TN TRB PG Assistant Mock Test- 3 (Mathematics) - TN TET MCQ

A discrete logarithm modulo of an integer to the base is an integer such that ax ≡ b (mod g). The problem of computing the discrete logarithm is a well-known challenge in the field of cryptography and is the basis of the cryptographic system i.e., the Diffie-Hellman key exchange.

The Spherical coordinates is of the form (r, θ, φ) and Cartesian coordinates is of the form (x, y, z) where x = r sin⁡θ cos⁡ϕ and y = rsin⁡θ sin⁡ϕ and z=rcos⁡θ. Now, substituting the values for r as 10, θ as 90, and φ as 60, substituting the values we get

x = 10 sin90 cos60 = 5

y = 10 sin90 sin60 = 8.66

z = 10 cos90 = 0.

The variance (V) for a Binomial Distribution is given by V = npq

Standard Deviation = 

Let T : R³ + R³

such that the matrix for T is given by 

First we check out the eigenvalues of A is given by 

The Eigen vectors for l =1 is u1 = 

as 

and Eigen vectors for λ = –2 are v2 = 

and if P = 

(a) Let  denote the zero polynomial, so  for every value of x. 

 ∈ W, since  Suppose f. g ∈ W. Then f(1) = 0 and g(1) = 0. Also, for scalars a and b, we have 

(af + bg) (1) = af(1) + bg(1) = a0 + b0 = 0 

Thus af + bg ∈ W, and hence W is a subspace. 

(b) . Suppose f, g ∈ W. Them f(3) = f(1) and g(3) g(1). Thus, for any scalars a and b, we have

(af + bg)(3) = af(3) + bg(3) = af(1) + bg(1) = (af + bg) (1) 

Thus af + bg ∈ W, and hence W is a subspace.

 (c) 0 ∈ W, . Suppose f . g ∈ W. Then f(–x) = –f(x) and g(–x) = –g(x). 

Also, for scalar a and b,

(af + bg)(–x) = af (–x) + bg(–x) = –af(x) – bg(x) = –(af + bg)(x) 

Thus ab + gf ∈ W, and hence W is a subspace of V. 

f₁g ∈ W

f(4) = 3 + f(2) 

and g(4) = 3 + g(2) 

If a, b ∈ ℝ, then 

af + bg)(4) = af (4) + bg(4) = a[3 + f(2)] + b[3 + g(2)], from (i) 

= af(2) + bg(2) + 3a + 3b = (af + bg) (2) + 3(a + b) 

≠ 3+( af bg )(2) 

∴ (af+ bg)( 4) ∉ W

Hence, W is not a subspace of V. 

Concept:
If a series ∑ aₙ converges then limₙ→∞ aₙ = 0.
In other words, if limₙ→∞ aₙ ≠ 0 then the series ∑ aₙ is not convergent.
Explanation: Given series ∑ (from n=3 to ∞) aⁿ / (nᵇ (logₑ n)ᶜ)
Option (3): a = 1, 1 ≥ b ≥ 0, c < 1.
Let a = 1, b = 0, c = -1.
Then the series becomes ∑ (from n=3 to ∞) 1 / (logₑ n)⁻¹ = ∑ (from n=3 to ∞) (logₑ n), which is not convergent as limₙ→∞ logₑ n ≠ 0.
Hence (3) is correct.

We are given 2 * x * 3 :
Apply the binary operation:

1. 2 * x
2. (2 * x) * 3

Using the definition of the binary operation:

1. 2 * x = 2 + x + 2x = 3x + 2
2. (2 * x) * 3 = (3x + 2) * 3 = (3x + 2) + 3 + (3x + 2) * 3

Given 2 * x * 3 = 7
⇒ (3x + 2) + 3 + (3x + 2) * 3 = 7
⇒ 3x + 2 + 3 + 9x + 6 = 7
⇒ 12x + 11 = 7
⇒ 12x = 7 - 11
⇒ 12x = -4
⇒ x = -4/12
⇒ x = -1/3
Therefore, the solution to the equation 2 * x * 3 = 7 is x = -1/3
Hence Option (1) is the correct answer.

Let's analyze each statement :
Statement 1: "There is a one-to-one correspondence between any two right cosets of the subgroup H in group G."
This statement is correct.
A right coset of a subgroup H in a group G is a set of the form gH = {gh | h ∈ H}, where g ∈ G.
Two right cosets gH and kH are either identical or disjoint.
This property ensures a one-to-one correspondence between any two right cosets.
Statement 2: "If H, K are the subgroups of G, then HK is a subgroup of G if and only if HK = KH."
This statement is also correct.
The condition HK = KH is a necessary and sufficient condition for HK to be a subgroup.
If HK = KH, then HK is closed under the group operation and contains inverses for each of its elements, making it a subgroup.
Statement 3: "If H, K are the subgroups of the abelian group G, then HK is a subgroup of G."
This statement is correct.
In an abelian group, the order of elements doesn't matter.
Therefore, HK = KH,
and from statement If H, K are the subgroups of G, then HK is a subgroup of G if and only if HK = KH.,
we know that HK is a subgroup.
Statement 4: "If H, K are the subgroups of G, then H ∩ K may or may not be a subgroup of G."
This statement is incorrect.
The intersection of two subgroups is always a subgroup.
This can be easily verified by checking the closure property and the existence of inverses within H ∩ K.
Therefore, the statement that is not correct is:
"If H, K are the subgroups of G, then H ∩ K may or may not be a subgroup of G."
Hence Option (4) is the answer.

