Question

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

• 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 $|x|\le \frac{\pi }{2}$
Here, $f(0)=\frac{1}{2}$, whereas
$\underset{x\to 0}{\mathrm{Lt}}=\underset{x\to 0}{\mathrm{Lt}}|\sin x|=0\ne f(0)$
$\therefore f$ 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.