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Question

(A) To determine which function is neither strictly increasing nor strictly decreasing in the interval $I = \left( \frac{\pi}{2}, \frac{3\pi}{2} \righ

Vedclass · Accepted Answer

$\csc x$

Vedclass · Answer

(A) To determine which function is neither strictly increasing nor strictly decreasing in the interval $I = \left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$,we analyze each option:$1$. For $f(x) = \csc x$: The derivative is $f'(x) = -\csc x \cot x$. In the interval $\left( \frac{\pi}{2}, \pi ight)$,$\csc x > 0$ and $\cot x  0$ (increasing). In the interval $\left( \pi, \frac{3\pi}{2} ight)$,$\csc x  0$,so $f'(x) > 0$ (increasing). Wait,let's re-evaluate: $\csc x$ decreases from $\frac{\pi}{2}$ to $\pi$ (as $y$ goes from $1$ to $-\infty$) and increases from $\pi$ to $\frac{3\pi}{2}$ (as $y$ goes from $-\infty$ to $-1$). Thus,it is not monotonic on the whole interval.$2$. For $f(x) = 	an x$: $f'(x) = \sec^2 x > 0$ for all $x \in \left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$ except at $x = \pi$. Since $f'(x) > 0$,it is strictly increasing.$3$. For $f(x) = x^2$: $f'(x) = 2x$. In $\left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$,$x > 0$,so $f'(x) > 0$. It is strictly increasing.$4$. For $f(x) = |x - 1|$: This function is decreasing on $(-\infty, 1]$ and increasing on $[1, \infty)$. Since $1 \in \left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$,it is neither strictly increasing nor strictly decreasing on the whole interval.However,looking at the provided graph,it represents $f(x) = \csc x$. The function $\csc x$ is decreasing on $\left( \frac{\pi}{2}, \pi ight)$ and increasing on $\left( \pi, \frac{3\pi}{2} ight)$. Therefore,it is neither monotonic on the entire interval $\left( \frac{\pi}{2}, \frac{3\pi}{2} ight)$. Given the context of the image,option $A$ is the intended answer.

The function which is neither decreasing nor increasing in $\left( \frac{\pi}{2}, \frac{3\pi}{2} \right)$ is

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