Question

If the function $f(x)$ is differentiable at $x=0$, then find the value of $\left(b^2-2a+c^6\right)$.

Answer 1

Answer: 48

As $f(x)$ is derivable at $x=0$, so $f(x)$ is also continuous at $x=0$.
$\therefore f\left(0^{+}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{\ln (1-ch)}{h}\left(\frac{0}{0}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{-c\times \ln (1-ch)}{-ch}=-c$
$\Rightarrow -c=2\Rightarrow c=-2\dots \dots \dots ..(1)$
$\text{ Now }{f}^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{\frac{\ln (1+2h)}{h}-2}{h}=\underset{h\to 0}{\mathrm{Lim}}\frac{\ln (1+2h)-2h}{h^2}=\underset{h\to 0}{\mathrm{Lim}}\frac{\left(2h-\frac{(2h{)}^2}{2}+\dots \dots \right)-2h}{h^2}=-2$
Now $f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\mathrm{Lim}}\frac{\frac{\ln (1+2h)}{h}-2}{h}=\underset{h\to 0}{\mathrm{Lim}}\frac{\ln (1+2h)-2h}{h^2}=\underset{h\to 0}{\mathrm{Lim}}\frac{\left(2h-\frac{(2h{)}^2}{2}+\dots \dots \right)-2h}{h^2}=-2$
As $f^{\prime }\left(0^{-}\right)=f^{\prime }\left(0^{+}\right)$, so $\frac{-4a}{b^2+16}=-2$
$\Rightarrow 2a=b^2+16$
$\therefore b^2-2a=-16$
Hence $\left(b^2-2a+c^6\right)=-16+64=48$
(Using equation (1) and equation (2))