CSIR NET Mathematical Science Mock Test - 3 - UGC NET MCQ

Given:

Consider the following four statements.

Statement 1: “Statement 3 is true.”

Statement 2: “Statement 1 is true”

Statement 3: “Statement 1 is true and Statement 2 is false”

Statement 4: “Statements 1, 2 and 3 are false”

Concept:

The concept at play in this set of statements is self-reference and logical consistency. These statements form a logical paradox similar to the well-known "liar paradox" or "Epimenides paradox."

Calculation:

Now, let's consider the implications:

If Statement 1 is true, then Statement 3 must be true.
If Statement 3 is true, then Statement 2 must be false (because Statement 3 claims that Statement 2 is false).
If Statement 2 is false, then Statement 1 must be false (because Statement 2 claims that Statement 1 is true).

This creates a logical contradiction. If Statement 1 is true, then it implies that Statement 1 is false. Therefore, there is no consistent solution, and the statements are mutually contradictory.

Hence the option (4) is correct.

Given:

At starting the debate:

Male participants : female participants = 2 : 1

Calculation:

Let the number of male participants at the start of the debate = 2x

The number of female participants at the start of the debate = x

According to the question:

⇒ (2x - 10)/(x + 4) = 3/2

⇒ 2 × (2x - 10) = 3 × (x + 4)

⇒ 4x - 20 = 3x + 12

⇒ x = 32

Total number of participants = (2x + x)

⇒ 3x = 3 × 32 = 96

∴ The correct answer is 96.

Concept:
Net Displacement: After calculating the movement components, the total displacement is the vector sum of the horizontal and vertical distances between the current and original positions.

Explanation:
Rajesh moves 1 km North-East (which is at a 45-degree angle) to Sunil’s house.
From Sunil's house, he moves 707 meters South to Arjun’s house.
Rajesh’s movement to Sunil's house forms a right-angled triangle, where the horizontal and vertical components of the movement are equal since it's in the North-East direction. The horizontal and vertical components of the 1 km (1000 m)
North-East move are 1000 × cos(45^∘) and 1000 × sin(45^∘), respectively. Both components are equal to .
From Sunil’s house, Rajesh moves 707 meters South, which directly affects the vertical component.
So, he is at the same latitude as his original house.
The horizontal component remains 707 meters to the East.
Since Rajesh is almost at the same latitude (vertical difference is 0.1 meters), the total distance from his house is approximately the horizontal displacement = Distance = 707meters
Thus, the correct answer is Option 3).

Concept:
Per capita GDP = Total GPD/Total population

Explanation:

City A:
In 2010, Per capita GDP = 115/150 = 0.767 billion USD
In 2020, Per capita GDP = 324/160 = 2.025 billion USD
So increase in per capita = 2.025 - 0.767 = 1.258 billion USD

City B:
In 2010, Per capita GDP = 1675/1200 = 1.39583 billion USD
In 2020, Per capita GDP = 2622/1400 = 1.87286 billion USD
So increase in per capita = 1.87286 - 1.39583 = 0.47703 billion USD

City C:
In 2010, Per capita GDP = 117/180 = 0.8167 billion USD
In 2020, Per capita GDP = 264/220 = 1.2 billion USD
So increase in per capita = 1.2 - 0.8167 = 0.3833 billion USD
Since 1.258 > 0.47703 > 0.3833 so required order from high to low is
A, B, C

Option (1) is correct

Concept:

A second order PDE of the form
(i) hyperbolic if S² - 4RT > 0
(ii) parabolic if S² - 4RT = 0
(iii) elleptic if S² - 4RT < 0

Explanation:

Given PDE is

Here S² - 4RT = (xy)² - 4x²(-2y²) = x²y² + 8x²y² = 9x²y2 = 9(xy)²
If y = 0 the S² - 4RT = 0 then PDE is parabolic.
(1) is false
If x > 0, y > 0, S² - 4RT > 0
PDE is hyperbolic

(2) is false
If x < 0, y > 0 then S² - 4RT > 0
PDE is hyperbolic

(3) is true
If x = 0, y > 0 then ^S2 - 4RT = 0
So, PDE can be parabolic
(4) is false

which is linear in y

IF = = e^-logx = 1/x

So, general solution is

Using y(0) = 1
⇒ 1 = 0 which is not possible.
Hence, the differential equation has no solution.
(4) is true.

