1. Let A(t) denote the area bounded by the curve y = e^–|x|, the x-axis and the straight lines x = –t and x = t, then A(t) is equal to

(a) 2

(b) 1

(c)

(d) 0

Ans. (a)

Sol.

2. If k is constant such that satisfies the differential equation

then k is equal to

(a) 1

(b) 0

(c) –1

(d) –2

Ans. (a)

Sol.

3. Which of the following functions is uniformly continuous on the domain as stated?

(a)

(b)

(c)

(d)

([x] is the greatest integer less than or equal to x)

Ans. (b)

Sol. For any > 0, we have

Clearly, x y as x, y 1. Hence, choosing a say we can say So by the definition, 1/x is uniformly continuous on

Another Method-1: Every Lipschitz function is uniform continuous so take the consider

where k = 1. So, is Lipschitz

f (x) is uniformly continuous.

Another Method-2: A function f such that is bounded for all x in domain then also f is uniformly continuous (Provided f is differential).

Hence f is uniformly continuous on

4. Let R be the ring of polynomials over and let I be the ideal of R generated by the polynomial x³ + x +1. Then the number of elements in the quotient ring R/I is

(a) 2

(b) 4

(c) 8

(d) 16

Ans. (c)

Sol. In Z₂, the given polynomial x³ + x + 1 is irreducible. Since.

for x = 0, (0)³ + 0 + 1 = 1 0

for x = 1, 1 + 1 + 1 = 1 0

and so,

(Since each coefficient of x³ + x + 1 have two choices).

Hence the number of elements in the quotient ring is 8.

5. Which of the following sets is the basis for the subspace

of the vector space of all real 2 × 2 matrices.

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

which gives y + t = 0 y = –t, x + 2y + t = 0 x – 2t + t = 0 x = t

so put value of x and y in W

So W can be expressed as linear combination of and it is also linear independent.

  Basis of W is

6. Let G be an Abelian group of order 10. Let S = {g G : g^–1 = g}. Then the number of non-identity elements in S is

(a) 5

(b) 2

(c) 1

(d) 0

Ans. (c)

Sol. There are exacted two group upto isomorphism of order 10. If it is abelian it will be z₁₀.

Now, S = {g G : g^–1 = g}

In each group identity always hold this. But in z₁₀ only one non-identity element 5 which is self inverse.

 5 + 5 = 0    order (5) = 2.

Hence    [5] = [5]^–1

7. Let (a_n) be an increasing sequence of positive real number such that series is divergent.

Let for n = 1, 2, … and for n = 2, 3, … Then is equal to

(a)

(b) 0

(c)

(d) a₁ + a₂

Ans. (a)

Sol. is divergent and a_k > 0 and a_{k + 1} > a_k for all k

8. For every function f : [0, 1] R which is twice differentiable and satisfies 1 for all x [0, 1], we must have

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

9. Let f : be defined by

Which of the following statements holds regarding the continuity and the existence of partial derivatives of f at (0, 0)?

(a) Both partial derivatives of f exist at (0, 0) and f is continuous at (0, 0)

(b) Both partial derivatives of f exist at (0, 0) and f in NOT continuous at (0, 0)

(c) One partial derivative of f does NOT exist at (0, 0) and f is continuous at (0, 0)

(d) One partial derivative of f does NOT exist at (0, 0) and f is NOT continuous at (0, 0)

Ans. (b)

Sol.

(along the path y = mx² if the limit exists)

Here, the limit value depends upon 'm', hence it is not unique. So, the limit doesn't exist. Therefore, f is not continuous at (0, 0).

10. Suppose (c_n) is sequence of real number such that exists and is non-zero. If the radius of convergence of the power series is equal to r, then the radius of convergence of the power series is

(a) less than r

(b) greater than r

(c) equal to r

(d) equal to

Ans. (c)

Sol.

11. The rank of the matrix is

(a) 3

(b) 2

(c) 1

(d) 0

Ans. (b)

Sol.

Here two row are non-zero    Rank is 2.

Another Method:

= 32 – 2.16 = 32 – 32 = 0

Rank 3 i.e rank < 3

Consider a 2 × 2 square sub matrix Rank = 2.

12. Let f : be a continuous function. If for all x then is equal to

(a) –1

(b) 0

(c) 1

(d) 2

Ans. (a)

Sol.

