1. Let for n Then the sequence {a_n} is

(a) Convergent

(b) Bounded but not convergent

(c) Diverges to

(d) Neither bounded nor diverges to

Ans. (a)

Sol.

2. The number of real roots of the equation x³ + x – 1 = 0 is

(a) 0

(b) 1

(c) 2

(d) 3

Ans. (b)

Sol. Let     f(x) = x³ + x – 1

   = 3x² + 1 > 0 f is increasing function.

Also,  

f is a continuous function, so range f =

Thus f attains each real value only. Hence f(x) = 0 for only one value of x.

3. The value of is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

4. Let V be the region bounded by the planes x = 0, x = 2, y = 0, z = 0 and y + z = 1. Then the value of integral is

(a)

(b)

(c) 1

(d)

Ans. (d)

Sol. (V : volume projected by the surface z = 1 – y on the xy-plane over R)

Alternatively:

V is the volume project by the surface x = 2 over G on the y-z plane

5. The solution y(x) of the differential equation satisfying the conditions is

(a) 4e^2x

(b) (16x + 4)e^–2x

(c) 4e^–2x + 16x

(d) 4e^–2x + 16x e^2x

Ans. (b)

Sol.

8 = (4 + 0) (–2) + c₂

c₂ = 16

So the solution is y = (4 + 16x)e^–2x

6. If y^a is the integrating factor of the differential equation 2xydx – (3x² – y²)dy = 0, then the value of a is

(a) –4

(b) 4

(c) –1

(d) 1

Ans. (a)

Sol. 2xydx – (3x² – y²)dy = 0

Let us multiply by y^a

7. Let and C be the positively oriented closed curve given by x² + y² = 1, z = 0.

If then the value of a is

(a) –1

(b) 0

(c)

(d) 1

Ans. (a)

Sol.

8. Consider the vector field where a is a constant. If then the value of a is

(a) –1

(b) 0

(c) 1

(d)

Ans. (c)

Sol.

9. Let G denote the group of all 2 × 2 invertible matrices, with entries from Let H₁ = {A G : det (A) = 1} and H₂ = {A G : A is upper triangular}.

Consider the following statements:

P : H₁ is a normal subgroup of G; Q : H₂ is normal subgroup of G

Then

(a) Both P and Q are true

(b) P is true and Q is false

(b) P is false and Q is true

(d) Both P and Q are false.

Ans. (b)

Sol. G = {group of all 2 × 2 invertible matrices with entries from R}.

H₁ = {A G : dt (A) = 1} and H₂ = {A G : A is upper triangular}.

Let A H₁ then P^–1 A P = A since |P^–1 AP| = |P^–1|.|A|.|P| = |A| = 1. So, H is normal.

But A H₂ need not be true that, P^–1 A P H₂ since P^–1 A P may or may not upper triangular.

10. For n let n = {nk : k }. Then the number of units of /11 and /12, respectively are

(a) 11, 12

(b) 10, 11

(c) 10, 4

(d) 10, 8

Ans. (c)

Sol.

And no. of units in is no. of elements prime to n.

Hence 10 elements prime to 11.

and 1, 5, 7, 11 four elements prime to 12.

Alternate method : Number of units in

11. Let A be a 3 × 3 matrix with trace (A) = 3 and det (A) = 2. If 1 is an eigenvalue of A, then the eigen-value of the matrix A² – 2I are

(a) 1, 2(i – 1), –2(i + 1)

(b) –1, 2(i – 1), 2(i + 1)

(c) 1,2(i + 1), –2(i + 1)

(d) –1, 2(i – 1), –2(i + 1)

Ans. (d)

Sol. Let other two eigen values are as we know

So eigen value of A is 1, 1 + i, l – i

So eigen value of A² is 1, 2i, –2i

12. Let T : be a linear transformation, where n 2. For k n, let E = and F =

Then

(a) If E is linearly independent, then F is linearly independent

(b) If F is linearly independent, then E is linearly independent

(c) If E is linearly independent, then F is linearly dependent

(d) If F is linearly independent, then E is linearly dependent

Ans. (b)

Sol.

