PYQ 2010 - VedPrep

IIT JAM Mathematics 2010
Previous Year Question Paper with Solution.

1. Which of the following conditions does NOT ensure the convergence of a real sequence of a real sequence {a_n}?

(a)

(b)

(c)

(d) The sequences {a_2n}, {a_{2n + 1}} and {a_3n} are convergent

Ans. (a)

Sol. Take an example of a_n such that

Clearly, But {a_n} does no converges as diverges.

2. The value of where G = is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

3. The number of element of S₅ (the symmetric group on 5 letters) which are their own inverses equals

(a) 10

(b) 11

(d) 26

Ans. (d)

Sol. In S₅ the number of elements which are own inverse equals to (number of elements of order 2 in S₅) + (identity elements (I))

So, In S₅, there once

be two cyclic decomposition

Which have order 2. And so,

for

So total 25 elements of order two and one element identity. so total number of element which are own inverse.

4. Let S be an infinite subset of such that S Which of the following statements is true?

(a) S must have a limit point which belongs to

(b) S must have a limit point which belongs to

(d) \S must have a limit point which belongs to S

Ans. (d)

Sol. S be an infinite set of s.t. S

(a) S must have a limit point which belongs to .

E.g. S = { is prime} then S

but S does not have a limit point. So not true

(b) Not true by same example

(d) /S, since S must have infinite point of irrational numbers and so,

R/S, must have all rationals. And for is limit point of and since x was arbitrary.

Hence, x S.

5. Let f : (1, 4) be a uniformly continuous function and let {a_n} be a Cauchy sequence in (1, 2).

Let for all n N. Which of the following statement is true?

(a) Both {x_n} and {y_n} must be Cauchy sequences in

(b) {x_n} must be a Cauchy sequence in but {y_n} need not be a Cauchy sequence in

(d) Neither {x_n} nor {y_n} needs to be a Cauchy sequence is

Ans. (d)

Sol. Again, is uniformly continuous in (1, 4). Since every continuous function be the uniformly continuous in a bdd interval.

Now, which is not cauchy.

Similarly for y_n. Hence neither {x_n} nor {y_n} needs to be cauchy sequence in .

6. Let be the gradient of a scalar function. The value of along the oriented path L from (0, 0, 0) to (1, 0, 2) and then to (1, 1, 2) is

(a) 0

(b) 2e

(d) e²

Ans. (b)

Sol. Given

7. Let denote the force field on a particle traversing the path L form (0, 0, 0) to (1, 1, 1) along the curve of intersection of the cylinder y = x² and the plane z = x. The work done by is

(a) 0

(b)

(c)

(d) 1

Ans. (c)

Sol. Work done

8. Let R[X] be the ring of real polynomials in the variables X. The number of ideals in the quotient ring R[X]/(X²–3X+2) is

(a) 2

(b) 3

(d) 6

Ans. (c)

Sol.

So, the ideals are, e, [x – 1], [x – 2] and [(x – 1) (x – 2)]

Hence 4 ideals

9. Consider the differential equation , where a, b > 0 and y(0) = y₀. As x then solution y(x) tends to

(a) 0

(b) a/b

(d) y₀

Ans. (b)

Sol.

Then x = 0, y = y₀

10. Consider the differential equation (x + y + 1)dx + (2x + 2y + 1)dy = 0. Which of the following statements is true?

(a) The differential equation is linear

(b) The differential equation is exact

(d) A suitable substitution transforms the differentiable equation to the variables separable form

Ans. (d)

Sol. (x + y + 1)dx + (2x + 2y + 1)dy = 0

It can be transform into variable separable form.

11. Let T: be a linear transformation such that T((1, 2)) = (2, 3) and T ((0, 1)) = (1, 4). Then T(5, 6) is

(a) (6, –1)

(b) (–6, 1)

(d) (1, –6)

Ans. (a)

Sol. Given that T(1, 2) = (2, 3), T(0, 1) = (1, 4)

Consider T(1, 0) = T[(1, 2) - 2(0, 1)] = T(1, 2) – 2T(0, 1)

= (2, 3) – 2(1, 4) = (2, 3) + (–2, –8) = (0, –5) = T(1, 0) = (0, –5)

T(5, 6) = T[5(1, 0) + 6(0, 1)] = 5T(1, 0) + 6T(0, 1)

= 5(0, –5) + 6(1, 4) = (0, –25) + (6, 24) = (6, –1)

T(5, 6) = (6, –1)

12. The number of 2 × 2 matrices over (the field with three elements) with determinant 1 is

(a) 24

(b) 60

(d) 30

Ans. (a)

Sol. = {0, 1, 2}.

