Range of values of $k$ in floor function

Question

If the function $f(x)=\lfloor 3.5+k\sin(x)\rfloor$ is an even function.Then set of values of $k$ is

Options :

$(a)\ (-0.5,0.5)$

$(b)\ [-0.5,0.5]$

$(c)\ (0,1)$

$(d)\ [-1,1]$

What I try : If function $f(x)$ is even function.

Then $\displaystyle f(-x)=f(x)$

So we have

$\displaystyle \lfloor 3.5+k\sin(x)\rfloor =\lfloor 3.5-k\sin(x)\rfloor $

$\displaystyle \lfloor 0.5+k\sin(x)\rfloor =\lfloor 0.5-k\sin x\rfloor $

(Because $\lfloor x+I\rfloor =\lfloor x\rfloor +I$ )

Now I did not understand how do I solve further, please help me

Answer 1

For $k=\frac 12$, we get $$f\bigg(\frac{\pi}{2}\bigg)=4\not=3=f\bigg(-\frac{\pi}{2}\bigg)$$

Therefore, $(b)(c)(d)$ are not correct.

For $-\frac 12\lt k\lt \frac 12$, since we have $$|k\sin(x)|=|k||\sin(x)|\lt \frac 12\times 1=\frac 12$$ we obtain $$-\frac 12\lt k\sin(x)\lt\frac 12\qquad\text{and}\qquad -\frac 12\lt -k\sin(x)\lt\frac 12$$ from which we get $$f(x)=3=f(-x)$$

Therefore, $(a)$ is correct.