If f is continuous at x = 1 then :


and
Since
⇒ k = 4
Hence Option (4) 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:
Given:
y = lim (n→∞) (n! / ((n+1)! + n!))
y = lim (n→∞) (n! / ((n+1)(n!) + n!))
y = lim (n→∞) (n! / (n!(n+1) + 1))
y = lim (n→∞) (1 / (n+2))
y = 1/∞ = 0

Concept:
A function is said to be continuous if

Calculation:
Given:
f(x) = (x + 1)cotx
Taking log on both sides
log (f(x)) = cot x log (x + 1)
For checking continuity at x = 0

= 0/0 form
Using L-Hospital rule


⇒ log (f(0)) = 1
⇒ f(0) = e¹ = e

Concept -

If Radius of convergence R of the power series is then

Then if the power series is given by then

Radius of convergence of the given power series is

And the interval of convergence is

Explanation -

Now let

Now ROC of is

Now the interval of convergence is

Hence the option (iii) is correct.

Given that A is a symmetric matrix. Therefore A = A'
Let A =
A² =
A² =
The diagonal entries of A² are a² + b² and b²+ c².
Sum of the diagonal entries of A² is 1.
Therefore, a² + b² + b² + c² = 1
⇒ a² + 2b² + c² = 1
∵ b² ≥ 0 and the coefficient of b² is 2, therefore the value of b should be 0 for the sum of entries of the
diagonal to be equal to 1.
∴ b = 0 and a² + c² = 1
The possible values of a and c can be
⇒ (a, c) ≡ (1, 0), (-1, 0), (0, 1), (0, -1)
Therefore, the possible number of such matrices is 4.
Hence, the correct answer is option (2).

Concept -

P - test - is convergent for p > 1

Explanation -

Given that be a sequence of positive real number such that is divergent.

For option (i) -

Let a_n = 1/n be a sequence of positive real number such that is divergent

Now

= is divergent series by p - test.

Hence option(i) is false.

For option (iii) -

Let an = 1/n be a sequence of positive real number such that is divergent

Now

= is divergent series by p - test.

Hence option(iii) is false.

Hence option(iv) is also false.

For option (ii) -

Let an = 1/n^k be a sequence of positive real number such that is divergent

Now

=

So clearly n² is least power in denominator so p - test, the series is convergent.

Hence option(ii) is true.

Concept:

To find the orthogonal trajectories of any curve

Step 1: Find the differential equation of the family of circles

Step 2: Replace dy/dx with -dx/dy to get the differential equation of the orthogonal trajectories

Step 3: Solve the new differential equation

Explanation:

We have , − 1 < h < 1 .

Differentiating both side with respect to x:
⇒

from given

⇒

⇒

⇒

Replacing dy/dx by -dx/dy:

⇒

⇒

Hence Option(1) is the correct Answer.

Concept:

For a vector field, V defined as Vx î + Vy ĵ + Vz k̂, the divergence (∇.V) is given by:

Also, for a Scalar function A, the gradient is defined as:

Calculation:

Given:

For the given vector field :

At the given point (1, 2, 3)

There are 3 distinct ring homomorphisms.
The homomorphism φ is determined by the image of 1 under φ because 1 generates the entire ring R.
Thus, we need to find the elements of S that, when multiplied by 4, yield 0 (the value to which 1 * 4 = 4 in R maps under φ), while also preserving the multiplicative identity.
The elements {0, 4, 8} of S satisfy the first property (0 * 4 = 4 * 4 = 8 * 4 = 0 in Z/12Z).
However, only the elements {1, 5, 9} of S serve as multiplicative identities,
Hence the total number of distinct ring homomorphisms that satisfy both properties is exactly 3.
Therefore, there are 3 distinct ring homomorphisms from R = Z/4Z to S = Z/12Z.
Hence option (3) is true.

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

f = y^x
ln f = x lny
⇒
⇒
⇒
= y^x−1 + xy^x−1lny
= 1⁽²⁻¹⁾ + [2×1⁽²⁻¹⁾ ln(1)] = 1

Concept (Comparison test):
We have two series Σ a_n and Σ b_n with a_n, b_n ≥ 0 for all n, and a_n ≤ b_n for all n. Then:

• If Σ b_n is Convergent, then Σ a_n is convergent.
• If Σ a_n is Divergent, then Σ b_n is Divergent.

P-Series test:
For the series Σ 1/n^k (if k > 1), the series is convergent; otherwise, it is divergent.
Calculation:
Given:
a_n = Σ 1/(2n-1)
b_n = Σ 1/n
We calculate the limit:
lim (n→∞) a_n / b_n = lim (n→∞) [(1 / (2n-1)) × n] = lim (n→∞) [n / (2n-1)] = 1/2
By the comparison test, both a_n and b_n are converging and diverging together.
Therefore, b_n = 1/n is divergent by the p-series test, and a_n is divergent.
Conclusion:
Therefore, a_n is divergent, but it does not imply that it is a Cauchy sequence.

Concept:

The total derivative is given by:

Calculation:

u = log xy where x2 + y2 = 1

x2 + y2 = 1

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.

Volume of the cylinder = πr²h

Hence, Option B is correct.

Let the initial number boys = 5x
The initial number of girls = 3x
(5x – 80) = (3x + 40)
2x = 120
x = 60
So the answer = 300
Hence, Option C is correct.