## IIT JAM MATEMATICS 2020Previous Year Question Paper with Solution.

1. Let sn = 1 + Then the sequence {sn} is

(a) monotonically increasing and is convergent to 1

(b) monotonically decreasing and is convergent to 1

(c) neither monotonically increasing nor monotonically decreasing but is convergent to 1

(d) divergent

Ans. (c)

Sol.

2. Let f(x) = 2x3 – 9x2 + 7. Which of the following is true?

(a) f is one-one in the interval [–1, 1]

(b) f is one-one in the interval [2, 4]

(c) f is NOT one-one in the interval [–4, 0]

(d) f is NOT one-one in the interval [0, 4]

Ans. (d)

Sol. f(x) = 2x3 – 9x2 + 7

For one-one ⇒ either strict increase or strict decrease.

(a) In the interval [–1, 1], f(x) is increase and decrease.

So the function is monotone.

(b) In the interval [2, 4], f(x) is increase and decrease.

So the function not one-one.

(c) In the interval [–4, 0], f(x) is strictly increase.

So function in one-one.

(d) In the interval [0, 4], f(x) is decrease and increase.

So the function is NOT one-one.

3. Which of the following is FALSE?

(a)

(b)

(c)

(d)

Ans. (d)

Sol.

4. Let g : be a twice differentiable function. If f(x, y) = then

(a)

(b)

(c)

(d)

Ans. (d)

Sol. We have

5. If the equation of the tangent plane to the surface z = 16 – x2 – y2 at the point P(1, 3, 6) is ax + by + cz + d = 0 then the value of |d| is

(a) 16

(b) 26

(c) 36

(d) 46

Ans. (b)

Sol. Given, z = 16 – x2 – y2.

Equation of tangent

6. If the directional derivative of the function z = y2e2x at (2, –1) along the unit vector is zero, then equals

(a)

(b)

(c)

(d)

Ans. (c)

Sol. We have z = y2e2x

7. If u = x3 and v = y2 transform the differential equation 3x5dx – y(y2 – x3)dy = 0 to , then is

(a) 4

(b) 2

(c) –2

(d) –4

Ans. (d)

Sol. Given differential equation 3x5 dx – y(y2 – x3)dy = 0.

This equation can be written as

8. Let be the linear transformation given by T(x, y) = (–x, y). Then

(a)

(b)

(c) the range of T2 is a proper subspace of the range of T

(d) the range of T2 is equal to the range of T

Ans. (d)

Sol.

Now this elements can be express as the linear combination of the basis of R2.

its characteristic polynomial is x2– 0x + (–1) = 0

x2 – 1 = 0

This is the characteristic equation for this [T] and every matrix satisfies its characteristic polynomial, so this is

9. The radius of convergence of the power series is

(a) e2

(b)

(c)

(d)

Ans. (d)

Sol.

Hence, radius of convergence of the given series is 1/e2.

10. Consider the following group under matrix multiplication:

H =

Then the centre of the group of isomorphic to

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

11. Let {an} be a sequence of positive real numbers. Suppose that L = . Which of the following is true?

(a)

(b)

(c)

(d)

Ans. (d)

Sol. We know that by theorems on limit

12. Define Which of the following is true?

(a) If then {sn} is monotonically increasing and

(b) If then {sn} is monotonically decreasing and

(c) If then {sn} is monotonically increasing and

(d) If then {sn} is monotonically decreasing and

Ans. (a)

Sol.

Now apply limit on both the sides of given recurrence relation.

13. Suppose that S is the sum of a convergent series Define tn = an + an + 1 + an + 2. Then the series

(a) diverges

(b) converges to 3S – a1 – a2

(c) converges to 3S – a1 – 2a2

(d) converges to 3S – 2a1 – a2

Ans. (d)

Sol. Given, S is the sum of a convergent series.

14. then

(a) does not exist at x = 0 for any value of a

(b) exist at x = 0 for exactly one value of a

(c) exists at x = 0 for exactly two values of a

(d) exists at x = 0 for infinitely many values of a

Ans. (a)

Sol.

This is not continuous at x = 0, so it is not differentiable at x = 0.

Hence, option (a) is correct.

15. Let f(x, y) = .

Which of the following is true at (0, 0)?

(a) f is not continuous

(b) is continuous but is not continuous

(c) f is not differentiable

(d) f is differentiable but both are not continuous

Ans. (d)

Sol. Given f(x, y) =

Now we see that

16. Let S be the surface of the portion of the sphere with centre at the origin and radius 4, above the xy-plane. Let is the unit outward normal to S, then equals

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Equation of the sphere with centre origin and radius 4.

i.e.     x2 + y2 + z2 = (4)2

Surface integral over the (S + D)

17. Let f(x, y, z) = x3 + y3 + z3 – 3xyz. A point at which the gradient of the function f is equal to zero is

(a) (–1, 1, –1)

(b) (–1, –1, –1)

(c) (–1, 1, 1)

(d) (1, –1, 1)

Ans. (b)

Sol. We have

18. The area bounded by the curves x2 + y2 = 2x and x2 + y2 = 4x, and the straight line y = x and y = 0 is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Given the curves

Curve (i) is a circle, so centre and radius of circle (i) are (1, 0) and r1 = 1.

