IIT JAM Mathematics Practice Test- 20 - Mathematics MCQ

Clearly the rule of the function is changing of x = 1 and 2.
So we shall test the differentiability of f(x) only at the points x = 1 and x = 2.
clearly Rf '(1) - Lf'(1)
and Rf*(2) ≠ Lf'(2)
hence f(x) is not differentiable at x = 2.

Given function F(x) = sin x

Given f is differentiable at one point of R So let f is differentiable at (x = a) ∈ R
Then Lf'(a) = Rf'(a)




Thus by (1).
Lf'(b) = Rf ’(b)
Since b be any arbitrary point of R at which f is differentiable
⇒ f is differentiable on R.


Thus the values of a₂,a₇,a₄,a₃ in equation (1) are

Hence by comparison test given series converge absolutely
⇒ option (d) is correct.

as least upper bound < greatest lower bound.

Let R(T) = Range of T
N(T) = null space of T
Let R(T)∩,N(T) = {0}
⇒ R(T) and N(T) both are disjoint subspaces of V.
Now if T(T(x)) = 0, where T(x) ∈ R(T)
⇒ T(x)∈N(T)
⇒ T(x) = 0

Here the eigenvalues of A are : 3 + i , 1 - i, 3i
⇒ The eigenvalues of A² are (3 + i)²,(1 - i)²,(3i)2
⇒ 8 + 6i, - 2i, - 9
Now the eigenvalues of 2A are: 6 + 2i, 2 - 2i, 6i
Thus eigenvalues of A² + 2A + 1 are
15 + 8i, 3 - 4i, - 8 + 6i

 proper closed subset of [0,1] but X not countable

 be a proper closed subset of [0,1] but it cannot contain an open interval.
⇒ None of given options are true.
⇒ (d) is correct.

The cross product  b = [0 1 0]^T and x = [x₁ x₂ x₃]^T is


By matching LHS and RHS, we get


Characteristic equation be,

So the eigen values are 0, i, - i

f(x, y, z) = Sin(x)Sin(y)Sin(z) - Cos(x) Cos(y) Cos(z) 
Since the function has 3 independent variables hence during differentiation we have to consider x and z as constant and differentiate it w.r.t. Y,
f’(x, y, z) = Sin(x)Cos(y)Sin(z) + Cos(x)Sin(y)Cos(z).

 be a homomorphism then in general, there are d
homomorphism from  where d = g.c.d(m,n).
Here m = 21 and n = 41 then g.c.d (21,41) = 1
So, there is only hom om orphism exists from 

By stoke's theorem, we know that 



 

Here (a) be a subspace but basis of it is {(0. 1, 0)} so it be a subspace of dimension 1.
(b), (c) are not subspace, so they don't have dim.
Now (d) be a subspace and its basis is {(0, 0. 1) (0, 1. 0)} so it be a subspace with dimension 2.

and W has real polynomials functions so this kind of function cannot be in W.
⇒ P(x) = 0 is the only element in W.
⇒ W = {0}
⇒ dim W = 0

So the number of such is clearly (n-2)l
[as σ (1) = 1 ,  will take remaining (n-2) letters among themselves and so σ will be a permutation on (n-2) letters]
O[N(12)] = 2(n-2)!
Here n = 5 .then [N(12)] = 2(5-2)!
= 2(3)!
=12

and the field is defined every where within and on the square of side unit length and centered at (x₀,y₀) with 
∴ By stoke's theorem
which is independent of (X₀, y₀)

A vector normal to the surface S is given by

Now we have

Where R is the projection of S on xy-plane.
The region R is bdd by x-axis, y-axis and x² + y² = 1, z = 0
So 

⇒ 
= 3/8

We choose the simplest parametrization we can think of  where 0 < t < 2. The component have continuous 1^st order derivative and  
The line integral of f over C is,



= 6√3

By rolle's theorem for
f(x) = x³ + bx² + cx in [1, 2], there exists 1 < α < 2 such that
f ‘(α) = 0 



⇒  8b + 3c = -16   .....(1)
also , f(1) = f(2) ⇒ 1 + b+ c = 8 + 4b + 2c
⇒ 3b + c = -7   ..... (2)
From (1) and (2)
b = -5 . c = 8

O(G) = 143
i.e. O(G) =11 x 13 (p < q)
Here.O(G) = pq , p < q and p X q -1, so G be a cyclic group of order 143 and every cyclic group is an abelian group.
Hence G is an abelian group, if G be an abelian group then centre of G is equal to group G.
G = Z(G) [∵ G is an abelian]
i.e. O[Z(G)] = 143 so Z(G) is isomorphic to Z₁₄₃.

Then

limit are 



then by under transformation we get

Now Surface integral becomes





and according to option correct answer is