CSIR NET Mathematical Science Mock Test - 8 - UGC NET MCQ

Let the cost price be Rsx 

S.P. of

New C.P. = 80% of

New gain 40%,

New S.P. of

Therefore, x = 250 

Cost price of the article = Rs250

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.

Let the wealth be = 100 x
wealth left for wife = 40x
wealth left for orphans = 100x - 40x wealth left for orphans = 60x

required = 20%

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-

6 + 6² = 42

42 + 11² = 163

163 + 16² = 419

419 + 21² = 860

Expansion of the product
(a₁ + a₂ + a₃) (b₁ + b₂ + b₃ + b₄) (c₁ + c₂ + c₃ + c₄ + c₅)
No. of terms = 3, 4, 5
∴ Product of terms = 3 × 4 × 5 = 60

Given:

Put

and

becomes,

Integrating both sides, we get:

Power set P ({0, 1, 2}) is the set of all subsets of {0, 1, 2}. Therefore, P({0, 1, 2}) = {null, {0}, {1}, {2}, {0, 1}, {0,2}, {1, 2}, {0, 1, 2}}.

∴ The series is convergent.

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

The level of significance

Option C is correct answer.

since f is twice differentiable, we have

By convexity with we have

which implies

We know that:

In

We have,

On Integrating both sides, we get:

......(1)

It is given that , i.e. at , putting this in (1),

Putting in (1), we get:

Given is Mdx

On Integrating, we get

⇒ It represent the function of straight lines

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.

A – B = {x : x ∈ A and x ∉ B}

= {x : x ∈ A and x ∈ Bc}

= A ∩ B^c

Since B is closed set, its complement Bc is open set.

∴ A – B is open set, because intersection of

two open sets is open set.

B – A = {x : x ∈ B and x ∉ A}

= {x : x ∈ B and x ∈ A^c}

= B ∩ A^c

Since A is closed, Ac is open and intersection

of two closed sets is closed, i.e., B – A is closed.

• Bounded above by 1 and maximal element is 1
• Bounded below by 0 and no minimal element

Choose ε > 0
By equicontinuity each is contained in an open set such that for all and all Form this it also follows that for all in
By the compactness of k there is a finite collection of these sets which covers k
Choose N so large that for all we have for each corresponding to this finite collection. Then for any there is an such that . Hence
for . Thus converges to f uniformly on k

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

Ans (D)

is a consistent estimator of

This statement is true regarding given statement.

We can see that V has dimension 5 . Looking at the minimal polynomial of T, we find the Jordan canonical form for T must be

which implies (1) is true and (2) is false.

since we have naturally isomorphic to
Thus the induced operator on the quotient space is nilpotent, so (3) is true.
However, if we look at , the induced operator on has as its minimal polynomial which implies (4) is false. We can see this by considering the Jordan form of  (It's the lower right submatrix of the above matrix.) Alternatively, let Then If we let , then 
As we again have that (4) is false.

If F₁ is a continuous function then so is G

Hence option A is correct.

Both B and D

Option C is correct answer.