JEE Mathematics Differential Coefficient MCQs Set 02

Answer: (a) 1

Explanation:
1. The product equals (1 − x^(2ⁿ⁺¹))/(1 − x) = 1 + x + x² + … + x^(2ⁿ⁺¹−1).
2. Differentiating, the constant term of the derivative is the coefficient of x, which is 1.
3. So dy/dx at x = 0 is 1.

Teacher's Note:
1. Expand only the lowest powers of x.
2. The product telescopes to a geometric series.

Answer: (c) \( \frac12(\sqrt3-1) \)

Explanation:
1. At x = 2π/3, cos x < 0 and sin x > 0, so y = −cos x + sin x near this point.
2. dy/dx = sin x + cos x = √3/2 − 1/2 = (√3 − 1)/2.

Teacher's Note:
1. Remove the modulus signs using the sign of each function at the given point.
2. cos(2π/3) = −1/2, sin(2π/3) = √3/2.

Answer: (c) \( \frac{1}{x\log_e x} \)

Explanation:
1. f(logₑx) = logₑ(logₑ x).
2. Its derivative is (1/logₑx)·(1/x) = 1/(x logₑx).

Teacher's Note:
1. Substitute first, then apply the chain rule.
2. d/dx ln u = u′/u.

Answer: (a) \( \sqrt2 \)

Explanation:
1. Multiply and divide by sin x: f(x) = sin 32x/(32 sin x).
2. f′(x) = [32·32 cos 32x sin x − 32 sin 32x cos x]/(32 sin x)².
3. At x = π/4: sin 32x = sin 8π = 0, cos 32x = 1, so f′ = (1024 · (1/√2))/(1024 · ½) = √2.

Teacher's Note:
1. The product of cosines of doubled angles collapses to a sine ratio.
2. At π/4 the numerator sine vanishes, which simplifies the quotient rule.

Answer: (b) −1

Explanation:
1. For x in [π, 2π], cos⁻¹(cos x) = 2π − x.
2. 5π/4 lies in this interval, so dy/dx = −1.

Teacher's Note:
1. cos⁻¹(cos x) equals x only on [0, π].
2. On [π, 2π] it is the reflected line 2π − x.

Answer: (a) \( \frac{1}{\sqrt2} \)

Explanation:
1. sec(tan⁻¹x) = √(1 + x²).
2. dy/dx = x/√(1 + x²), which at x = 1 is 1/√2.

Teacher's Note:
1. Put θ = tan⁻¹x so that sec θ = √(1 + tan²θ).

Answer: (a) 0

Explanation:
1. Taking logs: y ln x = x − y, so y = x/(1 + ln x).
2. dy/dx = [(1 + ln x) − 1]/(1 + ln x)² = ln x/(1 + ln x)².
3. At x = 1, ln x = 0, so dy/dx = 0.

Teacher's Note:
1. Take logarithms to separate y.
2. ln 1 = 0 kills the numerator.

Answer: (b) 1

Explanation:
1. At x = 0: 0 − y = 0, so y = 0.
2. Differentiating: e^(xy) + x e^(xy)(y + x y′) − y′ = 2 sin x cos x.
3. At x = 0: 1 − y′ = 0, so y′ = 1.

Teacher's Note:
1. Find y at the given x first.
2. Implicit differentiation then gives y′ directly.

Answer: (c) \( \frac{1-x}{x} \)

Explanation:
1. The exponent contains the same expression, so x = e^(y + x).
2. Hence ln x = y + x, i.e. y = ln x − x.
3. dy/dx = 1/x − 1 = (1 − x)/x.

Teacher's Note:
1. An infinite tower repeats itself, so substitute it by x.
2. Take logs to isolate y.

Answer: (a) \( \frac1x\log_2e\cdot\log_xe \)

Explanation:
1. y = ln(ln x)/ln 2.
2. dy/dx = (1/ln 2)·(1/ln x)·(1/x).
3. Since 1/ln 2 = log₂e and 1/ln x = logₓe, dy/dx = (1/x) log₂e · logₓe.

Teacher's Note:
1. Change the base to e before differentiating.
2. Reciprocals of natural logs are logs of e in the other bases.

Answer: (a) 2

Explanation:
1. dy/dx = f′(x²)·2x = 2x√(2x⁴ − 1).
2. At x = 1: 2·1·√1 = 2.

Teacher's Note:
1. Use the chain rule: the derivative of f(x²) is f′(x²) times 2x.
2. Evaluate f′ at x² = 1.

Answer: (a) 1

Explanation:
1. Write 5/13 = cos α and 12/13 = sin α, so the argument is cos(x + α).
2. For x in (0, π/2), x + α lies in (α, π/2 + α) ⊂ (0, π), so y = x + α.
3. Hence dy/dx = 1.