Curve (ii) also a circle, so centre and radius of circle (ii) are (2, 0) and r2 = 2.

19. Let M be a real 6 × 6 matrix. Let 2 and –1 be two eigenvalues of M. If M5 = aI + bM, where a, b R then

(a) a = 10, b = 11

(b) a = –11, b = 10

(c) a = –10, b = 11

(d) a = 10, b = –11

Ans. (a)

Sol. Given, M be a real 6 × 6 matrix = [M]6 × 6 and 2 and –1 be two eigen values of M, i.e. = 2 and = 1.

20. Let M be an n × n (n > 2) non-zero real matrix with M2 = 0 and let . Then

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

and eigenvalue of l is always 1.

i.e. if eigenvalue of M is then eigenvalue of P(M) is corresponding eigen vector should be same.

Here, , so eigenvalue of M is 0 and eigenvalue

(a) Therefore, eigenvalue of and eigenvalue of i.e.

(b)

(c)

(d)

Hence, option (b) is correct.

21. Consider the differential equation L[y] = (y – y2)dx + xdy = 0. The function f(x, y) is said to be an integrating factor of the equation if f(x, y) L[y] = 0 becomes exact.

If f(x, y) = , then

(a) f is an integrating factor and y = 1 – kxy, is NOT its general solution

(b) f is an integrating factor and y = –1 + kxy, is its general solution

(c) f is an integrating factor and y = –1 + kxy, is NOT its general solution

(d) f is NOT an integrating factor and y = 1 + kxy, is its general solution

Ans. (c)

Sol. Given differential equation

Hence, f is an integrating factor and y = 1 + kxy and y = 1 – kxy, is its general solution.

Option (c) is correct.

22. A solution of the differential equation that passes through the point (1, 1) is

(a)

(b)

(c)

(d)

Ans. (a)

Sol. Given differential equation

It equation can be written L.D.E. with constant coefficient equation as

c1 and c2 are arbitrary constant. So put c1 = 0 and c2 = 1 y = 1/x is solution of Eq. (i)

23. Let M be a 4 × 3 real matrix and let {e1, e2, e3} be the standard basis of R3. Which of the following is true?

(a) If rank(M) = 1, then {Me1, Me2} is a linearly independent set

(b) If rank(M) = 2, then {Me1, Me2} is a linearly independent set

(c) If rank(M) = 2, then {Me1, Me3) is a linearly independent set

(d) If rank(M) = 3, then {Me1, Me3} is a linearly independent set

Ans. (d)

Sol. If Rank (M) = 1 then M =

{Me1, Me2}

24. The value of the tripe integral where V is the region given by x2 + y2< 1, 0 < z < 2 is

(a)

(b)

(c)

(d)

Ans. (b)

Sol.

25. Let S be the part of the cone z2 = x2 + y2 between the planes z = 0 and z = 1. Then the value of the surface integral is

(a)

(b)

(c)

(d)

Ans. (b)

Sol. Given, equation of the cone

26. Which of the following is false?

(a)

(b)

(c)

(d)

Ans. (c)

Sol.

27. Let be a non-constant continuous function. Which of the following is true?

(a) The range of f is unbounded

(b) The range of f is a union of open intervals

(c) The range of f is closed interval

(d) The range of f is a union of at least two disjoint closed intervals

Ans. (c)

Sol.

Common region I, II, III, IV cases is

This is closed and bounded.

We know that, any subset of Rn which is closed and bounded, we way this is compact.

Therefore, D is compact.

Let

Since, D is compact is also compact.

Hence, the range of f is a closed interval.

28. Let be a continuous function such that and

|f(x) – f(y) – (x – y)| < sin (|x – y|2)

for all x, y [0, 1]. Then is

(a)

(b)

(c)

(d)

Ans. (a)

Sol.

On divided by |x – y| both sides

29. Let be the circle group under multiplication and i = Then the set is infinite} is

(a) empty

(b) non-empty and finite

(c) countably finite

(d) uncountable

Ans. (d)

Sol.

If we take any irrational number, then generate a infinite group corresponding to it irrational number.

Hence, we have uncountable many irrational in real line.

30. Let F = .

Consider the groups

G =

and H =

under matrix multiplication. Then the number of cosets of H in G is

(a) 1010

(b) 2019

(c) 2020

(d) infinite

Ans. (c)

Sol.

