Question

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

Answer 1

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