Given that Xi ~ LogNormal, then lnXi ~ N(μ, σ²). The log-normal distribution is positively skewed, not symmetric around zero, however, when we take the logarithm of a log-normal variable, it results in a normally distributed variable. This implies that the mean of lnX_i is 0.

In this case, it is known that for a set of independent and identically distributed normal random variables X_i, the sum of their squares (which matches our Zn) follows a Chi-squared distribution, with degrees of freedom equal to the count of the variables.

for option (i) -
Since the Chi-squared distribution is only defined for positive numbers, and given that the mean of

= (by properties of the Chi-squared distribution)
Hence option (i) is correct.

Concept:

(i) A complex function f(z) is entire function if it is analytic in whole complex plane.

(ii) If a complex function f(z) = u + iv is entire then it satisfy C-R equation i.e., u_x = v_y, u_y = - v_x

Explanation:

f : ℂ → ℂ is a real-differentiable function.

u, v : ℝ2 → ℝ by u(x, y) = Re f(x + i y) and v(x, y) = Im f(x + iy), x, y ∈ ℝ.

Also, ∇u = (ux, uy)

(1): Then "For c1, c2 ∈ ℂ, the level curves u = c1 and v = c2 are orthogonal wherever they intersect" this statement will satisfy only if f(z) is analytic function.

(1) is false

(3): f(z) is entire function so ux = vy, uy = - vx

then ∇u . ∇v = (ux, uy) . (vx, vy) = uxvx + uyvy = uxvx - vxux = 0 at every point.

(3) is true and (2) is false

(4): ∇u . ∇v = 0

⇒ (ux, uy) . (vx, vy) = 0

⇒ uxvx + uyvy = 0

⇒ uxvx = - uyvy

which does not imply ux = vy, uy = - vx

f is not an entire function.

(4) is false

Calculation:

Let z = x + iy & w = a + ib

⇒ |z + w| = |(x + a) + i(y + b)| =

⇒ & |z − w| = |(x − a) + i(y − b)| =

for |z + w| = |z−w| to be hold

⇒(x + a)² + (y + b)² = (x − a)² + (y − b)²

⇒ x² + a² + 2ax + y² + b² + 2by = x² + a² − 2ax + y² + b² − 2by

⇒ 4(ax + by) = 0 ⇒ ax + by = 0

Now, z ⋅ w̅ = (x + iy) (a − ib) = ax − ibx + iay + by = (ax + by) − i(bx − ay) = − i(bx − ay), {∵ ax + by = 0)

⇒ z ⋅ w̅ = − i(bx − ay) is purely imaginary.

The correct answer is option "4"

In the case of the normal distribution, a sufficient statistic for μ (the mean of the distribution) is the sum of the sample observations (∑X_i for i = 1 to n), because the mean value directly depends on the sum of all the observations.

On the other hand, the minimum value of the sample, min(X_i) for i = 1 to n, and the maximum value of the sample, max(X_i) for i = 1 to n, do not hold all necessary information to calculate μ, and thus they are not sufficient statistics.

In this question, the minimum value (min(X_i)) is listed as the statistic that is not sufficient for calculating μ.

Hence option (iii) is correct.

Concept:

Covariance Matrix Calculation:
Each  and since the X_i's are i.i.d., their variances and covariances can be computed easily.

The variance of Y_i is
The covariance between any two distinct Y_i and Y_j (where ) is

Thus, the covariance matrix of Y has 2's on the diagonal and 1's off-diagonal.

Explanation:

are independent and identically distributed (i.i.d.) random variables with mean 0 and variance 1.

