How to find whether $f(x)=\frac{2x(\sin(x)+\tan(x))}{2[\frac{x+2\pi}{\pi}]-3}$ is a many-to-one function or not?

Question

Question:

How to find whether the function $$f(x)=\frac{2x(\sin(x)+\tan(x))}{2[\frac{x+2\pi}{\pi}]-3}$$ is a many-to-one function or not?

where [.] represents greatest integer function or floor function.

Graph of $f(x)$:

Clearly, any horizontal line cuts the graph at more than one point, thus the function is many-to-one. I solved this without using the graph by taking an example, i.e., $f(0)=f(\pi)=0$, thus the function is many-to-one.

But, I wish to solve this problem algebraically(without taking any set of examples). Usually, we take derivative and see whether the sign of derivative changes or not to determine whether the function is many-to-one or not. But here due to the presence of floor function, I am unable to calculate the derivative. Is there any other way to prove the function is many-to-one, as solving by using graphing calculator is easier but can't be used in the exam.

score 1 · Answer 1 · answered Aug 11 '19 at 04:53

1

$\left[\frac{x+2\pi}{\pi} \right]= \left[\frac{x}{\pi} + 2\right] = \left[ \frac{x}{\pi} \right] +2 $

Now

$\left[\frac{x}{\pi} \right]=\begin{cases} \vdots \\ -1 & -\pi \leq x < 0 \\ 0 & 0\leq x< \pi \\ 1 & \pi \leq x < 2 \pi \\ \vdots \end{cases}$

answered Aug 11 '19 at 04:53

IrbidMath

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so the function is piecewise constant – IrbidMath Aug 11 '19 at 04:54
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Thanks for your reply. So we can consider the denominator to be a constant while differentiating, right? – Vishnu Aug 11 '19 at 05:09
Yeah and the derivative will be as a piecewise function also which is no defined at multiples of $\pi$ because it is not continuous – IrbidMath Aug 11 '19 at 05:13

How to find whether $f(x)=\frac{2x(\sin(x)+\tan(x))}{2[\frac{x+2\pi}{\pi}]-3}$ is a many-to-one function or not?

1 Answers1