Explanation -

ey(y′ − 1) = ex

⇒ dy/dx = ex−y + 1

Let x - y = t

1 – dy/dx = dt/dx

So, we can write

⇒ 1 − dt/dx = et + 1

⇒ −e^−t dt = dx

⇒ e^−t = x + c

⇒ ey−x = x + c

1 = 0 + c

⇒ e^y−x = x + 1

at x = 1

⇒ ey−1 = 2

⇒ y = 1 + log₂2

Hence Option (4) is correct.

Calculation:

Given, (h, k), (7, 3) and (–1, –3) lie on the line L₁.

Equation of line passing through (7, 3) and (–1, –3) is:

⇒ 4y + 12 = 3x + 3

⇒ 3x - 4y - 9 = 0

∴ L₁ : 3x - 4y - 9 = 0

Line perpendiclar to L₁ is L₂ : - 4x - 3y = λ 

It passes through (8, 10)

⇒ - 32 - 30 = λ 

⇒ λ = - 62

∴ L₂ : - 4x - 3y = - 62

⇒ L₂ : 4x + 3y = 62

∴ Intersection of L₁ and L₂ is (11, 6)

⇒ (h, k) = (11, 6)

⇒ hk = 66

∴ The value of hk is 66.

The correct answer is Option 2.

Formula used:

tan(α + β) = tan(π/3) = √3

tan(α - β) = tan(π/6) = 1/√3

cot(2β) = 1 / tan(2β)

Calculation:

From the given information, we can find the values of α and β:

α + β = π/3

α - β = π/6

Solving these equations simultaneously:

Adding the two equations:

⇒ (α + β) + (α - β) = π/3 + π/6

⇒ 2α = π/2

⇒ α = π/4

Now, substitute α = π/4 in α + β = π/3:

⇒ π/4 + β = π/3

⇒ β = π/3 - π/4

⇒ β = π/12

Now, calculate tan(α) and cot(2β):

tan(α) = tan(π/4) = 1

cot(2β) = 1 / tan(2β)

tan(2β) = tan(2 × π/12) = tan(π/6) = 1/√3

⇒ cot(2β) = 1 / (1/√3) = √3

Finally, calculate tan(α) × cot(2β):

⇒ tan(α) × cot(2β) = 1 × √3 = √3

∴ The correct answer is Option 3.

Concept:

Let R be a binary relation on a set A.

1. Reflexive: Each element is related to itself.

R is reflexive if for all x ∈ A, xRx.

2. Symmetric: If any one element is related to any other element, then the second element is related to the first.

R is symmetric if for all x, y ∈ A, if xRy, then yRx.

3. Transitive: If any one element is related to a second and that second element is related to a third, then the first element is related to the third.

R is transitive if for all x, y, z ∈ A, if xRy and yRz, then xRz.

Calculation:

For reflexive:

mRm = m – m = 0 and nRn = n – n = 0 which is not odd

thus, it is not reflexive.

For symmetric:

mRn = m – n is odd then nRm = n – m is also odd

∴ This relation is symmetric.

For transitive:

Let m = 5, n = 2 and third number (p) = 1

mRn = 5 – 2 = 3 (odd), nRp = 2 -1 = 1 (odd) and mRp = 5 – 1 = 4 (even)

i.e., mRn and nRp ≠ mRp

∴ This relation is not transitive.

Hence, option (1) is correct.

Explanation -

x² + y² − 8x − 4y + 16 = 0

(x − 4)² + (y − 2)² = 4

⇒ Centre (4, 2) , radius = 2

OA = 4 = OB

In ∆OBC

tan θ= 4/2 = 2 ⇒ sin θ = 2/√5

In ∆BDC

sin θ= BD/2 ⇒ BD = 4/√5

Length of chord of contact (𝐴𝐵) = 8/√5

Alternative

(𝑙) length of tangent = 4 and (𝑟) radius = 2

⇒Length of chord of contact = 2lr /(√(l2+ r2)

Square of length of chord of contact = 64/5

Hence the Correct option is (2)

Concept:

The characteristic equation of a square matrix A is given by |A - λI| = 0

Every square matrix satisfies its own characteristic equation.

Calculation:

⇒ (1 - λ)[(5 - λ)(2 - λ) - 14] - 2[4(2 - λ) - 21] + 3[12 - 2(5 - λ)] = 0

⇒ λ³ - 8λ² - 18λ - 7 = 0

Since, every matrix staisfies its own characteristic equation, we get:

A³ - 8A² - 18A - 7 = 0

Comparing with A³ - 8A² + αA + βI = 0, we gte:

α = - 18 and β = - 7