Question

Let $g\left(x\right)={\int }_0^{e^x}\frac{f\text{'}\left(t\right)}{1+t^2}dt$. Which of the following is true

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

Answer 1

The correct option is B

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

Explanation for correct answer Option(s)

Find the value of the given Equation.

The given Equation is:

$f:\left(-\infty ,\infty \right)\to \left(-\infty ,\infty \right)$ is defined as

$$\begin{gathered} f\left(x\right)=\frac{x^2-ax+1}{x^2+ax+1},\left(0<a<2\right) \\ g\left(x\right)={\int }_0^{e^x}\frac{f\text{'}\left(t\right)}{1+t^2}dt \end{gathered}$$

Let,

$$\begin{gathered} g\left(x\right)={\int }_0^{e^x}\frac{f\text{'}\left(t\right)}{1+t^2}dt \\ \text{differentiating with respect to}x \\ g\text{'}\left(x\right)=\frac{f\text{'}\left(e^x\right).e^x}{1+e^{2x}} \end{gathered}$$

Then,

$$\begin{gathered} f\left(x\right)=\frac{x^2-ax+1}{x^2+ax+1} \\ \text{differentiating with respect to}x \\ f\text{'}\left(x\right)=\frac{\left(x^2+ax+1\right)\left(2x-a\right)-\left(x^2-ax+1\right)\left(2x+a\right)}{{\left(x^2+ax+1\right)}^2} \\ f\text{'}\left(x\right)=\frac{2x^3+2ax+2x-a{x}^2-a^{2}x-a-\left(2x^3+2a{x}^2+2x+a{x}^2-a^{2}x+a\right)}{{\left(x^2+ax+1\right)}^2} \end{gathered}$$

To check positive solution, we will check only the numerator as the denominator is squared and thus will be positive.

$$\begin{gathered} f\text{'}\left(x\right)=2x^3+2ax+2x-a{x}^2-a^{2}x-a-\left(2x^3+2a{x}^2+2x+a{x}^2-a^{2}x+a\right) \\ f\text{'}\left(x\right)=2x^3+2ax+2x-a{x}^2-a^{2}x-a-2x^3+2a{x}^2-2x-a{x}^2+a^{2}x-a \\ f\text{'}\left(x\right)=4a{x}^2-2a{x}^2-2a \\ f\text{'}\left(x\right)=a\left(4x^2-2x^2-2\right) \\ f\text{'}\left(x\right)=a\left(2x^2-2\right) \\ f\text{'}\left(x\right)=2a\left(x^2-1\right) \\ \mathrm{Putting}\mathrm{the}\mathrm{inequality} \\ \Rightarrow 2a\left(e^{2x-1}\right)>0 \\ \Rightarrow \left(e^{2x-1}\right)>0 \\ \Rightarrow e^{2x}>0 \end{gathered}$$

Taking log on both sides we get,

$$\begin{gathered} \Rightarrow 2x>0 \\ \Rightarrow x>0 \end{gathered}$$

When $x$ is greater than zero $f\text{'}\left(e^x\right)$ will be positive and When $x$ is lesser than zero $f\text{'}\left(e^x\right)$ will be negative.

$g\text{'}\left(x\right)=f\text{'}\left(e^x\right)$

Both will have same sign

Case I : When $I:x\in \left(-\infty ,0\right)$

$$\begin{gathered} \Rightarrow f\text{'}\left(e^x\right)<0 \\ \Rightarrow g\text{'}\left(x\right)<0 \end{gathered}$$

Hence, $g\text{'}\left(x\right)$ is negative for $x\in \left(-\infty ,0\right)$

Case II : When $x\in \left(0,\infty \right)$

$$\begin{gathered} \Rightarrow f\text{'}\left(e^x\right)>0 \\ \Rightarrow g\text{'}\left(x\right)>0 \end{gathered}$$

Hence, $g\text{'}\left(x\right)$ is positive for $x\in \left(0,\infty \right)$

Therefore, the correct answer is Option B.