question bank for first year first sem mathematics - i regulation 2013

7/25/2019 QUESTION BANK FOR FIRST YEAR FIRST SEM MATHEMATICS - I REGULATION 2013

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UNIT-I (2marks questions)1 2

. n e c arac er s c equa on o e ma r x

2

.

Sol.

A I 0The characteristic equatin of A is

1 2 1 0

0 2

10

1 2

00 2

(1)(2 ) 0 0

2 220

232 0

The required characteristic equation is 2

32 0 .

1 2. a n e c arac er s c equa on o

5

.

Sol. 4

1 2e =

5

The characteristic equation of A is 2c

1c

20

c1sumof the maindiagonal elements

14 5

c2 A

2 1

5 4

4 10

6

Hencethecharacteristic

equationis 2(5)(6) 0

256 0

3. Find the sum and product of the eigenvalues of the

matrix 1 1 1

1 1 1 .

1 1 1

Sol.

sumof theeigenvalues sum ofthe diagonal elements

(1) (1) (1)

3

1 1 1product of thee

igenvalues

1 1 11 1 1

1(11) 1(11) 1(11)

1(0) 1(2) 1(2)

4

11 4 7

4. Two eigen values of the matrix 7 2 5 are 0 and 1,10 4 6

find the third eigen value.

Sol.

Given 10,2 1, 3?

sumof theeigenvalues sum of the main diagonal elements

123 11(2) (6)

0 13 3

3 2



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5. Verify the statement that the sum of the elements in the

diagonal of a matrix is the sum of the eigenvalues of the matrix

2 2 3

2 1 61 2 0

sol .sum of theeigenvalues sumof the maindiagonal elements

(2) (1) (0)

1

2 2 3

product of theeigenvalues 2 1 6

1 2 0

2(0 12) 2(0 6) 3(4 1)

24 12 945

6 2 2

2 36. The product of the eigenvalues of the matrixA 12 1 3

is16, Find the thirdeigenvalue. Sol.

let theeigenvalues of the matrix Abe 1,2,3.

Given1216

we knowthat 123A

26 2

2 3 1

2 1 3

6(9 1) 2(6 2) 2(2 6)

6(8) 2(4) 2(4)

32

16332

327. Two eigenvalues of the matrix

8 6 2

6 7 4 are3and0.what is the product of theeigenvalues of A?

2 4 3

sol . given13,20,3?

w.k .tThe sum of theeigenvalues sumof the main diagonal

elements 1238 7 3

3 0 318

315

productofeigenvalues 123(3)(0)(15)08. Find the sum and product of the eigen values of the matrix

2 0 1

0 2 01 0 2

sol .sumof theeigenvalues sum of the main diagonal elements

2 2 2

6

product of theeigenvalues A

12 0

0 2 01 0 2

2(4 0) 0(0) 1(0 2)

8 2

6



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9.Find the characteristic equation of the matrix

eigenvalues.Sol.

Given is a upper triangular matrix.

Hence the eigenvalues are 1,2

1 2and get its

0 2

11.Find the eigenvalues of A given

1 2 3

A 0 2 70 0 3

sol.1 2 3

W.k.t the chacteristic equation of the given matrix is

2(sumof theeigenvalues) (product of

theeigenvalues) 0 2(12)(1)(2) 0

232 0

10.Prove that if is an eigenvalues of a matrix A, then 1

is the

eigenvalue ofA1

proof ;

If X betheeigenvector corresponding to

then AX X

premultiplying bothsides by A1

,weget

A1

AX A1X

IX A1

X

X A1

X

1X A1

X

i.e,A1

X 1X

0 27A 0 0 3

clearly given Ais aupper triangular matrix

Hencetheeigenvalues are1,2,3theeigenvalues

of the given matrix Aare1,2,3

By the property theeigenvalues of the matrix A3are1

3,2

3, 3

3.

3 112.If and are ct e egen va ues o

1 5orm t e

matrix whose eigenvalues are 3and

3

1 7 5

0 2 9 0

0 0 5

Sol.(1)[(2 )(5 ) 0] 7[0 0] 5[0 0] 0(1)(2 )(5 ) 0

1, 2, 5

sumof theeigenvalues 122

25

2

30



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1 7 5

13.Sum of square of the eigenvalues of 0 2 9 is..0 0 5

Sol.

1

The characteristic equatin of A isA I

0

7 5

0 2 9 0

0 0 5

(1)[(2 )(5 ) 0] 7[0 0] 5[0 0] 0

(1)(2 )(5 ) 0

1, 2, 5

sumof theeigenvalues 12225230

4 6 6

1 3 214 .two eigenvalues of A= are equal and they are1 5 2

double thethird.Find the eigenvalues of

A. Sol.

Letthethirdeigenvaluebe

Theremainingtwoeigenvaluesare2,2

sumftheeigenvalues sumofthemaindiagonalelements

22(4) (3) (2)

55

1

theeigenvaluesofAare2,2,1

HencetheeigenvaluesofA2are 2

2,2

2,1

2

15.show that the matrix1 2

2

sa s es s own c arac er s c

equation. 1

Sol.

1 2LetA

2 1

The cha.equation of the given matrix is

A I 0

2S S

20

1

S1sumof main iagonal elements

11 2

S2A1 2 14 52 1

Thecharacteristicis 2250

Toprove A22A 5I 0

A2A.A

1 2 1 22 1 2 1

3 4

4 3

A22A 5I

3 42

1 2 1 05

4 3 2 1 0 1

0 0

0 0



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1 0. =

4 5expressA in terms of A and I using Cayley

Hamilton theorem.

A I 0Sol.The cha.equation of the given matrix is

1 0 1 0

4 5

1

1 0 0

4 5

(1)(5 ) 0

0 (1)(5 ) 0

265 0

By Cayley Hamilton theorem,

A26A 5I 0,A

26A 5I multiply

Aon both sides

A36A

25A 0

A36A

25A

6(6A5I) 5A

36A30I5A

31A30I

17.Write the matrix of the quadratic form

2x28z

24xy10xz2 yz.

