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Bi yani ' s Thi nk Tank

Concept based notes

MathsClass -XII

MS. VershaInformation Technology

Biyani Girls College, Jaipur

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Published by :

Think TanksBiyani Group of Colleges

Concept & Copyright :

Biyani Shikshan SamitiSector-3, Vidhyadhar Nagar,Jaipur-302 023 (Rajasthan)

Ph : 0141-2338371, 2338591-95 Fax : 0141-2338007E-mail : acad@biyanicolleges.orgWebsite :www.gurukpo.com; www.biyanicolleges.org

First Edition : 2009

While every effort is taken to avoid errors or omissions in this Publication, any mistake oromission that may have crept in is not intentional. It may be taken note of that neither thepublisher nor the author will be responsible for any damage or loss of any kind arising toanyone in any manner on account of such errors and omissions.

Leaser Type Setted by :Biyani College Printing Department

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Preface

Iam glad to present this book, especially designed to serve the needs of

the students. The book has been written keeping in mind the general weaknessin understanding the fundamental concepts of the topics. The book is self-explanatory and adopts the Teach Yourself style. It is based on question-answer pattern. The language of book is quite easy and understandable basedon scientific approach.

Any further improvement in the contents of the book by making corrections,omission and inclusion is keen to be achieved based on suggestions from thereaders for which the author shall be obliged.

I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay

Biyani,Director

(Acad.

) Biyani Group of Colleges, who are the backbones andmain concept provider and also have been constant source of motivationthroughout this Endeavour. They played an active role in coordinating the variousstages of this Endeavour and spearheaded the publishing work.

I look forward to receiving valuable suggestions from professors of variouseducational institutions, other faculty members and students for improvement ofthe quality of the book. The reader may feel free to send in their comments andsuggestions to the under mentioned address.

Author

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Unit 1

Chapter 1

Relations and functions

Q.1 Let A be set of firs t ten natural numbers. If R be a relation on A defined

by xRy x + 2y = 10 then

i. Express R and as set of ordered pairs1-

R

ii. Find domain of R and1-

R

iii. Find range of R and1-

R

Ans. Here

A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }

Relation R is defined as

xRy x + 2y = 10

10-xy =

2

Now for x = 1, y = 10 -1 9

= A2 2

Hence 1 is not related to any element of A.

Similarly use can observe that 3, 5, 7, 9 and 10 are also not related to anyelement of A.

Again use observe that

When 10-2x = 2,y = = 4 A 2R42

When 10-4

x = 4, y = = 3 A 4R32

When 6 10-6x = 6, y = = 2 A R22

When 8 10-8x = 8, y = = 1 A R12

Hence

i. R = {(2,4), (4,3), (6,2), (8,1) }

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and { }-1 = (4,2),(3,4),(2,6),(1,8)R

ii. Domain of R = {2, 4, 6, 8}

Domain of { }-1 = 4,3,2,1R

iii. Range of R = {4, 3, 2, 1}

Range of { }-1 = 2,4,6,8R

Q.2 Prove that the relation R on the set N x N defined by (a, b) R (c, d) a+d =b+

c fo r all (a, b), (c, d) N x N is an equivalence relation . Ans. To prove that the given relation is an equivalence relation we have relation to

show that it is reflexive, symmetric and transitive.

1) Reflexive Let (a, b) be an arbitary element of N x N. Then,

( a, b)N x N a+ b = b+ a [by commutativity ofaddition on N]

(a, b) R (a, b)

Thus (a, b) R (a, b) for all (a, b)N x NHence the given relation R is reflexive relation on N x N.

2) Symmetric Let (a. b), (c, d) N x N, Such that (a, b) R (c, d)Since (a, b) R (c, d) a + d = b + c

c + b = d + a [by commutativity ofaddition on N]

(c, d) R (a, b)Thus (a, b) R (c, d) (c, d) R (a, b) for all (a, b), (c, d)N x N .So R is symmetric relation on N x N

3) Transitive Let (a, b), (c, d) and (e, f)N x N.Such that (a, b) R (c, d) and (c, d) R (e, f)

Since (a, b) R (c, d) a + d = b + c (1)And (c, d) R (e, f) c + f = d + e (2)Adding equation (1) & (2), we get

a + d + c + f = b + c + d + e a + f = b+ e (a, b) R (e, f)

Thus (a, b) R (c, d) and (c, d) R (e, f) (a, b) R (e, f) for all (a, b), (c d), (e, f)

N x N.

So, R is transitive relation on N x N

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Hence R being reflexive, symmetric and transitivite is an equivalence relationon N x N.

Hence proved.

Q. 3 On the set N of natural numbers a relation R is defined asa R b a2-4ab+3b2=0 v (a, b N) . Prove that R is reflexive but no t

symmetric not transitivity.

