Q76 / 170 · Manipal (MET) 2017 · Mathematics
If $f(x)=\left\{\begin{array}{cc}3 x^2+12 x-1, & -1 \leq x \leq 2 \\ 37-x, & 2 \leq x \leq 3\end{array}\right.$, then
- A$f(x)$ is decreasing on $[-1,2]$
- Correct answerB$f^{\prime}(2)$ does not exist
- C$f(x)$ has the maximum value at $x=2$
- DNone of the above
Solution
Given, $f(x)=\left\{\begin{array}{r}3 x^2+12 x-1,-1 \leq x \leq 2 \\ 37-x, 2 \leq x \leq 3\end{array}\right.$
On differentiating, we get
$\begin{array}{ll}
& f^{\prime}(x)=\left\{\begin{array}{cc}
6 x+12, & -1 \leq x \leq 2 \\
-1, & 2 \leq x \leq 3
\end{array}\right. \\
\Rightarrow & f^{\prime}(x)\gt0 \text { for }-1 \leq x \leq 2
\end{array}$
$\therefore f(x)$ is increasing in $[-1,2]$
$\lim _{x \rightarrow 2^{+}} f^{\prime}(x)=-1 \text { and } \lim _{x \rightarrow 2^{-}} f^{\prime}(x)=24$
$\therefore f^{\prime}(2)$ does not exist.
Hence, the correct answer is option: 'B'.