.
The task is to find the first principal component based on the covariance matrix of .

Each where, X₀ is common across all Y_i's.
The covariance between any two different Y_i and Y_j depends on X₀.
The covariance matrix ∑_Y for Y will have entries:
, where is the Kronecker delta.

Since X₀ and X_i have variance 1, we get,

when (because of the common X₀).

when .

Thus, the covariance matrix ∑_Y is a matrix with diagonal entries 2 and off-diagonal entries 1. It is a symmetric matrix.
The first principal component corresponds to the eigenvector associated with the largest eigenvalue of the covariance matrix ∑_Y .
For a covariance matrix like this (with all off-diagonal elements equal and diagonal elements greater than off-diagonal elements), the first principal component will have equal weights on all components. Specifically, the eigenvector corresponding to the largest eigenvalue will be proportional to .
The first principal component can thus be expressed as ​​

This is a linear combination of the Y_i's, where each Y_i has an equal weight, scaled by to ensure unit length of the eigenvector.
From the available options, the correct representation of the first principal component is ​​

Thus, the correct answer is the first option.

Concept:

Limit of a Sequence of Functions:
1. Let {f_n} be a sequence of functions defined on a set D. We say that f_n converges pointwise to a function f on D if, for every x ∈ D

2. A stronger form of convergence is uniform convergence. The sequence {f_n} converges uniformly to a function f on D if 

Explanation: The problem gives a sequence of functions defined by

and asks about the limit of f_n(x) as n → ∞, denoted by f(x). We are tasked with determining which statement about f(x) is true.
We are asked to take the limit n → ∞ of the function:

As n → ∞, the term . So, for large n , the function f_n(x) approaches

Case 1:
For we have,

Case 2: x = 0
When x = 0 , the function becomes

Therefore, as n → ∞, we get f(0) = 0 .

The function f(x) , n → ∞ , is given by

This function is equal to |x| for all ,
Therefore, The correct option is 4).

Concept:

Linear Transformation: A function A between two vector spaces that preserves the operations of vector addition and scalar multiplication. In this case, A is a linear transformation from R^m (an m -dimensional space) to Rⁿ (an n -dimensional space).

One-to-One (Injective): A linear transformation is injective if distinct vectors in R^m are mapped to distinct vectors in Rⁿ . In terms of matrices, A is injective if its null space only contains the zero vector.

Onto (Surjective): A linear transformation is surjective if for every vector in Rⁿ, there is at least one vector in R^m that maps to it. In terms of matrices, A is surjective if its image spans the entire space Rⁿ (i.e., if ).

Bijective: A linear transformation is bijective if it is both injective (one-to-one) and surjective (onto).

A bijection implies that the linear transformation has an inverse, meaning A can map R^m to Rⁿ perfectly without losing or repeating information.

Explanation:

Option 1:
Consider the set X = {1, 2, 3} so m = 3 and set Y = {a, b, c, d} so n = 4.
Define the function A : X → Y by A(1) = a, A(2) = b and A(3) = c
This function is one-to-one (no two elements in X map to the same element in Y) but not onto (the element d \in Y is not mapped by any element of X).
In this case, m = 3 and n = 4, but m < n. Thus, the function is injective but not surjective, providing a counterexample to the condition m > n.

Option 2:
Consider the set X = {1, 2, 3, 4} (so m = 4) and set Y = {a, b, c} (so n = 3).
Define the function A : X → Y by
A(1) = a, A(2) = b, A(3) = c and A(4) = c
This function is onto because every element in Y is mapped by some element in X.
However, it is not one-to-one because two elements in X (3 and 4) map to the same element c in Y.

In this case, m = 4 and n = 3, but m > n. Therefore, the function is surjective but not injective, providing a counterexample to the condition m < n.

Option 3:
A transformation is bijective if it is both one-to-one and onto, meaning every element of Rⁿ has a unique preimage in R^m. This can only happen if m = n, so this is a correct statement.