Sol.1 1

coeff of x2

2coeff of xy 2coeff of xz

1 1coeff ofy2

Q=2coeff of xy 2coeff of yz

coeff of xz coeff of yz coeff of z

2 2 5

Q= 2 0 15 1 8

18.Determine the nature of the following quadratic form

f x1,x2,x3x122x2

2

sol .The matrix of Q.F is

1Q=

1coeff of x


1coeff of xy

1coeff ofcoeff ofy

2 yz

2 2

coeff of xz coeff of yz coeff of z

1 0 0

0 2=0 0 0

There for the eigenvalues are0,1,2. so find the eigenvalues oneeigenvalue is Zero another two eigenvalues are positive

.so given Q.F is positive semi definite.



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19. State Cayley Hamilton theorem.Every square matrix satisfies its own characteristic equation.

20. Prove that the Q.Fx22 y

23z

22xy2 yz2zx.

Sol.The matrix of the Q.F form,

coeff of x2 1 1


1 1coeff ofy

2Q=2coeff of xy 2coeff of yz

coeff of xz coeff of yz coeff of z 2 2

1

1

1

1 2 1

1 1 3

D1a1

D2

a1

a2

a1D3a2

a3

1 1(ve)

b1

1 1(2 1)

1(ve) b21 2

b1 c1

b2c21(61)1(31)1(12)

2(ve)b3c3

The Q.F is indefinite.


i i bl i


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UNIT II - SEQUENCES AND SERIESPart A

1. Given an example for (i) convergent series

(ii) divergent series (iii) oscillatory series

Solution:

(i) The series

+ is convergent

(ii) 1+2+3+.+n+ is divergent

(iii) 1-1+1-1+ is oscillatory

2. State Leibnitzs test for the convergence of an

alternating series

Solution:

The series a1-a2+a3-a4+. In which the terms are

alternately +ve andve and all ais are positive, is

convergent if

Solution:

(i) The converges or diverges of an infinite

series is not affected when each of its terms

is multiplied by a finite quantity

(ii) If a series in which all the terms are positiveis convergent, the series will remain

convergent even when some or all of its

terms are made negative

5. Define alternating series

Solution:

A series whose terms are alternatively positive and

negative is called alternating series

Eg: + is an alternating series

6. Prove that the series is convergent

(i) and

(ii)

3. State the comparison test for convergence of

series Solution:

Let anand bnbe any two series and let a

finite quantity 0, then the two series converges

or diverges together

4. State any two properties of an infinite series

Solution:

The nth term of the series is an=

Then an+1=

now = =

= =0(

Hence by DAlemberts test, an is convergent


it ti bl t i


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7. When is an infinite series is said to be (i)

convergent (ii) divergent (iii) oscillatory?

Solution:

Let an be an infinite series and let Sn be the

sum of the first n terms of an infinite series then

(i) If is finite the series is said to

be convergent

(ii) If If the series is said tobe divergent

(iii) If not tend to a definite limit or , then

the series is oscillatory.

8. State true or false

(i) If anis convergent, an2is also convergent.

(ii) If the nth

term of a series does not tend to zero

as n, the series is divergent.

The nth term of the series be an=

Then =1/n and =1/n+1

Since , n+1 n ,

anis decreasing and = =0

By Leibnitz test, the given series is convergent. Also the

series formed by the absolute value of its terms is

divergent. Hence the series is conditionally convergent.

10. For what values of p, the series + ++ + will be

(i) convergent (ii) divergent

Solution:

The p-series is convergent if p 1 and divergent if

(iii) The convergence or divergence of an infinite seies

is not affected by the removal of a finite number of

terms from the beginning

(iv) An absolutely convergent series is

convergent Solution:

All are true.

9. Prove that the series is conditionally

convergent

Solution:

UNIT-IIIDIFFERENTIAL CALCULUS

1) Find the curvature ofx

2

y

2

4x6 y1 0Solution:

f x3

2f y

22

f xxf y22f xyf xf yf yyfx

2

f =x2y

24x 6y 1


it ti bl t i


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f x2x 4 f y2y 6

f xx2 f yy2

xy 0

2 2

32 2

32 2

2x4 2 y6 2x4 2 y6

22 y620 22x4 2 2 2 y622x42

1 2 2 y62

(2x4)2

curvature 3 2 2 y62(2x4

2(2 y6)

2

2x42

1/ 2

2 2 y6 (2x4)2

1 2

2 y6 2(2x4)22) What is the formula of radius of

curvature in Cartesianform and parametric

form? Sol:

Cartesian form:(1

y12

)3/

2y2

Parametric form:x '2y '2 3/

2x 'y ''y 'x ''

3 Find the radius of curvature at x=0 on yex

Solution:

Given yex

3

Radius of curvature1

y12

2

y2

y ex

y ex y ]

x0e 1

1 1

y 2ex y 2]x0 e

0 1

1y123/ 2 113/ 2

2 2y2 1

4 Find the radius of curvature of the curvexyc2at( c, c)

Sol:

Given xy c2at (c ,c)

y c2

x

y c2 y c

2 1

x 2 c21 1

y 22c

2

y2 2c

2

2

x3 c3 c

1y1

23/ 2

113/ 2

c.23/ 2

y 2 2 / c 2

c 2.

5 What is the curvature of the curvex2

y2

25at thepoint (4,3) on

it. Sol:Since the given curve is a circle &

We know that radius of given circle is 5 unitsradius of curvature of a circle is equal to theradius of the given circle

5

curvature 1

1

5.


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6 Find radius of curvature of the curvexa

cos , y b sinat any point''

Sol:

x a cos y bsin

x ' a sin y 'b cosx '' a cos y '' b sin

x'2 y'23/ 2 a

2sin

2b

2cos

23/ 2

x 'y ''x ''y ' ab sin2ab cos

2

a2sin

2b

2cos

23/ 2(Qsin2cos21)

ab7 Find the radius of curvature at any point on the

curve r e.

