TN TRB PG Assistant Mock Test- 2 (Mathematics) - TN TET MCQ

It is a very well known theorem that if f and g be continuous real – valued function on the metric space M and A be the set of all x ∈ M s.t. f(x) < g(x) then A is open set. 

The definition of lipschitz function is given by 

Let A R ⊆ and let f : A → ℝ if there exists a constant k > 0 such that 

so we have lipschitz function in our question asked (as it satisfies lipschitz condition) now we will show uniform continuity of lipschitz function 

Since 

The given ε > 0 we can take δ = ε/k if x. y ∈ ℝ satisfy ��-�� < δ

then, 

Therefore f is uniformly continuous on R 

But for differentiability

Since 

⇒ f is differentiable at y ∀ ∈ x,y ℝ

⇒ f is differentiable function to 0

Let f(x) = Sin x ( 1 + cos x) 

⇒ f'(x) = cosx (1 + cosx) + sinx (–sinx) = cosx + cos²x – sin²x 

= cosx + cos²x 

f"(x) = – sin x – 2sin2x = – (sinx + 2sin2x) 

for maximum or minimum value of f(x), f'(x) = 0 

therefore cos x + cos2 x = 0 

⇒ cos x = – cos2x

⇒ cosx = –cos (π ± 2x) 

⇒ x = π ± 2x 

⇒ x = ��3 , –π 

Now 

Therefore f(x) is maximum at x =π/3. 

 Apply the logarithmic law, that is logax = xlog(a). Now the function is ln(1 + 5x)^-5 = -5log(1 + 5x). By taking the series = 

Consider

f_x = 2x

and

f_y = 2y

There is no a here. Thus, independent of a.

We know that

Now by differentiating equation (1) with respect to x, we get:

Closure of  Which is uncountable

Becausw x√2 is an irrational number for 

⇒ 24 – 6z = 0 

17 – 8x – 3y = 0 

11 – 26x – 5y = 0 

on solving the above eq.we get 

x = 1, y = 3 

Hence x = 1, y = 3, z = 4 

Differentiating f_xx = -y².sin(xy)

f_yy = -x².sin(xy)

f_xy = -yx.sin(xy)

Observe that f_xx. f_yy – (fxy)²

Hence, it is a saddle point.

Concept:

For a skew symmetric matrix

a_ij = a_ji

for diagonal element i = j

a_ii = -a_ii

2a_ii = 0

a_ii = 0

Hence diagonal element are zero.

Consider a 2 × 2 skew symmetric matrix

|A| = a²

Calculation:

Determinant of a skew symmetric even ordered matrix A is a perfect square.

AB = 0 ⇒ then A = 0 or B = 0 satisfies but it is not a mandatory case.
A =  and B =  then also = AB = 0
If AB = I then B = A⁻¹ ⇒ BA = A⁻¹ . A = I hence AB = BA = I

Option 3 is correct answer.

It is an optimization method for a linear objective function and a system of linear inequalities or equations.

The linear inequalities or equations are known as constraints. The quantity which needs to be maximized or minimized (optimized) is reflected by the objective function.

The fundamental objective of the linear programming model is to look for the values of the variables that optimize (maximize or minimize) the objective function.

Multiple techniques can be used to solve a linear programming problem. These techniques include:

Simplex method
Solving the problem using R
Solving the problem by employing the graphical method
Solving the problem using an open solver

Concept -

The curl of a vector field is given by the vector cross product of the del operator with , i.e., . It is computed as:

=

Explanation -

The form f corresponds to a vector field .

Now P = y+z , Q = z+x , and R = x+y into the formula for the curl:

1. Compute :

⇒ = 1 - 1 = 0

2. Compute :

⇒ = 1 - 1 = 0

3. Compute :

⇒ = 1 - 1 = 0

Put these values in the given formula we get -

Hence, the curl of is:

Now

Hence Option(3) is correct.

Concept:

Explanation:

If ϕ(x,y,z) is scalar homogeneous function og degree 4 and

Then, &

Now,

Use,

Since,

=

Given is homogenous function of degree 4,

Now, let λ=1

Concept:

If converges at x = x₀, then it also converges at all the values of x for which the following inequality is satisfied:

|x - a| < |x₀ - a|

Calculations:

converges at x = - 5

Here

a = 2, x₀ = - 5

|x - 2| < |- 5 - 2|

|x - 2| < 7

-7 < x - 2 < 7

-5 < x < 9

x = 5 lies inside (-5, 9)

Hence this series also converges at x = 5

Hence option (3) is correct.

Maximize z = 2x + y
subject to
 3x – 7y ≤ 21
 y – 2x ≤ 10
 x, y ≥ 0

Step 1. Rewrite the constraints:

• From 3x – 7y ≤ 21, subtract 3x: –7y ≤ 21 – 3x
 Multiply by –1 (and reverse the inequality): 7y ≥ 3x – 21
 So, y ≥ (3x – 21) / 7.

• From y – 2x ≤ 10, we have y ≤ 2x + 10.

Step 2. Identify the feasible region:

The variables must satisfy:  x ≥ 0,
 y ≥ 0,
 (3x – 21) / 7 ≤ y ≤ 2x + 10.

For example, at x = 0, the lower bound gives y ≥ –3, but y must be at least 0. The upper bound gives y ≤ 10. So the region is non-empty.

