If $f: R \in R$ is defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1}, 0<a<2$, then which of the following is true:

$(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$

We have,

$f(x)=\frac{x^2-a x+1}{x^2+a x+1}$

$\Rightarrow f^{\prime}(x) =\frac{\left(x^2+a x+1\right)(2 x-a)-\left(x^2-a x+1\right)(2 x+a)}{\left(x^2+a x+1\right)^2}$

$\Rightarrow f^{\prime}(x)=\frac{2 a\left(x^2-1\right)}{\left(x^2+a x+1\right)^2}$

$\Rightarrow f^{\prime \prime}(x) =\frac{4 a x\left(x^2+a x+1\right)^2-4 a\left(x^2-1\right)(2 x+a)\left(x^2+a x+1\right)}{\left(x^2+a x+1\right)^4}$

$\Rightarrow f^{\prime \prime}(x)=\frac{4 a\left\{x\left(x^2+a x+1\right)-\left(x^2-1\right)(2 x+a)\right\}}{\left(x^2+a x+1\right)^3}$

∴  $f^{\prime}(1)=0 $ and $f^{\prime \prime}(1)=\frac{4 a}{(2+a)^2}, f^{\prime \prime}(-1)=-\frac{4 a}{(2-a)^2}$

$\Rightarrow (2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=0$