of the point from which the chord contact of tangents drawn to the circle x2+ y2=16 subtends a right angle at the centre of the circle. 108. Find the locus of a point whose polar with respect to the circle r+y*=a* touches the circle

The equation of the circle which touches the cirel 2+y2+2x+2y-7 at (2,-1) and pasts through (1,0), is

0

find A.

system of circles is drawn through two fixed points. Tangents are drawn to these circles parallel to a given line. Find the equation of the locus of the points of contact. Find the locus of centres of the circles which touch the two circdes +y*=a2 and x2+y2=4ax

106. A line through

111.

The point on a circle nearest to the exterior P(1, 2), is 4 units away from P while the point Point farthest is (-3,-2). Then the equation of the circle

-0)+-b)*=

is

123. If (4, 1)

be an extremity of a diameter of the circie +6y 15 0 then the other extreu of the diameter is 124. The length of the chord 4x-3y = 5 of the circe The line 3x + 4y=k touches the circle x+y Then k = and the point of contact=126. The line 7y-r=5 is a tangent to the circle x+y-5x +5y =0. The other parallel tangen 125.

127. The length of

the tangent to the circle

x+y+6x-4y-3 0

+

c

(x-b) +(y-a)2=c2

and

a circle passes through the poinis of intersechon of thecoordinates axes with the lines Ar y + =0 and

131. 1f

r-2y+3=0 then the value of

is

through the point subtends an The chord whose equation is y-x=3, circle angle 30° in the major segment of the

+y=1passes 145.

x+y=k3 ifk=-

will touch each other if

Choose the corect option(s).

1

_

132. The area of the triangle fornmed by the positive r-axis, the normal and the tangent to the circle r2+y4 at (1, V3) is (2, 0) 133. The equation of the circle passing through (0, 4), and having the minimum radius is

146.

The radius of the circle

4x+ 4y-10x

+5y +5 0

is

(d) 5

(c) 3V5

-

and

of the circle which bisects the chord cut -2)+0+1)=16 line x - 2y-3 = 0, is offby the circle on the 135. The equation of the line passing through the points of intersection of the circles 3x+3y2x+ 12y-9 =0 + 2y -15 = 0 is. and x+y+6x 134.

136.

The equation of the diameter

The area bounded by the circles 2+y 4 and the rays

2x-3xy-2y=0,

y >0

is

x2+y=1 and given

.

by

137.The set of values of for which the circles -4y +a =0 have and +y2-4x +y2 exactly three real common tangents is 138. If the distances between the origin and centres ot r=1,2,3 the three circles x+y2-2,x-a=0; are in GP then the lengths of tangernts to the circles frpm any point on the circle y* =a are in -

z+

lines 4x-3y+2=0 and 8x = 6y +5 are two angents of the same circle thern the radius of the

139. 1f

x+y-2x

x+y+3x=y+ 10 is

the circle 16(x y)+48x-8y-43=0 nearest to the line 8x-4y +73 =0 is. tangents drawn 144. The chord of contact of the pair of + to the circle from each point on the line 2x y=4 143. The point on

+9 = 0 is

+ 6y

-

116.

104. A

105.

the circle r+y+4x

-

triangleisx+y-xy +Yr*+y*

(5, 1) is_ drawn from the point to the circle x*+y?=4 The equations of tangents which are inclined at 60° with the positive direction and of the x-axis are The pole of the line 3x + 5y + 17 = 0 with respect to

Fill in the blanks.

points whose parametric angles 8 differ by 103,

128,

147. The line 3x -4y =k will cut the circle

x+y2-4x - 8y -5 (a)k-35 ork> 15

= 0

at distinct points if (b)-35 15 oa-ycos =a touches the circle 148. The line xsin x+y=a, Then

a

(b) ae

[0,

T] (d) a is any angle (c)ae 149. The equation of a tangent to the circle = 0o +4y

T

-3

x*+y-6x

which is perpendicular to the liney

(a) x +2y +

=

27-1,is

t 4N5 = 0 (c)x +2yt (d) none of these 0 150. The equation of a tangent drawn to the circle x+y-2x +4y = 0 from the point (0, 1) is =0 4Y5 = 1

(a)2x-y+1 =0

(b) x+2y +

1

(b) 2x-y-2=0

c)x+2y-2=0 (d) x + 2y +1 =0 151. The centre of the circle passing through the origin and cutting off intercepts a and b on the r and y axes,is

circle will be 140. The

limiting

points

x+y+ 4x +9 = 0 are

of

the

coaxal

a(a) The circle r2+y2-2ax -2ay + a

circles

=0 touches

the axes of coordinates at. and. (6) The line (x-1)cos 0+(y- 1)sin 6 = 1 touches the circle whose equation is for all values of 0. 2

The equation of the circle passing through the Points of intersection of the circles xi+y'=6 and x+y6x+8= 0, and the point (1,1) is

(d) (a, b) 152. A square is

inscribed in the circle

x+y-2x+4y +3 = 0 and its sides are parallel to coordinate axes. Then one vertex of the square is

(a)(1+v2,-2) (b) (1- 2, -2) (c)(1,-2+v2) (d) none of these 153. Two circles (x-1)*+(y -3)= r and

2+y-8x

+ 2y +8=0

C-70 Problems Plus in IIT Mathematies

(a) 2 0.

i.e.,

,2a 20is

Selected Solved Examples

m

y mx+ 6. Condition for general equation of the second degree to represent a parabola

1.

.The line y= mx +c touches the parabola

The equation +2hxy + by+2gx+2fy +c =0 represents a parabola if h2= ab, ie., the second degree terms form a perfect square provided

The equation of the normal at (ri, y) is -2a

equation of a parabola into standard

form

equation (y B) = 4a(x a) can be reduced to the standard form by the transformations X, y-B=Y. The equation becomes Y= 4aX. which is the standard form in X, Y coordinates. (y-B)4a(ra) is the form of equation of a parabola whose axis of symmetry is parallel to the -

-

-a

:

V+T*

Note The lines ax +by+c= 0 and bx-ay perpendicular to each other]

X

The equation becomes Y'=J

+c'=0 are

axx +hry1+7y)+ byyi+8(r +X) +Sy+yi)+c=0 equation of the normal at (, yi) is The .Ash 1 (-x1)

y-(dy d

y1

daT

10. Chord of contact, polar line, pole

Let the equation of a parabola be y*-4ax = The chord of contact of tangents from the exterir point P(r, y) to the parabola is

TEyy-2a(a+z)=0

which is the

8. Parametric equations ofa parabola

T=yyh2a(x +X)=0 .The pole of a line L=0 with respect to the parabola

y=2at

are the parametric equations of a

the point (r y1) whose

parabola.

Any point on the parabola coordinates

y'=4ax

polar is the line L = 0.

has the 11. Chord with given

middle point

The equation of a chord of a second degree curve whose middle point is (z1, y1) is S, =T. So, for the parabola y- 4ax =0 it is 9.

Tangents and normals

i-4ax,= Yy1-2a( +1)

Let the equation of a parabola be y= 4ax.

.The equation of the tangent at (u y) to the parabola is

Y912a(x + x,) .The equation of the tangent at («t*, 2at) is

ty=x+at

**

-1)=-1, ie, a =B

Hence the equation of the parabola will be of the form

when a, rectum. (1)

S(1.1)

Solving (2), (3) we get Let

M-yi).

=

p=

So

12.

Diameter of a parabola

The locus of the middle points of parallel chord parabola is a line which is called a diameter

parabola.

=4a(4-B),

a= 4a(4 P)

.2

and

(1-a)" = 4a(9-B),

and

ie., 1-2a +a= 4a(9-B) (-2-a)*=4a(6- P),

V=f

i.e., 4+4a +a= 4a(6-B)

0, As Mis on

=0. So M=(0, 0). the directrix, (0, 0) satisfies (1).

Hence,=0. the equation of the directrixis x +y=0. Using focus-directrix properly, the equation of the parabola is

(3)-(2)

1-2a = 20a

(4)-(3)

3+6a = -12a

3+6a-12a

or

-3+6a

or

24a=-18;

(2)

equation of the parabola whose axis is to the y-axis and which passes through the points (0, 4),a,9) and (-2,6). Also find its latus rectum. ASthe axis is parallel to the y-axis, it will be x - a =0 for and the tangent to the vertex (which is

a

5(3+ 6a)

15

+30a

a-

a-

16 4-P or

-4xVS4V the ind parallel

=

1+-20a;

(5)

+4 0

The latus rectum

=

or-3(1-2a)

or 2( 1)+0-1))=(*+y) or 20-y-2-2y+2)-x+y?+ 2xy

or y-2xy-4x-4y or +y-1) y=4(x

4

.

1-2a204

(1-1-7

9-4-

ie,

a

51 40

8

4-p-9 L

from (1), the equation of the parabola is

Or

or

(1)

unknown constants, 4a being latus

a are

passes through (0, 4), (1, 9) and(-2,6). So

ie,

a

B,

(0-a)

As MV= VS, V is the middle point of

.

(-a)-4a-P)

(3)

MS.

The polar line of the point Pr yi) with respect to t parabola is

at

y-p=0

directrix

V=(a, p). Then a + B=1

78

standard form in X, Y coordinates.

r

Now, V is the foot of the perpendicular from S(1, 1) to

Let

some p.

=0 for

Y-0

theline x+y=1 and

B

.( (1)

directrix

Ser

a+b

the directrix will be of the form x+ y=a

-

x-a

directrix is parallel to the tangent at the vertex V. The edi

The equation of the tangent at (r, y) is

-axis.

The equation (ax+ by +c)'= bx-ay+c can be reduced to the standard form by the transformations ax +by+cy bx-ay +Cy

,

The equation of the normal at (at 2at) is y+ x 2at + at3 Let S ax2+2hxy + by+28x +2y+ c=0be a parab

y perpendicular to the axis) will be

Jatus rectum.

u

yma

abc+ 2/fgh-af-bg-ch0.

The

if

so any tangent to the parabola can be taken as

ax

7. Reduction of

Find the equation ot the parabola whose focus is (1, 1) x and the tangent at the vertex is +y=1. Also find its

-2=0

5)

C78

Mathematics Problems Plus in IIT

2x+3x- y+4 =0 and its latus rectum = 4a = 4

;

The point (E* rcos

6, Sin 8) is

on y= 4ax if

the vertex, focus, axis, directrix and latus rectum of the parabola y*+4x+4y-3 =0.

Now, if

y+4y +4 =-4x +7 or+2)4

Y, -1

ie,

o

there

are perpendicular lines

X, Y

vertex= (0, 0)x, v ie, in X, Y coordinates the vertex is (0, 0

when X-0-*=Xgivesxwhen Y

or and

=-43-

x

16x;

1,-x=X gives x

Xta=0,

when Y=0, y+2= Y gives y= -2

or

The equation of the axis is Y = 0, i.e.,y +2=0.

100

ie.,

thedirectrix is

= 4

x1 =4.

that

represents a parabola. Also find its focus and directrix.

.the

=

(-12)-9 16 144-144 0. Also,A#0

equation represents a parabola.

Now, the equation is (3x -4y)* = 5(4x + 3y + 12). Clearly, the lines 3x-4y =0 and 4x +3y + 12=0 are perpendicular to each other. So, let

v43=X

The equation of the parabola becomes

y-X4x

2ak

sin 6 fk=2a 4a2

IP

(

(2)

atj.2at,)

4x +3y

is the same for all positions of the chord

a(atž.2at2)

Solving (1) and (2) we get the coordinates of R. 6.

For what real values of a, the point (-2a, a + 1) will be an interior point of the smaller region bounded by the circle+y=4 and the parabola y? = 4x?

The point P(-2a, a + 1) will be

thecircle+y-4

53

+0.

0

and

an interior point of both

the parabola y-

4x

=

:

(1)-(2) or

chord PQ passes through K(k, 0) where :Suitably taken. Any point on PQ is (k + rcos where tan 0 is the slope of PQ. The equa parabola is y2=4ax

ie.,5a2+2

k

), Sin

*

a

(2),

0-xt-

h) +atzl2t1 -t)

x=at

f:th+t}

R= {at,t,

at

+ #)}

the area of APQR

(-20)+(a+1-4

positions of the chord. he

x (1)-

C:

0-x-4)+ alih-tšt)

y24

a is the same ur

a(t-)

y-t)=

y a(t, +)

From

0.

Prove that for a suitable point K on the axis of K, am parabola the chord PQ, passing through

and

b) cut

23.

Ellipse and Hyperbola

65.

is

The line 3x +5y = k touches the ellipse 16x+25y'= 400 if k is (a) + Vs

(b)t V15 (d) none of these 66. The equation of a tangent to the hyperbola 2-2y= 18 which is perpendicular to the line *-y=0 is (c) +25

(a) x+y=3

(6) x +y+

(c)X+y=312

(d) x +y+3v2 0

3= 0

C-110 Problems Plus in IIT MathematicCs

67. A

point on the ellipsex2+3y2 =37 where the normal is parallel to the line 6x 5y (a) (5,-2) 2,is -

(b) (5,2) (d) (-5,-2)

()-5,2) 68. If the

ordinate of the point of contact be 2 then the equation of the tangent x2+ 4y2 =25 is to (a) 3x +8y =25 (6) 8x + 3y = 25 (c) 8y 3x =25 (d) 3x-8y= 25 69. A tangent to the ellipse 16x2 +9y2 =144 making equal intercepts on both the axes is

(6) y=x-5

(a)y=x+5

(d) y=-x-5 ()y=-x+5 ellipsex*+ 4?1 6 to 70. If the tangent the to the circle + y' hep P' is a normal 8x to -

then o is equal (a)

n/2

b)

/4

(c) 0

d)-/ .

Das (3, 1) and (1, 1)passes An ellipse having foci thtog the point (1, 3) has the eccentricity

71.

(a)

v2-1(b) V3-1 ()

Answe1S 1.

5x+9y2+30x

2.

9x2+ 4y2 +36x -24y +36 0

3.

12x-4y2-24x+32y

5.

-

18y

41.v2 42. 5x2xy+ 5y76x

126 =0

-

127 =

0

vs+1,2)

6. (1,0), 23

6,23

33)(3v3)

2

10. a-2,

b=4

47.

hyperbola, (1, 2)

2x 3y -3v3 0 51. 2bva2-b2 unit2 49.

16. y-2

xt

tan

=0, 8x +5y -26 0;

aa 2a2-b7

b 26-a)

V2 + V13

y +2+ V13 =0

x+y2 =a2+ b2 29. c(x-a)-2xy = 0 30. 32. (a) 33.

-10a'y)+o*+a'y)-o 16x+ 10xy +y'=2

=4

37. (b) (x+y)*=16x2

8v2

5

espective

1

48.2 2x-v3y 50.

= 2v3

tanvb/a

52.4

53. 1

26.

(x2

82

= 0

(V61 45

5

24.

4

1

43.

88y+506

55.

a12-bm=n2

57.

rectangular hyperbola, v2.

58. 60.

59.0

25, V5

-92 40. hyperbola, (-2, 2)

61.(C)

(i), (a)

62. (a), (b),

34. 4x-y'=

56.

(), (d)

63. (c)

64. (a)

65. (c)

66. (a), (b)

67. (b), (d)

68. (a), (c)

69. (a), (b), (c),

70. (a), (c)

71.(a)

()

CalculuS

1.

Function

Recap of Facts and Formulae 1.

Function and its value

But F(x)-log x", x e Rand t(x)

BR is a real function then IEf:A B to which, corresponding to each fis a rule according a realf(x) € B. is unique xcA there fla) is the value of the function at x = a, i.e., x =a eA corresponds to f(a) c B.

4.

function may be defined in any one of the following ways: (1) uniform definition G1) piecewise definition (ii) general definition given by a property of the function.

B then x is the y=f(x), i.c., x e A independent variable and y is the dependent variable.

Domain and range of a real-valued function

Ify =f(r) be a real function then domain off= the set of real x for which fx) is real range of f the set of real values of f(x) for

For example Let (a)f(x) = x2+1

(b)g(r)= 2x -1,x (c) hx+

In (b), the definition is piecewise. For negative values of x, gtr) = 2x - 1 is to be used while for non-negative values of x, g(x) =X +3 is to be used.

¢(x)+ v(r) is D,nD

In (c), the definition is general. The function is described by no rules but by a property of h(x).

DnD-E the domain offt) =is y(x)

Clearly h(z) = e obeys the property, but there may be other functions satisfying the same property.

where E= the set of zeros of y(x). 5.

