CUET PG Mathematics Mock Test - 1 - CUET PG MCQ

Prime factorization of 36:
36 = 2² • 3²
Use the formula for Euler's totient function:


There are 12 generators of  
These are the integers k in {1, 2, ..., 35} such that gcd (k, 36) = 1
Hence Option (2) is the correct answer.

Given :-
w = u(x, y) + iv(x, y) is an analytic function of z = x + iy.
Concept used :-
If w is an analytic function then dw/dz = 0
Solution :-
As we know that
∂w/∂z = 1/2 (∂w/∂x − i ∂w/∂y) and ∂w/∂z̄ = 1/2 (∂w/∂x + i ∂w/∂y)
Subtracting both equations and putting dw/dz = 0, we get
dw/dz = −i ∂w/∂y

Concept:

Let f be a continuous function such that f '(p) = 0

• If f ''(p) > 0 then f has a local minimum at p.

• If f ''(p) < 0 then f has a local maximum at p.

Calculation:

f (x) = 3x4 + 4x³ - 12x2 + 12

⇒ f ' (x) = 12x³ + 12x² - 24x + 0 ----(1)

⇒ f ' (x) = 12x (x² + x - 2)

⇒ f ' (x) = 12x (x - 1)(x + 2)

Putting f ' (x) = 0

⇒ 12x (x - 1)(x + 2) = 0

⇒ x = 0, 1, -2 are the critical point

Finding f '' (x),

⇒ f '' (x) = 36x² + 24x - 24 [using (1)]

⇒ f '' (x) = 12 (3x² + 2x - 2)

Case 1: At x = 0,

f '' (x) = 12 (3(0)2 + 2(0) - 2)

⇒ f '' (x) = 12 (-2) = -24 < 0

Since, f '' (x) < 0 at x = 0

∴ x = 0 is the point of local maxima

Thus, f(x) is maximum at x = 0.

Case 2: At x = 1

f '' (x) = 12 (3(1)2 + 2(1) - 2)

⇒ f '' (x) = 12 (3 + 2 - 2) = 36 > 0

Since, f '' (x) > 0 at x = 1

∴ x = 1 is the point of local minima

Thus, f(x) is minimum at x = 1.

Case 3: At x = -2

f '' (x) = 12 (3(-2)2 + 2(-2) - 2)

⇒ f '' (x) = 12 (12 - 4 - 2) = 72 > 0

Since, f '' (x) > 0 at x = -2

∴ x = -2 is the point of local minima

Thus, f(x) is minimum at x = -2.

Hence, at point x = 0, f(x) is maximum

Concept:
(i) Every convergent sequence is bounded.
Explanation:
{xn} is a convergent sequence in ℝ. So it is bounded.
Then there exists a real number M such that |x_n| ≤ M.
{yn} is a bounded sequence in ℝ
Then there exists a real number L such that |yn| ≤ L.
Now, |xn + yn| ≤ |xn| + |yn| ≤ M + L
So, {xn + yn} is bounded.
Option (2) is true.
Let {xn} = {1/n} and {yn} = {(-1)ⁿ} then {xn} is a convergent sequence in ℝ and {yn} is a bounded sequence in ℝ.
But {xn + yn} = {1/n + (-1)n} which is not convergent and it has convergent and bounded subsequence.
Options (1), (3) and (4) are false

Concept:

Rolle's Theorem states:

If a function f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b),

and if f(a) = f(b), then there exists at least one point c in the open interval (a, b) such that f'(c) = 0

Explanation:

1. f(x) = sin(x), x ∈ [0, 2π]

Continuity: sin(x) is continuous everywhere, so it's continuous on [0, 2π]

Differentiability: sin(x) is differentiable everywhere, so it's differentiable on (0, 2π)

f(a) = f(b): f(0) = sin(0) = 0 and f(2π) = sin(2π) = 0, so f(a) = f(b)

Therefore, f(x) = sin(x) satisfies Rolle's Theorem.

2. f(x) = |x|, x ∈ [-1, 1]

Continuity: |x| is continuous everywhere, so it's continuous on [-1, 1]

Differentiability: |x| is not differentiable at x = 0. Therefore, it is not differentiable on the open interval (-1, 1)

Therefore, f(x) = |x| does not satisfy Rolle's Theorem.

3. f(x) = |x - 1|, x ∈ [-2, 2]

Continuity: |x - 1| is continuous everywhere, so it's continuous on [-2, 2].