Ans. Given set is N = {1, 2, 3,..}

Relation defined on N is

a R b a2-4ab+3b2=0 v a, b N.1) Reflexivity Let a N

2 2a - 4a.a + 3a

2 2 2= a - 4a + 3a2 2

= 4a - 4a

= 0

(a, a)R v a N. R is reflexive

2) Symmetry - Let a, b N such that (a, b) R2 2

2 2

4 3

4 3 0

( , )

( , ) ab b

b ba a

b a R

a b R a 0 + =

+

Hence (a, b) R but (b, a)R R is not symmetric relationEx. (3, 1) R because = 9-12+3 =12-12 = 023 4 3 1 3(1) + 2

But (1, 3) R because = 2 24(1)(3) 3(3)(1) + 1 12 27 0= +

3) Transitivity Let a, b, c N such that(a, b) R and (b, c) R

2 2( , ) 4 3 0a b R a ab b + =

and (b, c) R b2-4bc+3c2Then it is not necessary true that a2-4ac+3c2=0

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Ex. (9, 3) because = 81 108 + 27 = 02 4(9)(3) 3(3)9 + 2

and (3, 1) R because = 9-12+3 = 023 4(3)(1) 3(1) + 2

but (9, 1) R because = 81-36+32 24(9)(1) 3(1)9 + 0

R is not transitivity. Hence from (1), (2) & (3) it is clear that R is reflexive but notsymmetric and transitivity.

Q. 4 Let A = {1,2,3} then give examples of relations which are

1) Reflexive, symmetric and transitive

2) Symmetric and transitive but not reflexive

3) Reflexive and transitive bu t no t symmetric

4) Reflexive and symmetric but no t transitive

Ans. 1) is reflexive, symmetric and transitive1 {(1,1), (2, 2), (3,3)}R =

2) is symmetric and transitive but not reflexive2 {(1,1), (2,2)}R =

3) is reflexive and transitive but not

symmetric

3 {(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)}R =

4) is reflexive and symmetric but

not transitive

4 {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}R =

Q. 5 If a, b, {1, 2, 3, 4}, then check whether the following is function or not

f = {(a, b):b=a+1} also find it s range.

Ans. Here f = { (1, 2), (2, 3), (3, 4)}. Here we observe that an element 4 of the given setis not related to any element of the given set. So f is not a function.

Q. 6 If3

1

x

x

+

(x) =f then find [ { ( )}]f f f x

Ans. f (x) =3

1

x

x

+

Now ( ){ } =f x ( ) 3

( ) 1

f x

f x

=

+

3 31

31

1

xx

x

x

+

+ +

=3 3 3 2 6 3

3 1 2 2 1

x x x x

x x x x

+ = = =

+ +

Again

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f [f {f(x)}] = f3

1

x

x

+

=

33

1

31

1

x

x

x

x

+

+ +

3 3 3 43 1 4

x x x xx x

+ += =+ +

=

[ { ( )}]f f f x x =

Q. 7 Prove that the function f: R R, f(x) = cos x is many one function?Ans. Given f: R R such that f (x) = cos x

many one function: -

Let a, b R such that f (a) = f (b)

cos cos

2 ,

a b

a n b n

=

= I

f is many one function

Into function Let y R (Co domain)

If it is possible let f (x) =y

1

cos

cos y

x y

x =

=

x will exist if 1 1y When then pre image of y does not exist in R (Domain)[ 1,1]y R

Hence f is not on to function

f is in to function

Hence f is many one in to function.

Q. 8 If f: R R, f (x) = 2x 3 and g: R R, g(x) = 32

x +

then prove that fog =

gof =R

I

Ans. Given functions :f R R , f (x) = 2x-3

3: , ( )

2

xg R R g x

+ =

, so:gof R R

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(gof) (x) = g [f(x)] = g (2x-3) =(2 3) 3 2

2 2x

x x= =

+

gof (x) = x .. (1)

Again

fog (x) = f [g(x)] =

x + 3 x + 3= 2 - 3

2 2

= x + 3

= f

- 3 = x

fog (x) = x (2)

and : R R such that (R

I R

I x ) = x , vx R .(3)

from (1), (2) & (3) we get,

fog = gof =R

I

Hence proved

Q. 9 Let f: R R be defined by f (x) = 3x 7. Show that f is invertible and hence

find

1f

Ans. A function f is invertible if f is a bijection

1) Injectivity Let x, y R thenf(x) = f (y)

3 7 3 7x y

x y

= =

Thus f (x) = f(y) x=y for all x, y R. So, f is an injection2) Surjectivity Let y be an arbitary element of R, then f (x) = y

3 7

7

3

x y

yx

=+ =

Clearly7

3

yR

+

for all y R

Thus for all y R, there exists7

3R

yx

+= such that

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7

3

73 7

3

( )

( )y

y

f x y

f x f+

+ =

=

=

f is surjection. Hence :f R R is bijection. Consequently it is invertible

Let f(x) = y

1

7

3

7( )

3

3 7

y

yf y

x y

x

+

+ =

=

=

Therefore,1 :R Rf is given by

1 7( )3

xxf

+ =

Chapter 2

Binary Operation

Q. 1. Discuss the commutativity and associativity of the binary operation on Rdefined by

a * b =4

ab for all a, b R.