Equality of functions

flr)

function flx)

i.e.,

d(r) for all x e the common domain For example = x> 0 f(x) = log x2,x>0 and o(x) 2log x, same are equal functions because they have the +o«) we have domain (0, +oc) and for each xe (0,

f(r)

Some special piecewise functions

Modulus

Two functions f(x) and ox) are equal if () domain off= domain of and )

=

flx) =

=

=

hty) for all x, y eR.

xe

the domain off(r) = ¢(x) x y(x) is D, nDa

3.

y) = h(x)

In (a), the definition is uniform. For every R.f(x) = x*+1.

continuous functions. If the domain of ¢lx) be D, and the domain of y(x) be D then =

0 are not equal because they do not have the same domain.

where A c R,

=

Ixl,

x>0 0,x =0 x,

-X,

Signfunction

fx)

X

= 1,x

0

0,x =0

-1,x

Bthere is no a eA such thatf(a) =

function. Forexample

positive constant. If k is the least possible positive constant then k is the period of the function. For example cos x, Ix1,x2 1are even functions

Function Period

e

R to R

functionf(r) is evenif f-x)=f) for all x e domain functionf(x) is odd if for all x e domain f-)=f)

functionfx) is periodic if

A

B

If there is no such p e B thenfis an onto (surjectiyel

Even, odd and periodic functions A

Funcion

functions 8. Into and onto (surjective)

Ilx]-31llx]+2]>0.

Using sign-scheme,

Composite function

If fx) be a function whose domain is A and range

Is

and gy) be a function whose domain is B and the range is C then (g f) is a composite function whose domain A and the range is C such that (g z e Cwhere

is

f)=y

and g)-

toe

of))=

Forexample Lety=fa)-2+3 and gt)=sin then(g fM*) =glf(r)N=89) g(2x+3)= sin(2x +3).

2 ebefunction satisfying o

and f)-

xl-2 65 then find f6%

But

xl2

or

l>3.

x]= -3,-4,-5,..

x0

8a0,

= 3.

satisfies the relationf(r + y)=f)+f()

Iff)

from (1

FO=2-+N1+

2-212-1)50

0

and 9

6. Find the domain and range of the function

O

Now ain

Hence its

ie,a246)2-4(9+8a)s0

of)RE

20 32-(22-220 (25-3.2250 or

8.

This has to be true for all real y.

and1

range

Now 3-2-2120

or 91-2y+y9+8a +(a+64)y 8ay220 or 9+8ca)y+(a+46)y + (9 +8a)20.

(See Limit)

+3

or 5x-6

the function is not one-to-one for a

36(1-)+ 4(a +8y%8 + ay) 2 0+

forreal x,f() can aftain all real values except

and

D-7

ie., x=X

Asris real, D20

all real y(# 1), z is real, and

fO=lim

Let dom=D, domy=D Then domfsDoDa

So,

t8y)}+6(1y)x

3+1 for

- (8 + ay) = 0.

+3b (f()-ift)P13

for all x e R then prove thatf(r) is a periodic function.

Here,

Ifa + x) b3 -

=b+1-3bF)+3b1f(*}-{f(*)°

=1-fy°-3b

1fr)+362 f)-b}

=1-f)-b3

fla+x)-b+ f)-b=1 This is true for all x.

Putting a +x for x in (1), we get,

.()

Function Problems Plus in !IT

D-6

Mathematics

vsin.

fr)-v3-2*-21 and vr)=Vsin

for

D,, dom y = D2. Then domf=D,

Let dom

6.

or

3-2-(2)-220 (2)-3.23+2s0

or

(2-202-1) By

Find the domain and range of the function

s0

2-x20

rs2

1+z20

r2-1

or or

2e D

Now, sin

ie., 2°s2

s2;

0Srs1

r20

0ssin

zs

the principal value of sin"r is positive in the first quadrant}

When

domf

D,

5. Find the domain and range of

fa)

x-3x+2

+-6=0

or(r+3)Nx-2)=0,i.e., x =2,-3 off= R-(2,-31.

Now yf)

(*+3)(x-2)

f-3)=*

and

2

(-2-1-

6-)8(r-

=2

(6-8x

Or

-N(6

fis

Asf:R->R, the function will be onto if

OC

+6x8x2

-8)

-(-24x -30x? + 45x, +18)

a+6x-82

+6x-8

-8

N-24x-30x7+ 100x-48

a6x-8 Let

6(-) x)

(3x+6)03+6x1 -8x)-8x,(3x+6x-8) = -3x,03+ 6x -8x,)) 0

onto. Find the interval of values of a for which thefunction one-to-one for a=37Justify your answeh

for real values of x.

-(-66x, +55)] = 0

I55x66-z2(-66x,

or

55x,

2 5n

=

Now, putting x

81,N3x? + 6x, -8) +6(3 + 6x,-8x)) =0

-8x,)(3xj+6x

f(n)

1)

=5r

=

5(1 +2 +3+...

+ m)

+ 1)

{m

(3r+3

art6rR>R is defined byf)=a+6x-8r

can assume any real value

2

1363-)+6(- )(3+6x-8x)

7. A functionf:

X+3

=f(1)+5(r-1)=5 +5(r-

5 n=l

601)-8(«- xM3N? + 6r, Or

range offr) =[v3, v6].

x*+*-6

domain

x

the least value of flx) =3

Hene,f)--3x

flx) is not defined for those values of r for which

3t-+

I2

the greatest value off(r) = v6 and

*+X-66

-fr-3)+3.5

=f(r-r-1) + (7-1)5 because by (2), a

(1)

=...

(using ratio and proportion)

when-=5ie, =

+64)

3+6=9 R a-3,f)=

the least value of f(t)} =3 +0 3,

11.

. using (1)

=fr-2)+2.5=fr-3) +5}+2.5

37+6x-8 3+6x-8 3+6x8xi 3+6x-8x when

n D, = [0, 11n [O, 1] = [0,

fr-2)+5)+5,

=

fusing the given property)

1)+5

fr)=ftr-

Letfn)-fr). Then

thegreatest value of f(a)=3+2.5

D= 10, 1

sin).

"

0Sxss1

.

=fr-1)+f(1)

Bysign-scheme, 2 Sa s14.

-3+2

f:

8

a-16a +28) 50 (d+8)a 14) a-2)s0 (a16a

=f)+f(Y)

Heref(r)=fir-1+1)

... (2)

a-8

3+212+-

10, 1.

.

(a-14Xa -2) S0

domain off(x) = [-1,2]. Again, fz)}=(V2-x+ V +x)2 3+2N(2-rX1 +*)

21,

a>

1e,

+ y)

Also prove thatf(r) is odd.

from (1),

Domain of flr) = {x 12-x20 and 1 +r20).

sign-scheme,

1,

9+8a >0,

relation f(r

for all x, y e Randf(1) = 5 then find

s (a446+18+16a) a+46-18-16a) 0

fr)=2-r+vi+z.

Now,

8. 15 fxr) satisfies the

n-1

(a+46)-4(9 +8a)'so

ie,

Now,3-2-21-*20

3-2-20

the function is not one-to-one for a =3.

true for all real y.

Hence its D s0 and 9 + 8a>0

and

and

or

=

ay) 0.

+y)+

This has to be

rangeofi)-R-

={r eR1 sin z201.

(8 +

or(9+8a)y+(a*+46)y + (9 + 8a) 2 0.

(See Limt

real x.flt) can attain all real values except

for

-

36(1-y)+ 4(a +8y)(8 +ay) 20 +8ay 20 8a + (a +64)y 901-2y

or

1

= lim

nD

D, =(r eR I3-2%-2-*20} and D

all real y(* 1), x is real, and

fo)

v)

fr)-)+

So

So,

.

Ao y-3

)-V3-2-21-7

Let

at8y)}x* +6(1-y)r Asrisreal,D20

SeeLimit

4. Find the domain of the function

+55)] =0

-66 =x(55 -66x)

=

0

andy =0 in the given relation,

f0)=0. f0+0)=f0)+f0)% Also, putting-r for y in the given relation

for-)-f() +f-) +f(-r)

f0)=f) 0=fx)

+f(-x), ie.,f-x)=

-fx)

So, the function is odd.

9.

If a, b e R be fixed positive numbers such that

fla+

=b+ b°+1-3b fa)+3b (f*))*-1ftr)}°13 for all r e R then prove that f(r) is a periodic functio

fa +x)- b3 -b+1-3bf)+3biftr))2-1fN =1-f))-3b 1f)* +362.f) -b)

Here,

=1-ft)-b)

fa +)-b)3+ift«)-b)3=1

This is true for all x.

Putting a +x for x in (1), we get,

. .(1)

Problems Plus in

D-8

.(2)

f(2a +x)-b}+1fla+x)-b}=1

and glx)=Vx+1 forallr

f(2a +x)-b-iftr)-8}3=0 = or f(20+x)-b} lf«)-b13

f2+)-b=f)-b f2a+)=f)

or or

f)

is a periodic function.

MfetrM= hr)=x

because x20 lfrom definition of hril

Now,flst))=x for allx gK)=x for allx.

Asx20,(f8*)

the function h is the identity function.

Herefa)= -1

z

So

1

hat'm

ysin(-8)

18.f)=log-s3x-10)

F

is not

21.

Find the natural number a for which

fla

=xis an identity

the function f satisfies the relation +y)=f() fY) for all natural numbers x, y and alsof(1)= 2.

fx

Find the domain of the followingfunctions:

st)= log(1+v1+r)and ht)=f)-8)

then prove that h

7.

4. In the figure, BC = b and AH =h where AH L BC. If EF =x where EF BC then express 1 the area and

perimeter of the rectangle DEFG as functions..

9.f(x)=

VIxl-

+

INr-3

2.f)-snoa yscos4sin

14.f)-151xl--6

if f() is any function, f() +f-) is an even function while f(x)-f(-r) is an odd function. Hence, prove that every function can be expressed as the sum of an even function and an odd function.

fcos x).

27.f)=lo8103x4x +5) 28 f)

31.

x-2

is a bijective function.

product of two odd functions is an even function while their sum is an odd function.

41. Prove that the

42. (a)

f

1+

(b) Prove

Ix-11

is an even

thatf(r) = v1-x)

+

(1+)

43. If the functionfsatisfies

the relation

fr+y)+fr-y)= 2f)

,

f)

for all ye R and f0) + 0 then prove that f(r) is an even function.

2x

2 fX-3x+4 +4T 1)

and domain

=

R-11,4)

trenfind the range off(7). .

5If00,n eN then show thatfiftr) = x. 3. If fr)=log{a + v+z),

2

Check whether f(x)1+

onto or both or none.

function of r and indicate the domain of definition.

k=1

2

8x

+ Vsin(cos )

sin+2-x-

24.f)=cos

cx +d

Ifd=-4, show that ff}

+ 18 an one-one function. 37. Verify ifflr) iff{7) = + 37. =2 4x +30is

39.

23.

25, 1f log,y 6.

+2

all

y

(ii)f(2) *a.

(ü)f(y)=b

Find the functionf.

38.

1

22 fEcos

Exercises

1. Let fx)=

{x, y, z} > {a, b, c} be an one-one function. It is known that only one of the following statements is

true:

f)=3-+cos

But, by definition h(r) # x for x 1

)fz)

=

*z, 0 Srs1 -1.x>1

periodperiod

5

b) 15fx)- 1, x>0 0, x=0

.3/w2] espertively 72 R-2 spectively 74-1,1 73. (4, )

83. (c)

81 (c) 85. (b) 69.0(b)

92. false

3. true

84. (b), (c)

82.

(b), (c)

86. (c) 90. true

. true

then which of the tollowing are true?

-1, I0

[1.+)

80. (a), (c)

49.period

range

-.

there ate multupe optans Choe the correvt optiun(s) ne Nba perwdic function w hoe perioad is D none of these A. indeterninate B1 C.

9

47.period

off.

domain

of folkowinyg,

7

75.

5ten

inteu

bexotes true

(o) fu)1-bl*n,

ifr=-1,-350. Also find the set of

a is and a is a constant. cons and

8. Letybe a funetion of r givenby 101- B1y -2-4r- 2y Two-one and among the two eal values of

x

5. Prove that the function is that correspond to a real value of y, oneis

s-and the other is 2 9.1fr-41+3

0 where r

=

Vx*

+vi,

*e

R.

ye

R

then find the greatest and the least

values of r 10.Let ftr) =21 x,xS1. Prove that the function f"(r) exists and find the solution-set oftheequationf "() =f). -

Problems Plus in IUT Mathematics

11.

=x, *0. Find the domain and

a Letf be function defined by 2f(x-1)

range of

the functionf(r).

2. Differentiation

Recap of Facts and Formulae

1.Definition of

differential coefficient

d(tan

14

dx

1fy=f) be afunction then

differentfal coefficient ofy wrt. r is

o

f-

lim

+h-f)

or

fa)

= lim

a

_xl VE=1

d(sec

0

d

o diferential coeficient of y wrt. x atr =a is

+h-fla)

dcosecx ax

)dcos

Note

2. Differential coefficients of standard functions

(1/1) dx

using chain rule

de")

xV1-(1/**)

alOg

wherex>0 3. Rules

If a where a > 0

log

for differentiation

y=u() + v{r) t w()t... then dy du(a)+ dv(x) , dtu) (term-by-term differentiation)

cos Differentiation of a term:

dcos )

2sinx

1.

(a)2

2. (a) B

(b)[,41 ()R-rlx= nm+(-1)"2+sin (b)A,

B,

C (¢) B, C

6.

= domain ={b, al, range [Na-b, V2(a -

9.

greatest value = 3, least value =1

11.

R--11.

R-

b)1

respectively

-

10. 10, 11

ax

2) an2sec

Answers e

6ec d

dcot23

d cosec )

O

sec

du)v(r)}

tan x

COsec

osecX

=

d

du(?).

dx

d

D))

cot x

v(r) + ucr).

T-

otwr D-15

do(x) (product rule)

v()-

dv(x)

u()*dx

quotient rule)

tvcx))

dulx) lutvon dulvlr), dv(r)dx

dGin

ax

x2vx-1

(chain rule)

dulv{wtr)}1, dolw(r) dw(x) dofw(r)} dux) dx

-16

Problems Plus in lIT Mathematics

Diferentiation

of f)} *

dx

7.

wrt. x taking b) asa

constant+ d.c.of (ft)* * wtt. takingf(x) as aconstant

Higher derivatives of a function If y=fx) then the derivative of

x

i

D-17

wrt.

x is called tha

1

second derivative of y w.r.t. x and it is denoted by

0

lim

4. Differentiation of function represented parametrically

using

If yis a function of x given parametrically by ai y= 40), x=V) then

is a function of x such that x= 40), y = v() where t is the parameter then dy

1log

2.

im08 cos

5.

If

y,

.

andz = v() then

dzde

ie,

:fy=P()

1v-

a)

a function

given in the form

of a

d

=a"nx"-1-ny"-1 4 dx

P) A()

2h

w'x)

4) H(*)

rx) Vr)

u(x)tv(x) P ) 9'() A(r)

r 21-1

ieh wx)

r't v{T)

H(X)

u(r)vr)

w)

(r)

v '(1)

P(x)9(x) (r)

or

im

h0

r()

0g

cOS

los cos 4h-

Selected Solved Examples

los

at

25in

Cos f() in

h0

cos

Now, we have

2+2

or

4h213cos 4h 4h

sin 4h

or

-sin 2h+V3 cos 2h

lim

Now

Vi-x2n

V1-27

=y-1V-y V1-y2

sin4h

logcos 2hsin 2

m

*from the first principle.

x=is

-1,ta".

differentiation can also be done columnwise.

The

where h is a very small positive quantity, remains the same. So, not a turning point of definition.

2ny

fim V3 sin h

t sdtsis

u'()v'()

')

Clearly, the definition of the function

-y

-

hsin

0

r)then

A()H()v)|

fr)=logec z lcos 4xl +lsinxl then find

V1-y" =a"a"-y"» prove that dy=x"-1 dr

Differentiating both sides w.rt. X,

lim

u(X)(r)wr)

6. Relation betweenand

1. If

If V1-H +

2-1

Differentiation of one function wrt. another function

y=)

cos

Now, Hm 0 Differentiation of determinant

-log 2

d

ie, 8.

log,

definition of mod)

cos 3+4

Ify

dz ddxv

6v3

imsin

dx

+for)*).log ft).49) (rule of logarithmic diferentiation).

4v3+213

m 0

= lim h0

-E-1-in

cos 2h + sin 2h

22

2

in

sin 2h+ v3 cos 2h v3 +*-cos 1 2h sin 2h 2

(using L' Hospital's rule)

V1-x2"+V1-y" =a"x"-y") (1-)-(1-y) V1-y -a"-y"riy2-3a "(a"-y"V1-x-V-y

V--v-y or

+x")

a1-x+"=a" v1-y

-y"

Using (2) in (1),

V

Writing in the differential form,

V-T

dy=x"-

Vi-yd.

