Question

Consider the function $f:(-\infty ,\infty )\to (-\infty ,\infty )$ defined by $f\left(x\right)=\frac{x^2-ax+1}{x^2+ax+1}$ , $0<a<2$.

Let $g\left(x\right)={\int }_0^{e^x}\frac{f^{\prime }\left(t\right)}{1+t^2}dt$

which of the following is true?

• A. $g^{\prime }\left(x\right)$ is positive on $(-\infty ,0)$ and negative on $(0,\infty )$
• B. $g^{\prime }\left(x\right)$ is negative on $(-\infty ,0)$ and positive on $(0,\infty )$
• C. $g^{\prime }\left(x\right)$ changes $sign$ on both $(-\infty ,0)$ and $(0,\infty )$
• D. $g^{\prime }\left(x\right)$ does not change $sign$ on $(-\infty ,\infty )$

Answer 1

Answer: (B)

$g^{\prime }\left(x\right)$ is negative on $(-\infty ,0)$ and positive on $(0,\infty )$

$f\left(x\right)=\frac{x^2-ax+1}{x^2+ax+1}$ , $0<a<2$.
now, $g\left(x\right)={\int }_0^{e^x}\frac{f^{\prime }\left(t\right)}{1+t^2}dt$
$g^{\prime }\left(x\right)=\frac{f^{\prime }\left(e^x\right)e^x}{1+e^{2x}}=\frac{2a(e^{2x}-1)e^{2x}}{(1+e^{2x})(e^{2x}+a{e}^x+1{)}^2}$
Hence $g^{\prime }\left(x\right)$ is positive for $(0,\infty )$ and negative for $(-\infty ,0)$