Q76 / 150 · MHT CET 2011 · Mathematics
At the point $x=1$, the function $f(x)=\left\{\begin{array}{l}x^{3}-1,1 < x < \infty \\ x-1,-\infty < x \leq 1\end{array}\right.$ is
- Acontinuous and differentiable
- Correct answerBcontinuous and not differentiable
- Cdiscontinuous and differentiable
- Ddiscontinuous and not differentiable
Solution
LHL $=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x-1)=0$
$$
\begin{aligned}
\mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x) \\
&=\lim _{x \rightarrow 1}\left(x^{3}-1\right)=0 \\
\text { Also, } \quad f(1) &=1-1=0
\end{aligned}
$$
$\therefore f$ is continuous at $x=1$ Now, $L f^{\prime}(1)=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}$
$\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(1-h)-1-0}{-h} \\ &=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \\ \text { and } R f^{\prime}(1) &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \end{aligned}$
$$
\begin{array}{l}
\quad=\lim _{h \rightarrow 0} \frac{(1+h)^{3}-1-0}{h} \\
=\lim _{h \rightarrow 0} \frac{1+h^{3}+3 h+3 h^{2}-1}{h} \\
=\lim _{h \rightarrow 0} h^{2}+3+3 h=3
\end{array}
$$
$\therefore f(x)$ is not differentiable at $x=1$
Hence, the correct answer is option: 'B'.