Let the desired matrices ge of the form

det A = (ad – bc) mod 3.

a, b, c, d {0, 1, 2} ad, bc {0, 1, 2, 3, 4}. det A = 1mod 3 ad – bc = 1, 4, –2.

Possibilities of A are:

So, total number of A = 2 + 6 + 6 + 2 + 1 + 3 + 4 = 24.

13. The radius of convergence of the power series where a₀ = 1, a_n = 3^–n a_n–1 for n N, is

(a) 0

(b)

(d)

Ans. (b)

Sol.

14. Let T : be the linear transformation whose matrix with respect to standard basis {e₁, e₂, e₃} of is . Then T

(a) maps the subspace spanned by e₁ and e₂ into itself

(b) has distinct eigenvalues

(d) has a non-zero null space

Ans. (c)

Sol. Matrix of

T(e₁) = T(1, 0, 0) = (0, 0, 1) = e₃ T(e₂) = T(0, 1, 0) = (0, 1, 0) = e₂

Thus, T does not map subspace spanned by e₁ and e₂ into itself

(a) is not correct.

for (b) The eigen-value of T is |T – I| = 0

So T does not have distinct eigen value so (b) is not correct.

For (c) let us find eigen vector for

Vector for = 1 is (–1, 0, 1), Also T has one eigen value corresponding to = –1

eigen vector span R³

for (d) T(a, b, c) = (0, 0, 0) T(a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = (0, 0, 0)

aT(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = a(0, 0, 1) + b(0, 1, 0) + c(0, 0, 1) = (0, 0, 0)

a = 0, b = 0, c = 0 so (a, b, c) = (0, 0, 0), hence T has zero null space

(d) is not correct

15. Let T : be the linear transformation whose matrix with respect to the standard basis of is , where a, b, c are real numbers not all zero. Then T

(a) is one to one

(b) is onto

(d) has rank 1

Ans. (c)

Sol. T is

Then | T | = –abc + bac = 0

rank of T is 2.

(d) is incorrect

If rank of T is 2, then its nullity will be 1. Thus it will not be one to one.

It is not onto because range has tow vectors, so it will not generate R³.

The equation of line passing through the origin is

let (l, m, n) be the point on line.

Then T(l, m, n) = T[l(1, 0, 0) + m(0, 1, 0) + n(0, 0, 1)]

= lT(1, 0, 0) + mT(0, 1, 0) + n(T(0, 0, 1) = l(0, –a, –b) + m(a, 0, –c) + n(b, c, 0)

= (ma + nb, – al + nc, bl – cm)

(ma + nb, –al + nc, –bl – cm) does not lie on the line.

16. (a) Obtain the general solution of the following system of differential equations:

(b) Find the curve passing through and having slope at (x, y) given by differential equation 2(1 + y²)dx + (2x – tan^–1 y)dy = 0.

Sol. (a)

Multiply (D + 1) × (1) and 2 × (2)

CF = c₁e^3t + c₂e^-3t

So solution is

(b) 2(1 + y²)dx + (2x – tan^–1 y)dy = 0 …(1)

17. (a) Find the volume of the region in the first octant bounded by the surface x = 0, y = x, y = 2 – x², z = 0 and z = x².

(b) Suppose f : is a non-constant continuous function satisfying f(x + y) = f(x) f(y) for all x, y .

(i) Show that f(x) 0 for all x .

(ii) Show that f(x) > 0 for all x .

(iii) Show that there exists such that f(x) = for all x .

Sol. (a) Let S : z = x².

The required volume is the volume under the surface S over the region R in the xy plane where R is as shown in the figure.

Required volume,

(b) (i) f(x + y) = f(x) f(y)

f(x + 0) = f(x) f(0) f(x) (f (0) – 1) = 0 f(x) = 0 or f(0) = 1. But f is not a constant function, hence f(x) 0 . So, f(0) = 1 and f(x) 0 for all x .

18. (a) Let f (x) and g(x) be real valued functions continuous in [a, b], differentiable in (a, b) and let 0 for all x (a, b). Show that there exists c (a, b) such that

(b) Let 0 < < 4 and let {a_n} be a sequence of positive real numbers satisfying for n N. Prove that exists and determine this limit.

Sol.

Let h(x) = f(x) g(x) – g(b) f(x) – f(a) g(x)

f and g are continuous on [a, b], then so would be their algebra of f and g the function h in [a, b].

f and g are differentiable in (a, b), so their algebra function h would also be differentiable in (a, b).

