IIT Jam math 2016 solutions BY Trajectoryeducation

TRAJECTORY EDUCATION

1

JAM-2016 MATHS-SOLUTIONS

SECTION-A MULTIPLE CHOICE QUESTIONS (MCQ)Q.1 – Q.10 carry one mark each.

1. The sequence {sn} of real numbers given by

2sin sin sin2 2 2...

1 2 2 3 ( 1)n

nSn n

is(a) a divergent sequence (b) an oscillatory sequence(c) not a Cauchy sequence (d) a Cauchy sequence

Ans. (d)

Sn + 1–Sn =1sin

2 0( 1)( 2)

n

n n

So, <Sn> is monotonically increasing.

Sn =2sin sin sin

2 2 2...1 2 2 3 ( 1)

n

n n

1 1 1...1 2 2 3 ( 1)n n

= 1 1 1 1 11 ...2 2 3 1n n

=11 1

1n N

n

So, it is monotonically increasing & bounded above and hence, convergent. So, it is Cauchy.2. Let P be the vector space (over ) of all polynomials of degree 3 with real coefficients. Consider

the linear transformation :T P P defined byT(a0 + a1x + a2x2 + a3x3)= a3 + a2x + a1x2 + a0x3

Then the matrix representation M of T with respect to the ordered basis {1,x,x2,x3} satisfies(a) M2 + I4 = 0 (b) M2 – I4 = 0(c) M – I4 = 0 (d) M + I4 = 0

Ans. (b)Let ordered basis 2 3{1, , , }B x x x

3(1) (1) BP x P =

0001

2( ) ( ) BP x x P x =

0010


2 | JAM-2016 MATHEMATICS TRAJECTORY EDUCATION

2( ) ( ) BP x x P x =

0100

3 3( ) 1 ( ) BP x P x =

1000

M = [T]B= 3 1 2

0 0 0 10 0 1 0

( )( )0 1 0 01 0 0 0

a a a a

even permutation matrix.

So, M2 = I4

3. Let :[ 1,1]f be a continuous function. Then, the integral

0

(sin )xf x dx

is equivalent to

(a)0

(sin )2

f x dx (b)

0

(cos )2

f x dx

(c)0

(cos )f x dx

(d)0

(sin )f x dx

Ans. (a)

I=0

(sin )xf x dx

I=

0( ) (sin( ))x f x dx 0 0

as ( ) ( )a a

f x dx f a x dx =

0( ) (sin )x f x dx

Adding, 2I=0

(sin )f x dx

So, I=0

(sin )2

f x dx

4. Let be an element of the permutation group S5. Then the maximum possible order of is(a) 5 (b) 6(c) 10 (d) 15

Ans. (b)

1 2 3 4 5( )( )a a a a a has highest order 65. Let f be a strictly monotonic continuous real valued function defined on [a, b] such that f (a) < a and

f (b) > b. Then which one of the following is TRUE?(a) There exists exactly one ( , )c a b such that f (c) = c

(b) There exist exactly two points 1 2, ( , )c c a b such that f (ci) = ci, i = 1,2


JAM-2016 MATHEMATICS | 3

(c) There exists no ( , )c a b such that f (c) = c(d) There exists infinitely many points ( , )c a b such that f (c) = c

Ans. (a)Let g(x) = f(x) – xNow, g(a) < 0 at x = a and g(b) > 0 at x = bf (x) is strictly monotonic.For b > a, f (b) > f (a) So, it is monotonically increasing.In fact, g(x) is also monotonically increasing.So, it must intersect x-axis between x = a and x = b exactly once (Intermediate mean value theorem)

6. The value of( , ) (2, 2)

( ) 2lim

4x y

x yx y

is

(a) 0 (b)14

(c)13 (d)

12

Ans. (b)

( , ) (2, 2)

( ) 2lim

4x y

x yx y

=

( , ) (2, 2)

( ) 2 ( 2)lim

( 4)( 2)x y

x y x yx y x y

= ( , ) (2, 2)

