CUET PG Mathematics Mock Test - 3 - CUET PG MCQ

The transformation given is defined by:

T(x, y, z) = (x, y, 0)

Let α and β be scalars and (x₁, y₁, z₁) and (x₂, y₂, z₂) be two vectors. Then

T [α(x₁, y₁, z₁) + β(x₁, y₂, z₂)]

= T[(αx₁ + βx₂, αy₁ + βy₂, αz₁ + βz₂)]

= (αx₁ + βx₂, αy₁ + βy₂, 0)

= α (x₁ + y₁, 0) + β(x₂ + y₂, 0)

or equivalently

= αT(x₁, y₁, z₁) + βT (x₂, y₂, z₂)

= T[α(x₁, y₁, z₁) + β(x₂, y₂, z₂)]

= αT(x₁, y₁, z₁) + βT(x₂, y₂, z₂)

Rank:

The rank of a matrix is a number equal to the order of the highest order non-vanishing minor, that can be formed from the matrix.

For matrix A, it is denoted by ρ(A).

The rank of a matrix is said to be r if:

1) There is at least one non-zero minor of order r.

2) Every minor of matrix A having order higher than r is zero.

Property of Rank of Matrix:

​ρ(AB) ≤ min [ρ(A), ρ(B)]

Calculation:

Here A and B are 3 × 3 real matrices such that rank (AB) = 1

The determinant will therefore be 0, i.e.

|AB| = 0

Since |AB| = |A| |B|

|A| |B| = 0

Either |A| or |B| should be zero

|BA| = |B| |A| = 0

∴ BA is singular

Hence rank (BA) cannot be 3. (Because BA is 3 × 3 matrix)

A.

P.I = (e^x)/16
(A) matches with (II) which is in only Option(2)
Hence Option(2) is the correct answer.
D. (D² - 3D - 4)y = 2e^-x.

Put -1 in place of D then denominator becomes zero
In this case multiply numerator by x and differentiate denominator with respect to D

Now put -1 in place of D
P.I = x/(-5) • 2e^-x,
P.I = ((-2)/5) xe^-x
(D) matches with (IV)
Hence Option (2) is the correct answer.

Concept:

The dimension of a vector space V is the cardinality (i.e., the number of vectors) of a basis of V over its base field.

Explanation:

Fibonacci-like sequence is

an = an-1 + an-2 for n ≥ 3

For n = 3

a3 = a2 + a1

a4 = a3 + a2

..................

So basis B = {(a₁, 0, 0, 0, ...), (0, a₂, 0, ...)}

Hence the dimension of the ℝ-vector space of Fibonacci-like sequences

= cardinality of basis = 2

(2) is correct

Let y = mx


it depends on m.
⇒ limit does not exists
Hence Option (4) is the correct answer.

Given:
sin z = 10
where z is a complex number.
We need to determine whether this equation has a unique solution, exactly two distinct complex solutions, infinitely many complex solutions, or no solution.
The sine function in the complex domain is defined as:

where z = x + iy with x and y being the real and imaginary parts, respectively.
For real numbers, the sine function has a range of [-1, 1].
However, in the complex domain, there is no such restriction,
and the sine function can take on any complex value, including values with real parts greater than 1.
Since we are given sin z = 10 ,
we can conclude that there are complex numbers z for which the sine function can be equal to 10.
The sine function in the complex plane is periodic, and it achieves every possible complex value infinitely many times.
Therefore, the equation sin z = 10 has infinitely many complex solutions.
The correct option is (Option 3) .