Sol:3/ 2

r2r1

2r

2

rr12r22

Given r e

r e & re

1 2

e2e2

3/ 2

e23/ 2

2

e2ee2e2 e2e22e2

3/ 2 3 3

2 e

221e e2

e2

2.r

8 Find the radius of curvature at y=2a on the

curve y24ax

Sol:

Given y24ax 1

Formula1

y12

3/2

y2

diff 1w.r .to x,

2 yy14a

yy12a 2

y 2a1 y

y1 at y 2a 2a 1

diff 2w.r .to x,2a

yy2

y y 0 yy2

y1 1 1

y2y

2

y

y 2 at y 2a 1/ 2a

113/ 2

23/ 2 2a 23/ 2

1/ 2a1/ 2a

25/ 2ai.e.2

5/ 2.a4a2

9 Find the radius of curvature at (a,0) on y2

a3

x

3

x

Sol: Giveny2

a3

x

3

x




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2 a3 x

2

x

2 yya32x

x21

ya

3

x

2x2y y1

at (a ,0)y1

Hence we find dxdy

xy2 a

3 x

3

x. 2y y2.dy

dx03x

2

dydx

2xy( y23x

2) dy

dx0

dx 2xy

dy 3x2y

2

at (a,0)dx

0

dy


3x2y2

2 ydx

dx

2 yd

2x

dy

2x 2xy xdy

dy2 3x2y22

at (a,0)

d2x

3a200 2a0

6a

3

2

dy2 3a202 9a4 3a

1dx 2

3/ 2

dy 103/ 2

3

ad2x 2 2

dy2 3a

3

2a

10 Find the radius of curvature atx2on the curve

y 4sinx sin 2x .Sol:

y 4sinx sin 2x

y1dy

dx4cosx 2cos 2x

y 2d

2y

4sinx 4sin 2x dx

2

at x / 2,y14(0)2cos2at

x / 2,y2 4(1)4sin 43/ 2

1(2)2

3/ 2

1y1

y2 4

143/ 2 53/ 2 5.51/ 2 5 54 4 4 4

5 5 Qis ve

4




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11 Define the curvature of a plane curve and what isthe curvature of a straight line

Sol:

The curvature of a plane curve at K dds

The curvature of a straight line is zero.12 Find the radius of curvature at any point (x,y) on the

curve yclogx

sec

c

Sol:x

y c logsec

c

y1c.1 x x 1 tan xtan . sec c

x csec

c c c

c

y21

sec2 x

c c

tan 2x

3/ 2 2 x 3/ 2

2

3/ 2 1 secc c1y1

y2 1sec2x 1

sec2x

c c c c

sec3 x

xcc. c.sec

x csec

2

c

13 Find the radius of the curve given byx3 2cos,

y 4 2sinSol:

x 32cos y 42sin

dx 2sin 2cos

d2cos

ddy

cotdx 2sind y

d dy d

d

cot1

2

dx dx d 2sind dx

cos ec2

cos ec

3

2sin 2

1y3/ 2 3/ 2

1cot 3

1

cos ec

2y2

1 3 1 3

2 cos ec 2cos ec

2

14 Write the formula for centre of curvatureandequation of circle of curvature.Sol:

Centre of curvature:xxy1

1

y12

y2(1y )

y y &y2

Circle of curvature: xx2yy22

15 Find the centre of curvature ofyx2 of the origin.

Sol:




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The centre of curvature is given by

y1 1y12

X x 1( y1)2, Yy22

Given y x ;y 2x ; y2

2.1

at (0,0), y10

at (0,0), y22

x 021(0)

2 xX

y 102 y

1Y

2 2

at (0,0),X0

at (0,0),

1Y

2

0,1

Centreof curvatureis

216 Write properties of

evolutes. Sol:(i) The normal at any point of a curve touches the

evolute at the corresponding Centre of curvature.

(ii)The length of an of the evolute is equal to the of

curvature at the points on the original curve

corresponding to the extremities of the arc

(iii)There is only one evolute, but an infinite

number of involutes.

17 Find the envelope of the family of straight

lines y mx am2, m being the parameter.

Sol:

Given ymxam2

Diff. partially w.r.to m, we get,

0 x2am

m

2ax

y mx am2

x x 2y

2ax a

2a

x2

ax

2

2a 4a2

y x

2

x2

x

2

2a 4a 4a

x2 4ay is the required envelope

18 efine envelope of a family ofcurves. Definition:

A curve which touches each memberof a family of curve is called the envelope of that

family curves.

19 efine Evolute and Involute.

The locus of the centre of the given curve iscalled the evolute of the curve.

The given curve is called the Involute of its evolute.

20 Find the envelope of the family of lines

x

tyt2c,

t being theparameter. Sol:

Given family of lines can be written as,

yt22ct x 0 --------- (1)

The envelope ofAt2BtC0 is B

24AC

0 From (1) we get A = y, B= -2c, C =x

Putting these values in (2) we get,




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(2c)24 yx

0 4c24 yx

0 c2xy 0

(1)2(2)

2we get,

x y2 x y

2 2 2

cos sin sin cos 1 0a b ba

xy c2

This is required envelope.

21 Find the Envelope of the family of Straightlinesa

x2cos

2

2

sin2

a2

b2

x 2sin2y2cos2a2 b2

2xycossin

0 2xycossin

y mx m, where m is a parameter.Sol:

Given y mx a

(1)m

x2cos

2sin

2

y2

x2y

21

a2b

2

y2 sin

2cos

21

b2

ym m2x a

m2

x ym a0This is a quadratic in m

So the envelope is B24AC 0

Here A x ,B y ,c a

y24ax 0

y24ax

22 Find the Envelope of the family of lines ax

cos

b

ysin1,being the parameter

Sol:

Given,xcosysin1

1a b

'' we getdiffpartially (1)w.r .to

sin y cos0 (2)

a b

23 Find the envelope of the straight lines

x cosy sina sec,whereis the parameter.

Sol:

Givenxcosysinasec1

Dividing equation (1) by cos

we get,

x y tana cossec

a sec2

a(1tan2)a tan

2y tan

2(a x)0

Which is a quadratic equation in

tanHere A=a, B=-y, C = (a-x).

B24AC 0,

y24a (a x)0




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24 Find the envelope of ymx a2m

2b

2,where m is a

parameter.Sol:

y mx a2m

2b

2(

ymx)2a

2m

2b

2

y2m

2x

22mxy a

2m

2b

2

m2(x

2a

2)2mxy y

2b

20

Which is a quadratic equation in m.

