CSIR NET Mathematical Science Mock Test - 1 - UGC NET MCQ

Given:

Two mobile phones sold to Arjun at a profit of 11% each and rate of 13,875.

He also sells a laptop to seema at the rate of Rs. 38804 earning a 9% profit.

Formula used:

Profit or Gain = Selling price – Cost Price

Loss = Cost Price – Selling Price

Profit percentage = (Profit /Cost Price) x 100

Loss percentage = (Loss / Cost price) x 100

Calculations:

According to question,

If 109% of CP of laptop is equivalent to 38,804

Then, 9% of CP (i.e., profit) is equivalent to 38,804 × 9/109 = 3204

Similarly, if 111 % of mobile phone is equivalent to 13,875

Then, 11% is equivalent to 13,875 × 11/111 = 1375

Total profit

= profit on laptop+ 2 × profit on mobile

= 3204 + 2(1375)

= 3204 + 2750

= Rs.5954

∴ The selling price of a second mobile phone is Rs. 5954.

Given:
If 90 people are to be seated randomly in 15 rows of 6 seats each,

Concept:

Calculation:
In each row, there are 6 seats, and a person getting a seat at either end of a row corresponds to 2 out of those 6 seats.
For the first person, the probability of getting a seat at either end is  2/6 = 1/3

Hence the option (3) is correct..

Concept:

Mean :

= = 1

Standard Deviation :

Explanation:

Forest Patch A:

There are 25 quadrats, each containing exactly 1 tree.

Mean (μA): Total number of trees is 25. Mean is .

Standard deviation (σA): Since all values are the same (1), the standard deviation is .

Forest Patch B:

Quadrats contain the following numbers of trees:

Mean (μB): Total number of trees is 2 + 7 + 10 + 6 = 25 . Mean is .

Standard deviation (σB): Since values vary greatly, the standard deviation .

⇒ and are not equal.

⇒ The standard deviation in Patch B is higher due to the large variation in tree numbers.

Thus, Option 3) is correct.

Given that 540 people are to be seated in rows such that each row contains 4 fewer people than the previous row. We need to find which of the given numbers of rows is not possible.
Let n = number of rows and x = the number of people seated
The number of people in the second row is x−4 in the third row is x−8 and so on.
The total number of people seated is x + (x − 4) + (x − 8)+...+(x − 4(n − 1)) = 540
This is an arithmetic series with first term a = x, common difference d = -4 and number of terms n.
The sum of an arithmetic series is given by ​​​​​​

Checking Option 1: n = 5

⇒ x = 116

This is possible, so 5 rows is possible.

Checking Option 2: n = 6

This is possible, so 6 rows is possible.

Checking Option 3: n = 8
1080 = 8 × (2x − 4(8) + 4)
⇒ 35 = 2x − 28 35
⇒ x = 81.5 , which is not an integer, 8 rows is not possible.

Checking Option 4: n = 9
​1080 = 9 × (2x − 4(9) + 4)
⇒ 120 = 2x − 32
⇒ x = 76
This is possible, so 9 rows is possible.
The number of rows that is not possible is Option 3).

Since males and females pair randomly, we assume that the proportion of red, blue, and yellow is the same for both males and females (i.e., 50% of males are red, 30% of males are blue, and 20% of males are yellow and the same proportions apply to females).

To find the expected number of red male–yellow female pairings:
The probability of selecting a red male is 50% = 0.50
The probability of selecting a yellow female is 20% = 0.20.
The expected number of red male–yellow female pairings is the product of these probabilities and the total number of pairs =
Expected number of red male-yellow female pairs = 0.50 × 0.20 × 10,000 = 0.10 × 10,000 = 1,000
The expected number of pairings between red males and yellow females is 1,000
Pairing is random, meaning the probability of selecting any male or female of a specific color is proportional to their population distribution.
Hence, option 3) is correct.

