logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

$f(x)=\sin^{-1}\big(\large\frac{2x}{1+x^2}\big)$ is differentiable on

$(a)\;(-1,1)\qquad(b)\;R-[-1,1]\qquad(c)\;R-(1,1)\qquad(d)\;None\;of\;these$

Can you answer this question?
 
 

1 Answer

0 votes
$f(x)=\sin^{-1}\big(\large\frac{2x}{1+x2}\big)$
$\Rightarrow f'(x)=\large\frac{1}{\sqrt{1-\big(\Large\frac{2x}{1+x^2}}\big)^2}\frac{d}{dx}\big(\large\frac{2x}{1+x^2}\big)$
$\Rightarrow \large\frac{1+x^2}{\sqrt{(1+x^2)^2-4x^2}}\frac{(1+x^2).2-2x(2x)}{(1+x^2)^2}$
$\Rightarrow \large\frac{1}{\sqrt{(1-x^2)^2}}.\frac{2(1-x^2)}{1+x^2}$
$\Rightarrow \left\{\begin{array}{1 1}\large\frac{2}{1+x^2}&if\;|x| < 1\\\large\frac{-2}{1+x^2}&if\;|x| > 1\end{array}\right.$
$f'(x)$ does not exist at |x|=1 (i.e) $x=\pm 1$
$\therefore f(x)$ is differentiable on $R-[-1,1]$
Hence (b) is the correct answer.
answered Jan 2, 2014 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...