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The range of the function $f(x)=\sin [x],-\frac{\pi}{4} < x < \frac{\pi}{4}$, where $[x]$ denotes the greatest integer $\leq x$, is
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Verified Answer
The correct answer is:
$\{0,-\sin 1\}$
Given, $f(x)=\sin [x],-\frac{\pi}{4} < x < \frac{\pi}{4}$
$\begin{array}{ll}\text { Clearly, } & \sin 0=0 \\ \text { and } \quad & {\left[\frac{\pi}{4}\right]=\left[\frac{3.14}{4}\right]=0}\end{array}$
$\begin{aligned}
\therefore & \forall x \in\left[0, \frac{\pi}{4}\right], \sin [x]=0 \\
& \forall x \in\left[-\frac{\pi}{4}, 0\right],[x]=-1 \\
\therefore & \sin [x]=\sin (-1)=-\sin 1
\end{aligned}$
So, the range of $f(x)$ is $\{0,-\sin 1\}$.
$\begin{array}{ll}\text { Clearly, } & \sin 0=0 \\ \text { and } \quad & {\left[\frac{\pi}{4}\right]=\left[\frac{3.14}{4}\right]=0}\end{array}$
$\begin{aligned}
\therefore & \forall x \in\left[0, \frac{\pi}{4}\right], \sin [x]=0 \\
& \forall x \in\left[-\frac{\pi}{4}, 0\right],[x]=-1 \\
\therefore & \sin [x]=\sin (-1)=-\sin 1
\end{aligned}$
So, the range of $f(x)$ is $\{0,-\sin 1\}$.
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