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 \(f(x) = -|x + 2| + 4\) | (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:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
Each value follows from the bounds of the basic function: \(-1 \le \sin\theta \le 1\), \(-1 \le \cos\theta \le 1\), \(|t| \ge 0\) and \(t^2 \ge 0\).
(A) The maximum value of \(f(x) = \sin(3x) + 6\)
This matches (III) 7:
\(\sin(3x) \le 1\), so \(f(x) \le 1 + 6 = 7\)
(B) The maximum value of \(f(x) = -|x + 2| + 4\)
This matches (IV) 4:
\(|x + 2| \ge 0\), so \(f(x) \le 4\), with equality at x = −2
(C) The minimum value of \(f(x) = (3x + 1)^2 + 5\)
This matches (II) 5:
\((3x + 1)^2 \ge 0\), so \(f(x) \ge 5\), with equality at \(x = -\frac{1}{3}\)
(D) The minimum value of \(f(x) = 2\cos x + 4\)
This matches (I) 2:
\(\cos x \ge -1\), so \(f(x) \ge -2 + 4 = 2\)
So the matching is (A)-(III), (B)-(IV), (C)-(II), (D)-(I).