GATE MATHEMATICS 2021
Previous Year Question Paper with Solution.

1. Let A be a 3 × 4 matrix and B be a 4 × 3 matrix with real entries such that AB is non-singular.

Consider the following statement:

P: Nullity of A is 0

Q: BA is a non-singular matrix

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (d)

Sol. Take

Then AB = I3 × 3 is non-singular matrix.

and BA = is singular matrix.

We can easily conclude that

Rank (A) = 3 Nullity (A) = 4 – Rank (A) = 4 – 3 = 1

So, both statements P and Q are false.

So, option (d) is correct.

2. Let f(z) = u(x, y) + i v(x, y) for z = x + iy where x and y are real numbers, be a non-constant analytic function on the complex plane . Let ux, vx and uy, vy denote the first order partial respectively. Consider the following two functions defined on .

g1(z) = ux(x, y) – i uy(x, y) for z = x + iy ,

g2(z) = vx(x, y) + i vy(x, y) for z = x + iy .

Then

(a) both g1(z) and g2(z) are analytic in .

(b) g1(z) is analytic in and g2(z) is NOT analytic in .

(c) g1(z) is NOT analytic in and g2(z) is analytic in .

(d) neither g1(z) nor g2(z) is analytic in .

Ans. (b)

Sol. Given that f(z) = u(x, y) + iv(x, y) is non-constant analytic function on complex place .

= ux(x, y) + ivx(x, y)

= vy(x, y) + ivx(x, y)

= ux(x, y) – iuy(x, y)

= vy(x, y) – iuy(x, y)

and is also analytic function on .

Here g1(z) = ux(x, y) – iuy(x, y) = , which will be analytic.

g2(z) = vx(x, y) + ivy(x, y) = U + iV

Here U = vx(x, y), V = vy(x, y)

Ux = vxx, vx = vyx

Uy = vxy, Vy = vyy

Here Ux Vy and Uy –Vx. So C-R equations do not hold.

g2(z) is not analytic.

option (C) is correct.

Or take f(z) = z2 = (x + iy)2 = x2 – y2 + 2ixy

f(z) = u(x, y) + iv(x, y) where u(x, y) = x2 – y2, v(x, y) = 2xy

Consider g2(z) = vx(x, y) + ivy(x, y) = U2 + iV2 = 2y + i2x

Now, U2 = 2y U2x = 0, U2y = 2

V2 = 2x V2x = 2, V2y = 0

and Ux = Vy but Uy –Vx

g2(z) is not an analytic function.

Option (b) is correct.

3. Let T(z) = , ad – bc 0, be the Mobius transformation which maps the points z1 = 0, z2 = –i, z3 = in the z-plane onto the points w1 = 10, w2 = 5 – 5i, w3 = 5 + 5i in the w-plane, respectively. Then the image of the set S = {z : Re(z) < 0} under the map w = T(z) is

(a) {w : |w| < 5}

(b) {w : |w| > 5}

(c) {w : |w – 5| < 5}

(d) {w : |w – 5| > 5}

Ans. (c)

Sol. T(z) = , ad – bc 0 is Mobius map for which map for which

T(0) = 10, T(–i) = 5 – 5i and T() = 5 + 5i

Under Mobius transformation the cross ratio is preserved.

i.e. (z, z1, z2, z3) = (w, w1, w2, w3)

i.e. (z, 0, –i, ) = (w, 10, 5 – 5i, 5 + 5i)

=

=

z =

For mapping of Rez < 0,

i.e., < 0

|w – 5| < 5

Hence, Rez < 0 is mapped to |w – 5| < 5

option (c) is correct.

4. Let R be the row reduced echelon form of a 4 × 4 real matrix A and let the third column of R be .

Consider the following statements:

P: If is a solution of Ax = 0, then = 0.

Q: For all b rank [A\b] = rank [R\b].

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (d)

Sol.

Now Ax = 0

Q is incorrect.

Option (d) is correct.

5. The eigenvalues of the boundary value problem

are given by

(a) = (n)2, n = 1, 2, 3, ...

(b) = n2, n = 1, 2, 3, ...

(c) = kn2, where kn, n = 1, 2, 3, ... are the roots of k – tan(k) = 0

(d) = kn2, where kn, n = 1, 2, 3, ... are the roots of k + tan(k) = 0

Ans. (c)

Sol.

For non-trivial solution we must have c1 0

It has no non-trivial solution when < 0, so no negative eigenvalue.

Case-III: When = 0

y = c1 + c2x

c2 = 0

zero is not eivenvalues as no non-trivial solution obtained.

eigenvalues are given by = kn2, where kn = tan

Option (c) is correct.

