PYQ 2012 - VedPrep

IIT JAM Mathematics 2012
Previous Year Question Paper with Solution.

1. Let {x_n} be the sequence If

then the sequence {y_n} is

(a) monotonic

(b) NOT bounded

(d) convergent

Ans. (d)

Sol.

2. The number of distinct real roots of the equation x⁹ + x⁷ + x⁵ + x³ + x + 1 = 0 is

(a) 1

(b) 3

(d) 9

Ans. (a)

Sol. Let f(x) = x⁹ + x⁷ + x⁵ + x³ + x + 1

This is a polynomial of degree 9, an odd number. Hence, range f =

f(x) has at least one real root.

= 9x⁸ + 7x⁶ + 5x⁴ +3x² + 1 > 0

f is increasing f has at most one real root.

Hence, f(x) has exactly one real root as range f =

3. If f : is defined by

Then

(a) f_x(0, 0) = 0 and f_y (0, 0) = 0

(b) f_x (0, 0) = 1 and f_y (0, 0) = 0

(d) f_x(0, 0) = 1 and f_y (0, 0) = 1

Ans. (b)

Sol.

4. The value of

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

5. The differential equation (1 + x²y³+ x²y²)dx + (2 + x³y² + x³)dy = 0 is exact if equals

(a)

(b)

(d) 3

Ans. (b)

Sol. (1 + x²y³ + x²y²)dx + (2 + x³y² + x³)dy = 0

If Mdx + Ndy = 0

then it is exact if 3x²y² + 2x²y = 3x²y² + 3x²y

6. An integrating factor for the differential equation (2xy + 3x²y + 6y³)dx + (x² + 6y²)dy = 0 is

(a) x³

(b) y³

(d) e^3y

Ans. (c)

Sol. (2xy + 3x²y + 6y³)dx + (x² + 6y²)dy = 0

Multiply with e^3x

(2xy + 3x²y + 6y³)e^3xdx + (x² + 6y²)e^3x dy = 0

= 2x e^3x + 3x²e^3x + 18y²e^3x + 18y²e^3x = (2x + 3x² + 18y²)e^3x

e^3x is a integrating factor.

7. For c > 0, if is the unit normal vactor at to the cone then

(a) a² + b² – c² = 0

(b) a² – b² + c² = 0

(d) a² + b² + c² = 0

Ans. (a)

Sol.

8. Consider the quotient group of the additive group of rotional numbers. The order of the element in is

(a) 2

(b) 3

(d) 6

Ans. (b)

Sol. additive group of rational numbers.

Hence order of any type element will be equal to q, If (p, q) = 1 and p < q.

9. Which one of the following is TRUE?

(a) The characteristic of the ring 6 is 6

(b) The ring 6 has a zero divisor

(d) The ring 6 × 6 is an integral domain.

Ans. (c)

Sol.

6 has characteristic '0'. Hence

10. Let W be a vector space over and let T : W be a linear transformation such that S = {Te₂, Te₄, Te₆} spans W. Which one of the following must be TRUE?

(a) S is a basis of W

(b)

(b) {Te₁, Te₃, Te₅} spans W

(d) ker (T) contains more then one element

Ans. (d)

Sol. As we know

Rank (T) + nullity (T) = dim

Rank (T) = dim(Range of T)

Given that S = {Te₂, Te₄, Te₆} spans W then its basis less than or equal to 3.

Rank (T) 3, here dim = 6, then nullity (T) 3

Null space or kernel of T contains more than one element

11. Consider the following subspace of

W = {(x, y, z) 2x + 2y + z = 0, 3x + 3y – 2z = 0, x + y – 3z = 0}.

The dimension of W is

(a) 0

(b) 1

(d) 3

Ans. (b)

Sol. First solve the equations

2x + 2y + z = 0 …(1)

3x + 3y – 2z = 0 …(2)

x + y – 3z = 0 …(3)

On solving (2) – (3) × 3

we get –2z + 9z = 0 z = 0

then put z = 0 in above equation we get

2x + 2y = 0, 3x + 3y = 0, x + y = 0, let x =k than y = –k

(x, y, z) = (k, –k, 0) = k(1, –1, 0) W = {k(1, –1, 0)}

i.e. all the element of W spanned by (1, –1, 0) and it is basis

dim W = 1

Here dim (W) = nullity (A) = 1.

12. Let P be a 4 × 4 matrix whose determinant is 10. The determinant of the matrix – 3P is

(a) –810

(b) –30

(d) 810

Ans. (d)

Sol. We know if A be n × n matrix then |KA| = Kⁿ|A|

|–3P| = (–3)⁴ |P| ( |P| = 10)

|–3P| = 9 × 9 × 10 = 810

13. If the power series converges for x = 3, then the series

(a) converges absolutely for x = –2

(b) converges but not absolutely for x = –1

(d) diverges for x = –2

Ans. (a)

Sol. If the power series converges for x = 3, then the series

Converge absolutely on |x| < 3 atleast and uniformly converge on |x| r, r < 3.

