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

Given:

In two cylinders,

Ratio in radii of two cylinders = 4:7

Ratio in their heights = 7:5

Formula Used:

Surface area of the first cylinder =2πr₁h1

Surface area of the second cylinder =2πr₂h₂

Calculation:

Let r1 and r2 be the radii and h1, h2 be their heights

respectively.

⇒ r₁:r₂ = 4:7 and h₁: h₂ = 7:5

⇒ r1/r2 = 4 / 7 and h1/h2 = 7 / 5

⇒The ratio of surface area of the first and second cylinder is

⇒ 2πr₁h₁ / 2πr₂h₂

⇒r₁h₁ / r₂h₂

⇒4 × 7 / 7 × 5

⇒4 / 5

∴ The ratio of curved surface area of cylinder B to that of A = 5 : 4.

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.

Concept:

Uniform Distribution: In a uniform distribution, all outcomes within a given range are equally likely. Here, the length

of the fish is uniformly distributed between 2 and 4 inches, meaning every length in that range has the same probability.

Explanation:
The length of bristlemouth fish is uniformly distributed between 2 and 4 inches.
This means every length between 2 and 4 inches has an equal probability of being selected.
The total range of fish lengths is from 2 to 4 inches, so the range is 4 - 2 = 2 inches.
The range of fish lengths that are 3 inches or longer is from 3 to 4 inches. So the favorable range is 4 - 3 = 1 inch.
The probability of catching a single fish that is 3 inches or longer is the ratio of the favorable range to the total range
The probability of not catching a fish that is 3 inches or longer (i.e., catching a fish shorter ' than 3 inches) is P(Fish < 3  inches) = 1/2 = 0.5
If a fisherman catches 5 fish, the probability that none of the 5 fish are 3 inches or longer is P(None of the 5 fish ≥ 3 inches) = (0.5)⁵ = 0.03125
The probability that at least one fish is 3 inches or longer is the complement of the probability that none of the fish are 3 inches or longer = P(At least one fish ≥ 3 inches)=1−0.03125 = 0.96875
The probability that at least one of the 5 bristlemouth fish is 3 inches or longer is 0.96875.
Thus, the correct answer is Option 4).

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.

Concept -
Using Antilog Table

Explanation -
We have log₁₀X = 1 ⇒ X = 10
log10Y = 0.4 ⇒ Y = 2.5119
log10Y = 1.2 ⇒ Y = 15.849

For Option (4) -
At X = 10 the figure gives you approx 2.5 & 18.

Hence Option(4) is false.
For Option (3) -
At X = 10 the figure gives you approx 5 & 10.

Hence Option(3) is false.
For Option (2) -
At X = 10 the figure gives you approx 3.5 & 18.

Hence Option(2) is false.
For Option (1) -
At X = 10 the figure gives you approx 2.5 & 16.

Hence Option(1) is true.

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

Concept Used:

Rank: Rank of a matrix A, denoted as rank(A), is the maximum number of linearly independent rows or columns in A.

Singular Matrix: A singular matrix is a square matrix that is not invertible. In other words, a matrix is singular if its determinant is zero.

Invertible Matrix: A square matrix A is invertible if there exists another square matrix B (called the inverse of A), such that AB=BA=I, where I is the identity matrix.

Note: A matrix that is not invertible cannot have full rank.

Explanation:

If rank(AB) = 1,

AB has only one linear independent row or column, thus its determinant is 0 i.e., det(AB) = 0

⇒ det(A) ⋅ det(B) = 0

⇒ det(B) ⋅ det(A) = 0

⇒ det(BA) = 0

⇒ BA is a singular matrix, which is not invertible.

Thus, rank(BA) ≠ 3 [As only invertible matrix can have full rank]

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.

Concept:
If a function is continuous on a dense set S, it doesn't necessarily imply that the function is continuous on all of  , especially on , the complement of S in  .

Explanation:
Option 1: Continuity on a dense subset S does not imply continuity on the whole set R.
A function can be continuous on a dense subset but exhibit discontinuities on .
Therefore, this option is incorrect.

Option 2: g is defined only on S, so even if g is continuous on S, it says nothing about f's continuity on the rest of R. Continuity of g does not guarantee the continuity of f everywhere.
Hence, this option is incorrect.

Option 3: If for all (which is dense in R), and f is continuous on , by the density of S, f must be 0 everywhere on R, because a continuous function on a dense set that is 0 must be 0 on the entire set. Therefore, this option is correct.
Option 4: being identically 0 on S and f being continuous on S) does not imply f is identically 0 on . Continuity on S does not extend to the whole set without further conditions.
Therefore, this option is incorrect.
The correct answer is Option 3).

