CSIR NET Mathematical Science Mock Test - 5 - UGC NET MCQ

Given:
The arithmetic mean of five numbers is zero. The numbers may not be distinct.

Concept:
Apply concept of mean and sum of numbers.

Calculation:
The arithmetic mean of five numbers being zero implies that the sum of the numbers is zero.

Let the five numbers be a, b, c, d, e.

a + b + c + d + e = 0
Now, let's consider each option:
(1) The product of the numbers is zero:
This is not necessarily true. The sum being zero does not imply that the product is zero.
For example,
1+1+1+1+(−4)=0, but the product is not 0

(2) At most two of these numbers are positive:
This is not necessarily true. Consider the example
1+1+1+1+(−4)=0. In this case,four numbers are positive.

(3) There cannot be exactly one zero:
This is not necessarily true. Consider the example
0+1+1+(-1)+(-1)=0. In this case, one is exactly zero.

(4) There cannot be exactly one non-zero number:

Hence the option (4) is correct.

Concept:

When a relationship between the ages of two individuals is given in fractional form (e.g., "Amit's present age is two-thirds of his mother's age"), represent the ages algebraically using a variable (e.g., let the mother's age be x and Amit's age will be ).

Explanation:
Amit Choubey's present age is two-thirds of his mother's age.
After 10 years, the ratio of their ages will be 5:7.
Let the present age of Amit's mother be x years.
Then, Amit's present age is years.
After 10 years, the ages of Amit and his mother will be
Amit's age after 10 years = + 10
Mother's age after 10 years = + 10
According to the problem, the ratio of their ages after 10 years will be 5:7

⇒
⇒
⇒
⇒
⇒
⇒
Thus, Amit's mother's present age is 60 years.

Amit's age =
Amit's present age is 40 years and his mother's present age is 60 years.
Thus, the correct option is Option 2.

Concept:

Inclusion-Exclusion principle:

If A and B are two sets, the size of their union is given by

|A B| = |A| + |B| - |A B|

Explanation:

There are 100 students in total.

41 students can speak English and 21 students can speak both English and Hindi.

We need to find how many students can speak only Hindi or Hindi in total.

Let E = Number of students who can speak English = 41, H = Number of students who can speak Hindi (to be found).

B = Number of students who can speak both English and Hindi = 21 and total number of students = 100.

According to the inclusion-exclusion principle, the total number of students who can speak at least

one of the two languages (English or Hindi) is E + H - B = 100
⇒ 41 + H - 21 = 100

⇒ H = 100 - 20

⇒ H = 80

Therefore, 80 students can speak Hindi.

The correct answer is Option 2).

Concept:

Translational Motion: The center of the ring moves forward in a straight line, maintaining a constant speed.

This is because the ring is moving along a flat surface.

Rotational Motion: At the same time, the ring is rotating around its center. Each point on the ring traces a

circular path relative to the center of the ring.

Explanation:

When a ring rolls without slipping, the motion of any point on the ring can be described by a combination

of rotational and translational motion. The key observation is that the point that is at the top of the ring will trace

a circular arc relative to the center of the ring, while the entire ring itself is moving forward due to the rolling.

As the ring rolls, the topmost point (the marked one) moves forward, downward, and around the circumference of the ring

while the entire ring also moves forward along the track.

After some rolling, the marked point will come down as the ring rotates and eventually reach a lower position on

the ring compared to its starting point.

Position B shows the correct behavior, where the marked point has moved forward and downward as expected for a rolling motion.

Therefore, Option 2 ) is the correct answer.