6. The family of surfaces given by u = xy + f(x2 – y2), where f : is a differential equation, satisfies

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given that u = xy + f(x2 – y2)

Differentiating u w.r.t. x, = y + Df(x2 – y2). 2x

Differentiating u w.r.t. y, = x + Df(x2 – y2).(–2y)

Now, = y(y + Df(x2 – y2).2x) + x(x + Df(x2 – y2)(–2y)

= y2 + 2xy Df(x2 – y2) + x2 – 2xy Df(x2 – y2) = y2 + x2

Option (a) is correct.

7. The function u(x, t) satisfies the initial value problem

Then u(5, 5) is

(a)

(b) 1 – e100

(c)

(d) 1 – e10

Ans. (a)

Sol.

Its D'Alembert solution can be written as

Option (a) is correct.

8. Consider the fixed-point iteration

with = 3 + (x – 3)3, x (2.5, 3.5),

and the initial approximation x0 = 3.25.

Then, the order of convergence of the fixed-point iteration method is

(a) 1

(b) 2

(c) 3

(d) 4

Ans. (c)

Sol. Given fixed point iteration is xn + 1 = where = 3 + (x – 3)3, x (2.5, 3.5).

Now we know if α is fixed point of ,

then order of convergence is 'n' if

order of convergence is '3'

Option (c) is correct.

9. Let {en : n = 1, 2, 3, ...} be an orthonormal basis of a complex Hilbert space H.

Consider the following statements:

P: There exists a bounded linear functional f : H such that f(en) = for n = 1, 2, 3, ... .

Q: There exists a bounded linear functional g : H such that g(en) = for n = 1, 2, 3, ... .

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (b)

Sol. Given that {en : n = 1, 2, ...} is orthonormal basis of complex Hilbert space H.

In statement Q; we have given that a bounded linear function g : H defined as g(en) = for n = 1, 2, ... then by using Reiz representation theorem, there exists v, such that

g(x) =

Now using Bessel's inequality, we have

Then by p-test, L.H.S. of above is divergent

Hence it is contradiction

So (Q) is false statement.

In statement P, we have given that there exists a bounded linear function f : H such that

By proceeding the same as for statement, we found that

Hence statement P is true.

10. Let be given by Consider the following statements:

P : |f(x) – f(y)| < |x – y| for all

Q : f has a fixed point.

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (d)

Sol. P : Take x = 0 and y = 0.

then |f(x) – f(y)| = |f(0) – f(0)| = 0

0 < 0 not possible

Hence, Pis false,

Q : Put f(x) = x

such that

Hence, Q is also false.

Hence, option (d) is correct.

11. Consider the following statements

P: is metric on (0, 1).

Q: is a metric on (0, 1)

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (a)

Sol.

Consider L.H.S. of *.

= d1(x, z) + d1 (z, y)

Hence, d1 (x, y) < d1 (x, z) + d1 (z, y)

So, P is true

Q :

(1) d2 (x, x) = 0 (given)

(2) d2 (x, y) = d2(y, x) (clearly)

(3) d2 (x, y) > 0 [because mod is always a non negative number}

(4) To show d2 (x, y) < d2 (x, z) + d2 (z, y)

Consider L.H.S. we have d2 (x,y) = |x| + |y|

and R.H.S. is d2 (x, z) + d2 (z, y) = |x| + |z| + |z| + |y|.

Clearly |x| + |y| < |x| + |z| + |z| + |y|.

Hence, proved

So, Q is also true

12. Let f: be a twice continuously differentiable scalar field such that div = 6. Let Sbe the surface x2 + y2 + z2 = 1 and be unit outward normal to S. Then the value of is

(a)

(b)

(c)

(d)

Ans. (d)

Sol. From the Gauss divergence theorem, we have

Option (d) is correct

13. Consider the following statement:

P: Every compact metrizable topological space is separable.

Q: Every Hausdorff topology on a finite set is metrizable.

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (a)

Sol. Result: I Every compact metrizable toplological space is separable.

Result: II Every Hausdorff topology on a finite set is discrete topology and we know discrete topology is metrizable.

14. Consider the following topologies on the set of all real numbers:

Then the closure of the set {1} in (, T3) is

(a) {1}

(b) {0, 1}

(c)

(d) \{0}

Ans. (c)

Sol. Reason T3 = (Clearly)

Now closure of A is the intersection of all those closed sets which contains A. Now, here closed sets are and , but only contains {1}

closure of {1} = .