Ans. Commutati vi ty

a * b =4

ab and b*a =4

ba

We know that multiplication on R is commutative

4 4ab ba

= for all a, b R

for all a, b* *a b b a = RSo * is commutative on R.

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Associatively

Let a, b, c R then

(a * b) * c

( )4

4 1

abc

abc ==

6

.............................................(1)

and a * (b * c) = a *4

4 4 1

bc

bca

abc

= = 6

..(2)

From (1) & (2), we observe that a * (b*c) = (a*b) *c

Hence * is associative.

Q. 2. Let * be a binary operation on set Q {1} defined by a*b = a+ b a b , a, b Q - {1}

Find the identity element wi th respect to * on Q. Also prove that every

element of Q {1} is invertib le.Ans. Let the identity element e exist in Q {1} w.r.t * on Q - {1}, then

a *e = a = e*a for all a Q - {1} a*e = a for all a Q - {1} a+ e ae =a

e (1-a) = 0 e = 0

Thus o, is the identity element for * on Q {1}. Let a be an arbitary element ofQ {1} and

let b be inverse of a, then.

a * b = 0 = b *a [0 is identity element]

a * b = 0 a + b ab = 0

b (1 a) = - a

b =1

a

a

{1}

1 0

a Q

a

Since , therefore b={1}Qa {1}1

aQ

a

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Thus every element of Q {1} is invertible and the inverse of an element a is

1

a

a

Q. 3 Let * be an associative binary operation on a set S and a be an invertibleelement of

S then1 1)( aa =

Ans. Let e be the identity element in S for the binary operation * on S, then

1 1

1 1

*

* *

* e a a

a a e a a

a a

= =

= =

a is inverse of 1a

a = 1 1)(a

Hence proved

Chapter 3

Inverse trigonometric functions

Q. 1 Find the princ ipal values of the followings : -

1)

-1 -1cos2

(2) ( )-1 -sec 2

3) (4)( )-1cosec 1

-1

3

-1cot

Ans. (1) Let

-1 -1cos =2

cos = -1/2

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4) Let

-1 -1 = 3

cot

( )

1

3

cot = -

cot = -cot

3

cot = cot - /3

2cot = cot

3

2 =

3

-1 2-1cot =

33

Hence principal value of1 21cot

33is

.

Q. 2 Prove that1 11 1 1tan cos tan cos

4 2 4 2a a

b b

2ba

+ +

=

Ans. Let

1cos1

2

a

b

=

cos2a

b

=

LHS = ( ) ( )tan4 4

tan

+ +

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( )( )

( ) ( )

2 2

2

4

tan

4

=

= +

tan

+

tan - tan

1+tan 1-tan

tan = 11-tan 1+tan 4

+(1-tan)=

1-tan 1+tan

1+tan + 2 tan=

tan +

4

1-tan4

(1+tan)

tan

1+tan

2+1+tan -2tan

2

2

2

2

2

2=

)

1+tan = 2

1-tan

=

= = RHS2b

a

(1-tan

2

1-tan1+tan

2

cos

=

tanA+tanBtan(A + B)=

1-tanAtanB

tanA-tanBtan(A-B)

1+tanAtanB

Hence proved.

Q. 3 If1 15 12sin sin 90

x x

+ =

then find the value of x

Ans.1 15 12sin sin 90

x x

+ =

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-1 -1

-1 -1

-1 -1

-1 -1

2

2

5 1sin = 90 - sin

x x

5 12sin = cos

x x

sin x + cos x = 90

25 12cos 1- = cos

x x

25 121- =

x x

2

Squaring both sides, we get

2 2

2

2 2

2

2

25 1441- =

x x

x - 25 144=x x

x - 25 = 144

x = 169

x = 13

Since x = - 13 does not satisfy given equation . So x = 13 is correct solution.