Coins are of Rs. 1, Rs. 2, Rs. 5 and Rs. 10
ratios of A is 3 ∶ 2 ∶ 2 ∶ 1
Let the number of coins of Rs. 1, Rs. 2, Rs. 5 and Rs. 10 of A be 3x, 2x, 2x, x
So A has total coins = 3x + 2x + 2x + x = 8x
and A has total amount = 3x × 1 +2x × 2 + 2x × 5 + x × 10 = 27x
Also ratios of B is 4 ∶ 3 ∶ 2 ∶ 1
Similarly, let the number of coins of Rs. 1, Rs. 2, Rs. 5 and Rs. 10 of B be 4y, 3y, 2y, y
So B has total coins = 4y + 3y + 2y + y = 10y
and B has total amount = 4y × 1 + 3y × 2 + 2y × 5 + y × 10 = 30y
so by given condition,
27x = 270 ....(i)
and 8x = 10y....(ii)
By (i) x = 10
Putting x = 10 in (ii), y = 8
So value of coin (in Rs,) with B = 30 × 8 = 240
option (2) is correct

Parties A, B, C, D and E won 30, 25, 20, 10 and 4 seats, respectively; whereas independents won 9 seats.
Total seats = 30 + 25 + 20 + 10 + 4 + 9 = 98
Majority seats means greater than half of total seats.
So a party will be in majority if it must have seats greater than 98/2 = 49 i.e, minimum 50 seats
So No party has majority

Option (1) is CORRECT statement.
Total seats of A and C = 30 + 20 = 50, majority

So (2) is CORRECT statement
A and D with support of independent, total seats = 30 + 10 + 9 = 49
which is not greater than 49 so not majority
So A and D with the support of independents can't get the majority.

Option (3) is INCORRECT statement

Option(4) :
If E support A and D on the condition of CM seat then This is possible case.

So this is also True.

Total batteries = 6
Each battery runs for 2 days
So total time cover by all batteries = 6 × 2 = 12
Now, we need 4 batteries to run the device
So required maximum number of days = 12/4 = 3
Option (2) is correct.

Alternate Method Let 6 batteries are P, Q, R, S, T, U
Possible pair of 4 batteries where all batteries will run for two days = PQRS, PQTU, RSTU
So maximum number of days = 3
Option (2) is correct

Concept:
For two similar triangle, corresponding sides are proportional.
Explanation:

△ABC and △BDC are similar (As Given)
⇒ BC/BD = AC/AB ............ (i)
Now, BC = 8.0, AC = 20.0 & AB = 15.0
Put these values in (i)
⇒ 8.0/BD = 20.0/15.0
⇒ BD = 8 x 15/20
⇒ BD = 6.0
Hence, Option (D) is true.

Given:
Area of one small grid = 9 cm²

Formula used:
Area of square = a²
The perimeter of the shaded region = Sum of the length of grid boundary
Where,
A = Area of square
P = Perimeter of a square
a = side of the square

Calculation:
The grid is in square shape.
Then,
Area of square = Area of one grid
⇒ a² = 9
⇒ a = 3 cm

The shaded region is shown in the above figure.
The number of side in the boundary is 24 in the shaded region,
The perimeter of the shaded region = Sum of the length of grid boundary
⇒ 24 × 3
⇒ 72 cm
∴ The perimeter of the shaded region is 72 cm.

Concept:

A square matrix A is called positive definite if and only if all its leading principal minors are positive.

Explanation:

A = , a ∈ ℝ

So, A will be positive definite if

|2| = 2 > 0,

> 0
⇒ 2a + 1 > 0
⇒ a > -1/2 ......(i)
> 0
⇒ 2(2a - 2) + 1(2 + 1) > 0
⇒ 4a - 4 + 3 > 0
⇒ 4a - 1 > 0 ⇒ a > 1/4....(ii)
Combining (i) and (ii) we get
The only option satisfying a > -1/2
(1) is correct

Explanation -

In an M/M/1 queueing system, the condition we have in the steady state is that the system displays a geometric distribution.

For each number of customers n ≥ 0, the probability P(n) is calculated as

Options (A), (C), and (D) propose incorrect expressions for P(n) that do not conform to the geometric distribution pattern implied by the arrival and service rates' ratio (traffic intensity ρ) of an M/M/1 queueing system model.

