Question

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

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

Given that

$\Rightarrow f\left(g\right(x\left)\right)=x$

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

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

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

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

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

$\Rightarrow g\left(\frac{e^{2\times 501}-1}{2e^{501}}\right)=501=k$

$\Rightarrow g\left(\frac{e^{1002}-1}{2e^{501}}\right)=501=k$

Therefore, $k=501$

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