Further h(b) = f(b) g(b) – g(b) f(b) – f(a) g(b) = –f(a) g(b)

h(a) = f(a) g(a) – g(b) f(a) – f(a) g(a) = –f(a) g (b)

i.e., h(a) = h (b)

So, Rolle's theorem is applicable in h on [a, b]. Applying Role's theorem on h [a, b], there exists c (a, b) such that = 0

{a_n} is a decreasing sequence. And we have shown that 0 < a_n < 1. So, {a_n} is a monotonic bounded sequence hence convergent.

19. Let G be an open subset of R.

(a) If 0 G, then show that H = {xy : x, y G} is an open subset of R.

(b) If 0 G and if x + y G for all x, y G, then show that G = R.

Sol. (a) If x, y G, then there exists > 0 such that G and G as x, y are interior points of G, an open set.

Four possibilities arise about signs of x and y.

Case-I: Both x and y are positive real numbers.

We can always find such that x – > 0 and y – > 0, by taking < x, < y.

xy is an interior point of H₁

xy is an interior point of H as

Case-II: Both x and y are negative numbers.

We can find small enough such that x + < 0 and y + < 0.

xy is an interior point of H₁

xy is an interior point of H as H₁ H₁

Case-III: x > 0, y < 0.

We can find small enough such that x – > 0 and y + < 0.

xy is an interior point of H₁

xy is an interior point of H as H₁ H

Case-IV: x < 0, y > 0.

Proceeding as in case-III. we can say xy is an interior point of H.

Hence considering all the four cases, it has been proved that xy is an interior point for all x, y G, which proves further that H is an open set.

(b) 0 belongs to G and G is an open set > 0 such that (–2, 2) G.

Let a > 0, a be a real number.

x + y G for all x, y G ⇒ taking x [0,] and y = , [, 2] G

taking x[, 2] and y = , [2, 3] G.

Proceeding this way [n , (n + 1) ]

So for some n = N₁, a (N₁, , (N₁ + 1) ). (N₁ = [a/]. [ ] denotes the greatest integer).

Hence, any a , also belong to G. Therefore, G = .

20. Let p(x) be a non-constant polynomial with real coefficients such p(x) 0 for all x R. Define for all x R. Prove that

(i) For each > 0, there exists a > 0 such that |f(x)| < for all x R satisfying |x| > a, and

(ii) f : R R is a uniformly continuous function.

Sol. (i) In particular if we take p(x) = x² + where Then, p(x) = x² +

Now,

similarly we can extend this proof for other polynomials

(ii) As,

So for each x, N dependent on only (for which exist)

Hence, by the definition of uniformly continuous function, f : R R is a uniformly continuous function.

21. (a) Let M(K) and m(k) denote respectively the absolute maximum and the absolute minimum values of x³ + 9x² – 21x + k in the closed interval [–10, 2]. Find all the real values of k for which |M(k)| = |m(k)|. (6)

(b) Let = 1, and for n 3.

Prove that, for n N

Deduce that for any a, b R

Sol. Let f(x) = x³ + 9x² – 21x + k

= 3x² + 18x – 21 = 3(x² + 6x – 7) = 3(x + 7) (x – 1)

Critical points of f are x = 1, x = –7

f(–10) = –10³ + 9·10² + 210 + k = 110 + k

f(2) = 2³ + 9·2² – 21·2 + k = 4 + 36 – 42 + k = – 2 + k

f(1) = 1 + 9 – 21 + k = k – 11

f(–7) = k + 245

Hence, M(k) = k + 245 and m(k) = k – 11

|M(k)| = |m(k)| |k + 245| = |k – 11|

Hence, k = –117

So, by mathematical induction

Hence by mathematical induction, P(n) is true

22. (a) Let f(x, y) = ax² + xy + 1. Find sufficient conditions on such that (0, 0) is

(i) a point of local maxima of f(x, y)

(ii) a point of local minima of f(x, y)

(iii) a saddle point of f(x, y)

(b) Find the derivative of f(x, y, z) = 7x³ – x²z – z² + 28y at the point A = (1, –1, 0) along the unit vector What is the unit vector along which f decreases most rapidly at A? Also, find the rate of this decreases.

Sol.

(i) for (0, 0) to be a point of local maxima of f,

D (0, 0) > 0 and f_xx (0, 0) < 0

(ii) for (0, 0) to be a point of local minima of f,

D(0, 0) > 0 and f_xx (0, 0) > 0

(iii) for (0, 0) to be a saddle point of f,

(b) f(x, y, z) = 7x³ – x²z – z² + 28y

A directional derivative of f(x, y, z) along a direction given by unit vector

The direction along which maximum value directional derivatives occurs lies along .

since the unit vector gives the direction of maximum increase.