Option 4:
Consider the set X = {1, 2, 3} (so m = 3) and the set Y = {a, b, c, d} (so n = 4).
Define the function A : X → Y by A(1) = a, A(2) = b and A(3) = c
This function is one-to-one (injective) because no two elements of X map to the same element of Y.
However, m ≠ n, as m = 3 and n = 4.
Thus, the function is injective but m ≠ n, providing a valid counter example.
Thus, the correct statements is option 3).

Concept:

An entire function is a complex-valued function that is a complex differential in a neighborhood of each point in a domain in a complex coordinate space

Explanation:

f(z) = for z ∈ ℂ

singularities of f(z) is given by

1 - z - z² = 0 ⇒ z = =

So f(z) has a pole at z =

So options (1) and (2) both false

f has a Taylor series expansion so

f(z) = anzn = f(0) + z + ....

So a0 = f(0) = 1

and a1 =

Now, f'(z) = -(-1 - 2z)

So f'(0) = 1

So option (4) is correct and (3) false

Concept:

Let F(x) be a cumulative distribution function then
(i)
(ii) F is non-decreasing function

Explanation:

(2):

(Using L'hospital rule), not satisfying

Option (2) is false

Option (3) is false

f(0^-) = 1/2 and f(0⁺) = 1/4 and 1/2 > 1/4 so F(x) not satisfying the property "F is non-decreasing function"

Option (4) is false

Therefore option (1) is correct

Concept:

Explanation:
an = n + n-1 then

=  

Now,



Since the series , and are convergent series so is also convergent.
Hence the given series is absolutely convergent and hence convergent.
Now,

=
=
=
= e^-1 + 1 - e-1 + e-1 = e-1 + 1
Therefore the given series converges to e-1 + 1
(4) is correct

Concept:
Let Pp + Qq = R be a PDE where P, Q, R are functions of x, y, z then by Lagrange's method

= =

Explanation:
Given
uux + uy = 0, x ∈ ℝ, y > 0,
u(x, 0) = x, x ∈ ℝ.

Using Lagrange's method

= =

⇒ = =

Solving this we get
u = c₁...(i)
and putting u = c1 we get from first two terms
dx/c₁ = dy/1
dx = c₁dy
⇒ x - c₁y = c₂
⇒ x - uy = c₂...(ii)
From (i) and (ii) we get the general solution as
u = ϕ(x - uy)

Using u(x, 0) = x we get
x = ϕ(x) so ϕ(x - uy) = x - uy

Hence solution is

u = x - uy ⇒ u(1 + y) = x ⇒ u =

Therefore u(2, 3) = 2/4 = 1/2 

Option (3) is correct

Concept:
If xⁿ = 1 then o(x) divides n

Explanation:
ℤ/105ℤ ≅ ℤ₁₀₅
105 = 3 × 5 × 7

So
Given x2 = 1 so o(x) divides 2 Hence o(x) = 1 or 2
Element of ℤ2 of order 1 and 2 is 2
Element of ℤ4 of order 1 and 2 is 2
Element of ℤ6 of order 1 and 2 is 2
Hence total such elements = 2 × 2 × 2 = 8
Option (4) is correct

p(x) is a real polynomial of degree 3.
Let p(x) = a₀ + a₁x + a₂x² + a₃x³
So,

= (∞/∞ form)

= (Using L'hospital rule)

= (Again using L'hospital rule)

= (Again using L'hospital rule)

= 0

Option (1) is true.

When two independent Poisson random variables are added together, the resulting distribution is also a Poisson random variable whose parameter is the sum of parameters of the original variables. Therefore, the sum of Z and W is a Poisson random variable with parameter 4 + 5 = 9.

Hence option (i) is correct.

The difference between two Poisson random variables does not follow a Poisson distribution, so options (ii) and (iv) are incorrect. Option (iii) is also incorrect because the sum of two Poisson variables cannot result in a Poisson distribution with a negative parameter. A Poisson distribution's parameter must always be greater than or equal to zero.

