JEE Main 2017 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 2 April 2017 (offline), Maths Q71

Question 71 of 90 in this shift, Maths question 11 of 30.

If for

x\in\left(0,\frac{1}{4}\right)

, the derivative of

\tan^{-1}\left(\frac{6x\sqrt{x}}{1-9x^3}\right)

\sqrt{x}\cdot g(x)

, then

g(x)

equals :

(1) $\frac{3x\sqrt{x}}{1-9x^3}$
(2) $\frac{3x}{1-9x^3}$
(3) $\frac{3}{1+9x^3}$
(4) $\frac{9}{1+9x^3}$ Official answer

Official answer

Option 4

CBSE answer key (25/04/2017, used for result).

Chapter: Continuity and Differentiability
Topic: Inverse Trig Identities
Idea tested: tan^-1(2x/(1-x^2)) Piecewise Conversion

Same topic in other shifts

All Inverse Trig Identities questions

4 Apr 2026, Shift 1 · Q16If $y=\tan^{-1}\left(\frac{3\cos x-4\sin x}{4\cos x+3\sin x}\right)+2\tan^{-1}\left(\frac{x}{1+\sqrt{1-x^2}}\right)$ , then $\frac{dy}{dx}$ …MediumSingle correct
4 Apr 2026, Shift 2 · Q20The integral $\int_0^1\cot^{-1}\left(1+x+x^2\right)dx$ is equal to:HardSingle correct
5 Apr 2026, Shift 1 · Q24If $\frac{\pi}{4} + \sum_{p=1}^{11} \tan^{-1}\left(\frac{2^{p-1}}{1 + 2^{2p-1}}\right) = \alpha$ , then $\tan\alpha$ is equal to ________.MediumNumerical value
6 Apr 2026, Shift 1 · Q13Let $0<\alpha<1$ , $\beta=\frac{1}{3\alpha}$ and $\tan^{-1}(1-\alpha)+\tan^{-1}(1-\beta)=\frac{\pi}{4}$ . Then $6(\alpha+\beta)$ is equal to:MediumSingle correct
6 Apr 2026, Shift 2 · Q13If $\sin\left(\tan^{-1}\left(x\sqrt{2}\right)\right)=\cot\left(\sin^{-1}\sqrt{1-x^2}\right)$ , $x\in(0,1)$ , then the value of $x$ is :MediumSingle correct
8 Apr 2026, Shift 2 · Q12Let $\alpha = 3\sin^{-1}\left(\frac{6}{11}\right)$ and $\beta = 3\cos^{-1}\left(\frac{4}{9}\right)$ , where inverse trigonometric functions…HardTwo statements

Question text from the official JEE Main paper published by NTA (CBSE ran the 2017 exam); answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.