1. Let {a_n}, {b_n} and {c_n} be sequences of real number such that b_n = a_2n and c_n = a_{2n + 1}. Then {a_n} is convergent

(a) implies {b_n} is convergent but {c_n} need not be convergent

(b) implies {c_n} is convergent but {b_n} need not be convergent

(c) implies both {b_n} and {c_n} are convergent

(d) if both {b_n} and {c_n} are convergent

Ans. (c)

Sol. If (say), then any subsequence of it would have the same limit and hence converges.

2. An integrating factor of

(a) xe^3x

(b) 3xe^x

(c) xe^x

(d) x³ e^x

Ans. (b)

Sol. dividing by x, we get = e^–2x.

This above differential equation is equivalent to where integrating factor

3. The general solution of is

(a) (c₁ + c₂x)e^3x

(b) (c₁ + c₂ln x)x³

(c) (c₁ + c₂x)x³

(d)

Ans. (b)

Sol.

Above equation is a homogeneous differential equation with variable co-efficients.

   Let z = log x or x = e^z

    (D₁(D₁ – 1) –5D₁ + 9)y = 0  (D₁² – D₁– 5D₁ + 9)y = 0

    (D₁² – 6D₁ + 9)y = 0 (D₁ – 3)² y = 0

  General Solution    y = (c₁ + c₂z)e^3z

but z = log x y = (c₁ + c₂ ln x)e^{3log x} =

 y = (c₁ + c₂ ln x)x³

4. Let If is a solution of the Laplace equation then the vector field is

(a) neither solenoidal nor irrotational

(b) solenoidal but not irrotational

(c) both solenoidal and irrotational

(d) irrotational but not solenoidal

Ans. (d)

Sol. is solution of Laplace equation

and is irrotational but not solenoidal.

5. Let S be surface of the sphere x² + y² + z² = 1 and be the inward unit normal vector to S. Then is equal to

(a)

(b)

(c)

(d)

Ans. (d)

Sol. where is inward drain unit normal vector. Using Gauss Divergence theorem

6. Let A be a 3 × 3 matrix with eigen values 1, –1 and 3. Then

(a) A² + A is non-singular

(b) A² – A is non-singular

(c) A² + 3A is non-singular

(d) A² – 3A is non-singular

Ans. (c)

Sol. Eigen value of A calculated by |A – I| = 0 but = 1 is eigen value

|A – I| = 0 |A² – A| = 0 A² – A is singular because its determinant is zero.

Let = –1 is eigen value |A + I| = 0 |A² + A| = 0 A² + A is singular

Let = 3 is eigen value |A – 3I| = 0 |A² – 3A| = 0 A² – 3A is singular.

7. Let be a linear transformation and I be the identity transformation of . If there is a scalar c and a non-zero vector x such that T(x) = cx, then rank (T – cI)

(a) cannot be 0

(b) cannot be 1

(c) cannot be 2

(d) can not be 3

Ans. (d)

Sol. Dimension of T = 3.

consider   (T – CI)(x) = T(x) – CI(x) = Cx – Cx [ T(x) = Cx, I(x) = x] = 0

   (T – CI) x = 0 for non-zero x, (T – CI)x = 0

Nullity of (T – CI)x cannot zero.

As we know

    Rank + Nullity = Dimension

Rank of (T – CI) + Nullity of (T – CI) = 3

but Nullity of (T – CI) 0 ∴ Rank of (T – CI) < 3

8. In the group {1, 2, …………16} under the operation of multiplication modulo 17, the order of the element 3 is

(a) 4

(b) 8

(c) 12

(d) 16

Ans. (d)

Sol. This is U(n) group. And for each prime p, U(p) Z_{p – 1}

Hence   U(17) Z₁₆

and g.c.d (3, 16) = 1 means 3 is prime to 16

or     3¹⁶ 1 (mod 17).