(a) This statement is true. B follows a non-uniform distribution, specifically, it follows a chi-square distribution with 1 degree of freedom.

(b) This statement is not true. The expected value of B is not 4/3. For a random variable A uniformly distributed between -2 and 2, the resulting distribution B has an expected value of from -2 to 2, which equals 8/3, not 4/3.

(c) This statement is true. A can indeed take on values from -2 to 2, while B, being a square of A, is always non-negative.

(d) This statement is not true. The conditional probability P(A ≤ 1 | B ≤ 1) is not 1. Knowing that B ≤ 1 doesn't guarantee that A will be less than or equal to 1. For instance, A can be a value such as -1.5, which would make B > 1. Therefore, the conditional probability P(A ≤ 1 | B ≤ 1) doesn't equal 1, and varies depending upon the joint distribution of A and B.

The statements that are NOT true are (ii) and (iv).

Concept:
(i) A matrix A is diagonalizable if all its eigenvalues are real and distinct
(iii) Non-zero nilpotent matrix is not diagonalizable
(iii) Characteristic equation of a 2 × 2 matrix A is x² - tr(A)x + det(A) = 0

Explanation:
A ∈ M₂(ℝ)
Option 1): Characteristic equation
x² - tr(A)x + det(A) = 0
If (tr(A))² > 4 det(A) then all the roots are real and distinct
so eigenvalues of A are real and distinct
Then A is diagonalizable over ℝ.
Option (1) is true

Option 2): Let A =
Then tr(A) = 0 and det(A) = 0
So (tr(A))² = 4 det(A)
But A is a non-zero nilpotent matrix so not diagonalizable.
Option (2) is false
Option 3): If (tr(A))² < 4 det(A)

take A =
We get the Eigen Values in Complex Number
then all the roots of x² - tr(A)x + det(A) = 0 are complex which does not belongs to ℝ
Then A is diagonalizable over ℝ.
Option (3) is false
Therefore, Correct Option is Option 1).

Concept:
(i) Lebesgue measume μ*((x, y)) = x - y
(ii) Any countable set of real numbers has Lebesgue measure 0.

Explanation:
[x] denote the integer part of x for any real number x.

So [x] =

(1): {x ∈ [1, ∞) : limn→∞ [x]n exists }
limn→∞ [x]n exists then x ∈ [1, 2)
So, Lebesgue measure = 2 -1 = 1 ≠ 0
(1) is correct
(2): {x ∈ [1, ∞) : limn→∞ [xn] exists}
then x = {1} which is countable so Lebesgue measure = 0
(2) is false
(3): {x ∈ [1, ∞) : limn→∞ n[x]n exists}
then x = {0} which is countable so Lebesgue measure = 0
(3) is false
(4): {x ∈ [1, ∞) : limn→∞ [1 - x]n exists}
then x ∈ [1, 2)
So, Lebesgue measure = 2 -1 = 1 ≠ 0
(4) is correct

For the bilinear map ϕ(v, w) = v^T Aw there exist a 2 × 2 symmetric matrix B such that ϕ(v, v) = v^TBv for all v ∈ ℝ².

(3) correct

ϕ(v, w) = v^T Aw = v₁w₁ + 2v₁w₂ + 4v₂w₁ + 3v₂w₂

and ϕ(w, v) = w^TAv = v₁w₁ + 4v₁w₂ + 2v₂w₁ + 3v₂w₂

Hence ϕ(v, w) ≠ ϕ(w, v) for all v, w ∈ ℝ².

