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

Answer 1

The correct option is A $(2+a{)}^2{f}^{\prime \prime }(1)+(2-a{)}^2{f}^{\prime \prime }(-1)=0$

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

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

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