Question

Assertion :STATEMENT 1: Let $f\left(x\right)={\sin }^{-1}\left(\frac{2x}{1+x^2}\right)$, $f^{\prime }\left(2\right)=-\frac{2}{5}$ Reason: STATEMENT 2: ${\sin }^{-1}\left(\frac{2x}{1+x^2}\right)=\pi -2{\tan }^{-1}x$ $\forall x>1$

• A. Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1
• B. Both the statements are TRUE and STATEMENT 2 is NOT the correct explanation of STATEMENT 1
• C. STATEMENT 1 is TRUE and STATEMENT 2 is FALSE
• D. STATEMENT 1 is FALSE and STATEMENT 2 is TRUE

Answer 1

The correct option is A Both the statements are TRUE and STATEMENT 2 is the correct explanation of STATEMENT 1

Let $x=tan\frac{k}{2}$

Therefore $k=2ta{n}^{-1}\left(x\right)$

Then $si{n}^{-1}\left(\frac{2tan\frac{k}{2}}{1+ta{n}^2\frac{k}{2}}\right)$

$=si{n}^{-1}\left(sin\right(k\left)\right)$

$=\pi -2ta{n}^{-1}\left(x\right)$ for all $x>1$.

Hence

$f\left(x\right)=\pi -2ta{n}^{-1}\left(x\right)$

Therefore

$f^{\prime }\left(x\right)=\frac{-2}{1+x^2}$

Hence

$f^{\prime }\left(2\right)$ $=\frac{-2}{1+4}=\frac{-2}{5}$

Thus the reason is the correct explanation for the assertion.