Question

Let $f :\left(\frac{-1}{\sqrt{2}}, 1\right] \rightarrow(-\infty, \ln \sqrt{2}]$ be a function defined as $f ( x )=\ln \left( x +\sqrt{1- x ^2}\right)$ and $g ( x )$ $=\frac{ x ^2}{ f ( x )}$. If $f \left( x _0\right)=\ln \sqrt{2}$ then

• A. $g \left( x _0\right)=\frac{1}{\sqrt{2} \ln 2}$
• B. $g\left(x_0\right)=\log _2 e$
• C. $ g ^{\prime}\left( x _0\right)=2 \sqrt{2} \log _2 e$
• D. $g ^{\prime}\left( x _0\right)=\sqrt{2} \log _{ e } 2$

Answer 1

Answer: (B), (C)

$ f \left( x _0\right)=\ln \sqrt{2}=\ln \left( x _0+\sqrt{1- x _0^2}\right) \Rightarrow x _0=\frac{1}{\sqrt{2}} $
$g ( x )=\frac{ x ^2}{ f ( x )} \Rightarrow g ^{\prime}( x )=\frac{ f ( x ) \cdot 2 x - x ^2 f ^{\prime}( x )}{ f ^2( x )} $
$\Rightarrow g ^{\prime}\left( x _0\right)=\frac{2 x _0 f \left( x _0\right)- x ^2 f ^{\prime}\left( x _0\right)}{ f ^2\left( x _0\right)} \left\{\Theta f ^{\prime}\left( x _0\right)=0\right\} $
$ g ^{\prime}\left( x _0\right)=\frac{2 x _0}{ f \left( x _0\right)}=\frac{2 \cdot \frac{1}{\sqrt{2}}}{\ln 2}=2 \sqrt{2} \log _2 e \cdot$