Start with,

E is linearly independent. Since are linearly independent.

13. For n m let T₁ : and T₂ : be two linear transformations such that T₁T₂ is bijective. Then

(a) rank (T₁) = n and rank (T₂) = m

(b) rank (T₁) = m and rank (T₂) = n

(c) rank (T₁) = n and rank (T₂) = n

(d) rank (T₁) = m, and rank (T₂) = m

Ans. (d)

Sol.

Then T₁T₂(x) = T₁[T₂(x)] i.e. T₁T₂ : R^m R^m since T₁T₂ is bijective then it is non-singular so rank (T₁ • T₂) = m,

As we know rank (T₁T₂) min {rank (T₁), rank (T₂)} i.e. m min {rank (T₁), rank (T₂)} … (1)

Case-I: If m n then since T₁ is m × m matrix thus its rank not exceed n.

Similarly rank of T₂ cannot exceed n.

so from (1) m n so m = n

Case-II: If m n then rank of both T₁ and T₂ will be m. Thus Rank (T₁) = Rank (T₂) = m

14. The set of all x at which the power series converges is

(a) [–1, 1]

(b) [–1, 1]

(c) [1, 3)

(d) [1, 3]

Ans. (c)

Sol.

  R = 1, Power series convergent on –1 < x – 2 < 1

1 < x < 3 and convergent 1 < x 3

By leibnitz test this series will converge only for x – 2 = 1 not for 1.

15. Consider the following subsets of R:

   

Then

(a) Both E and F are closed

(b) E is closed and F is not closed

(c) E is not closed and F is closed

(d) Neither E nor F is closed

Ans. (c)

Sol.

16. (a) Let {a_n} be a sequence of non-negative real number such that converges, and let {k_n} be a strictly increasing sequence of positive integers. Show that also converges.

(b) Suppose f:[0, 1] is differentiable and f'(x) 1 at every x (0, 1). If f(0) = 0 and f(1) = 1, show that f(x) = x for all x [0, 1].

Sol.

17. (a) Suppose f is a real valued function defined on an open interval I and differentiable at every x I. If [a, b] I and then show that there exists c (a, b) such that .

(b) Let f : (a, b) be a twice differentiable function such that is continuous at every point in (a, b). Prove that

for every x (a, b).

Sol. (a) f is differentiable on I. Therefore, f is differentiable and hence continuous on [a, b] I. So, by the property of a continuous function on a closed interval [a, b], f attains minimum at least one point in [a, b].Out claim is that the point, say c, must be an interior point of [a, b], i.e., c (a, b). Let us assume that the point c is the interior point but the boundary point (one of the two) of [a, b], say c = a.

There exists an > 0 such that [a, a + ] [a, b].

18. Find all the critical points of the following function and check whether the function attains maximum or minimum at each of these points.

  u(x, y) = x⁴ + y⁴ – 2x² – 2y² + 4xy, (x, y)

Sol.

 u(x, y) = x⁴ + y⁴ – 2x² – 2y² + 4xy

 u_x(x, y) = 4x³ – 4x + 4y = 4 (x³ – x + y)

    u_y(x, y) = 4 (y³ – y + x)

u_xy(x, y) = 4

     u_xx(x, y) = 4 (3x² – 1)

u_yy(x, y) = 4 (3y² – 1)

 u_x(x, y) = 0 x³ – x – y = 0 ...(1)

 u_y(x, y) = 0 y³ + x – y = 0 ...(2)

