1. Let V be the vector space of all 6 × 6 real matrices over the field . Then the dimension of the subspace of V consisting of all symmetric matrices is

(a) 15

(b) 18

(c) 21

(d) 35

Ans. (c)

Sol. As we know dim means number of distinct non-zero element.

Consider 6 × 6 symmetric matrix i.e A^T = A

Note : Short Trick

Number of distinct element = 21 dim A = 21

2. Let R be the ring of all function from to under point-wise addition and multiplication. Let I = {f : R R | f is a bounded function}, J = {f : | f (3) = 0}.

Then

(a) J is an ideal of R but I is not an ideal of R

(b) I is an ideal of R but J is not an ideal of R

(c) both I and J are ideals of R

(d) neither I nor J is an ideal of R

Ans. (a)

Sol. If f I and g R.

Then |f g (x)| = |f (x) • g (x)| K • |g (x)| [since given that f is bounded say bounded by K].

But for g (x) we can not say it is bounded or not.

Hence I is not an ideal of R.

Addition of f₁ + f₂ is obviously belong to J.

Now, f J, g R, Then f g (3) = f (3) • g (3)

= 0 • g (3) = 0 f g (3) = 0

And so, J is an ideal of R.

3. Which of the following sequences of functions is uniformly convergent on (0, 1)?

(a) xⁿ

(b)

(c)

(d)

Ans. (c)

Sol.

is a uniformly convergent sequence (by M_n-test).

Applying the same criteria we can rule out option (a), (b) and (d).

4. Let T : be a linear transformation satisfy T³ + 3T² = 4I, where I is the identity transformation, then the linear transformation S = T⁴ + 3T³ – 4I is

(a) one-one but not onto

(b) onto but not one-one

(c) invertible

(d) non-invertible

Ans. (c)

Sol.

…(1)

…(2)

(from (2) + 3 × (1)) and using transformation definition

T is one-one.

Consider S = T⁴ + 3T³ – 4I = T(T³ + 3T²) – 4I = T(4I) – 4I

S = 4T – 4I

5. The number of all subgroups of the group (Z₆₀, + ) of integers modulo 60 is

(a) 2

(b) 10

(c) 12

(d) 60

Ans. (c)

Sol. In any Z_n number of subgroups equal to the number of divisors of n.

Hence number of division of n = (n)

(60) = (2² × 3 × 5) = (2 + 1) (1 + 1) (1 + 1) = 12

6. Let

Then the radius of convergence of the power series is

(a) 4

(b) 3

(c)

(d)

Ans. (b)

Sol.

radius of convergence of the power series is

if n is prime.

and R = 4 if n is not a prime

radius of convergence R = min(3, 4) = 3

7. The set of all limit points of the sequence is

(a) [0, 1]

(b) (0, 1]

(c) the set of all rational number in [0, 1]

(d) the set of all rational number in [0, 1] of the form where m and n are integers

Ans. (a)

Sol.

Elements of S_n increases from with a step size of any interval within [0, 1] of size would contain at least one element of S_n, hence of S. So, taking any point a [0,1] and its any neighbourhood (a , a + ) for some > 0, then it will contain at least one element of S_n by choosing n sufficiently large: > , n > log . Hence, any nbd of a has element of S, so a is a limit point of S. Therefore set of limit points of S = [0, 1]. The similar arguments hold for 0 and 1 as can be arbitrarily close to '0' and can be arbitrarily close to 1 by choosing sufficiently large value of n.

8. Let F : be a continuous function and a > 0. Then integral equals

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

9. The set of all positive values of a for which the series converges, is

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

which is non-zero and finite. So, u(n) and v(n) converge or diverge together

We know that converges iff p > 1

So, converge iff

Thus converge iff .

10. Let a be an non-zero real number. Then equals

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

11. Let T(x, y, z) = xy² + 2z – x²z² be the temperature at the point (x, y, z). The unit vector in the direction in which the temperature decreases most rapidly at (1, 0, –1) is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Temperature increases most rapidly in the direction of

Unit vector in that direction,

So temperature decrease most rapidly in the direction

12. Consider the differential equation 2cos(y²)dx – xysin(y²)dy = 0 then

(a) e^x is an integrating factor

(b) e^–x is an integrating factor

(c) 3x is an integrating factor

(d) x³ is an integrating factor

Ans. (d)

Sol. 2 cos(y²)dx – xy sin(y²)dy = 0 …(1)

Compare with Mdx + Ndy = 0

Then

Multiply equation (1) by x³, we get 2x³ cos(y²)dx – x⁴y sin(y²)dy = 0

Then

Here,

13. Suppose is a continuously differentiable vector field defined in a domain D in Which one of the following statements is NOT equivalent to the remaining ones?

