JEE Mathematics Differential Coefficient MCQs Set 03

Answer: (c) −2√(2π)

Explanation:
1. d/dx f(2x² + π/2) = 4x f′(2x² + π/2).
2. At x = √(π/2), 2x² + π/2 = 3π/2, and f′(3π/2) = −1 + sin 6π · cos(3π/2) = −1.
3. The value is 4√(π/2)·(−1) = −2√(2π).

Teacher's Note:
1. Apply the chain rule with the inner derivative 4x.
2. Evaluate f′ at the point 3π/2.

Answer: (b) \( \frac{488}{125} \)

Explanation:
1. t = x/(1 + x²), so at x = 2, t = 2/5 and dt/dx = (1 − x²)/(1 + x²)² = −3/25.
2. dy/dx = 2x + 2t·dt/dx = 4 + 2(2/5)(−3/25) = 4 − 12/125 = 488/125.

Teacher's Note:
1. Solve for t as a function of x.
2. Then differentiate y = x² + t² with the chain rule.

Answer: (a) AP

Explanation:
1. f(1) = f(−1) forces the x-term to vanish: f(x) = px² + q.
2. f′(x) = 2px, which is linear in x.
3. A linear function maps an AP to an AP.

Teacher's Note:
1. Use the condition to kill the linear coefficient.
2. A linear map preserves equal differences.

Answer: (a) 1

Explanation:
1. P(1) = P(0)² + 1 = 1, P(2) = P(1)² + 1 = 2, P(5) = 5, and so on.
2. So P(x) = x at infinitely many points, hence P(x) = x.
3. P′(0) = 1.

Teacher's Note:
1. Generate values from x → x² + 1 starting at 0.
2. A polynomial agreeing with x at infinitely many points is x.

Answer: (b) \( \frac78 \)

Explanation:
1. Replace x by 1/x and solve: f(x) = (5x + 4 − 3/x)/16.
2. y = x f(x) = (5x² + 4x − 3)/16.
3. dy/dx = (10x + 4)/16 = 14/16 = 7/8 at x = 1.

Teacher's Note:
1. Use the two equations in f(x) and f(1/x) as simultaneous equations.

Answer: (b) f′(x) = 0 for all x

Explanation:
1. The relation gives (1 + f(x))(1 + f(y)) = 2.
2. For fixed y, 1 + f(x) = 2/(1 + f(y)) is independent of x, so f is constant (f = √2 − 1).
3. Hence f′(x) = 0 for all x.

Teacher's Note:
1. Add 1 to both sides to factorise.
2. A function with f(x) independent of x is constant.

Answer: (a) cosec{g(x)}

Explanation:
1. For inverse functions g′(x) = 1/f′(g(x)).
2. f′(g(x)) = sin g(x), so g′(x) = 1/sin g(x) = cosec g(x).

Teacher's Note:
1. The derivative of an inverse is the reciprocal of f′ at the corresponding point.

Answer: (b) g(x) > 0

Explanation:
1. Let f = ax² + bx + c with a > 0 and b² < 4ac.
2. g = ax² + (b + 2a)x + (c + b + 2a), with discriminant (b + 2a)² − 4a(c + b + 2a) = b² − 4ac − 4a² < 0.
3. So g has no real root and, with a > 0, g(x) > 0 for all x.

Teacher's Note:
1. A positive quadratic has positive leading coefficient and negative discriminant.
2. Compute the discriminant of g.

Answer: (c) P(x) · P‴(x)

Explanation:
1. 2yy′ = P′ gives yy″ + y′² = P″/2, so y³y″ = y²(P″/2 − y′²) = P P″/2 − (P′)²/4.
2. Differentiating: (P′P″ + P P‴)/2 − P′P″/2 = P P‴/2.
3. Multiply by 2: P(x)·P‴(x).

Teacher's Note:
1. Express y³y″ through P and its derivatives.
2. Terms with P′P″ cancel.

Answer: (b) 0

Explanation:
1. Each derivative multiplies by 2 and shifts the phase by π/2.
2. d⁶y/dx⁶ = 2⁶ sin(2x + 3π) = −64 sin 2x.
3. At x = π/2, sin π = 0, so the value is 0.

Teacher's Note:
1. Use dⁿ/dxⁿ sin ax = aⁿ sin(ax + nπ/2).

Answer: (b) \( -\frac{\pi+4}{2} \)

Explanation:
1. dx/dt = cos t − t sin t, dy/dt = 1 + cos t.
2. dx/dy = (cos t − t sin t)/(1 + cos t), and d²x/dy² = [d/dt(dx/dy)]/(dy/dt).
3. At t = π/2 this evaluates to −(π + 4)/2.

Teacher's Note:
1. Differentiate dx/dy once more with respect to t, then divide by dy/dt.
2. Substitute t = π/2 at the end.