13. Let where a is constant. If the line integral over every closed curve C is zero, then a is equal to

(a) –2

(b) –1

(c) 0

(d) 1

Ans. (d)

Sol. Since for every closed curve

so, must be a conservative vector field

14. One of the integrating factor of the differential equation (y² – 3xy)dx+ (x² – xy) dy = 0 is

(a) 1/(x² y²)

(b) 1/(x²y)

(c) 1/(xy²)

(d) 1/(xy)

Ans. (b)

Sol. (y² – 3xy)dx + (x² – xy)dy = 0

Let's assume an integrating factor be x^ay^b. Then,

 (b + 2)x^ay^b+1 – 3x^a+1. y^b(b +1) = (a + 2)x^a+1. y^b – x^ay^b+1. (a + 1)

 (b + 2)x^ay^b+1 – 3(b + 1) x^a+1y^b = –(a + 1) x^ay^b+1 + (a + 2) x^a+1y^b

 b + 2 = (a + 1) and –3(b + 1) = a + 2

 b + 2 = –3 and a + 3b = –5 2b = –2 b = –1 and a = –2 x^ay^b = x^–2y^–1 =

15. Let C denote the boundary of the semi-circualr disk D = {(x, y) R² : x² + y²1, y 0} and let (x, y) = x² + y for (x, y) D. If is the outward unit normal to C, then the integral evaluated counter-clockwise over C, is equal to

(a) 0

(b)

(c)

(d)

Ans. (c)

Sol.

Let C consist of two piece wise curve C₁ and C₂

C₁ is circular part

C₂ is part of real axis

16. (a) Let Determine the eigen-values of the matrix B = M² – 2M + I.

(b) Let N be a square matrix of order 2. If the determinant of N is equal to 9 and the sum of the diagonal entries of N is equal to 10, then determine the eigenvalues of N.

Sol. (a) Eigen value can be found out by |M – λI| = 0

(1 + i – ) (1 – i – ) (i – ) = 0 l + i, i

Consider B = M² – 2M + I, As we know that is eigen value of M.

i.e.

eigen value of 2M is

eigen value of 2M is eigen value of I is 1,

Consider

eigen value of B is

    + 1 = (1 + i)² – 2(1 + i) + 1 = 1 + i² + 2i – 2 – 2i + 1

  = 1 – 1 + 2i – 2 – 2i + 1 ( i² = –1)

  = –1

for     = 1 – i

    + 1 = (1 – i)² –2(1 – i) + 1 = 1 + i² – 2i – 2 + 2i + 1

  = 1 – 1 – 2i – 2 + 2i + 1

= –1

For = i

   + 1 = i² – 2i + 1 = –1 – 2i + 1 = –2i

  eigen value of B are –1, –1, –2i

(b) Let Let N = ad – bc = 9 (given)

sum of diagonal element a + d = 10 (given)

Eigen value of N can be determine |N – I| = 0

but a + d = 10, ad – bc = 9

eigen value of N is 1, 9

Another Method: N be a square matrix of order 2. Then it has two eigen value say and .

As we know product of eigen values = determinant

Sum of eigen value = trace i.e. sum of diagonal

17. (a) Using the variation of parameters, solve the differential equation , given that x and are two solution of the corresponding homogeneous equations.

(b) Find the real number such that the differential equation

has a solution y(x) = for some non-zero real number a, b,

Sol.

Given x and are two integral of complementary function

So Wronkian  

Here,  u = x and

so x and are linearly independent solution.

For complete solution of differential equation is given by

where A and B are function of x compare y² + P. + Q·y = R where P, Q, R are function of x only or constant and

Here R = 1

The using variation parameter formula

 

Auxiliary equation is

[m² + 2( – 1)( – 3)m + (– 2)]y = 0 …(1)

It's s solution is y = a cos x + b sin βx means equation (1) has root which is purely imaginary

So – 2 > 0 > 2 …(2)

and   2( – 1)( –3) = 0 = 1 or 3 …(3)

from (2) and (3) we get = 3

18. (a) Let a, b, c be non-zero real number such that (a – b)² = 4ac. Solve the differential equation.

(b) Solve the differential equation dx + (e^ysiny – x) (y cos y + sin y)dy = 0.

