ExamHabit
Practise free

JEE Main 2017

JEE Main 2017 Mathematics question on Limits Continuity and Differentiability

For x∈(0,14)x \in \left(0, \frac{1}{4}\right), if the derivative of tan⁡−1(6xx1−9x3)\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right) is x⋅g(x)\sqrt{x} \cdot g(x), then g(x)g(x) equals
  1. A.
    91+9x3\frac{9}{1+9x^3}
  2. B.
    3xx1−9x3\frac{3x\sqrt{x}}{1-9x^3}
  3. C.
    3x1−9x3\frac{3x}{1-9x^3}
  4. D.
    31+9x3\frac{3}{1+9x^3}
Show answer and solution

Correct option: A

Solution

Rewrite the argument: 6x√x/(1-9x³) = 2·(3x^(3/2))/(1-(3x^(3/2))²), which matches the identity 2t/(1-t²) = tan(2·tan⁻¹(t)) — wait, actually use tan(A+B): 2t/(1-t²) corresponds to tan(2α) where t = tan α. So tan⁻¹(6x√x/(1-9x³)) = 2·tan⁻¹(3x^(3/2)).

Differentiating: d/dx[2·tan⁻¹(3x^(3/2))] = 2·(1/(1+9x³))·(9/2)·x^(1/2) = 9√x/(1+9x³).

Since this equals √x·g(x), we get g(x) = 9/(1+9x³).

[JEE Main 2017]

Practise Limits Continuity and Differentiability the way it is asked

A timed set of real JEE Main Limits Continuity and Differentiability questions, graded as the exam grades. No sign-up.

Start a Limits Continuity and Differentiability test

More: all JEE Main Limits Continuity and Differentiability questions · JEE Main 2017 · Mathematics chapter list