Question

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

$g^{\prime }\left(x_0\right)=2\sqrt{2}{\log }_{2}e$

$f\left(x_0\right)=\ln \sqrt{2}=\ln (x_0+\sqrt{1-x_0^2})$
$(\sqrt{2}-x_0{)}^2=1-x_0^2$
$2+x_0^2-2\sqrt{2}{x}_0=1-x_0^2$
$2x_0^2-2\sqrt{2}{x}_0+1=0$
$x_0=\frac{2\sqrt{2}\pm 0}{2\times 2}\Rightarrow x_0=\frac{1}{\sqrt{2}}$
So $g\left(x_0\right)=\frac{1}{2}\times \frac{1}{\ln \sqrt{2}}={\log }_{2}e$
$g^{\prime }\left(x\right)=\frac{f\left(x\right).2x-x^2{f}^{\prime }\left(x\right)}{f^2\left(x\right)}\Rightarrow g^{\prime }\left(x_0\right)=2\sqrt{2}lo{g}_{2}e$
Because, $f^{\prime }\left(x_0\right)=0.$