CSIR NET Mathematical Science Mock Test - 4 - UGC NET MCQ

Given:
A 360 ml aqueous solution contains 40% alcohol.And 3600 ml of water is added to the solution

Concept:
Apply concept of percentage.

Calculation:

Now,

The percentage of Alcohal in new solution

Hence the option (2) is correct.

Given:
On a track of 200 m length, S runs from the starting point and R starts 20 m ahead of S at the same time. Both reach the end of the track at the same time. S runs at a uniform speed of 10 m/s. If R also runs at a uniform speed,

Concept:

Calculation:
Total distance = 200 m
speed of S = 10 m/sec
Then Time taken by S = 200/10=20s
In 20 sec R cover only 180 m
then speed of R = 180/20 = 9 m/sec
Hence the extra time taken by R =20/9 = 2.2 sec

Hence the option (4) is correct,.

Given:
Consider a right angled triangle BAC with medians CM and BL having the same length.

Concept:
SAS similarity : If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

Calculation:
Consider a right angled triangle BAC with medians CM and BL having the same length.
In given figure,

AM / AB = AL / AC = 1 / 2
By SAS similarity,
△ MAL ≅ △ BAC
Then
ML / BC = AM / AB = AL / AC = 1 / 2
⟹ BC / ML = 2
Hence the option (1) is correct.

Let C be the total number of teachers who teach chemistry and P be the total number of teachers who teach physics.

Given -
Every second teacher who teaches chemistry also teaches physics, so the number of teachers who teach both subjects from the chemistry group is C/2.
Every third teacher who teaches physics also teaches chemistry, so the number of teachers who teach both subjects from the physics group is P/3.

Let x represent the number of teachers who teach both chemistry and physics. Since these two groups overlap, the number of teachers who teach both chemistry and physics must be the same in both cases.
Therefore, we can set up the equation:
C/2 = P/3

Cross-multiply to find the relationship between C and P :
3C = 2P

⇒ C = 2/3 P

Now, the number of teachers who only teach chemistry is the total number of chemistry teachers minus those who teach both subjects, which is:
Teachers who only teach chemistry = C - C/2 = C/2

Similarly, the number of teachers who only teach physics is:

Teachers who only teach physics = P - P/3 = 2P/3

To find the ratio of teachers who only teach chemistry to those who only teach physics, we calculate:

Teachers who only teach chemistry / Teachers who only teach physics =
Substitute C = into the equation:

=
Thus, the ratio of teachers who only teach chemistry to those who only teach physics is 1/2.

When tossing 4 coins, each coin has 2 possible outcomes (heads or tails). Therefore, the total number of possible outcomes is 2⁴ = 16 outcomes.
A favorable outcome occurs when the number of heads is greater than the number of tails.
In 4 coin tosses, we need 3 heads and 1 tail, or 4 heads and 0 tails.
The number of ways to choose 3 heads from 4 tosses is given by and there is only 1 way to get 4 heads, i.e., HHHH.
Thus, the total number of favorable outcomes is 4 (for 3 heads, 1 tail)+1 (for 4 heads) = 5 favorable outcomes. 
The probability is the ratio of favorable outcomes to total outcomes P(Favorable) = 5/16. 
The probability of a favorable outcome is Option 3).

The length of the road = 350 m

Each cycle consists of a painted line of 3.5 m and a gap of 3.5 m.
Therefore, the length of one full cycle is 3.5 m + 3.5 m = 7 m
Therefore, number of cycles = 350 / 7 = 50 cycles
Since, each cycle has a painted line of 3.5 m, and there are three painted lines on the road to make 4 lane.
Total painted line length = 50 × 3.5 × 3 = 150 × 3.5 = 525
Thus, the total length of the painted lines over the 350 m stretch of road is 525 meters.
Hence Option (3) is correct.