15. Let be differentiable. Let Du f(0, 0) and Dvf(0,)) be the directional derivatives of fat (0, 0) in the directions of the unit vectors and , respectively. If Duf(0, 0) = and then

Ans. 4

Sol.

Now solving (1) and (2), we get

16. Let denote the boundary of the square region R with vertices (0, 0), (2, 0), (2, 2) and (0, 2) oriented in the counter-clockwise direction. Then = _________.

Ans. 12

Sol. Let R be a region enclosed by

17. The number of 5-Sylow subgroups in the symmetric group S5 of degree 5 is _________.

Ans. 6

Sol. O(S5) = 120 = 5 × 23 × 3

Number of Sylow 5 – subgroups = n5 = 1 + 5k

such that 1 + 5k|24, then either k = 0 or 1, then n5 = 1 or 6.

If n5 = 1, then S5 has unique subgroup of order 5 which means there are only 4 elements of order 5 in S5, but it is absurd as number of elements of order '5' in S5 =

So, n5 must be 6.

Answer = 6

Or

Result: The number of Sylow p-subgroups in Sp is (p – 2)!, where p is a prime number.

Here, p = 5, the Sylow 5 – subgroups in S5 is (5 – 2)! = 3! = 6.

18. Let I be the ideal generated by x2 + x + 1 in the polynomial ring R = [x], where denotes the ring of integers modulo 3. Then the number of units in the quotient ring R/I is _________.

Ans. 6

Sol.

There are six elements in R/I which are unit (Unity of R/I is 1 + 2 + x + 1>)

Inverse of 1 + is 1 + .

Inverse of 2 + is 2 + .

Inverse of 2x + 2 + is x + and vice versa.

Inverse of x + 1 + is 2x + and vice versa.

But 2x + 1 + and x + 2 + has no inverse in R/I.

So, answer is 6.

19. Let T : be a linear transformation such that

Then the rank of T is _________ .

Ans. 2

Sol. Let T : be a linear transformation such that and .

Since T : is a linear transformation, then is also a linear transformation

So, we get and given that

dim (Rang (T2)) = 2

Rank (T2) = 2

We know that Rank (T) rank (T2) = 2

Rank (T) 2 ...(1)

Since Rank (T2) = 2 the Nullity (T2) = 1

then T2 is singular

T is singular

Nullity (T) 1 rank (T) ≤ 2 ...(2)

(Using Rank-Nullity theorem)

From (1) and (2), we get

rank (T) = 2.

So, Ans. = 2.

20. Let y(x) be the solution of the following initial value problem = 0, x > 0, y(2) = = 4. Then y(4) = _________.

Ans. 32

Sol.

y(2) = 0, = 4

The given equation can be transformed to

General solution is given by

y = c1x2 + c2x3

y(2) = 0 4c1 + 8c2 = 0

= 4 4c1 + 12c2 = 4

c2 = 1

c1 = –2

y(x) = x3 – 2x2

y(4) = 64 – 32 = 32.

21. Let f(x) = x4 + 2x3 – 11x2 – 12x + 36 for x

The order of convergence of the Newton-Raphson method

xn + 1 = xn

with x0 = 2.1, for finding the root = 2 of the equation f(x) = 0 is _________.

Ans. 1

Sol. Given f(x) = x4 + 2x3 – 11x2 – 12x + 36

and Newton Raphson method as

xn + 1 = xn

Now order of convergence of Newton Raphson method is 2 if for finding the root '' of f(x) with multiplicity 1 (i.e. simple root) and order of convergence is 1 if root of f(x), i.e., with multiplicity more than one (i.e. not a simple root).

Here f(2) = 24 + 2.(23) – 11(22) – 12(2) + 36 = 0

and f'(2) = 4(2)3 + 6(2)2 – 22(2) – 12 = 0

order of convergence is 1.

22. If the polynomial p(x) = (x + 1)x interpolates the data

then = _________.

Ans. 1

Sol.

23. Consider the Linear Programming Problem P:

Maximize 2x1 + 3x2

subject to 2x1 + x2 6

–x1 + x2 1

x1 + x2 3

x1 0 and x2 0.

Then the optimal value of the dual of P is dual to _________.

Ans. 8

Sol. We know by Fundamental Theorem of duality the optimal value of primal and dual is same. So here we find the optimal value of primal P given by

Maximize z = 2x1 + 2x2

subject to 2x1 + x2 6

–x1 + x2 1

x1 + x2 3

x1, x2 0

We solve the given problem graphically

The corner points of the given region are (0, 0), (3, 0), (1, 2) and (0, 1)

Extreme Points z = 2x1 + 3x2

(0, 0) 0

(3, 0) 6

(1, 2) 8

(0, 1) 3

optimal value of primal is 8.