Sol. …(A)

…(1)

…(2)

Differential equation (1) w.r.t. x again, we get

…(3)

Using equation (2), (3) and (A)

Auxiliary equation is am² + (b – a)m + c = 0

(b) dx + (e^{y sin y} – x)(y cos y + sin y)dy = 0

Compare with

Then

= e^{–y sin y + cos y – cos y} = e^{–y sin y}

= –y sin y + c

xe^{–y sin y} = –y sin y + c

   x = –y sin y e^{y sin y} + ce^{ysin y}

19. Let f (x, y) = x(x – 2y²) for (x, y) Show that f has a local minimum at (0, 0) on every straight line through (0, 0). Is (0, 0) a critical point of f ? Find the discriminant of f at (0, 0). Does f have a local minimum at (0, 0) ? Justify your answers.

Sol. Straight line through (0, 0): y = mx and x = 0 (m

on x = 0, f (0, y) = 0, a constant function so at each (0, y), fhas local minimum and local maximum.

on y = mx, f (x, y) = f (x, mx) = x(x – 2m²x²)

= x² – 2m²x³ = g(x) (say)

so, x = 0 is a critical point of g.

g has minimum at x = 0. That is, f has minimum at (0, 0) on y = mx.

f_x(x, y) = x – 2y² + x = 2x – 2y² f_x(0, 0) = 0

f_xy(x, y) = –4y f_xy(0, 0) = 0

f_yy(x, y) = –4x f_xy(0, 0) = 0

Discriminant of f at (0, 0), D(0, 0) = f_xx(0, 0). f_yy(0, 0) – (f_xy(0, 0)² = 0

f (x, y) = x(x – 2y²)

In any neighbourhood of (0, 0) there are points, say (, 0), at which f(, 0) > 0 and also there are points, say at which f < 0 and f (0, 0) = 0. So, f has neither minimum nor maximum at (0, 0). So, f has no minimum at (0, 0).

20. (a) Find the finite volume enclosed by the paraboloids z = 2 – x² – y² and z = z² + y².

(b) Let f : [0, 3] be a continuous function with

Evaluate

Sol. (a) 2 – x² – y² = x² + y² x² + y² = 1

So, the curve of intersection of the two surface is x² + y² = 1, z = 1, as shown in the figure.

The required volume,

(changing to the polar coordinates)

21. (a) Let S be the surface and let be the outward unit normal to S. If then evaluate the integral

(b) Let and If a scalar field and a vector field satisfy where f is an arbitrary differentiable function, then show that

Sol.

Let S : x² + y² + 2z = 2 x² + y² = –2(z – 1) is paraboloid with vertex (0, 0, 1)

Consider a closed surface which consist of two piece wise smooth, surface S and

22. (a) Let D be the region bounded by the concentric sphere S₁ : x² + y² + z² = a² and S₂ = x² + y² + z² = b², where a < b. Let be the unit normal to S₁ directed away from the origin. If = 0 in D and = 0 on S₂, then show that

Sol.

Let us consider a surface Consist of S and enclosing a volume D.

According to Gauss divergence

…(2)

from (1) and (2) and let is outwards drawn away from origin

Surface enclosed by C,

Using stroke's theorem,

23. Let V be the subspace of spanned by vector (1, 0, 1, 2), (2, 1, 3, 4) and (3, 1, 4, 6). Let T : V be a linear transformation given by T(x, y, z, t) = (x – y, z – t) for all (x, y, z, t) V. Find a basis for Null space of T and also a basis for range of T.

Sol. Part-I: Basis for null space. As we know N(T) = {X : T(X) = 0}. Let (x, y, z, t) belongs to V, such that

  T(x, y, z, t) = (0, 0, 0, 0) (x – y, z – t) = (0, 0) x = y, z = t

NN(T) = {(x, x, t, t)} (x, x, t, t) = x(1, 1, 0, 0) + t(0, 0, 1, 1)

{(1, 1, 0, 0), (0, 0, 1, 1)} is span N(T) and basis of null space.

Part-II: Basis of range space V is spanned by (1, 0, 1, 2), (2, 1, 3, 4) and (3, 1, 4, 6)

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

    T(3, 1, 4, 6) = (3 – 1, 4 – 6) = (2, – 2)

By definition range of T is spanned by T(1, 0, 1, 2), T(2, 1, 3, 4) and T(3, 1, 4, 6 ) i.e. (1, –1), (1, –1) and 2(1, –1).

Basis of range space = {(1, –1)}

24. (a) Compute the straight lines where D is the region in the xy-plane bunded by the straight lines y = x + 3, y = x – 3, y = –2x + 4 and y = –2x – 2.

