Application of Derivatives 4 Question 39
41. Let $g(x)=\int _0^{e^{x}} \frac{f^{\prime}(t)}{1+t^{2}} d t$. Which of the following is true?
(a) $g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$
(b) $g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$
(c) $g^{\prime}(x)$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$
(d) $g^{\prime}(x)$ does not change sign $(-\infty, \infty)$
Analytical & Descriptive Questions
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Answer:
Correct Answer: 41. $a=\frac{1}{4}, b=\frac{-5}{4} ; f(x)=\frac{1}{4} x^{2}-\frac{5}{4} x+2$
Solution:
- $\quad g^{\prime}(x)=\frac{f^{\prime}\left(e^{x}\right)}{1+\left(e^{x}\right)^{2}} \cdot e^{x}$
$$ =2 a \frac{e^{2 x}-1}{\left(e^{2 x}+a e^{x}+1\right)^{2}} \quad \frac{e^{x}}{1+e^{2 x}} $$
$$ g^{\prime}(x)=0 \text {, if } e^{2 x}-1=0 \text {, i.e. } x=0 $$
If
$$ x<0, e^{2 x}<1 \Rightarrow g^{\prime}(x)<0 $$