Hence, optimal value of dual is also 8.

24. Consider the Linear Programming Problem P:

Minimize 2x1 – 5x2

subject to 2x1 + 2x2 + s1 = 12

–x1 + x2 + s2 = 1

–x1 + 2x2 + s3 = 3

x1 0, x2 0, s1 0, s2 0 and s3 0.

If is a basic feasible solution of P, then x1 + s1 + s2 + s3 = _________.

Ans. 5

Sol. Minimize 2x1 – 5x2

subject to

2x1 + 3x2 + s1 = 12

–x1 + x2 + s2 = 1

–x1 + 2x2 + x3 = 3

x1 0, x2 0, s1 0, s2 0 and s3 0.

For basic feasible solution, we put 5 – 3 = 2 variables equal to zero

If we take s2 = s3 = 0, then

2x1 + 3x2 + s1 = 12

–x1 + x2 = 1

–x1 + 2x2 = 3

which gives x1 = 1, x2 = 2, s1 = 4

(1, 2, 4, 0, 0) is the basic feasible solution and given BFS is identical with the above

x1 + s1 + s2 + s3 = 1 + 4 + 0 + 0 = 5.

25. Let H be a complex Hilbert space. Let u, v H be such that = 2. Then

_________.

Ans. 2

Sol. Given u, v H and = 2

We have to find

Then

26. Let denote the ring of integers. Consider the subring R = of the field of complex numbers.

Consider the following statements:

P: 2 + is an irreducible element.

Q: 2 + is a prime element.

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (b)

Sol. First we find the units element of ring R = .

Let a + b is unit element in R

So, R such that (a + b) (c + d) = 1 ...(1)

Take conjugate on both sides, we get

(a – b) (c – d) = 1 ...(2)

Now multiply (1) and (2), we get

(a2 + 17b2) (c2 + 17d2) = 1

Above is true only if a2 + 17b2 = 1 and c2 + 17d2 = 1

a = ±1 and b = 0 as a, b

So, a + b = ±1

±1 are only units of R.

Let 2 + = (a + b) (c + d) ...(3)

for some a + b, c + d R

Take conjugate of equation (3) on both sides, we get

2 – = (a – b) (c – d) ...(4)

Multiply (3) and (4), we get

21 = (a2 + 17b2) (c2 + 17d2)

There are four possible cases

Case I: a2 + 17b2 = 3 and c2 + 17d2 = 7 any a, b, c, d such that above condition holds.

So, case-I is not possible.

Case II: a2 + 17b2 = 7 and c2 + 17d2 = 3 any a, b, c, d such above condition holds.

So, case-II is not possible.

Case III: a2 + 17b2 = 1 and c2 + 17d2 = 21

a = ±1, b = 0 and c = ±2, d = ±1 is only possible values of a, b, c, d such that above condition holds.

a + b = ±1 a + b is unit in R.

Case IV: a2 + 17b2 = 21 and c2 + 17d2 = 1

a = ±2, b = ±1 and c = ±1 and d = 0 is only possible values of a, b, c, d such that above condition holds.

c + d = ±1 c + d is unit in R.

In conclusion, either a + b is unit or c + d is unit in R

2 + is irreducible element in R.

Now 2 + | (2 + ) (2 – ) = 21

2 + | 3.7

Let 2 + |3, then a + b R

such that 3 = (a + b) (2 + )

3 = 2a – 17b 0 = a + 2b

After solving, we get values of a and b but do not belongs to

So, 2 + does not divide 3.

Similarly 2 + does not divide 7

2 + is not prime element.

So, option (b) is correct.

27. Consider the second-order partial differential equation (PDE) + (x2 + 4y2) = sin (x + y).

Consider the following statements:

P: The PDE is parabolic on the ellipse

Q: The PDE is hyperbolic inside the ellipse

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (a)

Sol.

Here R = 1, S = 4, T = x2 + 4y2

S2 – 4RT = 16 – 4(x2 + 4y2)

= 4(4 – (x2 + 4y2))

Equation is parabolic, when

S2 – 4RT = 0

i.e., 4 = x2 + 4y2

= 1.

Equation is Hyperbolic when

S2 – 4RT > 0

i.e., 4 – (x2 + 4y2) > 0

< 1

Option (a) is correct.

28. If u(x, y) is the solution of the Cauchy problem u(x, 0) = –x2, x > 0, then u(2, 1) is equal to

(a) 1 – 2e–2

(b) 1 + 4e–2

(c) 1 – 4e–2

(d) 1 + 2e–2

Ans. (c)

Sol.

