PYQ 2019 - VedPrep

GATE MATHEMATICS 2019
Previous Year Question Paper with Solution.

1. For a balanced transportation problem with three sources and three destinations where costs, a vailabilities and demands are all finite and positive, which one of the following statements is false?

(a) The transportation problem does not have unbounded solution

(b) The number of non-basic variables of the transportation problem is 4

(d) The transportation problem has at most 5 basic feasible solutions

Ans. (d)

Sol. The transportation problem has at most 5 basic feasible solutions.

2. Let f : [a, b] R (the set of all real numbers) be any function which is twice differentiable in(a, b) with only one root in (a, b). Let and denote the first and second order derivatives of f(x) with respect to x. If is a simple root and is computed by the Newton-Raphson method, then the method converges if

(a)

(b)

(c)

(d)

Ans. (a)

Sol. We know that f : (a, b) R be any function which is twice differentiable in (a, b) and only one root in (a, b). The function is convergent by Newton-Raphson method if and only if

3. Let f : C C (the set of all complex numbers) be defined by f(x + iy) = x³ + 3xy² + i(y³ + 3x²y), i = . Let denotes the derivative of f with respect to z. Then which one of the following statements is true?

(a) exists and

(b) f is analytic at the origin

(d) f is differentiable at 1

Ans. (d)

Sol. Given, f(x + iy) = x³ + 3xy² + i(y³ + 3x²y)

Let u = x³ + 3xy²

and v = y³ + 3x²y

f is differentiable at i and 1.

But f is not analytic at origin.

4. The partial differential equation (x² + y² – 1) is

(a) parabolic in the region x² + y² > 2

(b) hyperbolic in the region x² + y² > 2

(d) hyperbolic in the region 0 < x² + y² < 2

Ans. (d)

Sol. We have, (x² + y² – 1)

Here a = x² + y² – 1, b = 2

and c = x² + y² – 1

Now, b² – 4ac = (2)² – 4(x² + y² – 1)²

For hyperbolic b² – 4ac > 0

5. If u_n = n = 1, 2, 3,..., then which one of the following statements is true?

(a) Both the sequence and the series are convergent

(b) Both the sequence and the series are divergent

(d)

Ans. (c)

Sol. We have

6. Let = {(x, y, z) R³ : –1 < x < 1, – 1 < y < 1, – 1 < z < 1} and : R be a function whose all second order partial derivatives exist and are continuous. If satisfies the Laplace equation = 0 for all (x, y, z) , then which one of the following statements is true in ?

(R is the set of all real Numbers, and R³ = {(x, y, z) : x, y, z R}).

(a) is solenoidal but not irrotational

(b) is irrotational but not solenoidal

(d) is neither solenoidal nor irrotational

Ans. (c)

Sol.

Find vector point function

Here, we get div = 0

7. Let X = {(x₁, x₂,...) : x₁ R and only finitely many are non-zero} and d : X × X R be a metric on X defined by

d(x, y) = y = (y₁, y₂,...) in X.

(R is the set of all real numbers and N is the set of all natural numbers.)

Consider the following statements:

P : (X, d) is a complete matric space.

Q : The set {x X : d(0, x) 1} is compact, where, 0 is the zero element of X.

Which of the above statements is/are true?

(a) Both P and Q

(b) P only

(d) Neither P nor Q

Ans. (d)

Sol. P : Since every Cauchy sequence in X defined by d(x, y) = sup|x_i – y_i| for x = (x₁, x₂, ...), y = (y₁, y₂, ...) has not a limit point such that in space (X, d), sequence (x_n), does not exist.

x X such that x_n x as n so, (X, d) is not complate metric space.

Q : Set A said to be compact if it is complete and totally bounded. Since, the set {x X : d (0, x) 1} is not totally bounded so, it is not compact.

8. Consider the following statements:

I. The set Q × Z is uncountable.

II. The set {f : f is a function from N to {0, 1}} is uncountable.

III. The set { : p is a prime number} is uncountable.

IV. For any infinite set, there exists a bijection from the set to one of its proper subsets.

(Q is the set of the rational numbers, Z is the set of all integers and N is the set of all natural numbers)

Which of the above statements are true?

(a) I and IV

(b) II and IV

(d) I, II and IV

Ans. (b)

Sol. I. Cartesian product of Q × Z = pq, n(Q) = p = countable n(z) = q

II. F : N {0, 1} number of function = 2^N = uncountable

III. A = { is prime number}

Here, p is a infinite set,

But is a selected number.

