Question

Let $f:R\to R,f\left(x\right)=\ln (x+\sqrt{x^2+1})$ and $g:R\to R,g\left(x\right)=\{\begin{array}{cc}{x}^{\frac{1}{3}}& x\le 1\\ 2e^{1-x}& x>1\end{array}$, then the number of real solutions of the equation, $f^{-1}\left(x\right)=g\left(x\right)$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is C $4$

Given $f\left(x\right)=\ln (x+\sqrt{x^2+1})$
To find inverse of function $f\left(x\right)$, lets assume $f\left(x\right)=y$
$y=f\left(x\right)y=\ln (x+\sqrt{x^2+1})e^y=x+\sqrt{x^2+1}{e}^y-x=\sqrt{x^2+1}$

squaring on both sides

$e^{2y}-2x{e}^y+x^2=x^2+1e^{2y}-1=2x{e}^{y}x=\frac{e^{2y}-1}{2e^y}{f}^{-1}\left(y\right)=\frac{e^{2y}-1}{2e^y}{f}^{-1}\left(x\right)=\frac{e^{2x}-1}{2e^x}$
We have to find number of real solutions for $f^{-1}\left(x\right)=g\left(x\right)$
Case 1: when $x\le 1$, $g\left(x\right)=x^{\frac{1}{3}}$
$f^{-1}\left(x\right)=g\left(x\right)\frac{e^{2x}-1}{2e^x}=x^{\frac{1}{3}}$
By inspection, $x=0$ is a solution for this equation.
When $x\ne 0$, the equation can be written as $e^{2x}-2x^{\frac{1}{3}}{e}^x-1=0$,
The above equation can be written in a quadratic form whose $D>0$. Hence, we shall get $2$ more solutions.
Case 2: when x>1, $g\left(x\right)=2e^{1-x}$
$f^{-1}\left(x\right)=g\left(x\right)\frac{e^{2x}-1}{2e^x}=\frac{2e}{e^x}{e}^{2x}-1=4e{e}^{2x}=4e+12x\ln e=\ln (4e+1)x=\frac{1}{2}\ln (4e+1)$
for case 2, only possible solution is $x=\frac{1}{2}\ln (4e+1)$
So number of real solutions for $f^{-1}\left(x\right)=g\left(x\right)$ are $1+2+1=4$.