Auxillary equations are

Solving first two numbers, we have

Solving last two numbers dy = du, we have y = u + c2

At x = t and y = 0, u = –t2

ln t = c1 and c2 = t2

t = ec1 and put value in c2 = t2

we get c2 = e2c1

Solution to above problem is given by

y – u = e2(ln x – y)

u = y – e2(ln x – y)

= y – x2 e–2y

u(2, 1) = 1 – 4e–2

Option (c) is correct.

29. Let y(t) be the solution of the initial value problem = f(t), a > 0, b > 0, a b, a2 – 4b = 0, y(0) = 0, = 0, obtained by the method of Laplace transform. Then

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Let L(y(t)) = Y(s) and L(f(t)) = F(s)

Taking Laplace Transform on both sides

s2Y(s) – sy(0) – + a(sY(s) – y(0)) + bY(s) = F(s)

(s2 + as + b) Y(s) = F(s)

Y(s) =

Y(s) =

(Also given a2 – 4b = 0)

Taking inverse Laplace on both sides

Option (a) is correct.

30. The critical point of the differential equation > 0, is a

(a) node and is asymptotically stable

(b) spiral point and is asymptotically stable

(c) node and is unstable

(d) saddle point and is unstable

Ans. (a)

Sol.

the roots of auxiliary equation m2 + = 0 having negative real parts

every solution of given equation tends to zero as t tends to infinity

the critical point of the given equation is asymptotically stable

m is real

Hence both roots are real

The critical point behaves like node

Option (a) is correct.

31. The initial value problem

= f(t, y), t > 0, y(0) = 1,

where f(t, y) = –10y, is solved by the following Euler method

yn + 1 = yn + h f(tn, yn), n 0,

with step-size h. They yn 0 as n provided.

(a) 0 < h < 0.2

(b) 0.3 < h < 0.4

(c) 0.4 < h < 0.5

(d) 0.5 < h < 0.55

Ans. (a)

Sol.

and f(t, y) = –10y

By Euler's method yn + 1 = yn + hf(tn, yn)

Now for n = 0

y1 = y0 + hf(t0, y0) = 1 + h(–10(1)) = 1 – 10h

For n = 1

y2 = y1 + hf(t1, y1) = (1 – 10h) + h(–10y1)

= (1 – 10h) – 10h(1 – 10h)

= (1 – 10h)2

Similarly yn = (1 – 10h)n

0 < h < 0.2

Option (a) is correct.

32. Consider the Linear Programming Problem P:

Maximize c1x1 + c2x2

subject to a11x1 + a12x2 b1

a21x1 + a22x2 b2

a31x1 + a32x2 b3

x1 0 and x2 0, where aij, bi and and cj are real numbers (i = 1, 2, 3; j = 1, 2).

Let be a feasible solution of P such that pc1 + qc2 = 6 and let all feasible solutions of P satisfy –5 c1x1 + c2x2 12.

Then, which one of the following statements is NOT true?

(a) P has an optimal solution.

(b) The feasible region of P is a bounded set.

(c) If is a feasible solution of the dual of P, then b1y1 + b2y2 + b3y3 6.

(d) The dual of P has at least one feasible solution.

Ans. (b)

Sol. Given that –5 c1x1 + c2x2 12

LPP has bounded solution

Using fundamental theorem, we know that this LPP has optimal solution.

So, option (a) is correct.

Also, as the primal of LPP is bounded or optimal solution implies that dual also have optimal solution. Hence atleast one feasible solution of dual

So, option (d) is correct.

Now using weak duality theorem we have minimum of dual of LPP maximum of primal of LPP at feasible points.

b1y1 + b2y2 + b3y3 qc2

b1y1 + b2y2 + b3y3 6

Hence option (c) is correct.

option (b) is not true.

33. Let L2[–1, 1] be the Hilbert space of real valued square integrable functions on [–1, 1] equipped with the norm .

Consider the subspace M = {f L2[–1, 1] :

For f(x) = x2, define d = inf {||f – g|| : g M}. Then

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Firstly define a map T : H as T(x) = where I is constant function as I(t) = 1 [–1, 1]. It can be seen that T is linear functional

Hence T is bounded so it is continuous

i.e. M is kernel of our define linear map T and it is closed subspace of H.

Now, the distance from f(x) = x2 to the subspace M is the length of the projection vector of f(x) = x2 on

Hence option (a) is correct.

