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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
Each function is a non-negative quantity (a square, a modulus) or a sine added to a constant, so its bound comes from the least or greatest value of that part.
(A) The minimum value of \(f(x) = (2x-1)^2+3\)
This matches (III) 3. Since \((2x-1)^2 \ge 0\), with equality at \(x = \frac{1}{2}\):
\(f(x) \ge 3,\quad f\left(\frac{1}{2}\right) = 3\)
(B) The maximum value of \(f(x) = -|x+1|+4\)
This matches (I) 4. Since \(|x+1| \ge 0\), with equality at \(x = -1\):
\(f(x) \le 4,\quad f(-1) = 4\)
(C) The minimum value of \(f(x) = \sin(2x)+6\)
This matches (IV) 5. Since \(\sin(2x) \ge -1\), with equality at \(x = \frac{3\pi}{4}\):
\(f(x) \ge -1 + 6 = 5\)
(D) The maximum value of \(f(x) = -(x-1)^2+10\)
This matches (II) 10. Since \((x-1)^2 \ge 0\), with equality at \(x = 1\):
\(f(x) \le 10,\quad f(1) = 10\)
So the matching is (A)-(III), (B)-(I), (C)-(IV), (D)-(II).