Match List-I with List-II

List-I	List-II
(A) The maximum value of $f(x) = \sin(3x) + 6$	(I) 2
(B) The maximum value of $f(x) = -\vert x + 2\vert + 4$	(II) 5
(C) The minimum value of $f(x) = (3x + 1)^2 + 5$	(III) 7
(D) The minimum value of $f(x) = 2 \cos x + 4$	(IV) 4

Choose the correct answer from the options given below:

The range of $\sin(3x)$ is $[-1, 1]$, so:

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

Maximum of $f(x) = \sin(3x) + 6$ occurs when $\sin(3x) = 1$:

$f(x)_{\max} = 1 + 6 = 7 \longrightarrow \textbf{(III)}$

Since $|x + 2| \geq 0$, we have $-|x + 2| \leq 0$.

Maximum of $f(x) = -|x + 2| + 4$ occurs when $|x + 2| = 0$, i.e. $x = -2$:

$f(x)_{\max} = 0 + 4 = 4 \longrightarrow \textbf{(IV)}$

Since $(3x + 1)^2 \geq 0$, the minimum of $f(x) = (3x + 1)^2 + 5$ occurs when $(3x + 1)^2 = 0$, i.e. $x = -\dfrac{1}{3}$:

$f(x)_{\min} = 0 + 5 = 5 \longrightarrow \textbf{(II)}$

The range of $\cos x$ is $[-1, 1]$, so:

$-2 \leq 2\cos x \leq 2$

Minimum of $f(x) = 2\cos x + 4$ occurs when $\cos x = -1$:

$f(x)_{\min} = 2(-1) + 4$

$= -2 + 4$

$= 2 \longrightarrow \textbf{(I)}$

$(A) \rightarrow (III)$, $(B) \rightarrow (IV)$, $(C) \rightarrow (II)$, $(D) \rightarrow (I)$