TN TRB PG Assistant Mock Test- 5 (Mathematics) - TN TET MCQ

Concept Used:

A commutative group G is simple if and only if it has no nontrivial proper subgroups.

Prime order group has no nontrivial proper subgroups (except for the trivial subgroup and the group itself).

Explanation:

From both the statements in given above, we can conclude that any commutative Group G is simple if and only if O(G) is an prime number.

If R is the radius of convergence of then the series converges when |x| < R i.e., when - R < x < R
So interval of convergence is (- R, R)
(1) is correct

Explanation:

From properties of uniform continuity we can say that (1) and (2) are true.

Also product of two uniformly continuous need not be uniformly continuous.

For example f(x) = x, g(x) = x both are uniformly continuous on , but the product f(x)g(x) = x² is not uniformly continuous on .

(3) is also true

We know that sin x is uniformly continuous on [0, ∞] and and composition of two uniformly continuous function is uniformly continuous so f(x) = sin(sin x) is not uniformly continuous on [0, ∞].

(4) is NOT true.

Rewriting the constraints:
1. ⟶ equivalently:
2. (no change needed)
3. (Non-negativity constraints)
To check if the feasible region exists, we analyze the intersection of constraints:
Convert inequality to equality:
----------------(1)
This represents a line in the first quadrant
Convert to equality:

--------------(2)
This represents another line.
By checking whether these two lines enclose a bounded region in the first quadrant, we determine feasibility
Substituting x₁ = 0:
From x₂ = 3x₁ + 3, we get x₂ = 3
From x₂ = 0.25x₁ + 2.5, we get x₂ = 2.5 
Since both constraints are satisfied at some non-negative values, the feasible region exists
Since the constraints restrict x₁ and x₂ to a finite region, the LPP is bounded
Since the LPP is bounded and feasible, an optimal solution exists and is finite
⇒ The LPP has a finite optimal solution
Hence Option (4) is the correct answer.

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}

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 Used:-
The rate of change of a particular function with respect to the variable in that function is called the derivative of the function. If we differentiate a function with respect to the variable of the function, then the result we get is called derivative of the function.
A necessary condition for the complex-valued function to be differentiable that derivative w.r.t zbar is Zero
Given information:-
Given function is
f(z) = z2, Derivative of the function w.r.t z_bar is zero
On differentiating the above function with respect to z we get,

Now the value of the derivative at z = 4 needs to be found out. Put 4 in place of z in the above derivative,
⇒ f'(z) = 2z
⇒ f'(z) = 2×(4)
⇒ f'(z) = 8
So, for the function f(z) = z2, the value of the derivative at z = 4 is 8.
The correct option is 3.

Concept:

Method of finding the maxima & minima of z = f(x, y):

Step 1: Find for the given function

Step 2: Equate p & q to zero for obtaining stationary points (a, b).

Step 3: Calculate r, s, t at each stationary point (a, b).

Step 4:

i. If rt – s² > 0 & r > 0 then f (x, y) has a minimum at (a, b) & the minimum value is f(a,b).

ii. If rt – s² > 0 & r < 0 then f (x, y) has a maximum at (a, b) & the maximum value is f(a,b).

iii. If rt – s² < 0 then f (x, y) has neither a maximum nor a minimum value at (a, b) & (a, b) is a saddle point.

iv. If rt – s² = 0 & r < 0 then no conclusion by this method.

Calculation:

Given

f(x, y) =

Now, and

Putting and solving two equations we get

(x, y) = (a, a) or (-a, a)

Now at (a, a); ,

Hence, rt – s² = 3 > 0 and r > 0, hence it has a minimum value at (a, a)

Now, at (a, -a); ,

Hence, rt – s² = -5 < 0, hence it has neither minimum nor maximum at this point (a, -a)

Hence, (a, -a) is a saddle point

∴ Minimum value is, f(a, a) = 3a².

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

where C is the circle |z| = 2
The function has two singularities (poles) at z = 1 and z = 4
z = 1 is inside the contour |z| = 2
z = 4 is outside the contour |z| = 2
Since contour integration only depends on the singularities inside the contour, we only consider the residue at z = 1
To find the residue of at z = 1 , we express it in the form:
Residue = lim_z→1 (z - 1)f(z)
Multiplying by (z - 1) , we get:
Substituting z = 1 :

By the residue theorem,


Thus, the correct answer is (-2πi)/3 
Hence Option (3) is the correct answer.