2)

Problems Plus in IIT Mathenatics

-18 3. If

Diferentiation

Sinyo

sin x and g'() = cos *

u=fr*), v=gt*), f')=

cos

du

(osin yo)

sin(r) 2x

d=f')-2r

:f'l)=sin

f'r)=

x

sin(r*)

or

x'h" 7.

cos(r *)}

h)-2F) Now,

3x cos x3

3r2

(1

diff.coeff. of u" wrtxwhen v is

If fgh are differentiable functions of x and

A)= |(

a constant +diff.coeff. of u"wtt xwhen u is a constant.

8

(g

8. A functionf: R ->

in +(sin y)2.log

hyprove that

express y as an explicit function

6.If 2=y+ythen x

of

and prove that

dy

-1)

2

dx

A)

sin ycos

f+f) 2f+ 4sf'+*f"

g+8

Differentiatingftr + y) =f)

h

2h +4xh'+x*h"|

(=2f+xf ("=2(+ f)+21f'+x

+2))

2 sec log +2)

0

xh' 2xyf

2g+g"

sin yo

20

70

F)

xhRR-2R2

or ... (1)

wrt 2,

Nx-1)

taking the common factor x from R;

5y

fusing

-1

f(x)

But f0) 0

(1)

f0) =

0.

(given); sof0) =1.

Puttingx= 0 in(1),log 1 =0+c;

-25

f)

integrating, logf(«) =2x+c Again, putting x =0, y=0 in the given relation, fof0)-1)

()

=f'0)

f')-2

-

dfferentiaingagain

So=0. f)

t:f'0)=2

f')=2F)

f0+0) f0) R,

20.ysin yo)

fY) wxt. x,

Putting x = 0 and x for y, f')

2xth'+xh"

RR-2R2,R2 (sin yo) log

ye21 -yj-i-y) Differentiatng

=f)

Thus we get ft+y)

2x2-1

h

Puting r=-1, we get (assuming y =Yo when r=-1)

But y is independent of x.

4x-4fyi-2

2 2g+ 4xg +x'g"

4

+2Mog 2 tan log,

=0

Rsatisfies the equation given by

Fa+y)-1+-Fs)+f

+ xh

12x1

f")= -f)

fr+y)=f) fy), for all x, y in Randf() #0 for any x in R Let flr) be differentiable and f'(0) =2. Show that f')= f(*) for all x e R. Hence find f().

differentiating the given rela tion w.rt.z,

2cosy

t:

h 5.

y)

'()

h

hx)= 11 h(10) 11

We know,

sin(sin

.. (1)

'

t)

)=2f0) 80)+2g6) -f) hx)=c;buth(5) = 11. So 11 =c.

h'

+sec(2)+2"tanlog(a + 2y=0.

duax

2gt) g

-f)-8

Hence,d

Findatx=-1 when

(sin

f)+

f')=8)f")=8

h

dx

y

find h(10).

sinyo

cos(r*)

Letfbe twice differentiablesuch thatf"(a)--f) and f')=gta). If ho*) ={f*)}+{st>)* where h(5) 11,

Differentiating h(r) = 1f(x)}*+ lg(«)1* w.rt. x,

Jo

sinx

sin(),2x2

do

"

(sin yo)

Sin

8'(x)=

cos

8')= du

0forthenyis

e-2+tan log,1 =0 h'

83r=cos(r)-32

4.

-125

Now, putting r=-1 in the original relation,

Differentiating u =f(**) and v =g(*) wrt. x we get

do

D-19

or

h

then find

dudx

h'

c0.

logf)-2x

w.rt

fo)=e dx

Note If the differentiability of fx) is not given, thee correct method will be as given ahead.

Problems Plus in

D-20

ft)=

lim

f'(10)=1.

h

flrNf(h)- 1 10. Let

h0

=f(r)*lim

f0)

f(x) lim

: fr)

f29*

F0)= 1as obtained above)

f0)=2f(r)

fa

(takingx

lim lim

if x>0,f(r) 20

7Sigith

Differentiating w.rt.x,

2

Integrating

f-0

flr+ 1) =flr)

Thus (3)

-

=

f0)

bgiven) (1)

f) +f inf2 2e

Again, putting y = 0

f)+1)+3

-f)+4 1+4-5

2k+

-

(2k- 1)sin(2k-

sin(2k

-

1)sin(2k-

1)x

1)x

1}x -

1)xl]

(2k- 1)sin(2k + 1)r|.

-

cos 16.

a

Show that

v=+Bis asolution of the differential

.=

equation

Fndthederivatives w.r.t. x from the first principle atthetndtcated points: 7. xanr at a=1

18.

atx=2

lo

**erengusa

a)Cr

'

y= Va-r+log

20. sin

21.

b cos

-

Findif yisequal to:

l*,x*0 findf'. show that

ntati d

Cx

(x

ax showthatdi

atx=e log

-b)

-t

rthat

19. If

+b

f')af"(t)=0,f""()=0, F_0.

Ify =1 +

where A, B are constants

a VaT

log

ty

8 log xl atx-2

5.1 f)

0,

equatodr3rdr

e

12. sin(log )

2cas a, where

A, B are constants.

x

from (1),

r is a solution

dy of the differential equation y +

11.1)+ atr

fr)ax

+ 1)x

+

15. Show that y = Asin x + Bcos x+ Xsin

rcos

10. cos ata

we get

Hencef(1) = 1.

f5)=f(4) + 1 = {f3) +1)+1 f2)+1) +2

+.

dx

9x-1+1-3]

ie., fx) is identically zero.

fa)0.

sin 2kx sin *

prove that 2y*

wrt

f(l)=f(2)=f(3)=... =0, Sofrom (3),fr+1)=f)+1

1)x =

t cos(2-

7.

b=c

r

-Isin(2k+ 1)x + sin(2k

X

(given).

When 0,f(0) =0 +c, ie.,

Putting y 1 in (1), for all real x,

fa+ 1)=f)+ f1n"*1

5.

0)=a

sin

2kx

2sin

Vsin

r

ts. (2)

or 1, {from (2).

cOS KX

Diferentiate the following functions w.r.t. x from the first principle:

3. cos (eYan

=0

fu)-f)*"]

f)=0

2

Exercises

Taking = 0 and x in place of y we get

Puttingr=0,y= 1 in (1), 2n*1 Ta) =f0)+if(iy

12kcoS 2k Sin x- sin 2x cos

=x, x^ =y, 1 =2}

t

0

+(2k- 1)sin(2k 1xl1

x,

This holds for any real x, y. So y is independent of x, ie,

f0) =f(0) +f(0)

)-F0),,

1sinr +3sin 3x + 5sin 5x

et 2sinksin(2k X

Sin

1.

F)20

-

1)x. =

sin

cos r+cos 3x + cos 5x +.

f)+f2

Differentiating w.r.t.x,

+(2k-1)sin(2k- 1)x.

sin x +3sin 3x + 5sin 5x

sinRx

= b,

fo)=0.

or

S

From the question,

Letf) be a real function not identically zero such that +y)-f)+FYN**neN and xy are any real numbers andf (0) 20. Find the values off(5) and f'10). .. (1) Here,fr +y"* ')=flx) +1fty)) 7*

t..

D-21

sin

+ft,

where x are any real numbers and n eN. 11f)is diferentiableand f'(0)=aandf) find

9.

Putting x=0,y=0we get

+

Let S cos + cos 3x + cos 5x +..+Cos(2kHere the angles are in APwhose first term common diff. = 2.

f)+f(+f)+.

fh)-1

h+0

11. Find the sum of sin 3sin 3r + 5sin 5r

f=1

f)=*

f)-fh)-f

n0

Differentiation

Also (4) = ft)is a function whose value increases by 1 when variable r is increased by 1.

+h-fr) h

lim

IT Malhematics

dx2

(a

+b

cos )

VI -*

-va.vT-

cos+12N1-x

22. (xlog

x)og

log x

where

00 for all x then exists a functionftr) such that (a)ft) «0 for all x

b)-1f") (a0

)stt)=0

log a/1+

s+log

(d)

for any realx

for the cuve 2y=3-ris

37.

2yx V1-r

)-7

(a-

then (1)=-

90.

(a)1

+y-5=0

-2ry+5r

Thenaty=1is

Choese the correct option(s).

The value of

Dittrentiation

and at x=1,y=1. t

85. (

(a) 93. true

83.0

86.(

87. (a)

90.(

91.(a

94. true

95.

true

D-27

Diferentiation

Chapter Test Time: 60 minutes

resulting sentence becomes true. 1, In each of the following, fill in the blanks so that the (a) If (b) If

a

_.

andf)-sinz'then

,

f(x) = (ax + b)cos x + (cx + d)sir

()Iff(x) = og

b=_

and f"(x) = xcos x is an identity in x then

x

,d=

,C.

=and g(x) =fVx)) then g '(4)

|x-1|

r

(d) 2.

d(log x)

Findin terms of xift= 1+x y=x2+t2

function offandf'(x) 3. Ifg is the inverse 1

4.

Ify (1+x)(1 +x1+x)..

5. Ify =fx), p=and

q

6. Letf(r +y) =f(x).fy) xE

(1

=

+r)find

=then

inthesimplestform.

prove thatf "(x) can be obtained at all for allx, y and f(0) # 0 then

equation 7. Transform the differential

and z

prove thatg "() =1+{g(x)]°.

dx a function ofp and q. expressas dy2

Riff°(0) exists.

& Lety

-

dy

(1-x9+y=*by

=Provethat d

substitutingx = cost.

D-28

P'nWems

Pus in UT Mathematics

AnstveTS

-r) 2.sin

1. (a) (a) (c)

2.

4.

1

(d) 2log

rrg

b)0, 1,1,0 respectively *

2x1+x 2x

1-

A

3. Limnit, Indeterminate

Recap ofFacts and 1.

value. Indeterminante forms of values of a function at á indeterminate forms of

The

=ea

x+0

0, o -co; ,0,o 1

X

2.

Formulae

lim (1 +ax)

point are

Form

limits 3. Properties of 1f(x) + ¢(x)) =lim f(x)

lim

Standard limits

x>a

1 lim=0

.lim

f(x) x dlx)}=lim f(x)

x0

where n> 0 lim xl "=0,

lim(x)

x lim o (x)

f(x)

lim

lim

tlim d(r)

lim o(X)

ir

lim logf)

0

lim x"=0if lxl1 lim Ix1"=o if

lim fx))°=

1-

lim

lim o);

fx)*

X

a"

lim

na

1,

.lim

lim flo (r)) =f( li

r

where is in radian

lim

4.

If

x= 1

L'Hospital'srule

f(a)=0 and ¢(a)=0 then

x0

lim 0

lim

1;

where

is in radian

lim

measure

im f

"

X

then If limf(x)=o and lim (x)=0

-1.1

x0

a lim

is continuous

measure

0

cos

())iff()

f)

(x)

log

x0

D-29

lim f)

Problems Plus in lIT Mathematics

D-30

Lirmit, Indeterminate Form

Selected Solved Examples 1.

Evaluate lim +-Vs+r7-Ng +x+

limit

is a rational number. Xwhich n!

.from

(2), limit = 2 ifr is rational. Again, I cos (n! nx)l 0

2.

@X-a Sin

Evaluate:

.

+8-V10-x2

then find the

25.1Efr) is differentiable at x = a, show that

27.Ift,

lim

(vr-x+1+ax-b) =0

Objective Questions

cotx) an

24 l

Exercises Evaluate the following:

-

Fill in the blanks.

y+2y+3yy243

3a-2x4a +3x3a =12;

.(3)

along the

we get x =y+1. As (x, y) tends to (1,0)

When y= x-1,

along

2+a+C-2c=0,

()-0)

y=X-1.

b=a+C

(2)

_1as

Find lim y0

6

(1)

21. lim (2-)

lim

constants a and b.

(given)

12

34-2b+3c

33. If

+ce

x-2x sin x -x°cos x

2a+-20 *20T=2

D

Thus, a =3, b=12, c9.

(1+)

Then, limit = lim

Limit, Indeterminate Form

c3a=3x3 =9.

a(2+x)e+ae 4

E.(L + 2h)-2log(1 + h)

0

n+1abe

finite, find a e R and the

n'cos(,

o0then

lim

State whether the

(b) 0

(a)

1 2

(d) none of these I0

(a)

tncOS )-n sin x

54. If lim

,Tn-

Y)

Iff(2) = 2,f(2) = 1 then lim

a+ bxsin X+CCOS-2 then

56. If lim

(b) 3p

58. If 59.

lim

67.

(40

16.

(d) none of these

20

If lim

9r

=1 +p

(a)

a8 ()8

0

=m

and g ()

Ifftr)= cot then lim

and lim

-

=

x2

cos+

@,0a

3 2(1+x

94+p

(b)

201+a

1

(e) none of these

(a) none of these

59.

lim h0

61.

16fcx)=Vsthen lim f) cos Xt

(a) 0

(c)1

62 lim

x-log(l* 2

is

X (b) 3 (d) none of these

() 0.

lim 0

is equal to

22. e

23. 1

a

27.-

2in-

3h(N3cosh sin h)

(213

is equal to

a1,

34,4

limit

-1

35. 100

43. log3

45.

46. 0

47.-1

50. e

51.

S4

e/3

53.e2

52.

respectively

56. 48,-24,-48

respectively

55. 4

57. 4

58.24

59.5

61. (c)

62.((a)

63. (b)

64.(b)

65. (b)

66. (b)

67. (d)

68. (d)

70.(b)

71.

69. (d)

1)

1

49. e

b=-1

33.a=1,b=

2.

18.

to

)

32.

Choose the correct optiols).

where p, 4 #0 then Im is

21.e

39.2log 2

44. 15. e2 19.e

a

41.

-

12.8

18.1

a cos

+a'cos

40. 0

11.

31. a=-2,

68.

38. 2asin a

3

1

17.

26. 9sin

If fo)=2,ft)=1,ste)=-1,&to)=2 then

)-glaf)

10. co

9.,32

(b)-2

v2-cos-sin eaual

9/4

X-1

.2 14. 8 2(log 3)2

Vf)-3

Gr)s-V25-then lim G-G)

lim

60.

(d)-3p

37.12

36. then

equal to

is

n/2 Cos x

hmVr-3 Vr-3 r9

when (x, y)-> (0,0) along the curve yi=r

Answers

(a) 5 57. If f9)=9,f19)=4 then

=0

40 3-1log3

F)-3f(2)+f(4)

m X

na

y+0

(d) none of these

(a) 2p

X-2

I2

x-0

cot a

then

fr) is twice differentiable and f"0)=p

lim

75. lim

log(1 +)1

x-

1

(d) none of these

statements are true or false.

Xcos a

6)a+b-0

a=b

(c)2a=b 65. If

55.

cos )+bSin x

(1+a 64. If lim

Cot

lim

74.

(a)1(b)-1()0

t dr"+ex9-24h

ra

is equal to

im

(e) none of these x+1

73. lim S0S

3. true

74. false

.

(d)

true

60. (a)

72.

true

4.

Continuity,

Function of Graph Differentiability and

of Facts and Formulae

Recap

.Left-hand derivative off(x) 1.

Existence of limit

.lim h0 is

fla +h) is the

Sa -0) or f(a

and it right-hand limit off(x) at x =a

denoted byfla + 0) orf(a

+) or lim

a+0

r=a if

and it is

fa

fla-his orlim flx). -0 denoted by fla-0) orf(a-) a-0 im

exists iffla im fx) the value of lim

+ 0) =f(a - 0)

f(r) is

ie,

atx =a iff

Clearly, f(a) = lim

cannot does not exist then f(x)

Continuous at x = a

be

i

-

as from the'right). from the left as well

some standard Continuity and. differentiability of functions a), Polynomial functions (i.e., continuous and differentiable at sin x, cos x and é"are 5.

Differentiability of a function at a point

denoted by of flx) at x =a, lim

-a

and examining the continuity point at a differentiability of a function f(x) and diferentiability you start with the *=a, if conclude differentiable then you can you find find that f(x) is also continuous. But if is function the that have to differentiable at =a, you will also not is flx) separately. check the continuity continuity and find that the Instead, if you start with then you can conclude that function is not continuous nondifferentiable. But if you find the function is also check the you will also have to flx) is continuous, differentiability separately.

of

+), is the 0

fa)-fa)

fx)

0) does not arise. In So the question of getting f(a continuous at x=a it this case f) will be right-end a+0)=fla). Similarly, if x =a is the will be Point of the domain of definition then f(*) Continuous atr=a provided fla -0) =f@).

a9orfa

derivative offlx) at

While

domain the left-end point of the =ais then If definition from the left and cannot tend to a

h)-

finite

x=a

x = a.

