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

Given:

Given figure represents pH, partial pressure of CO2 (pCO2), and temperature (T) in an experiment conducted in a water sample over 20 days.

Concept:

Apply basic theory .

Calculation:

High CO₂ causes global warming (1):
This statement is not directly related to the information provided about pH, pCO₂, and temperature in a water sample. Global warming is a broader environmental issue influenced by various factors, including greenhouse gas emissions.

High temperature causes acidification (2):
Generally, high temperatures alone do not cause acidification. Acidification in aquatic systems is often associated with an increase in dissolved carbon dioxide (resulting in a decrease in pH) rather than temperature.

There is a decrease in pH and an increase in both T and pCO₂ over 20 days (3):
This statement could be plausible based on common trends in aquatic systems. An increase in pCO₂ could potentially lead to a decrease in pH. Additionally, temperature may influence these parameters.

pH and pCO₂ are positively correlated while pH and T are inversely correlated (4):
This statement could be true in some cases. A positive correlation between pH and pCO₂ means that as pCO₂ increases, pH decreases. The relationship between pH and temperature can be more complex and may depend on other factors.

Hence the option (3) is correct.

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.

Let first two digits are x and y . Then last digit is x+y
So,

Hence total pins are 55.

Hence the option (1) 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:

Net loss =

Explanation:
If a person gains x% on one item and loses x% on another item, the net result will always be a loss.

Net loss =

Here, the common percentage is 20%, so the net loss is

Therefore, the person incurs a net loss of 4%.
Thus, the correct option is Option 3).

All peacocks are bird
No insects are birds
Some birds extinct animals
Some insects are extinct animals
No peacocks are extinct animals
No peacocks are insects.

i.e.,

So option (3) is correct

Concept:
Number of ways of selecting r people from n people is ⁿC_r

Explanation:
Number of ways of selecting 2 people randomly from 7 people is

Now, Let 4 peoples are A, B, C, D
They have exactly one sibling so the pair of siblings can be (A, B), (C, D)
Also let 3 people be X, Y, Z
They have exactly two siblings so the pair of siblings can be (X, Y), (X, Z), (Y, Z)
So favourable outcomes are (A, C), (A, D) (B, C), (B, D),(A, X), (A, Y), (A, Z), (B, X), (B, Y), (B, Z), (C, X), (C, Y), (C, Z), (D, Y) (D, X), (D, Z)
Therefore probability that they are siblings = 16/21
Option (2) is correct

Let A: A tells the truth, A^c = A lies

B: B tells the truth, Bc = B lies

Similarly,

Therefore that the minimum probability that they would contradict each other i.e., one tell the truth and other lies

= P(A) × P(Bc) + P(B) × P(Ac)

Option (3) is correct

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.

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:
Supremum (sup): The supremum of a set is the least upper bound. For a sequence (a_n), is the smallest number that is greater than or equal to all the terms of the sequence.
Infimum (inf): The infimum is the greatest lower bound. For a sequence , is the largest number that is less than or equal to all the terms of the sequence.

Explanation:

Option 1:
The infimum and supremum of the sequence refer to its lower and upper bounds. If these two limits coincide, it implies that the sequence is squeezing towards a single point.
This is a true statement, as if the infimum and supremum converge to the same point, the sequence must converge to that point.

Option 2:
If the infimum of the sequence is equal to the limit of the sequence as n→∞, this implies that the sequence stabilizes at this value, suggesting that it is converging to that point.
This is a true statement, as the sequence is converging to its infimum, implying that it has a limit.

Option 3:
Counter example:
Consider the sequence .
1. As n→∞, lim_n→∞ an = 0.
2. The supremum of the sequence is

This sequence is clearly not constant because the values of a_n decrease as n increases.
However, we still have:

This shows that a_n is not constant.

Option 4:
The supremum of a sequence is the least upper bound of the values in the sequence.
It is the smallest number that is greater than or equal to every term in the sequence.
The infimum of a sequence is the greatest lower bound of the values in the sequence.
It is the largest number that is less than or equal to every term in the sequence.
Now, if , this means that the least upper bound and the greatest lower bound are the same. Let’s call this common value C.
Since the supremum C is an upper bound of the sequence, all terms in the sequence must be less than or equal to C.
Since the infimum C is a lower bound of the sequence, all terms in the sequence must be greater than or equal to C.
Therefore, for all n, the term a_n must satisfy , which implies that a_n = C for all n.

Hence, the required option is 3).

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:
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) Abel's test: Let is a convergent series, {bn} is a monotone sequence, and {bn} is bounded. Then is also convergent.
(ii) Alternating series test: Suppose that we have a series ∑a_n and either a_n = (−1)ⁿ b_n or a_n = (−1)^n-1b_n where b_n ≥ 0 for all n. Then if  b_n = 0 and, {b_n} is a decreasing sequence the series ∑a_n is convergent.

Explanation:
Given series
Let ∑a_n = and {b_n} =
Since ≥ 0 for all n,
= 0 and is a decreasing sequence so by Alternating series test ∑ an = is convergent.
Also, is a monotonic sequence and bounded as = e
Then by Abel's test is convergent

(1) is correct

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

We need to find units in

for Option 4)

it gives , a = -7, b= -4 and hence,

it implies given element is invertible.

Similarly, Option 1) is True

for Option 3)

Hence, not invertible

Therefore, Correct Options are Option 1 and Option 4.

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).

Concept: R² refers to the coefficient of determination, and refers to the adjusted coefficient of determination in a standard linear regression model.

Explanation:
Option 1: This is true. The adjusted R² accounts for the number of independent variables in the model and is usually less than R² unless the addition of variables explains the variation in the data better. Therefore, is typically less than or equal to R² .

Option 2: This is true. As you add more independent variables to the model, R² never decreases; it either increases or stays the same, since R² measures the proportion of the variance explained by the model, and adding more variables can only increase this or leave it unchanged.

Option 3: This is not necessarily true. The adjusted R² can decrease if the additional independent variables do not contribute enough explanatory power relative to the number of observations. However, it may also increase if the added variables improve the model significantly.

Option 4: This is false. The adjusted R² is typically either positive or zero, so it is not necessary to be positive.

Correct options: 1) and 2).