1.

(a) 3

(b) 2

(c) 1

(d) 0

Ans. (a)

Sol.

2. Let f(x) = (x – 2)¹⁷ (x + 5)²⁴. Then

(a) f does not have a critical point at 2

(b) f has a minimum at 2

(c) f has a maximum at 2

(d) f has neither a minimum nor a maximum at x = 2

Ans. (d)

Sol. = 17 (x – 2)¹⁶ (x + 5)²⁴ + (x – 2)¹⁷. 24(x + 5)²³

= (x – 2)¹⁶ (x + 5)²³ (17x + 85 + 24x – 48)

= (x – 2)¹⁶ (x + 5)²³ (41x + 37)

= 41(x – 2)¹⁶ (x + 5)²³ (x +37 / 41)

Clearly, does not change sign as we pass through x = 2 as (x – 2) has an even index. Hence, f has neither a minimum nor a maximum at x = 2.

3. Let

(a) 2f

(b) 3f

(c) 5f

(d) 7f

Ans. (d)

Sol.

Hence, f is homogeneous function of degree n = 7

So, by Euler's theorem,

4. Let G be the set of all irrational numbers. The interior and the closure of G are denoted by G⁰ and , respectively. Then

(a)

(b)

(c)

(d)

Ans. (c)

Sol. By the definition of the interior point, no point of G is interior point of G. By definition of the boundary point, of is a boundary point of g.

5. Let

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

6. Let C be the circle x² + y² = 1 taken in anti-clockwise sense. Then the value of the integral equals

(a) 1

(b)

(c)

(d) 0

Ans. (c)

Sol. Using green's theorem

7. Let r be the distance of a point P(x, y, z) from the origin O. Then is a vector

(a) orthogonal to

(b) normal to the level surface of r at P

(c) normal to the surface of revolution generated by OP about x-axis

(d) normal to the surface of revolution generated by OP about y-axis.

Ans. (b)

Sol. = grad always normal to level surface of at P.

8. Let T : be defined T(x₁, x₂, x₃) = (x₁ – x₂, x₁ – x₂, 0). If N(T) denote the null space and range space of T respectively, then

(a) din N(T) = 2

(b) din R(T) = 2

(c) R(T) = N(T)

(d) N(T) R(T)

Ans. (a)

Sol. Consider T (x₁, x₂, x₃) = (x₁ – x₂, x₁ – x₂, 0), for nullity T(x₁, x₂, x₃) = (0, 0, 0)

(x₁ – x₂, x₁ – x₂, 0) = (0, 0, 0) x₁ – x₂ = 0 (x₁ = x₂)

So null space N(T) = (x₁, x₁, x₃), consider x₁(1,1,0) + x₃(0,0,t) = 0 x₁ = x₃ = 0

(1, 1, 0) and (0, 0, 1) is linearly independent and span of N(T).

dim N(T) = 2

9. Let S be a closed surface for which Then the volume enclosed by the surface is

(a) 1

(b) 1/3

(c) 2/3

(d) 3

Ans. (b)

Sol.

10. If (c₁ + c₂ ln x)/x is the general solution of the differential equation

then k equals

(a) 3

(b) –3

(c) 2

(d) –1

Ans. (a)

Sol. This equation is homogeneous

Differential equation becomes

(D₁(D₁ – 1) + kD + 1) y = 0 (D₁² + (k – 1)D₁ + 1)y = 0 … (1)

It's solution is given y = (C₁ + C₂ ln x) /x, z = log x, x = e^z

y = (C₁ + C₂z)e^–z

So this solution tells that it have two equal real root that is –1.

Complementary function CF

(D₁ + 1)² = 0 D₁² + 2D₁ + 1 = 0

From (1) and (2)

(k – 1) = 2 k = 3.

11. If A and B are 3 × 3 real matrices such that rank (AB) = 1, then rank (BA) cannot be

(a) 0

(b) 1

(c) 2

(d) 3

Ans. (d)

Sol. A and B are 3 × 3 matrix, then AB is 3 × 3 matrix

but rank (AB) = 1 | AB | = 0 (by definition)

| A || B | = 0, so at least | A | or | B | should be zero so | BA | = | B || A | = 0

If | BA | = 0 and BA is 3 × 3 matrix rank (BA) cannot be 3.