Differentiability: |x - 1| is not differentiable at x = 1

Therefore, it is not differentiable on the open interval (-2, 2)

Therefore, f(x) = |x - 1| does not satisfy Rolle's Theorem

4. f(x) = 1/x, x ∈ [-1, 1]

Continuity: 1/x is not continuous at x = 0, so it's not continuous on [-1, 1]

Therefore, f(x) = 1/x does not satisfy Rolle's Theorem

Only f(x) = sin(x) satisfies all the conditions of Rolle's Theorem

Hence Option(1) is the correct answer.

Explanation -

(i) While some prime ideals may also be maximal, this is not the case for all prime ideals. There can be prime ideals which are not maximal. For instance, in the ring of integers, the ideal generated by the number 2 is a prime ideal but not a maximal ideal since it is contained in several larger ideals (like the ideal generated by the number 1).

Statement (1) is not always true

(ii) Every maximal ideal is indeed a prime ideal in these rings, since if an ideal is large enough that it cannot be properly contained in another ideal, it must also satisfy the property that if a product of two elements lies in the ideal then at least one of the factors is in the ideal..

Statement (ii) is true

(iii) There exist ideals which are neither prime nor maximal. For example, in the ring of integers, the ideal generated by the number 4 is not a prime ideal.

Statement (c) is not true.

(iv) As said above, ideals can exist which are not maximal. For example, in the ring of integers, the ideal generated by the number 2 is not a maximal ideal.

Statement (d) is not true.

We are given the set of vectors in :

Since S contains 5 vectors in , and the maximum number of linearly independent vectors in is 3, the set must be linearly dependent.
Thus, (A) is correct.
Choosing the Vectors v₁ = (1, 0, 0), v₂ = (0, 1, 0), v₃ = (1, 1, 0):
a(1,0,0) + b(0,1,0) + c(1,1,0) = (0,0,0)
a + c = 0 ⇒ c = -a
b + c = 0 ⇒ c = -b
Third component : 0 = 0 (always true)
Since we found c = -a and c = -b , we get a = b ,
and we can choose nonzero values for a, b, c such that this equation holds, meaning these three vectors are linearly dependent.
Thus, (B) is incorrect
Since S is a set of 5 vectors in  , any 4 vectors must be linearly dependent because the rank of the matrix formed by these vectors is at most 3
Thus, (C) is correct
Hence Option (3) is the correct answer.

The Graph of the region bounded by the curves y = ex and x = 1 in the first quadrant

x = 0 and x = 1
then y = 0 and y = ex
A =
A = e -1
Hence Option (4) is the correct answer.

Express the vector v as a linear combination of the vectors u₁, u₂, and u₃:
v = c₁u₁ + c₂u₂ + c₃u₃
(4, 9, 19) = c₁(1, -2, 3) + c₂(3, -7, 10) + c₃(2, 1, 9)
This gives us the following system of linear equations:
c₁ + 3c₂ + 2c₃ = 4 -------(1)
-2c₁ - 7c₂ + c₃ = 9 ------(2)
3c₁ + 10c₂ + 9c₃ = 19 ----(3)
Solve this system of equations :
Multiply equation (1) by 2 and add it to equation (2):
2(c₁ + 3c₂ + 2c₃) + (-2c₁ - 7c₂ + c₃) = 2(4) + 9
⇒ 2c₁ + 6c₂ + 4c₃ - 2c₁ - 7c₂ + c₃ = 8 + 9
⇒ -c₂ + 5c₃ = 17 --- (4)
Multiply equation (1) by -3 and add it to equation (3):
-3(c₁ + 3c₂ + 2c₃) + (3c₁ + 10c₂ + 9c₃) = -3(4) + 19
⇒ -3c₁ - 9c₂ - 6c₃ + 3c₁ + 10c₂ + 9c₃ = -12 + 19
⇒ c₂ + 3c₃ = 7 --- (5)
Now add equation (4) and equation (5):
(-c₂ + 5c₃) + (c₂ + 3c₃) = 17 + 7
⇒ 8c₃ = 24
⇒ c₃ = 3
Substitute c₃ = 3 into equation (5):
c₂ + 3(3) = 7
⇒ c₂ + 9 = 7
⇒ c₂ = -2
Substitute c₂ = -2 and c₃ = 3 into equation (1):
c₁ + 3(-2) + 2(3) = 4
⇒ c₁ - 6 + 6 = 4
⇒ c₁ = 4
So, we have c₁ = 4, c₂ = -2, and c₃ = 3.
Therefore, v = 4u₁ - 2u₂ + 3u₃
Hence Option (2) is the correct answer.