Right-uandderiva

0

B=fa. -h

continuous at fx) is not differentiable (finitely) at x = a. is not

thenf(x) is three are unequal t any two of the above

dlim.f()

h

(finitely) at

continuity and differentiability Relation between 4. x differentiable (finitely) at =a is f) x=a fx) is continuous at

fla+0)=fla -0) =f(a), f(a-h)=fla). ie, lim fla+h)=lim discontinuous at

lim Tat)=f@h

r >a

Continuity of a function at a point

Note

lim

=a,

value of equal to the common

continuous A functionf(x) is

h0

is called the and the common limit denoted byf "(a).

fla+0) andfta -0). 2.

Tla=h-f(a)

= finite, + 0) =f'la-0)

h0

i

and

-), is the lim

differentiable said to be A function f(r) is

f(x).

atx =a the left-hand limit off(x)

at x =a, denoted by

a"+ax"+.+

a) D-41

Problems Plus in IIT Mathematics

D-42

all points of the set R of real numbers.

is continuous and differentiable at all points of

log

,

+oc).

Continuity, Diferentiability and Graph of Function

all points of R -

10,

,continuous 3,...

and differentiable at

27,

is continuous everywhere and differentiable at all points except atr = a.

Ix-al

.Ixl

21

diocuso the existence of limit at r =1,-1.

Fromthe

=

or r2 Fromtheslgn-scheme, x S-1 Thus,the function is f(x) = x, xs-1

x

f) is continuous (or differentiable) and 4() is Ifdiscontinuous (or nondifferentiable) then f) + 4() and flr)x o(*) are discontinuous (or nondifferen tiable) at x =a.

Fig.

1

Fig. 2

f(x)=

X0. V16+ Vx -4 If possible, find the value of a so that the function may be continuous at x =0.

(1+h} =1

lim f(1 h) = lim

exists and it is equal to

lim()

fris

is continuous in an interval if it is continuous at each point of the interval.

i)

3. Let

VX

= lim

h0

f)

lim

a)

Ifa functionf) is differentiable at x = a, the graph of flx) will be such that there is only one tangent to the graph at the corresponding point. But if nondifferentiable at x=a, there will not be unique tangent at the corresponding point of the graph

fr)

Gi)

=

Whon discontinuouo alx

f(1+h)

f)=lim

202 (Whon contlnuous at x a)

¢(x) are continuous

r

f)

yX-1

x,*21

Algebraic property of continuity and differentiability and 4(x) are both continuous (or differentiable)

7. Continuity and

1.

'-10

fU)

y=f)=cos x-sin x, 0Sxs4

(1+gth) G(h), using given relation

sin xcosX

7

D-46

Problems Plus in

IT Mathematics 11.

*-o)-in

Continuity, Diferentiability and Graph of Function

The functionf) is defined as follows:

D-47

f)-ul+11-xl,-1sxs53

where ll is the greatest integer function. Det the points, if any, where fa) is not differe ntiable. Draw a graph of the function.

h

0

y=5 atr =3.

So, the graph is as below:

Hence, the required graph is as below:

3,5)

(1.2)

Here f(x) is the sum of two functions wh: re themselves piecewisely defined. In cases, at first break the domain of definition: subintervals keeping an eye on the turning o definition of the constituent functions. Here, the domain is the interval -1,3] which is equalto -1,0)u 0, 1)u [1, 2)u [2, 3) u 13).

lim h0

TIP

h

0

2 sin

lim h0

-h

l-N2

f() = r*-x*+x+1 and 8t)= max {f)0Stsxl, 0Srs1, 3-x, 1 0 in b) thenf'(x) is m.i. in (a, n

f)0

he

in (a, b) thenf(x)

.fis

sketch of monotonic functions are as follows:

y-txlis m. in f a.b]

a y

f(x) is m.d.

thatf is

Note A function may be invertible though it is nonmonotonic.

For example y = x,x is rational -x, x is irrational.

f(x) O

invertible is

a sufficient condition forf to be strictly m.i. (or m.d.) over the interval la, bl

is m.d. in (a, b)

yf(

Invertibility of a function

The inverse function is x = y, y is rational y is irrational.

-

in [ a,b]

D-83

Problems Plus in IIT Mathematics

D-84

are real such thatf(@) and f(P) are of opposite then ftr)=0 has a real root lying veen signs

6.

Use of Rolle's and Lagrange's theorems in equations

Iffx)=0 has two real roots a, a (repeated root) then fx)=0 willhave a root a.

functionfr) = 2sin x+ tan r-3r*

(-T/2, m/2).

-3

2cos x +

2cos-3cosr+1 cos x (cos-1x2cos-cosx-1)

t)(-)(+

So,fx)

1

>0for all x e (-z/2, a/2).

>0 for x e (- /2,0) and r e (0, 7/2)

f)=0

forr=0

Hence,fr) is increasing in (-/2, /2. 2. Find the interval of monotonicity of the function

fr)=2-1oglrl,

x*0.

Ifx0

and

(2r+1)(2x-1)>0

cos

cOS 1)*(2cos

"-o=1

d1+24x-3x2, x>0.

(-1/2,0)

0

x-sin x>0

axe,s0

f-" We

is monotonic increasing in fr) monotonic decreasing in (--1/2). If x>0,

or

f)>

So.,ftr) is differentiable at x = 0 also.

when-1/2 0

cos

(cos -1XN2c0s I +1Xcos x-1)

is a non-negative integer) As" (where n difterentiable everywhere,

f(0+0) [1 +2ax- 3

Butr 0 for all r>0, x * mi.

fr)is

*so

constant, Find the interval in where a is a positive which f' () is increasing.

(+)

fx)=x-

x>x-for all x >0.

sin x

flr)=1-cos

Letfr) =re

D-85

5. Prove that x> sin

Let

fa)

ae

is increasing in the interval (- T/2, Td2)

Here, f(x) =2sin x +tan x-3x. Clearly, the function flx) is defined at all points in

t

continuous and except perhaps at is differentiable everywhere

sign-scheme

from

R.

bel,) 4.

and B.

Selected Solved Examples 1. Prove that the

orcos -bs0,i.e, greatest value of cos x =1 b2 the

a

where a 2.

3

fO0,

ie.ft) is mi

f0,ie,f) 1

is

md

ismd

[leastf(), greatestf(7)]

Find the maximum and the minimum values of

-6-7x+10)-6-5-2

The sign-scheme for

=

the greatest and the least values of the

fo -1 rol215r-211 Hence find the range of fT,

unction

ru*

on is continuous and differentiable everywhere, Tis continuous and differentiable in 0 S rs2. Wenow that the maximum (or minimum) value or = the greatest (or least) among values at the critical points and the values at end pointsof definition.

='

,b-b+b-1,0sx ), xe (a, b) then prove thatf(r) > 0 ),xe (a, b).

37.

.

an

[a, b]

If flr) and

gtx) are differentiable functions 0SxS1 such that f(0) =2, g(0) =0, f(1) = 6, 8) then show that there exists c satisfying 00

1except (3)

Now, lim

=

0,i.e.,

a(x'+xi-I+2

stds

perhaps at the turning point of

-1+h-lim-1+)*+(1+b+1 h

.

andf-1) = ji so,f(r) is continuous at r y is an increasing function ofz.

+S g(x)dx increasesaasb-a

Again, lim

increases.

-f-) h

(-1

+h)+1-2

m 0

Let

f)=J

It+1|dt.

Examine the continuity and

diferentiability of f() in the interval

lim

-h1h+1

f-2sxg-1then

to-f-14-

polynomial

f-1-i)=lim-1-h3-(-1--

for all x, implies that g is

gla). dz

are

and

definition, i.e., x = -1.

h0

g(4-)> (3)

f-0

ax*+ax-ax + 2a

bothr

functions which are continuous and differentiable everywhere, flr) is continuous and differentiable in

-2.

-sb) _s(4-a)-gla)

aa.

3a-2b+c=0

f)=

+X+1,-1Sxs1

db

da

an increasing function.

(2)+(3) = Letf:0.-Rsuch that f0) 3 and

Differentiating (2) wrt.a,

As

f-1)=0

and

=s) +stb)

da

relative maximum/minumum).

and

dy

wrt.a. d

dydas(a)

-2SxS-1

fo)-

0.

As

+;

=0

+

Thus, the definition of f(r) is as follows

.(2)

so) +s) d4)=g(0)-gtb).

or

3ax*+2bx + c.

-8a+4b-2c

(1)

=4- 2a

(4-a)-a

dy

thenfr) From the question we havef(-2) =0

(3)

has

increases.

strndr

3+

function f(r) vanishes at r=-2

If>0 dr

a

for all x, prove

increases as (b-)

We have to prove that

Letfx)=ax*+bx*+cx

f

std+stde

=b-a=

26. A cubic

and let g(x) be

a+b=4 where a*as

intersects

y+)SOdt = 2.

l'uttingy= mr iny

J

as |*

o*

f(U)Jdt => **

and

inl is continuous and diferentiable

1f f(1) is

,

= 0

=

mx*

Provethat

xlog sin x dx =

m**+|

syA.

14.

ie.,

o(r)=2,

a +J

fNdt

+ Vcot x)Jdr

(Vtan x

Prove that

log sin x dr =

15.

Prove that

16.

Provethat

17.

Prove

=

18.

Prove that

19.

Prove that

20.

EvaluateJ

n/2

Ifx>

Evaluate tlhe following

log(tan x + cot x}dx = rlog

1 cos

2.

n

o

Odr +x)

=

sin -10n

0.

(1

, 1+ c .

2.

3.

r "a-x) "ur

P'rove that]2

Cos 0-sin 6

1+

sin

+

r"a-a) "dx.

cos d6

Evaluatef2sin COS

=

(2+

22. Prove

8E

24.

Prove that

y=|

dr=[ og0 +*).

X0,,is

sin Vtdt+J

cos

sin x

=

0.

Prove that

2x dr(2sin o Cos 'r+ sin "r

Evaluate the following:

Xsin a

1+cos

dr.

dx

r

4.

Prove that

5.

Prve that (

6.

fa f()+f(21-

Evaluate

log(l

tan a Nix.

t

1+cos a

25.

Nt

dt,

the equation of a straight line

39.

1+ cos

*x

sina,

J

Is

fr-chda.

sin

fxydx

r

cos "xdx = 0.

=J fdx

*

Jf2a- 1)dx.

function, prove that

f)it is odd.

fthdt, a 20 odd? Justify your answer.

1f-x)+fa)

= 0 in

and f)

the interval

periodic with period à, prove that

.11fft)=

periodic function of the period

-x|+ -2|.evaluate fmd.

If [xl is the greate

integer function then evaluate

42. If

S

Prove

thata]+ [-x]]dx =a -b,

nere (x] is the greatest integer S r

.

f(t)dt is

is a

fla + x) =f(x), prove that

fo= 43. Prove

xdr

Xdr

Show that

40. 16f0) is an even

41.

cos x-sin x|dr

26.(lx+ 1x-1|\dx

sin x

a cos *+b"sin *r

= 2b +

Evaluate the following

28.

xdr log(1+ cos xhds = -tlog2

x| dx

ftrdr=

Hence, show that 0

1+x

cos

Where

cos

where be Nand ae 38. Prove that

Prove that

|cos x |}dx

37. Show that

that

23.

+

1)xdx,

bos6 d0-a-Va*-b*).

Prove that

x|

cos x|dr

36.

cos

being an integer.

problem. will have real solutions. Hence, the

7.

2x dr

35.sinxdr

dx

Vtanx 12

parallel to the x-axis. Find its equation.

I'rove that

prove that

0

Exercises

1.

0,

log cos xdx=-log 2.

J

= 2,

ndt =2

lz]dx then define f(x) as polynomials in

x) =J

the interval [0, 3).

22

Ttlog 2.

thatj

+ that[Vxldr ="- Ddn 1)

sSr. where n e N and [r] is the greatest integer

2

32.

21. nm

-log

30. Prove

31.I

when |x| ->

o for all real values of r between -« and «, (r) atains values between 0 and « [Note If ¢ (a) = p, d (b) = q and d (x) is continuous then by evcry value between p and q will be attained from he (a) for x between a and b-this is clear graph of a continuous function] all x will also attain the value 2 for some real and (x)

ie. =

because

realm.

sndt we get m'r?.J fodt.

Consider the function o(1)

13,

discontinuous function cannot be continuous.So

-b

Iim

So, fta) is differentiable at x

xtan Provethat o tan X+ sec r

dx=m(T-2)

0

function. As the suma continuous, being a polynomial continuous functions is also continuous, o(r) is ale Continuous.

= lim

12.

Sndt=ft)

is continuousand

Asf)

(n-1»

thatf (x) =

andevaluate

-

four.

sin2x

is

sin 'r+ cos "x

sin 2x

sin'r+ cos'

a periodic

funetion

D-152

Problems

in l1T

Mathematcs

Prgertis

b

l,

nt

where n e N.

rdx=J ftrdr

that

Show

=J cos "r dx, n e Z, show that l,

Hence, evaluate

x-a

56. 1in

=l-

dt

sin

If ne

46.

sin"

2

dr zx =

is the middle term in the expansion

47. If

,

=}

tm

where t,

of4

s. 59.

tan "r da, n e N, show that 60.

48. If

-Hene, evaluatetan r

-*

l

-

(G

1)

-

in : +) 1"I10

62.

76 ForFor nEe Show that

63. 1

=

(1-x)".

prove that I,

g

Hence, evaluate

s

fy-

51.

f

Find

lag-z.

sin

yy

and

=0,

fo-)

=0,

12-

-

55.

f() at x

in!

0.

penodis

id)}-

=

)

-fia}

.

V

which

a, cOs kz

then find/r)

where 2-1>0.

sin

s an )fz)-žiz.

0. E fi)

kz

is

fnctior. F

henl:i seal

V)"dz

=

t

a

en e

even ctien

ot

where à is acorstart. is

(a)4

loglx\ds =

x,0sr a orr 0and the

28.

cu by Sketch the region enclosed the find

yla+**)=x'a2-r) and

byr--2,y

region. Find the area of this

ard bouni

its

the reg curves and identify 9. Sketch the 2 y= = logx and

area

the

Find

36.

bounded by the parabola

and the chord joining the points and (4,6).

=6+4x-x

cune

Also find its the xy area of the region in the Determine z5. a'y^=x(a*-x') enclosed by the curve 4y =

12.

15.

the

and y axes in the frs +y*s =a *, the quadrant. Find its area. area enclosed by the curve x=acos (Or Find the y= asin'0 in the first quadrant.) a 21. Find the area bounded by the curve y=sin 1 x-axis in the interval 0Sxs2T. boundel 22. Find the ratio of the areas of two regions 7-axis n te by the curve y = *"sin 2x and the interval 0 srsT. bounded by the cut 23. Determine the area y=x(X- 1), the y-axis and the line y=2 by the cu Sketch the region bounded

Sxs

and the xX-axis.

Find the area bounded by the normal at (1, 2) to the parabola y '=4x (y20), the curve itself and the axis of the parabola.

-6)

2

24.

144.

1

11.

.

Sketch

x

Find the area or the region bounded by the curves y= logX, y= Sin zx and the line x = 0. area of the region bounded by the curve 4. Find the - tan x, the tangent to itatx = q and the r-axis.

area

4r-a

=X*-x?

46. Find the area of the region in the Argand plane in which the point z will be located if

33.

35.

37.

V5s Iz| S2v3 and

Fill in the blanks.

38.

Find the area enclosed by the parabola y* =2x and two tangents to the parabola from the point (-2, 0).

39.

the Find the area bounded by the curve y x-axis and the two ordinates corresponding to the minima of the function.

=2x-x,

48.

The total

ordinate r

X-axis,

part of the curve y=1+and the ordinates

|X|

=

y=4r",x=0, y =1 and y=4 50. The area

is

the ordinate x = e

bounded by y=log,

and the x-axis is The area bounded by

51.

y=

X0 and the lines

yx(x-1)07-3) =0 is area bounded by the curve x lines |z] = 1 is-

52. The

= cos

y

and the

area of the region bounded by r* =y, y=X*+2 and the r-axis is

53. The

into two equal parts then find a.