Hence the envelope is B24AC0

Here A= (x2a

2), B=-2xy, C = y

2b

2

4x2y

24(x

2a

2)( y

2b

2) 0

x2(cos

2sin

2)y

2(sin

2

cos2)a

2x

2y

2a

2

26 Find the envelope of the family given byxmym1

,

m isparameter. Sol:

The given equation can be

written as m2y mx 10,

Which is quadratic equation inm

, Here,Ay, B x, c1

Hencethe envelopeis B24AC 0

x2y

2(x

2a

2)(y

2b

2)0

x y x y b x a y a b 0

i.e,b2x

2a

2y

2a

2b

2

x2

y2

12 2

x2

27 Find the envelope of

parameter.Sol:

4 y0x

24y

y mx 1m2 where m is a

a b

25 Find the envelope ofxcosysina, whereis

a parameter.

Sol:

Givenxcosysina (1)

Diff w.r.to

x siny cos0 (2)

Eliminate between (1) and (2)

1 2

2, we have

(xcosysin)2(xsinycos)

2a

20

2

x cos y sin 2xy sincos

2 2 2 2a

x sin y cos

2xysincos

Given ymx 1m2

y mx 1m2

Squaring onboth sides (y mx )21

m2y

22mxy m

2x

21m

2

m 2(x 21)2mxy y210.

Here A x21,B 2xy ,C

y21.B

24AC 0

(2xy)24(x

21)( y

21) 0

4x2y

24(x

21)( y

21) 0




16/44


UNITIV

FUNCTIONS AND SEVARAL VARIABLE

PART-A

1 x y u u 1. If ucos , . .x y cot u.

x y 2x y

Proof:

Given f (x ,y )cosu

x y

x y

As f is homogeneous function of degree n 1,2

it is satisfiesthe Euler 's equation.

x f y f nfx y

x(cos u)y(cos u) cos u.

x y 2

x (sinu )u y (sinu ) u cos u.

x y 2

x uyu 1 cos u.x y 2 sin u

x uyu cot u.

x y 2

As fis a homogeneous function of order n=2, it

satisfies the Eulers theorem.

x f y f nf

yx

x(tan u) y(tan u) 2 tan u.x y

x (sec2u )

uy (sec

2u )

u2 tan u.

x y

xu y u 2sin u 1 .x y sec

2ucos u

2sin ucos2u.

cos u

2sin u cos u.

xu

yu

sin 2u.x y

3. If ulogx

3 y

3 u

yu

2, P. .x

xy x y

2. If utan 1x

3y

3 u

yu

sin 2u., . .x

x yx y

Solution:

Given f (x ,y )tan u x

3y

3

x y




17/44


3 3 2

2 2

2 2

x y x y .1x 2xf y x

Solution: Given ulog

x y

2

x 2y 22

x 2y22

x

Similarly,

2f x

2y

2

x

2y

2

2

y

2

Let f eu x

3y

3

x y

As f is homogeneous functionof degree n 2,

it is satisfies the Euler 's equation.

xf

yf

nfx y

2f

2f

0x

2 y

2

5. If usin1x

tan1 y x

uy

u0s ow t at

y x x y

Solution: Hereuis a homogeneous function of degreen= 0.

x(e

u)

y(e

u)

2eu

x y

x (eu) uy (e

u) u 2e

u.

yx

x uy u 2.

yx

Hence the proof.

4. If f(x,y)log x2y

2 , show that

Solution: Givenf(x,y)log x2

y2f

12logx

2y

2

f 1 2x x

x 2x

2

y

2

x

2

y

2

2

2f

0 .x

2 y

2

Using Eulers theorem,xu

y

u

0

x y

6. If u x y z show thatxu y u z u0 .

y z x zx y

Solution: Given uxyz

y z x

u

1

z

x y x2

xu x z .........(1)x y x

u

x

1

y y2 z




18/44

q g p

yu

x

y

..........(2)y y z

z2 1 z 2

R.H .S r r

2

u y1z z2 x

zu

y

z

..........(3)z z x

Add eqn. (1),(2) & (3),we get

z z cos sin

x y

z2

2 z cos x y

2 z z

( sin) (cos)

x y2

z zsin 2 sincos

x y

2

xu

yu

zu

0.x y z

z

x

2 2 zsin

y

z zcos 2 sincos

x y

7. If zf(x,y)

z 2 z

wherex r cos,y r sin.Show that

2 z 2 1 z2

2

z2

x

2

z

y

x y

Solution:Wkt,

z

1 z

r

r r

z z x z yr x r y r

z cos zsinx y

z x z y

x y

z( rsin) z (rcos)x y

z

sin z

cos

yx

Thus, R.H.S = L.H.S

8. If zfx ,y ,x e ucosv , y e usinv show that

x

z

y

z

e

2u z

.v u y

Solution: Given z f x ,y ,xeucos v, yeusin v

z z x z y

u x u y u

z

eucosv

ze

usinv

x y




19/44

q g p

yz

z

yeucosv

zye

usinv

u x y

e2usinv cosv

ze

2usin

2v

z....(1)

x y

z

z x

z y

v x v y v

z u

sin v

z u

cos ve ex y

x z zxeusinv zxe ucos v

v x y

e2usinv cosv

ze

2ucos

2v

z....(2)

x y

(1) (2)

z z 2 u

sin2 2

z

x y e v cos vv u y

e2u zy

Hence proved.

9. If uxlog(xy)wherex3y

33xy 1find .

dx

Solution:

Given ,u x log (xy ) &x3y

33xy 1

du u u dy ....(1)

dx x y dx

ux

1( y) log (xy)

x xy

u

1log (xy)

x

u x 1x x

y xy y

consider ,x3y

33xy 1

Diff. w.r.to x,

3x2

3y2

y3y3x

y0

dx dx

3x23y3y23x

dydx0

dy

3x23y

x 2y

dx 3 y 3x y x

du x x 2y

(1) dx 1 log(xy) y y 2x10. Finddy whenx

3y

33axy

dx

Solution:

Let f (x ,y )x3y

33axy

f3x

23ay;

f3y

23ax

x y

dy f x 3x23ay x

2ay

dx f 3y 3ax y axy

11. Find dywhen y sinx x cosy

dx

Solution:

Given y sinx x cosy

y sinx x cosy 0




20/44

q g p

Let f (x ,y)x cosy y sinx

f cos yycosx& f x siny sinx

x y

dy f x cos yycosx

dx f y x siny sinx

dy cosy y cosxdx x siny sinx

12. If ux2y

2z

2and x e

t,y e

tsint ,z e

tcost find du

dt

with actual substitution.

