Which of the following is/are odd functions? (where $[\cdot]$ denotes the greatest integer function)
A) $f:\mathbb{R}\to\mathbb{R}$ such that $f(x) = \dfrac{1 + x \cdot |x|}{x + |x|}$ \hspace{1em}(wherever defined)
B) $f:\mathbb{Z}\to\mathbb{R}$ such that $f(x) = x \cdot \dfrac{x^4[x] + x^2 \tan x}{1}$
C) $f:\mathbb{R}\setminus\mathbb{Q}\to\mathbb{R}$ such that $f(x) = \sin\!\left(\dfrac{\pi}{2}\left[x + \dfrac{1}{2}\right]\right)$
D) $f:\left\{\dfrac{4k\pm 1}{2} : k\in\mathbb{Z}\right\}\to\mathbb{R}$ such that $f(x) = \sin\!\left(\dfrac{\pi}{2}\left[x+\dfrac{1}{2}\right]\right) - \dfrac{1}{2}$
Step-by-Step Solution
Key Concept: Check $f(-x) = -f(x)$ for each. For option C on $\mathbb{R}\setminus\mathbb{Q}$: if $x + \frac{1}{2} \notin \mathbb{Z}$, then $[-x+\frac{1}{2}] = -[x+\frac{1}{2}] - 1$, so $f(-x) = \sin\frac{\pi}{2}(-[x+\frac{1}{2}]-1) = -\sin\frac{\pi}{2}[x+\frac{1}{2}] = -f(x)$.
B: On integers, $[x]=x$, and $f(x) = x(x^4 \cdot x + x^2\tan x)$; verify $f(-x)=-f(x)$. C: $f(-x)=-f(x)$ shown by GIF property on irrationals. D: The domain is symmetric and $f(-x)=-f(x)$ holds. B, C, D are odd.
Correct Answer: BCD