Find the continuous function ftr) such that the a bounded by the x-axis, the curve y=f(7) and the ordinates x =1, x =a is equal to V1 +a* -v2 tor all

Choose the correct option(s). 54.

>1

ne

area bounded by the continuous curve y=Ju) and the linesy= 2, x =1 and = a, a> 1is equal to

x

l(20)3-3a+3 -212).

Find f)

if f) >2

for all

R, so that the area bounded by the curves

X-bx2 4. The (0,1) *na

the maximum.

r2is (0, 0), (2, 0), (2, 1) and s of a rectangle arehaving vertices at thne and y =

two parabolas

The area bounded by y= V4- x",y20 I-axis is (6) 27 (a) 4t

()T 55.

The area bounded and the x-axis is

(a)4

[1, a].

DE

area enclosed by the lines y= |7

The area in the region x>0, y>0 bounded by

49.

2

=2,x=4

1yl =1 is

and y =0 is

a divides the area bounded by

If the

=

The area bounded by the lines |x]+

47.

.

Curves.

z+z-227+8z +82>0.

Objective Questions

Find the area enclosed by the curves x*+y'= 4, + 8 =0 and a common tangent to the

x+y-6r

40.

-+1 and

Prove that the area bounded by y=

45.

eog

56.

and the

(d) none of these by the curve y= Sin x,

x E

[0, 27]

(6)0

(d) none of these c)2 The ratio in which the area bounded by the curves y=12r and x*=12y is divided by the line x = 3 is

a) 15:16 ()1:2

(b) 15:419

(d) none of these

Problems Plus in IIT

D-172

Mathematics

Answe7s 28. 1. 4

2aV1-e(ev1

3.

-

e2 + sin

area=a(T-21

'e)

s171.sin71 6

6.

8.-

43

10.

9. 12 12.

2

11.

log,2

areaog

13.(8t-V3)

V2

143 16.

29.

y

15. T+

V3 7+ 8n

17.3v3-)

V3

1.

19.121:4

20.

l(o.a)

34.log2 32

ra

(0-a)

35.

37.3/2+ sin

36. 36

39.2 120

38 22.

24

3

32. 1og2

area

21. 4t

logeX

30. 2

(a.0

(-a,0)

=

457-4

40. 22

41.1+X)

22x, x21 44. 32y9x, 4x*=

43. t1

42.

45.

4y 3(x-1)

area:

0-6)

d

9y

2-log3 +tan3 12sin

x»l

4v3

+5sinE+

64

49.

47. 2

48.

51. 2

52.2sin 153

55. (a)

56. (b)

1

50.1 54.(b)

26.5

27.a(2-

13.DifferentialEquation of the First Order

Recap ofFacts and Formulae .Differential equation, its degree and order An

equation involving derivatives (i.e.,

,

The differential equation of 2% dx

differential equation. etc.) in x and y is a the highest order derivative in the The order of equation is the order of the equation. ,Thedegree of the highest order derivative (when put degree of the equation. in rational form) is the For example

dy

is a differential equation of the first

y+=2

order and the first degree (linear).

=0

G

is a differential equation of the

dx2 second order and the first degree.

mv-(

being order and the second degree, it =x in therationalised form.

first 1

=x3 is a differential equation of the

ax

differential equation is the We know x2+y2=a2 is a circle whose centre will it parameter, Ongin anda is the radius. If a is a epresent a family of concentric circles with the 2 Formation of

common

centre (0, 0).

Dilferentiating

2+y?=a dy 2x+2y=0 dx

equation of the second order, obtained by eliminating the parameters by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameters. differential equation the The general solution of a differential equation is integrating relation in the variables x, y obtained by (removing derivatives) where the relation contains as many arbitrary constants as the order of the equation. the The general solution of a differential equation of first order contains one arbitrary constant while that constants. of the second order contains two arbitrary of equation differential So, the general solution of a curves. the first order gives one-parameter family of of the In the general solution, if particular values solution arbitrary constants are put, we get a particular which will give one member of the family of curves. 3. Solution of

differential equation of the first order and the first degree Simple standard forms of differential equation of the first order and the first degree are as follows. 4. To solve

=

Method Make a suitable substitution so that the

equation becomes variable separable in the new variables.

sis and constant).

Forexample

curves of one h tferential equation of a family ofthe first order, is a differential equation. of eliminating

1) of two parameters is a differential

.Reducible into variable separable.

a differential equation for all the members of the it does not contain any parameter (arbitrary

ned by differentiation.

family of curves (like

.Variable separable. Form: flr)dx + oy)dy 0 Method Integrate it, i.e., find J ftx]dx +] d(yldy =c

1e,xdx + ydy = 0 (in differential form).

eter

=

a

the

parameter

Take the.equation (r

-y)2=1

This is not in the form of variable separable because we cannot write it in the formf(x)dx + ¢(y)dy = 0.

by

D-175

Problems Plus in lT

D-176

Dferential Equation of the First Orde

Mathenatics

xdy ydr

T'utr-y

1

1-

1

(r-y)

ax

To

Homogeneous equation. Form: d variablesX,

=

vx and solve he equation

the new

in

U.

Nonhonmogeneous equations. OV *

dy Fom

Solvable forp.

..

*

*.

*2 put r=X

If 2

a,h+b,k +C

0,

+

h, y= Y+k such that

+ c2 =0. In this way, the homogeneous in X, Y. Then use the =v

ar

or

+

by = v. The

variable separable equation changes in the form of

in

Solvable for x.

*

Q)

called by e", Method Multiply the equation becomes integratin8 factor. Then the equation

P

Q).

integrating, y

PirMix

e'rN = } Q(x) e"aM

dx

pc

Method Put S(y)

y = z;

or fp)

+

x

cx +f©) which

y

dx

length of the cartesian subnormal

=

AD =y =

1.

dtan dtan

yus y

tan)u

in the equation. The resulting

equation

y-x

is

the equation of the orthogonal

trajectories.

Find the differential equation of the family of curves arbltrary A,B

equation has two arbitrary constants. So its diferential equation will be of the second order. Here, the

We

.

have y me"(Acos x+ Bsin x)

Differentiating () wrt.

(1)

X,

+

orJedy=Je"dx;

.ca. 3.

Determine the equation of the curve passing through the origin, in the form y=f®), which satisfies the

sin(10x

differential equation

z; then

10

+6

10

Differentiating (2) w.t.t. X,

+e(-Asin

x + Bcos

the +e(-Acos

-y,using

or

This is

2.

'

X

tangent at P(x, y) he equation of the

y-f) is Y-y-(X-).

to to the

curv

6y).

+e'(-Asin x+ Bcos x)

or

(2) and (1)

X

- Bsin X

or

10

equation becomes

6sin z + 10;

6sinz+ 10

or3sinz+5"d

1

O

+

x +Bsin x)+ e"(-Asinx + Bcos x)

- (x)

4y =ydx xdy

place of in differential

are Acosx+ Bsin ) where ye constants. Also write its order and degree.

3

the differential equation.

1+tan"

xdy-

-

y.--0

Method To find the orthogonal trajectories of a lamily of curves whose differential equation is known, put

normais Some results on tangents and

which is in the linear form. Form 2flotx, y)hdo=0. Exact differential equations. cquation in this Method In order to put a differential following ditferentials. form, one has to remember the ydx d(ry) » xdy + ydx dlog dlog

=y

Selected Solved Examples

a

singular This solution is called the

Qr):

P(rz

The equation becomes

1s

+X =0 all the u andf'P) any arbitrary constant) touching (without = y cr +f(C) given by solution

Sy) d5(y) dy dy

0

curveswhose differential equation isflx.

P'ut 10x +6y

=0.

which we get a solution

dx

1

CP =y

or the cartesian subtangentt=CA

e(Acos

solution is When p= the general lines, gives a family of straight = p from y=pxI When f'p) +x 0, eliminating cu

R(y) hen

wrt

c,

P) Sty)-Q)

such that

orthogonal trajectories of a family of curves form another family of curves such that each cunve ot one tamily cuts all the curves of the other family at nge angles. differential cquation of the orthogonal The is the family of trajectories of the curvesf|x, y.

.The initial ordinate of the tangent -OB

x+fP)-0.

Reducible into linear equation. Form R(y)

Form x =f(y.P)

y which gives Method Differentiate x =f(y. P) in p and y. Solve first order and first degree equation p from x =f(y, P) and it to get vy. p, c) = 0. Eliminate VyP, C)=0. +f P) Clairaut'sequation. Form y=px x, we get Method Differentiating w.r.t.

X, v.

y= Linear cquations. Form+P(X)

=

7. Orthogonal trajectory

The

pP)

a Method Differentiatey=f%.p) W.rt. which gives in p and x. Solve cquation degree first and order first 0. it to get é(,p, c) = ©)=0. Eliminate p from y =f(X, P) and ¢(z, P.

azlh + bak

put a,r+ b,y

If

the curve

length ot the normal= PD

.The length . The

to

Tu,y)

the solution.

Solvable for y. Form y=f(x,

equation becomes method tor homogeneous equations.

R)

. The

x

p-Snlx. y) =0 Form p-fX, yNip -fT, order and first deg Method Solve the rst etc. If d,x, y ,c)=0, ete equations p -fixy) =0, same arbitrary constant c for the (taking solutions y, C) 'nz, y. ¢) =0 i each) then 4(X, y, e) **, y))

at

x).

ot the tangent The length

*

dx a+by

Method

dx

equation ot the first order bus solve differential

equao Is -yd(X-

-f()

v

equation of the f Standard forms of dilterentual (here wed follows are as and higher degree

separable. which is in the form of variable

Method 1lut y

The

"(ay)=-

higher degree

dz

=

dx

=z

-l

dsin

az

then 1-

2;

Solve

log,

Here=e at

see

d

ar+by. +

by.

edy=e

5

ax

5tan+6tan5+5

sin2

.

D-178

or

5t2+6t+5

Mathematics Problems Plus in IIT

x+C

puting

tan

ifferential Equatiom the of First Order

-

tfc)-y= z; thenf'()-d f'lt)

becomes,f'(r) -

(1)

esin

=X*

d(1-2f

v do

"sin

=

=0

v dv = c

I=Je "sin

-Ir

do

vdv = 0

"sin

or

Now,

or

v+

=

sin

2) = cC fo)+log(1 +y-f(a)) f)+ logil

=-sin

-

5.

tan

or

It is a

5t+3

4 or

an

5tan

Solve xdy-ydx

5+3=4tan 4(r

5tan(5x

+

3y)

+ 3

Je"cos

v dv

+c)

.. (1) =

y

UX.

=v+x. dx Vr+y

(sin v

I=e

"(sin v+cos

or

or

=x v1+v

x

Putx

or

cos v) =

c

+

Vo+

1)= log x+ log

y

e

+y*=

0; or

v+ Vo*+1 = cX

y+

log I+log(v* + 2o +2)

or

los+2yx+ 2)-an

8. Solve

-

tan

(o+

1)

=c

1

=.

(y-3+3)y--4

42y-X-4 iet dx y-3x+3 Putx = X+h, y=Y+k; then dx = dx, dy = dY the

equation becomes

a_20*-X+h)-4

dY.

or

Y+k-3(X

+ h)+3

X+(2*--4) dx Y-3X+(k-3h+3)

Select h, k such that 2-h-4=0,k-3h +3 0.

Vx* + y

=

Solve

xe-

to solve the equation,

xe"-

.

or

ysin

equation becomes,

y°+I+y°=0

ysind+rsinž dy=0

nenU+1So, the equation

fl)

v+x

cx,

puty Ths is a homogeneous equation. So,

where fl) is a given function ofx.

Y+X+Y=0

+ xsin

xe"-vxsin v + xsin

on

=0

v-+

becomes

substitution of r

DEcomes dY

=

X+2,y=Y+3,

2Y-X

dxY-3X hich

which is a homogenous equation

*he

25tan 4x 4x)

tan4-3tan 4x")

Here-f)-y)f'lu)

or

dX

Then dx = dX, 2ydy =

put y'= vx; then 2y

5(4-3tan4r)55(4-3tan 5tan 4r

f)=fr)

log

lere

ay. dY Using this in the given equation,

So, in order

tan 4x

4(3+4tan 4x)3

4do x5Í (+ 1)2+1 2+20+2 -fd=o

or

Solving these, h = 2,k=3.

54tan 47+tan

1-

Solve

**2 * (sin +

***

3y)=

0

2+2+2

v)

xc+s sincos

X, y*=Y.

+2

2+2)-2 du=

v)

*2 dx

tan 4r

4.

od

20

r

or

solve it.

1x

loglv

y

+

n

that the differential equation y 'dy + (* + y)dr = 0 can be reduced to a homogeneous equation. Hence,

5tan(5x+3y) +3 - 4tan |4r + tan" tant5r

v.

7. Show

Putting in (),

or

log

log

Putting in the equation from (1), we get

v1+

+

cos

21-e

trom(),

-y=

4tan 4c

tan-4c

4c;

vx* +y* dr.

x

v + cos

=-* "(sin v+ cos v)-I

equation. I'ut

We have the equation

- 4tan 4(x+c)

t passes through (0, 0). So, 5tan 0+3

tan tan

homogeneous

Then Then

4(7 +c)

=

=C.

v do

v-Jcos

or-og(1-2)=f)-c 5

D179

e"+x'sin v 0

is

PutY- VX; thenxV+XK

2X-X . V+X xVX-3XK V+X or

2V-1 orXXV-3

v-+3V-1 -5V-4V

Integrating).5V-dV

Xlv+*+x+dx

+x *1++

pX =0

1+0 2 dr

0

the equation

homgeneous.

= log

X+ log

-2V+5)av =log excx 0F-2-+5V-14log V or og -V 5V-1)*zv-5/2(21/2 =

log cX

Probems Plus

in iIT

that Select h, ksuch

+21 or

oF-io

X

. =

or

5-2

=2

log ex

+dg. 9. Solve (r +2yNdr-dy)=dr

Putx-2y=;

the

=0

-

14 + 24

nes 3u + 2

+ =

c(r

Solve

11.

or

=0

2-1+2y+1

or

Integrating factor=

euation becomes

y=-

()

uls

or

becomes-5*r=r

in the linear form. So,

integrating factor =

e"

Multiplying by it

+log =log u- log

c

or

-lo lo 4 u-v

or

X-2+=1 lo x-2-(-1) -log2+Y-1XX-2-Y-D

2ez=-2

sin

y-Isin Solve (1+y

2

Here.

(1y

dr= (tany-1dy. tan "y-

a--2. or

(putting*=9

J:*#

=log(X-2)+kg

orlogy31. (r-2i-8)y

-2 log(z*-2)+hg

or

-( =

I+y

s

is in

1+3

the linear form.

Cleari

grating factor= e

1

yi=1

Maxltiplying

žy=

dz, dt

-yi-3a*g

dy.

Put= X. =Y; thern 2zdr= X,2y theeuaion becomes (3X -2Y-3dr 2X 3Y-7K =

thea l= PutX=-k,Y=t-c

4

then

-2=-2x.

or

log u + log

Itis

-2or6-1) Here (23y-7)ziz

=;

Put

(X-2)

=

Hereyx'y

cos

or

3-1d

is in the linear form.

e

rr log

+c= (tan"y-1)

factor, Nfultiplying by the integrating

oF3osl--og

then 1 +2

=

r'y-ry.

13. Solve

**" du

=

xean

.y-costhis

3+2

11-- :

where tan "y

=- cos

Here,

where X=ut2,Y=+1

3u+2uz 3+2

" du

1+y

eds,

=y-cos

1.

-2)

- 4log c = log c°=log k (say)

--jeid=-1

**)

= 1.

becone du

Integrating we get

.2)

3h+2k-8 =0

x(5)., -5h

and so k

c

-3=k{x* -yi-1)3.

2k-8)

2h+ 3k-7

Put r= u; then

d

(7 -2¥* 1M

(1)

+

+

2-y2-1)

or

+ 3k- 72

-1/2

2NI2-3)2-3)-(5-121 XF -2))

=

From2 x (2)-3

o2Y-5+ 121X

2N-3)-(1=--3)-

(a-2y-1ir

*k)-8 3(4 h)+2("

(i 2+3+ 3u+2+ (3/h

log ex

er

2(u+)* 5("+ )-7

ddu

(-v5V- 1).5-

logty"-)-žbst-y-1) log

becomes,

(1)

oF-o

Diferential Equation of the irst Order

NMatkemaiths

bogy-9-ost-1) log(-y-1)+

by it

-1

Here-2-1-2

logta*-

loglz*-2)kg

=

+y

z

roblems Plus in irT

Ditferential Equation of the First Order

Mathemath

D182

dy dr sin

2

or

the

xdy

dy

-2lincarfom log

integrating factor =

rcos Bd0) or

,

-yax = TcOs

the

=r'd6 equation becomes

-

y Now, y

rdr

+

rdr + de

0;

or

yax

ydx = xy °(1 +log x)dx

=xy(1+ log )4x

xy(1+ "(1+ log xjdx -d

We know, d(x

"

+

xuy

ydy +-

=0.