Concept:
A function is written in the form of the ratio of two polynomial functions is called a rational function.
Rational functions are continuous at all the points except for the points where the denominator becomes zero.
f(x) = P(x) / Q(x)
where P(x) and Q(x) are polynomials and Q(x) ≠ 0.
f(x) will be discontinuous at points where Q(x) = 0.
Calculation:
Given:
f(x) = (4 - x²) / (4x - x³)
This is a rational function, so it will be discontinuous at points where the denominator becomes zero.
4x - x³ = 0
x(4 - x²) = 0
x(2² - x2) = 0
x(2 + x)(2 - x) = 0
x = 0, x = - 2 and x = 2
Hence the function f(x) = (4 - x²) / (4x - x³) will be discontinuous at exactly three points 0, - 2 and 2.
Mistake Points
There may be a doubt that some factors are eliminating each other so first, we have to simplify this.
Note that,
If (4 - x2) = 0 then the f(x) will come indeterminant or 0/0 form.
So, the function will not have any value for x = ± 2. So, these will also be the points of discontinuity.
Also, if (4 - x2) ≠ 0
⇒ f(x) = (4 - x²) / (4x - x³) = 1/x
Here, x = 0 is also the point of discontinuity.
There will be exactly three points 0, - 2 and 2.

If f is continuous at x = 1 then :


and
Since
⇒ k = 4
Hence Option (4) is the correct answer.

Concept -

(i) n³ + 1 = (n+1)(n² + 1 - n)

(ii) and

(iii)

Explanation -

We have the series

=

=

=

=

=

= e + (e - 1) = 2e - 1

Hence the given series is convergent and cgs to finite limit 2e - 1.

Hence option(iii) is correct.

Concept:
The radius of convergence of the series
is
Explanation:
Here in the series
, and k = 2
The radius of convergence of the series is
1/R =
⇒ 1/R =
⇒ 1/R =
⇒ R = ∞
(2) is correct

Concept:

1. Determinant of a n x n Matrix is the Product of its Eigen values.
2. If Eigen-values of matrix A are a, b ,c then eigen values of its transpose are same and eigen values of A-1 are 1/a , 1/b ,1/c.

Explanation:
Let A is the 3 × 3 Matrix whose eigen values are 6, 5 and 2,
Eigen values of A-1 will be respectively.
Eigen values of transpose of A-1 will be .
Now,
Determinant of (A-1)T = Product of Its eigen values
=
= 1/60
= 0.016.
Hence Option(4) is the correct answer.

To determine which of the given sets is a subspace of ℝ³, we need to check if each set satisfies three conditions for being a subspace:

1. It contains the zero vector.
2. Closed under scalar multiplication: For any vector u in W and any real scalar c, c·u must be in W.
3. Closed under addition: For any vectors u, v in W, u + v must be in W.

Let’s analyze each option:

Option A: W = {(x, y, z) ∈ ℝ³ : x + 4y – 10z = –2}

• Zero vector (0, 0, 0) does not satisfy the equation 0 + 4(0) – 10(0) = –2 ⇒ 0 ≠ –2.
• Since W does not contain the zero vector, it cannot be a subspace.

Option B: W = {(x, y, z) ∈ ℝ³ : x·y = 0}

• Zero vector (0, 0, 0) satisfies 0·0 = 0.
• Closure under addition: Consider u = (1, 0, 0) and v = (0, 1, 0). Both satisfy x·y = 0 individually. However, u + v = (1, 1, 0), and 1·1 = 1 ≠ 0, so u + v is not in W.
• Therefore, W is not closed under addition and is not a subspace.

Option C: W = {(x, y, z) ∈ ℝ³ : 2x + 3y – 4z = 0}

1. Zero vector: (0, 0, 0) satisfies 2·0 + 3·0 – 4·0 = 0, so it is in W.
2. Closure under addition:
   – Let u = (x₁, y₁, z₁) and v = (x₂, y₂, z₂) be in W. Then 2x₁ + 3y₁ – 4z₁ = 0 and 2x₂ + 3y₂ – 4z₂ = 0.
   – Adding these, 2(x₁ + x₂) + 3(y₁ + y₂) – 4(z₁ + z₂) = 0, so u + v is in W.
3. Closure under scalar multiplication:
   – Let r be any real scalar. For u = (x, y, z) in W, 2x + 3y – 4z = 0.
   – Then for r·u = (r·x, r·y, r·z), 2(r·x) + 3(r·y) – 4(r·z) = r(2x + 3y – 4z) = r·0 = 0, so r·u is in W.

