GATE MATHEMATICS 2015
Previous Year Question Paper with Solution.

1.    Let T : be a linear map defined by

T(x, y, z, w) = (x + z, 2x + y + 3z, 2y + 2z, w).

Then, the rank of T is equal to _________.

Ans.

Sol. We have, T : defined by

T(x, y, z, w) = (x + z, 2x + y + 3z, 2y + 2z, w)

Let {e1, e2, e3, e4} be the basis of . Then,

T(e1) = T(1, 0, 0, 0) = (1, 2, 0, 0)

= 1(1, 0, 0, 0) + 2(0, 1, 0, 0) + 0(0, 0, 1, 0) + 0(0, 0, 0, 1)

T(e2) = T(0, 1, 0, 0) = (0, 1, 2, 0)

= 0(1, 0, 0, 0) + 1(0, 1, 0, 0) + 2(0, 0, 1, 0) + 0(0, 0, 0, 1)

T(e3) = T(0, 0, 1, 0) = (1, 3, 2, 0)

= 1(1, 0, 0, 0) + 3(0, 1, 0, 0) + 2(0, 0, 1, 0) + 0(0, 0, 0, 1)

T(e4) = T(0, 0, 0, 1) = (0, 0, 0, 1)

= 0(1, 0, 0, 0) + 0(0, 1, 0, 0) + 0(0, 0, 1, 0) + 1(0, 0, 0, 1)

[MT]4 × 4 =

Rank of T = 3

Alternative Method:

Let (x, y, z, w) Ker T

T(x, y, z, w) = (0, 0, 0, 0)

(x + z, 2x + y + 3z, 2y + 2z, w) = (0, 0, 0, 0)

x + z = 0 ...(i)

2x + z ...(ii)

2y + 2z = 0 ...(iii)

w = 0 ...(iv)

On solving above equations, we get

x = 0, y = z and w = 0

Ker T = (0, z, z, 0)

*****

So, rank of T = 4 – 1 = 3.

2.    Let M be a 3 × 3 matrix and suppose that 1, 2 and 3 are the eigenvalues of M.

If M–1 = for some scalar , then is equal to _________.

Ans.

Sol.

3.    Let M be a 3 × 3 singular matrix and suppose that 2 and 3 are eigenvalues of M. Then, the number of linearly independent eigenvectors of M3 + 2M + I3 is equal to _________.

Ans.

Sol.

Now, L = M3 + 2M + I3

L has three distinct eigenvalues 1, 13 and 34.

Hence, L has three eigenvectors.

4.    Let M be a 3 × 3 matrix such that M and suppose that m3 for some Then, is equal to _________.

Ans.

Sol.

On comparing, we have

–2a = 6

b = –3

c = 0

a = –3, b = –3 and c = 0

5.    Let f : be defined by f(x) = Then, the function f is

(a) uniformly continuous on [0, 1) but not on (0, )

(b) uniformly continuous on (0, ) but not on [0, 1]

(c) uniformly continuous on both [0, 1) and (0, )

(d) neither uniformly continuous on [0, 1) nor uniformly continuous on (0, )

Ans. (c)

Sol. The function f is uniformly continuous on both [0, 1) and (0, ).

6.    Consider the power series where

an =

The radius of convergence of the series is equal to _________.

Ans.

Sol. Given, power sereis where

an =

Let R be the radius of convergence,

When, n is odd, then

Now, R = min(3, 5) = 3

So, radius of convergence is 3.    

7.    Let C = {z C : |z – i| = 2}. Then, is equal to _________.

Ans.

Sol. We have, C = {z C|z – i| = 2}, which represent the following circle.

8.    Let X ~ B and Y ~ U(0, 1). Then is equal to _________.

Ans.

Sol. It is equal to 6.

9.    Let the random variable X have the distribution function,

F(x) =

Then, is equal to _________.

Ans.

Sol. The cumulative distribution function is

F(x) =

The only points that receive non-zero. Probability are 0, 1, 2 and 3.

The probability mass function at each point is the change in the cumulative distribution function at the point.

10.    Let X be a random variable having the distribution function,

F(x) =

Then, E(X) is equal to _________.

Ans.

Sol. The comulative distribution function is

F(x) =

The only points that recieve non-zero probability are 0, 1, 2 and .

