JEE Main 2015 · 4 Apr · Shift 1 · Mathematics Q89

MathematicsSingle correct+4 / -1Question 89 of 90

Let
$\tan^{-1}y=\tan^{-1}x+\tan^{-1}\left(\frac{2x}{1-x^2}\right),$
where $|x|<\dfrac1{\sqrt3}$ . Then a value of $y$ is :

\dfrac{3x-x^3}{1-3x^2}

\dfrac{3x+x^3}{1-3x^2}

\dfrac{3x-x^3}{1+3x^2}

\dfrac{3x+x^3}{1+3x^2}

Answer and worked solution

Official answer: A

For $|x|<\frac1{\sqrt3}$ , $\tan^{-1}\dfrac{2x}{1-x^2}=2\tan^{-1}x$ . So $\tan^{-1}y=3\tan^{-1}x$ and
$y=\tan(3\tan^{-1}x)=\frac{3x-x^3}{1-3x^2}.$

JEE Main 2015 · 4 Apr · Shift 1, Mathematics question 89