Question

If $g$ is inverse function of $f$ where
$f\left(x\right)={\int }_0^{\pi }\frac{1}{\sqrt{1+t^2}}dt$ and
$\int g{\left(g^{\prime }\right(x\left)\right)}^{2}dx=\frac{{[1+{\left(g\right(x\left)\right)}^{\alpha }]}^{\beta }}{\gamma }+c$. Then the value of $\alpha \beta \gamma$ is equal to [where $\alpha ,\beta ,\gamma \in R$]

• A. $9$
• B. $6$
• C. $3$
• D. $2$

Answer 1

The correct option is A $9$

Given: $f\left(x\right)={\int }_0^{\pi }\frac{1}{\sqrt{1+t^2}}dt$

Since, $g$ is inverse function of $f$,

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

Integrating the equation,

$\Rightarrow \int g{\left(g^{{}^{\prime }}\left(x\right)\right)}^{2}dx=\frac{[1+{g\left(x\right)}^3]}{3}+C$

$\Rightarrow \alpha =3,\beta =1,\gamma =3\Rightarrow \alpha \beta \gamma =9$