Since AB and DE are perpendicular to AE so AB and DE are perpendicular to BD
Hence ∠CBD = 90^∘ - 30^∘ = 60^∘
similarly, ∠CDB = 90∘ - 30∘ = 60∘
We know that sum of angles of triangle is 180∘
So, ∠BCD = 180∘ - (60∘ + 60∘) = 60∘
Hence ΔBCD is equiangle triangle so equilateral triangle,
Now AE||BD so AE = BD = 15 cm
So BC= CD = BD = 15 cm
Also given AB = DE = 15 cm
hence perimeter of ABCDE = 15 + 15 + 15 + 15 + 15 = 75 cm
Option (2) is correct

Concept:

The extremal of the functional

I(y) = , y(x₁) = y₁, y(x₂) = y₂ is give by the Euler equation

= 0

Explanation:

I(y) = , y(0) = 1, y(1) = 2

Here F = , so using Euler's equation

= 0

⇒ 0 - = 0

⇒ y'' = 0

Integrating

y = ax + b

y(0) = 1 ⇒ b = 1

y(1) = 2 ⇒ a + b = 2 ⇒ a + 1 = 2 ⇒ a = 1

So, extremal is

y = x + 1

(1) is true.

By using a Binomial expansion we get,

and Hence ,
So ut = f'(x + t) -g'(x - t) - 3t²
Applying Initial Conditions ,
u(x,0) = sin x
we get f(x) + g(x) + e^x = sin x ...........(i)
and ut(x, 0) = 0 ⇒ f'(x) - g'(x) = 0 ..........(ii)
From (ii), f'(x) = g'(x) and by integrating we get f(x) = g(x)
Putiing this in (i) we get

u(x, t) = π /4 - e^π /2
Therefore, Correct Option is Option 3).

Concept:

The order of permutation group S_n is n!

Explanation:

Order of S₇ is 7!.

One element in S₇ is (1 2 3 4 5 6 7) which is of order 7.

Option (2) is false

(1 2 3 4 5 6)(7) ∈ S₇ which is of order lcm(6, 1) = 6

So, option (1) is false

(1 2) (3 4 5 6 7) ∈ S7 which is of order lcm(2, 5) = 10

So, option (4) is false

But we will not get any element of order 8 in S7

Hence option (3) is true

Concept:

Rank-nullity theorem: Let T: V → W be a linear map rank(T) + nullity(T) = dim V.

Explanation:

T : ℝ3 → ℝ3 is the linear transformation defined by T(x, y, z) = (x + y, y + z, z + x) for all (x, y, z) ∈ ℝ3.

Standard basis of ℝ3 is {(1, 0, 0), (0, 1, 0), (0, 0, 1)}

T(1, 0, 0) = (1, 0, 1)

T(0, 1, 0) = (1, 1, 0)

T(0, 0, 1) = (0, 1, 1)

Hence matrix representation of T is

T =

det(T) = = 1(1 - 0) + 1(1 - 0) = 2 ≠ 0

So, rank(T) = 3 and nullity(T) = 3 - 3 = 0

Option (4) is true

Concept:

1. Power Set: The power set of a set S, denoted P(S), is the set of all subsets of S.
If S has n elements, then (S) contains 2ⁿ elements.
2. Countably Infinite Set: A set is countably infinite if its elements can be put into a one-to-one correspondence with the natural numbers (i.e., it has the same cardinality as N).
3. Uncountably Infinite Set: A set is uncountable if it is not countably infinite (for example, the real numbers R).
4. Power Set and Cardinality: If the power set (S) is countably infinite, then S cannot be finite.
This is because for any finite set S, its power set (S) has 2ⁿ elements, where n is the number of elements in S, and 2ⁿ is always finite. Additionally, if S is uncountably infinite, then its power set P(S) will be uncountably infinite.

Explanation:
Option 1: This cannot be true because if S is finite, its power set will also be finite, not countably infinite.
Option 2: This is incorrect option because if S is any countably infinite set then its power set must be uncountable.
Option 3: This cannot be true because if S were uncountable, its power set would be uncountable as well.
Option 4: This is true, as there is no countably infinite set whose power set is countably infinite, so C is empty.
The correct option is 4).

Concept:
If and  are cyclic groups, the number of group homomorphisms from to  is given by
Number of homomorphisms = gcd(n, m), where gcd(n, m) is the greatest common divisor of n and m.