IV. A function is bijective

function is one-one and onto

For any infinite set, there exists a bijection from the set to one of its proper subset.

9. Let f : R² R be defined by

(R is the set of all real numbers

and R² = {(x, y) : x, y R})

Which one of the following statements is true?

(a) f has a local maximum at origin

(b) f has a local minimum at origin

(d) The origin is not a critical point of f

Ans. (c)

Sol. We have, f(x, y) = x⁶ – 2x²y – x⁴y + 2y²

It has no local minima and maxima.

f has a saddle point at origin.

10. Let by any sequence of real numbers such that if the radius of convergence of is r, then which one of the following statements is necessarily true?

(a) r³ 1 or r is infinite

(b) r < 1

(c)

(d)

Ans. (a)

Sol. Radius of convergence,

11. Let T₁ be the co-countable topology on R (the set of real numbers) and T₂ be the co-finite topology on R.

Consider the following statements:

I. In (R, T₁), the sequence converges to 0.

II. In (R, T₂), the sequence converges to 0.

III. In (R, T₁), there is no sequence of rational numbers which converges to

IV. In (R, T₂), there is no sequence of rational numbers which converges to

Which of the above statements are true?

(a) I and II

(b) II and III

(d) I and IV

Ans. (b)

Sol. II. In confinite topology, only closed subset are finite sets. Since

III. The countable topology on any set X consists of the empty set and all countable subsets of X. Rational sequence (where q_n 0) in T₂ space is convergent if it is eventually constant.

12. Let X and Y be normed linear spaces, and let T : X Y be any bijective linear map with closed graph. Then which one of the following statements is true?

(a) The graph of T is equal to X × Y

(b) T^–1 is continuous

(d) T is continuous

Ans. (c)

Sol. According to closed graph theorem (functional analysis), any continuous linear bijection T : X Y from one normed linear space to another is necessarily an isomorphism in the sense that the inverse map is also continuous and linear.

This implies that the projection from to X, which is continuous linear bijection conversely, to deduce this claim. From closed graph theorem, we conclude that the graph of the inverse T^–1 is the reflection of the graph o T hence, the graph of T^–1 is closed.

13. Let g : R² R² be a function defined by g(x, y) = (e^x cos y, e^x sin y) and (a, b) = g (R is the set of all real numbers and R² = {(x, y) : x, y R})

Which one of the following statements is true?

(a) g is injective

(b) If h is the continuous inverse of g, defined in some neighbourhood of (a, b) R², such that

h(a, b) = then the Jacobian of h at (a, b) is e²

h(a, b) = then the Jacobian of h at (a, b) is e^–2.

(d) g is surjective

Ans. (c)

Sol. We have, g : R² R² be a function defined by g(x, y) = (e^x cos y, e^x sin y) and (a, b) = g suppose f : Rⁿ R^m is a function which taken as input the vector X Rⁿ and produces as output the vector f(x) R^m, then the jacobian matrix J of f is an m × n matrix denoted as

From the above information, option (c) can be concluded as correct.

14. Let u_n = (the set of all natural numbers). Then is equal to ......... .

Ans. 0

Sol.

15. If the differential equation is solved using the Euler's method with step-size h = 0.1, then y(1.2) is equal to ......... (round off to 2 places of decimal).

Ans. 2.47

Sol. We have,

= 2 + (0. 1)(2.236)

= 2 + 0.2236

= 2.2236

So, the approximation to the solution at

x₁ = 1.1 is y₁ = 2.2236

At the next step we have,

So, the approximation to the solution at x₂ = 1.2 is y₂ = 2.47.

y(1.2) = 2.47.

16. Let f be any polynomial function of degree at most 2 over R (the set of all real numbers). If the constants a and b are such that = af(x) + 2f(x + 1) + bf(x + 2), for all x R, then 4a + 3b is equal to ......... (round off to 2 places of decimal).

Ans. –7.5

Sol. Let f(x) = x² + ax + b; = 2x + a

Given, = af(x) + 2f(x + 1) + bf(x + 2)

2x + a = a(x² + ax + b) + 2(x² + 2x + 1 + ax + a + b) + b(x² + 4x + 4 + ax + 2a + b)

2x + a = x²(a + 2 + b) + x(a² + 4 + 2a + 4b + ab + ab + 2a + 2b + 2 + 4b + 2ab + b²

Equating the coefficient of x², x and constant terms

We get, a + b + 2 = 0 ...(i)

a² + 2a + 4b + ab + 2 = 0 ...(ii)

b² + 2a + 6b + 2ab + 2 = a ...(iii)

Solving Eqs. (i), (ii) and (iii), we get

17. Let L denotes the value of the line integral , where C, a circle of radius 2 with centre at origin of the xy-plane, is traversed once in the anti-clockwise direction. Then is equal to ......... .