From (1) + (2), x³ + y³ = 0 (y + x) (x² – xy + y²) = 0

x + y = 0 y = –x

Putting y = –x into (1),

   x³ – x – x = 0

x(x² – 2) = 0

So,   

Critical points of u(x, y) are P₁ (0, 0),

Discriminant of

     D(0, 0) = 16 – 16 = 0

So, u has minima at

At P₁(0, 0), the derivative test fails as D(0, 0) = 0

 u(x, y) = x⁴ + y⁴ – 2x² – 2y² + 4xy

= x⁴ + y⁴ – 2(x – y)²

= (x⁴ + y⁴) – 2x²y² – 2(x – y)²

On the line y = x, u(x, y) = (2x²)² – 2x² – x² – 0 = 4x⁴ – 2x⁴ = 2x⁴ > 0

On the line y = 0, u(x, y) = x⁴ – 2x² = x²(x² – 2) < 0 for

In any neighbourhood of size, radius at (0, 0) we get points on x-axis at which u(x, y) < 0 and points on y = x at which u(x, y) > 0.

Hence, u(x, y) has neither maximum nor minimum at (0, 0).

19. (a) Let : [a, b] be differentiable and [c, d] = {(x): a x b}, and let f :[c, d] be continuous. Let g : [a, b] be defined by for x [a, b]. Then show that g is differentiable and = f ((x) (x) for all x [a, b].

(b) If f : [0, 1] is such that for all x R the find

Sol. (a) Firstly we will introduce here "Leibnitz Rule":

are function of x, t and g(x), h(x) has first order its derivative, then,

  

given that

g : [a, b] differentiate (1) w.r.t. 'x'

  

Since f(t) is function of 't' only. Hence, Also, hence c is constant term. So, So above expression be in the form, Hence proved.

20. Find the area of the surface of the solid bounded by cone and the paraboloid z = 1 + x² + y².

Sol.

Solving the two equations for intersecting curve:

Hence, intersecting curve: x² + y² = 1, z = 2

The surface area, S = S₁ S₂

(R is as shown in the figure)

21. Obtain the general solution of each of the following differential equations:

(a)

(b)

Sol.

Here h and k choosen such that

  h + 2k + 8 = 0

  2h + k + 7 = 0

So h = –2 and k = –3 on solving, so

Put    X = x – h = x + 2,  y = y – k = y + 3

22. (a) Determine the value of b > 1 such that differential equations

satisfying the conditions y(1) = 0 = y(b) has a nontrivial solution

(b) Find γ(x) such that y(x) = e^4x(x) is a particular solution of the differential equation

Sol.

Given    y(b) = 0

 0 = c₁ sin log b  c₁ 0 otherwise solution become trivial

so   sin log b = 0, so log b =

Since log b > 0, n = 1, 2 … ∴ b =

Consider

On integrating, we get

23. (a) Change the order of integration in the double integral

(b) Let Using Green's theorem, evaluate the line integral where C is the positively oriented closed curved which is the boundary of the region enclosed by the x-axis and the semi-circle in the upper half plane.

Sol.

24. (a) If then evaluate the surface integral where S is the surface of the cone lying above the xy-plane and is the unit normal to S making an acute angle with

(b) Show that the series converges uniformly on for p > 1.

Sol. (a) Consider a closed surface conisting of S and

Hence given series converges uniformly on

25. (a) Find a value of c such that the following system of linear equations has no solution:

   x + 2y + 3z = 1,

   3x + 7y + cz = 2,

  2x + cy + 12z = 3.

(b) Let V be a vector space of all polynomials with real coefficients of degree at most n, where n ≥ 2. Considering element fo V as functions from to , define . Show that W is a subspace of V and dim(W) = n.

Sol. (a) x + 2y + 3z = 1

3x + 7y + cz = 2

2x + cy + 12z = 3

 

Augumented matrix can be [A : B]

For no solution r(A) r([A : B]) make last row of A is zero

We get –c² + 13c – 30 = 0 c = 3, 10

if c = 3 then all element of last row is zero.

  r(A) = r([A : B]), If c = 10, then the given system of linear equations has no solution.

Hence W is subspace of V.

so basis of vector space is

26. (a) Let A be a 3 × 3 real matrix with det (A) = 6. Then find det (adj A).

