Question

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

• A. $\frac{2}{3}$
• B. $\frac{3}{2}$
• C. $\frac{3}{4}$
• D. 3

Answer 1

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

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

$f^{{}^{\prime }}\left(x\right)=\frac{1}{\sqrt{1+x^3}}$

$f^{{}^{\prime }}\left(h\right(x\left)\right)=\frac{1}{\sqrt{1+h^3\left(x\right)}}$

$h^{{}^{\prime }}\left(x\right)=\sqrt{1+h^3\left(x\right)}$

$h^{\prime \prime }\left(x\right)=\frac{3h^2\left(x\right)}{2\sqrt{1+h^3\left(x\right)}}\times h^{{}^{\prime }}\left(x\right)$

$\frac{h^{\prime \prime }\left(x\right)}{h^2\left(x\right)}=\frac{3}{2}\frac{h^{{}^{\prime }}\left(x\right)}{\sqrt{1+h^3\left(x\right)}}=\frac{3}{2}$