Question

Let $f\left(x\right)={\int }_0^x\frac{dt}{\sqrt{1+t^3}}$ and g(x) be the inverse of f(x), then which one of the following holds good?

• A. $2g^{\prime \prime }=g^2$
• B. $2g^{\prime \prime }=3g^2$
• C. $3g^{\prime \prime }=2g^2$
• D. $3g^{\prime \prime }=g^2$.

Answer 1

The correct option is B $2g^{\prime \prime }=3g^2$

Here $f\left(g\right(x\left)\right)=x$ ($\because g\left(x\right)=f^{-1}\left(x\right)$)

i.e., ${\int }_0^{g\left(x\right)}\frac{dt}{\sqrt{1+t^3}}=x$ .....(1)

Let, $g\left(y\right)=t$

$\Rightarrow g^{{}^{\prime }}\left(y\right)dy=dt$

Substituting above equation in (1) and changing the limits...

${\int }_0^x\frac{g^{{}^{\prime }}\left(y\right)dy}{\sqrt{1+{g\left(y\right)}^3}}=x$

But, ${\int }_0^{x}dy=x$

Therefore, $\frac{g^{{}^{\prime }}\left(y\right)}{\sqrt{1+{g\left(y\right)}^3}}=1$

$\Rightarrow g^{{}^{\prime }}\left(y\right)=\sqrt{1+{g\left(y\right)}^3}$

$\Rightarrow g^{{}^{\prime }}(y{)}^2=1+{g\left(y\right)}^3$

Taking derivatives on both sides,

$\Rightarrow 2g^{{}^{\prime }}\left(y\right)g^{{}^{\prime \prime }}\left(y\right)=3g\left(y{)}^2{g}^{{}^{\prime }}\right(y)$

Cancelling $g^{{}^{\prime }}\left(y\right)$ on both sides,

$2g^{{}^{\prime \prime }}\left(y\right)=3g(y{)}^2$

Hence, option B is correct.