The probability mass function at each point is the change in cumulative distribution function at the point.

11.    In an experiment, a fair die is rolled untill two sixes are obtained in succession. The probability that the experiment will end in the fifth trial is equal to

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The fifth trial is equal to .

12.    Let x1 = 2.2, x2 = 4.3, x3 = 3.1, x4 = 4.5, x5 = 1.1 and x6 = 5.7 be the observed values of a random sample of size 6 from a U( – 1, + 4) distribution where is unknown. Then, a maximum likelihood estimate of is equal to

(a) 1.8

(b) 2.3

(c) 3.1

(d) 3.6

Ans. (a)

Sol. A maximum likelihood estimate of is equal to 1.8.

13.    Let = {(x, y) R2 |x2 + y2 < 1} be the open unit disc in R2 with boundary If u(x, y) is the solution of the Dirichlet problem

uxx + uyy = 0 in

u(x, y) = 1 – 2y2 in

Then u is equal to

(a) –1

(b)

(c)

(d) 1

Ans. (c)

Sol. It is equal to .

14.    Let c Z3 be such that is a field. Then, c is equal to _________.

Ans.

Sol. Take c = 2. Then,

f(X) = X3 + cX + c = X3 + 2X + 1

f(0) = 1

0 is not root.

f(1) = 13 + 2 × 1 + 1 = 4

1 is not root.

f(2) = 23 + 2 × 2 + 1 = 13

2 is not root.

So, X3 + 2X + 1 is irreducible over Z3[X].

is field.

Hence, correct value of c is 2.

15.    Let V = C1[0, 1], X = (C[0, 1], and Y = (C[0, 1], || ||2). Then, V is

(a) dense is X but not in Y

(b) dense in Y not in X

(c) dense in both X and Y

(d) neither dense in X nor dense in Y

Ans. (c)

Sol. V is dense in both X and Y.

16.    Let T : (C[0, 1], ) be defined by T(f) = for all f C[0, 1]. is equal to _________.

Ans.

Sol. is equal to 1.

17.    Let be the usual topology on R. Let be the topology on R generated by

B =

Then, the set is

(a) closed in (R, ) but not in (R, )

(b) closed in (R, ) but not in (R, )

(c) closed in both (R, ) and (R, )

(d) neither closed in (R, ) nor closed in (R, ).

Ans. (c)

Sol. The given set is closed in both (R, ) and (R, ).

18.    Let X be a connected topological space such that there exists a non-constant continuous function f : X , where is equipped with the usual topology. Let f(X) = {f(x) : x X). Then,

(a) X is countable but f(X) is uncountable

(b) f(X) is countable but X is uncountable

(c) both f(X) and X are countable

(d) both f(X) and X are uncountable

Ans. (d)

Sol. Let f(X) = {(X): X X}, then both f(X) and X are uncountable.

19.    Let d1 and d2 denote the usual matric and the discrete matric on , respectively.

Let f : (, d1) (, d2) be defined by f(x) = x, x . Then

(a) f is continuous but f–1 is not continuous

(b) f–1 is continuous but f is not continuous

(c) both f and f–1 are continuous

(d) neither f nor f–1 is continuous

Ans. (b)

Sol. f–1 is continuous but f is not continuous.

20.    If the trapezoidal rule with single interval [0, 1] is exact for approximating the integral Then, the value of c is equal to _________.

Ans.

Sol. Given, interval is [0, 1].

Take h = 1,

y = f(x) = x3 – cx2

y(0) = f(0) = 0

and y(1) = f(1) = 1 – c

By trapezoidal rule,

21.    Suppose that the Newton-Raphson method is applied to the equation 2x2 + 1 – = 0 with an initial approximation x0 sufficiently close to zero. Then, for the root x = 0, the order of convergence of the method is euqal to _________.

Ans.

Sol. Given, f(x) = 2x2 + 1 –

Take initial approximation at x0 = 0.1

Then, f(0. 1) = 0.0099

f'(0. 1) = 0.1979

= 0.0499

f(0.0499) = 0.0024

= 0.0257

f(0.0257) = 0.0006

= 0.01400.

= 0.9988

1.

22.    The minimum possible order of a homogeneous linear ordinary differential equation with real constant coefficients having x2 sin sin(x) as a solution is equal to _________.

Ans.

Sol. Solution is equal to 6.