Teacher's Note:
1. Convert 5cos x − 12 sin x into a single cosine.
2. Check that x + α stays in [0, π] so that cos⁻¹(cos θ) = θ.

Answer: (a) \( \frac{-1}{2|x|\sqrt{x^2-1}} \)

Explanation:
1. Let u = √((x+1)/(x−1)); then dy/dx = u′/(1 + u²), 1 + u² = 2x/(x − 1).
2. u′ = −1/[(x − 1)²u] = −1/[(x − 1)^(3/2)(x + 1)^(1/2)].
3. Simplifying gives dy/dx = −1/(2|x|√(x² − 1)); the modulus appears because the domain includes x < −1 as well as x > 1.

Teacher's Note:
1. Differentiate inverse tangent of a root with the chain rule.
2. Both branches x > 1 and x < −1 give the same formula with |x|.

Answer: (b) −1

Explanation:
1. Taking logs: y ln x + x ln y = ln 16.
2. Differentiating: y′ ln x + y/x + ln y + x y′/y = 0.
3. At (2, 2): y′ ln 2 + 1 + ln 2 + y′ = 0, so y′(1 + ln 2) = −(1 + ln 2) and y′ = −1.

Teacher's Note:
1. Use logarithmic differentiation for variable exponents.
2. Factor (1 + ln 2) on both sides.

Answer: (b) 0

Explanation:
1. ln y = (ln x)/x, so y′/y = (1 − ln x)/x².
2. y′ = x^(1/x)(1 − ln x)/x².
3. At x = e, 1 − ln e = 0, so y′ = 0.

Teacher's Note:
1. y = x^(1/x) has its maximum at x = e.
2. Hence the derivative is zero there.

Answer: (d) 1

Explanation:
1. For |x| < 1, both functions equal 2 tan⁻¹x.
2. So the ratio of their derivatives is 1.

Teacher's Note:
1. Substitute x = tan θ in both expressions.
2. Equal functions have derivative ratio 1.

Answer: (b) \( \frac25 \)

Explanation:
1. dx/dt = 5t⁴ − 15t² − 20 = −30 at t = 1.
2. dy/dt = 12t² − 6t − 18 = −12 at t = 1.
3. dy/dx = (−12)/(−30) = 2/5.

Teacher's Note:
1. For parametric forms divide dy/dt by dx/dt.
2. Substitute t = 1 only after differentiating.

Answer: (a) 0

Explanation:
1. For 1 < x < 2, [x] = 1, so f(x) = cos(π/2 − x³) = sin x³.
2. f′(x) = 3x² cos x³.
3. At x³ = π/2, cos(π/2) = 0, so f′ = 0.

Teacher's Note:
1. Replace the greatest integer by its constant value on the interval.
2. cos(π/2 − θ) = sin θ.

Answer: (d) none of these

Explanation:
1. Put x = tan θ: (√(1 + x²) − 1)/x = (sec θ − 1)/tan θ = tan(θ/2).
2. So the function is θ/2 = ½ tan⁻¹x.
3. Its derivative with respect to tan⁻¹x is 1/2, which is not among the first three options.

Teacher's Note:
1. The substitution x = tan θ turns the expression into a half-angle tangent.
2. Differentiate with respect to θ = tan⁻¹x.

Answer: (a) −4

Explanation:
1. cosec⁻¹(1/u) = sin⁻¹u, so the function is sin⁻¹(2x² − 1).
2. Put x = cos θ: sin⁻¹(cos 2θ) = π/2 − 2θ, and √(1 − x²) = sin θ.
3. The required ratio is (−2)/cos θ = −2/(1/2) = −4 at θ = π/3.

Teacher's Note:
1. Use cosec⁻¹(1/u) = sin⁻¹u.
2. With x = cos θ, check that π/2 − 2θ is in the principal range at θ = π/3.

Answer: (a) \( \frac32x\cos x^3\operatorname{cosec}x^2 \)

Explanation:
1. du/dx = 3x² cos x³ and dv/dx = 2x sin x².
2. du/dv = (3x² cos x³)/(2x sin x²) = (3/2) x cos x³ cosec x².

Teacher's Note:
1. Divide the two derivatives with respect to x.
2. Cancel one factor of x.

Answer: (b) \( \frac{2(1+x-x^2)}{(x^2+1)^2}\sin\left(\frac{2x-1}{x^2+1}\right)^2 \)

Explanation:
1. dy/dx = f′(u)·du/dx with u = (2x − 1)/(x² + 1).
2. du/dx = [2(x² + 1) − 2x(2x − 1)]/(x² + 1)² = 2(1 + x − x²)/(x² + 1)².
3. So dy/dx = [2(1 + x − x²)/(x² + 1)²] sin u².

Teacher's Note:
1. Chain rule: outer derivative sin(u²) times the quotient-rule derivative of u.