(a) There exists a function such that for all (x, y)

(b) holds al all points of D

(c) for every piecewise smooth closed curve C in D

(d) The differential pdx + qdy is exact in D

Ans. (c)

Sol.

p dx + q dy is exact is as there exists a function as defined above, such that

14. Let f, g : [–1, 1] , f (x) = x³|x|, then

(a) f and g are linear independent on [–1, 1]

(b) f and g are linearly dependent on [–1, 1]

(c) is NOT identically zero on [–1, 1].

(d) There exist continuous functions p(x) and q(x) such that f and g satisfy + qy = 0 on[–1, 1]

Ans. (b)

Sol. Given that f(x) = x³, g(x) = x² |x|

(i) and (ii) simultaneously holds if c₁ = c₂ = 0

f(x) and g(x) are linear independent in [–1, 1].

Note: f(x), g(x) are linear dependent in [–1, 0] or [0, 1]. So, correct answer is (b).

(c) is wrong because

Note: (1) If y₁y₂ are linear dependent w(y₁y₂) = 0, but converse is not true.

(2) If w(y₁ y₂) 0 y₁ and y₂ are linear independent but converse is not true.

Note: Converse of (1) and (2) are true. If y₁ and y₂ are solutions of differential equation. So, (d) is wrong.

15. The value of c for which there exists a twice differentiable vector field with curl is

(a) 0

(b) 2

(c) 5

(d) 7

Ans. (c)

Sol.

16. Consider A contains 100 cc of milk and container B contains 100 cc of water. 5 cc of the liquid in A is transferred to B, the mixture is thoroughly stirred and 5 cc of the mixture in B is transferred back into A. Each such two-away transfer is called a dilution. Let a_n be the percentage of water in container A after n such dilutions, with the understanding that a₀ = 0.

(a) Prove that and that, in general for n = 1, 2, 3, …..

(b) Using (a) prove that for n = 1, 2, 3, …..

Find and explain why the answer is intuitively obvious.

Sol. (a) W = Water, M = Milk

First dilution:

So,

After (n – 1) dilutions:

After 5 cc transfer from A to B, B has

Than after (n) dilutions, % of water in (A):

Also,

Hence, for n = 1, 2, 3,… and are the same sequence.

This intuitively true as after infinite number of dilutions, milk and water would be equally distributed in the two containers. Same result would also be obtained after infinite dilutions if we take the dilution from B to A, i.e., if we take 5 cc first form B to A and then 5 cc back form A to B and counting it as one dilution.

17. (a) Let f : be a non-negative function. Assume that for every the series is convergent and has sum a_m and further that the series is also convergent and has sum L. Prove that for every n, the series is convergent and if we denote its sum by b_n then the series is also convergent and has sum L.

(b) Define f : by

Show that

Sol.

…(i)

Since, (a) and (b) are convergent, so equation

(i) is also convergent. So it can be written as

[since given that ]

Thus, is convergent and converge to L.

= [f(1, 1) + f(1, 2)] + f(1, 3) + …] + [f(2, 2)] + f(2, 3) + ….]

+[f(3, 1) + f(3, 2) + f(3, 3) + f(3, 4) + ….] + …….

so by g.p infinite sum,

18. (a) Evaluate where R = [0, 1] × [0, 1].

(b) Let Show that I I + 100.

Sol. (a)

Therefore,

19. Let D = {(x, y): x 0, y 0}. Let f (x, y) = (x² + y²)e^–x–y for (x, y) D.

Prove that f attains its maximum on D at two boundary points.

Deduce that

Sol. f(x, y) = (x² + y²)e^–x–y

f_x(x, y) = (2x – x² – y²)e^–x–y

f_y(x, y) = (2y – x² – y²)e^–x–y

f_x(x, y) = 0 x² + y² = 2x ...(1)

f_y(x, y) = 0 x² + y² = 2y ...(2)

Solving (1) and (2), x = y.

Putting y = x into (1), 2x² = 2x x = 0, x = 1

So, (x, y) (0, 0), (1, 1) · (0, 0) and (1, 1) satisfies (1) and (2)

So, critical points are P₁(0, 0) and P₂(1, 1)

f(0, 0) = (0² + 0²)e^–0–0 = 0 ...(A)

f(1, 1) = (1² + 1²)e^–1–1 = 2e^–2 ...(B)

i.e., on boundary, x = 0 line, f(x, y) = y²e^–y = g(y) (say)

(using L' Hospital's Rules twice)

Hence, f_max (x, y) on x = 0 line is f(0, 2) = 4e^–2 ...(3)

Similarly, f_max (x, y) on y = 0 line is f(2, 0) = 4e^–2 ...(4)

...(5)

Similarly, ...(6)

Boundary of D : x = 0, x and y

So, the greatest value of f using (A), (B), (3), (4) and (5), (6) is 4e^–2 = f(0, 2) = f(2, 0)

Hence, f(x, y) < 4e^–2

(x² + y²)e^–x–y < 4e^–2

f attains maximum at two boundary points (0, 2) and (2, 0).