Explanation:
The number of group homomorphisms from is given by the greatest common divisor (gcd) of the orders of the two groups. That is Number of homomorphisms = gcd(150, 90)
Prime factorization of 150:
Prime factorization of 90:
Now, the gcd of 150 and 90 is the product of the lowest powers of the common prime factors:

The number of group homomorphisms from  is 30.
Thus, option 1) is correct.

Concept:
The supremum sup A of the set is the least upper bound for x. The infimum inf A is the greatest lower bound of x

Explanation:
We are given the set } as a subset of , and we need to determine the correct supremum (sup) and infimum (inf) of this set.

We are asked to solve for x in the inequality .

Multiply the numerator and denominator by

So, the inequality becomes
The set A consists of rational numbers  that satisfy this inequality. Therefore,
The supremum sup A of this set is , which is the least upper bound for x.
The infimum inf A is 0 , since x approaches but never reaches 0 from the positive side.
The correct statement is sup A = 3 + 2√2
Hence option 2) is correct.

Concept:
Conditional Probability:
For two events A and B, the conditional probability of A occurring given that B has occurred is denoted as

, provided that P(B) > 0.
P(A₃∣A₂)
Explanation:

Option 1:
Counter example:



Left-hand side:  0.625 × 0.6667 = 0.4167 
Right-hand side: 0.1667
Here, the left-hand side 0.4167 is greater than the right-hand side 0.1667, which violates the inequality.
Thus, this provides a counter example where

So, option 1) is false.
Option 2:
Let us define three events in a probability space with the following probabilities:

Left-hand side:
Right-hand side:
Hence,
So option 2) is false.
Option 3:
This statement looks plausible because the probability of the union of two events is generally less than or equal to the sum of the individual probabilities. This would make the product of the conditional probabilities potentially greater than or equal to the union probability. Hence, this option is true.

Option 4:


Left-hand side:
Right-hand side:
The correct answer is Option 3).

Concept:
Translation of Functions: The solution represents a translation  of the function along the x-axis as time t progresses. This is important when analyzing the properties of the solution because many properties, such as integrability and the Lebesgue measure of sets, are preserved under translation.

Lebesgue Measure: The Lebesgue measure of a set is a way of assigning a "size" to the set.

The key property relevant here is that the Lebesgue measure of a set is invariant under translation.
In other words, if a set  is translated by some fixed amount, its measure remains the same.

Explanation:

where  is an arbitrary C¹ (continuously differentiable) function. Two statements S₁ and S₂ are given about the solution u(x, t), and we are to determine whether both statements are true or false.
This is a first-order linear PDE. The standard method to solve this type of equation is by using the method of characteristics. The PDE has the following characteristic equations

This tells us that the solution is constant along the lines x − 2024t = constant, meaning the solution takes the form:
Thus, the solution to the PDE is
Statement S₁:

If and |A_t| denotes the Lebesgue measure of , then for every , .
The solution to the PDE is , which means that u(x, t) is a translation of the initial condition.
Translation does not change the Lebesgue measure of a set. Therefore, if we define the set A_t as, the measure |A_t| will remain the same for all t ≥ 0 because the function is simply a translated version of u₀(x).
Thus, Statement S₁ is true.

Statement S₂:
If u₀ is Lebesgue integrable, then for every t ≥ 0, the function u(x, t) is Lebesgue integrable.
The solution is a translation of the initial condition u₀(x).
The translation of a Lebesgue integrable function is still Lebesgue integrable.
Therefore, if u₀(x) is integrable, u(x, t) will also be integrable for all t ≥ 0.
Thus, Statement S₂ is also true.
Both statements S₁ and S₂ are true.
Therefore, the correct option is Option 1).

Concept:

Zero of a Quadratic Form:
In mathematical terms, a quadratic form is an expression of the form:
 where a_ij are constants and x_i, x_j are variables. The zeros of the quadratic form are the values of the variables (x₁,x₂,…,x_n) that make Q(x₁,x₂,…,x_n) = 0

Explanation:

Option 1: x² + 2y² + 3z² 
This is a sum of squares. Since the terms are all positive for any nonzero values of x, y, z this quadratic form cannot be zero unless x = y = z = 0. Therefore, this form does not have any zeros other than (0, 0, 0).