Hence, W in option C is a subspace of ℝ³.

Option D: W = {(x, y, z) ∈ ℝ³ : x ∈ ℚ}

• Zero vector (0, 0, 0) is in W since 0 is rational.
• Closure under scalar multiplication: Let u = (q, y, z) with q in ℚ, and let c be an irrational number (e.g., √2). Then c·u = (c·q, c·y, c·z). Because c·q is irrational if q ≠ 0, this new vector is not in W.
• Therefore, W is not closed under scalar multiplication and is not a subspace.

Conclusion: Only option C is a subspace of ℝ³.

(an)n ≥ 1,
aₙ = cos( (-1)ⁿ * (nπ/2) + (nπ/3) )
a₁ = cos( (-π/2) + (π/3) ) = sin( (π/3) ) = √3/2
a₂ = cos( (π + 2π/3) ) = cos( (π/3) ) = 1/2
a₃ = cos( (-3π/2 + π) ) = 0
a₄ = cos( (2π + 4π/3) ) = - cos( (π/3) ) = -1/2
a₅ = cos( (-5π/2 + 5π/3) ) = -√3/2
a₆ = -1
...
a₁₂ = 1
Hence limsup aₙ = 1
Option (1) is false.
a₂ₙ = cos( (nπ + 2nπ/3) ) = cos( (5nπ/3) )
a₁ = cos( (5π/3) ) = 1/2
a₂ = cos( (10π/3) ) = -1/2
a₃ = cos( (5π) ) = -1
a₄ = cos( (20π/3) ) = -1/2
a₅ = cos( (25π/3) ) = -1/2
a₆ = cos( (10π) ) = 1
lim sup a₂ₙ = 1
Option (2) is true and (3) is false.
a₃ₙ = cos( (-3π/2 + nπ) )
Similarly, we can show option (4) is false.

To find : Dimension of Solution Space for Differential Equation
Given: Consider the differential equation:

The question asks for the dimension of the solution space W .
Explanation:
The dimension of the solution space for a linear differential equation is determined by the order of the equation.
Here:

• The differential equation is third-order (the highest derivative is (d³y/dx³)).
• Therefore, the solution space will have a dimension equal to 3.

This means there are three linearly independent solutions to this third-order differential equation.
Then dimension of the solution space W is 3.
Hence Option (1) is the correct answer.

Concept:
1. Cauchy-Riemann (C-R) Equations:
The C-R equations are a pair of equations involving the partial derivatives of a complex function
f(z) = u(x, y) + iv(x, y), where z = x + i y.
The equations are:
∂u/∂x = ∂v/∂y
∂u/∂y = -∂v/∂x
2. Differentiability of a Complex Function:
A complex function f(z) is said to be differentiable at a point z₀ if the limit of the difference quotient exists as h → 0:

Connection between C-R Equations and Differentiability:
It turns out that the C-R equations are a necessary condition for a complex function to be differentiable.
In other words, if a function is differentiable at a point, its partial derivatives must satisfy the C-R equations at that point.
However, the C-R equations are not sufficient for differentiability.
We also need an additional condition known as the Cauchy-Riemann condition, which ensures that the partial derivatives are continuous.
Explanation:
Option 1: The C-R equations are only satisfied by constant functions:
This is not true.
Many non-constant functions, such as polynomials, exponential functions, and trigonometric functions, satisfy the C-R equations and are differentiable in their domains.
Option 3: The C-R equations are only used for real-valued functions:
The C-R equations are specifically for complex-valued functions.
Option 4 : The C-R equations are only applicable to polynomials: Again, the C-R equations apply to a broader class of functions, including analytic functions, which encompass polynomials but also many other functions.
Therefore, the only correct statement is:
C-R equations are a necessary condition for the differentiability of a complex function.
Hence Option (2) is the correct answer.