Let unit digit be y and tens digit be x then the number is 10x + y
If we reverse the digits then the number will be 10y + x
Difference of squares of these two number
= (10x + y)² - (10y + x)²
= 100x² + 20xy + y² - 100y² - 20xy -x²
= 99x2 - 99y2
= 99(x2 - y2) = 9 × 11(x2 - y2)
So the difference of square is always divisible by 1, 9, 11, 99
Option (4) is correct

1. Rate of Deviation per Hour:
- Clock A gets faster by 8 minutes per hour.
- Clock B gets slower by 12 minutes per hour.
2. Time shown on each clock relative to the true time:
- After t hours, the time deviation for Clock A is +8t minutes (it's fast).
- After t hours, the time deviation for Clock B is -12t minutes (it's slow).
Let t be the true time in hours since the clocks were synchronized at 05:00 hrs.
3. Clock Readings:
- Clock A shows 15:12 hrs.
- Clock B shows 12:12 hrs.
4. Expressing Time on Each Clock:
- Time on Clock A (fast by 8t minutes): 5:00 + t should be adjusted to show 15:12.
Thus, 300 + 60t + 8t = 912 minutes.
- Time on Clock B (slow by 12t minutes): 5:00 + t should be adjusted to show 12:12.
Thus, 300 + 60t - 12t = 732 minutes.
5. Solving for t:
- From Clock A: 68t = 612 or t = 612/68 = 9 hours.
- From Clock B: 48t = 432 or t = 432/48 = 9 hours.
Both calculations show t = 9 hours, meaning 9 hours have passed since 05:00 hrs.
6. True Time Calculation:
- The true time is 05:00 hrs + 9 hours
= 14:00 hrs.
Therefore, the correct time at that instant, when Clock A shows 15:12 and Clock B shows 12:12, is 14:00 hrs.
Hence Option (B) is correct.

y(x) = 
⇒ y(x) = 
⇒ y(x) = λxe^xc ....(i)
where c 

Putting the expression of y(x) from (i) in (ii) we get


(2) is correct

(1) If f is one-one then f ∶ U → f(U) is bijection so that g ∶ f(U) → U be an analytic function.
⇒ g is continuous and U is open so inverse image of U must be open.
⇒ f(U) is open.

Hence Option (4) is false and Option (1) is correct.
(2) f(z) = sin z, U = {z ∶ 0 < Im z < 4π}
(3) Take f and U as in (b) then
The correct answer is option (1).

Concept:

Let the system of ODE be

then if the eigenvalues of the Jacobian matrix is
(i) real and both positive then the critical point is unstable node
(ii) real and both negative then the critical point is asymptotically stable node

Explanation:
Given system of ODE is


f(x, y) = -5x + 3y, g(x, y) = -4x + 2y

Jecobian (J) = =

So eigenvalues of J are given be
x² - tr(J)x + det(J) = 0
x2 + 3x + 2 = 0
(x + 2)(x + 1) = 0
x = -1, -2
Both eigenvalues are negative and distinct.
So critical point is a stable node
(1) is correct

The numerator of the conditional probability expression should be the binomial probability of having exactly two allergic reactions from m trials, which is given by

The denominator should be the probability that at least one person has an allergic reaction. This is one minus the probability that no person has an allergic reaction, or

So, the correct answer is:which corresponds to Option A.

Note:  ^mC₂ stands for "m choose 2". The term, , derives from the property of the binomial distribution where the probability of "at least one success" can be computed as 1 minus the probability of "no success".

Hence the option(i) is true.

Concepts Used:
1. Cumulative Distribution Function (CDF):
The CDF F(x) gives the probability that the random variable X takes a value less than or equal to x . That is, F(x) = P(X ≤ x) .

2. Finding Probability Using the CDF:
The probability that the random variable X lies within a certain interval (a, b] is given by:

P(a < X ≤ b) = F(b) - F(a)

3. Probability at a Specific Point (Jump Discontinuity):
The probability at a specific point x = c is the difference in the CDF just to the right and just to the left of c :
P(X = c) = F(c⁺) - F(c^-)

Explanation -
We are given a cumulative distribution function (CDF) F(x) of a random variable X as:

F(x) =

From the definition of the probability from the CDF: P(a < X b) = F(b) - F(a)

In this case, we need to calculate .

Since and , we use the formula for both 1/3 and 3/4 :

Thus, the probability is:

=

The probability at a point is the jump in the CDF at that point. We need to calculate F(0⁺) - F(0^-) .

From the CDF definition: F(0⁺) = F(0) =

⇒ F(0^-) = 0

Thus,
Now, we add the two results:

Thus, the final answer is 17/36.

Concept:

Lipschitz condition: A function f(x, y) satisfies the Lipschitz condition with respect to y in a domain

 if there exists a constant L > 0 such that for any (x, y₁) and (x, y₂) in D, the following inequality holds

The constant L is called the Lipschitz constant.

Explanation:

, where ∈ > 0 is a constant, and  .
S1: There is an ∈ > 0 such that for all  , the IVP has more than one solution.
S2: There is a such that for all ∈ > 0 , the IVP has more than one solution.
The differential equation is . Since ∈ > 0 , the right-hand side of the equation is always positive and well-defined for any .
In general, the uniqueness of solutions to IVPs can often be determined by the Lipschitz condition.
For a function , we need to check if the function satisfies the Lipschitz condition with respect to y.