23.    The Lagrangian of a system in terms of polar coordinates is given by

L =

where, m is the mass, g is the acceleration due to gravity and denotes the derivative of s with respect to time. Then, the equations of motion are

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given, Lagrangian of a system in terms of polar coordinates (r, 0) is

L =

The Lagrangian equation of motion are

From Eq. (i), we have

From Eq. (i), we have

Hence, Lagrangian equation of motion are

24.    If y(x) satisfies the initial value problem

(x2 + y)dx = xdy, y(1) = 2. Then, y(2) is equal to _________.

Ans.

Sol. Given, initial value problem is

(x2 + y)dx = xdy with y(1) = 2

Using y(1) = 2, we have

2 = 1 + c c = 1

Now, y(x) = x2 + x

y(2) = 22 + 2 = 6

25.    It is known that Bessel functions Jn(x), for n 0, satisfy the identity

for all t > 0 and x

The value of is equal to _________.

Ans.

Sol. The correct answer is 1.

26.    Let X and Y be two random variables having the joint probability density function

f(x, y) =

Then, the conditional probability is equal to

(a)

(b)

(c)

(d)

Ans. (d)

Sol. The correct answer is 8/9.

27.    Let = (0, 1] be the sample space and let be a probability function defined by

P(0, x] =

Then, P is equal to _________.

Ans.

Sol. Let = (0, 1] be the sample and be a probability function defined by

28.    Let X1, X2 and X3 be independent and identically distributed random variables with E(X1) = 0 and if is defined through the conditional expectation

Then, is equal to _________.

Ans.

Sol. The correct answer is 2.5.

29.    Let X ~ Poisson where > 0 is unknown. If is the unbiased estimator of then is equal to _________.

Ans.

Sol. The correct answet is 9.

30.    Let X1, ... , Xn be a random sample from N(µ, 1) distribution, where . For testing the null hypothesis H0 : µ = 0 against the alternative hypothesis H1 : consider the critical region

where, c is some real constant. If the critical region R has size 0.025 and power 0.7054, then the value of the sample size n is equal to _________.

Ans.

Sol. The value of sample size n is equal to 25.

31.    Let X and Y be independently distributed central chi-squared random variables with degrees of freedom and respectively. If and m + n = 14, then is equal to _________.

Ans.

Sol. The correct answer is .

32.    Let X1, X2, X3 ... be a sequence of independent and identically distributed random variables with P(X1 = 1) = and P(X1 = 2) = . If for n = 1, 2, 3, ..., then is equal to _________.

Ans.

Sol. It is equal to 1.

33.    Let u(x, y) = 2 f(y) cos (x – 2y), (x, y) be a solution of the initial value problem

2ux + uy = u

u(x, 0) = cos(x).

Then, f(1) is equal to

(a)

(b)

(c) e

(d)

Ans. (b)

Sol. Given, equation is

2ux + uy = u

u(x, y) = eyg(x – 2y)

Using u(x, 0) = cos x

But given that,

u(x, y) = 2f(y) cos(x – 2y)

On comparing, we get

34.    Let u(x, t) x t 0, be the solution of the initial value problem

utt = uxx

u(x, 0) = x

ut(x, 0) = 1.

Then, u(2, 2) is equal to _________.

Ans.

Sol. Given, initial value problem is

utt = uxx

u(x, 0) = x

ut(x, 0) = 1

Hence, c = 1, f(x) = x and g(x) = 1

The D'Alembert's solution is

35.    Let W = Span be a subspace of the Euclidean space , Then, the square of the distance from the point (1, 1, 1, 1) to the subspace W is equal to _________.

Ans.

Sol.

v = (1, 1, 1, 1) = (1, 1, 0, 0) + (0, 0, 1, 1)

Then, (1, 1, 0, 0) is the component of v, which is orthogonal to W i.e.

= (1 – 1)2 + (1 – 1)2 + (1 – 0)2 + (1 – 0)2

= 2

36.    Let T : be a linear map such that the null space of T is {(x, y, z, w) : x + y + z + w = 0} and the rank of (T – 4I4) is 3. If the minimal polynomial of T is , then is equal to _________.

Ans.

Sol. T : be a linear map such that the null space of T is {(x, y, z, w) : x + y + z + w = 0}

Rank of T = 3.

and given that, rank of (T – 4l3) = 3

Geometric multiplicity of 4 = 4 – 3 = 1

= 1.

