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. $f\left(x\right)$ is decreasing on $(1,1)$ and has a local minimum at $x=1$
• B. $f\left(x\right)$ is increasing on $(1,1)$ and has a local maximum at $x=1$
• C. $f\left(x\right)$ is increasing on $(1,1)$ but has neither a local maximum nor a local minimum at $x=1$
• D. $f\left(x\right)$ is decreasing on $(1,1)$ but has neither a local maximum nor a local minimum at $x=1$

Answer 1

The correct option is A $f\left(x\right)$ is decreasing on $(1,1)$ and has a local minimum at $x=1$

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

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

$f^{\prime }\left(x\right)<0$ for $x\in (-1,1)$

therefore, $f\left(x\right)$ is decreasing in $x\in (-1,1)$

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

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

and thus has minima at $x=1$