This function is continuous and bounded for all because ∈ > 0 ensures no singularity at y = 0.

Therefore, the function satisfies the Lipschitz condition, ensuring that the IVP has a unique solution for each when ∈ > 0 .
S1: This statement claims that there is some ∈ > 0 for which the IVP has more than one solution for all .
From our uniqueness analysis (Lipschitz condition), the IVP actually has a unique solution for all ∈ > 0 and .

Therefore, S1 is false.
S2: This statement claims that for some and for all ∈ > 0, the IVP has more than one solution.
Based on the same reasoning (Lipschitz condition and continuity of the derivative), the solution remains unique for all ∈ > 0 and for any y₀ . Therefore, S2 is also false.
Both statements S1 and S2 are false.
Thus, the correct option is 4).

Concept:

The moment of inertia (in kg meter2) of the cylinder about a diameter of its base is

Explanation:

Radius of the cylinder, , Height of the cylinder, , Density of the cylinder,

, where

Substituting the given values,

Thus,

Now, substitute the values , and into the formula

⇒

⇒

⇒

Hence correct option is 1).

Concept:
Even functions are symmetric about the y-axis, satisfying f(-x) = f(x).
Odd functions are symmetric about the origin, satisfying f(-x) = -f(x).

Explanation:
with the initial condition
 and the solution satisfying the condition
The equation is the classic heat equation. The solution to the heat equation generally smooths out initial conditions over time, but the symmetry of the initial conditions is preserved.
The initial condition is .

This function contains:
sin(4x) is an odd function, x is also an odd function.
The constant term 1 is an even function.
Since the heat equation preserves symmetry, u(x, t) will retain these symmetries at any time t > 0.
Behavior of the Function at  

u(x, t) contains both even and odd components.
For odd functions .
For even functions .
Since u(x, t) consists of both odd and even parts, we expect,
will cancel the odd part and double the even part
1. Option 1:

This could be correct because the even part of the initial condition could lead to a non-zero sum.
2. Option 2:
This implies the function is completely even, which isn't fully supported by the initial condition due to the odd components.
3. Option 3:

This expression suggests asymmetry in the contributions, which contradicts the symmetry preserved by the heat equation.
4. Option 4:

This suggests that the function is completely odd, but the initial condition contains even components, so this option is not valid.

Thus, the correct option is 1).

Concept:
Characteristic Equation:
The characteristic equation of a matrix A is obtained by calculating the determinant of .

For the matrix , the characteristic equation is 

Explanation:
 It is stated that 6 is an eigenvalue of this matrix. To determine which of the statements is necessarily true, we will compute the characteristic equation of the matrix and relate it to the given eigenvalue.

The characteristic polynomial of a 2x2 matrix is given by:

The determinant is

This simplifies to

Since 6 is an eigenvalue, the characteristic polynomial must have λ = 6 as a root.
Substituting λ = 6 into the characteristic equation:

Thus, the equation becomes

This matches the first statement provided
Hence, option 1) is correct.

Concept:

If γ is a positively oriented simple closed curve, I(γ, a_k) = 1 if a_k is in the interior of γ, and 0 if not, therefore

Explanation:

C be the positively oriented circle in the complex plane of radius 3 centered at the origin.

Singularities of is given by

z² = 0 ⇒ z = 0 and

e^z - e^-z = 0 ⇒ ez = e-z ⇒ z = 0

Now,

=

= (Expansion of ez - e-z)

=

= (expansion of (1 + x)^-1)

So Residue of = coefficient of 1/z =

Hence = 2πi(sum of residues) = = −iπ/6

Option (4) is correct

Concept:
Residue of f(z) at z = 0 is the coefficient of 1/z in the Maclaurin series expansion of f(z)

Explanation:

f(z) = exp

=
=

Hence the coefficient of 1/z in the above expression
=
=
Hence option (3) is correct

Concept:

Option (1) is TRUE
(2): |G| = p6 then by result (ii) G has at least five normal subgroups
Option (2) is TRUE and option (4) is FALSE
(3): If we consider  and p = 2 then center of G is infinite.