37.    Let M be an invertible Hermitian matrix and let x, y be such that x2 < 4y. Then,

(a) both M2 + xM + yI and M2 – xM + yI are singular

(b) M2 + xM + yI is singular but M2 – xM + yI is non-singular

(c) M2 + xM + yI is non-singular but M2 – xM + yI is singular

(d) both M2 + xM + yI and M2 – xM + yI are non-singular

Ans. (d)

Sol.

det (M2 + xM + yl) = (1 + x + y) (4 + 2x + 1) (9 + 3x + 1)

Let x = y = 1

det (M2 + xM + yl) = 3 × 7 × 13 = 273 0

M2 + xM + yl is non-singular.

det (M2 – xM + yl) = (1 – x + y) (4 – 2x + y) (9 – 3x + y)

Let x = y = 1

det (M2 – xM + yl) = 1 × 3 × 7 = 21 0

M2 – xM + yl is non-singular.

38.    Let G = {e, x, x2, x3, y, xy, x2y, x3y} with O(x) = 4, O(y) = 2 and xy = xy3. Then, the number of elements in the centre of the group G is equal to

(a) 1

(b) 2

(c) 4

(d) 8

Ans. (b)

Sol. We have, G = {e, x, x2, x3, y, xy, x2y, x3y} with O(x) = 4, O(y) = 2 and xy = yx3 is a dihedral group of order 8.

Centre of G, here only two elements.

39.    The number of ring homophrism from Z2 × Z2 to Z4 is equal to _________.

Ans.

Sol. Let Z4 = {0, 1, 2, 3}

0 + 0 = 0

1 + 1 = 2

2 + 2 = 0

3 + 3 = 2

0 is only independent element.

Number of ring homorphism from Z2 × Z2 Z4 is 1.

40.    Let p(x) = 9x5 + 10x3 + 5x + 15 and q(x) = x3 – x2 – x – 2 be two polynomials in Q[x]. Then,
over Q,

(a) p(x) and q(x) are both irreducible

(b) p(x) is reducible but q(x) is irreducible

(c) p(x) is irreducible but q(x) is reducible

(d) p(x) and q(x) are both reducible

Ans. (c)

Sol. We have, p(x) = 9x5 + 10x3 + 5x + 15

Here, 5|15, 5|5, 5|10

But 5|9 and 25|15.

Hence, by Eisenstein's criterion, p(x) is irreducible over Q.

Now, q(x) = x3 – x2 – x – 2

q(2) = 23 – 22 – 2 – 2 = 0

x – 2 is a factor of q(x).

q(x) is reducible over Q.

41.    Consider the linear programming problem

Maximize 3x + 9y,

Subject to 2y – x 2

3y – x 0

2x + 3y 10

x, y 0.

Then, the maximum value of the objective function is equal to _________.

Ans.

Sol. We have,

Maximize z = 3x + 9y

Subject to 2y – x 2

3y – x 0

2x + 3y 10

x, y 0

The graph is as follows:

z(0, 0) = 3 × 0 + 9 × 0 = 0

z(0, 1) = 3 × 0 + 9 × 1 = 9

z(2, 2) = 3 × 2 + 9 × 2 = 24

=

maximum value = 24.

42.    Let S = and T = S {(0, 0)}. Under the usual metric on

(a) S is closed but T is NOT closed

(b) T is closed but S is NOT closed

(c) both S and T are closed

(d) neither S nor T is closed

Ans. (d)

Sol. The correct answer is (d).

43.    Let H = Then, H

(a) is bounded

(b) is closed

(c) is a subspace

(d) has an interior point

Ans. (b)

Sol. It is closed.

44.    Let V be a closed subspace of L2[0, 1] and let f, g L2[0, 1] be given by f(x) = x and g(x) = x2.
If = Span {f} and Pg is the orthogonal projection of g on V, Then, (g – Pg) (x), x [0, 1], is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The correct option is .

45.    Let p(x) be the polynomial of degree atmost 3 that passes through the points (–2, 12), (–1, 1),
(0, 2) and (2, –8). Then, the coefficient of x3 in p(x) is equal to _________.

Ans.