Option 2: x² + 2y² + 3z² − 2xy
This includes the cross term −2xy. We would need to test if there are any nontrivial solutions to this form being zero. However, the squared terms dominate, and the cross term alone is unlikely to reduce the entire expression to zero unless x = y = z = 0 .
Thus, no nontrivial zeros exist.

Option 3: x² + 2y² + 3z² − 2xy−2yz
This has two cross terms −2xy − 2and −2yz. Even with the cross terms, the squared terms are positive, and reducing the expression to zero would still likely require x = y = z = 0. Therefore, no nontrivial zeros are expected.

Option 4: x² + 2y² − 3z²
This is different because it contains both positive and negative squared terms. Specifically, z² has a negative coefficient, which can allow for cancellation between the positive and negative terms. It is possible for some combination of nonzero x, y, z values to satisfy the equation x² + 2y² − 3z² = 0. For example, take x = 0, y = 0 and z = 0. Then, f(0, 0, z) = 0 when z = 0  , providing a solution other than (0, 0, 0).
Thus, Option 4) is correct.

Concept:

Leibnitz rule for differentiation:

=

Explanation:

(1):

So
y(x) = 2 K(x, t) y(t) dt
⇒ y(x) = 2 t(1 - x)ydt + 2 x(1 - t)ydt....(i)
⇒ y' = 2(-t)y(t)dt + x(1 - x)y(x) - 0 + 21(1-t)y(t)dt + 0 -x(1 - x)y(x).1
⇒ y' = 2(-t)y(t)dt + 2(1 - t)y(t)dt
⇒ y'' = 2[0 - xy(x).1 - 0] + 2[0 + 0 -(1 - x)y(x)]
⇒ y'' = -2y(x)
⇒ y''(x) + 2y(x) = 0
Also by (i)
y(0) = 2 t(1 - 0)ydt + 2 0(1 - t)ydt = 0 and
y(1) = 2 t(1 - 1)ydt + 2 1(1 - t)ydt = 0

hence satisfies

Therefore option (1) is correct

Concept:

(a) Let V be a vector space. A function β : V × V → ℝ, usually denoted β(x, y) = , is called an inner product on V if it is positive, symmetric, and bilinear. That is, if
(i) <x, y> ≥ 0, <x, x> = 0 only for x = x

(ii) <x, y> = <y, x>
(iii) <rx, y> = r<x, y> 
(iv) <x + y, z> = <x, z> + <y, z>
(b) If Q = ax₁y₁ + bx₁y₂ + cx₂y₁ + dx₂y₂ be a quadratic form then the necessary condition of being it inner product space is a > 0, d > 0 and ad - bc > 0
i.e., if these condition does not satisfy then it will not be an inner product space.

Explanation:

x = (x1, …, xn) and y = (y1, …, yn) ∈ ℝn
For n = 2

(1)  = x1y1 + x1y2 + x2y1 + x2y2

So, A =

Here a > 0, d > 0 but ad - bc = 1 - 1 = 0

So it does not define inner product space.

Option (1) is false

(4)

So, A =

Here a = 0, d = 0 but ad - bc = 0 - 1 = -1 < 0

So it does not define inner product space.

Option (4) is false

(2)

=

=

Consider x = (1,1) and y = (1,1)

then <2x,y> = 2(2)² + 2 (2)² + 2 (1)² + 2(1)² = 20

But 2 = 2( 2 (1)2 + 2(1)2 + 2 (1)2 + 2(1)2 ) = 16

Thus <2x,y> ≠ 2 and so is not an inner product.

Option (2) is false

Hence option (3) is true.

maximize x + 3y, subject to A ≤ b,

where A = and b =

So constrains are
-x - y ≤ -1...(i)
y ≤ 5...(ii)
-x + y ≤ 5...(iii)
x + 2y ≤ 14...(iv)
-y ≤ 0 ...(v)

Let z = x + 3y....(vi)
We know that maximum solution always satisfy all the constrains.

(1): The objective function attains its maximum at in the feasible region.
Putting x = 0, y = 5 we can verify that it satisfy all the constraints from (i) to (v)
-0 - 5 < -1 so (i) satisfies. Similarly all constrains satisfies.
Here z = 0 + 3 × 5 = 15

So it can be a maximum solution.