Ans. 32

Sol.

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

18. The temperature T : at any point T(x, y, z) in inversely proportional to the sequare of the distance of T from the origin. If the value of the temperature T at the point R(0, 0, 1) is then the rate of change of T at the point Q(1, 1, 2) in the direction of is equal to ......... (round off to 2 places of decimal).

(R is the set of all real numbers, R³ = {(x, y, z): x, y, z belongs to R} and denotes R³ excluding the origin)

Ans. 0.22

Sol. Given, T (x, y, z) where k is the constant of proportionality.

Also, we have,

Now, rate of change of T is given by

Now, the rate of change of T in direction of , i.e. in the direction of is

19. Let f be a continuous function defined on [0, 2] such that f(x), x = 0, y = 0 and x = b is where b [0, 2], then f(1) is equal to ......... (round off to 1 place of decimal).

Ans. 0.5

Sol.

20. If the characteristic polynomial and minimal polynomial of a square matrix A are ( – 1) ( + 1)⁴ ( – 2)⁵ and ( – 1) ( + 1) ( – 2), respectively, then the rank of the matrix A + I is ........., where I is the identify matrix of appropriate order.

Ans. 6

Sol. A is diagonalisable if its minimal polynomial is a product of distrinct linear monic factors.

M_A() = ( – 1) ( + 1) ( – 2)

– 1 = 0 = 1

+ 1 = 0 = – 1

– 2 = 0 = 2

P as such that |P| 0 and A = P^TDP

D = diag (1, –1, –1, –1, –1, 2, 2, 2, 2, 2)

A + l = P^–1DP + l = P^–1(D + l)P

P(A + l) = P(D + l)

= P (diag (2, 0, 0, 0, 0, 3, 3, 3, 3, 3)

= 6.

21. Let be a primitive complex cube root of unity and i = Then the degree of the field extension over Q (the field of rational numbers) is .........

Ans. 4

Sol. E(f) is a f-vector space.

have two intermediate field having degree of extension, 2.

22. Let C : cos t + i sin t, 0

Then the greatest integer less than or equal to is ......... .

Ans. 2

Sol.

|x| = 1

|cos t + i sin t| = 1

2z² – 5z + 2 = 0

Using Cauchy's Residue theorem,

= 2.09

Greatest integer = 2.

23. Consider the system:

3x₁ + x₂ + 2x₃ – x₄ = a,

x₁ + x₂ + x₃ – 2x₄ = 3,

x₁, x₂, x₃, x₄ 0

If x₁ = 1, x₂ = b, x₃ = 0, x₄ = c is a basic feasible solution of the above system (where a, b and c are real constants), then a + b + c is equal to ......... .

Ans. 7

Sol. We have,

3x₁ + x₂ + 2x₃ – x₄ = a

x₁ + x₂ + x₃ – 2x₄ = 3

x₁, x₂, x₃, x₄ 0

x₁ = 1, x₂ = b, x₃ = 0, x₄ = c

∴ 3 + b + 0 – c = a

and 1 + b + 0 – 2c = 3

b – c = a – 3 ...(i)

b – 2c = 2 ...(ii)

Solving (i) and (ii)

b = 2c – 8, c = a – 5

Since, x₁, x₂, x₃, x₄ 0

b 0, c 0

2a – 8 0, a 5

a 4, a 5

a 5, a = 5

b = 2(5) – 8 = 2

c₂ = 5 – 5 = 0

a + b + c = 5 + 2 + 0 = 7.

24. Let f : C C be a function defined by f(z) = z⁶ – 5z⁴ + 10. Then the number of zeros of f in {z C : |z| < 2} is ......... . (C is the set of all complex numbers)

Ans. 4

Sol. Let h(z) = –5z⁴, g(z)

|h(z)| = 5|z⁴| = 80 at z = 2

and |g(z)| = |z⁶ + 10|

= 64 + 10 = 74 at z = 2

|g(z)| 74 < 80 = |h(z)| on z = 2

By the Rouche's theorem, the function h and h + g f have same number of zeroes inside |z| = 2, since h has only a zero of order 4 at z = 0 inside |z| = 2, the function h + g f has four zeros in |z| < 2.

