Question

If $f\left(x\right)$ is an odd function, then

• A. $\frac{f(-x)+f\left(x\right)}{2}$is a constant function
• B. $\left[|f\right(x)|+1]$ is even where, $\left[x\right]=$ greatest integer $\le x$
• C. $\frac{f\left(x\right)-f(-x)}{2}$ is odd function
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\frac{f(-x)+f\left(x\right)}{2}$is a constant function
B $\left[|f\right(x)|+1]$ is even where, $\left[x\right]=$ greatest integer $\le x$
C $\frac{f\left(x\right)-f(-x)}{2}$ is odd function

Given: $f(-x)=-f\left(x\right)$
$f\left(x\right)+f(-x)=0$.
Hence, $\frac{f\left(x\right)+f(-x)}{2}$ is a constant function.
Now, let $g\left(x\right)=\left[|f\right(x)|+1]$
$\Rightarrow g(-x)=\left[|f\right(-x)|+1]=\left[|f\right(x)|+1]$ [Given: $f(-x)=-f\left(x\right)$], which is an even function.
Let $P\left(x\right)=\frac{f\left(x\right)-f(-x)}{2}=f\left(x\right)$ an odd function.
Hence, options 'A' , 'B' and 'C' are correct.