1 1lim42x y x y

7. Let ˆˆ ˆ( )r xi yj zk and | | .r r If f(r) = ln r and 1( ) , r 0,g r

r satisfy 2 ( ) 0,f h r g

then

h(r) is

(a) r (b)1r

(c) 2r (d)2r

Ans. (c)

ˆ ˆ (log )ff i i rx x

= 1ˆ xir r

= 2

1 rr

g =1 1ˆ, g ir x r

= 2 3

1ˆ x rir r r

2 f h g = 0

2 3

2r hrr r

= 0

h= 2r



8. The nonzero value of n for which the differential equation2 2 2 3 2(3 ) ( 3 ) 0, 0xy n x y dx nx x y dy x

becomes exact is(a) –3 (b) –2(c) 2 (d) 3

Ans. (d)

Mdx + Ndy = 0 is exact ifM Ny x

2 2 26 3 6xy n x nx xy 2 3 0,3n n n

9. One of the points which lies on the solution curve of the differential equation( ) ( ) 0y x dx x y dy

with the given condition y(0) = 1, is(a) (1, –2) (b) (2, –1)(c) (2, 1) (d) (–1, 2)

Ans. (c)(y – x)dx + (x + y)dy= 0

(ydx + xdy)–xdx + ydy= 0Integrating

2 2

2 2x yxy = c

(0, 1) lies on the curve

So, c =12

So,2 2

2 2x yxy =1

(2, 1) lies on curve10. Lest S be a closed subset of ,T a compact subset of such that .S T Then S T is

(a) closed but not compact (b) not closed(c) compact (d) neither closed nor compact

Ans. (c).S T T

Since, T is bounded, So, S T is also bounded

S T is closed also, since, intersection of two closed set is closed.So, S T is closed and bounded.So, S T is compact

Q.11 – Q.30 carry two marks each.11. Let S be the series

(2 1)1

1(2 1)(2 )k

k k

and T be the series( 1)

3

2

3 43 2

k

k

kk



of real numbers. Then which one of the following is TRUE?(a) Both the series S and T are convergent(b) S is convergent and T is divergent(c) S is divergent and T is convergent(d) Both the series S and T are divergent

Ans. (b)

For S, ak = 2 1 2 1

1 1(2 1)2 2 kk k b

k

kb will form geometric series and hence convergent.So, from comparision test it is convergent.

For T,1

33 43 2

k

kk

=1

3 3613 2

k

k

=

213

2 31 61 11 3 22 3

k

k k

As k

12333 4

3 2

k

k ek

So, T is divergent.12. Let {an}

1

4na

=3

13 , 1, 1

81n

n

a n aa

Then all the terms of the sequence lies in

(a)1 3,2 2

(b) [0, 1]

(c) [1, 2] (d) [1, 3]Ans. (d)

a1=1

2

4a

=3 1 2441 81 81

a2 =8161

3

4a

=32

2

3 3 64 1381 4 81 27a

a

3

1a = 3

9 16 16 1.716 2187 9

a

When ,n an and 1na

4 =

3381



1 =

3

81

4 81 3

13. The largest eigenvalues of the matrix1 4 164 16 1

16 1 4

is

(a) 16 (b) 21(c) 48 (d) 64

Ans. (b)

Looking at the matrix111

X

is the eigenvector for 21

AX= 21X

det =1 4 164 16 1

16 1 4

= 1 1 2 3

21 4 1621 16 121 1 4

C C C C

=1 4 16

211 16 11 1 4

=1 0 0

211 12 15 21.( 189)1 3 12

det = 1 2 3 21.( 189)

trace = 1 2 3 21

If, 1 2 321,

and 2 3& can be 17 or –17So, largest eigenvalues is 21

14. The value of the integral1

2

1

(2 )! (1 ) ,2 ( !)

nn

n x dx nn

is

(a)2

(2 1)!n (b)2

(2 1)!n

n

(c)2( !)2 1

nn (d)

( 1)!2 1nn



Ans. (c)

I =1

22

1

(2 )! (1 )2 ( !)

nnn x dxn

=1

22

0

(2 )! 2 (1 )2 ( !)