9. A ring R has maximal ideals

(a) if R is infinite

(b) if R is finite

(c) if R is finite with at least 2 elements

(d) only if R is finite

Ans. (c)

Sol. If R is finite with at least 2 elements

10. The integral is equal to

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

11. Let f: be continuous and g, h : be differentiable. Let where u = g (x, y) and = h (x, y). Then

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

12. Let y = f(x) be a smooth where such that 0 < f (x) < K for all x [a, b]. Let

L = length of the curve between x = a and x = b

A = area bounded by the curve, x-axis and the lines x = a and x = b

S = area of the surface generates by revolving the curve about x-axis between x = a and x = b Then

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

…(2)

13. Let f : R R be defined by f (t) = t² and let U be any non-empty open subset of R. Then

(a) f(U) is open

(b) f^–1(U) is open

(c) f(U) is closed

(d) f^–1(U) is closed

Ans. (b)

Sol. Given f : R R by f (t) = t³.

So, f is a continuous function. By the property of a function which is continuous that each open set has open pre-image under inverse of continuous function.

Ans so, f^–1 (U) is open

14. Let f : (–1, 1) be such that f⁽ⁿ⁾(x) exists and |f⁽ⁿ⁾ (x)| 1 for every n 1 and for every x (–1, 1). Then f has a convergent power series expansion in neighbourhood of

(a) every x (–1, 1)

(b)

(c) no x (–1, 1)

(d)

Ans. (a)

Sol. We can Maclaurin's series of f as

    

Remainder of this series if we writer upto a polynomial of degree n

15. Let a > 1 and f, g, h : [–a, a] R be twice differentiable functions such that for some 'c' with 0 < 0 < 1 < a,

f (x) = 0 only for x = –a, 0, a;

= 0 = g (x) only x = –1, 0, 1;

g'(x) = 0 = h (x) only for x = –c, c.

The possible relations between f, g, h are

(a)

(b)

(c)

(d)

Ans. (b)

Sol. As shown in the figure satisfying all the given conditions, the relationship between f, g and h can be

or  f, g, h : [–a, a] twice differentiable functions such that for some c with 0 < c < 1 < a < 1

f(x) = 0 only for x = –a, 0, a

= 0 = g(x) only x = –1, 0, 1

= 0 h(x) only for x = –1, c

so the possible relation are .

16. (a) Solve the initial value problem

(b) Solve the differential equation (2y sin x + 3y⁴ sin x cos x)dx – (4y³ cos² x + cos x)dy = 0

Sol.

It's auxiliary equation is m² – 1 = 0 m = ± 1

Complementary function CF = c₁e^x + c₂e^–x

Particular integral   

Applying the formula

Consider

So      y = CF + PI

   

Now at    x = 0, y = 1

…(1)

Now consider

Now at    x = 0, = 1

    1 = C₁ – C … (2)

On solving equation (1) and (2), we get

(b) (2y sin x + 3y⁴ sin x cos x)dx – (4y³ cos² x + cos x)dy = 0

Compare with Mdx + Ndy = 0

 M = 2y sin x + 3y⁴ sin x cos x

So differential equation is not exact

Now (2y sin x cos x + 3y⁴ sin cos²x)dx – (4y³ cos³x + cos²xdy) = 0 is exact D.E. then sol is

17. Let G be a finite abelian group of order n with identity e. If for all a G, a³ = e, then by induction on n, show that n = 3^k for some non-negative integer k.

Sol. If G = {e}, then it is of type as desired as

So,     n = 1 = 3⁰ = 3^k for k = 0. … (1)

Assuming the G is such that and |G| = n = 3^k for some non-negative integer k.

Let one additional element in G be . Then e. So, G must also include .

For G to be a group and would get multiplied by remaining elements of G to give new elements of G.

So, new elements of G are 3^k + 3^k. So, now total elements in G is 3^k + 3^k = 3^k+1. … (2)

So, from (1) and (2), by induction, G has elements of order equal to n = 3^k for some non-negative integer k.

18. (a) Let f: [a, b] be differentiable function. Show that there exist two points c₁, c₂ (a, b) such that 2f(c₁)

(b) Let

Find a suitable value of such that f is continuous. For this value of , is f differentiable at (0, 0)? Justify your claim.