Sol. Let p(x) = a0 + a1x + a2x2 + a3x3.

where, a0, a1, a2 and a3 are to be determinded.

Since, p(x) passes through the points

(–2, 12), (–1, 1), (0, 2) and (2, –8). Therefore,

a0 – 2a1 + 4a2 – 8a3 = 12 ...(i)

a0 – a1 + a2 – a3 = 1 ...(ii)

a0 = 2 ...(iii)

a0 + 2a1 + 4a2 + 8a3 = –8 ...(iv)

Putting a0 = 2, we get

–2a1 + 4a2 – 8a3 = 10 ...(v)

–a1 + a2 – a3 = –1 ...(vi)

2a1 + 4a2 + 8a3 = –10

On solving Eqs. (iv), (v) and (vi), we get

a1 = 3, a2 = 0 and a3 = – 2,

Thus, we have p(x) = 2 + 3x – 2x3

Coefficient of x3 is –2.

46.    If for some the integration formula

=

holds for all polynomials p(x) of degree atmost 3, then the value of is equal to _________.

Ans.

Sol. We have,

Let p(x) = x

Let p(x) = x2

Let p(x) = x3

From Eqs. (i) and (ii), we get

47.    Let y(t) be a continuous function on [0, ) whose Laplace transform exists. If y(t) satisfies

then, y(1) is equal to _________.

Ans.

Sol. y(1) is equal to 28.

48.    Consider the initial value problem

If y(x) 0 as x 0+, then is equal to _________.

Ans.

Sol. Given, initial value problem is

Let x = ez

Auxiliary equation is

The solution of the given equation is

y(x) = c1e3z + c2e–2z

On putting z = log x, we get

y(x) = c1x3 + c2x–2 ...(ii)

Using y(1) = , we get

= c1 + c2 ...(iii)

Using 6, we get

6 = 3c1 – 2c2 ...(iv)

On solving Eqs. (iii) and (iv), we get

c1 = 6 – 2

c2 = – c1

= – 6 + 2

= 3 – 6

The solution of the given equation is

y(x) = (6 – 2)x3 + (3 – 6)x–2

49.    Define f1, f2 : [0, 1] by f1(x) =

Then,

(a) f1 is continuous but f2 is not continuous

(b) f2 is continuous but f1 is not continuous

(c) both f1 and f2 are continuous

(d) neither f1 nor f2 is continuous

Ans. (a)

Sol. f1 is continuous but f2 is not continuous.

50.    Consider the unit sphere

S = {(x, y, z) : x2 + y2 + z2 = 1}

and the unit normal vector = (x, y, z) at each point (x, y, z) on S. The value of the surface integral is equal to _________.

Ans.

Sol. It is equal to 4.

51.    Let D = . Define

f(x, y) =

Then, the minimum value of f on D is equal to _________.

Ans.

Sol. We have, D =

Putting, fx = 0 and fy = 0, we get

On solving Eqs. (ii) and (iii), we get

x = 10, and y = 10

fxx(10, 10) fyy(10, 10) – [fxy(10, 10)]2

=

Further, fxx(10, 10) = 1 > 0 and fyy(10, 10) = 1 > 0

(10, 10) is the minimum point.

Note: f(a, b) is an extreme value of f(x, y), if fx(a, b) = 0 = fy(a, b) and fxx(a, b) fyy(a, b) – [fxy(a, b)]2 > 0.

This extreme value is a maximum or a minimum accordingly as fxx(a, b) or [fyy(a, b)] is negative or positive.

Minimum value = f(10, 10)

52.    Let D = {z C : |z| < 1}. Then, there exists a non-constant analytic function f on D such that for all n = 2, 3, 4, ...

(a)

(b)

(c)

(d)

Ans. (c)

Sol. The correct answer is .

53.    Let be the Laurent series expansion of f(z) = in the annulus . Then, is equal to _________.

Ans.

Sol. We have, = ... a–2z–2 + a–1z–1 + a0 + a1z + a2z2 ...

54.    The value of is equal to _________.

Ans.

Sol. It is equal to 2.

55.    Suppose that among all continuously differentiable functions y(x), x with y(0) = 0 and y(1) = , the function y0(x) minimizes the functional

Then, is equal to

(a) 0

(b)

(c)

(d)

Ans. (b)

Sol. Given function is

Using y(0) = 0, d = 0