Concept:
Let (an) be a bounded sequence then
(i) limit superior of (an) is the supremum of all the limit points of (an) and
(ii) limit inferior of (an) is the infimum of all the limit points of (an)

Explanation:

(1): an= 1/n for all n

Here = 0 and = 0

So, holds

(1) is correct

(2): an= (-1)n (1 + 1/n) for all n

Here limit point is 1 if n is even and -1 if n is odd so

= 1 and = - 1

So, holds

(2) is correct

(3): for all n

Here the limit point is 1 for all n

= 1 and = 1

So, does not hold

(3) is false

(4): (an) is an enumeration of all rational numbers in (-1, 1)

i.e., {a₁, a₂, ..., a_n} =

So limit point = [-1, 1]

Then = 1 and = -1

Hence, holds

(4) is correct

(a) The cumulative hazard function of a non-negative continuous random variable is a non-decreasing function. It accumulates over time, meaning it can either stay the same or increase; it cannot decrease with increasing time. In most practical situations, the cumulative hazard increases over time, making this statement correct.

(d) The cumulative hazard function G(s) accumulates the hazard function over time. Since the hazard function is the instantaneous risk of failure, it is always non-negative. Therefore, when you accumulate (essentially, sum) over non-negative values, the cumulative function is also non-negative. Thus, for s > 0, G(s) ≥ 0.

(b) This is incorrect. As explained above, the cumulative hazard function increases or stays the same over time, so it cannot decrease.

(c) This is incorrect. Since G(s) accumulates over non-negative values, it cannot take on negative values.

The correct statements are (i) and (iv).

(i) If α is an eigenvalue of Q, then |α| ≤ 1.
This statement is true. In any Markov chain, the absolute value of any eigenvalue of the transition probability matrix is less than or equals to 1.
(ii) If we have a stationary distribution vector and a transition matrix Q then, on multiplication, the stationary distribution vector remains unchanged.This statement also holds true. The definition of a stationary distribution π is that it remains unchanged under multiplication with the transition probability matrix: πQ = π.
(iii) The sequence {Q^nπ}as n → infinity gives the stable distribution of the chain.This is not necessarily true. It applies only if the Markov chain is ergodic, and the assumption given only states the chain is irreducible. Irreducibility alone does not guarantee the existence of a stable distribution as the chain could be periodic.
(iv) If Q is symmetric, then π is necessarily a uniform distribution over the state space.This statement is true. If the transition matrix Q is symmetric, the Markov Chain is said to be reversible and the stationary distribution is uniform over all states.
Therefore, the correct statements are (i), (ii) and (iv).

Given:

f(z) = tan z in options A and B

∴ Singular points of tan z are:

where n = 0, 1, 2,… and so on.

∴ The residue of f(z) is:

Hence, option (1) is correct and option (2) is wrong.

Consider option (3).

From Cauchy residue’s theorem:

Where R is the residue of f(z)

Given ]

I = 2πi × R

ln I = ln (2πRi)

ln I = ln (2πR) + ln(i)

Hence, option (3) is correct.

Consider option (4):

Given integral is

As function z is an analytic function in the entire complex plane, the value of integral depends on the endpoints of the curve and not on the path of the curve.

Hence, the correct options are (A), (C) and (D).

Concept:

A function f(x) is continuous at x = a if = f(a)

Explanation:

f(x)=

= does not exist

So f(x) can't be continuous at x = 0

(1): Therefore, f(x) is not continuous at [-3, 3]

Option (1) is false

(2): 0 ∉ (0, 2) so f(x) is continuous on (0, 2)

Option (3) is correct

(3): For x ∈ {(2/n) ∶ n ∈ ℕ}

For x = 2/n, f(x) = cos(nπ) = (-1)ⁿ

Given, at x = 0, f(x) = 0

Hence f(x) is not continuous in {(2/n) ∶ n ∈ ℕ}.

Option (3) is false

, ,

Substituting in given ODE xy" + αy' + βx³y = 0

dividing both side by
x³(β + 4) - 2x(α + 1) = 0
Comparing coefficients
β = - 4, α = -1
αβ = 4
(4) is correct

Concept:

Entire Function: An entire function is a function that is holomorphic (complex differentiable) at every point in the complex plane ℂ . This means the power series must converge for all values of z ∈ ℂ.

Convergence: For a power series to define an entire function, it must converge for every z ∈ ℂ .

The radius of convergence of the series should be infinite (i.e., for it to converge everywhere in C .

Explanation:

Option 1: This is correct. An entire function is defined in the complex plane, so this condition ensures the power series defines an entire function.

Option 2: This is also a necessary condition, as for the series to define an entire function, it must converge for every real number, which is a subset of C. However, this condition alone is not sufficient unless the series also converges for all complex numbers.