34. Let C[0, 1] be the Banach space of real valued continuous functions on [0, 1] equipped with the supremum norm. Define T : C[0, 1] C[0, 1] by

Let R(T) denote the range space of T. Consider the following statements:

P: T is a bounded linear operator.

Q: T–1 : R(T) C[0, 1] exists and is bounded.

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (b)

Sol. Given T : C[0, 1] C[0, 1]

So T is bounded operator

First to check that T is one-one

On differentiating both sides, we get

f1(t) = f2(t)

Hence T is 1 – 1

Now if we choose

f(x) = xn

But ||f(t)|| = ||xn|| = 1

Fot T–1 is to be bounded or continuous, we have

||T(f)|| K||f||

But for choosen f(x) = xn, we have

= K

But this is not true always.

Hence it T–1 is not bounded.

35. Let l1 = {x = (x(1), x(2), ..., x(n), ...)| be the sequence space equipped with the norm Consider the subspce and the linear transformation T : X l1 given by (Tx) (n) = n x(n) for n = 1, 2, 3, ... . Then

(a) T is closed but not bounded

(b) T is bounded

(c) T is neither closed nor bounded

(d) T–1 exists and is an open map

Ans. (a)

Sol. Correct option is (a)

36. Let be given by fn(x) = n x3 e–nx for n = 1, 2, 3, .....

Consider the follwoing statements:

P : (fn) is equicontinuous on [0, 10],

Q: does NOT convere uniformly on [0, 10].

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (b)

Sol. P : (fn) is equicontinuous on [0, 10]

Lemma : If K is compact metric space, if fn belongs to C (K) for n = 1, 2, .... and if {fn}converges uniformly on K, then {fn} is equicontinuous on K.

Here fn = nx3e–nx

Apply Mn – test

Now, z = nx3 e–nx

Hence, {fn}is uniformly convergent on [0, 10].

So, by lemma {fn} is equicontinuous on [0, 10].

Therefore, P is true.

Q: does not converge uniformly on [0, 10]

fn = nx3e–nx

as R.H.S. is converent series, so by Weierstrass's M-test

is uniformly convergent on [0, 10].

So, Q is false

37. Let is given by

Consider the following statements:

P : f is continuous at (0, 0) but f is NOT differentiable at (0, 0).

Q : The directional derivative Du f(0, 0) of f at (0, 0) exists in the direction of every unit vector u .

Then

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (a)

Sol.

For continuity,

Put x = r cos .

y = r sin

= 0

Hence, f is continuous at (0, 0)

To check differentiability

= 0

be definition of differentiability in two variables we have f differentiable t (x0, y0) whenever we have

f(x0 + h, y0+ k) – f(x0, y0) = fx (x0, y0) h + fy(x0, y0) k +

We have

Now,

Hence, f is not differentiable

Q : Definition of directional derivative at point a = (a1, a2) along the direction u = (u1, u2) is

always exists

So, Q is true

Hence, option (a) is correct

38. Let V be the solid region in bounded by the paraboloid y = (x2+ z2) and the plane y = 4. Then the value of is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We have to find

39. Let f : be given by f(x, y) = 4xy – 2x2 – y4. Then f has

(a) a point of local maximum and a saddle point.

(b) a point of local minimum and a saddle point.

(c) a point of local maximum and a point of local minimum.

(d) two saddle points.

Ans. (a)

Sol. f(x, y) = 4xy – 2x2 – y4

fx= 4y – 4x, fxx= – 4

fy= 4x – 4y3, fyy = –12 y2, fxy = 4

Now, A = fxx fyy – (fxy)2

= (–4) (–12 y2) –16

= 48y2 – 16

Put fx= 0 4x = 4y

x = y ... (1)

and fy = 0 4x = 4y3

x = y3 ... (2)

On solving (1) and (2) we et, the critical points are P(0, 0), Q (1, 1), and R (–1, –1).

At P (0, 0), A = – 16 < 0

P (0, 0) is a saddle point.

At Q (1, 1) A = 48(1) – 16 = 32 > 0 and fxx < 0

(–1, –1) is also a point of local maxima

Option (a) is correct

40. The equation xy –z log y + exz = 1 can be solved in a neighbourhood of the point (0, 1, 1) as y = f(x, z) for some continuously differentiable function f. Then

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

41. Consider the following topologies on the set of all real numbers.

T1 is the upper limit topology having all sets (a, b] as basis.

T2 =

T3 is the standard topology having all sets (a, b) as basis.

Then

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Result: Let B and be basis for the topologies T and , respectively, on X. Then the following are equivalent

(1) is finer than T

(2) For each x X and for each B B containing x

Claim: T3 T1,

i.e., T1 is finer than T3

(where B is Basis of T3 and is Basis for T1}

Hence T3 T1

So, options (B) and (C) are rejected.

