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

Answer 1

Answer: (B)

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

$(x^2+ax+1)f\left(x\right)=x^2-ax+1$a
Differentiating we obtain $(2x+a)f\left(x\right)+(x^2+ax+1)f^{\prime }\left(x\right)=2x-a\left(i\right)$
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 (ii)
$2f\left(x\right)+(2x+a)f^{\prime }\left(x\right)+(2x+a)f^{\prime }\left(x\right)+(x^2+ax+1)f^{\prime \prime }\left(x\right)=2$
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 $f(-1)=\frac{2+a}{2-a}(-2+a)f(-1)+(2-a)f^{\prime }(-1)=-(2+a)\Rightarrow f^{\prime }(-1)=0$
Putting $x=-1$ in (ii), we obtain $f^{\prime \prime }(-1)=\frac{-4a}{(2-a{)}^2}$
so $(2+a{)}^2{f}^{\prime \prime }(1)+(2-a{)}^2{f}^{\prime }(-1)=0$
$\left(i\right)\Rightarrow f^{\prime }\left(x\right)=\frac{1}{x^2+ax+1}$
$\left[\right(2x-a)-(2x+a)\times \frac{x^2-ax+1}{x^2+ax+1}]=\frac{2a(x^2-1)}{(x^2+ax+1{)}^2}$
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 $(-1,1)$
$g^{\prime }\left(x\right)=\frac{f^{\prime }\left(e^x\right)e^x}{1+e^{2x}}<0$ if $f^{\prime }\left(e^x\right)<0$
But
$f^{\prime }\left(e^x\right)<0$ if $e^x\in (-1,1)$ For $x>0e^x>1$ but for $x<0,0<e^x<1$ Thus $f^{\prime }\left(e^x\right)<0$ if $x\in (-\infty ,0)$
and is positive for $x\in (0,\infty )$