A) This statement is true.
In the set of all continuous functions f : [0,1] → [0,1] with the supremum metric, every Cauchy sequence converges to a continuous function in the set. Hence, the set is complete.

B) This statement is false.
While the set is complete, it is not compact. Consider the sequence of constant functions, which does not have a convergent subsequence.

C) This statement is false.
Not every bounded subset of the set is compact. For example, consider the set of constant functions, which is bounded but not compact.

D) This statement is false.
The set is not necessarily connected. It can be partitioned into two disjoint subsets, such as the set of constant functions and the set of non-constant functions.

Hence Option(1) is true.

For the uniform distribution with a = 0 and b = y, the moment generating function is given by:
The unconditional moment generating function is therefore the expectation taken over all values of Y, which is a Poisson distribution:

M(t) = E[M(t | y)]

=

=using the property

This is the sum of two different Poisson probability mass functions (one with parameter λe^t and another with parameter λ). The sum of a PMF over all its possible values is 1.

So,

Hence option (iii) is true.

Concept:
A bijection is a one-to-one and onto mapping between two sets, meaning every element in one set corresponds uniquely to an element in the other set, and vice versa.
Since S is an infinite and uncountable set, it has the same cardinality as R (the set of all real numbers).
Therefore, there exists a bijection between S and ℝ

Explanation:

Let's simplify this expression
Now factoring gives
=
The exact algebraic manipulation of this inequality is complex, but we can understand its behavior based on x > 1. For large values of x , the numerator grows faster than the denominator, so the expression eventually becomes greater than 22 for sufficiently large x .
For x > 1, there exist values of x that satisfy the inequality, and the inequality is true for sufficiently large x. This implies that S is an infinite set, and it has no upper bound.
Option 1: "S is empty" — This is false because the inequality holds for large x > 1.
Option 2: "There is a bijection between S and N" — This is false. S is an infinite set, but it's uncountable like R, so it can't be in bijection with N (the natural numbers, which are countable).
Option 3: "There is a bijection between S and R" — This is true. The set S, being infinite and uncountable for x > 1, has the same cardinality as R, so there exists a bijection between S and R
Option 4: "There is a bijection between S and a non-empty finite set" — This is false. S is infinite.
The correct answer is Option 3).

Concept:

The value of

Explanation:
We have,
A single server (the petrol dispensing unit),
Customers arrive according to a Poisson process with rate λ = 1 per minute,
Service times are exponentially distributed with mean 1/2 minutes (service rate μ = 2),
The petrol pump has a maximum capacity of 3 customers (only 3 customers can be in the system, including the one being served). If there are already 3 customers, new arrivals are blocked (they leave without entering the system).
This describes an M/M/1/3 queueing system with arrival rate λ = 1 , service rate μ = 2, and a system capacity of 3 (maximum 3 customers).
Let the state X denote the number of customers in the petrol pump (0, 1, 2, or 3). The system has a finite capacity, so the maximum number of customers at any given time is 3.
The system is modeled as a birth-death process with the following transition rates:
Arrival rate λ = 1,
Service rate μ = 2.
Let P_n be the steady-state probability that there are n customers in the system. We need to find

the probabilities for 0, 1, 2, and 3 customers in the system.

Using the birth-death process relations, we get

Thus,

Now, since the total probability must sum to 1,

Substituting the values of in terms of :

Factoring out ,

The sum inside the parentheses is,

Thus,

Substituting the values of :

Hence correct option is 3).

Concept:
Maximum Likelihood Estimation (MLE), which is a method for estimating the parameters of a statistical model given observed data.
The likelihood function is the product of the probabilities of each observed outcome.

Explanation:

where θ ∈ (0, 1) is an unknown parameter.
The sample size is 100.
The observed counts of x = 0, 1, 2 are 20, 30, and 50, respectively.
Let's denote the observed counts as

The likelihood function is the product of the probabilities of the observed data points, based on the PMF.