We can parameterize the surface using x and y as parameters:
r(x, y) = <x, y, x² - y²>
Find the Partial Derivatives
∂r/∂x = <1, 0, 2x>
∂r/∂y = <0, 1, -2y>
∂r/∂x × ∂r/∂y = < -2x, 2y, 1>
||∂r/∂x × ∂r/∂y|| = √((-2x)² + (2y)² + 1²) = √(4x² + 4y² + 1)
The surface area (A) is given by the double integral over the region D in the xy-plane, which is the unit circle x² + y² ≤ 1:

Since the region of integration is a circle
Let x = rcosθ , y = rsinθ and dA = r dr dθ
The limits of integration are 0 to 1 for r and 0 to 2π for θ.

Let u = 4r² + 1, so du = 8r dr
When r = 0, u = 1, and when r = 1, u = 5

Now,

Therefore, the surface area is (π/6)(5√5 - 1)
Hence Option 3 is the correct answer.

Concept:
If Pde is of the form ϕ(D, D') z = e^(ax + by).
Then PI = 1 / ϕ(D, D') (e^(ax + by)) = 1 / ϕ(a, b)
Explanation:
(D² - D² + D - D') Z = e^(2x + 3y)
PI = 1 / (D² - D² + D - D') e^(2x + 3y)
PI = 1 / (4 - 9 + 2 - 3) e^(2x + 3y)
PI = -1 / 6 e^(2x + 3y)
Hence, (1) option is true

Concept use:

The Mean Value Theorem for Integrals states that if f is continuous on the closed interval [a, b], then there exists at least one number c in the open interval (a, b) such that:

∫from a to b f(x) dx = f(c) · (b - a)

This theorem guarantees the existence of at least one point c in the interior of the interval [a, b] where the value of the function f(c) multiplied by the length of the interval (b - a) gives the integral of f(x) over that interval

Explanation:

In the given statement, if f is the derivative of some function on the interval [a, b], then the Mean Value Theorem for Integrals applies.

According to this theorem, there exists a number c in the open interval (a, b) such that the integral of f(x) over the interval [a, b] is equal to f(c) multiplied by the length of the interval (b - a).

Hence, the correct option is: f(c) · (b - a)

Hence, The Correct Option is 2.

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:

Stoke's Theorem:

The surface integral of the curl of a function over a surface bounded by a closed surface is equal to the

The line integral of the particular vector function around a closed curve C is equal to the surface integral of the curl of a function over a surface S bounded by C.

, where

Solution:

Using Stoke's theorem,

Let

= 0

⇒

∴

The correct answer is option 3.

Concept:

Find the stationary points.

By solving the above two equations, we will get the values of x and y.

Putting these values in F(x,y) will get the minimum value.

Calculation:

Given,

........(i)

............(ii)

Solving (i) and (ii), we get:

x = 1 and y = 3

Minimum value = 9

Concept:

Consider an exact differential equation:

M dx + N dy = 0

The condition of an exact differential equation:

Calculation:

Given,

and

α = -1, β = -2

Concept:
In the original integral, the integration order is dx.dy. This integration order corresponds to integrating first with respect to x and afterward integrating with respect to y. Here, we are going to change the integration to be dy.dx, which means integrating first with respect to y.
Calculation:

Then,
y = 0, y = 1
x = 1, x = e^y
According to the given limit, we draw the required diagram,

Here,
We integrate the given integral with respect to x.

After that,
We integrate the given integral with respect to y.

So, we have to change the limit
x = 1, = e
y = log x, y = 1
Then, we arrange like this

After,
Comparing the obtained integral to the given integral
=
Then,
a = 1, b = e, c = log x, d = 1.
So,
e(a - d) = e^{(1 - 1)}
= e⁰
= 1

We have given that G/Z is a cyclic group, so let Zg is a generator of the cyclic group G/Z, where g ∈ G.
W e now show that G is an abelian group i.e.,
Since a ∈ G, so Za ∈ G/Z. But G/Z is a cyclic group which is gen era ted by Zg. Thus there exists an integer m such that

Za = (Zg)^m = Zg^m [∵ Z is a normal subgroup of G]

Again 

a ∈ Za and Za = Zg^m ⇒ a ∈ Z g^m.

Now

a ∈ Z g m ⇒ z₁ ∈ Z such that a ∈ z₁ g^m

Similarly, for b ∈ G, b = z₂ gⁿ, where z₂ ∈  z and n is any integer.

Now

Again

Supose that  then we are to prove that N₁ = N₂ we have  but  therefore  that is N1 is equal to some coset of N₂ in G. But two cosets of N₂ in G are either disjoint or identical since e ∈ N₁ and e ∈ N₂ th e re fo re N₁ and N₂ are not disjoint so we must have N₁ = N₂.

The given equation can be re-written as 

Integrating 

Putting y² = v so that 2y (dy/dx) = (dv/dx). We then get

which is linear. Its I.F =