Question

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

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

Answer 1

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

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