So, the function decrease most rapidly and rate of decrease

23. Using x = e^u, transform the differential equation

to a second order differential equation with constant coefficients. Obtain the general solution of the transformed differential equation.

Sol.

Its complementary function m² + 3m + 2 = 0

Solution of differential equation

y = CF + PI

24. Let G be a group and let A(G) denote the set of all automorphism of G, i.e. all one-to-one, onto, group homomorphisms from G to G. An automorphism f : G G of the f(x) = axa^–1, x G (for some a G) is called an inner automorphism. Let I(G) denote the set of all inner automorphism of G.

(a) Show that A(G) is a group under composition of functions and that I(G) is a normal subgroup of A (G).

(b) Show that I(G) is isomorphic to G/Z(G), where

(G) = {g G : xg = gx for all x G} is the centre of G.

Sol. (a) A(G) set of all automorphisms of G.

I (G) set of all inner automorphism of G.

f A(G) f is an automorphism of G.

By the def. of an automorphism it is one-one onto homomorphism from G to G. So f^–1 exist.

f(x) = x identity also, f·g A(G), so A(G) is group.

g_f f I(G) Ʌ g A(G). Then

(g f g^–1) (x) = g f (g^–1(x)) = g = [a(g^–1(x))a^–1]

By the property of inner automor.

So, I(G) is a normal subgroup of A(G).

(b) Given Z(G) = {g G : (x) (g) = (g) (x) x G} is the centre of G.

x g = g x x^–1 xg = x^–1 g x

g = x^–1 g x ...(i)

Number of different isomorphic function will be quotient of G by elements of the form in (i)

25. (a) Given an example of a linear transformation T : such that for all R².

(b) Let V be a real n-dimensional vector space and let T : V V be a linear transformation satisfying for all V.

(i) Show that n is even.

(ii) Use T ot make V into complex vector space such that the multiplication by complex numbers extends the multiplication by real numbers.

(iii) Show that, with respect to the complex vector space structure on V obtained in (ii), T : V V is a complex linear transformation.

Sol. (a) T :

let define T(a, b) = (–b, a), then T[T[a, b]] = (–a, –b) = –(a, b)

but if n is odd then not satisfied.

Let u = a + ib C

Then T(u₁, u₂….u_2m) = (–u_m+1…u₁…u_m)

26. Let W be the region bounded by the planes x = 0, y = 0, y = 3, z = 0 and x + 2z = 6. Let S be the boundary of this region. Using Gauss' divergence theorem, evaluate where and the outward unit normal vector to S.

Sol. Using Gauss divergence theorem

27. (a) using strokes' theorem evaluate the line integral where L is the intersection of x² + y² + z² = 1 and x + y = 0 traversed in the clockwise direction when viewed from the point (1, 1, 0).

(b) Change the order of integration in the integral

Sol. (a) Using stroke's theorem

28. In a group G, x G is said be conjugate to y G, written x ~ y, if there exists z G such that x = zyz^–1.

(a) Show that ~ is an equivalence relation on G, Show that a subgroup N is G is normal subgroup of G if and only if N is a union of equivalence classes of ~.

(b) Consider the group of all non-singular 3 × 3 real matrices under matrix multiplication. Show that (i.e. the two matrices are conjugate).

Sol.

(i) reflexivity : since exe^–1 = x (where e is an identity element) so x ~ x.

(ii) symmetry : x ~ y x w x w^–1 = y[say w = z^–1]

y ~ x

(iii) Transitivity also obviously hold.

Hence, '~' is an equivalence relation on G.

And hence '~' divides G into distinct equivalence classes. [By equivalence partition theorem].

And also let N be a normal subgroup of G. Let y an element z N s.t.

y = x z x^–1 so, y ~ z. Hence N is union of equivalence classes of ~.

Again if N is a union of equivalence classes of ~.

Then, a z N and x G s.t. y = x z x^–1 N.

N is a normal subgroup in G.

(b) Firstly find out the E.V. of this matrix.

The eigen values are 1, 1, 3.

A·M·of 1 = 2.

Now,

x + 2y = 0 which gives G·M = 2

So, A·M· = G·M· of each eigen value. Hence matrix is diagonalizable.

Similarly, we can prove that is also diagonalizable. So, both matrices are conjugate to a diagonal matrix 'D'. Hence

P A P^–1 = P B P^–1 = D

29. Let S denote the commutative ring of all continuous real valued function on [0, 1], under pointwise addition and multiplication. For a [0, 1], let M_a = {f S|f(a) = 0}.