Finding range of $f(x)= \frac{\tan(\pi\lfloor x^2-x \rfloor)}{1+\sin(\cos x)}$

Question

Range of
$$f(x)= \dfrac{\tan(\pi\lfloor x^2-x \rfloor)}{1+\sin(\cos x)},$$ where $\lfloor x \rfloor$is greatest integer function.

Options-

• (A). R- $\lfloor 0,\tan 1 \rfloor$
• (B). R-$ \lfloor \tan 2,1 \rfloor$,
• (C). $\lfloor \tan 2,\tan 1 \rfloor$
• (D). $\{0\}$

My attempt-- I tried by going with the options and eliminating and checking, it was complicated.

Can you please help me find in a better theoretical or geometrical way.

Answer 1

Realize that the greatest integer function will only give you integers. So this:

$$\require{mathtools} \tan(\pi{\lfloor}{x^2 - x}{\rfloor})$$

Will only ever give you integer multiples of $\pi$. Then remember that the tangent of all integer multiples of $\pi$ is $0$ because $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$ and $\sin \theta$ is always $0$ when $\theta$ is an integer multiple of $\pi$. Thus the whole function is always $0$ in its domain as the numerator is always $0$—and the range is $\{0\}$.

You can go even further to say the function is defined for all real numbers because $\cos x$ in the denominator oscillates between $-1$ and $1$. That means:

$$1 + \sin -1 \leq 1 + \sin(\cos x) \leq 1+\sin 1$$

Since $\sin 1$ and $\sin -1$ are never less than $\approx -0.842$ the denominator is never zero thus the function is defined for all real numbers and is always zero.