Concept:
The argument of $$a a can be any real number, as it represents the angle that the line connecting the origin to the point $$a
$\mathrm{makes\; with\; the\; positive\; real\; axis.}$

Explanation:
Given, and is a complex number.
We are to find which statement is true.

Option 1: This implies that for a complex number z, if is less than 1, the modulus of is less than . To verify if this is true, we can try specific examples for z and a under the condition that |z| < 1 and .
Let and First, check the moduli:

This satisfies the condition and |z| < 1 .

Clearly, , so the inequality does not hold for this example.
The statement is false because we found a counterexample where || is not less than |z - a| .

Option2: This implies that if the modulus of z - a is equal to 1 - , then = 1 . This looks like a symmetry condition related to distances in the complex plane. For points on the unit circle (i.e., |z| = 1 ), this is true because the modulus of z would satisfy this equation. Therefore, this statement is true.

Option 3: This implies that for a point on the unit circle (i.e., |z| = 1 ), the modulus of z - a is less than |1 - | .
To verify this, we will try specific examples for z and a under the condition that |z| = 1 and .

Let z = 1 (since |z| = 1 ) and

Since z = 1 and (because a is real),

So the inequality does not hold in this example because both sides are equal.

Therefore, the statement is not true.

Option 4: This suggests that if the modulus of 1 - is less than z - a, then z is inside the unit circle.

Let z = 1/2 and a = 1/3.
Since a = 1/3 , we have (because is real).


In this case, and .

Clearly, , so the condition does not hold for this example, and hence it does not verify the statement.
The statement seems to not hold.
The option 2) is correct.

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:

The solution of utt − c²uxx = 0 with initial condition, u(0, t) = u(L, t) = 0 for all t and boundary condition y(x, 0) = f(x), y_t(x, 0) = g(x) for 0 < x < L is

u = [f(x + ct) + f(x - ct)] + (D'alembert solution)

Explanation:

Given

utt − uxx = 0, 0 < x < 2, t > 0

u(0, t) = 0 = u(2, t), ∀ t > 0,

u(x, 0) = sin(πx) + 2sin(2πx), 0 ≤ x ≤ 2,

ut(x, 0) = 0, 0 ≤ x ≤ 2.

Here f(x) = sin(πx) + 2sin(2πx) and g(x) = 0, c = 1

So u(x, t) = [f(x + t) + f(x - t)] +

= [sin(π(x+t)) + sin(π(x-t))] + sin(2π(x+t)) + sin(2π(x-t))

So u(1, 1) = (sin 2π + sin 0)+ sin4π + sin 0 = 0

Option (1) is false

u(1/2, 1) = [sin (3π/2) - sin(π/2)] + sin(3π) - sin π = - 1

Option (2) is false

u(1/2, 2) = [sin (5π/2) - sin(3π/2)] + sin(5π) - sin 3π = 1

Option (3) is correct

Also

ut(x, t) = [cos(π(x+t)) - cos(π(x-t))] + 2π cos(2π(x+t)) + 2π cos(2π(x-t))

So ut(1/2, 1/2) = [cos π - cos0] + 2π cos 2π + 2π cos0 = - π + 2π + 2π = 3π

Option (4) is false

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

For option (1):
Any Hausdorff topology on a finite set is discrete. So it cannot have 3 different topologies.

For option (2):
Since X is finite and Hausdorff. So X × X is also finite and Hausdorff. So X × X has discrete topology. Hence every function f : X × X → ℝ is continuous.

For option (3):
Since f: ℝ → (X, discrete) is continuous.
So f is constant map and hence gof has any one element.

For option (4):
Suppose A is finite and dense i.e. A̅ = X. Now since A is finite set in a metric space.
⇒ A is closed set i.e. A̅ = A
⇒A = X so X is finite metric space and hence topology on X is discrete. So every f : X → ℝ is continuous.

Hence option (2) is true.

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

The radial part of the Lagrangian reads as follows:

The Euler-Lagrange equation for the coordinate r gives the equation of motion:

which when plugged into the Lagrangian gives:

Therefore, the correct answer is (i).

(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).

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

x² - x + 1 = 0 ⇒ x = = = ω, ω², where ω = is cube root of unity

So x² - x + 1 = (x - ω)(x - ω²)

Hence splitting field of the polynomial x² - x + 1 are Q(ω), Q(ω²)

(1), (2) & (3) are correct

Concept :

The order of an element in a direct product of a finite number of finite groups is the lcm of the orders of the components of the element.

Suppose φ is an isomorphism from a group G onto a group G'.Then G is Abelian iff G' is Abelian.