(b) Let and be non-zero vectors in 3, such that is not a scalar multiple of . Prove that there exists a linear transformation T : such that T³ = T, T= and T has at least three distinct eigenvalues.

Sol. (a) det (A) = 6

We know A(adj A) = | A |ⁿ I_n | A || adjA |=| A |ⁿ | adj A | = | A |^n–1 = 6² = 36

(b) T : let T(x₁, x₂,…,x_n) = (x₂, x₁, 0, 0, 0,…, 0)

let T²(x₁, x₂…..x_n) = T(x₂, x₁, 0, 0,…., 0) = (x₁, x₂, 0, 0,…., 0) = T(x₁, x₂,…., x_n)

T³(x₁,x₂….x_n) = T²(x₂,x₁, 0,…., 0) = T(x₁, x₂,…., x_n)

we see that (x₁, x₂….x_n) and (x₂, x₁, 0,…., 0) are independent

The eigen value of T i.e. = 0

so distinct eigen value 0, 1, –1,

27. (a) If E is a subset of that does not contain any of its limit points, then prove that E is a countable set.

(b) Let f : (a, b) be a continuous function. If f is uniformly continuous, then prove that there exists a continuous function g:[a, b] such that g(x) = f(x) for all x (a, b).

Sol. (a) If E be subset of that does not contain any of its limit point. To show that E is a countable set. E can be either a finite or an infinite subset of .

Case (I) If E is finite countable

Case (II) If E is infinite, then no point a E s.t. a is limit point of E. So each nbd of a must have only finite member of E. Hence, we can partition the set E in countably infinite subsets which have only finite members of e. And so, E itself be a countable subset of .

Example: (1) No limit point

   (2) limit point '0' but not in the set.

(b) f(a, b) R is uniformly continuous function. So, for every > 0,

 |f(x₁) = f(x₂)| < |x₁ – x₂| depends only not on 'x'. Also,

let f(x₂) = f(c) = g(x) [say]

And

  a < c < b

and at x = c, g(x) = f(c), then

   g(x) = f(c) for all x (a, b)

28. (a) On define a binary operation * as follows: For (x,y,t), in

Then show that (*) is a group, and find its centre.

(b) let For any m, show that I = is an ideal of Further, find the generators of I.

Sol. (a) (i) Closure property is obvious

(ii) Associativity :- [(x, y, t) *

holds

(iii)

= 0. So (0, 0, 0) is the identity element of the above set.

Now, (x, y, t)* = (0, 0, 0)

 

So,   = (–x, –y, –t). Hence inverse exist.

By the def. of centre of a group,

(x, y, t)* (x', y', t') = (x', y', t') * (x, y, t)

  (x, y) = (0, 0) Then it will always hold. Hence centre is {(0, 0, t) ; t R}.

(b) For any K N

   K Z = {Kn : n Z} is an ideal of Z.

Now,  I = mZ n Z. To show I is an ideal.

Let a, b I   a m Z ^ a n Z

So, a is a multiple of m and n both.

Thus a is a multiple of l·c·m of m and n say 'l'.

Similarly b is multiple of l.

a – b also multiple of l.    a – b I.

Now, r Z, a · r = r · a = (l · p)r [if a = l · p]

[Since Z is Commutative]·

  a r Z. Hence I is an ideal. And generator of I is l · c · m (m, n).

29. Let G be a group of order p², when p is a prime number. Let xG. Prove that {y G : xy = yx} = G.

Sol. By class equation we have,

where  Z(G) = centre of group.

  N(a) = normalizer of 'a'.

We have, if 0(G) = pⁿ, p being a prime number.

Then     0(Z(G)) > 1. Also, Z(G) N(a).

Since

Now, given O(G) = p². Then we have two choices of order of Z(G).

(i) If O(Z(G)) = p² = O(G) G is abelian group.

(ii) If O(Z (G)) = p

Then is cyclie G is abelian.

So, every group of order p² abelian.