Also (AB)^T=B^TA^T

Taking transpose of XTCX:

i.e. ⇒ Let XTCX = A

Transpose of A ⇒ A^T = (X^TCX)^T

= X^TC^T(X^T)^T (using (AB)T=BTAT)

= X^TC^TX

= X^T(-C)X = - X^TCX = - A

X^TCX + X^TCX = 0

2X^TCX = 0

X^TCX = 0

Therefore, XTCX is a null matrix.

Concept:
f(p,q) = 0;
Calculation:
The given PDE is ∂z/∂y = ln(∂z/∂x)
Writing in terms of p and q,
⇒ q = ln (p)
⇒ p = e^q;
Let p = a,
⇒ q = ln a;
The complete solution will be
z = ax + y ln a + c where a and c are arbitrary constants.

Dirichlet Conditions in Fourier Transformation are as follows:

• f(x) must absolutely integrable over a period, i.e., ∫ from t₁ to ∞ |x(t)| dt < ∞
• f(x) must have a finite number of extrema in any given interval, i.e. there must be a finite number of maxima and minima in the interval.
• f(x) must have a finite number of discontinues in any given interval, however, the discontinuity cannot be infinite.
• f(x) must be bounded.

The Gamma function is defined as:

t = x²
dt = 2x dx
dx = dt / (2√t)
Our integral becomes:


This integral now matches the form of the Gamma function with z = 1/2:

Γ(1/2) = √π
Therefore,
Hence, option (3) is the correct answer.

An element x of a ring R is called nilpotent if there exists some positive integer n, called the index (or sometimes the degree), such that
xⁿ = 0.
i.e., an element α of a ring is said to be nilpotent if αm = 0 for some m > 1
(3) is correct

Here: uₙ = (-1)ⁿ⁻¹ (1+2+3+...+n) / (n+1)³
= (-1)ⁿ⁻¹ n(n+1) / 2(n+1)³
= (-1)ⁿ⁻¹ n / 2(n+1)²
Let:
aₙ = n / 2(n+1)²
aₙ₊₁ = 1/2 [n / (n+1)² - (n+1) / (n+2)²]
= 1/2 [n²-1 / (n+1)²(n+2)²] > 0
aₙ₊₁ < aₙ
lt n → ∞ aₙ = 1/2 lt n → ∞ n / (n+1)² = 0
Thus, by Leibnitz's rule, ∑aₙ and therefore ∑uₙ is convergent.
|uₙ| = 1/2 n / (n²+1)
Taking vₙ = 1/n
lt n → ∞ |uₙ| / vₙ = 1/2 lt n → ∞ n² / (n²+1) = 1/2 ≠ 0
Since ∑vₙ is divergent, therefore ∑|uₙ| is also divergent.
Thus, the given series ∑uₙ is conditionally convergent.

Statement (A): If the real part of f(z) is constant, then f(z) is a constant function
If the real part u(x, y) is constant, then its partial derivatives u_x = 0 and u_y = 0
From the Cauchy-Riemann equations:
u_x = v_y, u_y = -v_x
Since u_x = 0 and u_y = 0, it follows that v_y = 0 and v_x = 0, meaning the imaginary part v(x, y) is also constant
If both u and v are constant, then f(z) is a constant function
(A) is correct
Statement (B): If |f(z)| is a nonzero constant in D , then f(z) is a constant function in D
If |f(z)| is constant, then f(z) lies on a circle of fixed radius in the complex plane
By the open mapping theorem, a non-constant analytic function is an open map, meaning it would map open sets to open sets, which contradicts the assumption that |f(z)| is constant
Hence, f(z) must be constant (B) is correct
Statement (C): If f'(z) = 0 everywhere in D , then f(z) is a constant function in D
If f'(z) = 0 for all z in D , then the function is locally constant everywhere
This means f(z) does not change, implying it is a constant function (C) is correct
Statement (D): If |f(z)| is a nonzero constant in D , then f(z) is constant only for some z in D
This statement contradicts (B), which we already proved as true
If |f(z)| is constant, then f(z) must be constant everywhere, not just for some z (D) is incorrect
The correct statements are (A), (B), and (C)
Hence Option (1) is correct

Explanation:

[A] is a square matrix, real and symmetric.

