Question

Consider the function
$f ( x )= \begin{cases} |\sin x | & \text { for } 0<| x | \leq \frac{\pi}{2} \\ \frac{1}{2} & \text { for } x =0\end{cases}$
Assertion: $f$ has a local maximum value at $x =0$.
Reason: $f '(0) = 0$ and $f ''(0) < 0$

• A. Assertion is correct, reason is correct; reason is a correct explanation for assertion.
• B. Assertion is correct, reason is correct; reason is not a correct explanation for assertion
• C. Assertion is correct, reason is incorrect
• D. Assertion is incorrect, reason is correct.

Answer 1

Answer: (C)

We draw the graph of $ f ( x ) $ for $x \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$

i.e., for $| x | \leq \frac{\pi}{2}$
Here, $f (0)=\frac{1}{2}$, whereas
$\underset{ x \rightarrow 0}{Lt} =\underset{ x \rightarrow 0}{Lt}|\sin x |=0 \neq f (0)$
$\therefore f$ is discontinuous at 0
$\Rightarrow f$ is not derivable at $0 .$
Thus, Reason is false.
However, we note that for all
$x \in\left(\frac{-\pi}{6}, 0\right) \cup\left(0, \frac{\pi}{6}\right)$
$|\sin\, x|<\frac{1}{2}$ i.e., $f(x) < f(0)$
$\Rightarrow $ fhas a local maximum value at $0 .$
So, Assertion is true.