Let u = 2ax - x² 
so
Now, 2ax - x² = u implies du = (2a - 2x)dx  , or du = -2(x - a) dx
Using the substitution u = 2ax - x² :

1. When x = 0 : u = 0 ,
2. When x = 2a : u = 0 (as 2ax - x² = 0 at x = 2a ).

The limits of integration become symmetric about x = a
Rewriting x in terms of u , we evaluate the integral using symmetry :

Thus p = 103, q = 6
p² + q² = 103² + 6² = 10609 + 36 = 10645
Hence Option (1) is the correct answer.

∇Φ = (∂Φ/∂x) î + (∂Φ/∂y) ĵ + (∂Φ/∂z) k̂
∂Φ/∂x = 2xyz + 4z² , ∂Φ/∂y = x²z and ∂Φ/∂z = x²y + 8xz
∇Φ = (2xyz + 4z²) î + (x²z) ĵ + (x²y + 8xz) k̂
∇Φ(1, -2, 1) = [(2(1)(-2)(1) + 4(1)²) ] î +[ (1²)(1)] ĵ + [(1²(-2) + 8(1)(1))] k̂ = ĵ + 6 k̂
The unit vector in the direction of 2î - ĵ - 2k̂ is:
(2î - ĵ - 2k̂) / |2î - ĵ - 2k̂| = (2î - ĵ - 2k̂) / √(2² + (-1)² + (-2)²) = (2î - ĵ - 2k̂) / 3
Directional Derivative = ∇Φ(1, -2, 1) • (2î - ĵ - 2k̂) / 3
= ( ĵ + 6 k̂) • (2î - ĵ - 2k̂) / 3
= [0 + (1)(-1) + (6)(-2)] / 3
= (0 - 1 - 12) / 3
= -13/3
Hence Option (1) is the correct answer.

Concept:
Transportation problem is crucial part of linear programming problem which can be connected for required sources of supply to corresponding destination of demand, with the end goal that the aggregate transportation cost to be limited.
Explanation:
The essential phase of any transportation problem as initial basic feasible solution.

• Initial basic feasible solution must be feasible i.e. it must satisfy all the supply and demand constraints.
• The number of positive allocations must be equal to m+n-1 where m is number of rows and n is number of columns.

Nondegenerate basic feasible solution: A basic feasible solution is non – degenerate if it has exactly m+n-1 positive allocations in individual positions. If the allocations are less than required number, then it is known as degenerate basic feasible solution. This solution is not easy to modify, because it is impossible to draw a closed loop for each occupied cell.

Concept:
Equation of the form y = x × g(p) + f(p) is called Lagrange's form.
When g(p) = p, then the equation, y = px + f(p) is called Clairaut's equation and the solution of such type of equation is given by: y = Cx + f(C).
Calculation:
Given:
y = px + √(4 + p²) where f(p) = √(4 + p²)
The above equation represents Clairaut's form so the solution is y = Cx + f(C).
∴ y = Cx + √(4 + C²)
∴ (y - Cx)² = 4 + C²
∴ (y - Cx)² - C² = 4 is the required solution.

