Match List-I with List-II

List-I	List-II
(A) The minimum value of f(x) = $(2x-1){}^2+3$	(I) 4
(B) The maximum value of f(x) = $-|x+1|+4$	(II) 10
(C) The minimum value of f(x) = $\sin \left(2x\right)+6$	(III) 3
(D) The maximum value of f(x) = $-(x-1){}^2+10$	(IV) 5

Choose the correct answer from the options given below:

The correct answer is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).

To find the correct match for each function, let us evaluate the minimum or maximum value of each function step-by-step.

Step 1: Evaluate (A)
The given function is:
$f\left(x\right)=(2x-1{)}^2+3$
Since the square of any real number is always non-negative, we have:
$(2x-1{)}^2\ge 0$
Adding 3 to both sides:
$(2x-1{)}^2+3\ge 3$
Thus, the minimum value of f(x) is 3, which occurs when $2x-1=0$ (i.e., $x=1/2$).
Therefore, (A) matches with (III).

Step 2: Evaluate (B)
The given function is:
$f\left(x\right)=-|x+1|+4$
Since the absolute value is always non-negative:

Multiplying by -1 reverses the inequality:

Adding 4 to both sides:

Thus, the maximum value of f(x) is 4, which occurs when $x=-1$.
Therefore, (B) matches with (I).

Step 3: Evaluate (C)
The given function is:
$f\left(x\right)=\sin \left(2x\right)+6$
We know that the sine function oscillates between -1 and 1 for any real argument:
$-1\le \sin \left(2x\right)\le 1$
Adding 6 to all parts of the inequality:
$-1+6\le \sin \left(2x\right)+6\le 1+6$
$5\le f\left(x\right)\le 7$
Thus, the minimum value of f(x) is 5.
Therefore, (C) matches with (IV).

Step 4: Evaluate (D)
The given function is:
$f\left(x\right)=-(x-1{)}^2+10$
Since the square term is non-negative:
$(x-1{)}^2\ge 0$
Multiplying by -1:
$-(x-1{)}^2\le 0$
Adding 10 to both sides:
$-(x-1{)}^2+10\le 10$
Thus, the maximum value of f(x) is 10, which occurs when $x=1$.
Therefore, (D) matches with (II).

Combining all the results:

• (A) → (III)
• (B) → (I)
• (C) → (IV)
• (D) → (II)