Question

If $f:[-2,2]\to R$ defined by $f\left(x\right)=x^3+\tan x+\left[\frac{x^2+1}{p}\right]$ is an odd function, then the least value of $\left[p\right]$ is
($[.]$ represents the greatest integer function)

Answer 1

$f\left(x\right)=x^3+\tan x+\left[\frac{x^2+1}{p}\right]$
$f$ is an odd function.
$\therefore f(-x)=-f\left(x\right)\Rightarrow -x^3-\tan x+\left[\frac{x^2+1}{p}\right]=-x^3-\tan x-\left[\frac{x^2+1}{p}\right]\Rightarrow \left[\frac{x^2+1}{p}\right]=0$

Now, $x\in [-2,2]$
$\therefore x^2+1\in [1,5]$
So, for $f$ to be an odd function, $p\in (5,\infty )$
So, the least value of $\left[p\right]$ is $5$.