Solution: Givenux2y

2z

2,xe

t,ye

tsint,ze

tcos t

du u dx u dy u dz

dt x dt dt z dt

2x et2 y(e

tsin te

tcos t) 2z(e

tcos te

tsin t)

2 etxy(sin tcos t) z(cos tsin t)

2 et e

te

tsin

2te

tsin tcos te

tcos

2te

tsin tcos t

2e e e sin tcos t

2e 2e

13. Find if usin (x/y), wherexet,yt

2.

dt

Solution:

duu .dx u.dydt t dt y dt

x 1cos . e cos x x 2t

2

y ydu e e 2e

dt

cos

t

2 2

t

3

u

u

u14. If u = f( y z , z x , x y ) find .x z

Solution: Given ufyz,zx,xy

Let r y z,s z x and t x y

u u r u s u tx r x s x t x

u

(1) u

(1) .....(1)s t

u u r u

y r y s y t y

u (1) u (1) .....(2)

r t

u u r u

z r z s z t z

u (1) u(1) .....(3)

r s

(1) (2) (3)u

u

u

0x y z

15. Find the minimum value of F = x2+y

2subject to the

Constraint x=1.

Solution: GivenF = x2+y

2

= square of the distance from the

origin The minimum of F is 1.




21/44

q g p

16. Define Jacobian.

If uand vare functions of the two independent variables

u u

x and y, then the determinant x y is called the Jacobian

v

v

y y

18. If u2xy,vx2y

2 and x r cos,y r sin,

evaluate(u, v)

(r,)

Solution:

u,v u,v x,y

r, x,y r,

of u ,vwith respect tox,y.

(x, y)

17. Find the Jacobian (r,)

It is denoted by

x,y

.u,v

if x r cos,y r sin.

u u x x

x y

r

v v y yx y r

Given u 2xy v x2y

2

Solution: Givenxrcos

xcos

r

y r sin

r

x,yx x

rr, yy

r

x,y

r,

y

r sin

ysin

r

yr cos

cos r sin

sin r cos

r cos2r sin

2

r cos2sin2

r

u

2 y v

2xx x

u2x

v 2 y

y y

Given xrcos y r sin

xcos

ysin

r r

y r sin y r cosr

u,v

2 y 2x cos r cos

r, 2x 2 y sin r cos

4 y24x2r cos2r sin2

4x2y2r cos2sin2




22/44

4 r2 r

u ,v

4r

3

r,

19. Ifu2, v

x2then find

(u, v).

x y (x, y)

Solution:

Given u y

2

v x

2

x y

u

y2 v

2x

x x2 x y

u

2 y v

x2

y x y 2

u ,v

u u

2 2 y

x y x

x

x ,y v v 2x

x2

x y y y

y2 x

2 2 y 2x

x

2

y

2

x y

14 3

u ,v

3 x ,y

20. Ifxu(1v), yuvcomputeJ&J, and proveJ.J1.

Solution: Given xu1vand y uv

x

1vy

vu u

x u y uvv

x ,yx x

J u vu ,v y y

u v

1vv u

u (1v )(uv)

u uv uvu

x ,y ' u ,v 1

J u& J

uu ,v x ,y

To prove: J .J= 1

' x ,y u ,v 1J J u

u ,v x ,y u

J J'1

21. Ifxrsincos, yrsinsin, z r cos.Find J.

Solution:

Given x r sincos,y r sinsin,z r cos

x x x

x ,y ,z

r

J

y y y

r ,, r y y y

r




23/44

sincos r coscos r sinsin

sinsin r cossin r sincos

cos r sin 0cos(r

2cossincos

2r

2cossinsin

2)

r sin(r sin2cos

2r sin

2sin

2)

r2sincos

2sin2cos2r2sin3(sin2cos2)

r2

sinsin2

cos2

J r

2sin

22. Expand f(x, y) exy

in Taylors series at(1, 1)up to

second degree.

f x ,y f a ,b 1 fxa ,b x a f ya ,b y b1

1 fxxa ,b x a 22fxya ,b y b x a

...2

f yya ,b y b1x 1 y 1

exy

1e x 1

24 x 1 y 1 y 1

2

2

23. Find the Taylors series expansion ofexs in ynear the

up to the first degree terms.point 1,

4

Solution:

Solution:

Given f x ,y exy

and the po inta 1,b 1

f

x ,y

exy f

1,1 e

f

xx ,y

e

xy y f

x1,1 e

f yx ,y exy

x f y1,1 e

f x ,y exsin y 1,f4

fxx ,y ex

sin y 1,

x

4

f yx ,y ex

cos y 1,

y

4

e1

sin

4

e1

sin

4

e1

cos

4

e1

2

e1

2e

12

fxxx ,y exy

y2 f xx1,1 e

fxyx ,y exy

(1)y exy

(x ) fxy1,1e e 2e

f yyx ,y exyx2 f yy1,1 eThe Taylors series is

The required expansion is

f x ,y f a ,b 1f xa ,b x a f ya ,b y b1


24/44

fx ,y

f 1,

x 1 f

f

x , y y ,4 4 4 4

ex

sin y 1

1 x 1 y

4e 2

24. Write condition for finding maxima and minima. Necessary Conditions:

The necessary conditions for f(x, y)to have a maximum

f f

or minimum at (a, b)are that 0 and 0 at(a,b)x y

Sufficient Conditions:

Let r fxx a,b;s fxya,band t f yya,b

(i) If rts20 and r0 at (a, b) , then fis maximumand

f (a, b) is maximum value

(ii) If rts20 and r0 at(a,b) , then fis minimumand f(a, b) is minimum value.

(iii) If rts20 , then fis neither maximum norminimum

at (a, b).

(iv) If rt s2= 0 , in this case further investigation are

required.

25. Find the stationary points of

f (x,y)x3y

33x 12y 20.


