CSIR NET Mathematical Science Mock Test - 6 - UGC NET MCQ

The patient who is being operated by Mrs. Jain is the elder brother of the son of paternal grandfather of Mrs. Jain, the doctor. So, the patient is the paternal uncle of Mrs. Jain.

Strong nuclear force is the strongest force. It is present inside the atom responsible for binding the protons and neutrons and also inside the proton and neutron in binding up the quarks.

The series is pqp/pqp/pqp/pqp. Thus, the pattern 'pqp' is repeated.

Rise of mercury indicates increase in atmospheric pressure. As air descends, it warms and contracts, which reduces or prevents the formation of clouds. Because of this effect, areas of high pressure often create clear, dry weather.

16.66% = 1/6
let Actual length be 6m Madhvi wears share = 6 x 5/6 = 5m Actual length 6m is greater than the 5m by

Sum of the score of 24 students = 24 x 54

Actual Sum of the scores of 24 students = 24 x 54 + 88 - 64 = 1320

Actual Average = 1320/24 = 55

Total age of 750 boys
Total age of 700 boys
Average age of 50 boys

Sum of ages of 7 members of the family = 37 × 7 = 259

Sum of ages of 6 members of the family = 31 × 6 = 186

Hence, age of the head of family = 259 – 186 = 73 years

(a) Expected time of customer in queue is

hour.

(b) Probability (at least two customers in the system

(c) Probability (the server is idle)

We know that,

1. Combination: If we have to select r objects out of n objects where the order of the selection is not important then it can be done in ⁿC_r ways.

Mathematically it is given by:

2. Addition and Multiplication principal:

If there are m ways to choose an object one and n ways to choose an object two then the number of ways of selecting objects one and two is given by m x n.

If there are m ways to choose an object- say object one and n ways to choose an object- say object two, then the number of ways of selecting objects one or two is given by m + n.

Case-I: We can select one boy and one girl from the given children.

The number of ways of doing so is:

Case-II: We can also select both boys.

The number of ways of doing so is: ³C₂ = 3

Therefore, the number of ways of selecting either one girl and one boy or both boys are given by:

6 + 3 = 9

Therefore, the number of ways of selecting two children such that there is at least one boy is 9.

So unique solution

We know



So choice ( 2 ) is answer

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

The inverse of a matrix is unique

Given the integral equation

where

Substituting the value of from

(i) in C, we get

or i.e.,

Hence from (i),

which shows that the given integral equation has only trivial solution. Thus it does not contain any eigen values or eigen functions.

Given differential equation is
Put

On integrating, we get



Integrating both sides, we get

Applying log on both sides, we get

Let sequence 〈 x_n 〉 is a Cauchy sequence, for

we have

The sequence is bounded.

Let A = {a1, a2,…} is a countable set

{a₁}, {a₂},… is a collection of countable subsets of A.

∴ m*(∪a_n ) ≤ m* a _n = 0 (Lebesgue outer measure of single element set is equal to zero),

⇒ m*A = 0.

Since L is non-empty, it follows from (b) that the constant functions belong to L.
Given let For  we have Choose a positive real number . since is continuous, the set  is closed. since is compact, is bounded on say by have a function such that 
since (g f ∈ L

For


∴ Finite sub-collection does not covers (0, 1)

Choose ε > 0.

By equicontinuity each x ∈ K is contained in an open set Ox such that σ[fn(x), fn(y)] < ε/3 for all y ∈ Ox and all n. Form this it also follows that σ[f(x), f(y)] ≤ ε/3 for all y in Ox. By the compactness of K there is a finite collection {Oxi,…,Oxn} of these sets which covers K.

Choose N so large that for all n ≥ N we have σ[fn(xi), f(xi)] < ε/3 for each xi corresponding to this finite collection. Then for any y ∈ K there is an i ≤ k such that y ∈ Oxi. Hence σ[fn(y), f(y)] ≤ σ[fn(y), fn(xi)] + σ[fn(xi), f(xi) ] + σ[f(y), f(xi)] < ε for n ≥ N. Thus 〈 fn 〉 converges to f uniformly on K

Take A = −I= −I. Then A³ = −1 but A does not have distinct eigenvalues.
The minimal polynomial of a matrix A may be defined as the polynomial of smallest degree that is satisfied by A and has highest coefficient equal to 1.

Ans (D) n = 3000 will ensure | p – θ | ≤ 0.02 with probability at least 0.95.

3

Hence option D is correct.

Both A and C

Hence option D is correct.