Question

Let $g\left(x\right)={\int }_0^{c^x}\frac{f^{\prime }\left(t\right)}{1+t^2}dt$, which of the following is true?

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

Answer 1

The correct option is B $g^{\prime }\left(x\right)$ is negative on $(-\infty ,0)$ & positive on $(0,\infty )$

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

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

$f^{\prime }\left(x\right)$ is positive for $x\in (-\infty ,1)\cup (1,\infty )$

$f^{\prime }\left(e^x\right)=\frac{2a(e^x+1)(e^x-1)}{{(e^{2x}+a{e}^x+1)}^2}=0$

$f^{\prime }\left(e^x\right)$ is negative for $(-\infty ,0)$

$f^{\prime }\left(e^x\right)$ is positive for $(0,\infty )$

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

Applying Leibitz rule we get

$g^{\prime }\left(x\right)=\frac{f^{\prime }\left(e^x\right)e^x}{(1+e^x)}=0g^{\prime }\left(x\right)=\frac{20(e^x+1)(e^x-1)e^x}{\left({1+e}^{2x}\right)}$

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