Calculation :
G = z₃⊕z₃⊕z₃
|G|=27
The possible order of elements in z₃ is 1 and 3, so the possible order of elements in G is 1 and 3.
All the non-identity elements are of order 3.
Thus number of elements of order 3 in G is 26.
|H| = 27
Every non-identity element in H is also of order 3 since,

Thus, number of elements of order 3 in H are 26.
We know that direct sum is abelian iff are abelian.
Thus, G is abelian.
But H is not abelian. (Take a = 1, c=1 in one element and a=1,c=0 in another element and check they do not commute.)
So G and H are not isomorphic.
Thus options (1), (2) and (4) are correct.

Concept:
Abel's test: Suppose the following statements are true:
∑a_n is a convergent series,
{bn} is a monotone sequence, and
{bn} is bounded.
Then ∑a_nb_n is also convergent.

Explanation:

By Abel's test

converges on -1 < x ≤ 1

∑un is uniformly convergent

is also convergent uniformly

∑v_n is also convergent uniformly

(1), (2), (4) 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:

Explanation:

We are given the function , defined as

We are tasked with determining the truth of various statements about the partial derivatives, continuity, and differentiability of this function at (0, 0) .

Option 1: To check whether the partial derivative with respect to x exists at (0, 0), we calculate it as follows

When y = 0 , the function reduces to

Thus,

Hence, exists and is equal to 0.

Option 2: We calculate the partial derivative with respect to y at (0, 0) using the limit definition:

When x = 0 , the function is defined as f(0, y) = 0 . Thus,

Therefore, also exists and is equal to 0.

Option 3: To check continuity at (0, 0) , we evaluate the limit
For , . Along the path y = x , we have:

As , . Along the path , we have

As x →0, this also tends to 0. However, the function is 0 when x = 0 , and the limit approaching from different paths gives inconsistent behavior, indicating that the function is not continuous at (0, 0) .
Thus, f is not continuous at (0, 0) .

Option 4: A function is differentiable at (0, 0) if it is continuous and its partial derivatives exist and are continuous in a neighborhood of (0, 0) . Since the function is not continuous at (0, 0) , it cannot be differentiable there. Thus, f is not differentiable at (0, 0) .
Hence correct options are 1), 2), 3) and 4).

Concept:

Bounded Sequence: A bounded sequence means that all the elements of the sequence lie within some fixed interval [-M, M] for some positive M. This ensures that the sequence does not diverge to infinity or negative infinity.

Non-existence of a Limit: The condition implies that the sequence oscillates between multiple values and does not converge to a single value.

Bolzano - Weierstrass Theorem: This theorem states that every bounded sequence has a convergent subsequence.
Therefore, there exist subsequences of (a_n) that converge to different limit points.

Structure of Set S: The set S contains all the limit points of the subsequences of (a_n).

Explanation:

Option 1: This cannot be true because, by the Bolzano - Weierstrass theorem, there must be at least one convergent subsequence. Therefore, S contains at least one element.
False.

Option 2: This cannot be true because if the sequence does not converge, it must oscillate, which implies that there are multiple limit points. Hence, S will contain more than one element.
False.

Option 3: Since the sequence does not converge to a single value, there must be at least two distinct subsequential limits. Therefore, S must have at least two elements.
True.

Option 4: This is not necessarily true because a bounded sequence can have an infinite number of subsequential limits. For example, the sequence could oscillate between an infinite set of values.
False.

The correct answer is 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:

Intermediate Value Theorem:
Let f : [a, b] →  :  be a continuous function on the closed interval [a, b]. If N is any value between f(a) and f(b) (i.e., f(a) < N < f(b) or f(b) < N < f(a)), then there exists at least one c∈ (a, b)  such that f(c) = N

Explanation:

Let  be non-empty and f : K → K be continuous such that |x − y| ≤ |f(x) − f(y)| for all x, y ∈ K.
Let's evaluate each statement based on the properties of the function f given the condition |x − y| ≤ |f(x) − f(y)|.

Option 1:The condition |x − y| ≤ |f(x) − f(y)| guarantees that  f is injective (as we can show |f(x₁) − f(x₂)| = 0 implies x₁ = x₂. However, it does not ensure that f covers all points in K. Thus, f can be injective without being surjective.

Option 2: If K = [0, 1] and f is continuous and injective (from the earlier point), then by the Intermediate Value Theorem, f must map K [0, 1] onto itself, implying that f must be surjective.

Option 3: Since f is continuous and injective, we can deduce that f^-1 is continuous as well, provided that f(K) is compact. This is a consequence of the property that continuous injective functions on compact intervals have continuous inverses.

Option 4: Since we have established that f^-1 is continuous when f is continuous and injective on a compact interval K.

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