25/44

Solution: Given f(x,y)x3y

33x12y20

fx3x23fy3y

212For stationary points fx0, fy0

3x230x

21x1 3y

212 0 y

21 y2

The stationary points are (1,2), (1,-2),(-1,2) & (-1,-2).

26. Find the stationary points of zx

2

xyy

2

2xy.

Solution: Givenzx2xyy

22xy

zx2x y 2 ,zy x 2y 1For stationary points fx0, fy0

2xy2 and x2 y1 Solving x =1, y =0

The stationary point is (1,0)

27.Find the maximum and minimum values ofx2xy y

22x y

Solution: Given f(x,y)x2xyy22xy

fx2x y 2 f y x 2y 1

fxx2 f yy 2

fxy 1

At maximum and minimum point: fx= fy= 0(1,0) may be maximum point or minimum point. At (1,0): fxx. fyy( fxy)2= 4-1 =

3 > 0 & fxx=2 > 0

(1,0) is a minimum point

Minimum value = f(1,0)= -1incseitquestions.blogspot.in

28. A flat circular plate is heated so that the temperatureat

any point (x,y) is u(x,y) = x2+2y

2-x. Find the coldest

point on the plate.

Solution: Given u x22y

2x

u x 2x 1 u y 4

xx2 uyy4u

xy0

For stationary points ux0 2x1 0 x

2

u y0 4y 0 y 0

The point is

1,0


26/44

At 1,0 u u uxy 2 80&u 2 0xx yy xx

2

The point1

,0

2

is the minimum point.

1,0Hence the point

2

is the coldest point.

29. Find the shortest distance from the origin to the curve

x28xy 7y

2225.

Solution:

Let f x 2y 2&x 28xy 7y2225

f x 2 y 2x 2 8xy 7y2 225

fxx0 1x4y0 (1)f

y

y

0 4x1 7

y 0 (2)

Solving (1) & (2) = 1, = 1

9

If 1x 2y &5y2225

(no real valueof y)

If y 2x x 9 5, y 20

1. y3x1x2

x3

(12. (i) Find the Jacobian

u ,v ,w

, ifx ,y ,z x y z u ,y z u v ,z u v w

(ii) If ux2y

2, v2xy. f(x, y) (u, v) show that

2f

2f 2 2

2

2

x y

x2

y2

u

2 2

v




27/44

UNIT-V

PART-A

1. Evaluate1dx

xe

yx

dy0 0

Sol:

=1 e

y xdydxLet I

00

= ey xx axdx eax

0

dydx Q e

0 a1

xey x x

dx0

0

1(xe

x xxe0)dx

0

1(xex) dx

0

1x (e 1)dx

0

(e1)1x dx

0




28/44

(e1) x2 12 0

(e1)1

0

2

e 1

2

2. Evaluateb a

dxdy

1 1 xy

Sol:b adxdy dx

Let I logx

1 1 xy x

adx dy

y1 x 1

logx1alog y1

b

log alog1log blog1(log a0)(log b0) (Qlog1 0)

log alog b

3. Evaluatea a

2x2dydx

0 0

Sol:

Let I a a 2x2dydx

0 0

a

y0a2

x2 dx

0

a 0dxa2x20

a a

2x

2dx

0

xax a x a 1

sin2 2

0

2

a2 2 2

a a sin1 0 sin1(0) (1)

2 2 2

sin1

(0)0,Qsin 0 0

sin 1 sin1 (1)

2 2

a2

0 00 2 2

a

2

4

4. Evaluate1 xy(x y)dxdy

00

Sol:




29/44

Let I 1 x

xy(xy)dxdy

0 0

1 x x2yxy2dxdy0 0

1 x(x2yxy2) dydx(correct form)00

xy3

1 x y

x

dx2 30 0

x2 2 x 31 x x

dx

2 30

1 x x

5 2

Qx3 2 5 2

dx x x2 30

1x

3dx

1x

5 2 dx

0 2 0 3

1 x1

1x7 2

1

2 34

0 7 2 0

1 10 1 2 02 4 3 7

1

8212

2116168

16837

in

2 y dxdy.

0 0x2 2Sol:

y2 dxdye

0 0x2 22 y

0 0x2

dx 2 dy

2 1tan 1 x

1y y

21 1 y

1

tan

y y

21tan1(1) dy

1 y

21

dy

1y 4

21dy

41y

log 2 log1

4 2 acos

6. Evaluate r2drd

0 0

Sol:

y

dy

0tan

1(0) dy

4log 2 (Qlog1 0)




30/44

2 acos

Let I r2drd

0 0

2 r3a

cos

d

0 0

2a

3cos

3d

0 3

a32

cos d

3 0

a3 3 1

3 3

2a

3

9

sin

7. Evaluate

2

r ddr0 0

Sol:

Let I 2 sin

r ddr

0 0

2 sin

r dr d correct form0 0

sin

2 r n 1n 3 d

2 ......1,if nis odd 0 2 0n n 2

Q cosn

n 1n 3

2sin

20 ...... ,if nis even

0 d

n n 2 2 20

n 1n 3

1

2 2 ......1,if nis odd

0sin

2

0sin

n n n 2

2 d n 1n 3 ...... ,if niseven

n n 2 2

1.1 . 2 2 2 8

cos

8. Evaluate r dr d0 0

Sol:




31/44

cos

Let I r dr d0 0

r2rcos

d

20 r 0

0 d cos220

cos

2d

0 2

1 2

2cos d0

11cos 2d

2 0 2

1 sin 2

4

2 0

1 sin 2

0

4

2

2

1 0

4 4

ey dxdy is difficult to solve

0x

y

But by changing the order we get,

yeydxdy0 0 y

ey

y x0

ydy

0

ey( y0) dy

0

eydy

0

9. Why do we change the order of integration in multipleintegrals? Justify your answer with an example?

Sol :

Some of the problems connected with double

integrals,which seen to be complicated,can be made easy

to handle by a change in the order of integration.

Example:

ey

0

ey0

(e e

0) (0 1) 1

10. Expressa a

2x2 2

)dxdyin polar co-ordinates

0 y ySol:




32/44

The region of integration is

bounded by y 0,y a ,x y ,x a.

