Match List-I with List-II

$\begin{array}{|l|l|} \hline \rule{0pt}{2.8ex}\text{List-I} & \text{List-II} \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(A) The minimum value of } f(x) = (2x - 1)^2 + 3 & \text{(I) } 4 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(B) The maximum value of } f(x) = -|x + 1| + 4 & \text{(II) } 10 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(C) The minimum value of } f(x) = \sin(2x) + 6 & \text{(III) } 3 \\[1.2ex] \hline \rule{0pt}{2.8ex}\text{(D) The maximum value of } f(x) = -(x - 1)^2 + 10 & \text{(IV) } 5 \\[1.2ex] \hline \end{array}$

Choose the correct answer from the options given below:

For $f(x) = (2x - 1)^2 + 3$:

A squared term $(2x - 1)^2$ is always greater than or equal to $0$.

The minimum value of $(2x - 1)^2$ is $0$, which occurs when $2x - 1 = 0$, i.e., when $x = \frac{1}{2}$.

$f(x) = (2x - 1)^2 + 3$

Minimum value $= 0 + 3$

Minimum value $= 3$

(A) matches with (III)

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For $f(x) = -|x + 1| + 4$:

The absolute value $|x + 1|$ is always greater than or equal to $0$, so $-|x + 1|$ is always less than or equal to $0$.

The maximum value of $-|x + 1|$ is $0$, which occurs when $x + 1 = 0$, i.e., when $x = -1$.

$f(x) = -|x + 1| + 4$

Maximum value $= 0 + 4$

Maximum value $= 4$

(B) matches with (I)

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For $f(x) = \sin(2x) + 6$:

The sine function has range $[-1, 1]$.

The minimum value of $\sin(2x)$ is $-1$.

$f(x) = \sin(2x) + 6$

Minimum value $= -1 + 6$

Minimum value $= 5$

(C) matches with (IV)

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For $f(x) = -(x - 1)^2 + 10$:

The squared term $(x - 1)^2$ is always greater than or equal to $0$, so $-(x - 1)^2$ is always less than or equal to $0$.

The maximum value of $-(x - 1)^2$ is $0$, which occurs when $x - 1 = 0$, i.e., when $x = 1$.

$f(x) = -(x - 1)^2 + 10$

Maximum value $= 0 + 10$

Maximum value $= 10$

(D) matches with (II)

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The correct matches are:

(A) - (III), (B) - (I), (C) - (IV), (D) - (II)