Consider the function f(x)=[ x], x [0, ]. Assertion (A) : f(x) is… - Vidyadip

MCQ

Consider the function $f(x)=[\sin x], x \in[0, \pi]$. Assertion $(A) : f(x)$ is not continuous at $x=\frac{\pi}{2}$.
Reason $(R) : \lim _{x \rightarrow \frac{\pi}{2}} f(x)$ does not exist.

A
Both $(A)$ and $(R)$ are true and $(R)$ is the correct explanation of $(A).$
B
Both $(A)$ and$ (R)$ are true but $(R)$ is not the correct explanation of $(A).$
✓
$(A)$ is true but $(R)$ is false.
D
$(A)$ is false but $(R)$ is true.

✓

Answer

Correct option: C.

$(A)$ is true but $(R)$ is false.

We know that for all
$x \in\left[0, \frac{\pi}{2}\right) \cup\left(\frac{\pi}{2}, \pi\right], 0<\sin x<1$
$\Rightarrow[\sin x]=0$
$\therefore \lim _{x \rightarrow \frac{\pi}{2}}[\sin x]=0$
Thus, we see that the Reason is not true.
Also, $f\left(\frac{\pi}{2}\right)=\left[\sin \frac{\pi}{2}\right]=1$
$\Rightarrow \lim _{x \rightarrow \frac{\pi}{2}} f(x) \neq f\left(\frac{\pi}{2}\right)$
$\therefore f$ is not continuous at $x=\frac{\pi}{2}$

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