Characteristic equation,

(A - λI) = 0 …………(i)

Solving for λ, eigen value

Let symmetric and real matrix A

…………(ii)

⇒ (a - λ)² – b² = 0

⇒ (a - λ)² = b²

⇒ (a - λ) = ± b

⇒ λ = a ± b

We are given the differential equation:
3xy + y² + (x² + xy)dy/dx = 0
(3xy + y²) + (x² + xy) dy/dx = 0
(x² + xy) dy/dx + (3xy + y²) = 0
(x² + xy) dy + (3xy + y²) dx = 0
(3xy + y²) dx + (x² + xy)dy = 0
which is of the form:
M(x, y) dx+ N(x, y) dy = 0
where: M = 3xy + y² and N = x² + xy
A differential equation is exact if:



Since , the equation is not exact, and we need an integrating factor (IF)
An integrating factor μ(x, y) makes the equation exact by multiplying both M and N by μ(x, y)
(A) x as an Integrating Factor:
Multiplying by x :
x(3xy + y²)dx + x(x² + xy)dy = 0
New M' = x(3xy + y²) = 3x²y + xy² 
New N' = x(x² + xy) = x³ + x²y 
Check exactness:


⇒ x is an integrating factor
(B) x² as an Integrating Factor:
Multiplying by x² :
x²(3xy + y²)dx + x²(x² + xy)dy = 0
New M' = x²(3xy + y²) = 3x³y + x²y² 
New N' = x²(x² + xy) = x⁴ + x³y'
Check exactness:


⇒ x² is not an integrating factor
(C) 3x as an Integrating Factor:
Multiplying by 3x :
3x(3xy + y²)dx + 3x(x² + xy)dy = 0
New M' = 3x(3xy + y²) = 9x²y + 3xy²
New N' = 3x(x² + xy) = 3x³ + 3x²y 
Check exactness:


⇒ 3x is not an integrating factor
(D) as an Integrating Factor:
Multiplying by :



We now check if the equation becomes exact, meaning:

which confirms exactness
Thus, is a valid integrating factor
⇒ (A), (C), and (D) are correct only
Hence Option (2) is the correct answer.

Concept:
L-Hospital Rule: Let f(x) and g(x) be two functions
Suppose that we have one of the following cases,
I. lim (x→a) f(x) / g(x) = 0/0
II. lim (x→a) f(x) / g(x) = ∞/∞
Then we can apply L-Hospital Rule ⇔ lim (x→a) f(x) / g(x) = lim (x→a) f'(x) / g'(x)
Calculation:
lim (n→∞) (n² / 2ⁿ)
Apply L'Hospital;
lim (n→∞) (2n / 2ⁿ log 2)
Again apply L'Hospital rule;
lim (n→∞) (2 / 2ⁿ (log 2)²)
= 2/∞ = 0

Concept:
1. Loxodromic Transformation:
A Möbius transformation with two distinct, non-real fixed points
It exhibits rotation and scaling simultaneously
It maps circles and lines into spirals or other geometric shapes with a constant angle of intersection
2. Hyperbolic Transformation:
A Möbius transformation with two distinct real fixed points
It maps lines and circles to hyperbolas
The transformation involves scaling along the real axis and rotation in the complex plane
Explanation:
1.
This is a Möbius transformation of the form , where a = 3i , b = 4 , c = 1 , and d = -i
To determine the nature of the transformation, we examine the fixed points and their properties
The fixed points of a Möbius transformation can be found by setting w = z :

Multiplying both sides by z - i , we get:
z(z - i) = 3iz + 4
⇒ z² - iz = 3iz + 4 
⇒ z² - 4iz - 4 = 0 
This is a quadratic equation, and solving for z , we can determine that w₁ has two distinct fixed points
Given that w₁ has two distinct fixed points and the transformation involves rotation and scaling, w₁ is a loxodromic transformation
2.
This is another Möbius transformation
Finding Fixed Points:
We set w₂ = z:

Multiplying both sides by z - 7 , we get:
z(z - 7) = z
⇒ z² - 7z = z
⇒ z² - 8z = 0 
⇒ z(z - 8) = 0
Thus, the fixed points are z = 0 and z = 8 , both real and distinct
Since w₂ has two distinct real fixed points, w₂ is a hyperbolic transformation
Conclusion:
w₁ is loxodromic, as it has two distinct fixed points and exhibits both scaling and rotation
w₂ is hyperbolic, as it has two distinct real fixed points and maps lines to hyperbolas

We can answer this through counter-examples.

