Question

If $y={\sin }^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$ and $\frac{dy}{dx}=\frac{2^{x+1}\log 2}{f(x)}$ 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 $2^x=\tan \theta$
$={\sin }^{-1}\left(\frac{2\tan \theta }{1+{\tan }^2\theta }\right)$
$\Rightarrow {\sin }^{-1}(\sin 2\theta )=2\theta$
$\Rightarrow y=2{\tan }^{-1}\left(2^x\right)$
$\therefore \frac{dy}{dx}=\frac{2\cdot 2^x\log 2}{1+4^x}=\frac{2^{x+1}\log 2}{4^x+1}$
$\therefore f(x)=4^x+1$
$\Rightarrow f(0)=2$