12. The differential equation representing the family of circles touching y-axis at the origin is

(a) linear and of first order

(b) linear and of second order

(c) non-linear and of first order

(d) non-linear and of second order

Ans. (c)

Sol. Equation of circles touching y-axis is given by

(x – a)² + y² = a², where a is radius … (1)

x² + a² – 2ax + y² = a² x² + y² – 2ax = 0 … (2)

differentiate w.r.t. x, we get,

… (3)

From (2) and (3)

so, differential equation is non-linear and of first order.

13. Let G be a group of order 7 and (x) = x⁴, x G. Then is

(a) not one-one

(b) not onto

(c) not a homomorphism

(d) one-one, onot and a homomorphism

Ans. (d)

Sol. |G| = 7, a prime number, So G is a cyclic group, say, G = {e, a, a²,……, a⁶}.

So, clearly G is one-one and onto.

Also, (x, y) = (xy)⁴ for any x, y G.

= x⁴y⁴ (as G is cyclic, hence abelian)

= (x) (y)

is a homomorphism.

14. Let R be the ring of all 2 × 2 matrices with integer entries. Which of the following subsets of R is an integral domain?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

. Hence commutative

If x = 1, then and so identity also exists.

Hence no zero divisor so, it is integral domain.

15. Let f_n(x) = n sin²ⁿ⁺¹ x cos x. Then the value of is

(a)

(b) 0

(c)

(d)

Ans. (a)

Sol.

16. (a) Test the convergence of the series

(b) Show that

Sol.

So, the series converges (by the Ratio Test).

(b) Let f (x) =

f (0) = ln (2) – ln 2 – 0 = 0.

17. Find the critical points of the function f (x, y) = x³ + y² – 12x – 6y + 40. Test each of these for maximum and minimum.

Sol.

For critical points f_x (x, y) = f_y (x, y) = 0 x = 2 or –2 and y = 3

So, the critical point are P₁(2, 3), P₂ (–2, 3)

f_xx (x, y) = 6x, f_xy (x, y) = 0, f_yy (x, y) = 2

Discriminant of f,

D(2, 3) = 24 > 0

D(–2, 3) = –24 < 0

So, f has no extremum at P₂(–2, 3)

f_xx (2, 3) = 6 × 2 = 12 > 0 f has minimum at P₁(2, 3).

18. (a) Evaluate where R is the region bounded by the lines x = 0, y = 1 and the parabola y = x².

(b) Find the volume of the solid bounded above by the surface z = 1 – x² – y² and below by the plane z = 0.

Sol.

(b) The required volume,

19. Evaluate the surface integral where the surface S is represented in the form

Sol.

Surface area for S,

20. Using the change of variables evaluate where the region R is bounded by the curves xy = 1, xy = 3, y = 3x and y = 5x in the first quadrant.

Sol.

21. (a) Let u and be the eigenvectors of A corresponding to the eigenvalues 1 and 3 respectively. Prove that u + is not an eigenvector of A.

(b) Let A and B be real matrices such that the sum of each row of A is 1 and the sum of each row of B is 2. Then show that 2 is an eigenvalue of AB.

Sol. (a) u is eigen vector of A corresponding to eigen-value

Au = lu, is eigen vector of A corresponding to eigen value 3 A = 3

Consider A(u + ) = Au + A = 1u + 3, so (u + ) is not an eigen value of A.

(TIPS: Ax = is eigen value then where is eigen vector of .

(b) Let A be matrix whose sum of all terms in row is 1.

Let Sum of all terms in row is 1

Here one of the eigen value of A is 1.

One of eigen value of B is 2. ( is eigen vector) multiply both side by A

( because 1 is eigen value of A)

So, 2 is eigen value of AB.

22. Suppose W₁ and W₂ are subspace of spanned by {(1, 2, 3, 4), (2, 1, 1, 2)} and {(1, 0, 1, 0) (3, 0, 1, 0)} respectively. Find a basis of W₁ W₂. Also find a basis of W₁ + W₂ containing {(1, 0, 1, 0), (3, 0, 1, 0)}.

Sol. Part-I: Basis of W₁ W₂, W has a basis {(1, 2, 3, 4), (2, 1, 1, 2)} and W₂ has a basis {1,0,1,0}, (3,0,1,0)}, so W₁ is the linear combination of = a(1, 2, 3, 4) + b(2,1,1,2)

W₂ the linear combination of = c(1,0,1,0) + d(3,0,1,0)

but W₁ W₂ contain element which is in both W₁ and W₂ i.e. both element present in W₁ and W₂.

