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If $F(x)=\int_{0}^{x} \frac{\cos t}{\left(1+t^{2}\right)} d t, 0 \leq x \leq 2 \pi .$ Then
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The correct answer is:
$F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
Given function is
$$
F(x)=\int_{0}^{x} \frac{\cos t}{\left(1+t^{2}\right)} d t, 0 \leq x \leq 2 \pi
$$
On differentiation w.r.t. $x$. (apply Leibnitz rule) $F(x)=\frac{\cos x}{1+x^{2}} \times 1=\frac{\cos x}{1+x^{2}},$ where $\left(1+x^{2}\right)>0$
Here, $\cos x>0 \Rightarrow x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$
and $\cos x < 0 \Rightarrow x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
So, $F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
$$
F(x)=\int_{0}^{x} \frac{\cos t}{\left(1+t^{2}\right)} d t, 0 \leq x \leq 2 \pi
$$
On differentiation w.r.t. $x$. (apply Leibnitz rule) $F(x)=\frac{\cos x}{1+x^{2}} \times 1=\frac{\cos x}{1+x^{2}},$ where $\left(1+x^{2}\right)>0$
Here, $\cos x>0 \Rightarrow x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$
and $\cos x < 0 \Rightarrow x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
So, $F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
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