• To discard Option 1 and Option 3 we take R = (Z, +, .) where pZ is maximal for any prime p.
• To discard Option 4 we take R = (F,+, .), where F is a field. And, a field is a commutative ring with unity where {0} is the only prime ideal.
• So, Option 2 is the only correct option. In fact, R = (Z₆, +₆, ·₆) has exactly two maximal ideals, M = {0,2,4} and N = {0,3}

Order: The highest derivative present is the second derivative (d²y/dx²)
⇒ it's a second-order differential equation.
Linearity: A differential equation is linear if it can be expressed in the form:

where , and f(x) are functions of x only.
In our equation, the term -2tan(x + y) (dy/dx) involves both the dependent variable (y) and its derivative (dy/dx)
⇒ the equation is non-linear
Homogeneity:
A differential equation is homogeneous if f(x) = 0
In our equation, , which is not zero
⇒ it's a non-homogeneous equation
The given differential equation is second-order . It is non-linear. It is non-homogeneous
Hence Option (4) is the correct answer.

f(x) = (1/2) [ (x³ / 3) ]² | from 0 to 2
f(x) = (1/2) ( 2³ / 3 - 0 )
f(x) = 4/3

Statement (A):
If P^-1 XP  is a diagonal matrix for some invertible matrix P, then X has a basis of eigenvectors 
P^-1 XP being diagonal means X can be written in diagonal form
A matrix is diagonalizable if and only if there exists a basis of eigenvectors
Since P is invertible, the columns of P are eigenvectors of X
So (A) is correct
Statement (B):
If X is a diagonal matrix with different diagonal values and XY = YX , then Y must also be diagonal
Let's say X is:

where (diagonal values are different)
If XY = YX , it means Y must be forced to have zeroes outside the diagonal, otherwise multiplication would not work properly
So Y must also be diagonal
So (B) is correct
Statement (C):
If X² is diagonal, then X must be diagonal
Let's take an example:

Here, X² = 0, which is diagonal
But X itself is not diagonal
So (C) is incorrect
Statement (D):
If X is diagonal and it commutes with every matrix Y , then X must be a multiple of the identity matrix
If X is diagonal but not a multiple of I , there will be some matrices Y for which 
Only scalar multiples of identity X = λI commute with all matrices
So (D) is correct
Correct statements are (A), (B), and (D)
Hence Option (1) is the correct answer

Concept:
The order of a differential equation is determined by the number of independent arbitrary constants present in its general solution
Explanation :
A. y = mx -----------(1)
differentiating with respect to x
dy/dx = m = y/x
⇒ x dy = y dx
⇒ x dy - y dx = 0
Correct match for A is II
(B) (x - a)² + 2y² = a², a is an arbitrary constant
Differentiate with respect to x:
2(x - a) + 4y(dy/dx) = 0
x - a = -2y(dy/dx)
a = x + 2y(dy/dx)
Substitute a back into the original equation:
(x - (x + 2y(dy/dx)))² + 2y² = (x + 2y(dy/dx))²
(-2y(dy/dx))² + 2y² = x² + 4xy(dy/dx) + 4y²(dy/dx)²
4y²(dy/dx)² + 2y² = x² + 4xy(dy/dx) + 4y²(dy/dx)²
2y² = x² + 4xy(dy/dx)
2y² - x² = 4xy(dy/dx)
which matches (I)
(C) y² = 4ax, a is an arbitrary constant
Differentiate with respect to x:
2y(dy/dx) = 4a
y(dy/dx) = 2a
a = (y/2)(dy/dx)
Substitute a back into the original equation:
y² = 4x(y/2)(dy/dx)
y² = 2xy(dy/dx)
y = 2x(dy/dx)
which matches (III)
(D) y = a cos(x + b), a and b are arbitrary constants
Differentiate with respect to x:
dy/dx = -a sin(x + b)
d²y/dx² = -a cos(x + b)
Since y = a cos(x + b), we have:
d²y/dx² = -y
d²y/dx² + y = 0
which matches (IV)
Matching Results:
(A) y = mx matches with (II) ydx - xdy = 0
(B) (x - a)² + 2y² = a² matches with (I) 2y² - x² = 4xy(dy/dx)
(C) y² = 4ax matches with (III) y² = 2xy(dy/dx)
(D) y = a cos(x + b) matches with (IV) d²y/dx² + y = 0
Therefore, the correct answer is: (A) - (II), (B) - (I), (C) - (III), (D) - (IV)
Hence Option (3) is the correct answer.