nnn x dxn

Let 2 ,x y x y

dx =1

2dy

y

I =1 1

22

0

(2 )! 12 (1 )2 ( !) 2

nnn y y dyn

= 2

(2 )! 1 , 12 ( !) 2n

n B nn

2

(2 )! ½ ( 1)2 ( !) 3( )

2

n

n nn

n

=2

(2 )! ½ !1 1 3 12 ! ..... ½2 2 2 2

n

n n

n n n n

=1

2

(2 )!22 (2 1)(2 1)...3.1

n

nn

n n

=1

2

(2 ) (2 1)(2 2)...2.1 22 (2 1)(2 1)...3.1

n

n

n n nn n

=1

2

2 2 ! 2( !)2 (2 1) 2 1

n n

nn n

n n

15. If the triple integral over the region bounded by the planes

2x + y + z = 4, x = 0, y = 0, z = 0given by

( ) ( , )2

0 0 0

,x µ x y

dz dy dx

then the function ( ) – µ( , )x x y is(a) x + y (b) x – y(c) x (d) y

Ans. (d)( ) ( , )2

0 0 0

x µ x y

dz dy dx

=4 22 4 2

0 0 0

x yx

dz dy dx

So, µ(x,y)= 4 – 2x – y

( )x = 4 – 2x( ) ( , )x µ x y = 4 – 2x – 4 + 2x + y = y

16. The surface area of the portion of the plane y + 2z = 2 within the cylinder x2 + y2 = 3 is

(a)3 5

2 (b)

5 52

(c)7 5

2 (d)

9 52



Ans. (a)

S =2 2

1 z z dxdyx y

=114

dxdy

= 25 5 ( 3)2 2

dxdy

=3 5

2

17. Let 2:f be defined by

f (x, y) =

2

if 0

0 if 0

xy x yx y

x y

Then the value of2 2f fx y x y

at the point (0, 0) is

(a) 0 (b) 1(c) 2 (d) 4

Ans. (b)At (x,y)= (0, 0)

0

( ,0) (0,0) 0 0lim 0y yxy h

f h ff

h h

0 0

(0, ) (0,0) 0lim lim 1x xyx k k

f k f kfk k

0

(0, ) (0,0) 0 0(0,0) lim 0y k

f k ffk k

0

( ,0) (0,0) 0 0(0,0) lim 0x h

f h ffh h

0 0

( , ) ( ,0)( ,0) lim lim 0y k k

f h k f h hkf hk h k

2

0 0

( , ) (0, )(0, ) lim limx h h

f h k f k kf k kh h k

the value of2 2f fx y x y

at the point (0, 0) is 1

18. The function 2 2 2 2( , ) 3 4 3 12 1f x y x y y x y has a saddle point at(a) (0, 0) (b) (0, 2)(c) (1, 1) (d) (–2, 1)

Ans. (d)

xf =6xy – 6x

yf =3x2 + 12y2 – 24y



r = 6 6xxf y

t= yyf =24y – 24

s = xyf = 6xrt – s2 =144 (y – 1)2 –36x2

=36[4(y–1)2 – x2] < 0 at (–2, 1)

19. Consider the vector field ˆ ˆ( ),F r yi xj

where ˆ ˆ,r xi yj and | | .r r If the absolute value of

the line integralc

F dr along the closed curve c : x2 + y2 = a2 (oriented counter clockwise) is 2 , then

is(a) –2 (b) –1(c) 1 (d) 2

Ans. (a)

F = ˆ ˆ( )r yi xj

= ˆ ˆ(sin cos )r r i j

F dr

= ( 1) ˆr e dr

= ( 1) ( 1)r dl r rd = ( 2)r d

=2

( 2) 2

0

2 2a d a

If 2

20. Let S be the surface of the cone 2 2z x y bounded by the planes z = 0 and z = 3. Further, let C be

the closed curve forming the boundary of the surface S. A vector field F

is such that ˆ ˆ.F xi yj

The absolute value of the line integral ,c

F dr where

(a) 0 (b) (c) 15 (d) 18

Ans. (d)