Q. 4 Prove that

21 1

2

1cos tan {sin(cot )}

2

xx

x

+ = +

Ans. We have

1 1

2

2

2

1 1

2 2

2 2

22

1sin(cot ( )) sin sin

1

1

1

1 1cos tan cos cos1 2

1 1

22

xx

x

xx x

x x

xx

=

+

=+

+ =

+ +

+ += =

++

Hence proved

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Q. 5 If1 1cos cos

x y

a b

+

= then prove that

2 22

2 2

2cos sin

x xy y

a ab b + =

Ans. Given1 1cos cos

x y

a b

+ =

{ }

2 2-1

2 2

-1 -1 -1 2 2

2 2

2 2

2 2 2

2 2

2 2

2 2

x y x y= cos - 1- 1- =

a b a b

cos x +cos y = cos xy + 1- x 1- y

xy x y- 1- 1- = cos

ab a b

xy x y-cos = 1- 1-ab a b

x y

a b

22 2 2 2

2 2 2 2

2xy x y x y- cos +cos = 1- - +

ab a b a b

2 22

2 2

2 22

2 2

x 2xy y- cos + = 1-cos

aba b

x 2xy y- cos + = sin

aba b

Hence proved

Q. 6 Solve the following equation

-1 -1 -1

2

1 1tan + tan = tan

1+ 2x 4x +1 x

2

Ans. Given

1 1

2

- - -1 1tan + tan = tan1+ 2x 4x +1 x

1 2

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( )( )

-1 -2

2

-1 -1 -1

2

1 1+

21+ 2x 1+ 4xtan = tan1 1 x

1-

1+ 2x 1+ 4x

x + ytan x + tan y = tan

1-xy

4x +1+1+ 2x 2=

1+ 2x 4x +1 -1 x

6x+2

1 2+ 4x + 8x + 2x - 1( )

2

2=

x

2 ( )3x+1

2 ( )

( ) ( )

( )( )

22

3 2 2

3 2

2

2

2

= x4x + 3x

3x + x = 8x + 6x

3x - 7x - 6x = 0

x 3x - 7x - 6 = 0

x 3x - 9x + 2x - 6 = 0

x 3x x - 3 + 2 x - 3 = 0

x 3x + 2 x - 3 = 0

-2x = 0,3,

3

Practice Problems

1) Find principal values of the followings

i)1 1sin

2

ii) ( )1tan 3

2) Solve the following equation

( ) ( )

-1 -1 -1 -1x xsec - sec = sec b - sec aa b

3) If ( ) ( )1 1tan 3 tan 24

x x + = then find value of x.

4) Solve the following - ( ) ( ) (1 1 1tan 1 tan ( ) tan 1 tan 3 )1x x x + + + = x .

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Chapter 2

Determinant

Q. 1 Find determinant of A =

-1 6 -2

2 1 1

4 1 -3

Ans. |A| = (-1)1 1 2 1 2 1

- 6 - 21 -3 4 -3 4 1

= -1 (-3-1) -6 (-6-4) -2 (2-4) = 4+60+4

|A| = 68

Q. 2 Find determinant of A =

1 2 -1 3

2 1 -2 3

3 1 2 1

1 -1 0 2

Ans. |A| =

1 -2 3 2 -2 3 2 1 3 2 1 -2

1 2 1 - 2 3 2 1 -1 3 1 1 - 3 3 1 2

-1 0 2 1 0 2 1 -1 2 1 -1 0

1

= 1{1(4-0) +2(2+1) +3(0+2) {-2{2(4-0) +2(6-1) +3(0-2)}-1{2(2+1)-1(6-1)

+3(-3-1)}-3{2(0+2)-1(0-2)-2(-3-1)}= 1{4+6+6}-2{8+10-6}-1{6-5-12}-3{4+2+8}= 1{16}-2{12}-1{-11}-3{14}= -39

Q. 3 Check whether the following matrix is s ingular or not

A =

1 0 2

1 2 3

2 4 6

Ans.A matrix is singular if A =0

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1 0 2

A = 1 2 3 = 1[12 -12]- 0[6 -6]+ 2[4 -4] = 0

2 4 6

A = 0

Hence A is singular matrix.

Q. 4 Find the mines and cofactors of elements of the

determinant

2 -3 5

6 0 4

1 5 -7

Ans. We have0 4

M = = 0 - 20 = -2011 5 -7

6 4M = = -42 - 4 = -46

12 1 -7

6 0M = = 30 - 0 = 30

13 1 5

-3 5M = = 21 - 25 = -4

21 5 -7

2 5M = = -14 - 5 = -19

22 1 -7

2 -3M = = 10 + 3 = 1323 1 5

-3 5M = = -12 - 0 = -12

31 0 4

2 5M = = 8 - 30 = -22

32 6 4

2 -3M = = 0 + 18 = 18

33 6 0

Co- factors

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( )

( ) ( )

( )

( )