25. Let l² = {x = (x₁, x₂, ...): x_i be a normed linear space with the norm

||x||₂ =

Let g : l² C be the bounded linear functional defined by

g(x) = for all x = (x₁, x₂, ...) l².

Then (sup{|g(x)| : ||x||₂ ≤ 1})² is equal to ......... (round off to 3 places of decimal)

Ans. 0.125

Sol. For 1 p < for x = (x₁, x₂,..., x_n) lⁿ,

We have ||x||_p =

Since, g : l² C be the bounded linear functional defined by

g(x) = for all x = (x₁, x₂,...) l²

sup {|g(x)| : ||x||₂ 1} = 0.35 [ sup{|g(x)| : ||x||₂ 1} defines a norm on l² (0, 1)]

(sup{|g(x) : ||x₂|| 1})² = 0.125.

26. For the linear programming problem (LPP):

Maximise Z = 2x₁ + 4x₂

Subject to –x₁ + 2x₂ 4, 3x₁ + x₂ 6,

x₁, x₂ 0, R,

(R is the set of all real numbers)

Consider the following statements:

I. The LPP always has a finite optimal value for any 0.

II. The dual of the LPP may be infeasible for some 0.

III. If for some , the point (1, 2) is feasible to the dual of the LPP, then Z 16, for any feasible solution (x₁, x₂) of the LPP.

IV. If for some , x₁ and x₂ are the basic variable in the optimal table of the LPP with x₁ = , then the optimal value of dual of the LPP is 10.

Then which of the above statements are true?

(a) I and III

(b) I, II and IV

(d) II and IV

Ans. (b)

Sol. Given LPP is

...(i)

The dual of above LLP is

Minimise Z = 4w₁ + 6w₂

...(ii)

For LPP (i), we have the following graph

It can be observed from the graph that for 0, LPP always has a bounded feasible region (OABC) and so it always has a finite optimal solution for any 0. Hence, the statement I is true. So, either option (a) or (b) is true.

Clearly, from the option (a), (b), we can say statement III will be correct.

Now, let us check the statement IV.

Since, x₁ = , therefore x₂ =

[ optimal solution occurs at corner point of feasible region and according to given value of x₁, it occurs at point of intersection of given lines].

Now, at x₁ = and x₂ = ,

z =

= 1 + 9 = 10

Also, we know that the optimal value of the primal objective function is equal to the optimal value of the dual objective function.

Thus, the optimal value of dual of the LPP is 10.

Hence, the statement IV is also correct.

27. Let f : R² R be defined by

f(x, y) =

Consider the following statements:

I. The partial derivatives exist at (0, 0) but are unbounded in any neighbourhood of (0, 0).

II. f is continuous but not differentiable at (0, 0).

III. f is not continuous at (0, 0).

IV. f is differentiable at (0, 0).

(R is the set of all real numbers and R² = {(x, y) : x, y R}

Which of the above statements is/are true?

(a) I and II only

(b) I and IV only

(d) III only

Ans. (b)

Sol. We have,

f(x, y) is continuous at (0, 0).

exist at (0, 0) but are unbounded because sin and cos functions are oscillating in (–1 to 1).

Also, f is differentiable at (0, 0).

28. Let K = be an infinite matrix over C (the set of all complex numbers) such that

(i) for each i N (the set of all natural numbers), the ith row (k_{i, 1}, k_{i, 2,...}) of K is in and

(ii) for every x = (x₁, x₂,...) l¹, is summable for all i N, and (y₁, y₂,...) l¹,

where y_i =

Let the set of all rows of K be denoted by E.

Consider the following statements:

P : E is a bounded set in

Q : E is a dense set in .

Which of the above statements is/are true?

(a) Both P and Q

(b) P only

(d) Neither P nor Q

Ans. (b)

Sol. If K = be an infinite matrix, then by Schur's test, there exists positive numbers p_i > 0, g_j > 0 (i, j N), > 0 and > 0 satisfying and then the matrix K defines an operator l² with ||K||²

On matrix A with pi = qj = 1 and = = the norm of the operator which defines A is E_i|a(i)|. This argument shows that the map. A : E() which assigns to a sequence Hence, the set E is bound in .

29. Consider the following heat conduction problem for a finite rod

= – xe^t – 2t, t > 0, 0 < x <

with the boundary conditions

u(0, t) = – t², u( t) = – – t², t > 0 and the initial condition u(x, 0) = sin x – sin³ x – x, 0 x If v(x, t) = u(x, t) + xe^t + t², then which one of the following is correct?

