If $f(x)=\frac{a^2-1}{a^2+1} x^3-3 x+2 \log _e 5$ is a decreasing function of x for all $x \in R$, then the set of possible values of a, is

$[-1,1]$

We have,

$f(x)=\frac{a^2-1}{a^2+1} x^3-3 x+2 \log _e 5$ for all $x \in R$

$\Rightarrow f'(x)=3\left(\frac{a^2-1}{a^2+1}\right) x^2-3$ for all $x \in R$

For f(x) to be decreasing on R, we must have

$f'(x)<0$ for all $x \in R$

$\Rightarrow 3\left(\frac{a^2-1}{a^2+1}\right) x^2-3<0$ for all $x \in R$

$\Rightarrow \left(\frac{a^2-1}{a^2+1}\right) x^2-1<0$ for all $x \in R$

$\Rightarrow \frac{a^2-1}{a^2+1} \leq 0 \Rightarrow a^2-1 \leq 0 \Rightarrow a \in[-1,1]$