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JEE Main 2019 · 12 Apr · Shift II

JEE Main 2019 Mathematics question on Limits Continuity and Differentiability

The derivative of tan⁡−1(sin⁡x−cos⁡xsin⁡x+cos⁡x)\tan^{-1}\left(\dfrac{\sin x - \cos x}{\sin x + \cos x}\right), with respect to x2\dfrac{x}{2}, where (x∈(0,π2))\left(x \in \left(0, \dfrac{\pi}{2}\right)\right) is
  1. A.
    1
  2. B.
    23\dfrac{2}{3}
  3. C.
    12\dfrac{1}{2}
  4. D.
    2
Show answer and solution

Correct option: D

Solution

The expression tan⁻¹((sin x - cos x)/(sin x + cos x)) simplifies by dividing numerator and denominator by cos x to get tan⁻¹((tan x - 1)/(tan x + 1)) = tan⁻¹(tan(x - π/4)) = x - π/4 (for x ∈ (0, π/2)).

So the derivative of (x - π/4) with respect to x is 1. Now we need the derivative with respect to x/2, not x. Using the chain rule: d(x - π/4)/d(x/2) = [d(x - π/4)/dx] / [d(x/2)/dx] = 1 / (1/2) = 2.

[JEE Main 2019 · 12 Apr · Shift II]

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