1+1 2A = -1 .M = (-1) (-20) = -20

11 11

1+2 3A = -1 .M = -1 (-46) = 46

12 12

1+3 4A = -1 .M = (-1) 30 = 30

13 132+1 3A = -1 .M = (-1) (-4) = 421 21

2+2 4A = (-1) .M = (-1) (-19) = -19

22 22

2+3 5A = (-1) .M = (-1) (13) = -1323 23

3+1 4A = (-1) .M = (-1) (-12) = -12

31 31

A3

3+2 5= (-1) .M = (-1) (-22) = 22

2 32

3+3 6A = (-1) .M = (-1) 18 = 1833 33

Q. 5 If W is one of the imaginary cubs root of unity, find thevalue of

2

2

2

1 w w

w w 1

w 1 w

=

Ans. Given

2

2

2

1 w w

w w 1

w 1 w

=

givesC C +C + C1 1 2 3

2 2

2 2

2

1+ w + w w w

w + w +1 w 1

w + w +1 1 w

=

Since 21+ w + w = 0

So

2

2

0 w w

= 0 w 10 1 w

Now finding its determinant & expending along first column

= 0 3 2 2 4w -1 - 0 w -w + 0 w-w = 0

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Q. 6 Prove that ( )2 2a b ax + by

b c bx + c2

y = (b - ac) ax + 2bxy + cy

ax + by bx + cy 0

Ans. Let A =a b ax + byb c bx + cy

ax + by bx + cy 0

Appling , we get3 3 1

c - xc -ycc2

A =

a b ax + by - ax - by

b c bx + cy - bx - cy

ax + by bx + cy 0 - x(ax +by) - y(bx +cy)

A = 2 2

a b 0

b c 0ax + by bx + cy -(ax + 2bxy + cy )

Now expending along we get3

c

( ) 2 2

b bx + cy - c(ax + by) - 0[a(bx + cy) - a(ax + by)] + 2bx2

y + cy )[ac - b ]A = 0 -(ax

b

(a

(

2 2

2 2

+ 2bx 2

2

y + cy )(ac - b )

ac)(ax + 2bxy + cy )

A = - x

A =

Hence proved

Q. 7 Find the area of tr iangle with vert ices at the points (3, 8), (-4,2) and (5,-1).

Ans. Let A (3, 8), B (-4, 2), C (5,-1) are three given vertices oftriangle. So the area ofABC is

given by

12

=3 8 1

-4 2 1

5 -1 1

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{ }

[ ]

[ ]

1

3 2+1 -8{-4-5}+1{4-10}2

133+89+1(-6)

2

19+72-62

75

2

=

=

=

=

Q. 8 If the points (a, 0), (0, b) and (1, 1) are coll inear, Prove that a+ b = ab

Ans. The given points are collinear soa 0 1

0 b 1 = 0

1 1 1

a[b -1]+1[0- b] = 0

a b - a - b = 0

a b = a + b

Hence proved.Q. 9 using determinant find the equation of the line joining the

points (1, 2) and (3, 6).Ans. Let P (x, y) be a point on line AB i.e. the points A (1, 2), B (3, 6)and P(x, y) are collinear so

ABP = 0

1 2 11

3 6 12

x y 1

= 0

0=

1 2 1

3 6 1

x y 1

1[6 - y]-2[3 - x]+ 1[3y - 6x] = 0

6

- y - 6

+2 x +3 y - 6 x =0-4x+ 2y = 0

y = 2 x

Which is the required equation of line.Q. 10 Find the adjoint of the matrix

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1 3 3

A = 1 4 3

1 3 4

Ans. First we have to find cofactors

1+111

1+2

12

1+3

13

2+1

21

2+2

22

2+3

23

3+1

31

3+2

32

3+3

33

c = (-1) 7 = 7

c = (-1) (1) = -1

c = (-1) (-1) = -1

c = (-1) (3) = -3

c = (-1) (1) =1

c = (-1) (0) = 0

c = (-1) (-3) = -3

c = (-1) (0) = 0

c = (-1) (1) =1

T7 -1 -1

adjA = -3 1 0

-3 0 0

7 -3 -3

adjA = -1 1 0

-1 0 1

Q. 11 Find the inverse of matrix

2 -1

A = 3 4 .

