Derivatives, Tangents, Normals and Implicit Functions (JEE Advanced)

Question 1

Let f(x)=sin⁻¹((2x)/(1+x²))+cos⁻¹((1−x²)/(1+x²)) for x∈mathbbR, where sin⁻¹ and cos⁻¹ take principal values. The value of f'(−2)+f'(−1/2)+f'(1/2)+f'(2) is ______. Exactly one option is correct.

Accepted Answer

C. Put x=tanθ with θ∈(−π/2,π/2). Then (2x)/(1+x²)=sin2θ and (1−x²)/(1+x²)=cos2θ, where 2θ∈(−π,π).
sin⁻¹(sin2θ) equals 2θ for lvert xrvert≤1, π−2θ for x>1, and −π−2θ for x<−1.
cos⁻¹(cos2θ) equals 2θ for x≥0 and −2θ for x<0.
So f(x)=4tan⁻¹x on (0,1), f(x)=0 on (−1,0), f(x)=π on (1,∞), and f(x)=−π−4tan⁻¹x on (−∞,−1).
Hence f'(1/2)=4/(5/4)=16/5, f'(−1/2)=0, f'(2)=0 and f'(−2)=−4/5.
The sum is 16/5−4/5=12

Question 2

The equation y³−3y+x=0 defines three differentiable functions y(x) near x=0. Consider the one with y(0)=√3. For this branch, the value of y''(0) is ______. Exactly one option is correct.

Accepted Answer

A. Differentiate: 3y²y'−3y'+1=0, so 3(y²−1)y'=−1. On this branch y²−1=2≠0 at x=0, so the slope exists. (The curve has vertical tangents only where y²=1, i.e. at x=±2, away from x=0.)
At x=0: y'(0)=−1/(3·2)=−1/6.
Differentiate 3(y²−1)y'+1=0 again: 6y (y')²+3(y²−1)y''=0, so y''=−(2y (y')²)/(y²−1).
At x=0: y''(0)=−(2√3·1/36)/2=−(√3)/36.
Difficulty: Very hard

Question 3

Let b∈mathbbR. The set of all values of b for which exactly two distinct lines through the point (2,b) are tangent to the curve y=x³ is ______. Exactly one option is correct.

Accepted Answer

D. The tangent at (t,t³) is y=3t²x−2t³. It passes through (2,b) if and only if b=g(t), where g(t)=6t²−2t³.
Different values of t give different lines: equal slopes need t'=−t, and then the intercepts −2t³ and 2t³ differ unless t=0. So the number of tangents equals the number of distinct real roots of g(t)=b.
g'(t)=12t−6t², so g has a local minimum g(0)=0 and a local maximum g(2)=8.
There are three ro

Derivatives, Tangents, Normals and Implicit Functions: questions with solutions

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