31. Let such that a < b < c. Which of the following is/are true for any continuous function satisfying f(a) = b, f(b) = c and f(c) = a?

(a)

(b)

(c)

(d)

Ans. (a), (c), (d)

Sol.

This function is well define because composition of two function is continuous f is continuous

f2 is continuous x be a polynomial of continuous.

Hence, g(x) is continuous.

32.

(a)

(b)

(c)

(d)

Ans.

Sol.

But be an alternative series which is Cgt by Lebnitz's test.

33. Let and a < b. Which of the following statements(s) is/are true?

(a) There exists a continuous function such that f is one-one

(b) There exists a continuous function such that f is onto

(c) There exists a continuous function such that f is one-one

(d) There exists a continuous function such that f is onto

Ans. (a), (c), (d)

Sol. One-One (Injective)

Function F : A  B is one-one if image of distinct element of A are distinct under f.

For example,

Onto function

Function F : A  B is onto if every element of B is image of some element of A.

For example,

Now for any f(x) = x is a continuous function then option (a), (c) and (d) will be correct options.

34. Let V be a non-zero vector space over a field F. Let be a non-empty set. Consider the following properties of S

I. For any vector space W over F, any map extends to a linear map from V to W.

II. For any vector space W over F and any two linear maps, satisfying f(s) = g(s) for all we have f(v) = g(v) for all v ∈ V.

III. S is linearly independent.

IV. The span of S is V.

(a) (I) implies (IV)

(b) (I) implies (III)

(c) (II) implies (III)

(d) (II) implies (IV)

Ans. (b), (d)

Sol.

35. Let where p, q are real constants. Let y1(x) and y2(x) be two solutions of L[y] = 0, x > 0, that satisfy y1(x0) = 1, for some x0 > 0. Then,

(a) y1(x) is not a constant multiple of y2(x)

(b) y1(x) is a constant multiple of y2(x)

(c) 1, ln x are solutions of L[y] = 0 when p = 1, q = 0

(d) x, ln x are solutions of L[y] = 0 when

Ans. (a), (c)

Sol.

We know that, by Wronskian formula,

Hence, option (a) is correct and option (b) is wrong.

36. Consider the following system of linear equations

x + y + 5z = 3,  x + 2y + mz = 5  and  x + 2y + 4z = k.

The system is consistent if

(a)

(b)

(c)

(d) k = 5

Ans. (a), (d)

Sol. Given linear equation x + y + 5z = 3

37.

Which of the following is/are true?

(a) a > b

(b) a < b

(c)

(d)

Ans. (b), (c)

Sol.

38. Let S be that part of the surface of the paraboloid z = 16 – x2 – y2 which is above the plane z = 0 and D be its projection on the xy-plane. Then the area of S equals

(a)

(b)

(c)

(d)

Ans. (a), (d)

Sol. Given z = 16 – x2 – y2.

Differentiate w.r.t. x and y, we get

39. Let f be a real valued function of a real variable such that where K > 0. Which of the following is/are true?

(a)

(b)

(c)

(d)

Ans. (a), (d)

Sol.

40. Let G be a group with identity e. Let H be an abelian non-trivial proper subgroup of G with the property that If then

(a) K is a proper subgroup of H

(b) H is a proper subgroup of K

(c) K = H

(d) there exists no abelian subgroup L ⊆ G such that K is a proper subgroup of L

Ans. (c), (d)

Sol. Any group G consisting of only the identity element is a subgroup of G and being the trivial group. The group consisting of only the identity element is a subgroup of G being the trivial group.

If H is a subgroup of G and let .

Then H is a normal subgroup of the subgroup K is G, the

41.

Ans. 1

Sol.

42. Let and S be the sphere given by (x – 2)2 + (y – 2)2 + (z – 2)2 = 4. If is the unit outward normal to S then is _________.

Ans. 32

Sol. Given equation of sphere is

It can be written as by guass divergence theorem

43. Let be such that f, are continuous functions with f > 0,

is _________.

Ans. 0

Sol. Given that

Now, we calculate

44.

Then is _________. (rounded off to two decimal places)

Ans. 0.08

Sol. Given that

45. Let f(x, y) = ex sin y, x = t3 + 1 and y = t4 + t. Then at t = 0 is _________. (rounded off to two decimal places)

Ans. 2.72

Sol. Given F(x, y) = ex sin y, x = t3 + 1 and y = t4 + t

Here, f is function of x, y and x, y is function of t. So by chain rule.

The above value put in the Eq. (i), we get

46. Consider the differential equation

+ 10y = f(x), x > 0,

where f(x) is a continuous function such that

Then the value of is _________.

Ans. 0.1

Sol. Given differential equation

47.