= a(1,2,3,4) +b(2,1,1,2) for some a,b

i.e., linear combination of basis of

Similarly, for some c, d = c(1,0,1,0) + d(3,0,1,0)

i.e., linear combination of basis of W₂.

from (1) and (2), we get a(1,2,3,4) +b(2,1,1,2) = c(1,0,1,0) + d(3,0,1,0)

Comparing both side, we get

a + 2b = c + 3d …(3)

2a + b = 0 …(4)

3a + b = c + d …(5)

4a + 2b = 0 …(6)

From (4) and (6), we get 2a + b = 0 b = –2a

From (5) 3a – 2a = c + d a = c + d …(7)

From (3) a – 4a = c + 3d –3 = c + 3d …(8)

From (7) and (8), we get d = –2a,c = 3a

= a(1,2,3,4) + b(2,1,1,2) = a(1,2,3,4) –2a(2,1,1,2)

= a(1 – 4, 2 – 2, 3 – 2, 4 – 4) = a(–3,0,1,0)

i.e. W₁ W₂ span by (–3,0,10) dim (W₁ W₂) = 1

and basis of W₁ W₂ = {(–3,0,1,0)}

Part-II: Find basis of W₁ + W₂. As we know

dim (W₁ + W₂) = dim(W₁) + dim(W₂) – dim(W₁ W₂) = 2 + 2 – 1 = 3

Basis of W₁ + W₂ contains 3 elements by definition

W₁ + W₂ = {α + α; α belongs to W₁. β belongs to W₂}

So, W₁ + W₂ contain both W₁ and W₂. ∴ W₁ W₂, W₂ W₁ + W₂

So basis contain three element but {1,0,1,0}, {3,0,1,0} are given then other element belongs to W₁.

Basis of W₁ + W₂ is = {(1,0,1,0), (3,0,1,0), (1,2,3,4)} or

= {(1,0,1,0), (3,0,1,0), (2,1,1,2)}

23. Determine y₀ such that the differential equation – y = 1 –e^–x, y(0) = y₀ has a finite limit as x

Sol.

24. Let (x, y, z) = e^x sin y. Evalute the surface integral where S is the surface of the cube 0 x 1, 0 y 1 and is the directional derivative of in the direction of the unit outward normal to S. Verify the divergence theorem.

Sol.

There are six surfaces

x = 0 x = 1

y = 0 y = 1

z = 0 z = 1

Hence Gauss divergence theorem is verified.

25. Let y = f(x) be a twice continuously differential function on satisfying f (1) = 1 and Form the second order differential equation satisfied by y = f (x), and obtain its solution satisfying the given conditions.

Sol. Consider differential equation w.r.t. x, we get

This is homogeneous equation Let z = log x

Equation charge into

Then auxiliary equation 4m² – 4m + 1 = 0

Consider f (1) = 1 i.e x = 1, f (x) = 1

1 = (c₁ + c₂ log1) ( log1 = 0)

…(2)

26. Let be the group under matrix addition and H be the subgroup of G consisting of matrices with even entries. Find the order of the quotient group G/H.

Sol.

Since is of order 2, so as 2 × 2 matrix of dimension 4.

actually each entry of G/H has exactly two choices either 0 or 1.

So, the 2 × 2 matrices has 4 elements and each element has two choices.

2 × 2 × 2 × 2 = 16 elements.

27. Let

Show that f is uniformly continuous on

Sol.

28. Find and hence prove that the series.

is uniformly convergent on .

Sol.

Since, is a convergent series, by P-test given series converges series converges uniformly on

29. Let R be the ring of polynomials with real coefficients under polynomial addition and polynomial multiplication. Suppose

I = {p R : sum of the coefficients of p is zero}.

Prove that I is a maximal ideal of R.

Sol. Given R be the ring of polynomials with real coefficients under polynomial addition and polynomial multiplication.

I = {p R : sum of the coefficients of p is zero}.

It is easy to prove that I be an ideal

Now let us assume that U be an ideal of R s. t. I C U R. Then a polynomial r(x) U.t. sum of coefficients of r(x) is non zero.

Let coefficient of r(x) is and since U contains I, then a non-zero number l and a polynomial S(x) in I s.t. sum of coefficient of s(x) is zero say + l in I. And since U is ideal so, s(x) – r(x)

Hence l is maximal ideal.