CSIR NET Mathematics Mock Test - 9 - CSIR NET Mathematics MCQ

Let the number of girls be x.

∴ Number of boys = 2x 

Total number of students = x + 2x = 3x 

135 = 3x

Number of girls = 45

Number of boys = 45 x 2 = 90 

Number of girls who failed in the examination = 45 x 1/3 = 15 

Number of boys who failed in the examination = 90 x 1/6 = 15 

Total number of students who failed in the examination = 15 + 15 = 30 

Percentage of students who failed in the examination = 30/135 x 100 = 22.22%

∴ Percentage of students who passed the examination = (100 - 22.22)% = 77.8% 

So named because it resists (or inhibits) the flow of current.

The IEEE (Institute of Electrical and Electronics Engineers) was formed in 1963 by the merger of the Institute of Radio Engineers (IRE, founded 1912) and the American Institute of Electrical Engineers (AIEE, founded 1884).

Nithya is Sam’s Sister and Mogan is Sam’s Father → Nithya is Mogan’s Daughter.

Selvan is Rajan’s Son and Rajan is Mogan’s Brother → Selvan is Mogan’s Nephew.

So, Nithya is Selvan’s Cousin.

Lara > Smith > Waqar > Warner

The white pigment in old painting turns black due to formation of PbS. This white pigment is restored by using hydrogen peroxide.

As Anjali is the sister of Vishal, Vishal is the son of Manoj. Manoj and Rohit are brothers because both are the sons of Rakesh. It means Rohit is the paternal uncle of Vishal and Anjali.

Freezing point of alcohol is −115⁰C while freezing point of mercury is −39⁰C Hence, to measure below −39⁰C alcohol thermometer is used.

The four main lakes of nainital are : Nainital lake, sattal lake, bhimtal lake and Naukuchiyatal lake.

The Charpit's equations are

First and second gives ,

and

Given

Compare it with

We get

in the second and fourth quadrants

Given

By given condition

If are intersect then

Which is not possible

So are never intersect

Consider the polynomial

If is odd, and has a root, then there must exist a real number with the property .

To prove this, we assume that the set of contains only integers and If not, we can easily multiply the values to make the assumption true. This means that is continuous over the set of real numbers.

Also, the limit of as goes to either infinty or minus infinity.

This means and have the same sign at extreme values of but since is odd, must have opposite signs for and

So there must be real values of with the property and

The Intermediate Value Theorem states that there must be a real number on the interval with the property .

Where f and g are twice differentiable functions

(bounded below)
(bounded above)
is bounded sequence

7/9

Option C is correct answer.

X + Y and X – Y are uncorrelated for all values of ρ

Option D is correct answer.

{(1, 0, 1), (0, 1, 0)}

Option A is correct answer.

The integral equation 

is a Volterra integral equation of second kind.

The νth order approximation is given by


In general, we have

Hence, the solution of the integral equation is given by

Given differential equation 

Integrating both sides of the given differential equation, we get

The required integral equation

If A is a set containing ‘n’ elements, then number of subset of B=2ⁿ

Then, number of subset of a set of 6 elements is 2⁶

So choice (4) is answer

Here
(i)
(ii)
(iii)

(a) has jump at
(b) has discontinuity of first kind.
(c) Measure of discontinuity is 2

A. Sup R = + ∝

B. Inf R = – ∝

C. + ∝, – ∝ ∈ R

Since f is a measurable real valued function, then for real number α the {x : f(x) < ∞} is measurable.

{x : f(x) + c < α} = {x : f(x) < α – c} so f + c is measurable.

{x : cf(x) < α} = {x : f(x) < α/c }so cf is measurable.

Since f is a real valued function such that for each let,

Also let,

So we have forms a decreasing sequence of measurable sets.
For some and we have

Hence given so that, i.e.

The first term of the sequence

Second term of the sequence

Third term of the sequence

Fourth term of the sequence

The th term of the sequence

Ans (B) States 1, 2, 3, 4 are recurrent and state 5 is transient

The singularities are z = 2 and z = 2nπ, n = ±1, ±2, . . . .

The singularities z = 2nπ, n = ±1, ±2, . . . , are simple poles since they are simple zerosof z²(1- cosz.)

All of the options