Concept:
If z = f(x, y) and x, y is the function of t.
Then
dz/dt = (∂z/∂x) × (dx/dt) + (∂z/∂y) × (dy/dt)
Calculation:
Given:
z = e^x siny ⇒ (∂z/∂x) = e^x siny and (∂z/∂y) = e^x cosy
x = log t ⇒ (dx/dt) = 1/t
y = t² ⇒ (dy/dt) = 2t
dz/dt = (∂z/∂x) × (dx/dt) + (∂z/∂y) × (dy/dt)
∴ dz/dt = (e^x siny / t) + 2t e^x cos y
∴ (e^x / t) (siny + 2t² cosy)

CONCEPT :
If roots are real and different then complimentary solution is given by F(x) = c1 em x + c2 en x
CALCULATION :
Differential equation f"(x) = f(x)
f"(x) - f(x) = 0
[D2-1] f(x) = 0
The corresponding auxiliary equation D2 - 1 = 0
Root of auxiliary equation D2 = 1
⇒ D = ±1
Here, m = 1 and n = - 1
Roots are real and different. So complimentary solution is given by
⇒ f(x) = c1 ex + c2 e-x
Put initial condition f(0) = 2 in the above equation we get,
⇒ 2 = c1 e0 + c2 e0
⇒ 2 = c1 + c2 -----(1)
∵ f(x) = c1 ex + c2 e-x
⇒ f`(x) = c1 ex - c2 e-x
Put initial condition f`(0) = 3 in the above equation
⇒ 3 = c1 - c2 -----(2)
By adding equation (1) and equation (2) we get,
⇒ 2c1 =5 ⇒ c1 = 5/2
Put value of c₁ in equation (1)
⇒ 2 = 5/2 + c2 ⇒c 2 = - 1/2
Solution of the given differential equation f(x) = 5/2ex - 1/2e-x

So, the value of f(4)

Hence, option B is the correct answer.

By using Gaussian elimination method:

⇒ R₂ - R₁ and R₃ - 2R₁ ⇒
= 
⇒ R₃ - R₂

1x + 1y + 1z = 150
1y + 2z = -50
0 = -50
Here, two equations and 3 unknowns as third equation is invalid
and also rank of [A] < rank of [A : B]
So inconsistent and having no solution

Concept:
For metrix then |P| = (a × d) - (b × c)
Inverse of matrix P is given by
P⁻¹ = adj P / det.(P)
P⁻¹ = adj P / |P
Multipllication of imaginary number i × (-i) = 1
Calculation:
Given:
Let,
|P| = (4 + 3i) × (4 - 3i) – (i) × (-i) = 16 + 9 – 1 = 24

Statement I: Every infinite group has infinitely many subgroups.
An infinite group typically has subgroups corresponding to all its elements and subsets.
Since there are infinitely many elements in an infinite group, you can create infinitely many subgroups.
Example:
Consider the infinite cyclic group Z (integers under addition).
For each integer n ≠ 0 , the set nZ = {nk | k ∈ Z} is a subgroup of Z .
These subgroups are distinct for different values of n ,
and since there are infinitely many integers n ,
there are infinitely many subgroups of Z.
⇒ Statement(I) is correct.
Statement II: There are only finitely many non-isomorphic groups of a given finite order.
For any given finite order n , the classification theorem of finite groups ensures
that there are only finitely many group structures (up to isomorphism) for that order.
Example:
Consider groups of order n = 4 :
By Lagrange's theorem and group theory classification:
1. The cyclic group Z₄.
2. The Klein four-group  .K₄ = Z₂ x Z₂
These are the only two non-isomorphic groups of order 4.
Thus, the number of non-isomorphic groups of a given finite order (here n = 4 ) is finite.
⇒ Statement(II) is correct.
Hence Option(1) is the correct answer.

Concept:

If A is any square matrix of order n, we can form the matrix [A – λI],

where

I is the nth order unit matrix.

The determinant of this matrix equated to zero i.e. |A – λI| = 0 is called the characteristic equation of A.

The roots of the characteristic equation are called Eigenvalues or latent roots or characteristic roots of matrix A.

Eigenvector (X) that corresponding to Eigenvalue (λ) satisfies the

equation AX = λX.