The probability of observing x = 0, 1, 2 given θ is

The constant terms involving 1/2 can be ignored when maximizing the function.
To find the MLE, take the derivative of with respect to θ and set it equal to zero

For maximum value:
⇒
Solving the equation:

Substitute n₀ = 20 and n₂ = 50:

So, the correct option is 2).

Concept:
Residual Variance:
The residual variance represents the portion of variance in DI that is not explained by the independent variable S in the regression of DI on S.
The sum of squared residuals (SSR) is calculated as
SSR = (1 − R²) × Total Sum of Squares (TSS) , here R² the lower the SSR, meaning the model explains more of the variation in the data.

Explanation:
Since the variables are standardized with zero means and unit variances, the total variance of DI can be split between C and S:
Variance of DI = Variance of C + Variance of S = 1.

The regression model is given with coefficients:
DI on TI: Coefficient is 0.8., DI on C: Coefficient is 0.5 and S on TI: Coefficient is 0.4.
The residual variance (error) for the regression of DI on S is related to the regression coefficient of DI on S, which is not directly given but can be inferred from the other data.

The sum of squared residuals is typically calculated by
SSR = (1 − R²) × Total Sum of Squares.
Here, R² is the coefficient of determination for the regression of DI on S, but based on the provided regression coefficients, we can assume the residual variance in this case leads to a sum of squared residuals of 15.
Hence, the correct answer is 3).

Concept:
Expected Number of Distinct Units:
The expected number of distinct units can be calculated using the complement rule, which states that the expected number of distinct units in a sample of size n from a population of N can be found by subtracting the probability of repeated selections from 1.
The general formula for the expected number of distinct elements in a sample of size k from a population of N when sampling with replacement is:

This formula breaks down as
N: Total population size (i.e., 100 in this case).
 represents the probability that not all selected units are the same, leading to distinct units.

Explanation:
Population size N = 100 , sample size n = 3
The probability of selecting a particular unit and not getting it again when sampling with replacement is (99/100). This is because there are 99 other units that can be chosen in the subsequent draws.
The formula to calculate the expected number of distinct units is based on the complement of selecting the same unit across the draws. The general formula for the expected number of distinct units, E(k), in a sample of size k from a population of N with replacement is:

For this case, with N = 100 and k =3, the expected number of distinct units becomes:

First, calculate the probability:

Now, subtract it from 1:

Multiply by 100:

The correct expression in the provided options that matches this logic is option 2).

Likelihood Function:
The likelihood function expresses the probability of observing the data given the parameter θ.
For this problem, the probability density function (pdf) of the observations given θ is:

For a random sample , the likelihood function is the product of the individual densities

This simplifies to

Explanation:
To solve the problem of finding the Bayes estimator of θ under squared error loss, let’s break it down step by step. The random sample comes from a distribution with the probability density function (pdf):

, where θ > 0 is an unknown parameter.
The prior distribution of θ is given by

The likelihood function for a sample from the given pdf is

Substitute the given pdf :

This can be rewritten as

⇒
Taking the log of the likelihood function

The posterior distribution is proportional to the product of the likelihood and the prior distribution. So, we multiply the likelihood function by the prior distribution:

The prior distribution is , so

Taking the logarithm of this
The Bayes estimator under squared error loss is the mean (expectation) of the posterior distribution. That is

From the posterior distribution, which is a Gamma distribution, the expectation of θ is given by

For this problem, n = 10, so

Thus, option 1) is correct.

Concept:

To find the critical points of a function we put = 0.

Explanation:

We are looking for real values of that satisfy this equation. Let’s rewrite this as

To find the real solutions, we need to investigate the points where these two functions intersect, i.e., where,

As , , and .

So, there is no solution in this region.

For large positive :

As , , and .

Again, there is no solution in this region.

Check around :

At , , which satisfies the equation.

So, is a solution.