Let us transform this integral in polar co-ordinates by taking

x r cos,y r sin,dxdy rdrd.

Consider the limitsx y ,x a ,y 0.

If y 0 r sin 0 r0,sin 0

r0,0

If x y r cos r sin cos 1sin

tan 1

a4

If x a r cos a r

cosra sec

a a

asecx2 4 (rcos) rdrd

2

y

2 2

r sin2 3 2

0 y 0 0 r cos 4 asec r cos drd

r 2(cos2sin2 3 20 0 )

4 asec

cos2drd

0 0

11 .Find dxdyover the region bounded byx 0, y 0,x y 1

Sol:

Givenx 0, y 0 &x y 1

The region of integration is the triangle.

Herex

varies fromx0 to x1y

y varies from y 0to y 1

I dxdy

R

1 1y

dxdy0 01

x 10ydy

0

1(1y) dy

0

y y2 12 0

11

21

2

12.Find the area of a circle of radius a by doubleintegration in polar

Co-ordinates

Sol:

The equation of circle whose radius is a is given by




33/44

r 2a cos

The limits for

r :r 0to r 2a cos

:0 to 2

Area 2 upper area22 acos

2 rdrd0 0

2 r2 2 acos

2

0 2

d

0 2

4a2cos

2d

0

2

4a

2

cos

2

d

0

4a2 2 1

2 2

4a21

a2

2 2

13. Define Area in polar Co-

ordinates Sol:

Area=rdrd

R

14. Express the Volume bounded

byx 0,y 0,z 0and x y z 1

in triple integration.

Sol:For the given region

z varies from 0to 1x2y

2

y varies from 0to 1x2

x varies from 0to 1

1x y

I 11

2

dzdydx0 0 0

15. Evaluate 2 3 2xy2z dzdydx

0 11

Sol:2 32

Let I xy zdxdydz

0 1 1

2 3 2dy

2 x x

0 1 1

x 2 y z2 3

02

11

40

27

1 4

1

2 3 3 22

(2)26 3

26 3

2

16. Find the volume of the region bounded by the

surface y x2,y x

2and the planesz 0,z 3

Sol:




34/44

y2x (1)

x2y (2)

Substituting (2)in (1)we

get x4x

x4 x 0

xx

3 1 0

x 0,1

1 3x

Re quired volume dzdydx

0x20

1

x z 3

0dydx

0x2

1x

3dydx0x

2

1

3 yx2xdx

0

31 x

2 dxx

0

x3 2 x3

1

3

3 2 3 0

2x1

x3

3

0

32(1) 13

2 1 1

17. Sketch roughly the region of integration for1x

f(x, y) dy dx.

0 0

Sol:

The region of integration is bounded byx0,x1, y0, yx

Here x varies from x 0to x 1y

varies from y 0to y x

a a x

18. Sketch the region of integration dydx.0 ax x

Sol:

Given x varies from x = 0 to x = a

y varies from y a2x

2 to y ax x

2

i.e., y2x

2awhich is a circle with centre (0,0)

andradius a.




35/44

x 2y 2ax a

2 a

2

y 20

2

4

a2 2

a2

i.e., x

2

y

4

This is a circle with centre

(a/2,0) and radius a/2.

19. Change the order of integration ina x

f(x, y) dydx

0 0

Sol:

Givena x

f(x, y) dydx

0 0

The region of integration is

bounded byx 0,x a ,y 0,y x

i.e.,x varies from x 0to x a represents Vertical path

y varies from y 0to y x represents Vertical strip

Now changing the order of integration we get

x varies from x y to x a represents Horizontal strip

y varies from y 0to y a represents Horizontal path

a x

f (x ,y )dydx a a

f (x ,y )dx dy

0 0 0 y

20. Sketch roughly the region of integration for the following

double integrala a 2x2f(x,y)dxdy

0 0

Sol:

Given that x varies from x 0to x a

y varies from y 0to y a2x

2i.e.,y

2x

2a

2

x2y

2a

2

Which is a circle with centre (0,0) and radius a




36/44

11y

21.Change the order of integration in f (x ,y )dxdy0 0

Sol:

Given x varies from x 0to x 1y i .e.,x y 1represents

Horizontal y varies from y 0to y 1represents Horizontal path

The region of integration is bounded by y0, y1,x0,xy1

x varies from x 0to x 1represents Vertical path

y varies from y 0to y 1x represents Vertical strip

After changing the order of integration limits of x and y

becomesx 0,x 1,y 0and y 1x .11y 1 1x

i.e., f (x ,y )dxdy f (x ,y )dydx0 0 0 0


37/44


UNIT-IPART-B

= Find all the eigenvalues and eigenvectors of the matrix



2 2 1

1 3 1

1 2 2

2 2 3

2 1 6

1 2 07 2 0

2 6 2

0 2 5

2 1 2

1 2 14. s ng ayey am on eorem n w enA=

1 1 21 2 2

5. Using Cayley Hamilton theorem find A1

1 3 0When A

0 2 11 0 3

6. Using Cayley Hamilton theorem find A1

2 1 1n

1 1 1

6. Using Cayley Hamilton theorem find the inverse of the matrix

1 1 4

7.Find aA1

3 2 1 , Using Cayley Hamilton theorem.

2 1 1

1 0 3

8 17A

3 0 8


38/44


3 1 1

1 3 1 by means of an orthogonal transformation.8. Diagonalise the matrix A=

1 1 310 2 5

2 2 39.Reduse the matrix to diagonal form.

5 3 53 1 1

10. Diagonalise the matrix 1 5 1 by means of an orthogonal

1 1 36 2 2

2 3 111. Diagonalise the matrix by an orthogonal

2 1 3transformation.

= Reduce the quadratic form Q6x23y

23z

24xy2yz

4zxinto canonical form by an orthogonal transformation.

= Reduce the quadratic form 8x127x2

23x3

212x1x28x2x34x3x1to

the canonical form by an orthogonal transformation and hence show

that it is positive semi-definite.