Properties of Eigenvalues:

The sum of Eigenvalues of a matrix A is equal to the trace of that matrix A

The product of Eigenvalues of a matrix A is equal to the determinant of that matrix A

Calculation:

Let,

Given eigen vector

⇒

⇒ (4 - λ) (101) + 2 × 101 = 0

⇒ λ = 6

Concept:
Integral test: This test is used for series of the form ∑∞n=1 f(n), where f(x) is a continuous, positive, and decreasing function for x ≥ 1. The series converges if and only if the corresponding improper integral ∫∞1 f(x)dx converges.
Explanation:
limn→∞ (1/n+1 + 1/n+2 + ⋯ + 1/2n)
Sn = ∑∞n=1 1/n+k
limn→∞ Sn = limn→∞ ∑∞n=1 1/n+1+k/n
= ∫10 1/(1+x) dx
= loge(1 + x)|0 to ∞
= log2

Given :
V = {(a, b, c, d) : b - 5c + 2d = 0}
W = {(a, b, c, d) : a - d = 0, b - 3c = 0}
From W: a - d = 0 ⇒ a = d
From W: b - 3c = 0 ⇒ b = 3c
From V: b - 5c + 2d ⇒ 0
Substitute the values of a and b from the equations of W into the equation of V:
3c - 5c + 2d = 0
⇒ -2c + 2d = 0
⇒ -2c = -2d
⇒ c = d
Now we have:
a = d , b = 3c and c = d
So, the vectors in V ∩ W can be represented as:
(d, 3d, d, d) = d (1, 3, 1, 1)
The intersection V ∩ W is spanned by a single vector: (1, 3, 1, 1)
Therefore, its dimension is 1
The dimension of V ∩ W is 1
Hence Option (1) is the correct answer.

(I) FALSE: But trace of LHS = RHS.
(II). TRUE
(III). FALSE: Inverse of a matrix exists only iff determinant of the matrix in non- zero.
(IV). TRUE

Concept:
The classification of partial differential equations (PDEs) of the form
aU_xx + bU_xy + cU_yy = 0 depends on the discriminant D = b² - 4ac:

• Elliptic if D < 0
• Parabolic if D = 0
• Hyperbolic if D > 0

Given:
Tricomi's equation is given by:
U_xx + xU_yy = 0.
We need to determine the type of this equation for different values of x .
Explanation:
The classification of partial differential equations (PDEs) of the form aU_xx + bU_xy + cU_yy = 0 depends on the discriminant D = b² - 4ac:

• Elliptic if D < 0
• Parabolic if D = 0
• Hyperbolic if D > 0

After comparing In Tricomi's equation U_xx + xU_yy = 0:
here a = 1 , b = 0 , and c = x .
The discriminant D = 0² - 4 • 1 • x = -4x.
Thus:
When x < 0 , D > 0 , so the equation is hyperbolic.
When x = 0 , D = 0 , so the equation is parabolic.
When x > 0 , D < 0 , so the equation is elliptic.
The correct answer is that Tricomi's equation is hyperbolic for x < 0 and elliptic for x > 0.
Hence Option (2) is the correct answer.

Concept:

A function f(x,y) is defined for (x,y) = (a,b) is said to be continuous at (x,y) = (a,b) if:

i) f(a,b) = value of f(x,y) at (x,y) = (a,b) is finite.

ii) The limit of the function f(x,y) as (x,y) → (a,b) exists and equal to the value of f(x,y) at (x,y) = (a,b)

Note:

For a function to be differentiable at a point, it should be continuous at that point too.

Calculation:

Given:

For function f(x,y) to be continuous:

and finite.

f(a,b) = f(0,0) ⇒ 0 (given)

= 0

f_x(0, 0) = {f(h, 0) - f(0, 0)} / h = 0

fy(0, 0) = {f(0, k) - f(0, 0)} / k = 0

∵ the limit value is defined and function value is 0 at (x,y) = (0,0), ∴ the function f(x,y) is continuous.

Hence, Option 2, 3 & 4 all are correct

Hence, Option 1 is not correct

Hence, The Correct Answer is option 1.