+y

or

+y) =2(xdx + ydy)

theequation becomes

dxy+

dtan

0

+

log

+P)-y

or

y=

curve passing throug Find the equation of the

whose differential equation is yr+y ydr=zy-x)dy.

xtan -

P(x)

42 = QP)

v(X)

+p(x) v(x) = g(7) +p(xMu(x)- v(x)) =f)-g{r)

(1) dx .

- otr)-e'PMJ= {fx)-s(»)) e'Pa

orlu(x)

2

(it

. 8)

or

=0 y(ydx -xdy) + x(ydx + zdy)

or

xdxy)=0

)+

dx

P(x)

y-y)=0

.

Let

(4)

(2)-(3)

-x" **

+r(-sin 8jd0 and

in

linear torm)

d

-2)+

From (4)

and (5),

P)

(y1-y»)=0

.

,

(u(t)-

(5)

are positive. v{T)}%

-

: u(x)-v(X)}A

y-

{u(z;)-vtz,)la

> 0

f)>gla) for * > > {u(71)

x,

and eJFaM*

> 0}

- v{T)})u> 0 because u(X,)> Ar,)

u(r)-v(r) -1

(1)

(eod,eo

Clearly

:

tan 0

was

F)-s(r) e""dx .

P--y)-0

Here, (ry+y ydr (iy3-x dy

ie,-x+yi, dr=

=Q)

..

+p(x) u(x) =fr)

=

= rcos 8 and y = rsin 6 Second Method Put x

drcos 6

1y2

and

solutions of the differential equation

Pr)y-Q)

g -3

17.

.

Here,

From (1)- (2),

Integrating.a+y)+ tan"

=c

then prove that y=y4 +Ch -y)

general solution

y, are two

rom

+y+ 2tan

respectively

ux)-v(x)Me'PRM

logš-*

(exact equation)

satisfy the differential equations

and glr) are continuous functions. If where p)f() u(z)> v{z) for some x, and f) > gt*) for all x>ru prove that any point (, Y), where x>*y does ot satisfy the equations y=u(7) and y =v(r),

of the equation where c is any constant. For what relation between the constants Bya also be a a,P will the linear combination ay+ solution? is the

xd

a + B =1.

d+PM=f andp)p =g)

are two solutions of the differential equation

+Pry=Q)

and

xdy-ydx xdy-yax +y

dtan

-14-/r*a

y +2x-5x y= 0.

curve is

If yu ya

As y

Q), using (2) and (3)

19. Let u(r) and v()

he

log r)dx

15. Solve xdx

if

Qr) = Q).

+B)

Hence

c

tan"

2)} dx=0. Solve xdy-ly +xy°(1+ log -

o

(ay* Py») =Qr)

aQt)+p-Q)-

(a

2+2+2 2=0

2+0=c

18.

-2log y-

or

y+2x +2cx 'y = 0

= 0

(1, 2). So tpasses through

Here, xdy

y+Cy -a) ay+ By2 will bea solution

ay+By)+P)-

y= an 16.

*

or

Jrdr+J de =G

y

y)

=log c(y1 -y)

log-V)

rsin B(cos bdr- rsin0 de)

e

ze*y) or

+

integrating, log (y-yi)= log(Y1 or

b{Sin dar + rcos 6d0) -

equation becomes

Multiplying by

or

rdr

then 2xd

d

rsin B(sin bdr

dividing by x'y

0

rcos b(cOs Bdr- rsin Bde) t

yi--

Put-z;

0 +

xdx + ydy=

2y d-x--(d+2)

or

rcos Bde

u(r)

* D(X)

> 0

for x >xi

when

x > X1.

Hience the problem.

Mathematics Pnvlems Plus in IIT

2py-c= x

20. Solve

+(y

dr

1)+ yp

pp-

2v

1) 0

-

1Mxp + y)=0

(

=

4xp*- 3p*x

=

xp*

y Au= -2x* +X

x(y- 2xp)

=

y = 2xp

+p*

.the

methods

equation

to nna

+ 4y(y+)= 4x(Ty + ©XY+*)+ (ry c).

***"

Put

dy

xdy + ydx

is (y

-

X -

c\ay

-

d

dy

Where p

Here, xp=l+p"

C

general solution

dy.

2. Solve 1

=0

=0

d(ry)

xy .the

4y+x*=0.

a tamily of straight lines

But the

singular ssolution 4y tr*=0. vith a of

tne

solving first order differential general solution of

2y+22y+2

Now, p-1 =0

the temperature of the substance at time r anu :beT.The rate of cooling of the substance kT-30)C/minute (from the question).

Let

solutior tions are y=cr +C

+ 2x

Putting this in

or P*y*.

PIO

Dyferential Equation the o First Order

p)-3p * +2p)

P(2y+2x )=xy +c Xy* p =

yp - y=0

-ap+

or

=0.

-xp - y =0where p dx

Here ap(y or

x)y

-

+

29|2xp

= P p. Then

ddx

dx

equation is

the

-

3 dp

=

rate of cooling rate of decrease of temperature-d

from the question,-=k(T

Hog(T-30))=kll

logk15;

:which is of the formr =f0.

y=2" dr *dx

Solve

Here, y 2xp +p* where p

p 2p+

dr

dp

=

solvable

or

dy

0

Here, y= xp

px)= or

lo%P

byp',

+2px= dp

aax,

=

+

++ b)

Thus, we have y

and 3p 'r+ 2pC

=

2xp + p*

+2p=0

Odx

0

The ra

?A3*

y=xp +p*gives y = cx

0

+c

p=-

nce

ha

15 cooling of a substance in moving air to the difference of temperatures or tue and the air. A substance cools irom minutes. Find when the substance wi"

the it being known that of air is 30°C

emperature 32°C,

constant temperature

Solving

V (from

(1),-k-dt;

=e**i t=

0, u = 2V

(1)

... (2)

kp

Att= 0, u = 2V and v=

When

.

-ku

=

r

onal

x+2p

and B are two separate reservoirs of water. The capacity of A is double that of B. Both the reservoirs are filled completely with water. Water is released simultaneously from both the reservoirs. For each of the reservoirs, the rate of flow out at any instant is proportional to the quantity of water left in the reservoit. After one hour, the quantity of water in A is 1.5 times the quantity of water in B. After how many A hours from the time of release of water, do both and B have the same quantity of water? B u Let at timet hours, the volume of water in A and be and v respectively. From the question,

and

d(p r)= -2p dp

ior

2

Y-2ax

0;

2p)

minutes.

t

-20

2p

Pp

p'r-2pdp--

dx

y=

this is of the form y = p +fP). Differentiating w.rt. x,

(x+

+b

P=V-2arh

+ b)=p

p p° where

p?.

-2p

2

4x

2p

20r

log3

26. A

integrating

y=*

Hog(T-30),,lo

e °%(Op= ap*

from ()

.(1)

"=e

log a =

=3logP*

eliminating p the p-eliminant, obtained by and (2), is the general solution.

The integrating factor

(1)

a being an arbitrary constant

y+c=5-logP

23. Solve

integrating, we get

dp= 3log p + log a.

p1-d

yse--

dxx-2 dpp 2log

dp

dlos log

for y)

2x+2p=0

Multiplying

wheret minutes is the required timne.

or

Differentiating (1) w.r.t. y.

or

nn"E+p

,

isolvable for x)

dx

4*P dx *P Or

p)

=

It is in the form y =f, P) Differentiating w.r.t. X,

klog

dx

c)= 0.

dy

2x

30)

T-30dt

.(1)

21.

-

the question).

logu-k,t +c

. (3) 2V=e

Trovlems 1lus in

)gives,

u

=

Simiiariy, from (2) we get From the question,

if p

r-

..

Ve m I

then

u =

z

u

(p

4

Using (o),2-G1.ie

Tos-log 2 ie.

and

be

los2

T-

same will have the

The

quation

curve

(*

+

- 1). So it

Now, the line is 2x -y-4

y

cx

t2V-.

(1,

1

1

is a

Hence it

=0,

or

ie,+1.

dy+(-yMr-0.

Puty vx; thenr**

this in td

.

()

or

y-

ie,

But in the first

quadrant x, y are positive.

ay1.

.

Solving y 2= 4(r+y)-8

of the equation and

*y* (or lines) are tne requíred curves

ay Y = 0

and

X

46. A

(X- x), i.e., X=- " OA)

Curve passing

point(

through

y) equal to

r(OB)

(1, 2)

Find

has

Here, siope* dx

y-2

its Sio

the area

the line Dounded by the curve and dy

-

and 2r-y-4=0, we get

e ta

o

2r-y-4

or

24)4y-8 -6y=0, ie. y=0,b Then y = 6, * =

dx

ie, Y=y

*

l.

=0 successively, we get

2

1-0

12

1y- 1,-3.

solution This is the singular (l, a curve and it passes through

-n,

2xy

It is a homogeneous equation.

pomt (r, y) of t..e of the tangent at the

Solving (1) with

a

place of Puttingin dy or

p-.Putting

=0

0.

2r+

75S,6)

So r+y quadrant x, y are positive.

*

r=t

=0

4r y This is the differential equation of the circles The equation of orthogonal trajectories is

-2 0.

lne is y = -r t2,ie, r *y=t2

But in the tirst

y

passes through A(2,0) and (0,-4).

=

the

d

+y-ay

parabola.

vertex C is (1,2) and the axis is y

Its

dy

2x+2

y-4y +4= 4r-4

y)-8

12 =9.

af().

T-y(X-r)

and Y-y

=

(-2)=4(*

or

-p

y*

-

0, differentiating w.rt. x,

-8

the curve does not cut the y-axis.

= 21

Find the orthogonal trajectories of the circles x"+y°-ay=0where a is a parameter.

29.

Here,y2-ay=

its roots are imaginary.

ct 2V-¢, ie, (l-c)*-4e. cm-1 (1+c) -0

then

NOw,

is

y

y-4y+8=0;

or

in (2, If it passes through This gives a tamily of lines.

ditferential possible curves

0)-8, i.e., x=2

When x= 0,y'= 4(0 +y)

i,

Pc

d putting

0 = 4(T *

*4y

urve dy = f

(1), the required area

from

A(2,0) only. the curve cuts the x-axis at

or

O

,

When y =

Again,

=0

ar(BCAODB)-S

2+c,ie., c=-4

find the area, we have to draw a rough sketch of the

To

y-x=t2V-p

-4

thecquation of the curve is-2y 2-4 y-4y= 4x - 8, ie, y '=4(x + y) -8.

curve

-P

of water

y

=p

taking

p

2

kSheurs the nservoirs

curve be

apP),

-

Pp*dt2

(, 1) such that the A cure (or line) passes through coondinate aves and the triangle tormed by the curve is in the first tangent at any point of the area equal to 2. Form the quadrant and has its ot the equation and find the equations

Let the

y)g

or y=ap+2N-p Differentiating w.rt.

q =1

2e

i.e.,

A

B

-

=- (y-ap);

(6)

27,

So passes through (1,2).

2

or

of water in Let, atter 7 hours, the volume yual. 2V¢*= Ve ** » (5) )and

241-21

-2y 2c.

S-2dy2J dr;

Vo

=c-i **

z

=*,

-2)ady = 2dx

.

OB

or

Dividing these,32

qantity

5OA

%= Ve

and (5)

atter

= 2 =

dy

t, whent

=

the first quadrant

the axes in

(5)

D17

Diferential Equation of the First Order

NMathematics

by the tangent a rea of the triangle formed nd

.. (9

2Ve *

In

log*log(l+*)-bg al+r)-c n5. So B= (5,

or

learly, the required area

ar{OABDO)-ar(BCAODB)

arDABDO)

-

d=0: antegrating we get

dy-

(1)

1-c.

ie,a+y*-a

orm the ame Note It the orthogonat trajectories amy ot Curves tamily ot curves as the given curves is caled o then the gIven system seforthago.

IIT ProbiemshPlus in

42.

Exercises to e

equation corresponding 1. Find the differential arbitrary -c), where c is an tamily of curves y = dr constant. conic sections difterential equation of al 2 Obtain the a, b are parameters. where ax by*1, family of circles differential equation of the 3. Find the origin. touching the I-axis at the 4. 5.

Solve

xdy +y*dx= zy{zdy yáx).

Solve1+r*tan "x +y=0. sec

6. Solve sec'r. tan ydx 7.

Solve

8.

Solve

9.

10.

Solve

Solve

-

13. Solve

yMy =

+

26.

cos Solve the equaion

+ dx

27. Solve

sin

yCos(x dx

16.

.

+|1-je

it. homogeneous form. Hence, solve = 0. Solve (4r-y+3)dy + (2r -3y-1dx

in the

21. Solve

dy 3r-4y +2

22. Solve a7 23. Solve

1dx.

(x-1)dy

- y)dx. =

36.

y

-.

(1+y)+(r-e-an

+

ydy +

(r*+y)dy =0.

1x

sin =cos x sinx-

:

56.

ve where v is a function of

equation when y

= 1

and=0, dr

= 0.

smallest integer bigger than or equal to

log 10- Iog9 log (1.04) -0.03 47. Show that the equation of the curve passing throug (,0) and satisfying the differential equation *o*

(1+y jdx- xy dy = 0 will be x"-y'=1. whose slope at TOVe that the equation of a curve (,y) is

ify=1 when x =2

24y_ +2 d =**

and

58.

60.

t

curves for which the cartesian equation of the constant lengn. ength of the tangent is of portion ot the tangen Find the curves for which the between the axes o

xy

aany

p=

point intercepted

rectangular hyperbola. Find the equation of curves for which the cartesian subtangent varies as the abscissa. Find the equation of curves for which the cartesian subnormal is constant Find the orthogonal trajectory of the family of parabolas y°s 4ax where

a is a

parameter

Prove that the family of curves A

Ab+a

isa parameter, is self orthogonal.

objective Questions

Fill

in the blanks.

62. For

the differential equation

and

degree 3.

65.

order

Vis,

the

=

equation is general solution of a differential then (y+c)° cx, where c is an arbitrary constant the order of the differential equation is it the

:64. The general

(

solution of

The particular solution where y(0) = 1, 15 =

of

lE

66.

Let

67.

If y() is a solution of (1+)

>0.

=0is_

+

log y)dy

sin -dx

=

+

ydx = 0

F(t)- F(1)

then one of the possible values of kis-

equal to the radius vector

*nd

Find the curves for which the distances of the origin from the tangent and the normal at any point (¥, y) e equal. ne normal at any pont P(z, y) of a curve meets the be X-axis at G. It the distance of G from the origin a twice the abscissa of P prove that the curve is

where

which passes through the point

(2,1), is r3+21y =8. on the square of the intercept cut by any tangent y-axis is equal to the product of the coordinates he suchn of the point of contact then find the equation of curves. normal is the curves for which the length of the Find

*".

-+e)+1-0.

y=2px + yp°where

the point ot contact.

-2+y=xe",

the average food requirement per person remains constant, prove that the country will become self-sufficient in food after n years where n is the

differential equation

solution of the equaton Find the particular

41. Solve

55.

4%61.

40. Solve

=0.

.

the59.

y(1)=5

ylr+y*+1)

y

A radioactive Substance decays with time such that at any moment the rate or decay of volume is proportional to the volume at that time. The half-life of the substance is the time it takes for half substance to disappear. Calculate the half-life of the substance if 20%% ot it disappears in 15 years. Its (b) A country has a food deficit of 10%. S7% per population grows contunuously at a rate ot year. Its annual food production every year is more than that of the previous year. Assuming that

dx.

Solve

39. Solve

order equations,

du

x+20.

Reduce the differential equation

forr

+xy)dy =0. + 1)dx.

38. Solve

Using methods ot sOlving first

46. (a)

x(t -1)

- x*)+x log x.

y-=r+ Vi-s

reference, is of constant length. Find the equation of the curve for which the length or the normal is constant (a) and the curve passes through the point (1, 0) 54. Find the family of curves which are such that the area of the rectangle constructed on the abscissa of any point and the initial ordinate of the tangent at that point 1s a constant. Find the curves for which the portion of the tangent included between the coordinate axes is bisected at 53.

