Question

If $y=\sin ^{-1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$ and $\frac{d y}{d x}=\frac{2^{x+1} \log 2}{f(x)}$ then $f(0)=$ _______.

• A. $0$
• B. $-2$
• C. $2$
• D. $2 \log 2$

Answer 1

Answer: (C)

Now $y=\sin ^{-1}\left(\frac{2 \cdot 2^{x}}{\left(2^{x}\right)^{2}+1}\right)$
let $2^{x}=\tan \theta$
$=\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right)$
$\Rightarrow \sin ^{-1}(\sin 2 \theta)=2 \theta$
$\Rightarrow y=2 \tan ^{-1}\left(2^{x}\right) $
$\therefore \frac{d y}{d x}=\frac{2 \cdot 2^{x} \log 2}{1+4^{x}}=\frac{2^{x+1} \log 2}{4^{x}+1} $
$\therefore f(x)=4^{x}+1$
$\Rightarrow f(0)=2$