Hence, option (a) is correct.

42. Let denote the set of all real numbers. Consider the following topological spaces.

X1 = (, T1), where T1 is the upper limit topology having all sets (a, b] as basis.

X2 = (, T2), where T2 =

Then

(a) both X1 and X2 are connected.

(b) X1 is connected and X2 is NOT connected.

(c) X1 is NOT connected and X2 is connected.

(d) neither X1 nor X2 is connected.

Ans. (c)

Sol. Result: A topological space is connected iff the only sets which are both open as well as closed are and X.

X2 is connected

Let U X2 which is both open as well as closed.

Also Uc is finite, i.e., Uc = {x1, x2, ..., xn}

but now U is not closed because U is not finite.

X1 is not connected

Let U = (a, b]

Claim: U is both open as well as closed because Uc =

as is open and is open, so is their union.

So, Uc is open, i.e., U is closed.

So, X1 is not connected.

Hence, option (c) is correct.

43. Let be an inner product on the vector space over Consider the following statements:

(a) both P and Q are TRUE

(b) P is TRUE and Q is FALSE

(c) P is FALSE and Q is TRUE

(d) both P and Q are FALSE

Ans. (a)

Sol.

[by using Cauchy Schwartz inequality,

Now, we have

P is True.

Take u = v,

Option (a) is correct.

44. The G be a group of order 54 with center having 52 elements. Then the number of conjugacy classes in G is _________.

Ans. 145

Sol. G be group of order 54 = 625 and O(Z(G)) = 52 = 25.

There are 25 conjugacy classes and each contain single element as we know that a O(Cl(a)) = 25.

For element 'a' which does not belong to Z(G), the O(Cl(a)) > 1 and O(Cl(a)) =

O(N(a)) < O(G) and O(Z(G)) < O(N(a)) and O(N(a))|O(G)

O(N(a)) = 53

If a Z(G), then O(Cl(a)) =

So, 600 elements of G (which does not belong to Z(G)) divided into 120 conjugacy classes and each conjugacy class contain '5' elements.

So, total conjugacy classes of G are 25 + 120 = 145.

45. Let F be a finite field and F× be the group of all non-zero elements of F under multiplication. If F× has a subgroup of order 17, then the smallest possible order of the field F is _________.

Ans. 103

Sol. Given F is finite filed such that F*(= F\{0}) has subgroup of order 17.

By using Lagrange's theorem, O(F*) = 17K for some K

O(F) = 17K + 1

We just only think such smallest integer 'K' such that O(F) = (prime)n for some n

( Order of finite field is pn for a prime and n

If K = 1, then O(F) = 18 p prime

So, K = 1 is not possible.

If K = 2, then O(F) = 35 p prime

So, K = 2 is not possible.

If K = 3, then O(F) = 52 p prime

So, K = 3 is not possible.

If K = 4, then O(F) = 69 p prime

So, K = 4 is not possible.

If K = 5, then O(F) = 86 p prime

So, K = 5 is not possible.

If K = 6, the O(F) = 103 = (103)1, as 103 is a prime number.

Smallest possible order of such field is 103.

46. Let R = {z = x + iy : 0 < x < 1 and and be the positively oriented boundary of R. Then the value of the integral is _________.

Ans. 11

Sol. Let f(z) = ez – 2

By Argument principle

where N = Number of zeroes of f(z) inside counting multiplicity

P = Number of poles of f(z) inside counting multiplicity

Zeroes of f(z) are ez – 2 = 0

ez = 2

z = log 2 = ln 2 + out of which ln 2, ln 2 ± ln 2 ± ln 2 ± ln 2 ± ln 3 ± lies inside .

N = 11 and f(z) has no poles

P = 0.

Hence,

47. Let D = and f : D be the function defined by f(z) = .

If f(z) = for z D, then 6a2 = _________.

Ans. 3

Sol.

Given f(z) is analytic at z = 0 so in Taylor series expansion

48. The number of zeroes (counting multiplicity) of P(z) = 3z5 + 2i z2 + 7i z + 1 in the annular region {z : 1 < |z| < 7} is _________.

Ans. 4

Sol. P(z) = 3z5 + 2iz2 + 7iz + 1

Let F(z) = 7iz and G(z) = 3z5 + 2iz2 + 1

On |z| = 1, |f(z)| = |7iz| = 7|z| = 7 and |G(z)| = |3z5 + 2iz2 + 1| 3|z|5 + 2|z|2 + 1 |G(z)| 6

Hence on |z| = 1, |G(z)| < |F(z)|, therefore F(z) and F(z) + G(z) has same number of zeroes inside region |z| < 1.