Obtain all the solutions of (y+ 1)+2=z

x. Hence, solve the

cos (T-y)-sin x sin 2

y(zy°

+3

x(1+x)y(1 (1-x

+ y) +

xdy =

37. Solve

=(2y +X19. Solve (2r+4y+ 1dy 20. Solve

45.

ax

d y-3y+2y

4x5y

44.

nas a solution which is a

solve the differential equation

dy=0

ž+ tan

dx

cos'

ycos(xdy-ydx)+ Xsin (xdy + ydr) =0

put the equation By making suitable substifufions,

18. 5ove

43.

dy

34.

ay2x*2ry-4 17.

2r)

1t

= using the substitution y

Solve (1+*°4y y(T

when 15. Solve

Td-(tan

3x where y(0)=4

vx +4

33. Solve xdx

-3y

"M) dr 14. Solve (1+e

+2* y -y-5r dr =0.

Solve

35. Solve the

dr

straight lhnes ana parabola.

ytr).

and y

where Ix

32. Solve

z

x(x

Solve x(1 -x"dy

+ 31. Solve ydx x(1

vr+y+1 dx =r+ y-1. xdy= Vr^-y dx. +

25.

+ 30. Solve ydx

dy

Solve y'dx

Ify+ay)= x{sinx* log X), find

29.

(r+y*+ 2ry)a I+y= sin

24.

tan xdy =0.

+y)+1 =0.

+rtan(r

11. Solve ydr 12.

"y

Prove that tne geriera 50uton of the differential form equation y= (r+ of

1)z- dx

.

129

Diferential Equation of the First Orde

Mathematics

D-188

then y1)-

y=1 and y(0) =-1

Problems Plus in l1T Mathena

D-190

AuswersS

dy

(dy

dy

2.y dx2

28. cos

29. 3.

(-y)2xy

5.

X= tan

7. log

4.log

C

6.

sin(r+y)+r=c

9. tan(x

ay

tan r tan y

8.r+y=

atan

2vr +y+1+log(v

+

y+1 -1)

13.

ye

sin=c =

y+1+2)

12. kxy= x+2y

kr2

14. x + yeY = c

15. cx = sin =

X+1,y2= Y+1;

6Vy*-1+vr-1 tan 23 V-1

23

log 12x+3y3-5+Vr-1-1))=c 17. (x +y+2)*=cy-2x-1)2

33. 2y +

34. sin

19. 20.

y 22.

+ceN1

1-

y=c+ tan"y

32.3x4

(2x2 +2y2-1) 2tan

=z

(y-e+ c\y+e+c)

25.

26. ycos 2x 27.

= sin

X

cos°x

yr+ Nx2+4) =r*+ (x2+4)32

+c

38.

(2y-x-c)2y +3x-

39.

(xy-c(x-y2-c) =0 eliminant of xp

x

= 0

c) = 0

= ce P

and y=x(p +p^)

y?= 2cx +c 43. (y +1)c+2 =cx, (y + 1)2 + 8x=0

41.

44. xy + c= kx

y=e

1

46. nearly 46.6 years

49.

x= ce2 Ny/*

51.

X=alog

52.

x+y

53.

y2+(x-1 ta)2 =a254.y=cx*

55. xy

50. xty2=a2

a-va-y+ Na-y +c

2 =

a2

=c

58. cx=y

y 5x + cx v1 -x

y

36.y sin

37.

x =1+ y*+cv1 +y

23. xetan

1) +c(r -1)-2a

log(r*+y*)=c

35. 2xy sin

18. (y-x-1)(5y

-3x -1) = c 3log(4x +8y-1)= 4(x-2y +C) y(1 +x)=xlog x-x*+ cx

+**)+ cy

xY

40. p

16. Substitutions: x2

V1

y

-

-log( log x+

V1+x =ylog(x +

y+C

+ y) sec(x + y) = x +C

10. x+c=

11.

C

-

20--1)+ 31. log y=+c 30.

=

sin 'x sin x ++ce sin

y

56.

x+y=e

59.y

-2tan

2ax +c

60.

2x+y2=c2

62.1,2 respectively

63.

one

64. yx*= o(y + 2)

65.

y(x- 1+log y) + 1 =0

66. 16

67

Probability

1. Elementary

Probability

Recap of Facts and Formulae

.

1.

Sample space and event

Here n(S)

of all possible outcomes of a random is experiment called the sample space or probability space. space is an event. Every subset of a sample example In throwing a dice, the sanmple space

setS

The

n(E)

S={1,2,3, 4, 5, 6} and n(S) = 6. = E 1,3,5} c S. So E, is an event and n(E,) 3. The event E = {1,3, 5} is also expressed as the event of getting an odd number in throwing a

the number of numbers of four distinct digits beginning with 4 that can be made with 1,2, 3 and 4

=

=

dice.

1x°P =3!

required probability =

and S c

cS

P=4!.

=

For

S. So o and S are also events for the S. ¢ is the impossible event (null event)

total length

Probability of an event is finite) then

Ifthe sample space S is discrete (i.e. n(S) by the probability P(E) of the event E is given P(E)

or

nE)

or

=*

n(S)

number of favourable outcomes total number of outcomes permutation or The problems of restricted problems of combination are convertible into work out such probability. Students are advised to

favourable area total areaa

favourablevolume total volume

according as the sample points (outcomes) are distributed over a length (one dimension) or an area dimensions). (two dimensions) or a volume (three from 4 PM to For example A football match is played match (not before 6 PM. A boy arrives to see the probability that he the match starts). What is the which takes will miss the only goal of the match match? place at the 15th minute of the has to reach between To arrive at the match, the boy total possible length of time 4 PM and 6 PM. So, the will miss the goal if he = 120 minutes. He hours 2 6 PM. So, the favourable arrives between 4.15 PM and = 105 minutes length of time = 1 hour 45 minutes goal probability of his missing the

corresponding chapter roblems by using exercises in the in Algebra.

=

probability can If n(S) and n(E) are both infinite, the below: be estimated by geometrical method as PE)=avourable length

sample space and S is the certain event.

2.

total number of numbers of four distinct 4 digits that can be made with 1,2, 3 and (without restriction)

=

For such problems,

n(E) number of ways under the restriction n(S) number of ways without restriction distinct For example How many numbers of four 1, 2, 3 and 4 digits can be made with the digits which begin with 4? tis a problem on restricted permutation. A similar Sum in probability will be as the following: P(E) =-

the

A four-digit number of distinct digits is written 1, 2,3 and 4. What is Wn at random by using digits number begins with the

105. 7 120

Probability (chance) that

8

0sP(E)S1.

the digit 4?"

F-3

Problcms Plus in

Multiplication theorem) P(E,) P(E = P(E;): P(E;) P(E3). PE, events E, Ez Cannot be utually .Tiwo non-null at the same time. exclusive and independent independent events then E, E, :Ei. If E, E are also independent. and E,, Ez' are

PEnE)=

nE E)

Complementary event

3.

erent E ' (or E or E") of The complementary happening and E is

the event

the event of not + P(E') = 1 = 1- PE), i.e., PE) E

P(E )

The The

If

odds in favour ot the event

P(E) PE')

E

PE)

odds against the event E

E

=

a

:

b then P(E) =

Union and intersection of events E, is the event of at E E, The union E, u of events and

given by P(E2/l

-

LAddition theorem)

u E,) =E P(E)

PE EuE,U...

-

2sitjsn

PE;n E)

.the number of favourable selections = r{E) =7x

(5, 6), (6, 1), (6, 2),

..,

..,

..

x

31 50

determinant of the second order is made with the elements 0 and 1. What is the probability that the determinant made is non-negative?

The number of determinants of the second order that canbe made with 0,1

each

Repeated trial is p and that o the probability of success in one trial p then failure is q so that +g =1 trials = "C, p9 the probability of successes inn expansion o binomial the in term 1e., (r+1)th

(q+P)

PE,)+ P(E>) + UEuE) = PCE;) + PE) P(E)

successes in n trials the probability of at least r +"CGP

+..

="Cp'q"-+"C,.1p*'q"-ve

.Two events E, and E, are independent iff = P(E,n E) P(E,) P(E). Thus, for independent events:

successes in n trlais the probability of at most r

="C9+"C, pq"-l+... +"C,p'4

n(S)

=

16

ofthefour places ofelements can

16.

=the event of gettingnon-negative determinants. Then E' = the event of getting negative determinants. Clearly, negative determinants of the second order that can be made with 0, 1 are

1-19-11

10) with replacement, determine the probability that the 0ots of the 11, 2, 3, 4, 5, 6, 7, 8, 9,

equationx+px +q=0 are real.

:

Roots of x?+ px +g=0

or

pz 44,i.e.,

where 1 Sx, s6,1

and x, then

sx

6,.1

Sr,

$ 6.

The number of solutions of this equation

in (r+x*+r'+r'+*+)

coefficient of x =coefficient of

coeficient of

r" in (1+x+x+r*+x'+r555 x "

coefficient of x

"

in in

(1-x°)-1-x)*

tC-cr*+C"..

x('C+Cr+ C?..+SCa"... to «)

required probability=

13

16

1514 2413:12.9-8-7.6 15 7 13

are real if p-4q2

Xy, Xy

+X+Is =16

PE)=1-P(E') =1-;

Tw

set

If the numbers shown be integers r, X

=coefficient of x " in

Selected Solved Examples If two numbers p and q are chosen at random from the

Five ordinary dice are rolled at random and the sum of the numbers shown on them is 16. What is the probability that the numbers shown on each is any one from 2,3, 4 or 5?

=

"(E) 3

P(E)=

.

+

Let E

Hence, the

1.

c64x6314

(10, 10)

2. A

2 x2x2 x 2

r

PEE)=

n(S)

(8, 10),

to have the

9.

if

nE)7x147x14x2.

(6,9),

be filled in 2 ways-by 0 or 1).

If

14.

the required probability

1+2+4 +6 +9+ 10 10+10+10 62.

i=1

Mutually exclusive and independent events

PE

corner = 2x7=14.

10.

62 required probabinty(S) 100

the

x;Pi

E(x)=

PEnE,nEx). 1siejsksn .+(-1)"- P(E, n E2... En) exclusive .Two events E and E are mutually events PE, nE)= 0. So, for mutually exclusive

10 ,

2), (4, 1), (4, 2), (4,3), (4,4), ,

(10, 1), (10, 2),

+

6.

10

7,1), (7, 2), ..., 7, 10), (8, 1), (8, 2), (9, 1), (9,2),.. 9, 10),

n(E)=

z, isp value x is pPix if the probability of the value variable r Xn of the random Iz, values x, the X,... If Pn respectively have the probabilities pi,P2» P3 E(x), is given by then the expectation of x, denoted by

PE3)-P(E, nE) PEnEs)- P(Es n E) + P(E, nEz n E)

5, 1), (5, 2),..

PEE)

.The expectation of the random variable

+

..,

= 10 x 10 = 100

{(2, 1), (3, 1), (3,

and E which

E)PE

are selected (or The number of ways to select two consecutive rows columns) =7, because there are 8 rows (or columns) of small squares. For each pair of two consecutive rows (or columns), the number of pairs of squares having exactly one common

ifp=6thenq=1,2,3,..9 ifp 7 thenq=1,2, 3,.., 10

Expectation

8.

P(E, n Ez)

PE,) + PE)

common if they Two squares selected can have a comer from two consecutive rows (or columns).

ifp=3

ifp= 10 then q= 1,2, 3,..

g

least one of the events E, E, happening. E, is the event E, The intersection E, n E, of events and of both the events E,, E, happening.

PE uEE)=

ifp =4 thenq=1,2,3, 4 3,4,5,6 ifp 5 then q=1,2,

nS)="C

ifp=9 then q= 1,2, 3,.

nB).

total number of ways to select 2 squares

the

occur in conjunction place (i.e, E, i and E, takes place after E has taken E takine dependent on E,) then the probability of denoted by P(E2/ E,), is place after E, has taken place,

E, E are two events

1

ifp =8 then q= 1,2,3,

we have

Conditional probability

7.

5.

PE uE)= P(E,) + P(E)-

PMA) =P(AoB)+ P(A

B

q=

ifp =2 then thenq= 1,2

Now, (S)

= a:b then

. If

.If odds against the event

-

-

any two events A and

P(E)

odds in favour of the event

E

then are independent events P(E)N1 P(Ez).

PE UE={1

For

PEa+b

.

E, E

If

Odds in favour, odds against

4.

e

Tt

are chosen at random from the small C chance drawn on a chessboard. What is the two squares chosen have exactly one corner in common?

uares

the

otal number of

= small squares on a chessboard 64

10.9.7= 1365

- 630

=

735

n(S)=735. Now, n(E) = the number of integral solutions of

16 where X X Ss5 2xS5, 2s2s5, 2

+2+=

Problems Plus in IT

cocificient ofr

"

in

+X*+x*+r*'

coefficient of r " in coefficient of r ° in

(1

+

x

in which the remaining n 1 The number of ways two places on two Sides can take places keeping ne of

xP'Co+C,x+°C^x*+.. +Csx") 10 x 10

+

n(S)"Cm-1

3)

n(E) 15249

quired probaDintyn(S)

(m-1)! (n-m -2) !

735

-

consecutively, three Out of (2n + 1) tickets numbered that the are drawn at random. Find the chance numbers on them are in AP + , Let the tickets be numbered 1, 2, 3, ... 2n 1 m(S)= the number of ways to select three numbers 2n

Now, possible Al's with common ditference 4), ...,(2n-1, 2n, 2n + 1),

1

possibilities. 1.e, Possible Al's with common difference 2 are (1,3, 5), (2,4,6), .,(2n-3, 2n-1, 21+ i.e., 2n -3 possibilities.

2-

(n -1)(n-2)

divisible by 10? x, y. Let the two non-negative integers be if the sum of digits Nowx+y' will be divisible by 10 10. y is 0 or x in the units places of and x 10 ways units place of can be filled in

The

1

1, 2,3,.. ,9 can be used). in 10 ways. Also, the units place of y can be filled x, y to have a the number of ways for the numbers

(:any one of 0,

:

1),

places= 10 x 10 = 100. digit each in their units can be x* as well as y digits in the units place of can square of any number from 1,4,5,6,9 because the in the units place. digits one of these have only be places of x' and yshould Sum of the digits in units

***********************************************************

The

Possible AP with common difference n is 1), ie., possibility (1,n+ 1, 1

2+

n{E)

=

(27

-

1) +

(2n

1+3+5t...

-

3)

+

(Zn - 5) +

..+1

+(2n--1)

-12x1(n- 1)2)

=

(S)

= C,

or

:

n?

So,

number of ways of drawing 2n cards of which are white and n are black; n = 1, 2,... 26

"C,"C,

if x

have 0 in unt has 0 in units place, y must

1)

in a row, his man parks his car among n cars standing On his return he finds car not being parked at an end. cars are still there. What is the that exactly m of the n on two sides of probability that both the cars parked

6. A

his car, have left? crosses (x). Clearly, his car is at one of the

Xxx *

X

*

*

5

un must have 5 in in units place, y

x has 4 or 6 in units in units place

place, y must have

the

1+2x2+2* 1x1+2 x2+2x 2+1x

3-2

2

+7.6+7

A

(

number is divisible by

11

if the difference of the sum in even

of the digits in odd places and that of the digits

places is divisible by 11. As the number is of seven digits we must have (for favourable cases), sum of four digits in odd places- sum of three digits in even places 33, 44, 55.

X+y=* X -y=ll

X+y=59

"Co+°C X+ "C2*r'+...+Czsx

or

X-y

-y=

+y=5*

x+y=5

or

33

or

I-y=44

(Gii)

55

I-y=

(vi)

(

(iv

:Clearly, ),

+y= 59 X-y=2 (ii)

or

and Gv) do not give integral values of

x

and y.

constant term in

(1

+X)

(i)

But

(vi)

+"C

c+c}+*c+. of x

(1+)

in ln25

Coetficient of x

in (1

y =24;

»

iv)

46, y = 13;

=57,y=2

.

+x)="C26

C=C3+*c}+"C}.*C nE)="C2»-"C6= 52 (26!3 therequired probaolty

I= 35,

=

Obviously, from (a), (b), (C), (d) and (e) we get, sum of three digitsy cannot be 2 or 13. Hence, only favourable case takes place when the sum of four digits in the odd places = 35 and the sum of the three aigits in even places= 24. in the favourable numbers we will get, 9,9,9,8 in odd places and 8, 8, 8 in even places places 9,9,9, 8 in odd places and 9, 8,7 in even or even places or 9,9,9,8 in odd places and ,9,6 in by 11 the number of numbers divisible

constant term in (1+

coefficient

3'3*31*3*2

52 (26 ) .