Ans. A = 8 + 3 = 11 0

So, A is non singular matrix and therefore it is invertible. Nowfinding co-factors

1+1

11

1+2

12

2+1

21

2+2

22

c = (-1) 4 = 4

c = (-1) 3 = -3

c = (-1) (-1) = 1

c = (-1) 2 = 2

T

4 -3adj A =

1 2

4 1adjA =

-3 2

Hence -11

A =A

.adj(A)

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=

4 1 4/11 1/111=

-3 2 -3/11 2/1111

=

4/11 1/11

-3/11 2/11

1

Q. 12 Find the inverse of

1 3 3

A = 1 4 3

1 3 4

and verify that -13A A = I

Ans. From question No. 10 we find that

adj A =

7 -3 -3

-1 1 0

-1 0 1

A = 1[16 - 9] - 3[4 - 3] + 3[4 - 3] = 7 - 3 - 3 = 1 0

So A is invertibleHence

-1

-1

1A = adjA

A

7 -3 -3 7 -3 -31

A = -1 1 0 = -1 1 01

-1 0 1 -1 0 1

Now

-1

-1

-1

-1

3

7 -3 -3 1 3 3

A A = -1 1 0 1 4 3

-1 0 1 1 3 4

7 -3 - 3 21-12 - 9 21- 9-12

A A = -1+1 + 0 -3 + 4 + 0 -3 + 3 + 0

-1+ 0+1 -3 + 0+ 3 -3 + 0+ 4

1 0 0

A A = 0 1 0

0 0 1

A A = I

Hence verified

Q. 13 Solve the following system of equations by using cramersrule.

x + 2 y = 3

4x+ 8y = 12

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Ans. We have

1

2

1 2D = = 8 - 8 = 0

4 8

3 2D = = 24 - 24 = 0

12 8

1 3D = = 12 -12 = 0

4 12

Since D, and all are equal to zero so the given system of

equations has infinitely1D 2D

many solutions.Let y=k then from equations x+2y=3x+2k=3x= 3-2k

Hence, x=3-2k, y= k is the solution of the given system ofequations, where k is arbitrary

real number.

Q. 14 solve the following system of equations by cramers rulex 2y = 4- 3x +5y = -7

Ans. We have

1

2

1 -2D = = 5 - 6 = -1 0

-3 54 -2

D = = 20 -14 = 6-7 5

1 4D = = -7 +12 = 5

-3 -7

So, by Cramers rule, we have

1

2

D 6x = = = -6

D -1

D 5

y = = = -D -1 5

x= -6, y= -5 is required solution.

Q. 15 Solve the following system of equations2x+3y+4z=0x+ y+ z = 02x-y+3z=0

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Ans. We have2 3 4

D = 1 1 1

2 -1 3

= 2(3 +1)- 3(3 - 2) + 4(-1- 2)=8 - 3 - 1 2

D = -7 0

So, the given system of equations has only the trivial solutionsi.e x=0, y=0, z=0

Q. 16 Solve the following homogeneous system of equationsx+y-2z=0..............................................................(1)

2x+y-3z=0...(2)5x+4y-9z=0.(3)

Ans. We have,1 1 -2

D = 2 1 -3

5 4 -9

= 1(-9 +12) -1(-18 +15) - 2(8 - 5)

= 3 + 3 - 6

D = 0

So, the system of equations has infinitely many solutions.Consider eq. (1) & (2). Put z=k in equations(1) and (2), we get

x+y =2k2x+y=3k

Solving these equations by cramers rule

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1

2

1

2

1 1D = = 1- 2 = -1

2 1

2k 1D = = 2k - 3k = -k

3k 1

1 2kD = = 3k - 4k = -k

2 3k

D -kx = = = k

D -1

D -y = =

D

k

-= k

1

x=k, y=k and z=k gives the solution for each value of k.

Q. 17 Use matrix method to solve the following system ofequations

5x 7y =27x 5y =3

Ans. The given system of equations can be written as

5 -7 x 2=

7 -5

y 3

Or A x = B, where

=

5 -7 x 2A = , X , B

7 -5

y 3

So, the solution is given by . So the find we have tofind co factors

-1X = A B

-1A

1+1

11

1+2

12

2+1

21

2+2

22

T

-1

-1

C = (-1) (-5) = -5

C = (-1) (7) = -7

C = (-1) (-7) = 7

C = (-1) (5) = 5

-5 -7 -5 7adj A = =

7 5 -7 5

A = -25 + 49 = 24

-5 71 1A = adj(A) =

-7 5A 24

X = A B

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-5 7 21X =

-7 5 324

-10 + 21 111 1X = =

-14 +15 124 24

11/24X =

1/24

Hence11

x =24

and y=1

24

Q. 18 Show that the following system of equations is consistent

2x y + 3z = 53x + 2y z = 7

4x + 5y 5z =9Ans. The given of equation can be written as

2 -1 3 x 5

3 2 -1

y = 7

4 5 -5 z 9

A X =B

Where

= =

2 -1 3 x 5

A 3 2 -1 , X

y ,B = 7

4 5 -5 z 9

Now

=

2 -1 3

A 3 2 -1

4 5 -5

= 2[-10 + 5]+ 1[-15 + 4]+ 3[15 - 8]

= -10-11+ 21 = 0

So A is singular. So the given system of equation is eitherinconsistent on consistent with

infinitely many solutions according as (adj A) B 0 or (adj A)B= 0 respectively.