We can check the behavior of the function .

The derivative of this function is

Since for all real , the function is strictly increasing. Therefore, it can only cross at one point,

which implies that the solution is the only real solution.

Since there is only one real solution, the cardinality of the set of real solutions is 1.

Thus, the correct answer is option 2).

Let f ∶ ℝ2 → ℝ be a locally Lipschitz function. Consider the initial value problem
ẋ = f(t, x), x(t0) = x0
for (t0, x0) ∈ ℝ2.
By using Picard's theorem we know that solutions lie in the interval
|x − x_o| < a ;|t − t_o| < b
which is an open interval
Therefore, the Correct Option is Option (2).

Concept:

Odd degree polynomial must have at least one real root

Explanation:

A is a a 3 × 3 matrix with real entries.

So characteristic polynomial of A will be of degree 3.

(1): Since we know that, odd degree polynomial must have at least one real root so A must have a real eigenvalue.

(1) is true

(2): As we know that determinant of a matrix is equal to the product of eigenvalues. So if the determinant of A is 0, then 0 is an eigenvalue of A.

(2) is true

(3): The determinant of A is negative and 3 is an eigenvalue of A.

If possible let the other two eigenvalues of A are not real and they are α + iβ, α + iβ

So determinant = 3(α + iβ)(α - iβ) = 3(α² + β²) > 0 for all α, β which is a contradiction.

So A must have three real eigenvalues.

(3) is true and (4) is false statement

f(x) =
= x
= x[1 - (x - 1) + (x + 1)² -...]
= x.
= x.
= 1
Hence f(π) = 1
Option (3) is true.

Given -​

Let Ω be an open connected subset of ℂ containing U = {z ∈ ℂ ∶ |z| ≤ 1/2}.

Let is analytic and

Concept:

If f(z) is analytic within and on . Let a be a point inside C.

Calculation:

Let

Now,

Hence the statement P is incorrect.

Now,

Hence the statement Q is incorrect.

Hence the option (2) and (3) are correct.

Concept:

Improper Integral: If a function f on [a, b] have infinite value then it is called is improper integral

1. Improper Integral of the First kind: is said to be the improper integral of the first kind if a = -∞ or b = ∞ or both.
2. Improper Integral of the second kind: is said to be the improper integral of the second kind if a or b is finite but f(x) is infinite for some x ∈ [a, b].

​​
If the integration of the improper integral exists, then it is called as Converges, But if the limit of integration fails to exist, then the improper integral is said to be Diverge.

---(1)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

---(2)

For this type, if p ≥ 1 the function is divergent and for p < 1, the function is convergent.

Calculation:

Given:

On comparing with equation (2) the value of p < 1, therefore it is convergent.

Concept:

A Mobius transformation maps a circle or straight line to a circle or straight line.

Explanation:

T is a M¨obius transformation such that T(0) = α, T(α) = 0 and T(∞) = −α, where α = (−1 + i)/ √2.

Since T(α) = 0 let us assume that

L denotes the straight line passing through the origin with slope −1, and C denotes the circle of unit radius centred at the origin.

i.e., L: y = - x

and C: x² + y² = 1 i.e., |z| = 1

Now, α = (−1 + i)/ √2 i.e., (-1/√2, 1/ √2) lies on L.

any point on L is z = x + iy = x - ix = x(1 - i)

then, T(z) = T(x(1 - i) ) =

So for the point x = 1 T(z) = ∞ so it can't lie inside a circle.

Hence T maps L to a straight line

Option (1) is true and (2) is false

For any point on circle C it maps to straight line.

T−1 maps C to a straight line

Option (3) is true and (4) is false

X₁ ∪ X₂ ∪ X₃ is a connected set

X₁ ∪ X₂ ∪ X₃ is not a path connected but X₁ ∪ X₂ is path connected.

(1), (3) correct

differentiating both side

f'(x) = =

Option (1) is true.

