Question

Let $f:R\to R$ be defined by $f\left(x\right)=\ln (x+\sqrt{x^2+1})$, then number of solution of $|f^{-1}x|=e^{|x|}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $Infinite$

Answer 1

The correct option is A $1$

ln $(x+\sqrt{x^2+1})=f\left(x\right)=y$

$e^y=x+\sqrt{x^2}+1$

$e^y-y=\sqrt{x^2+1}$

$e^{2y}+x^2-2e^{y}x=x^2+1$

$e^{2y}-2e^{y}x-1=0$

$a^2-2ax-1=0$

$x=\frac{e^{2y}-1}{2e^y}$

$f^{-1}\left(x\right)=\frac{e^{2x}-1}{2e^x}=e^{1\times 1}$

$e^{2x}-1=e^x.2e^x=2e^{2x}$

$e^{2x}=1$

$x=0$ only one solution