(2): The objective function attains its maximum at in the feasible region.
Putting x = -2, y = 3 we can verify that it satisfy all the constraints from (i) to (v)
Here z = -2 + 3 × 3 = 7, which can't be maximum solution as 15 > 7
So option (2) can't be true

(3): The objective function attains its maximum at in the feasible region.
Putting x = 1, y = 0 we can verify that it satisfy all the constraints from (i) to (v)
Here z = 1 + 3 × 0 = 1, which can't be maximum solution as 15 > 1
So option (3) can't be true

(4): Putting x = 14, y = 0 we can verify that it satisfy all the constraints from (i) to (v)
Here z = 14 + 3 × 0 = 14, which can't be maximum solution as 15 > 14
Hence The objective function does not attain its maximum at in the feasible region.

So option (4) is correct option

Concept:

Leibnitz rule of differentiation

+ -

Explanation:

y(x) =

Differentiating both sides

y'(x) = 0 + .1 - 0

⇒ y'(x) =

⇒ y'(1) = e

Option (4) is true.

(i) By definition of independence between two events, the probability of the intersection of E and F equals the product of the probabilities of E and F.
Hence, ℙ(E ∩ F) = ℙ(E) x ℙ(F).

(ii) This statement is known as the inclusion-exclusion principle for two events. It's correct irrespective of whether the events E and F are independent or not.

For independent events, it simplifies to this form because ℙ(E ∩ F) = ℙ(E) x ℙ(F).

(iii) This statement is in general true, but in this problem statement, it is given that E and F are independent. So, this does not apply. Note: If two events were dependent, then the conditional probability of the events could differ from their marginal probabilities, which could validate this statement. However, in our case, it contradicts the given information of independence.

(iv) This statement is false. Actually, if ℙ(E | F) = ℙ(E), then E and F are independent. It's based on the definition of independence, where the conditional probability of an event given another event is simply the probability of the event itself.

The correct options are (i) and (ii).

a) This is incorrect. The sum of a normal and a Bernoulli distributed variable does not follow a normal distribution. As an intuition, the Bernoulli variable introduces a discontinuity (it can only be 0 or 1), which is not characteristic of normal distributions.

b) This is incorrect. The mean of a sum of independent random variables is the sum of their means. Therefore, the mean of W is E[U] + E[V] = 0 + 0.5 = 0.5.

c) This is incorrect.

The variance of a sum of independent random variables is the sum of their variances. V being a Bernoulli variable, its variance is (0.5)(1 - 0.5) = 0.25.

Hence, the variance of W is 1 (variance of U) + 0.25 (variance of V) = 1.25. The standard deviation is then the square root of the variance, √(1.25) = 1.12 (approximately).

d) This is correct. The probability distribution of the sum of two or more independent random variables is the convolution of their individual distributions.

S1: ecos(t) = e2022sin(t)
Taking log both sides
⇒ logecos(t) = log e2022sin(t)
⇒ cos t = 2022 sin t
⇒ tan t = 1/2022
⇒ t = 0.00049455 ∈ (0, π)
Hence ecos(t) = e2022sin(t) for t = 0.00049455 ∈ (0, π)
S1 is false
S2: x = loge (1 + xet)
⇒ e^x = 1 + xet
⇒ ex - 1 = xet

Since log function is positive and 
so t ∈ (0, x)
S2 is true
S3: We know that
|sin x| ≤ |x| for all x
and exponential function is increasing function so
e|sin (x)| ≤ e|x| for all x ∈ ( -1, 1)
S3 is true
(4) is correct

Option 1, 2, and 3 are correct.

From the option, we can see that we are only concerned with function values at 0 ≤ x < 1 and 1 ≤ x < 3.

Let us find f'(x) at this range,

Let, f'(x) = g(x)

To find the differentiability of g(x), we need to find g'(x)

g'(1⁺) =-4, g'(1^-) = 2

∴ g'(1⁺) ≠ g'(1^-)

∴ g(x) is NOT differentiable at x = 1

To find maxima and minima,

Let us try to draw f'(x)

2x - 1 = 0 is a straight line.