-4+24+12

5

. )

= 40

from (1) and (2), *

place, y musthave it x has 3 or 7 in units in units place to have numbers T, number of ways for the sum is 0 or whose places digit in their units

02,

X+y=

ne

or or9

5!2! 5! 6

4!2!

62 210

or

s place, y must have 0r if x has or 9 in units in units place 0r place, y musthave 4 if x has 2 or 8 in units

has place

7

3!4!

seven-dig

7.7,7

:If the two sums are denoted by x and y respectively then

in units place

(21+1)X2n-1) 4n-1

7

0,11,22,

1

(2m+1)2n(2n

total number of ways to form a whose sum of digits is 59

the

number

E("C,

Multiplying these and equating the constant terms on both sides,

,

10.

=

n

-c+C+*c3+... "c Now, (1 +)"=

place

the required probability

nE)

nE) =

-2

-1 (using property of binomial coefficients)

(-1)1

Two non-negative integers are chosen at random from replacement. the set of non-negative integers with What is the probability that the sum of their squares is

7.

are

(1,2, 3), (2,3,

(n-m)!

C+"C6t+"Ca=2*'-1

-2

11

+

m-1)

C

7- m- 1)

5.

,

total numberof ways of drawing even numberof cards

S)=

135.

5 x5=

from 1,2,3,..

n ordinary pack of 2 cards an even number of are drawn at random. Find the probability of getting equal numoe Or DIack and red cards. an ordinary pack ot 2 cards, 26 are black and 26 are

red.

-

n(E)

:CC,+C2 C2+ *C°C

In

(d)9,9,9,9,9,8,6

c)9,9,9,9,9,7,7 (e) 9,9,9,9,9,9,5

8.ds

his car vacant=" Cm-1

therequired probability

Probubility

probability= 9 the required 0050

="-"Cm-1:

)°-(1+*)

rC+Cr+Cr*..+Cr 10+

Element

ways in which the remaini The number of the -1 (excluding the car of cars can take their places man) there are n -1 places for them-

+x*+x'+x*

in (x* (1

coefficient of

MatheaticS

sum of the digits of a seven-digit numbers is divisible by ne probability that this number is 59, clearly at

As 7 x least

digits and the sum of seven ast three of the digits must be number will be as the seven digits of the b

y,

1Ollows: ta)

9,9,9, 8, 8, 8, 8

(b)9,9,9,9,8,8,7

the required probability

210 21 0.3, PMB)=04

that PA)= 10. A, B, C are events such = PC)= 0.8, P{AB) = 0.08, PLAC) 0.25 and 0.75 then PMABC)=0.09. If

P{AuBU)2

0.23 Sx P{BC) lies in the interval

:We know, P{A

FEiiillItiE

U

BuC)S1.

s048.

show that

Problems Plus in

sP(A UBUC)Ss1 = But, PlAUBuC) PA) + P(B) +

075

0.4 +0.8 0.08-P(BC) 159-0.36 PBC)

-

P(Å

-

PLA

n B)

(HH..

m times) xx

T(HH..

m times) xx

...

028 +0.09

PA)=P(C) or P(B)=

xT(HH...

... x

times) xx ..

But0 ) P(D/E;) + P{E) PDE)

C

PE)

P(E/E)+ P(Ey) P(E/E,) +P(Ep) P(E/ED) + P(Ec) P(E/E-}

10 50

+ As PE)+ PEz) P(E;)= Ez, E

PEs) P(E/E;)

P(Es/E)=

9

20C1

above).

because after drawing 2 black balls in the first draw there are 4 black balls and 2 white balls for the next draw. (1), the required probability

B

spades.

x9,111 20 x 19 9190

10025

C

Let A,

Similarly, P(E/ED) =P(EEc)=78 1275

the last defective in the 12th drawv

P(D/E)

events E

1312.

150125

We have to find P(Es/E), ie., the probability of the missing card being spade when the two cards drawn are

= 2x 10!8I*20

PE)

C

PE/E,)= probability of getting two defectives and nondefectives in the first 11 draws and 2

from

draw

F-29

probability of drawing 2 spades when one heart is missing

PEE)= .. (1)

PE/E,)= Probability of getting one defective and 10 nondefectives in the first 11 draws and the last defective in the 12th

C

PB/E)

Conditional Probability and BayesS Thcorenm

the required probability + PE)- P(E)- P(E/E2) P(Eg) P(E/E,)

a

PE)21

PLE)P(B/E,)

+

Total

roduced event ot a product being producedat A event of product being produced at B event of a product being produced

=PC)

As the question is answered by either guessing or copying or knowing,

PU-1Now, PA/G) = probability of answering correctly by

8uessing

=:

there are 4 choices and only one of them is correct

Inlems

pobubility

A=

of

answering correctly

Plus n

In

ATatiNmaatics

otal Conditional Prabability and Bayes' Thevrem

1Le, the probabi have to tind rb/T), spoke truth. ball was drawn when they By Bayes' theorem,

We

by

copying

given) PA/K)

=

probability ot answering corrextiy

P(B/1)=

by

t

ab

PB/E) rEDP(B/E,)+ P¢B/E,)+ P(B/E,)+ P{B/E,) 35

he knows he certainiy

1.3Z5.13 10 6 64

21+15

36

4

8. An

10

times out of 6. A speaks truth 3 times out of 4 while 10. A ball is drawn at random from a bag containing one black ball and five other balls of different colours. Both A and B report that a black ball has been drawn from the bag Find the probability of their assertion B, 7

six bals

=

3

bag containing balls ot other colours

the event of a black ball drawn from the bag

Obviously, exclusive. Now, P(T/B)

speaking truth speaking lies.

Let

B

=

B

E

ball of another colour

a

= P(Ez) =

P(E) = P(Eg)=

P(L/B)

=

balls. Let B= the event of drawing 3 black 3 black bals Now, PIB/E,) = probability of drawing when E, happens

balls and 5 white balls the event of the urm containing 6 black E, are equiprobable and they are

exhaustive. Now, PW/E,) =

probability of 3 balls drawn being white in case of

11

2, 66 x 70 280

ever Bayes theorem for equiprobable

Let

B

909

909

303

=the event of drawing a black ball in the next draw.

Now, the ball in the next draw may be black after the 3 balls were dravn from the bag containirng 6 black white and 3 white balls. Its probability is PE/W) PIBAE,/W), etc. by the theorem on total conditional probability P(B) =

PE, /W)

PIBAE, /w

-

+P(E,/W)-PIBAE, /W}. Now, PLB/E,/W)}

already drawn

C =

=

C

C3 3

PIBAE,/W)l=77

Bayes' theorem for equiprobable events,

PIBAE/W

P(W/E;) P(W/E)+ P(W/E,)

+

P(W/ES) + P(W/E)

C

PEAE,/W))=

5

C using (,

3b

T12,1 84

x1

66x70 55

84

909

140

0 90**30

P(B) PB)-9051*07

30 33 909

(55132+210+280)=c0

909

(1)

1 the probability of drawing black in case ot E when 3 whites are

=

10C.

P(W/E) C 12c

after draws The bag will contain no black balls with By balls if B took place in conjunction

Dy

x70 420 14

66 11

3

PE,/W)=

So, we have to find P(E/B).

84 3033*

+PE, /W)-PLBAE, /W)

E

6

Similarly, P(WIEA)

by

C1. PB/E,)=g

11

PE/W)1.1

PEs/W)-PIBAE, /W)}

7

C Similarly, P(BJEg)= 6

=

probability of both telling lies when a black ball is not drawn

33

1,2+ 33 30

84

ball will be drawn in the

balls and 6 white balls.

(:their sum 1

10

10

black

the event of the um containing 6 black balls and 4 white balls

P(W/E)=

probability of both telling truth when black ball is drawn

probability of B telling truth

909

P(E,/W)=1,

contains at least 3 white balls. the event of the urn containing 6 black

Each of Ex Ey Es

are exhaustive and mutually

:probability of A telling truth

a

Es=the event of the urn containing 6 black

and P(B) 1- 6 and

unknown number

balls and 3 white balls

the event of the bag containing 4 black balls and 2 balls of other colours 1

Es

E

=

=

and

Clearly, the urn

3 black bals

E,= the event of the bag containing 6 black balls Clearly, they are equally probable and no cbhe are blaà possibilities exist as 3 balls drawn at random means the bag has at least 3 black balls.

PE,)

being true?

urn ontains 6 black balls

that the chance that

12

E=the event of the

and

Clearly, PB)

15

909

30

white balls. 1hree balls are drawn successively and not replaced and are all found to be white. Prove

Es=the event of the bag containing 5 black bals

B

b6 x 70

next draw is

there are

and

B

22 11

1

(S 6) of

or unknown colours; th a balls are drawn at random and found to be Find the probability that no black ball is left in theb (Or Find the probability that the bag contained erad 3 black balls.) ag

E

T- the event of both A and L= the event of both A and

35

10

Let

B =

1

20

I.

ansiwers cormxtiy) probability that he knew We have to find PiK/A,ie., the when he answered corrctly. By Baves' theorem, PK) PA/K) PK/4A) PMG)- PMAG)+ P(C)- PiA/C) P(K) P(A/K)

Let

0 =

to

+ P(B) P(L/B)

P(B) P(T/8)

nowing 1

imilarly, P(E, /W)

P(B)- PCIJ8)

677

PrWems Plas

contains

Apue

5 coins, each either a

in

ilT Mathematics Total Cenditional Probability anal Bayes' "Theorem

PE/E,)

titty-paisa coin

PE/E

and One-rupee coin. wo coins are drawn at random expected at found to be one-rupee coins. kind the coins. remaining of the total value pobable) Clearly. at least two of the coins are one-rupee coins. Let coin and f stand for tifty-paisa oneruper tor stand or

rNE/E)+P(E Eg)

* PEE)P(EE

I0

+

3

1

10

20

10

coin

following: he coins in the purse can be any one of the 27. 3 3. : ir, If 5. Let E, = the event ot the purse containing 2r and 3f E the event ot the purse containing šr and 25 etc. Clearly, the events are E;, Es. E, and Es, and they are

Similarly, P(E3/E) =

3

10 20

PE/E)=1

PE)=

Also E, E. E, and exclusive. Now, P(E/E;)

=

Es

-+ 0

are exhaustive and mutually

PEsE)1. +

probability of drawing wo one-rupee coins when E happens =

10

++1

20

(value

of Ir and 2f)

+

P(E,/E) (value of 2r and 1f)

+

P(Es/E) (value of 3r) -

20(Rs 1.50)2 Rs 2)10 (Rs 2.50) R

C P(E/ES)==1 probability of the purse containing 2r and 3f when 2r are drawn, etc. By Bayes theorem for equiprobable events, Now, PE/E)

3

P(E2/E) (value of 3f)

P(E,/E)

PE/E)C1 E)

0

expected total value of the remaining coins

the

Similarly,

+15 200

3

3

3

=

=

**|D'0*Í*Rs

2.625.

. There are

three bags each containing 5 white and 2 black balls. AlSo, there are 2 bags each containing 1 white and black balis.A ball is drawn at random from a bag and it is found to be black. What is the chance that the black ball came from the first group? 9. A can hit a target 4 times in 5 shots; B can hit 3 times in 4 shots and C twice in 3 shots. They fire once each. f two of them hit, what is the chance that C has missed it? 10. A bag contains 5 balls of unknown colours. A ball is drawn and replaced twice. On each occasion it is found to be red. Again, two balls are drawn at a time. What is the probability of both the balls being red ? 11. A letter is known to have come from either MAHARASTRA or MADRAS. On the postmark only consecutive letters RA can be read clearly. What is the chance that the letter came from MAHARASTRA? 12. A bag contains 10 coins of which at least 2 are onerupee coins. 1wo coins are drawn and both are Tound to be not one-rupee coins. Whatis the probabilhty ot the bag to contain exactly 2 one-rupee coins?

There are two bags, one of which contains three black and four white balls while the other contains four black and three white belis. A dice is cast. If the face 1: or 3 turns up, a ball is ta ken out from the first bag. But if any other face turns up, a ball is taken from the second bag. Find the probability of getting a black

3.

respectively. A ball is drawn from an urn chosen at random. What is the probability that a white ball is drawn if the choices of urns are equiprobable?

:

5.

e

not i probability that a certain eiec does n used is 0.10. If it o ais when first for lasts it that probability the pon Edately, a new co s U.9. What is the probability that one wll last year? A factory A produces 10% detecvealves defective another factory B produce 20% nd 5 valve ron bag contains 4 valves of factory A a randu at actory B. If two valves are drawn

4. The

ball. 3 2. Three urns contain 2 white and 3 black balls, white 1 black ball and 2 black balls; and 4 white and

An urn contains two balls each of which i5e um. Wnat A white ball is added to the tne u probability of drawing a white ball from mpone

1

2

. 0.29 3

0.891

train and the probabilities

of these are

late if he takes car, scooter, bus or train

Given

that he reached office in

:and is5

time,

fhnd the

the 22nd

century

probability that he travelled by a car

Objective Questions Fillin

the blanks

15. The probability that in a chosen

-

at random

there

year of will

be 53

Sundays,

is

There are two bags X and Y; X contains 3 white balls and 2 black balls, and Y contains 2 white balls and

16.

4 black balls. A bag and a ball out of that bag are picked at random. The probability that the ball is

white,

coupors each with a word written on them. On one of them the word PATNA is written and on the other PLATE. A coupon is taken at random and 3 letters are selected at random tem the letters of the word on the coupon probability that the selection contains two vowels, is

17. There are two

he

15.

1wo persons A and I throw two dice each. 1he probability that Y throws a sum greater than X s it is knoWn that X hrows a sum of at when

leasto.

13

12

1615

19

2

respectively:. The probability that he reaches office

6.

10.

C

A heads and 1 was chosen. 4. A person goes to ofice either by car, scooter, bus or

AnswersS

or black

coin

random and tossed 3 times giving ta'l Find the probability that the coin

man has three coins A, B, C. The coin A is unbiased. The probability that a head will show

Exercises

in case of the

A coin is chosen at

5. A

1.

.35

when B is tossed is while it is

o

20

equaly probable. PrE,) = P\E;) = PME) =

find the probability that at least valve is defective. unbiasecd coim is tossed. If the result is a 6. An head, a s pair ot unDiasea ace is rolled and the number btained by adding the numbers shown on them is noted. It the resuit is a tal, àa card trom a well-shuffled erea 2, 3, 4,., pack of eleven ca 12 is picked and the number ot the card is noted. What is the chance that the noted number is either 7 or 8? . A bolt factory has three machines A, B and manufactunng respectvely z7o, 35% and 40% of the total production. hese the machines produce 5%, 4% and 27o aerectve bolts respectively. A bolt is selected at random and it is tound to be defective. Find the probability that it was manutactured by the machine (a) A (b) B (c)C. bag,

18.

108

14 17.

Total Conditional Probabilitya

Bays Theovem

Chapter Test Time 75 reseztes Abag contains 3 white and 3 black balls. A person draws 3 ballsat rardom fon t He en drops 3 red balls into the bag and azain draws out 3 bals at rardom. What is te arce that the latter 3 balls will be of different colour? 2. Abag contains n coins of unknon values. A con is drawmat rardom and it is ford o a rupee. What is the chance that it is te the only rupee con in the baz? 3. In a city filmland there are 3 male superstars and 3 ferrale sperstarn Ad r make a film with 3 superstars. If the film deides has 2 rale saperstas yestz te chance of the film to be a hit iswhile the ae chance is here 1 mae zd 2 superstars. f the 3 superstars are from the same sex, he carce oé te begats the time of making the film, only 3 superstars arearalabe 1.

rd1eale

ai

At

toigtrte

te

director takes them all then what is the probablity of his flbeigak? 4. Three groups A, B and C are competing for positions an he Boardé Des cf a company. The probabilities of their winning are 05, 03 and 02 spesrey The probabilities of introducing a new product ifa particular gop ws 2219erey, 0.6 and 0.5. Find the probability that a new product will be intodacd 5. A purse containing 16 notes, four each of denominations Rs 10, Rs 20, Ps 50 znds 100, is left on a table. A maid servant steals two of the notes. Afer at ie za akas ot wo notes at random from the purse and finds that they anont to Rs 200 What is te probability that an amount of Rs 200 was stolen by the aid servart Écn is pse? 6. An um containing 6 white balls and 4 black balls. A ball is drawn at randoaadisp back into the un along with 5 balls of the same colour as that of the bal dza Aball is drawn again at random. What is the probability that the ball drawa now is white? 7. Abag contains p white and q red balls. m(m are