Question

If $f\left(x\right)=\frac{e^x-e^{-x}}{2}$ and if $f\left(g\right(x\left)\right)=x$ then $g\left(\frac{e^{1002}-1}{2e^{501}}\right)$ is equal to 'k' then sum of digits of 'k' is

Answer 1

Given that $f\left(g\left(x\right)\right)=x$

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

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

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

Substitute $g\left(x\right)=501$, we get

$\Rightarrow \frac{e^{2\times 501}-1}{2e^{501}}=501=k$

$\Rightarrow \frac{e^{1002}-1}{2e^{501}}=501=k$

$\therefore k=501$

Hence sum of digits of $k=5+0+1=6$