Function Analysis

Qn: Show that the function given by f(x) = sin x is 

A) Strictly increasing in $(0, \frac{\pi}{2})$

B) Strictly decreasing in $(\frac{\pi}{2}, \pi)$

C) Neither increasing nor decreasing in $(0, \pi)$

[How to solve]

 

Ans: The given function is $f(x) = \sin x$.

Therefore, $f′(x) = \cos x$

A)

Since for each $x \in (0, \frac{\pi}{2})$, $\cos x > 0$, we have $f′(x) > 0$.

Hence, $f$ is strictly increasing in $(0, \frac{\pi}{2})$.

B)

Since for each $x \in (\frac{\pi}{2}, \pi)$, $\cos x < 0$, we have $f′(x) < 0$.

Hence, $f$ is strictly decreasing in $(\frac{\pi}{2}, \pi)$.

C)

It is clear from the results obtained in (A) and (B) that $f$ is neither increasing nor decreasing in $(0, \pi)$.