F dr

= ˆF ndS

= ˆ ˆ ˆ( )xi yj ndS =- ˆ ˆrre ndS

=-2 3

0 0

r rdr d

ˆ ˆ(as 1& )re n dS rdrd

=-18

18F dr



21. Let y(x) be the solution of the differential equation

1; (1) 0, 0.

x

d dy dyx x ydx dx dx

Then y(2) is

(a)3 1 ln 24 2 (b)

3 1 ln 24 2

(c)3 ln 24 (d)

3 ln 24

Ans. (d)d dyxdx dx

= x

dyxdx =

2

12x c

dydx = 1

2x c

x

At x = 1 11

102x

dy cdx

dydx

=1

2 2x

x

y=2 1 log

4 2x x c

1(1) 04

y c

y=2 1 1log

4 2 4x x

y(2)=3 1 log 24 2

22. The general solution of the differential equation with constant coefficients2

2 0d y dyb cydx dx

approaches zero as ,x if(a) b is negative and c is positive (b) b is positive and c is negative(c) both b and c are positive (d) both b and c are negative

Ans. (c)(D2+bD + c)y= 0

D=2 4

2b b c

The solution will be exponentially decaying if b > 0 and c > 0



23. Let S and S denote the set of point x in such that every neighbourhood of x contains somepoints of S as well as some point of complement of S. Further, let S denote the closure of S. Thenwhich one of the following is FALSE?(a) (b) ( \ ) ,T T T

(c) ( ) , , ,T V T V T V T V

(d) ( \ ),T T T Ans. (c)

We can take example If T = (0,3), {0,3}T

If V= (2,6), {2,6}V

T V =(0, 6)

( )T V ={0, 6}

T V ={0, 2, 3, 6}

So, ( )T V T V 24. The sum of the series

22

( 1)2

n

n n n

is

(a)1 5ln 23 18

(b)1 5ln 23 6

(c)2 5ln 23 18

(d) 2 5ln 23 6

Ans. (c)

22

( 1)2

n

n n n

=

2

( 1)( 2)( 1)

n

n n n

=2

1 1 1( 1)3 1 2

n

n n n

=1 1

1 1 1 1 13 2 1 2 2 2 2 3m mm m m m

=1 1

1 1 1 1 13 2 1 2 2 2 2 2 3m mm m m m

=1 1 1 1 1 1 1 1 1 11 ... ...3 2 3 4 4 5 6 7 8 9

=1 1 1 1 5 2 52 1 ... log 23 2 3 4 6 3 18



25. Let 1 1( )1 | | 1 | 1|

f xx x

for all [–1,1].x Then which one of the following is TRUE?

(a) Maximum value of f (x) is 32

(b) Minimum value of f(x) is13

(c) Maximum of f(x) occurs at12 (d) Minimum of f(x) occurs at x = 1

Ans. (a)

f (x) =1 1

1 1 1x x

=1 1

1 2x x

for 1 0x

For 0 1,x f (x) =1 1

1 2x x

f '(x) = 2 2

1 1 01 2x x

(2 – x)2 =(1 + x)2

x =12

f "(x) = 3 3

2 2 10 at21 2

xx x

So, f (x) is minimum at 12

x

Minimum value is2 23 3 =

43

Let us check the value at end point

At x = 1, f (x) =1 312 2

At x = –1, f (x) =1 1 52 3 6

So, maximum value is32

26. The matrixcos sinsin cos

Mi i

is unitary matrix when is

(a) (2 1) ,2

n n (b) (3 1) ,3

n n

(c) (4 1) ,4

n n (d) (5 1) ,5

n n

Ans. (a)For unitary matrix, the columns must be orthonormal

<C1, C2>= 1 20 0C C



sincos sin

cosi

i

=0

sin cos sin cos =0sin 2 =0

2 = n

= 2n

27. Let0 1 11 0 , \{0}

2 0M

and b a non-zero vector such that Mx = b for some 3.x Then

the value of xTb is(a) (b) (c) 0 (d) 1

Ans. (c)MX =b

XTMX = 0T T TX b X b X MX 28. The number of group homomorphisms from the cyclic group 4 to the cyclic group 7 is

