JEE Main 2015 · 10 Apr · Shift 1 · Mathematics Q89

MathematicsSingle correct+4 / -1Question 89 of 90

If $f(x)=2\tan^{-1}x+\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)$ , $x>1$ , then $f(5)$ is equal to :

\dfrac\pi2

\pi

4\tan^{-1}(5)

\tan^{-1}\left(\dfrac{65}{156}\right)

Answer and worked solution

Official answer: B

For $x>1$ , $\sin^{-1}\dfrac{2x}{1+x^2}=\pi-2\tan^{-1}x$ . So $f(x)=\pi$ for all $x>1$ , and $f(5)=\pi$ .

JEE Main 2015 · 10 Apr · Shift 1, Mathematics question 89