Question

If $f^{\prime }\left(x\right)=\frac{1}{-x+\sqrt{x^2+1}}$ and $f\left(0\right)=-\frac{1+\sqrt{2}}{2}$, then $f\left(1\right)$ is equal to

• A. $-\log (\sqrt{2}+1)$
• B. $1$
• C. $1+\sqrt{2}$
• D. None of these

Answer 1

The correct option is D None of these

$f^{\prime }\left(x\right)=\frac{1}{-x+\sqrt{x^2+1}}$
$f^{\prime }\left(x\right)=x+\sqrt{x^2+1}$
$\Rightarrow f\left(x\right)=\int (x+\sqrt{x^2+1})dx$
$f\left(x\right)=\frac{x^2}{2}+\frac{x}{2}\sqrt{x^2+1}+\frac{1}{2}\log |x+\sqrt{x^2+1}|+C$
Put $x=0$ in $f\left(x\right)$, we get
$f\left(0\right)=C$. $\Rightarrow C=-\frac{1}{2}-\frac{1}{\sqrt{2}}$
Hence, $f\left(1\right)=\frac{1}{2}+\frac{1}{2}\sqrt{2}+\frac{1}{2}\log |1+\sqrt{2}|+(-\frac{1}{2}-\frac{1}{\sqrt{2}})$
$=\frac{1}{2}\log (1+\sqrt{2})$