(b) Evaluate

Sol.

25. (a) Does the series converge uniformly for x [–1, 1] ? Justify.

(b) Suppose (f_n) is a sequence of real-valued function defined on and f is a real-valued function defined on such that for all n N and a_n 0 as n Must the sequence (f_n) be uniformly convergent on R? Justify.

Sol. (a)

We can break into two series

First one is conditionally convergent by liebnitz test and second one is convergent.

Both does not have uniform convergence in [–1, 1].

So, converges uniformly is not true.

(b) Yes, it must the sequence (f_n) be uniformly convergent on R.

As we define, for a uniformly convergent sequence,

 

uniformly if

And also, given that

Hence, M_n and so convergence is uniform.

26. (a) Suppose f is a real valued thrice differentiable function defined on such that > 0 for all x . Using Taylor's formula, show that

for all x₁, x₂ in with x₂ > x₁.

Sol. (a) By Taylor's formula for f on about x = x₁,

Again, using the Taylor's formula for f' about x = x₁ in the interval ,

So, from (1) and (2).

(b) (a_n) is an increasing sequence with a upper bound. Also, (b_n) is a decreasing sequence with a lower bound. Then (a_n) and (b_n) are convergent sequences. Let their limits be lim a_n = a and lim b_n = b. Also, (a_n) an increasing sequence, so (b_n) is a decreasing sequence, so Also, Hence, So, there exists a real number x such that x may belong; to [a, b].

27. Let G be the group of all 2 × 2 matrices with real entries with respect to matrix multiplication. Let G₁ be the smallest subgroup of G containing and G₂ be the smallest subgroup of G containing and . Determine all elements of G₁ and find their orders. Determine all elements of G₁ onto G₂? Justify.

Sol. Given G be the group of all 2 × 2 matrix with real entries with respect to matrix multiplication. G₁ be the smallest subgroup of G containing

Thus be closure property of a group,

  

and since G₁ is subgroup, then identity

Thus now we will check the order of each element.

 

similarly  

Thus each non identity elements has order 2 and total element in G₁ is 4.

Thus it is isomorphic to K₄.

Now, G₂ the smallest subgroup of G containing

again by the property of subgroup, closure property holds. Thus,

Similarly,

and also    

Thus G₂ is isomorphic to Z₄. Since it has two elements of order 4 and one element of order 2. Hence there does not exist a one to one homorphism. since between two finite groups of same order if any one-one map exist, then it will be isomorphism. But K₄ and Z₄ are not isomorphic since in K₄ no element of order 4.

28. (a) Let p be a prime number and let be the ring of integers. If an ideal J or Z contains the set pZ properly, then show that J = . (Here pZ = {px : x Z}).

(b) Consider the ring R = {a + ib : a, b Z} with usual addition and multiplication. Find all invertible elements of R.

Sol. (a) For any prime p, pz be the maximal ideal in and so why if any ideal J contains pz. Then it must be J = .

Now we will show that pz is maximal.

If pz is contained in J then an element x J s.t. x is not a multiple of p. And since p is prime so (x, p) = 1 and so, an element y J s.t. xy = 1 (say inverse of x) and hence J = Z.

Other pf, if a is not a multiple of p. Now, ap, a²p, a³p …… will given integers which with respect to module p will go in all chasses [0], [1],…… [p – 1]. Hence J = z.

(b) R = {a + ib: a, b }

Any element (a + ib) R be invertible if an element c + id s.t. (a + ib)(c + id) = 1

   ac – bd = 1

So there are four possibilities of choices of a, b, c and .

(i) a = c = 1 and b = d = 0

(ii) a = c = –1 and b = d = 0

(iii) a = c = –1 and b = 1, d = –1

(iv) a = c = –1 and b = –1, d = 1

So, inverse of four elements 1, –1, i and and –i exists in R.

29. (a) Suppose E is a non-empty subset of which is bounded above, and let = sup E. If E is closed, then show that a E. If E is open, then show that E.

(b) Find all limit of the set

Sol. (a) E subset of , which is bounded above = sup E.

Since supremum of a bounded above set always be it's limit point.

And if E is closed to E contains all it's limit point and so E.

Again suppose that E is open, so interior points of E will be contained in E but it's boundary point one not contained in it. And for any non-empty open set supremum of the set will always be the boundary point. Hence, E.

(b) contains all the natural numbers. Since

E contains

Hence the set of limit points of E =