Consider $f(x) = \sin(3x) + 4, \forall x \in \mathbb{R}$. Then

(A) Maximum value of $f(x)$ is 5

(B) Minimum value of $f(x)$ is 3

(C) Maximum value of $f(x)$ is attained at $x = \frac{\pi}{6}$

(D) Minimum value of $f(x)$ is attained at $x = 0$

Choose the correct answer from the options given below:

The function is $f(x) = \sin(3x) + 4$

The sine function satisfies $-1 \leq \sin(3x) \leq 1$ for all $x \in \mathbb{R}$.

Adding 4 throughout:

$-1 + 4 \leq \sin(3x) + 4 \leq 1 + 4$

$3 \leq f(x) \leq 5$

Therefore:

Maximum value of $f(x) = 5$ (when $\sin(3x) = 1$)

Minimum value of $f(x) = 3$ (when $\sin(3x) = -1$)

Statement (A) is TRUE - Maximum value is 5

Statement (B) is TRUE - Minimum value is 3

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For statement (C), the maximum occurs when $\sin(3x) = 1$.

$\sin(\theta) = 1$ when $\theta = \frac{\pi}{2}$

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

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

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

$f(\frac{\pi}{6}) = \sin(3 \times \frac{\pi}{6}) + 4$

$= \sin(\frac{\pi}{2}) + 4$

$= 1 + 4$

$= 5$

Statement (C) is TRUE - Maximum is attained at $x = \frac{\pi}{6}$

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For statement (D), at $x = 0$:

$f(0) = \sin(3 \times 0) + 4$

$= \sin(0) + 4$

$= 0 + 4$

$= 4$

The minimum value is 3, not 4.

The minimum occurs when $\sin(3x) = -1$, which happens when $3x = \frac{3\pi}{2}$, giving $x = \frac{\pi}{2}$.

Statement (D) is FALSE - Minimum is not attained at $x = 0$

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Statements (A), (B), and (C) are correct.