Statement A:
U = x² - y² is the real part of an analytic function f(z)
For a function to be analytic, it must satisfy the Cauchy-Riemann equations:

Here, u = x² - y², so:

Let v be the imaginary part of f(z) = u + iv
Solving , we get:
v = 2xy
Hence, f(z) = x² - y² + i(2xy) = z²
The function f(z) is not of the form f(z) = z + c , so Statement A is incorrect
Statement B:
The zeros of cosz occur when cosz = 0
For cosz = 0, we have:

This matches the given condition in the statement
Hence, Statement B is correct
Statement C:
If f(z) is entire and bounded for all z in the complex plane,
then by Liouville's theorem, f(z) must be a constant function
Hence, Statement C is correct
Statement D:
Evaluate the integral:

The integrand has a singularity at z = 0 and z = -1 . Only z = 0 lies within the contour |z| = 1
Using the residue theorem:
Residue at z = 0:
The integral equals 2πi x Residue = 2πi  x 1 = 2πi  
Hence, Statement D is incorrect
The correct statements are B and C Only
Hence Option (1) is the correct answer.

(A) A closed set either contains an interval or else is nowhere dense.
This statement is correct
This is a consequence of the perfect set theorem and the Baire category theorem
A closed set in a complete metric space (like the real numbers) can be decomposed into a perfect set (which contains an interval) and a nowhere dense set.
(B) The derived set of a set is closed
This statement is correct
The derived set (the set of limit points) of any set is always closed
(C) The union of an arbitrary family of closed sets is closed.
This statement is incorrect
The union of an arbitrary family of closed sets is not necessarily closed
For example, consider the closed intervals [-1/n, 1/n] for n = 1, 2, 3,
Each of these sets is closed, but their union is (-1, 1) which is open, not closed.
(D) The set R of real numbers is open as well as closed.
This statement is correct
In any topological space, the entire space and the empty set are always both open and closed
Therefore, the correct statements are (A), (B), and (D).
Hence Option (1) is correct.

Given that, AA^T = A^TA and B = A^-1 A^T

BB^T = (A^-1 A^T) (A^-1 A^T)^T

= A^-1 A^T (A^T)^T (A^-1)^T

= A^-1 (A^TA) (A^-1)^T ---------------(Matrix multiplication is associative)

= A^-1 A A^T (A^-1)^T ----------------(∵ AAT = AT A)

= (A-1 A) AT (A-1)^T ---------------(Matrix multiplication is associative)

= I A^T (A^-1)^T-----------------------(A-1 A = I)

= (A^-1 A)^T -------------------------(∵ (A-1 A)^T = AT (A-1)T )

= (I)^T = I

(A).Let G = <a> be a cyclic group of order n, then G = <aᵏ> if and only if gcd(k, n) = 1
This statement is TRUE
The element aᵏ generates the entire group G if and only if the greatest common divisor (gcd) of k and n is 1
This is a fundamental property of cyclic groups.
(B). Let G be a group and let a be an element of order n in G. If aᵏ = e, then n divides k.
This statement is TRUE
If aᵏ = e, it means that k is a multiple of the order of 'a' (which is n)
In other words, k = nr for some integer r
This is a direct definition of the order of an element.
(C). The center of a group G may not be a subgroup of the group G.
This statement is FALSE
The center of a group G, denoted as Z(G) = {z ∈ G | zg = gz for all g ∈ G}, is a subgroup of G
It always contains the identity element, is closed under the group operation, and is closed under taking inverses.
(D). For each 'a' in a group G, the centralizer of 'a' is a subgroup of group G.
This statement is TRUE
The centralizer of 'a' in G, denoted as C(a) = {x ∈ G | xa = ax}, is the set of all elements in G that commute with 'a'
It is always a subgroup of G
It contains the identity, is closed under the group operation, and is closed under taking inverses.
Therefore, the correct statements are (A), (B), and (D)
Hence Option (1) is the correct answer.

Concept:
Homogeneous system AX = 0
Non – homogeneous solution AX  0
Let A is an m × n matrix

‘m’ = number of equations
‘n'= number of unknowns
Trivial solution (or) zero solution (X) = 0
Non – trivial solution (X)   0
Note:
1. Every homogeneous equation is always consistent with the trivial solution.
2. A non – trivial solution may or may not exist. But if exists infinite number only.
Linear dependence: Two vectors X, Y are said to be linearly dependent if X = α Y or Y=α X. Otherwise they are linearly independent.
Null-space:
The nullspace of the matrix A is denoted by N(A). It is the set of all n-dimensional column vectors x such that Ax = 0.
Nullity is defined as the dimension of null space
= number of linearly independent solutions
Nullity = no.of unknowns – the rank of A = n – r