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)
|
List-I |
List-II |
|
(A) The maximum value of $f(x)=\sin(3x) + 6$ |
(III) 7 |
|
(B) The maximum value of $f(x) = -|x+2|+4$ |
(IV) 4 |
|
(C) The minimum value of $f(x) = (3x+1)^2+5$ |
(II) 5 |
|
(D) The minimum value of $f(x) = 2\cos x+4$ |
(I) 2 |
Matching List-I with List-II:
(A) The maximum value of \( f(x) = \sin(3x) + 6 \)
The maximum value of \( \sin(3x) \) is 1, so maximum value of \( f(x) = 1 + 6 = \mathbf{7} \)
⇒ (A) → (III)
(B) The maximum value of \( f(x) = -|x + 2| + 4 \)
The minimum value of \( |x + 2| \) is 0 (when \( x = -2 \)), so maximum of \( f(x) = -0 + 4 = \mathbf{4} \)
⇒ (B) → (IV)
(C) The minimum value of \( f(x) = (3x + 1)^2 + 5 \)
The minimum value of a square is 0, so minimum of \( f(x) = 0 + 5 = \mathbf{5} \)
⇒ (C) → (II)
(D) The minimum value of \( f(x) = 2\cos x + 4 \)
Minimum of \( \cos x \) is -1, so minimum of \( f(x) = 2(-1) + 4 = \mathbf{2} \)
⇒ (D) → (I)