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Question: Find the range of $f(x)=sin[x] $where $(-π/4)<= x <= (π/4) $and$ [x] $is the greatest integer function of $x$

Since$ π=3.141..., π/4 < 1$ so $[π/4]=0 $so upper limit of $f(x)$ is $0 $ also$ -π/4 > -1$ so$ [-π/4] is -1$. Therefore the range should be$ {sin(-1),0}$. Though I think I'm right, the solution in the answer key is given as ${sin(-1),0,sin(1)}$. How is it possible to get$ sin(1) $as a result? Please verify, Thanks!

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    $\sin (1)$ shouldn't be there. – 2017-02-11
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    try to write in math mode – 2017-02-11
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    yeah $\sin(1)$ shouldn't be there. – 2017-02-11
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    What you mean it shouldn't be there? For $0 <= x < 1$, $floor(x)=0$ and for $1 <= x <= \frac{\pi}{4}$, $floor(x)=1$? So the domain consists of $x \in \{-1, 0, 1\}$ and the range thus becomes $\sin(-1), \sin(0), \sin(1)$? – 2018-05-10

2 Answers 2

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The floor function of a float/real returns the integer right below of the number.

For $0<=x<1$, $floor(x)=0$ and for $1<=x<=\frac{\pi}{4}$, $floor(x)=1$? So the domain consists of $x \in \{-1, 0, 1\}$ and the range thus becomes $\sin(−1),\sin(0),\sin(1)$.

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You have determined that [x] takes on the values -1 (when x < 0) and 0 (when x $geq$0 )

The range is the set of output values of sin[x] so when x < 0, f(x) = sin[x] = sin(-1)

and when when x $geq$0, f(x) = sin[x] = sin(0) = 0.

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    Expect when $1<= x <= \frac{\pi}{4}$ and then $floor(x)=1$? – 2018-05-10