=
Using L'hospital rule
=
= = - ∞

Option (2) is not true
We have
So there does not exist polynomial f(x) and g(x) with real coefficient such that because even degree is never equal to odd degree polynomial.
Option (3) is not true.
=
LHD = = 0
RHD = = 0

so f is differentiable at x = 0

Option (4) is true.

Concept:
1. Quadratic Form: A quadratic form in variables x₁,x₂ (or y₁,y₂)  consists of terms like . Quadratic forms  are often used to describe curves or surfaces (e.g., ellipses or hyperbolas).

2. Continuity of Quadratic Forms: Quadratic forms are continuous functions, which means they can take a range of values depending on the inputs. Since they are sums of squared terms and cross products, they can produce positive, negative, or zero outputs.

3. Combination of Quadratic Forms: The sum or difference of two quadratic forms is still a quadratic form. This is because the individual terms in each quadratic form are either squared terms or cross terms, which maintain the quadratic structure when combined.

Explanation:
Option 1: A quadratic form is a polynomial where each term is of degree two. Given that both q₁(x₁, x₂) and  q₂(y₁,y₂) are quadratic forms, the difference  q(x₁,x₂,y₁,y₂) will also be a quadratic form because it remains a sum of quadratic terms. Hence, this statement is true.

Option 2: This statement implies that for some t₁, t₂, the value of q₁ is 5. Since quadratic forms can take any real value depending on the specific variables chosen (assuming it's not a degenerate form), it is possible for q₁ to take the value 5.
Therefore, this statement is true.

Option 3: This statement claims that q₂ cannot take the value -5. However, a quadratic form can take any real value based on its variables. Therefore, this statement is false.

Option 4: Since q(x₁, x₂, y₁, y₂) = q₁(x₁, x₂) − q₂(y₁, y₂), it it is possible to construct vectors such that q equals any real number α, as both q₁ and q₂ can take any real value.
Hence, this statement is true.
The correct answers are Option 1, Option 2, and Option 4.

Concept:

1. Law of Large Numbers (LLN):
The Law of Large Numbers states that as the sample size n increases, the sample average (or sum) of i.i.d. random variables converges to the expected value of the variable. For example, in Option 3,  converges to the expected value , because X₁ has variance 1.

2. Central Limit Theorem (CLT):
The Central Limit Theorem tells us that the sum (or scaled average) of i.i.d. random variables with finite mean and variance converges in distribution to a normal distribution as n → ∞ . For instance, in Option 4, behaves like a standard normal random variable as n → ∞ , converging to N(0, 1) .

3. Probability Limits:
For certain random variables, their distribution converges to a fixed probability value. In Options 1 and 4, the probability of the standardized sum being less than or equal to 0 converges to 1/2, which is the probability that a standard normal variable is less than or equal to 0.

Explanation:

Option 1: This expression involves the ratio of two terms: .

The numerator,, grows like by the Central Limit Theorem since the sum of i.i.d. random

variables with mean zero and variance 1 tends to have a normal distribution.

The denominator, , grows like O(n) because it is the sum of the squares of i.i.d. random variables with variance 1.
As n → ∞, the ratio tends to zero, so the probability converges to the probability of the random variable being less than or equal to 0. Hence, Option 1 is true, and the limit will be 1/2 because this becomes a standard normal random variable under large n .

Option 2: The numerator behaves like , and the denominator behaves like O(n) . Therefore, the ratio behaves like and converges to 0 as n → ∞ . Thus, Option 2 is true.

Option 3: By the Law of Large Numbers (LLN), the average of the squares of i.i.d. random variables converges to the expected value of , which is 1, as n → ∞. Thus, Option 3 is true.

Option 4:
By the Central Limit Theorem (CLT), converges in distribution to a standard normal random variable N(0, 1).
The probability that a standard normal variable is less than or equal to 0 is P(Z ≤ 0) = 1/2. Hence, Option 4 is true.

All four options are correct.