Mathematics - Differentiation Question with Solution | TestHub

\because f(x)= & \frac{\left(x^2+a x+1\right)-2 a x}{x^2+a x+1}=1-\frac{2 a x}{x^2+a x+1} \\ \therefore \quad f^{\prime}(x) & =-\left[\frac{\left(x^2+a x+1\right) \cdot 2 a-2 a x(2 x+a)}{\left(x^2+a x+1\right)^2}\right] \\ & =-\left[\frac{-2 a x^2+2 a}{\left(x^2+a x+1\right)^2}\right]=2 a\left[\frac{\left(x^2-1\right)}{\left(x^2+a x+1\right)^2}\right] \end{aligned} \begin{aligned} & \text { and } \begin{aligned} f^{\prime \prime}(x) & =2 a\left[\frac{\left(x^2+a x+1\right)^2(2 x)-2\left(x^2-1\right)\left(x^2+a x+1\right)(2 x+a)}{\left(x^2+a x+1\right)^4}\right] \\ & =2 a\left[\frac{2 x\left(x^2+a x+1\right)-2\left(x^2-1\right)(2 x+a)}{\left(x^2+a x+1\right)^3}\right] \end{aligned} \\ & \text { Now, } f^{\prime \prime}(1)=\frac{4 a(a+2)}{(a+2)^3}=\frac{4 a}{(a+2)^2} \end{aligned} and $f(-1)=\frac{4 a(a-2)}{(2-a)^3}=-\frac{4 a}{(a-2)^2}$ \therefore(2+a)^2 f^{\prime \prime}(1)+(2-a)^2 f^{\prime \prime}(-1)=4 a-4 a=0 $$