(a) 7 (b) 3(c) 2 (d) 1

Ans. (d)Number of group homomorphism from 4 7 to is gcd(4, 7) = 1

29. In the permutation group ( 5),nS n if H is the smallest subgroup containing all the 3-cycles, thenwhich one of the following is TRUE?(a) Order of H is 2 (b) Index of H is Sn is 2(c) H is abelian (d) H = Sn

Ans. (b)The 3-cycles are all even permutation, so they are contained in An. Thus, .nH H We can prove thatevery element of An can be written as product of three cycles.Any element nA can be written as product of an even number of transpositions.

= T1T2 ...T2k–2 T2k = (T1 T2)(T3 T4) ...(T2k–1 T2k)Thus, if we can show that product of any pair of transpositions must be in H then must be in H.Let us consider such a product,= (a b)(c d)Case 1: a and b are same as c and d, then = 1 which is certainly in H.Case 2: (a b) and (c d) has one element in common then is itself a 3 cycle and therefore is in H. Forexample (1 2)(2 3) = (1 2 3)In general (x y)(y z)= (x y z) if x, y, z are all differentCase 3: a, b, c and d are all differentthen, (a b)(c d)= (a b c)(b c d)So, H= An

and hence, index of H in nS is 2.



30. Let :f be defined as

f (x) = 21 sin ln if 0.

0 if 0x x x x

x

Then, at x = 0, the function f is(a) continuous and differentiable when 0 (b) continuous and differentiable when 0 (c) continuous and differentiable when 1 0 (d) continuous and differentiable when 1

Ans. (b)

f (x)= 1 2sin ln if 0x x x x

It is continuous if 1 0 and differentiable if 1 1

SECTION-B (MULTIPLE SELECT QUESTIONS (MSQ))31. Let {sn} be a sequence of positive real numbers satisfying

2sn+1 = 2 3 , 1.4ns n

If and are the roots of the equation 21

32 0 and ,4

x x s then which of the following

statement(s) is(are) TRUE?(a) {sn} is monotonically decreasing (b) {sn} is monotonically increasing

(c) lim nns

(d) lim nn

s

Ans. (a, c)

2 324

x x = 0

4x2 – 8x + 3= 0 4x2 – 6x – 2x + 3= 0 2x(x – 3) – (2x – 3)= 0 (2x – 3)(2x – 1)= 0

x=3 1,2 2

So, =1 3,2 2

Let s1 = 1;

2s2 = 21

34

s

s2 = 178

s

2s3 = 22

3 7 3 134 8 4 8

s

s3 = 21316

s

So, {sn} is monotonically decreasing and converges at .



32. The value(s) of the integral

cos , 1x nx dx n

is (are)(a) 0 when n is even (b) 0 when n is odd

(c) 2

4n when n is even (d) 2

4n when n is odd

Ans. (a, d)

I=0 0

cos 2 cos 2 cosx nx dx x nx dx x nx dx

=12 sin sinx nx nx dx

n n

= 20

12 sin cosx nx nxn n

= 2

2 cos 1nn

So, I =0 for even n

= 2

4n for odd n

33. Let 2:f be defined by

f (x, y) =if 0

.0 elsewhere

xy xx

Then at the point (0, 0), which of the following statement(s) is(are) TRUE?(a) f is not continuous (b) f is continuous(c) f is differentiable (d) Both first order partial derivatives of f exist

Ans. (b)

f (x, y)=for 0for 0

0 for 0

y xy x

x

x

y

zz y

x = – (plane)

at < 0

zx

= 0at = 0z y

x = (plane)

at > 0

O

f (x, y) is continuous at (0, 0) and both first order partial derivatives of f exist.



34. Consider the vector field ˆ ˆF xi yj

on an open connected set 2.S Then which of the followingstatement(s) is(are) TRUE?

(a) Divergence of F

is zero on S

(b) The line integral of F

is independent of path in S

(c) F

can be expressed as a gradient of a scalar function on S(d) The line integral of F

is zero around any piecewise smooth closed path in S

Ans. (b, c, d)

F

is radial and hence conservative.35. Consider the differential equation

sin 2 dyxdx

= 2 2cos , 1 2.4

y x y

Then which of the following statement(s) is(are) TRUE?

