CSIR NET Mathematics Mock Test - 5 - CSIR NET Mathematics MCQ

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Sum of age of leaving person = 48

total increase of of age = 8 x 2 = 16 years

increased total age due to addition of two men = 48 + 16 = 64;

Average age of two new men = 64/2 = 32 years

According to the given condition,

Therefore, person is standing in the South-East direction from his house.

According to the question, Average of
Put




Now check from the option Option: (C)

(satisfied)

Hence (B) is the correct answer.

Let the given complex number be z.

On rationalizing the denominator,

Here, given partial differential equation may be written as:

------(1)

Charpit's auxiliary equations for (1) are

Or,

So choice (2) is false

So choice (3) is false

So choice (4) is false

Reciprocal of

So, C is Answer

[0, 1] is closed set

⇒ Every point [0, 1] is limit point of [0, 1]

⇒ So there are uncountable limit point of [0, 1]

also has a uncountable limit point

The Lagrange auxiliary equation is

By (2) & (3) fraction of (1), we get

By first and IInd fraction of (1), we get



It is bounded as at
So choice (3) is answer

Now, By Leibnitz Rule

Now, By Leibniz Rule,

So choice (2) is answer

And


∴ A is Hermitian.

A and B are idempotent if

A² = A, B² = B

(A + B) is idempotent if

(A + B)² = A + B

Here (A + B)² = A² + AB + BA + B²

= A + AB + BA + B

∵ (A + B) is idempotent, it is possible only

when AB + BA = 0.

Transformation does not alter the rank of matrix.

By Cramer's Rule

.

= 
= 

is a convergent sequence is convergent
is convergent and converges to zero.

is bounded


decreasing sequence
is bounded and monotone sequence

and the radius of convergence

No extremals

Hence option A is correct.

Ce^–t/2

Hence option B is correct.

Trick: dim v = No. of element - No. of L. I. condition

= 100 - 3

= 97

Given:

At ,

Therefore, is continuous at

At ,

Therefore, is continuous at

At ,

Therefore, is continuous at

If X is sequentially compact and U is an open covering of X. since X is compact, X is totally bounded, we can find a finite number of balls of radius which cover  since each ball is contained in a set the collection is a finite sub-collection of U which covers X

Let x ∈ A ⇒ 0 < x ≤ a

y ∈ B ⇒ 0 < y ≤ b

x > 0 and y > 0 ⇒ xy ≤ ab

⇒ ab is an upper bound for C.

⇒ C is bounded above

⇒ By completeness axiom, C has a sup C = c

(say)

c is a sup C ⇒ c ≤ ab …(1)

Now we have to prove c ≥ ab.

Let x ∈ A and y ∈ B

⇒ xy ≤ ab

Choose ε > 0 : xy > ab – ε

Then we have c = sup C ≥ xy > ab – ε

⇒ c > ab – ε

⇒ x + ε > ab

⇒ c > ab …(2)

By (1) and (2) c = ab = sup C.

Both B and C

Hence option D is correct.