Question

If $f\left(x\right)={\int }_0^x(1+t^3{)}^{\frac{-1}{2}}$ and g (x) is the inverse of f, then the value of $\frac{g"\left(x\right)}{g^2\left(x\right)}$ is

• A. $\frac{3}{2}$
• B. $\frac{2}{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{3}{2}$

$f\left(x\right)={\int }_0^x(1+t^3{)}^{\frac{-1}{2}}dti.e.f[g\left(x\right)]={\int }_0^{g\left(x\right)}(1+t^3{)}^{\frac{-1}{2}}dti.e.x={\int }_0^{g\left(x\right)}(1+t^3{)}^{\frac{-1}{2}}dt$ [$\therefore$g is inverse of f $\Rightarrow$ f[g(x)] = x]
Differentiating again with respect to x, we have
$2g^{\prime }g"=3g^2{g}^{\prime }$
gives $\frac{g"}{g^2}=\frac{3}{2}$