20. (a) Let a₁, b₁, a₂, b₂ . Show that the condition a₂b₁ > 0 is sufficient but not necessary for the system

to have two linearly independent solutions of the form and with

(b) Show that the differential equation representing the family of all straight lines which have an intercept of constant length L between the coordinate axes is

Sol.

Consider (D – b₂) × (1) × b₁ × (2)

So

Then auxiliary equation

m² – (a₁ + b₂)m + (a₁b₂ – b₁a₂) = 0

But solution are given,

It show that it has real and distinct root.

Consider discriminant of equation (3)

If a₂b₁ > 0 then D > 0 and the differential equation will have linearly independent solution and auxiliary equation has real and distinct root.

But if a₂b₁ < 0 and smaller than (a₁ – b₂)² then the differential equation will have two linearly independent solution.

a₂b₁ > 0 is sufficient but not necessary condition for two independent solution.

(b)

Equation of straight line

Length of intercept L

So equation of straight line becomes

Differential equation w.r.t. x, we get

putting value of a and b in equation (1), we get

21. Let A, B, k > 0. Solve the initial value problem

(a) Also show that then the solution y(x) is monotonically increasing on and tends to

(b) if then the solution y(x) monotonically decreasing on and tends to as

Sol.

22. (a) Evaluate along the portion from (1,0,1) to (3,4,5) of the curve C, which is the intersection of the surface z² = x² + y² and z = y + 1.

(b) A particle moves counterclokwise along the curve 3x² + y² = 3 from (1,0) to a point P, under the action of the force Prove that there are two possible locations of P such that the work done by is 1.

Sol.

Given W = 1

Squaring both side, we get

So, there are two value of i.e two possible location of P that work done by is 1.

23. Verify Stokes theorem for the hemisphere x² + y² + z² = 9, z 0 and the vector field

Sol. The bounding curve C of S is given as, x = 3 cos , y = 3 sin , z = 0

Consider

put x = 3 cos , y = 3 sin , z = 0, dx = –3 sin d, dy = 3 cos d, dz = 0

Now consider

Consider a surface consisting of piecewise smooth surface S and

Hence stokes theorem is verified.

24. (a) Let T : be the linear transformation defined by T(x, y, z) = (x + 2y, x – z). Let N(T) be the null space of T and Find a linear transformation S : W such that TS = I, where I is the identity transformation on

(b) Suppose A is a real square matrix of odd order such that A + A^T = 0. Prove that A is singular.

Sol.

Now let find linear transformation S : W such that

TS = I or S = T^–1 let a = x + 2y, b = x – z

(b) A + A^T = 0 A = –A^T, As we know det(A) = det(A^T) = det(–A) ( A^T = –A) = (–1)ⁿ det A n is odd then (–1)ⁿ det A = –det A

2det(A) = 0 det(A) = 0 A is singular

25. (a) Find all pairs (a,b) of real numbers for which the system of equations

x + 3y = 1, 4x + ay + z = 0, 2x + 3z = b

has (i) a unique solution, (ii) infinitely many solution, (iii) no solution.

(b) Let A be a n×n matrix such that Aⁿ = 0 and A^n–1 0. Show that there exist a vector such that forms a basis for

Sol. (a) x + 3y = 1

4x + ay + z = 0

2x + 3z = b

Then Augumented matrix

(1) Unique Solution: rank [A] = ran[A:B] = 3 if rank (A) = 3 then a – 10 10

(a, b) where a 10

(2) Infinite many solution: for infinite many solution r[A] = r[A : B] < 3

last element of row must be zero.

a – 10 = 0,

(a, b) is (10, –10)

(3) For no solution r(A) r([A : B]) let r(A) < 3 and r([A : B]) = 3

For r(A) < 3 a – 10 = 0 a = 10

For r([A : B]) = 3

(a, b) is a = 10 and b –10

(b) Let and it is given that Aⁿ = 0 and A^n-1 0

To prove is basis of Rⁿ

Let a₀, a₁….a_n+1

Then consider

multiply A^n–1 both side.

Again multiply (1) by A^n–2

which means is linearly independent and any x R³ by spanned by these vector.

Basis of Rⁿ is

26. (a) In which of the following pairs are the two groups isomorphic to each other? justify your answers.

(i) and S¹, where is the additive group of real numbers and S¹ = {z : |z| = 1} under complex multiplication.

(ii) (, +) and (Q, +).

(b) Prove or disprove that if G is a finite abelian group of order n, and k is positive integer which divides n, then G has at most one subgroup of order k.