(a) The solution is unbounded when 0x (b) The solution is unbounded when2

x

(c) The solution is bounded when 0x (d) The solution is bounded when2

x

Ans. (c, d)

sin 2 dyxdx =2y + 2 cos x

1sin cos

dy ydx x x =

1sin x

If1

sin cosdx

x xe =

2 cosec2x dxe

=12 log tan2 1

tanx

ex

1tan

yx

=1 1

sin tandx c

x x

=1

sinc

x

y =–sec x + c tan x

at x = , 1 2 1 2 24

y c

c =1So, y =tan x – sec x

0, 1x y

2

lim tan secx

x x

=

2

sin 1limcosx

xx

0 form0

=2

coslim 0sinx

xx

y is bounded when 0 and2

x x



36. Which of the following statement(s) is(are) TRUE?(a) There exists a connected set in which is not compact(b) Arbitrary union of closed intervals in need not be compact(c) Arbitrary union of closed intervals in is always closed(d) Every bounded infinite subset V of has a limit point in V itself

Ans. (a, b)

If An =1 ,1n

1n

nA

= (0, 1] which is not closed

Hence, arbitrary union of closed intervals in need not be closed and hence need not be compact.(a, b) is connected set in which is not compact.

37. Let 5 12 113 13

x x

P x

for all .x Then which of the following statement(s) is(are) TRUE?

(a) The equation P(x) = 0 has exactly one solution in (b) P(x) is strictly increasing for all x(c) The equation P(x) = 0 has exactly two solutions in (d) P(x) is strictly decreasing for all x

Ans. (a, d)

P'(x) = 5 5 12 12log log 013 13 13 13

x x

P(x) is strictly decreasing.

Let5

13 = sin , then12 cos13

P(x)= 0 sin cos 1x x x= 2

38. Let G be a finite group of o(G) denotes its order. Then which of the following statement(s) is(are)TRUE?(a) G is abelian if o(G) = pq where p and q are distinct primes(b) G is abelian if every non identity element of G is of order 2

(c) G is abelian if the quotient group G

Z G is cyclic, where Z(G) is the center of G

(d) G is abelian if o(G) = p3, where p is primeAns. (b, c)

39. Consider the set 3 , , , .x

V y x y zz

For which of the following choice(s) the

set V becomes a two dimensional subspace of 3 over ?(a) 0, 1, 0 (b) 0, 1, 1

(c) 1, 0, 0 (d) 1, 1, 0 Ans. (a, c, d)

Two dimensional subspace of 3 is a plane passing through origin.



40. Let 1 1 , .3 7n mS n m

Then which of the following statement(s) is(are) TRUE?

(a) S is closed (b) S is not open(c) S is connected (d) 0 is a limit point of S

Ans. (b, d)0 is limit point which is not contained in S.So, S is not closed. S is not open because it does not contain nbd of

1 13 7 =

1021

SECTION-C (NUMERICAL ANSWER TYPE (NAT))Q. 41 – Q. 50 carry one mark each.41. Let {sn} be a sequence of real numbers given by

sn = 1 12 1 sin , .2

n n nn

Then the least upper bound of the sequence {sn} is ________Solution :

Ans. 0.5

42. Let {sk} be a sequence of real numbers, where

sk = / , 1, 0.a kk k Then

1/1 2lim n

nns s s

is ________Solution :

Let us consider a sequence whose nth term is given byan = 1 2 ns s s

1/Lt nnn

a

= / 111lim lim lim 1 nn

nn n nn

a s na

=1 Ans. 1

43. Let1

32

3

xx x

x

be a non-zero vector and .T

TxxAx x Then the dimension of the vector space

3y A 0 y over is ________

Solution :

A=T

Txxx x

A is a projection matrix and has rank 1So, dimension of null space of A is 2. Ans. 2