(a)

(b)

(c)

(d)

Ans. (a)

Sol. …(i)

u(0, 1) = – t² u (, t) = – e^t – t², t > 0

u(x, 0) = sin x – sin³ x – x

v(x, t) = u(x, t) + xe^t + t²

Now, v(0, t) = u(0, t) + t² = 0, v(, t) = 0

v(x, 0) = u(x, 0) + x = sin x – sin³ x

= + e^t + 2t = (from 1) ...(ii)

From the above equations, option (a) can be concluded as right answer.

30. Let f : C C be non-zero and analytic at all point in Z.

If F(z) = f(z) cot (z) for z then the residue of F at n Z is ......... .

(C is the set of all complex numbers, Z is the set of all integers and denotes the set of all complex numbers excluding integers)

(a)

(b) f(n)

(c)

(d)

Ans. (b)

Sol.

31. Let the general integral of the partial differential equation

(2xy – 1) + (z – 2x²) = = 2(x – yz)

be given by F(u, v) = 0, where F : R² R is a continuously differentiable function.

(R is the set of all real numbers and R² = {(x, y) : x, y R}

Then which one of the following is TRUE?

(a) u = x² + y² + z, y = xz + y

(b) u = x² + y² – z, v = xz – y

(d) u = x² + y² – z, v = yz – x

Ans. (a)

Sol. Given,

Using Langre method,

Second solution, zdx + xdz + dy = 0

Integrate, xz + y = c₂

32. Consider the following statements:

I. If Q denotes the additive group of rational numbers and f : Q Q is a non-trivial homomorphism, then f is an isomorphism.

II. Any quotient group of a cyclic group is cyclic.

III. If every subgroup of a group G is a normal subgroup, then G is abelian.

IV. Every group of order 33 is cyclic.

Which of the above statements are true?

(a) II and IV

(b) II and III

(d) I, III and IV

Ans. (a)

Sol. I. No, f(x) = 0 = Q

II. G = < a >, = < aN > its true.

III. Q₈ = {1} = {1, – 1} = z(G) = < j >, < k > is 4 = Q₈ not true.

IV. O(G) = 33 = 3 × 11 its true.

33. A solution of the Dirichlet problem = 0, 0 < r < 1, – is given by

(a)

(b)

(c)

(d)

Ans. (c)

Sol. = 0, r < a

The general solution

Where a₀, a_n and b_n are constant.

As it is an even function

Now, from Eq. (i), we get

34. Consider the subspace Y = {(x, x) : x C} of the normed linear space If is a bounded linear functional on Y, defined by (x, x) = x, then which one of the following sets is equal to is a norm preserving extension of to

(a) {1}

(b)

(c)

(d) [0, 1]

Ans. (d)

Sol. First, in order for subspace to be well-defined operator acting on the function needs to be in C²[0, 1] for all f C²(0, 1]. In particular is measurable, and taking f = 1, it follows that is measurable function on [0, 1].

35. Consider the following statements:

I. The ring is a unique factorisation domain.

II. The ring is a principal ideal domain.

III. In the polynomial ring Z₂[x], the ideal generated by x³ + x + 1 is a maximal ideal.

IV. In the polynomial ring Z₃[x], the ideal generated by x⁶ + 1 is a prime ideal.

(Z denotes the set of all integers, Z_n denotes the set of all integers modulo n, for any positive integer n)

Which of the above statements are true?

(a) I, II and III

(b) I and III

(d) II and III

Ans. (b)

Sol. I. Since z[i] satisfies all the conditions of unique factorisation domain so, the ring is a unique factorisation domain.

II. An ideal of the Euclidean ring R is maximal off is generated by some prime element of R. So, the polynomial ring z₂[x], the ideal generated by x³ + x + 1 is a maximal ideal.

36. Let M be a 3 × 3 real symmetric matrix with eigenvalues 0, 2 and a with the respective eigenvectors u = (4, b, c)^T, v = (–1, 2, 0)^T and w = (1, 1, 1)^T.

Consider the following statements:

I. a + b – c = 10.

II. The vector x = satisfies Mx = v + w.

III. For any d span {u, v, w}, Mx = d has a solution.

IV. The trace of the matrix M² + 2M is 8.

(y^T denotes the transpose of the vector y)

Which of the above statements are true?

(a) I, II and III

(b) I and II

(d) III and IV

Ans. (b)

Sol. Real symmetric: Eigenvalues 0, 2 and a with eigenvectors u, v and w respectively.