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1+111

1+212

1+313

2+121

2+222

2+323

3+131

3+232

3+333

c = (-1) (-10 + 5) = -5

c = (-1) (-15 + 4) = 11

c = (-1) (15 - 8) = 7

c = (-1) (5 - 15) = 10

c = (-1) (-10 -12) = -22

c = (-1) (10 + 4) = -14

c = (-1) (1 - 6) = -5

c = (-1) (-2 - 9) = 11

c = (-1) (4 + 3) = 7

-5 11 7

adj A = 10 -

5

11

7

-5 10 -5

22 -14 = 11 -22 11

-5 11 7 -7 -14 7-5 10 5

adj A) (B) = 11 -22 7

7 -14 9

-25 + 70 - 45 0

= 55 - 154 + 99 = 0

35 - 98 + 63 0

(adj A) (B) = 0

(

Thus AX=B has infinitely many solutions and the given systemof equation is consistent.

Unit III

Chapter 1

Continuity and Differentiability

1. Check the continuity of the function f(x) at the origin :

( ) ; 0

1; 0

xf x x

x

x

=

=

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Ans. We have to show that the given function is continuous at x= 0,so

LHL lim ( ) lim (0 )0 0

f x f hx h

=

lim ( )

0

lim lim 1( )0

1

f h

hh h

h hh h o

LHL

=

= =

=

=

RHL lim ( ) lim (0 )00

f x fh

hx

= ++

lim ( )0

lim lim 1( )0

1

f hh

h h

h hh h o

RHL

=

= =

=

=

Now f(0) = 1Since LHL RHL, so the function f(x) is not continuous at the

origin.

2. Test the continuity of the function at x= 0

sincos , 0

( )

2, 0

xx when x

f x x

when x

+ =

=

Ans. LHL lim ( ) lim (0 )0 0

f x f hx h

= lim ( )

0f h

h=

sin( )lim cos( )

( )0

hh

hh

= +

sin( )lim lim cos( )

0 0

hh

hh h= +

1 1 2

2LHL

= + =

=

RHL lim ( ) lim (0 )00

f x fh

hx

= ++

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( )lim cos( )

( )0

( )lim lim ( )

0 0

1 1 22

Sin hh

hh

Sin hCos h

hh h

RHL

= +

= +

= + = =

And f(0) = 2Since f(0) = LHL = RHLSo the given function f(x) is continuous.

3. Find the values of a and b for which the following

function is continuous at x = 1.

2 1

( ) 1

5 2 1

x a when x

f x b when x

x when x

+ >

= =

0. So f

(x) is continuous

on [1,2] and differentiable on (1,2). Thus both the conditions of

langranges mean value

Theorem is satisfied. Hence there exist some ( )1,2c such that :

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( )

( )

2

(2) (1)'( )

2 1

( ) log

1'( )

(2) log 2, (1) log 1 0

(2) (1)'( )

2 1

log 2 01

11

log 2

1log

2log 22 4

log 2 log log 42 2 2

1 log 2

log 1,22

(2) (1)'

2 1

e

f ff c

f x xe

f x

xf fe e

f ff x

ex

ex

x ee

Now e

e

c e

f ff c =

=

=

=

= =

=

=

=

=

= =

<

<

< >

+ < >

< < >

< < Hence after 2 second and before 5 second velocity will be negative

and acceleration will

be positive.

3. Find the points on curve 2 2 2 3 0x y x+ = where tangent is

i)Parallel to x axis

ii)Perpendicular to x axis

iii)Makes equal angle with both axes

Ans. Given curve is _________________(1)2 2 2 3x y x+ = 0

Differentiating with respect to x we get

2 2 2 0

1

1_______________(2)

dyx y

dx

dyx y

dxdy x

dx y

+ =

+ =

=

i)Tangent is parallel to x axis

tan 0 0

10

1

dy

dx

x

y

x

= =

=

=

Substituting x = 1 in equation (1)21 2 3

2 4

2

y

y

y

+ =

=

=

0

Hence required points are (1,2) and (1, -2).

ii)Tangent is perpendicular to x - axis

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tan90dy

dx =

=

o

So from equation (2)1 1

0

0

x

y

y

= = =

Substituting y = 0 in equation (1)

( )( )

2 0 2 3 0

2 2 3 0

3 1 0

1,3

x x

x x

x x

x

= + =

= =

= + =

= =

Hence required points are (-1, 0) and (3, 0).iii) Making equal angle with both axes

tan 45 1dy

dx = =o

From equation (2)1

1

1

x

y

y x

= =

=

Substituting y = (1 x) in equation (1)

( )

( )

22 1 2 3 02 21 2 2 3 0

22 4 2 0

2 2 1 0

1 2, 1 1 2

x x x

x x x x

x x

x x

x y

= + + == + + =

= =

= =

= = = m 2

Hence required points are ( ) ( ),1 2 2 1 2, 2and+

4. Find tangent and normal of the curve2 2

12 2x ya b

+ = , on

( )cos , sina b .

