Question

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 \( f(x) = \sin(2x) + 6 \)	(III) 3
(D) The maximum value of \( f(x) = -(x - 1)^2 + 10 \)	(IV) 5

Choose the correct answer from the options given below:

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

Hint:

To find the max/min of simple quadratic and absolute value functions, find the value that makes the squared or absolute part zero. For trigonometric functions, use their known range (e.g., [-1, 1] for sine and cosine).

Answer 1

The Correct Option is C

Step 1: Conceptual Understanding: Determine the minimum or maximum value for each function in List-I by analyzing their properties. This includes understanding the ranges of basic functions like squares, absolute values, and trigonometric functions.

Step 3: Detailed Explanation:

(A) f(x) = $(2x - 1)^2 + 3$: The term $(2x - 1)^2$ is a square, with a minimum value of 0 occurring when $2x - 1 = 0$. Thus, the minimum value of the function is $0 + 3 = 3$. This matches (III).

(B) f(x) = $-|x + 1| + 4$: (Assuming a typo in the OCR, likely missing the negative sign before the absolute value). The term $|x + 1|$ has a minimum value of 0. Consequently, $-|x + 1|$ has a maximum value of 0. The maximum value of the function is $0 + 4 = 4$. This matches (I).

(C) f(x) = sin(2x) + 6: The range of sin(2x) is [-1, 1]. The minimum value of sin(2x) is -1. Therefore, the minimum value of the function is $-1 + 6 = 5$. This matches (IV).

(D) f(x) = $-(x - 1)^2 + 10$: The term $(x - 1)^2$ has a minimum value of 0. Therefore, $-(x - 1)^2$ has a maximum value of 0. The maximum value of the function is $0 + 10 = 10$. This matches (II).

Step 4: Final Answer: The correct matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II). This corresponds to option (3).