Question

Let $f(x)=12\left(\frac{e^{3x}-3e^x}{e^{2x}-1}\right)$ be defined for $x>0$ and $g(x)$ be the inverse of $f(x)$ If $\underset{8}{\overset{27}{\int }}g(x)dx=a\ln 3-b\ln 2-c$, then the value of $a-(b$ $+c)$ is

Answer 1

Answer: 7

$\underset{\ln 2}{\overset{\ln 3}{\int }}f(x)dx+\underset{8}{\overset{27}{\int }}g(y)dy=27\ln 3-8\ln 2$
and $\underset{\ln 2}{\overset{\ln 3}{\int }}f(x)dx=12-12\ln 3+12\ln 2$
$\therefore \underset{8}{\overset{27}{\int }}g(y)dy=39\ln 3-20\ln 2-12$
Hence, $a=39;b=20;c=12$
$\therefore a-(b+c)=39-32=7.$