CSIR NET Mathematics Mock Test - 2 - CSIR NET Mathematics MCQ

the number is in the format of 221x + 64. where x is the quotient.

dividing by 13 gives 221x/13 + 64/13

whatever the x value 13 completely divides 13 * 17 = 221

so you need to find the remainder only 64/13 = 13 * 4 + 12

so 12 is the remainder here.

A collection of atoms or molecules that can be excited to a higher energy state is called an active medium. Before lasing can occur, the active media is "pumped". The process of raising the atoms in the active media from a lower energy state to a higher state is like pumping water up from a well.

The correct sequence of teaching is:

Fixing goals and content-Decision about strategy- Diagnosis of the learner- Actions and Reactions- Appropriate testing devices- Feedback to teaching.

On December 12, 1901, a radio transmission received by Guglielmo Marconi resulted in the first transmission of a transatlantic wireless signal (Morse Code) from Poldhu, Cornwall, to St. John's, Newfoundland

On September 12, 1958, Jack Kilby demonstrated the first working IC while working for Texas Instruments, although the U.S. patent office awarded the first patent for an integrated circuit to Robert Noyce of Fairchild.

The only daughter of the father of my (Priyankesh) sister’s brother will be Priyankesh's sister only.

The father of Anjali's sister's brother means father of Anjali. Only son of Anjali's father means brother of Anjali

This uses the 'haversine' formula to calculate the great-circle distance between two points that is, the shortest distance over the earth's surface - giving an 'as-the-crow-flies' distance between the points (ignoring any hills they fly over, of course!).

The extreme points of the feasible region ABCD are A(0, 0), B(60,000; 40,000), C(75,000; 25,000) and D(75,000; 18,750).

The value of objective function at extreme points are:

Z(C) = 7500 + 5000 = Rs. 12500

Z(D) = 7500 + 3750 = Rs. 11250

Z(B) = 6000 + 8000 = Rs. 14000

Since the maximum value of Z occurs at extreme point B, the optimal solution the problem is : x₁ = 60,000; x₂ = 40, 000 and max . Z = Rs 14,000

Given word is

MATHEMATICS

Then the number of words that can be formed by permuting the letters of "MATHEMATICS' is

Let

satisfies the PDE

Also

So All condition are satisfied

Now

So choice (4) is answer

Here Lagrange's Auxiliary equation are given by

Taking first two fractions of (1), we get

Taking first and third fraction of (1), we get

Given condition is

using (2) & (3), (4) reduces to

So, (A), (B), (C) is Answer

We know Arg

So D is Answer

By graphically

Let

Given integral Equation is

Apply Laplace transform on both sides, we get

Then



Apply Inverse laplace Transform on both sides we 

choice (4) is answer

A and B are idempotent

is Symmetric if

∴ adj A is Symmetric.

Interchanging of two rows (or columns) changes the sign of determinant.

The partial
{s_n} is increasing sequence and unbounded from above
∴ It is divergent sequence
The series is divergent series.

I. A function f : z → z, defined by f(x) = x + 1, is one-one as well as onto.

Calculate

Calculate

Now,

So, is one-one function.

Consider

is onto.

II. A function , define by , is one-one but not onto.

,

Calculate

Calculate

Now,

So, is one-one function.

Clearly, for all

So, does not assume values

is not an onto function.

So, Both I and II are correct.

Option D is correct answer.

We know that

By the iterated kernels, we have

or

or

or

or

Hence the resolvent kernel is determined as

or

Substituting the function y(x) = 1 – x in the given equation, we have

Thus is a solution of the integral equation.

The eigen values of A are 1,1,2,2

The characteristic polynomial of A is

The minimal polynomial of A is of the form

Now A. M. of 1 is 2

And A.M. of 2 is 2

So, By Rank-Nullity Theorem

Nullity

Now partition of AM. of 1 i. e. 2 is

G. M. of 1 is 1

Trick G. Mecide the partition of A.M.

The highest no. of partition of 2 is power of in minimal polynomial

So is factor of minimal polynomial

Now A.M. of 2 is 2

Then By Rank-Nullity Theorem

Nullity

So A. M. of 2 is 2

Now partition of A.M. of 2 i.e. 2 is and G.M. of 1 is 1.

Trick

The highest no. of partition of 2 is power of in minimal polynomial So is factor of minimal polynomial

So minimal polynomial of matrix is

Hermitian matrix: A square matrix is said to be Hermitian if for all .

• The necessary and sufficient condition for a matrix to be a Hermitian is that
• The diagonal element of a Hermitian matrix is purely real.

Example:

is a hermitian matrix.

The eigenvalue of a real symmetric (or Hermitian) matrix is always real and the eigenvalues of a real skew-symmetric (or skew Hermitian) matrix are either zero or purely imaginary.

Given

Then power set of A is

So choice (3) is answer

(A) A set of integer is countable

(B) A set of rational number is countable

is differentiable at x = 0