Question

Let $g$ be a real valued differentiable function on $R$ such that $g\left(x\right)=3e^{x-2}+4{\int }_2^x\sqrt{2t^2+6t+5}dt\forall x\in R$ and let $g^{-1}$ be the inverse function of $g$. If $\left(g^{-1}{)}^{\prime }\right(3)$ is equal to $\frac{p}{q}$ where $p$ and $q$ are relatively prime, then find $\frac{p+q}{6}$

Answer 1

$g\left(x\right)=3e^{x-2}+4{\int }_2^x\sqrt{2t^2+6t+5}dtg\left(x\right)=y\Rightarrow g^{-1}\left(y\right)=x$

Differentiating the above equation w.r.t $x$ gives

$g^{{-1}^{{}^{\prime }}}\left(g\left(x\right)\right)g^{{}^{\prime }}\left(x\right)=1$

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

When $x=2,g\left(x\right)=3e^{2-2}+4{\int }_2^2\sqrt{2t^2+6t+5}dt=3+0=3$

$g^{\prime }\left(x\right)=3e^{x-2}+4\sqrt{2x^2+6x+5}$

$g^{\prime }\left(2\right)=23$

$x=2g\left(x\right)=3$ implies $\left(g{)}^{-1}\right(3)=2$

Now,

$g^{-1^{\prime }}\left(g\right(x\left)\right)=g^{-1^{\prime }}\left(3\right)=\frac{1}{g^{\prime }\left(2\right)}$

$g^{{-1}^{{}^{\prime }}}\left(3\right)=\frac{1}{g^{{}^{\prime }}\left(2\right)}=\frac{1}{23}=\frac{p}{q}$

$\therefore p+q=24\left(componendo\right)\therefore \frac{p+q}{6}=4$