CSIR NET Mathematics Mock Test - 8 - CSIR NET Mathematics MCQ

The steps which are required to design a questionnaire includes the aim of the study i.e. writing primary and secondary aims of the study, to prepare a draft of questionnaire in which number of questions will be asked related to the topic, review the literature so as to frame the relevant questions and revision of the draft for making it error free.

The best example of changing into electric energy from chemical energy is primary cells or batteries, the dry cell is also made up in this phenomenon.

Increase in manufacturing cost = 10% of 40% of total

= 4% of total

and tax = 0.1% of total

4.1% of total = Rs. 82

Total = Rs 2000

So, profit = 50% = Rs. 1,000

A knot is one nautical mile per hour (1 knot = 1.15 miles per hour ). The term knot dates from the 17th century, when sailors measured the speed of their ship by using a device called a "common log."

Pattern is-

8 × 3 + 3 = 27

27 × 5 + 6 = 141

141 × 7 + 9 = 996

996 × 9 + 12 = 8976

Pattern is-

6 + 6² = 42

42 + 11² = 163

163 + 16² = 419

419 + 21² = 860

Father of my son’s father -> Radhika’s Father in law

Radhika 's Father-in-law brother -> Radhika 's Uncle

Let the cost price = 100

Marked Price = Rs 125

To gain 10 %, Selling Price = Rs 110

Required Discount Percentage = (125 − 110)/125 × 100 = 12%

Given:

Put

and

becomes,

Integrating both sides, we get:

Last digit is 8

Clarification:

Note how we used the fact that 6 multiplied any number of times with itself would always produce last digit of 6. That is {6 (mod

10) }ⁿ = 6 (mod 10)

Since it is a decimal system, therefore mod 10 has been chosen

For example, take any number like 2018. Last digit of this number is the remainder when this number is divided by 10.

In general this is a very useful technique for solving missing digit problems

If and then

If m > n

Then no. of on-to-functions are

Now, here m = 4, n = 3

Then no, of onto functions are

(Chinese Remainder Theorem): - Let n be positive integer such that ged for . Then the system of linear congruences

Has a simultaneous solution, which is unique modules the integer
Given is and is b (r od 37 ) and
⇒ The given congruence has unique solution (By above result)
∴ For any

diag (1,……1) = I and I² = I

∵ Transformation does not alter the rank of matrix

The sequence {–n} is bounded above since there exist a real number –1 : –1 ≥ –n, ∀n ∈N

Here f(x) = x²m is an even function

f'(x) = 2m x^2m–1 is an odd function

Here f(x) = sin nx is an odd function

f'(x) = cos nx, which is an even function

since f is twice differentiable, we have

By convexity with we have

which implies

Given integral equation

and since is one of the solution

Thus (i) is identically satisfied by Hence is a required solution.

Given is Mdx

On Integrating, we get

⇒ It represent the function of straight lines

Let

Given, U is a harmonic function

Treating z and as independent variable and

is harmonic function. Then

Here, and

1. Relation is not reflexive since,

2. Since, so, belong to

But, and so, is not symmetric

3. Since, , so belong to

Also, , so belong to

But, so is not transitive

So, R satisfies none of the reflexivity, symmetry and transitivity.

Here,

Thusis a proper integral if , and improper if being the only point of infinite discontinuity of the integrand in this case.

Take


Converges if and only if


Therefore, the integralconverges if and only if , which also includes the case when the integral is proper.

Given in and we can find an open set containing such that for all since D is dense, there must be a point and  converges, so it must be a Cauchy sequence, so for we have

Thus, for all

Thus is a Cauchy sequence and converges by the completeness of Y
Let . To prove is continuous at
let be given. By equicontinuity there is an open set containing such that for all and all in
Hence for all in we haveand f is continuous at x

Both B and D

Option C is correct answer.