Question

Suppose f: R -- R is given by $f\left(x\right)={\log }_e(x+\sqrt{x^2+1})$ . Then the number of solution $\left|f^{-1}\left(x\right)\right|=e^{-\left|x\right|}$ is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is A 1

Let $f^{-1}\left(x\right)=g\left(x\right)⟹f\left(g\right(x\left)\right)=x$

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

$g\left(x\right)+\sqrt{\left(g\right(x){)}^2+1}=e^x$

$\left(g\right(x)-e^x{)}^2=(g\left(x\right){)}^2+1$

$-2g\left(x\right)e^x+e^{2x}=1⟹g\left(x\right)=\frac{1}{2}(e^x-e^{-x})$

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

$x>0⟹e^x=3e^{-x}⟹2x=\ln 3⟹x=\frac{1}{2}\ln 3$

$x<0⟹e^x-e^{-x}=2e^x⟹e^x=-e^{-x}$ which is not possible

No of solutions is $1$