The function $f(x) = \sin 3x$, $x \in \left[0, \frac{\pi}{2}\right]$

(A) is increasing on $\left[0, \frac{\pi}{6}\right]$

(B) is decreasing on $\left[\frac{\pi}{6}, \frac{\pi}{2}\right]$

(C) is increasing on $\left[0, \frac{\pi}{2}\right]$

(D) is decreasing on $\left[0, \frac{\pi}{2}\right]$

Choose the correct answer from the options given below:

Given: $f(x) = \sin 3x$, $x \in \left[0, \frac{\pi}{2}\right]$

To determine where the function increases or decreases, find the derivative using the chain rule:

$f'(x) = \cos 3x \cdot 3$

$f'(x) = 3\cos 3x$

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The function increases when $f'(x) > 0$, which means $\cos 3x > 0$

The function decreases when $f'(x) < 0$, which means $\cos 3x < 0$

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For $x \in [0, \frac{\pi}{2}]$, the range of $3x$ is:

$3x \in [0, \frac{3\pi}{2}]$

The cosine function is positive in $[0, \frac{\pi}{2})$ and negative in $(\frac{\pi}{2}, \frac{3\pi}{2}]$

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Finding the critical point where $\cos 3x = 0$:

$3x = \frac{\pi}{2}$

$x = \frac{\pi}{6}$

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For $x \in [0, \frac{\pi}{6}]$:

$3x \in [0, \frac{\pi}{2}]$

In this range, $\cos 3x \geq 0$, so $f'(x) \geq 0$

Therefore, the function is increasing on $[0, \frac{\pi}{6}]$

Statement (A) is correct.

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For $x \in [\frac{\pi}{6}, \frac{\pi}{2}]$:

$3x \in [\frac{\pi}{2}, \frac{3\pi}{2}]$

In this range, $\cos 3x \leq 0$, so $f'(x) \leq 0$

Therefore, the function is decreasing on $[\frac{\pi}{6}, \frac{\pi}{2}]$

Statement (B) is correct.

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Statement (C) claims the function is increasing on the entire interval $[0, \frac{\pi}{2}]$, but the function decreases after $x = \frac{\pi}{6}$

Statement (C) is incorrect.

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Statement (D) claims the function is decreasing on the entire interval $[0, \frac{\pi}{2}]$, but the function increases before $x = \frac{\pi}{6}$

Statement (D) is incorrect.

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The correct answer is: (A) and (B) only