Sol. (a) Let (f (x))² = g(x) for differentiable function g : [a, b] . Apply Lagrange's theorem to g in the interval [a, b],

…(1)

Again, applying Lagrange's theorem to f in the interval [a, b],

…(2)

Here, Lagrange's theorem are applicable for f (x) and (f (x))² as both are continuous and differentiable on [a, b].

From (1) and (2)

(b)  f (x, y) = (x² + y²) [ln (x² + y²) + 1]

Converting it to the polar coordinates

f (x, y) = r²[ln (r²) + 1]

So, for f to be continuous at (0, 0),

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

19. (a) Let S be the surface x² + y² + z² = 1, Use Stoke's theorem to evaluate where C is the circle x² + y² = 1, z = 0, oriented anticlockwise.

(b) Show that the vector is conservative. Find its potential and also the work done in moving a particle from (1, 0) to (2, 1) along some curve.

Sol. (a) By stroke's theorem

so  

S is part of sphere x² + y² + z² = 1 above xy plane. Let us consider surface made up two piecewise smooth surface S and S', where S' is the base of hemisphere whose equation x² + y² = l, z = 0

By divergence theorem

     ds = dxdy

(b) vector field is conservative if By stroke's theorem.

Here

  = x²y – xy⁴

so the potential is given by =x²y – xy⁴ + 3x

consider work done

so work done in moving a particle from (1, 0) to (2, 1) is equal to 5.

20. Let T : is defined by T(x, y, z) = (y + z, z, 0). Show that T is a linear transformation. If is such that = 0, then show that B = forms a basis of . compute the matrix of T with respect to B. Also find a such that 0

Sol. Part-I: Show T(x, y, z) = (y + z, z, 0) is linear transformation.

Let   = (x₁,y₁, z₁), = (x₂, y₂, z₂)

Then = T[a(x₁, y₁, z₁) + b(x₂, y₂, z₂)] = T[a(x₁, y₁, z₁) + (bx₂, by₂, bz₂)]

= T[(ax₁ + bx₂, ay₁ + by₂, az₁ + bz₂)] = (ay₁ + by₂ + az₂, az₁ + bz₂, 0)

= (ay₁ + az₁, az₁, 0) + (by₂ + bz₂, bz₂, 0)

= a(y₁ + z₁, z₁, 0) + b(y₂ + z₂, z₂, 0) = aT[(x₁, y₁, z₁)] + bT[(x₂, y₂, z₂)]

=

So T is linear transformation

Part-II:      Show B =

Let   = T(x, y, z) = (y + z, z, 0)

B = form a basis if it is linearly independent.

Let a,b, c R. Then a(x, y, z) + b(y + z, z, 0) + c(z, 0, 0) = 0

(ax + by + bz + cz, ay + bz, az) = 0

Since z 0, az = 0 a = 0, ay + bz = 0 b = 0 and c = 0

is linearly independent and has dimension 3

is basis ( standard basis of is {e₁, e₂, e₃} then by theorem if any linearly independent has same dimension then it is basis)

Part-III: Compute matrix T w.r.t. B.

Part-IV: Find

21. (a) For each N, define f_n : [–1, 1] R by

Compute for each n. Analyze pointwise and uniform convergence of the sequence of function {f_n}.

(b) Let f : be a continuous function with |f (x) – f(y)| |x–y| for every x, y . is f one-one? Show that there cannot exist three point a, b, c with a < b < c such that f(a) < f(c) < f(b).

Sol. (a) For each n N, define f_n : [–1, 1] R by

Thus f_n(x) is pointwise converge in [0, 1] convergence is also uniforms as.

But x and y are any two arbitary number. Hence c is any real number.

Since, f is one one and continuous function so

In each case f(x) is a monotomic function

Hence, f (a) < f (b) < f (c) for a < b < c or

f (a) > f (b) > f (c) for a < b < c

But not f (a) < f (c) < f (b) for a < b < c.

22. Find the volume of the cylinder with base as the disk of unit radius in the xy-plane centred at (1, 1, 0) and the top being the surface, z = {(x – 1)² + (y – 1)²}^3/2.

Sol. Required volume,