Converges absolutely for x = –2 since if |x| < R converges then for | x₀ | < R converges absolutely where |x₀| < |x|.

14. If , then the set of all limit points of Y is

(a) (–1, 1)

(b) (–1, 1]

(d) [–1, 1]

Ans. (d)

Sol.

15. If C is smooth curve in from (0, 0, 0) to (2, 1, –1), then the value of

(a) –1

(b) 0

(d) 2

Ans. (c)

Sol.

16. (a) Examine whether the following series is convergent:

(b) For each x let [x] denote the greatest integer less than or equal to x. Further, for a fixed (0, 1), define for all x Show that the sequence {a_n} converges to .

Sol. (a)

Hence, by the ratio test the series converges.

(b) (using L' Hospital rule)

(again using L' Hospital rule)

17. (a) Evaluate

(b) For a, b with a < b, let f : [a, b] be continuous on [a, b] and twice differentiable on (a, b). Further assume that the graph of f(x) intersects the straight line joining the points (a, f(a)) and (b, f(b)) at a point (c,f(c)) for a < c < b. Show that exists a real number (a, b) such that = 0.

Sol.

(b) f is continuous on [a, c] and differentiable on (a, c), so by Lagrange's men value theorem, there exists such that slope of the line segment joining A and B.

Similarly, applying Lagrange's Mean value theorem on f on [c, b]:

Applying Roll's theorem on in the interval is continuous on as is differentiable hence continuous on [a, b] and so there exists such that = 0.

i.e. there exists (a, b) such that = 0.

18. (a) Show that the point (0, 0) is neither a point of local minimum nor a point of local maximum for the function f : given by f(x, y) = 3x⁴ – 4x²y + y²for (x, y)

(b) Find all the critical point of the function f : given by f(x, y) = x³ + y³ – 3x – 12y + 40 for (x, y) Also, examine whether the function f attains a local maximum or a local minimum at each of these critical points.

Sol. (a) f(x, y) = 3x⁴ – 4x²y + y²

= (3x² – y) (x² – y)

f(0, 0) = 0

For a neighbour x² + y² < 4² of (0, 0), we have points, such as P₂(, –), at which f(x, y) > 0, (f(,–)) = (3² + ) (² + ) > 0), and have points, such as P₁(, 2²), at which f(x, y) < 0 (f(, 2²) = (3²– 3 ^ ) (² – 2²) = –⁴ < 0). So, every neighbourhood of (0, 0) contain points which has values higher as well as lower than f(0, 0).

Hence, f(0, 0) is neither a point of local minimum nor a point of local maximum.

(b) f(x, y) = x³ + y³ – 3x – 12y + 40

f_x(x, y) = 3x² – 3 = 3(x² – 1) = 3(x – 1) (x + 1)

f_y(x, y) = 3y² – 12 = 3(y – 4) = 3(y – 2) (y + 2)

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

Discriminant of f, D(x, y) = f_xy(x, y) f_yy(x, y) – (f_xy(x, y))² = 12x

f_x(x, y) = 0 x = 1 or –1

f_y(x, y) = 0 y = 2 or –2

Hence, critical points are (1, 2) (–1, 2), (1, –2) and (–1, –2).

D(1, 2) = 12 > 0

f_xx (1, 2) = 6 × 1 = 6 > 0

f has minimum at (1, 2)

D(–1, 2) = –12 < 0

f has saddle point at (–1, 6). So f has neither minimum nor maximum at (–1, 6).

D(1, –2) = 12 > 0

f_xx (1, –2) = 6 > 0

f has minimum at (1, –2)

D(–1, –2) = –12 < 0

19. (a) Evaluate

(b) Using multiple integral, find the volume of the solid region in bounded above by the hemisphere and bounded below by the cone

Sol.

(b) Sphere, S : (z – 1)² = 1 – x² – y², i.e., S : (z – 1)² + x² + y² = 1

Cone,

Solving the two equations

(z – 1)² + z² = 1

(z – 1)² = (1 – z)(1 + z) (z – 1)(z + 1 + 1 + z) = 0

z = 1, –2 z = 1

x² + y² = 1

Volume is enclosed between the cone and the sphere. Therefore, the required volume,

(R is as shown in the figure)

20. Find the area of the portion of the surface z = x² – y² in which lies inside the solid cylinder x² + y² 1.

Sol. Required surface area.

21. Let y(x) be the solution of the differential equation such that y(0) = 2 and = 2. Find all values of [0, 1] such that infimum of the set {y(x)|x is greater then or equal to 1.

Sol. then auxiliary equation m² – 1 = 0 m ± 1

y = c₁e^x + c₂e^–x, Given y(0) = 2

2 = c₁ + c_{2 ..(1)}

For finding value of x s.t. y is minimum

but infimum is greater than 1

than we get infimum equal to or greater than 1.

