Question

Consider the function $f:(-\infty ,\infty )\to (-\infty ,\infty )$ defined by$f\left(x\right)=\frac{[x^2-ax+1]}{[x^2+ax+1]};0<a<2$. Which of the following is true

• A. ${(2+a)}^{2}f\text{'}\text{'}\left(1\right)+{(2–a)}^{2}f\text{'}\text{'}(-1)=0$
• B. ${(2–a)}^{2}f\text{'}\text{'}\left(1\right)–{(2+a)}^{2}f\text{'}\text{'}(-1)=0$
• C. $f\text{'}\left(1\right)f\text{'}(-1)={(2–a)}^2$
• D. $f\text{'}\left(1\right)f\text{'}(-1)={-(2+a)}^2$

Answer 1

The correct option is A

${(2+a)}^{2}f\text{'}\text{'}\left(1\right)+{(2–a)}^{2}f\text{'}\text{'}(-1)=0$

Explanation of the correct option :

Step1. Given function :

$f\left(x\right)=\frac{[x^2-ax+1]}{[x^2+ax+1]};0<a<2$

Discriminant of denominator $a^2–4<0$

Denominator $\ne 0,\forall x\in R$

Step 2. Differentiate with respect to x :

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{\left[\right(x^2+ax+1)(-2a)+2ax(2x+a\left)\right]}{{[x^2+ax+1]}^2}\\ & =& \frac{[2a{x}^2–2a]}{{[x^2+ax+1]}^2}\\ & =& \frac{\left[2a\right(x^2–1\left)\right]}{{[x^2+ax+1]}^2}\\ f\text{'}\left(1\right)& =& f\text{'}(-1)=0\\ f\text{'}\text{'}\left(x\right)& =& \frac{[2a{(x^2+ax+1)}^2(2x)–2a(x^2–1)2(x^2+ax+1)(2x+a\left)\right]}{{(x^2+ax+1)}^4}\\ f\text{'}\text{'}\left(1\right)& =& \frac{2a[2(2+a\left)\right]}{{(2+a)}^3}=\frac{4a}{{(2+a)}^2}\\ f\text{'}\text{'}(-1)& =& \frac{-4a}{{(2-a)}^2}\\ f\text{'}\text{'}\left(1\right){(2+a)}^2+f\text{'}\text{'}(-1){(2–a)}^2& =& 0\end{array}$

Hence, the correct option is A.