Step 3. Examine the objective function:

The objective is z = 2x + y. Notice that for any fixed x, the maximum possible value of y is given by y = 2x + 10 (from the second constraint). Plugging this in gives:  z = 2x + (2x + 10) = 4x + 10. Since x is unbounded above (there’s no constraint limiting x to a maximum value), as x increases, z increases without bound.

Thus, the LPP is unbounded.

The correct answer is:

(a) The LPP is unbounded.

(A). The unit normal to the surface ?³ − ??? + ?³ = 1 at (1, 1, 1)
Let F(x, y, z) = x³ - xyz + z³ - 1
The gradient of F gives a normal vector to the surface:
∇F = (∂F/∂x, ∂F/∂y, ∂F/∂z) = (3x² - yz, -xz, -xy + 3z²)
At the point (1, 1, 1), the normal vector is:
∇F(1, 1, 1) = (3(1)² - (1)(1), -(1)(1), -(1)(1) + 3(1)²) = (2, -1, 2)
The unit normal vector is:
? = ∇F / |∇F| = (2, -1, 2) / √(2² + (-1)² + 2²) = (2, -1, 2) / √9 = (2/3, -1/3, 2/3) = (1/3)(2?̂ - ?̂ + 2?̂)
So, (A) matches with (IV).
(B). If φ = y / (x² + y²), (∇φ)|_(1,0) =
∇φ = (∂φ/∂x, ∂φ/∂y)
∂φ/∂x = -2xy / (x² + y²)²
∂φ/∂y = (x² + y² - 2y²) / (x² + y²)² = (x² - y²) / (x² + y²)²
At (1, 0):
∂φ/∂x = -2(1)(0) / (1² + 0²)² = 0
∂φ/∂y = (1² - 0²) / (1² + 0²)² = 1
∇φ(1, 0) = (0, 1) = ?̂
So, (B) matches with (II).
(C). If F = x²?̂ + 2z²?̂ - 3y²?̂, (∇ × F)|(0,-1,-1) =

= ?̂(∂(-3y²)/∂y - ∂(2z²)/∂z) - ?̂(∂(-3y²)/∂x - ∂(x²)/∂z) + ?̂(∂(2z²)/∂x - ∂(x²)/∂y)
= ?̂(-6y - 4z) - ?̂(0 - 0) + ?̂(0 - 0) = (-6y - 4z)?̂
At (0, -1, -1):
∇ × F = (-6(-1) - 4(-1))?̂ = (6 + 4)?̂ = 10?̂
So, (C) matches with (III).
(D). If φ = y / (x² + y²), (∇φ)|(0,1) × ?̂ =
From part (B), we found ∇φ = (∂φ/∂x, ∂φ/∂y)
∂φ/∂x = -2xy / (x² + y²)²
∂φ/∂y = (x² - y²) / (x² + y²)²
At (0, 1):
∂φ/∂x = -2(0)(1) / (0² + 1²)² = 0
∂φ/∂y = (0² - 1²) / (0² + 1²)² = -1
∇φ(0, 1) = (0, -1) = -?̂
(∇φ) × ?̂ = (-?̂) × ?̂ = -(-?̂) = ?̂
So, (D) matches with (I).
Final Answer:
(A) - (IV)
(B) - (II)
(C) - (III)
(D) - (I)
Therefore, the correct option is 1. (A)-(IV), (B)-(II), (C)-(III), (D)-(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.

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.

Explanation:

G is a simple group.

A simple group is a nontrivial group whose only normal subgroups are the trivial group and the group itself.

By this definition, if G has only two normal subgroups {e} (the trivial group containing only the identity element) and G itself, then G is a simple group

Concept:
Gradient / Normal vector of surface / Slope of surface / Rate of maximum increase or decrease:
Let ϕ(x, y, z) = C represents any scalar point function, then its gradient is defined as
grad ϕ = ∇ϕ = î ∂ϕ/∂x + ĵ ∂ϕ/∂y + k̂ ∂ϕ/∂z = Normal vector = Σ î ∂ϕ/∂x

1) Direction of (grad ϕ) = Direction of Normal vector = Direction of surface
2) Rate of maximum increase = |grad ϕ|
3) Unit normal on ϕ' = ȳ̂ = grad ϕ / |grad ϕ|
Calculation:
Given,
Surface S : x2 + y2 + z2 - 48 = 0
We know that; grad ϕ = ∇ϕ = î ∂ϕ / ∂x + ĵ ∂ϕ / ∂y + k̂ ∂ϕ / ∂z
∴ grad (S) = 2x î + 2y ĵ + 2z k̂
At point (4,4,4) we get:
grad (S) = 8 î + 8 ĵ + 8 k̂
∴ Unit normal on the sphere will be:
∴ grad S / |grad S| ⇒ (8î + 8ĵ + 8k̂) / √(8² + 8² + 8²)
= (î + ĵ + k̂) / √3

Concept:

• Conditionally convergent:
  If the series Σ b_n is convergent and Σ |b_n| is divergent, then the series Σ b_n is called conditionally convergent.

• Absolutely convergent:
  The series Σ b_n with both positive and negative terms (not necessarily alternating) is called absolutely convergent if the corresponding series Σ |b_n| with all positive terms is convergent.

• If a series is convergent but not absolutely convergent, it is called conditionally convergent.
  An example of a conditionally convergent series is the alternating harmonic series.

Thus, a conditionally convergent series is a series that is convergent but not absolutely convergent.
Hence, the correct answer is option 2.

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

Hence, Option C is correct.