22. (a) Assume that y₁(x) = x and y₂(x) = x³ are two linearly independent solutions of the homogeneous differential equation Using the method of variation of parameters, find a particular solution of the differential equation

(b) Solve the differential equation subject to the condition y(1) = 1

Sol.

but it is given that y₁(x) = x and y₂(x) = x³

As we know y = C.F. + P.I. but it is given y₁(x) = x and y₂(x) = x³

but it is given that y(1) = 1

23. (a) Let be the position vector field in and let f : be a differentiable function. Show that

(b) Let W be the region inside the solid cylinder x² + y² 4 between the plane z = 0 and the paraboloid z = x² + y². Let S be the boundary of W. Using Gauss's divergence theorem, evaluate where and is the outward unit normal vector to S.

Sol.

(b) Shaded region is the volume enclosed by S.

By Gauss's divergence theorem,

(changing to cylindrical coordinates)

24. (a) Let G be a finite group whose order is not divisible by 3. Show that for every g , there exists an h such that g = h³.

(b) Let A be the group of all rational numbers under addition, B be the group of all non-zero rational numbers under multiplication and C the group of all positive rational numbers under multiplication. Show that no two of the groups A, B and C are isomorphic.

Sol. (a) Given that G be a finite group say 0(G) = m.

Then m = 3n – 1 or 3n + 1 [ m is not divisible by 3].

In A, there is only one element of finite order itself identity.

In B, –1 has order 2 which is of finite order other than identity.

Hence A and B not isomorphic.

Similarly B and C are not isomorphic.

Now, suppose on isomorphism of between A and C. Then 2 f is onto, = 2.

But that is a contradiction as there is on rational no. x s.t. x² = 2. Hence the result follows.

25. (a) Let I be an ideal of a commutative ring R. Define

A = {r R | rⁿ I or some n

Show that A is an ideal of R.

(b) Let F be a field. For each p(x) F[x] (the polynomial ring in x over F) define :F[x] F × F by (p(x)) = (p(0), p(1)).

(i) Prove that is a ring homomorphism

(ii) Prove that the quotient ring F[x]/(x² – x) is isomorphic to the ring F×F.

Sol. (a) I be an ideal of a commutative ring R. Given A = {r R : rⁿ I same n

Now, r, s A

Hence, is ring homomorphism.

(iii) By first fundamental theorem of ring isomorphism,

So, (f [x]) is equivalent to product of two fields i.e.

26. (a) Let P, D and A be real square matrices of the same order such that P is invertible. D is diagonal and D = PAP^–1. If Aⁿ = 0 for some n then show that A = 0.

(b) Let T : V W be a linear transformation of vector spaces. Prove the following:

(1) If spans V and T is onto, then spans W.

(2) If is linearly independent in V and T is one-one, then is linearly independent in W.

(3) If is a basis of V, and T is bijective, then is a basis of W.

Sol. (a) P is invertible, D is diagonal

D = PAP^–1

Consider D.D = (PAP^–1) (PAP^–1) = PA(P^–1P) AP^–1 = PA²P^–1

Similarly Dⁿ = PAⁿP^–1

If Aⁿ = 0 then PAⁿP^–1

D is diagonal matrix then Dⁿ is also the diagonal matrix.

We know if Dⁿ = 0 only if D = 0 Dⁿ = 0 D = 0

Now consider D = PAP^–1

P^–1DP = P^–1 PAP^–1P = A

If D = 0 then p^–1DP = 0 A = 0

(b) T : V W is a linear transformation T is one-one i.e.

T is onto i.e. if an element such that

T is bijective i.e. T is one-one and T is onto.

(i) If span V i.e. if then can be written as linear combination of .

Let Given T is onto i.e. s.t . Consider

is a linear combination of vector

spans W.

(ii) If is linearly independent in V.

i.e. c₁ = c₂ = ……= c_k = 0

Given T is one-one i.e.

Consider = T(0) ( T(0) = )

Here, c₁ = c₂ =……= c_k = 0

is linearly independent in W

(iii) Given is basis of V and T is bijective i.e. one-one and onto.

From definition of basis span V and T is onto

span W …(1)

If basis of V then is linearly independent and T is one-one.

is linearly independent …(2)

from (1) and (2)

is basis of w.

27. (a) Let be a basis of vector space Find the matrix of the transformation T with as the basis of both the domain and co-domain of T.

(b) Let W be a three dimensional vector space over and let S:W W be a linear transformation. Further, assume that every non-zero vector of W is an eigen-vector of S. Prove that there exists an such that S = where I:W W is the identity transformation.

Sol.

matrix of transformation T with as basis

(b) Let is non-zero vector of W correspond to eigen value

28. (a) Show that the function f : , defined by f(x) = x² for x is not uniformly continuous.

(b) For each n let f_n : be a uniformly continuous function. If the sequence {f_n} converges uniformly on to a function f : , then show that f is uniformly continuous.