CSIR NET Mathematical Science Mock Test - 10 - UGC NET MCQ

The median of six numbers:

Formative evaluation helps students in the following ways:

• It provides constant feedback to both teacher and student concerning learning successes and failure while instruction is in process.
• It often happens during the course of instruction. Teacher clarifies the doubts of students in the class itself.
• Feedback to candidates reinforces successful learning and locates the particular learning errors that need correction.
• The teacher raises learner’s motivation through a question-answer session.

The Hungarian brothers, Laszlo and George Biro, made the first ball point pen in 1894. It followed the first workable fountain pen which was invented by L.E. Waterman in 1884.

Indian rivers can be divided into West flowing (Arabian Sea) rivers and East flowing (Bay of Bengal). Chenab river does nor flow into the Arabian Sea.

Stronger intermolecular force in solid Explanation: In solids, the particles are arranged in a regular pattern, touching each other. They attract each other with a strong force (because they are so small and so close). This means that they cannot change places. So solids cannot change its shape.

• North Circar: Coastal plain from Odisha to Andhra Pradesh
• Malabar Coast: Coastal plain from Mangalore to Kanyakumari
• Coromandal Coast: Coastal plain from Andhra Pradesh to Tamil Nadu
• Konkan Coast: Coastal plain from Gujarat to Goa

P + Q → P is the father of is brother of R R - S → R is mother of S

Q is maternal uncle of S

Two liquids of densities 3 kg upon metre cube and 4 kg per metre cube.
Let mass of liquid 1 be "m" and mass of liquid 2 be "m"
Formula: Density
Volume of liquid 1,
Volume of liquid 2,
After mixing two liquid
Volume of mixture is
Mass of mixture would be
Let density of mixture be
Thus,

The problem can be formulated as follows:

Minimize (total cost)

Subject to the constraints

since each of them happened to be of 'greater han or equal to type', the points satisfying hem all will lie in the region that falls towards the ight of each of these straight lines.

The solution space is the intersection of all hese regions in the first quadrant. This is shown shaded in the following figure.

The solution spaces open with and is extreme points.

Value of objective function at the extreme points can be obtained as follows:

Optimal value of the objective function. Since the minimum cost of Rs. 160 is found at point Q(4, 2), i.e., x₁ = 4 and x₂ = 2, the optimal solution for the firm is to purchase 4 units of product A and 2 units of product B in order to maintain a minimum cost of Rs. 160.

Also y = 0 satisfy the condition
So two solutions

And

Here

∴ A is Skew-Hermitian.

By Cramer's Rule

If A² = I, then A is diagonalisable over real numbers

Hence option B is correct.

Let,

A Volterra's integral equation of second kind.

Given a

= 

and are L. I.

and has no common zeros and and has no common point of extrema

(A) and (B) are not true

Now, If or has repeated root

are

If are Neither nor has repeated root

Now union of two open set is open

In option (3)

(2,3) and (4,5) are open set

⇒ (2, 3) ∪ [4, 5] are open set

Also (2, 3) ∪ [4, 5] are not interval

So choice (3) is answer

Here
is converges for

Trick [No. of element - no. of L. I condition = dimension
Now
. No. of element of
No of L.I condition of
so dim

Given:

We have to find the value of if

Consider:

After comparing, we get,

Adding equation (1) and (2),

If x_n = 1/n
then

(Bounded below)

(Bounded above)

(Bounded below and above)

Since D is a closed set, it contains all its points of closure, and , a complement of closed set is open, therefore, is a open set.

Let with open. Let
be a metric for which is bounded by 1 equivalent on to and for which is complete set

Then is a metric on which is equivalent to
Let be a Cauchy sequence in Then, since must be a Cauchy sequence for Hence converges in to a point since and are equivalent on in since limits of sequences in metric spaces are unique, we have for all
Let this common value be Then in
X. since The equi-
valence of and implies that converges to
in and so is complete.

According to given question

Option (D) i.e There exist an unbounded open subset B of C and an entire function f such that ∂(f (B)) ⊆ f (∂B) is true .

Both B and C

Hence option D is correct.