= Reduce the quadratic formx125x2

2x3

22x1x22x2x3

2x3x1to the canonical form by an orthogonal transformation

= Reduce the quadratic formx2y

2z

22xy2 yz2zxto

canonical form by an orthogonal transformation

8 6 2

16. Find all the eigenvalues and eigenvectors of the matrix 6 7 4

2 4 3

Cseitquestions.blogspot.in


39/44

17.Obtain the orthogonal transformation whish will transform the

Quadratic form Q2x1x22x2x32x3x1into sum of squares.

UNIT-II

PART-B

1. Show that converges to 0

2. The series is convergent and its sum is 1.

3. Prove that the series 1-2+3-4+. Oscillates infinitely

4. Prove that the geometric series 1+r+r2+.+rn-1+ converges, if 0 and divergesto .

5. Examine the convergence of the series +

6. Test the convergence of the series

7. Using the integral test, discuss the convergence of the series



10. For what values of x are the following series convergent.




40/44

UNIT-III

PART-B

1. Show that the radius of curvature at any point (x , y) on theasteroid x

2/ 3y

2/ 3a

2/3is3(axy)

1/3

3a,

3aon2. Find the radius of curvature at

2 2

x3y

33axy

3. Find the radius of curvature at the (a, 0) on the

curvexy2a

3x

3

4. (i) If

is the radius of curvature at any point (x ,y) on

y

ax 2 2/ 3 x ythe curve , P.T

a x a y x

(ii)Find the radius of curvature of the curve

r a (1cos)at 2

5. Find the equation of the centre of curvature of the

rectangular hyperbolaxy=12 at the point (3,4)

6. Find the equation of the circle of curvature at

(c,c) on xy = c2

7. Find the circle of curvature of the curve x y a

= Find the envelope of the family of

ax

by

1, where a and b are connected by the

relation a2b

2c

2

(ii)x

y

1, where a and b are connected by the a b

relation abc2

(iii) x2

2

1, where a and b are arbitrary constants

a b given by a

2b

2c

2

10. Find the evolute of Parabola, Ellipse,

Hyperbola, Rectangular hyperbola, Astroid.

11. Show that the evolute of the cycloid

x a(sin),y a(1cos)is another

cycloid12. Find the evolute of Parabola, Ellipse, Hyperbola,

Rectangular hyperbola, Asteroid. considering it

as the envelope of its normals.




41/44

x3

a3cos

a,a

at the point

4 4.

8. Find the envelope of family of curves

3y

3 1,

b sin

being the parameter.

(i) sinxsinysin(xy) (ii) sinxsinysin(xy)

6. Find the volume of the largest rectangular parallelepiped

that can be inscribed in the ellipsoid. (OR)

Find the volume of the largest parallelepiped and which can

inscribed in the ellipsoidx2

2

z21

a2 b

2 c

2

UNITIV

PART-B

1. (i) If Z = f(x, y)and u ,v other two variables such that

u lx my ,v ly mx show that

2z

2z 2 2

2z

2z

(l m

x y u v

If u sin1x

3y

3 u

yu

2 tan u, . .x

xy x2. Given the transform u = e

xcos yand v = e

xsinyand that

is a function of uand vand also ofxand y, prove that

2 2 2 2 2 2 (u v

x

2

y

2

u

2 2

v

3. Find the maxima and minima value of

(i) f (x,y)x3y

33axy

(ii) f (x,y)x

3y

2(1x y)

4. In a plane triangle ABC find the maximum

value of cosAcosBcosC.

5. Find the maximum and minimum value of

7. (i).Find the dimensions of the rectangular box with out top of

maximum capacity whose surface is 108 sq.cm?

(ii) A rectangular box open at top to have the volume of

32 cube feet .find it dimensions, if the total area is

minimum.

8. If x r cos,y r sin.Prove that the Jacobian

J (x, y)r and J (r,) 1

r(r,) (x, y)

9. Find the Jacobian( y1, y2, y3) if, y1

x2x3 , y2x1x3

x x(x,x ,x)1 2 3 1 2

y x1x23

3

(OR)

Find the Jacobian(u, v, w), if u yz , v

zx,w

xy

(x, y, z) x y z

10. (i) Expand ex

cosyabout 0, up to third using Taylors

2

series.




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(ii) Expand ex

log(x + y)in powers of x and y up to terms of third

degree using Taylors theorem.

(iii) Expand ex

sin yabout (x ,y) up to third using Taylors

series

1 x y u u 1. If ucos , . .x y cot ux y 2x y

(ii) If ux2y

2z

2 andx e

t, ye

tsin t, ze

tcos t, find

du

dt

12. (i) Find the Jacobian

u,v,w

, ifx,y,z x y z u ,y z u v ,z u v w

(ii) If ux2y

2,v2xy. f(x, y) (u,v) show that

2f

2f 2 2

2

2

4(x y

x2

y2

u

2 2

v

UNIT-VPART-B

1. Evaluate e(x

2y2

)dxdy

0 0

Evaluate by changing the order of integrationa

a y

3. y dxdy0 a y

12y

4.Change the order of integration in the integral xy dydx and0 2

evaluate it.

5.Change the order of integration and evaluatea a

(x2y

2) dydx

0x

6.Chan e the order of inte ration and evaluatea2a

x

xy dydx

0x 2 a

7.Change the order of integration and evaluate1 2x

xy dydx0 x

8.Evaluate by changing into polar co-ordinates the integrala a

x2 dxdy

2 2

0 y x

9. Change into polar co-ordinates and evaluate e

(x

2y2

)dxdy

0 0

10.Evaluateaa

2x

2 by changing into polar co-ordinates

11. Evaluate

for which

2. Change the order of integration in the double integral

2 a 2ax

dxdy

2axx2

0y

xy dxdy over theregion over the positive


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quadrantx y 1

5. Find by double integration the area enclosed by the curve

x2 3 y2 3 a2 3

6. Find the area common to y24axandx

24ayusing

double integration

7. Find using double integration the area of the cardioid

r a(1cos)log 2x x y

15. Evaluate ex

yzdxdydz

0 0 0

16. Find by Triple integral the volume of Tetrahedronbounded by the

planesx0, y0, z0 and axb

ycz1

17.Find the volume of that portion of the ellipsoidx

2

y2

z21

a2 b2 c2

which lies in the first octant


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question bank for first year first sem mathematics - i regulation 2013

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