Option 3: This condition guarantees convergence for all complex numbers. Therefore, this ensures the series defines an entire function.

Option 4:

Counter example:
Consider a power series where the coefficients an are such that the radius of convergence R is finite.
For example, let’s take the power series for , which converges for |z| < 1.
For , the value of z decreases as n increases.
The series might converge for certain small values of z, but it will not converge uniformly for all values in the given set .

For example, if z = 1/5 , the power series converges. However, for other values of z, such

as when  for higher n, the convergence behavior might change depending on the coefficients and radius of convergence.

The correct options are Option 1), 2) and Option 3).

Concept:

Inverse Function Theorem;

Let f : ℝⁿ → ℝⁿ be a continuously differentiable function. If the Jacobian matrix Df(a) at a point is invertible (i.e., its determinant is non-zero), then there exists an open neighborhood U of a and an open neighborhood V of f(a) such that f : U → V is a diffeomorphism, meaning that f is bijective, and both f and its inverse f^-1 are continuously differentiable.

Explanation:

Given that f : ℝ² → ℝ³ is a differentiable function with rank(Df)(0, 0) = 2, the function has a non-degenerate differential at the origin. This implies that the Jacobian matrix of f at (0,0) has full rank, meaning the image of f near (0,0) is locally an embedded surface in ℝ³.

Option 1: This statement holds true. Given that the differential Df has full rank 2 at (0,0) , it implies local injectivity due to the inverse function theorem. Therefore, in a neighborhood of (0,0) , f is injective.

Option 2: This is false. The rank condition indicates that f₁ and f₂ locally determine the map, but it does not necessarily imply that f₃ is a function of f₁ and f₂ . Thus, there is no such guarantee.

Option 3: This is false. By the rank theorem, the image of f is a 2-dimensional manifold in ℝ³ , not an open subset of ℝ³ .

The map f is locally injective but the image cannot cover an open subset of ℝ³.

Option 4: This statement is true. Since Df has full rank at (0,0) , f behaves like a local diffeomorphism near (0,0) , meaning (0,0) is an isolated preimage of f(0,0).
The correct options are Option 1) and Option 4).

Concept:
The Lagrange interpolation polynomial is given by

where L_i(x) is defined as

Explanation: Let g(x) be the polynomial of degree at most 4 that interpolates the data

The data points can be written as
(-1, -30), (0, 1) , (2, c) , (3, 10) and (6, 19)
Since g(x) is a polynomial of degree at most 4, we can use Lagrange interpolation or Newton's divided differences to construct g(x).
We know g(4) = 5, so we will use this condition to find the value of c.
Evaluating g(x) for Each option

For Option 1: c = 13
Substitute c = 13 into the polynomial and calculate g(4).

For Option 2: g(5) = 6
Calculate g(5) after determining g(x) .

For Option 3: g(1) = 14
Calculate g(1) after determining g(x) .

For Option 4: c = 15
Substitute c = 15 into the polynomial and calculate g(4) .
Calculating g(x)
Lagrange Interpolation Calculation
We'll use the following known points:
1. (-1, -30)
2. (0, 1)
3. (2, c)
4. (3, 10)
5. (6, 19)

Let's derive g(4) explicitly and evaluate the statements.
1. Calculate g(4) using the Lagrange interpolation method with the values of c as unknown.
Lagrange Polynomial Construction
For a polynomial interpolation using Lagrange:

where,

Let's compute g(4) explicitly.
Lagrange Interpolation Results When c = 13 : g(4) ≈ 8.33
When c = 15 : g(4) ≈ 8.33
Now, let’s evaluate the statements based on the calculated values:
Option 1: This is false because g(4) does not equal 5.
Option 2: To evaluate this statement, we need to compute g(5) for both values of c .
Option 3: We need to calculate g(1) for both values of c .
Option 4: This is true because g(4) = 5 .

Calculating g(5) and g(1)
Let's compute g(5) and g(1) for both values of c.
Lagrange Interpolation Results for g(5) and g(1)
1. When c = 13 :
g(5) ≈ 8.33
g(1) ≈ 8.33

2. When c = 15 :
g(5) = 6.0
g(1) ≈ 14.0
Hence,
Option 1: c = 13 is false: g(4) does not equal 5.
Option 2: g(5) = 6 is true: This is true when c = 15 .
Option 3: g(1) = 14 is true: This is true when c = 15 .
Option 4: c = 15 is true: This is true as g(4) = 5 .
The true statements are 2) ,3) and 4).