Now, for f'(x) = 2x2 - 8x + 7

4x - 8 = 0, x = 2

∴ we have minima at x = 2

From,

∴ we have local minima at x = 2 and local maxima at x = 1

Concept:

Leibniz's rule of differentiation:

Explanation:

f(x) = k(x, y) ϕ (y) dy

Differentiating both sides using the Leibnize rule of differentiation

f'(x) = k(x, y)ϕ(y)dy + k(x, x)ϕ(x).1 - 0

f'(x) = k(x, y)ϕ(y)dy + k(x, x)ϕ(x) ....(i)

If we compare (i) with

ϕ(x) + G(x, y) ϕ (y)dy = g(x)....(ii)

we get k(x, x) = 1, g = f' and G(x, y) =

Option (1) is correct

Also dividing both side of (i) by k(x, x) weget

= ϕ (y)dy

so comparing with (ii) we get
G(x, y) = and

Option (3) is correct

Explanation:

T2 = λT for some λ ∈ ℝ

So T satisfies the polynomial t2 - λt = t(t - λ)

minimal polynomial of T is t or t - λ or t(t - λ)

if minimal polynomial is t then T = 0. In particular, T is singular. (2) is true by vacuity.

(3) is false as T satisfies T2 = kT with k≠0, and we may not be free to choose k ; however, if T=λI for some λ then it would be true with λ = 0 only

is true only for |λ| = 1

So (1) false

Concept:

The sum of squares for the error α can be calculated as

Total Sum of Squares = Sum of Squares for Treatments+Sum of Squares for Blocks + Sum of Squares for Error

Explanation:
Total Sum of Squares:
The total sum of squares is given as 144.
The sum of squares for the treatments is 48, and the sum of squares for the blocks is 72.
The sum of squares for the error α can be calculated as:
Total Sum of Squares = Sum of Squares for Treatments + Sum of Squares for Blocks + Sum of Squares for Error
Substituting the known values:
144 = 48 + 72 + α
⇒ α = 144 − 48 − 72 = 24
Thus, α = 24.
The total degrees of freedom are given as 19.
The degrees of freedom for treatments are 4, and for blocks, they are 3.
The degrees of freedom for error m can be calculated as
Total Degrees of Freedom = Degrees of Freedom for + Degrees of Freedom for Blocks + Degrees of Freedom for Error
Substituting the known values,
19 = 4 + 3 + m
⇒ m = 19 - 4 - 3 = 12
Thus, m = 12.
Mean Squares:
The mean square for treatments is already given as 12.
The mean square for blocks is given as 24.
The mean square for error γ is calculated as:

Thus, γ = 2.
F-Ratio for Treatments: The F-ratio for treatments β is calculated as

Thus, β = 6, α = 24
β = 6, m = 12 and γ = 2
Thus, the correct options 2) and 4).

Concept:

The right-hand derivative (for x > 0) at x = 0 is
The left-hand derivative (for x > 0) at x = 0 is

Explanation:
The function f(x) = x|x| can be rewritten as:

Both pieces are continuous (since x² and -x² are continuous functions), and at x = 0 , the function evaluates to 0 from both sides. Therefore, f(x) is continuous for all real values of x .

Option 1: f is continuous on  is true
Now, let's check the differentiability.
For x > 0 , the function is f(x) = x², which is differentiable with derivative f'(x) = 2x .
For x < 0 , the function is f(x) = −x², which is differentiable with derivative f'(x) = -2x .
At x = 0 , we need to check the left-hand derivative and right-hand derivative:

The right-hand derivative (for x > 0) at x = 0 is

The left-hand derivative (for x < 0) at x = 0 is .
Since both the left-hand and right-hand derivatives at x = 0 are 0, the function is differentiable at x = 0 , and the derivative at x = 0 is 0.
Thus, the function is differentiable everywhere on .
Option 2: f is differentiable on  is true
Option 3: f is differentiable only at 0 is False (it is differentiable everywhere)
Option 4: f is not differentiable at 0 is False (it is differentiable at 0)
Hence, correct options are 1) and 2).