MHT CET 2024 3 May Shift 2 Question Paper: Download PCM Question Paper with Answers PDF

Step 1: Let y = sec(tan⁻¹ x).

Let θ = tan⁻¹(x), so tan(θ) = x.

Step 2: Differentiate y.

We know sec²(θ) = 1 + tan²(θ) = 1 + x², so sec(θ) = √(1 + x²).

Thus, y = √(1 + x²).

Step 3: Differentiate with respect to x:

dy/dx = (1/2√(1 + x²)) × 2x = x/√(1 + x²).

Step 4: Evaluate at x = 1:

dy/dx = 1/√(1 + 1²) = 1/√2.

Step 1: Simplify the logarithmic expression.

y = log_e(e^3x) + log_e(((x - 4)/(x + 3))^3/2).

Using properties of logarithms, this simplifies to:

y = 3x + (3/2) × log_e((x - 4)/(x + 3)).

Step 2: Differentiate y with respect to x.

dy/dx = 3 + (3/2) × d/dx [log_e((x - 4)/(x + 3))].

Using the quotient rule for log differentiation:

d/dx [log_e((x - 4)/(x + 3))] = [(x + 3)(1) - (x - 4)(1)] / [(x - 4)(x + 3)].

This simplifies to 7 / [(x - 4)(x + 3)].

Step 3: Substitute back:

dy/dx = 3 + (3/2) × [7 / ((x - 4)(x + 3))] = 3 + 21 / [2(x - 4)(x + 3)].

The equation of a circle with center at (h, 0) and radius h is given by:

(x - h)^2 + y^2 = h^2.

Expanding and simplifying, we get:

x^2 + y^2 + h^2 - 2hx = h^2,

or x^2 + y^2 - 2hx = 0.

Differentiating with respect to x:

2x + 2y dy/dx - 2h = 0.

Rearranging for h gives:

h = x + y dy/dx.

Substitute h back into the original equation:

x^2 + y^2 - 2x(x + y dy/dx) = 0.

Expanding and simplifying:

x^2 + y^2 - 2x^2 - 2xy dy/dx = 0,

or y^2 - x^2 - 2xy dy/dx = 0.

Finally, rearranging gives:

2xy dy/dx = y^2 - x^2.

We are given f(x) = 3x + 6 and g(x) = 4x + k. The condition f ∘ g(x) = g ∘ f(x) means:

f(g(x)) = g(f(x)).

Compute f(g(x)):

f(g(x)) = f(4x + k) = 3(4x + k) + 6 = 12x + 3k + 6.

Compute g(f(x)):

g(f(x)) = g(3x + 6) = 4(3x + 6) + k = 12x + 24 + k.

Equate the two compositions:

12x + 3k + 6 = 12x + 24 + k.

Cancel out 12x and solve for k:

3k + 6 = 24 + k.

3k - k = 24 - 6,

2k = 18,

k = 9.