Ans. 0.25

Sol.

48. Let f(x, y) = 0 be a solution of the homogeneous differential equation

(2x + 5y)dx – (x + 3y)dy = 0.

If f(x + , y – 3) = 0 is a solution of the differential equation

(2x + 5y – 1)dx + (2 – x – 3y)dy = 0,

then the value of is _________.

Ans. 7

Sol. Given differential equations

Put x = x – a and y = y + 3 in Eq. (ii), we get

On comparing Eqs. (i) and (iii), we get

49. Consider the real vector space P2020 = .

Let W be the subspace given by W = .

Then, the dimension of W is _________.

Ans. 1011

Sol.

50. Let be a non-trivial non-injective group homomorphism. Then, the number of elements in the kernel of is _________.

Ans. 3

Sol. Given

51.

Ans. 0.25

Sol. Given series

It is telecopies series

52. Consider the expansion of the function f(x) = in powers of x, that is valid in |x| < . Then the coefficient of x4 is _________.

Ans. 33

Sol.

We know that,

(1 – x)–1 = 1 + x + x2 + x3 + ...

( 1 + x)–1 = 1 – x + x2 – x3 ...

∴    3[(1 – x)–1(1 + 2x)–1] = 3[(1 + x + x2 + x3 + x4 + ...)

(1 – 2x + 4x2 – 8x3 + 16x4 + ...)]

= 3[16 – 8 + 4 – 2 + 1] coefficient of x4

= 3(11) = 33

53. The minimum value of the function f(x, y) = x2 + xy + y2 – 3x – 6y + 11 is _________.

Ans. 2

Sol. Given, F(x, y) = x2 + xy + y2 – 3x – 6y + 11.

For minimum value, fx = 0 and fy = 0

difference w.r. to 'x' of Eq. (i), we get

by solving Eqs. (ii) and (iii), we get

by solving Eqs. (ii) and (iii), we get

x = 0 and y = 3

x = 0 and y = 3 are extreme points

Now, r = fxx = 2 > 0    r > 0

s = fxy = 1

t = fyy = 2

and rt – s2 = 2(2) – (1)2 = 3 > 0

Hence, r > 0 and rt – s2 > 0

So, minima exist.

Now, put x = 0, y = 3 in Eq. (i), we get minimum value

(0)2 + (0)(3) + (3)2 – 3(0) – 6(3) + 11 = 2

Hence, minimum value is 2.

54. Let x > 0 and g(x) = a0 + a1(x – 1) + a2(x – 1)2 be the sum of the first three terms of the Taylor series of f(x) around x = 1. If g(3) = 3 then is _________.

Ans. 0.33

Sol. Given

This is taylor expansion upto two terms.

Put x = 1 in Eq. (i), we get

These values put in the Eq. (ii), we get

55. Let C be the boundary of the square with vertices (0, 0), (1, 0), (1, 1) and (0, 1) oriented in the counter clockwise sense. Then, the value of the line integral is _________. (rounded off to two decimal places)

Ans. 0.67

Sol.

Using green theorem, we get

Limit x = 0 to 1, y = 0 to 1

56. Let be a differentiable function with for all x. Suppose that are two non-zero solutions of the differential equation

satisfying

Then, the value of p is _________.

Ans. 8

Sol. Given differential equation

Now, from Eq. (i)

(4D2 – pD + 3)y = 0

A.E. is    4m2 – pm + 3 = 0

We consider three cases :

Case I. 6 × 2 = 12, 6 + 2 = 8 = p

Case II. 12 × 1 = 12, 12 + 1 = 13 = p

Case III. 4 × 3 = 12, 4 + 3 = 7 = p

Case I, taken p = 8, i.e. 4m2 – 8m + 3 = 0

Hence, p = 8 satisfying the given equation and condition.

57. If x2 + xy2 = c where is the general solution of the exact differential equation M(x, y)dx + 2xy dy = 0 then M(1, 1) is _________.

Ans. 3

Sol. Given differential equation x2 + xy2 = c.

On differentiate

58. Let M = . Then the value of det(8I – M)3 is _________.

Ans. –216

Sol. M =

M =

59. Let be a linear transformation with Nullity(T) = 2. Then, the minimum possible value of Rank(T2) is _________.

Ans. 3

Sol. We know that

Rank (T) = dim (domain space) – Nullity (T)

Let A is m × n and B is n × p matrix.

60. Suppose that G is a group of order 57 which is not cyclic. If G contains a unique subgroup H of order 19 then for any , o(g) is _________.

Ans. 3

Sol. Given O(G) = 57 which is not cyclic

We know that by Langrange theorem,

because any finite group is cyclic some element whose order is same, is same as the order of group.

(iv) Hence, o(g) = 3.