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JEE Advanced · Mathematics · 25. AOD

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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. A
    \(g^{\prime}(x)\) is positive on \((-\infty, 0)\) and negative on \((0, \infty)\)
  2. B
    \(g^{\prime}(x)\) is negative on \((-\infty, 0)\) and positive on \((0, \infty)\)
  3. C
    \(g^{\prime}(x)\) changes sign on both \((-\infty, 0)\) and \((0, \infty)\)
  4. D
    \(g^{\prime}(x)\) does not change sign on \((-\infty, \infty)\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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