I. a + b – c = 10, 2 + 2 + 6 = 10 its true.

II. v + w = (–1 + 1, 2 + 1, 1 + 0) = (0, 3, 1)

IV. M² + 2M

These eigenvalues 0 0² + 29 = 0

2 2² + 2.2 = 8

Sum of the eigen values = 0 + 8 + 8 = 16 its not true.

Hence, I and II are true.

37. Consider the region

in the complex plane. The transformation x + iy e^{x + iy} maps the region onto the region S C (the set of all complex numbers). Then the area of the region S is equal to

(a)

(b)

(c)

(d)

Ans. (a)

Sol. The mapping

u + iv = e^{x + iy}

u + iv = e^x .ev^iy

u + iv = e^x (cos y + i sin y)

Now eliminate y, we get, u² + v² = e^2x

Line x = –1 is mapped to the curve, u² + v² = e^–2

and the line x = 2 is mapped to the curve, u² + v² = e⁴

The graph of the curve

38. Consider the sequence of functions, where g_n(x) N and is the derivative of g_n(x) with respect to x.

(R is the set of all real numbers, N is the set of all natural numbers).

Then which one of the following statements is true?

(a) does not converge uniformly on R

(b) converge uniformly on any closed interval which does not contain 1.

(d) converge uniformly on any closed interval which does not contian 0.

Ans. (d)

Sol.

39. Consider the boundary value problem (BVP R (the set of all real numbers), with the boundary conditions y(0) = 0, = k (k is a non-zero real number).

Then which one of the following statements is true?

(a) For = 1, the BVP has infinitely many solutions

(b) For = 1, the BVP has an unique solution

(d) For = –1, k > 0, the BVP has a solution y(x) such that y(x) > 0 for all x

Ans. (d)

Sol.

40. Conisder the ordered square , the set [0, 1] × [0, 1] with the dictionary order topology. Let the general element of be denoted by x × y, where x, y [0, 1].

Then the closure of the subest

S =

(a) S ((a, b] × {0}) ([a, b) × {1})

(b) S ([a, b) × {0} ((a, b] × {1})

(d) S ((a, b] × {0})

Ans. (a)

Sol. Dictionary order topolgy for points (x, y) and (u, v) in the square is defined as (x, y) < (u, v)

(x < u) (x = u y < v)

Here, set [0, 1] × [0, 1] is the dictionary order topology and the general element of l² be denoted by x × y, where x, y [0, 1]. We know that the closure of (0, 1) in R equals to [0, 1].

Closure of the subset S = S ((a, b] × {0}) ([a, b) × {1})

41. Let P₂ be the vector space of all polynomials of degree at most 2 over R (the set of real numbers). Let a linear transformation T : P₀ P₂ be defined by

T(a + bx + cx²) = (a + b) + (b – c)x + (a + c)x².

Consider the following statements:

I. The null space of T is { (–1 + x + x²):  R}.

II. The range space of T is spanned by the set {1 + x², 1 + x}.

III. T(T( 1 + x) = 1 + x².

IV. If M is the matrix representation of T with respect to the standard basis {1, x, x²} of P₂, then the trace of the matrix M is 3.

Which of the above statements are true?

(a) I and IV

(b) I, III and IV

(d) II and IV

Ans.

Sol. We have a linear transformation T : P₂ P₂ defined by T(a + bx + cx²) = (a + b) + (b – c) x + (a + c)x²

I. The null space of T is defined as

{a + bx + cx² : T + (a + bx + cx²) = 0 P₂}

= {a + bx + cx² : (a + b) + (b – c) x + (a + c)x² = 0}

= {a + bx + c²; a + b = 0; b – c = 0; a + c = 0}

= {a + bx + cx²; a = –b = –c = 0; a + c = 0}

= {(–1 + x + x²);  lR}

Thus, the statement I is true.

II. Range space of T is defined as

T(P₂) = {T(a + bx + cx²), a, b, c lR}

= {(a + b) + (b – c)x + (a + c)x², a, b, c lR}

= {a(1 + x²) + b(1 + x) + cx² – cx, a, b, c lR}

= {a(1 + x²) + b(1 + x) + cx² – cx – c + c, a, b, c lR}

= {a(1 + x²) + b(1 + x) + c(x² + 1) – c(x + 1), a, b, c lR}.

42. Let T₁ and T₂ be two topologies defined on N (the set of all natural numbers), where T₁ is the topology generated by B = {(2n – 1), 2n}: n N} and T₂ is the discrete topology of N.

Consider the following statements:

I. In (N, T₁), every infinite subset has a limit point.

II. The function f : (N, N₁) (N, T₂) defined by

f(n) =

is a continuous function.

