KCET · Maths · Functions
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
- A \(\{0\}\)
- B \(\{0,-1\}\)
- C \(\{0, \pm \sin 1\}\)
- D \(\{0,-\sin 1\}\)
Answer & Solution
Correct Answer
(D) \(\{0,-\sin 1\}\)
Step-by-step Solution
Detailed explanation
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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