Paragraph: Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by…

Paragraph: Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2$
Question: Let $g(x)=\int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} d t$. Which of the following is true ?
  1. $g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$
  2. $g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$
  3. $g^{\prime}(x)$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$
  4. $g^{\prime}(x)$ does not change sign on $(-\infty, \infty)$

Solution

$\because g^{\prime}(x)=\frac{f^{\prime}\left(e^x\right)}{1+\left(e^x\right)^2} \cdot e^x=2 a\left[\frac{e^{2 x}-1}{\left(e^{2 x}+a e^x+1\right)^2}\right]\left(\frac{e^x}{1+e^{2 x}}\right)$ $\begin{aligned} & g^{\prime}(x)=0 \text {, if } e^{2 x}-1 & =0, \\ \text { i.e., } & & x=0 \\ \text { If } & & x < 0, e^{2 x} < 1 \Rightarrow g^{\prime}(x) < 0 .\end{aligned}$

Asked in: JEE Advanced 2008 (Paper 2)

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