Sol.

Then a + Z has finite order q.

And if a Q^c then 'Q'. This sign is of order, not is '0' zero (a) is infinite. So has infinite element of infinite order and simultaneously infinite element of infinite order.

Similarly, any root of unity on |Z| = 1 has finite order and also there infinite element of infinite order.

Hence both

(ii) (Z, +) and (Q, +).

Not isomorphic since (Z, +) is a cyclic group and the generator of (Z, +) are + 1 and +1 and –1. But (Q, +) is not cyclic since no element x (Q, +) s.t.

(b) In any K₄ type group.

2/4 means 2 divides 4 but K₄ has exactly three subgroup of order 2.

Example : U(8) = {1, 3, 5, 7}

0 (3) = 2

0 (6) = 2

0 (7) = 2

So, {1, 3}, {1, 5} are three subgroup of order 2.

(b) Proof: a³ = e

{a, a², e} are three members of G.

So for K = 1.

3^K = 3¹ = 3. It is true.

Let us assume that b is not any power of a, then b, b² are distinct elements. How, by closure property of the group, ab, ab², a²b, a²b² will also be another four elements of the group.

So, number of elements = 9 = 3².

{e, a, a², b, b², ab, ab², a²b, a²b²}

Now, let O(G) = 3^K–1

then one element 'c' (say) other G, we have a new group say G with elements CG, C²G and C.G as the elements of new group with each element satisfying c³ =

| G₁ | = 3 | G | = 3.3^K–1 = 3^K

Hence proved.

27. Let I and J be ideals of a ring R. Let IJ be the set of all possible sums where a_i J for i = 1, 2, ….., n and

(a) Prove that IJ is an ideal of R and

(b) Is it true that IJ = I J ? Justify your answer.

Sol. (a) Since I and J be ideals of a ring R and where a_i I, b_i J for

i = 1, 2,……, n and n N

Also, r R, then

So, I J is an ideal

(b) For (Z, +, ·) if

I = set of all even integers and

J = set of all integers divisible by 4.

Then I J = set of all integers divisible by 8.

But I J = set of integers divisible by 4.

Thus I J I n J but I J I J [always].

28. A sequence {f_n} of functions defined on an interval I is said to be uniformly bounded on I if there exists some M such that |f_n (x)| M for all x I and for all n N.

(a) Prove that if a sequence of functions {f_n} converges to a function f on I and {f_n} is uniformly bounded on I, then f is bounded on I.

(b) Suppose the sequence {f_n} and {g_n} of functions converge uniformly to f and g respectively on I and both are uniformly bounded on I. Prove that the product sequence {f_ng_n} converges to fg uniformly bounded. on I.

Sol. (a) {f_n} is uniformly bounded |f_x(x)| M for all x I and n for some M. Assume that |f(x)|> M for some x = x₀ I, i.e., |f(x)| M. Say f(x₀) = M + m (m > 0, m Since, {f_n(x)} converges to f(x₀), for any > 0, there exists a N such that |f_n(x₀) – f(x₀)| < n > N.

Here |f_n – M| may go sufficiently close to zero but |m| remains a constant. So. for some > 0. Hence, from (1),

which is not true for any > 0 bot only > which contradicts that {f_n(x₀)} converges to f(x₀) emplying that our assumption is wrong. So, |f(x)| M, hence bounded.

(b) Let and for some M_f, M_g

For any > 0,

for some N , and for all x I. … (1)

This is possible because {g_n} and {f_n} are uniformly convergent, so and can exists with any values greater than zero. Then we can take the values of and such that say = = e/4. So, from (1) {f_ng_n} converges to fg uniformly on I.

29. (a) Prove that if f is a real-valued function which is uniformly continuous on an interval (a, b), then f is bounded on (a, b).

(b) Let f be a differentiable function on an interval (a, b). Assume that is bounded on (a, b). Prove that f is uniformly continuous on (a, b).

Sol. (a) Let f is unbounded on (a, b), i.e, f goes to at a or b. Then, for uniform continuity: for any > 0 there exists a such that for all x, y (a, b).

Here, |x – y| (0, b – a) and |f(x) – f(y)| can be infinitely large as f is unbounded on (a, b). So, is arbitrarily infinitesimally small, i.e., it tends to zero. So, we won't get a > 0 such that so, f would not be uniformly continuous on (a, b) which is a contradiction. Hence, our assumption that f is unbounded on (a, b) is wrong.

Therefore, f is bounded.

(b) For any

For x, y at a finite distance, is finite and for which is finite as f' is bounded. Hence, is bounded as (x - y)² (0, (b – a)²).

So, does not go to zero, So, there exists a > 0 such that

Hence, f is uniformly continuous on (a, b).