Q104 / 200 · WBJEE 2013 · Mathematics
If $f(x)=\left\{\begin{array}{ll}x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2\end{array}\right.$ then
- A$\lim _{x \rightarrow 2} f(x)$ does not exist
- B$f$ is not continuous at $x=2$
- Correct answerC$f$ is continuous but not differentiable at $x=2$
- D$t$ is continuous and differentiable at $x=2$
Solution
Given. $f(x)=\left\{\begin{aligned} x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2 \end{aligned}\right.$
LHL
$$
\begin{array}{l}
=f(2-0)=\lim _{h \rightarrow 0}(2-h)^{3}-3(2-h)+2 \\
=(2)^{3}-6+2=8-6+2=4 \\
\begin{aligned}
LH L=f(2+0) &=\lim _{h \rightarrow 0}(2+h)^{3}-6(2+h)^{2} \\
&=(2)^{3}-6(2)^{2}+9(2)+2 \\
&=8-24+18+2=4
\end{aligned}
\end{array}
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 2} f(x)$ exist
and $\begin{aligned} f(2)=(2)^{3} &-6(2)^{2}+9(2)+2 \\ &=8-24+18+2=4 \end{aligned}$
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(2)$
So, $f(x)$ is continuous at $x=2$
Now, $\quad f^{\prime}(x)=\left\{\begin{array}{cc}3 x^{2}-3, & x < 2 \\ 3 x^{2}-12 x+9, & x \geq 2\end{array}\right.$
$\begin{array}{ll}\therefore \quad & L f^{\prime}(2)=3(2)^{2}-3=12-3=9 \\ \text { and } \quad & R f^{\prime}(2)=3(2)^{2}-12(2)+9 \\ & =12-24+9=-3\end{array}$
$\because \quad L f^{\prime}(2) \neq R f^{\prime}(2)$
$\therefore f(x)$ is not differentiable at $x=2$
Hence. $f$ is continuous but not differentiable at $x=2$
Hence, the correct answer is option: 'C'.