44. Let f be a real valued function defined byf (x, y) =2 ln (x2y2ey/x), x > 0, y > 0.

Then the value of f fx yx y

at any point (x, y), where x > 0, y > 0, is ________

Solution :f (x, y)= 2 log x2y2 ey/x = u

x2y2ey/x = eu/2 = v

v =2

2 2 / 4 /2

y x y xyx y e x ex

v is a homogeneous function of degree 4v vx yx y

=4v from Euler’s theorem

/2 /21 12 2

u uu dux e y ex dy

=4eu/2

u dux yx dy

=8 Ans. 8

45. Let 3ˆ ˆF xi x y j

be a vector field for all (x, y) with ˆ ˆ0 and .x r xi yj Then the value of the

line integralC

F dr from (0, 0) to (1, 1) along the path 2 3: , , 0 1c x t y t t is ________

Solution :On c, x = 3 2t dx tdt

y = 3 23t dy t dt

F dr = 3xdx x y dy

= 9 22 3t tdt t t t dt =(2t2 + 3t4 + 3t11)dt

F dr

= 1 2 4 11

02 3 3t t t dt

=112

3 5

0

2 3 33 5 12

tt t

=2 3 1 91 1.5163 5 4 60

=1.52 Ans. 1.52

46. If : 1,f defined by 1

xf xx

is expressed as

f (x) =

2

32 1 22 ,3 9 1

c xx

where lies between 2 and x, then the value of c is ________Solution :

f (x) = f (2 + (x – 2))



= 212 2 2 22!

f f x x f

f (x) =11

1 1xx x

f (2) =23

f '(x) =

2

1 1291

fx

f "(x) =

3 3

2 2,1 1

fx

f (x) =

2

32 1 223 9 1

xx

c =–1 Ans. -1

47. Let y1(x), y2(x) and y3(x) be linearly independent solutions of the differential equation3 2

3 26 11 6d y d y dy ydx dx dx =0

If the Wronskian W(y1, y2, y3) is of the form kebx for some constant k, then the value of b is ________Solution :

(D3 – 6D2 + 11D – 6)y= 0(D3 – D2 – 5D2 + 5D + 6D – 6)y = 2

(D – 1)(D2 – 5D + 6)y= 0(D – 1)(D – 2)(D – 3) =0

y1 =ex, y2 = e2x, y3 = e3x

Wronskian

W=

2 31 2 3

1 1 1 2 31 2 3

2 33 2 21 2 3

2 34 9

x x x

x x x

x x x

y y y e e ey y y e e e

e e ey y y

= 2 3

1 1 11 2 31 4 9

x x xe e e

=ke6x

So, b = 6 Ans. 648. The radius of convergence of the power series

21

4 21

nn

nx

n n

is ________

Solution :

R= 1/2

1/2

1

lim limn n

nn nn

aaa

=

1/2

14 1 2lim

1 4

n

nn

n nn n



=1/21 1 0.5

4 2

Ans. 0.5

49. Let : 0,f be a continuous function such that

0

x

f t dt =2

2 4 sin 2 2cos 2 .2x x x x

Then the value of 14

f

is ________

Solution :