Which of the above statements is/are true?

(a) Both I and II

(b) only I

(d) Neither I nor II

Ans. (a)

Sol. We have, T₁ is the topology generated by B = {{2n – 1,2n} : n N} and T₂ is the discrete topology on N.

I. Let, x B and A B, then for every G T₁ with x G such that

(G – {x}nA

g_n (N, T₁), every infinite subset has a limit point

II. The function f : (N, T₁) (N, T₂) defined by

f(n) =

is an onto function.

f^–1(V) is open in N for every open set V in N.

Given function is continuous.

43. Let 1 ≤ p < q < . Consider the following statements:

where l^p = {(x₁, x₂,...}: x_i }

and L^p[0, 1] = {f : [0, 1] R : f is µ-measurable,

< , where µ is the Lebesgue measure}

(R is the set of all real numbers.)

Which of the above statements is/are true?

(a) Both I and II

(b) only I

(d) Neither I nor II

Ans. (b)

Sol. We know that if p [1, ) and f_n L^p([0, 1])

Then f_n converges to f.

i.e. ||fn||_p ||f||_p

By triangle inequality

This solution does not seem to generalise to the p > 1

Now, by counter example for

44. Consider the differential equation

If Y(s) is the laplace transform of y(t), then the value of Y(1) is ......... (round off to 2 places of decimal).

(Here, the inverse trigonometric functions assume principal values only)

Ans. 0.83

Sol. We have,

Taking laplace transform on both sides of the equation,

Eq. (i) is now reduced to

Taking inverse of laplace transformation,

y(t) = t sin t

y(1) = sin(1) = 0.83.

45. Let R be the region in the xy-plane bounde by the curves y = x², y = 4x², xy = 1 and xy = 5.

Then the value of the integral is equal to ......... .

Ans. 12

Sol. Given equation of curves are

y = x²...(i)

y = 4x²...(ii)

xy = 1 ...(iii)

xy = 5 ...(iv)

On drawing the graph for above curves, we get the following figure

Clearly, shaded region is region R.

Let us compute the integral by changing the variables.

For this subsitute u = xy and v = where x, y > 0

Now, let us compute the Jacobian.

Let S be the region in the uv-plane, then the region S is bounded by the curves

Now the integral

46. Let V be the vector space of all 3 × 3 matrices with complex entries over the real field. If W₁ = {A V : A = A^–T} and W₂ = {A V : trace of A = 0}.

Then the dimension of W₁ + W₂ is equal to ......... .

(A^–T denotes the conjugate transpose of A.)

Ans. 17

Sol.

a_ij + b_ij= a_ji – ib_ji

On comparing,

A^–T = A

dim W₁ = 6 + 3 = 9

dim W₂ = 18 – 2 = 16

dim(W₁ W₂) = 9 – 1 = 8

dim(W₁ + W₂) = dim(W₁) + dim(W₂) – dim(W₁ + W₂)

dim(W₁ + W₂) = 9 + 16 – 8 = 17.

47. The number of elements of order 15 in the additive group Z₆₀ + Z₅₀ is ......... (Z_n denotes the group of integers modulo n under the operation of addition modulo n, for any positive integer n)

Ans. 48

Sol. Let an element (a, b) Z₆₀ × Z₅₀ such that 0(a, b) = 15.

We know that order of (a, b) is |(a, b)| = lcm (|a|, |b|).

Now, as a Z₆₀, therefore |a| must divide 60, hence |a| = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

Similarly, as b Z₅₀, therefore |b| must divide 50, hence |b| = 1, 2, 5, 10, 25, 50

The only possibilities for the order of (a, b) to be 15, i.e. lcm (|a|, |b|) = 15 are

|a| = 15, |b| = 1 or |a| = 15, or |a| = 3, |b| = 5

Hence, the number of elements of order 15

where is an Eqler-function

= (8 × 1) + (8 × 4) + (2 × 4)

= 8 + 32 + 8

= 48.

48. Consider the following cost matrix of assigning four jobs to four persons :

Then the minimum cost of the assignument problem subject to the constraint that job J₄is assigned to person P₂, is ......... .

Ans. 27

Sol. Since job J₄ is assigned to person P₂, therefore our problem is to assign three jobs (J₁, J₂, J₃) to three person (P₁ P₃, P₄) at a minimum cost. So, consider the following cost matrix.

Now, reducing the matrix row-wise, we get the following matrix

Now, reducing the matrix column-wise, we get the following matrix

Now, examine the rows one-by-one until a row containing only one zero element is found.

