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

$(x^2+ax+1)f\left(x\right)=x^2-ax+1$
Differentiating we obtain $(2x+a)f\left(x\right)+(x^2+ax+1)f^{\prime }\left(x\right)=2x-a$ ...(i)
Putting $x=1$ and noting that $f\left(1\right)=\frac{2-a}{2+a},$ we obtain
$(2+a)f\left(1\right)+(2+a)f^{\prime }\left(1\right)=2-a\Rightarrow f^{\prime }\left(1\right)=0$
Differentiating (i) again
$2f\left(x\right)+(2x+a)f^{\prime }\left(x\right)+(2x+a)f^{\prime }\left(x\right)+(x^2+ax+1)f^{\prime \prime }\left(x\right)=2$ ...(ii)
Putting $x=1$, we have $2\cdot \frac{2-a}{2+a}+(2+a)f^{\prime \prime }\left(1\right)=2$ $\Rightarrow f^{\prime \prime }\left(1\right)=\frac{4a}{(2+a{)}^2}$
Putting $x=-1$ in (i) and noting that $f(-1)=\frac{2+a}{2-a}\Rightarrow (-2+a)f(-1)+(2-a)f^{\prime }(-1)=-(2+a)\Rightarrow f^{\prime }(-1)=0$
Putting $x=-1$ in (ii), we obtain $f^{\prime \prime }(-1)=\frac{-4a}{(2-a{)}^2}$

So $(2+a{)}^2{f}^{\prime \prime }(1)+(2-a{)}^2{f}^{\prime }(-1)=0$
(i) $\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2+ax+1}$
$\left[\right(2x-a)-(2x+a)\times \frac{x^2-ax+1}{x^2+ax+1}]=\frac{2a(x^2-1)}{(x^2+ax+1{)}^2}$
So $f^{\prime }\left(x\right)<0$ if $x\in (-1,1)$ and $f^{\prime }\left(x\right)>0$ for $x\in (-\infty ,-1)\cup (1,\infty )$

i.e. f increases in$(\infty ,-1)\cup (1,\infty )$ and decreases on $(-1,1)$