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 minimum at $x=1$
• C. $f\left(x\right)$ is decreasing on $(-1,1)$ and has a local maximum 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 $x=1$

Explanation for the correct option:

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

$\Rightarrow f\text{'}\left(x\right)=\frac{\left[2a\left(x+1\right)\left(x-1\right)\right]}{{\left(x^2+ax+1\right)}^2}$

$\Rightarrow f\text{'}\left(x\right)>0$ in $(-\infty ,-1)\cup (1,\infty )$and $f\text{'}\left(x\right)<0$ in $(-1,1)$

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

and as $f\text{'}\left(x\right)<0$for $x<1$and $f\text{'}\left(x\right)>0$ for $x>1$

i.e. $f$decreases before $x=1$and increases after it.

So, by first derivative test, $f$has a local minimum at $x=1$

Hence, the correct option is A.