Paragraph: Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 Question: Which of the following is true ?

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Aicharya · Accepted Answer

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

Paragraph: Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by…

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