Ans. Given curve2 2

12 2

x y

a b+ =

Differentiating w. r. t. x, we get

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( )cos , sin

220

2 2

2

2

a b

y dyx

dxa b

dy xb

dx ya

dy a

dx

= + =

=

=

2cos b

b 2sin a

( )cos , sin

cos

sin

1 sin

cosa b

b

a

a

bdy

dx

=

= =

Hence equation of tangent at ( )cos , sina b is

( )

( )

cossin cos

sin2 2sin sin cos cos

2 2cos sin cos sin

cos sin

by b x a

aay ab bx ab

bx ay ab

bx ay ab

= =

= = +

= + = +

+ =

Equation of normal at ( )cos , sina b

( )

( )( )

sinsin cos

cos2 2cos sin cos sin sin cos

2 2sin cos sin cos

2 2sec cos

ay b x a

b

by b ax a

ax by a b

ax by ec a b

= =

= = +

= =

=

5. A balloon, which always remain spherical has a variable

diameter (3

2 32

x + ) . Determine the rate of change of volume with

respect to x.

Ans. Diameter of balloon = ( )3

2 32 x +

( )3

2 34

Radius x = +

Volume of a sphere (V) =4 33

r=

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( )

( )

34 3

2 33 4

392 3

16

x

V x

= +

= +

Rate of change of volume ( )29 3 2 3 216

dv xdx

= +

( )227

2 38

dvx

dx

= +

6. A stone is dropped in to a quite lake and waves move in a circle

at a speed of 3.5 cm/sec. At the instant when the radius of the

circular wave is 7.5 cm, how fast is the enclosed area increasing?

Ans. Let r be the radius and A be the area of the circular wave at any time t

then

( )

( )

2 3.5 / sec.

2

2

2 3.5

7

7 7.5

252.5 /sec.

drA r and cm

dt

dA dr r

dt dt

dA dr r

dt dt

dAr

dt

dA rdt

dA

dt

cm

= =

=

=

=

=

=

=

7. Sand is pouring from a pipe at the rate of The

falling sand forms a cone on the ground in such a way that the

height of the cone is always one sixth of the radius of the base.

How fast is the height of the sand cone increasing when the

height is 4 cm?

312 /sec.cm

Ans. Let r be the radius, h be the height and v be the volume of the sand

cone at any time t.

Then.

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( )

( )

1 23

1 2363

312

236

212 36

1

23

1 1

2 484 3 4

V r h

V h h

V hdV dh

hdt dt

dhh

dt

dh

dt h

dh

dt h

=

=

= =

=

=

= =

=

Thus, the height of the sand cone is increasing at the rate of1

/sec.48

cm

8. Find all the points of local maxima and minima of the function

( ) 3 26 9 8f x x x x= +

Ans. Given y = ( ) 3 26 9 8f x x x x= + . Then 23 12dy

9x xdx

= +

For maxima and minima 0dy

dx=

( ) ( )( )( )

23 12 9 0

2 4 3 0

2 3 3 0

3 1 3 0

1 3 0

1,3

x x

x x

x x x

x x x

x x

x

+ =

+ =

+ =

=

=

= Now we have to check that whether these points are the points of

maxima or minima. So

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( )

26 12

2

26 1 12 6 0

2 1

d yx

dx

d y

dx x

=

= =

=

Hence the function has minimum value at x = 3. The minimum

value of the function at x = 3 is

( ) ( ) ( ) ( )( )

3 23 6 3 9 3

27

f x

f x

= + =

8

54 27+

( )

8

8f x

=

9. Find the maxima and minima of the function ( ) ( )2

1 xf x x e= .

Ans. Given ( )2

1 xy x e=

( ) ( )2

2 1 1dy x xx e x edx

= +

For maxima and minima 0dydx

=

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( ) ( )2

2 1 1 0

2

xx x e

x

+ =

22 2x x +

( )

1 0

2 1 0

20, 1 0

1

xe

xx e

xe x

x

+ =

=

=

=

Now ( ) ( ) ( )2 2

2 2 1 2 1 12

d y x x xe x e x e xdx

= + + + xe

( ) ( )2 2

2 4 1 12

d y xe x xdx

= + +

Now for x= 1

2' 2 4 0 0

21

22 0

21

d ye

dx x

d ye

dx x

= + +

=

=