Question

If $f\left(x\right)=\frac{\cos x}{\left[\frac{x}{2\pi }\right]+\frac{1}{2}}$ $\forall$ $x$ $\in$ $[-2\pi ,2\pi ]$ and $[\cdot ]$ represent greatest integer. Then which of the following is true

• A. $f\left(x\right)$ is an odd function
• B. $f\left(x\right)$ is an even function
• C. $f{}^{\prime }\left(x\right)=2\sin x\forall -2\pi \le x<0$
• D. $f{}^{\prime }\left(x\right)=-2\sin x\forall 0<x<2\pi$

Answer 1

Answer: (A), (C), (D)

The correct options are
A $f\left(x\right)$ is an odd function
C $f{}^{\prime }\left(x\right)=2\sin x\forall -2\pi \le x<0$
D $f{}^{\prime }\left(x\right)=-2\sin x\forall 0<x<2\pi$

Case 1: $-2\pi \le x<0$
$f\left(x\right)=\frac{\cos x}{-1+\frac{1}{2}}=-2\cos x$
case 2: $0\le x<2\pi$
$f\left(x\right)=\frac{\cos x}{0+\frac{1}{2}}=2\cos x$
Clearly $f(-x)=-f\left(x\right)\Rightarrow f\left(x\right)$ is an odd function.
Option 'C' and 'D' are obviously correct.