(1) false

ϕ(v, w) = 0

will be true for for all w ∈ ℝ²

if (v₁ + 4v₂ 2v₁ + 3v₂) = 0
⇒ v₁ = v₂ = 0 ⇒ v = 0

(2) false

Similarly by calculation we can show that (4) is also false

Given,

and B

clearly, eigen value of A and B are 2,2,3

for matrix A

A.M of eigenvalue 2 is 2

G.M of eigen value 2 is 1 ( G.M = n - rank (A-2I))

since number of jordan block corresponding to 2 is 1

then jordan form of A is

(say B)

Any matrix is similar to its jordan form

Therefore Option 1) and 3) is correct

Since A.M not equal to G.M

this implies not diagonalizable

Therefore, Option 2) is False

For Option 4)

Option 4) is Correct

Therefore, Correct Options are Option 1), 3) and 4.

Concept:

To minimize the functional, we use the Euler-Lagrange equation for the functional, which is

Explanation:

subject to the conditions:
1. (i.e., y(x) is continuously differentiable).
2. = 0
3.

Euler-Lagrange Equation

Substitute into the Euler-Lagrange equation

This is a second-order linear differential equation,

The general solution to this differential equation is:

where A and B are constants to be determined using the boundary conditions.

Apply y(1) = 1 :

⇒    ... (1)

2. Apply y(-1) = 1 :

⇒  ... (2)

From equations (1) and (2), we can solve for A and B . Add both equations

Thus,   ...(3)

Now subtract equation (2) from equation (1):

Since, , it must be that  A = B      ...(3)

Substitute A = B into equation (3):

This matches option 3).

Concept:
The optimal value is the value of f at the point where it achieves its maximum or minimum, often subject to a set of constraints.

Explanation:
Maximize x₁ + x₂ + x₃
Subject to the constraints
1) x₁ + x₂ - x₃ ≤ 1
2) x₁ + x₃ ≤ 2 
3) 0 ≤ x₁ ≤ 1/2
4) x₂ ≥ 0,
5) 0 ≤ x₃ ≤ 1

We are maximizing x₁ + x₂ + x₃ under the given constraints. The feasible region is defined by the intersection of the constraints, which will give the possible values for x₁ + x₂ and x₃ .

Option 1: Suppose

Check the first constraint which violates .

Try (x₁ + x₂ + x₃) = (0, 2, 1) : 0 + 2 - 1 = 1 , and 0 + 1 = 1 satisfy the constraints. Thus, x₁ + x₂ + x₃ = 0 + 2 + 1 = 3

Therefore, the optimal value of the objective function is 3.
Option 1 is true.

Option 2: From the previous step, we already found that the maximum value of x₁ + x₂ + x₃ is 3, which is greater than 3/2.
Option 2 is false.

Option 3: We found that the point (0, 2, 1) satisfies all the constraints and gives the maximum value of the objective function. Extreme points are vertices of the feasible region, and since this point lies on the boundary, it is an extreme point. Option 3 is true.
Option 4: Let’s check if satisfies all the constraints:
First constraint: , which satisfies ≤ 1.
Second constraint:

Thus, this point satisfies all the constraints. However, the value of , which is less than the optimal value of 3.
Option 4 is false.

The correct options are Option 1) and Option 3).

Concept:

A point z₀ is an essential singularity if the Laurent series expansion of f(z) around z₀ has infinitely many negative power terms. Specifically, it can be written as:

where a_n for n < 0 are not all zero.
A point z₀ is a removable singularity if  where L is a complex number. If we define f(z₀) = L, then f(z) becomes analytic at z₀.

Explanation:

For and g(z) = f(z)sin (z) .
1. Function :

To analyze the behavior of f(z) as z → 0, consider the Taylor series expansion of sin(x):

Therefore,

As z → 0, f(z) exhibits infinitely oscillatory behavior, indicating that f(z) has an essential singularity at z = 0.

2. Function g(z) = f(z)sin(z):
Near z = 0 , sin(z) behaves like z, hence

Since is nearly equal to when z → 0
Hence g(z) has essential singularity at z = 0.
Option 1: f has an essential singularity at 0: True
Option 2: g has an essential singularity at 0: True
Option 3: f has a removable singularity at 0: False
Option 4: g has a removable singularity at 0: False
The true statements are option 1) and 2).