Question

Let $f:\left(\frac{-1}{\sqrt{2}},1\right]\to (-\infty ,\ln \sqrt{2}]$ 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)

$f\left(x_0\right)=\ln \sqrt{2}=\ln \left(x_0+\sqrt{1-x_0^2}\right)\Rightarrow x_0=\frac{1}{\sqrt{2}}$
$g(x)=\frac{x^2}{f(x)}\Rightarrow g^{\prime }(x)=\frac{f(x)\cdot 2x-x^2{f}^{\prime }(x)}{f^2(x)}$
$\Rightarrow g^{\prime }\left(x_0\right)=\frac{2x_{0}f\left(x_0\right)-x^2{f}^{\prime }\left(x_0\right)}{f^2\left(x_0\right)}\left\{\Theta f^{\prime }\left(x_0\right)=0\right\}$
$g^{\prime }\left(x_0\right)=\frac{2x_0}{f\left(x_0\right)}=\frac{2\cdot \frac{1}{\sqrt{2}}}{\ln 2}=2\sqrt{2}{\log }_{2}e\cdot$