P(z) has 1 zero inside |z| < 1.

Let Q(z) = 3z5 and R(z) = 2iz2 + 7iz + 1

On |z| = 7, |Q(z)| = |3z5| = 3|z|5 = 3.75 and |R(z)| = |2iz2 + 7iz + 1| 2|z|2 + 7|z| + 1 |R(z)| 2.72 + 7.7 + 1 = 148

Hence on |z| = 7, |R(z)| < |Q(z)|, therefore Q(z) and R(z) + Q(z) has same number of zeroes inside region |z| < 7.

P(z) has 5 zeroes inside |z| < 7

Thus total number of zeroes inside |z| < 7

Hence, P(z) has 4 zeroes inside annulus 1 < |z| < 7.

49. Let A be a square matrix such that det(xI – A) = x4(x – 1)2 (x – 2)3, where det (M) denotes the determinant of a square matrix M.

If rank (A2) < rank (A3) = rank (A4), then the geometric multiplicity of the eigenvalue 0 of A is _______ .

Ans. (*)

Sol. Given that A is a square matrix such that det (xI – A) = x4(x – 1)2 (x – 2)3

If rank (A2) < rank (A3) = rank (A4)

Above statement is wrong because

(as we know that

So, question is wrong.

50. If y = is the power series solution of the differential equation – 24x2y = 0, then = _________.

Ans. 2

Sol.

2a2 + 6a3x + (12a4 – 24a0)x2 + ... + = 0

Comparing terms, we have 12a4 – 24a0 = 0

51. If u(x, t) = Ae–t sin x solves the following initial boundary value problem

Then A = _________.

Ans. (*)

Sol. (Marks to all)

52. Let V = {p : p(x) = a0 + a1x + a1x + a2x2, a0, a1, a2 } be the vector space of all polynomials of degree at most 2 over the real field Let T : V V be the linear operator given by

T(p) = (p(0) – p(1)) + (p(0) + p(1)) x + p(0)x2.

Then the sum of the eigenvalues of T is _________ .

Ans. 1

Sol. T : V V be a linear transformation given by

T(p(x)) = (p(0) – p(1)) + (p(0) + p(1))x + p(0)x2

Let B = {1, x, x2} be basis of V

T(1) = 0 + 2x + x2 = 0.1 + 2.x + 1.x2

T(x) = (0 – 1) + (0 + 1)x + 0.x2 = –1.1 + 1.x + 0.x2

T(x2) = (0 – 1) + (0 + 1)x + 0.x2 = –1.1 + 1.x + 0.x2

Matrix representation of T w.r.t. basis 'B' is given by

Sum of eigenvalues of T = Trace (T) = 0 + 1 + 0 = 1.

53. The quadrature formula

is exact for all polynomials of degree 2, then = _________.

Ans. 2

Sol.

As formula is exact for all polynomials of degree 2

Take f(x) = 1

54. For each x (0, 1], consider the decimal representation x = .d1d2d3... di ... . Define f = [0, 1] by f(x) = 0 if x is rational and f(x) = 18 n if x is irrational, where n is the nuber of zeroes immediately after the decimal point upto the first non zero digit in the decimal representation of x.

Then the Lebesue integral

Ans. 2

Sol. We have x = .d1d2d3...

Where n = number of zeroes immediately after the decimal point upto first non-zero digit in the decimal representation of x

= 0 + 18 [0.1 – 0.01] + 18 × 2 [0.01 – 0.001] + 18 × 3 [0.001 – 0.0001] + ....

55. Let be an optimal solution of the following Linear Programming Problem P:

Maximize 4x1 + x2 – 3x3

subject to

2x1 + 4x2 + ax3 10

x1 – x2 + bx3 3

2x1 + 3x2 + 5x3 11

x1 0, x2 0 and x3 0, where a, b are real numbers.

If is an optimal solution of the dual of P, then p + q + r = _________. (round off to two decimal places).

Ans. 3.17

Sol. maximize z = 4x1 + x2 – 3x3

subject to

2x1 + 4x2 + ax3 + s1 = 10

x1 – x2 + bx3 + s2 = 3

2x1 + 3x2 + 5x3 + s3 = 11

We solve it by using Simplex Algorithm

As given is the optimal solution, so above table will be optimal simplex table.

From above table we can obtain variable as y1 = zj for s1, y2 = zj for s2, y3 = zj for s3

p + q + r = 3.17 (upto two decimal places).