Then, we make assignment and mark cross 'X' over all zero lying in the column containig the assigned zero. Now, continue in this manner until all the rows have been examined, we get the following matrix.

Now, similarly examine the columns one-by-one, we get the following matrix.

In the above matrix, we observe that row P₃ and column J₃ have on assignment, therefore required solution cannot be obtained at this stage.

Now, let us draw minimum number of lines to cover all zeroes atleast once.

We observe that the number of lines drawn in this manner is 2 which is less than 3 (order of cost matrix).

Now, the smallest element among all uncovered elements is 1. Subtracting this element from the uncovered elements and adding to every element that lies at the intersecting of two lines and leaving the remaining elements unchanged, we get the following matrix.

Now, make the assignments as shown in the table

Thus, the optimal assignment is

P₁ J₃, P₃ J₂

P₄ and P₂ J₄

and the minimum total cost is 6 + 8 + 7 + 6 = 27

49. Let y:[–1, 1] R with y(1) = 1 satisfy the Legendre differential equation

(1 – x²) + 6y = 0 for |x| < 1.

Then the value of is equal to ......... (round of to 2 places of decimal.)

Ans. 0.27

Sol.

which is in the form of

Here, adjusted value of a₀ and a₁ are and 0, respectively.

50. Let Z₁₂₅ be the ring of integers modulo 125 under the operations of addition modulo 125 and multiplication modulo 125. If m is the number of maximal ideals of Z₁₂₅ and n is the number of non-units of Z₁₂₅, then m + n is equal to ......... .

Ans. 26

Sol. Z₁₂₅= Z₅₃

m = number of maximal ideals = 1

n = number of non-units = 125 –

= 125 – (5³ – 5²) = 25

Therefore, m + n = 1 + 25 = 26.

51. The maximum value of the error term of the composite trapezoidal rule when it is used to evaluate the definite integral with 12 sub-intervals of equal length, is equal to ......... (round off to 3 places of decimal).

Ans. 0.021

Sol. Here, a = 0.2, b = 1.4 and = 12

Now, nh = b – a

The given function is f(x) = (sin x – log_e x)

Max. = more than 20 in the (0.2, 1.4)

So, error in the trapezoidal rule

52. By the simplex method, the optimal table of the linear programming problem:

Maximize Z = x₁ + 3x₂

Subject to x₁ + x₂ + x₃ = 8, 2x₁ + x₂ + x₄ = , x₁, x₂, x₃, x₄ 0,

where are real constants, is

Then the value of is ......... .

Ans. 15

Sol. According to question, we have

Here, key element,

53. Consider the inner product space P₂ of all polynomials of degree at most 2 over the field of real numbers with the inner product

Let {f₀, f₁, f₂} be an orthogonal set in P₂, where f₀ = 1, f₁ = t + c₁, f₂ = t² + c₂f₁ + c₃ and c₁, c₂, c₃ are real constants. Then the value of 2c₁ + c₂ + 3c₃ is equal is ..........

Ans. –3

Sol. Given,

Since, {f₀, f₁, f₂} be an orthognal set in P₂

C₂ = –1 …(ii)

54. Consider the system of linear differential equations

= 5x₁ – 2x₂,

= 4x₁ – x₂,

with the initial conditions x₁(0) = 0, x₂(0) = 1.

Then log_e (x₂(2) – x₁(2) is equal to ......... .

Ans. 2

Sol.

We take the Laplace transform to both differential equations

sX₁(S) – x₁(0) = 5X₁(s) – 2X₂(s)

sX₂(s) – x₂(0) = 4X₁(s) – X₂(s)

Now take the initial condition

(s – 5) X₁(s) + 2X₂(s) = 0 ...(i) [ x₁(0) = 0]

–4X₁(s) + (s + 1)X₂(s) = 1 ...(ii) [ x₂(0) = 1]

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

Taking the inverse Laplace transform, we get

x₁(t) = e^t – e^3t

Now, 2x₂(t) = 5x₁ –

= 5(e^t – e^3t) – (e^t – 3e^3t)

x₂(t) = 2e^t – e^3t

log_e(x₂(2) – x₁(2) = log_e(2e² – e⁶ – e² + e⁶)

= log_e e² = 2

55. Consider the differential equation

x(1 + x²)

The sum of the roots of the inidical equation of the Frobenius series solution for the above differential equation in a neighbourhood of x = 0 is equal to ......... .

Ans. 10

Sol. …(i)