0

x

f t dt =2

2 4 sin 2 2cos22x x x x

Differentiating wrt. xf (x) =x + 4 sin 2x + 8x cos 2x – 4 sin 2x

4f

= 4 0 44 4

14

f

=1 0.254 Ans. 0.25

50. Let G be a cyclic group of order 12. Then the number of non-isomorphic subgroups of G is ________Solution :

12= 2 12 3Number of non-isomorphic subgroups of G is 6. Ans. 6

Q. 51 – Q. 60 carry two marks each.

51. The value of

21

1lim 8

n

n

nn

n

is equal to ________

Solution :1/lim nnn

u

= 1lim nn

n

uu

un = 1

18

n

nn

n

1n

n

uu =

11 11 111 1

2

1 11 1

2

11 8 1 18 18 11

1 18 8 18

nnn n

n nn n

nnnn

n nn n

1lim nn

n

uu

=

11

1 1 21 1

1 1 1

2

118 1 8 1lim 1

88 18

n

n n

n n n

n nn

n

Ans. -1



52. Let R be the region enclosed by 2 2 2 24 1 and 1.x y x y Then the value of

R

xy dx dyis ________

Solution :y

x

|xy| is positive in all the four quadrant.In the first quadrant

xy dx dy =2

2

1 1 12

0 012

3 (1 )2 4

x

x

xxy dy dx x dx

= 1

3

0

38

x x dx

=3 1 18 2 4

=3 2 1 38 4 32

R

xy dx dy =34

32

=3 0.3758 Ans. 0.375

53. Let

M =1

32

3

1 11 1 , 1, , , and .1 1

xxx

x

Then Mx = 0 has infinitely many solutions if trace(M) is ________Solution :

Whatever be the answer it will be true in generalSo, if 1

trace(M) =3 Ans. 354. Let C be the boundary of the region enclosed by y = x2, y = x + 2, and x = 0. Then the value of the line

integral 2 3 ,

Cxy y dx x dy

where C is traversed in the counter clockwise direction, is ________Solution :



y

x

(2, 4)

The question should have been more clear because said region is in first and second quadrant both.Taking first quadrant.

2 3xy y dx x dy =N M dx dyx y

= 2

2 22

0

3 2x

x

x x y dx dy

=2

222 2

0

3x

x

x y xy y dx

= 2

2 2 2 2 4

0

3 2 2 4 4x x x x x x x x x dx

= 2

4 3 2

0

2 2 6 2 4x x x x dx

=25 4 3 2

0

2 2 6 2 45 4 3 2x x x x x

=64 8 16 4 8 0.85 Ans. 0.8

55. Let S be the closed surface forming the boundary of the region V bounded by x2 + y2 = 3, z = 0, z = 6.

A vector field F

is defined over V with 2 1.F y z

Then the value of

1 ˆ ,S

F n ds

where n̂ is the unit outward drawn normal to the surface S, is ________.Solution :

Applying Gauss Divergence theorem

ˆS

F nds =

VFds

= 6

0

2 1y z dz dx dy

=62

02

2zyz z dx dy

= 12 24y dx dy= 24 3 72



So, 1 ˆF n ds

=72 Ans. 72

56. Let y(x) be the solution of the differential equation2

2 5 6d y dy ydx dx =

00, 0 1, 1.

x

dyydx

Then y(x) attains its maximum value at x = ________Solution :

(D2 + 5D + 6)y= 0y= C1e–2x + C2e–3x

y' = –2C1e–2x – 3C2e–3x

At x = 0, y = –1 –2C1 – 3C2 = –1

2C1 + 3C1 = 1At x = 0, y = 1 C1 + C2 = 1

C1 = 2, C2 = –1y =2e–2x – e–3x

y' = 2 3 3 34 3 0 log4 4

x x xe e e x

y" = 2 3 9 64 9 648 9 8 9 ve16 27 2 3

x xe e

So, y(x) is maximum at x = log (0.75) Ans. log(0.75)57. Then value of the double integral

0 0

sinx y dy dxy

is ________

Solution :Region of integration is bounded by

0, , 0,y y x x x y

x

( , )

x =

Changing the order of integration

0

sin

y

y dx dyy

= 0 0

sin sinyy x dy y dyy



= 0cos 2y Ans. 2

58. Let H denote the group of all 2 × 2 invertible matrices over 5 under usual matrix multiplication. Then

the order of the matrix2 31 2

in H ________

Solution :

2 3 2 31 2 1 2

=2 24 2

32 31 2

=2 2 2 3 1 04 2 1 2 0 1

I

order =3 Ans. 3

59. Let 1 2

1 2 0, 1 0 ,

1 5 23 1

A B N A

the null space of A and R(B) the range space of B. Then the

dimension of N A R B over is ________Solution :

Both the columns of B are not lying in the null space of A as1 21 0, 0 0

3 1A A

N A R B =000

dim N A R B =0 Ans. 0

60. The maximum value of f (x, y) = x2 + 2y2 subject to the constraint y – x2 + 1 = 0 is ________Solution :

Use method of lagrange’s multipliers

Ans. 0.875

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