Match List-I with List-II
Consider the function \(f(x) = 2x^3 - 21x^2 + 36x + 80, x \in [0, 6]\). Then
| List-I | List-II | ||
|---|---|---|---|
| (A) | one of its critical points is at \(x\) = | (I) | −28 |
| (B) | Its absolute maximum value is | (II) | −42 |
| (C) | Its absolute minimum value is | (III) | 97 |
| (D) | Its second derivative at \(x = 0\) is | (IV) | 6 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
The critical points are where f'(x) = 0:
\(f'(x) = 6x^2 - 42x + 36\)
\(f'(x) = 6(x^2 - 7x + 6) = 6(x - 1)(x - 6)\)
\(x = 1\) or \(x = 6\)
On a closed interval, the absolute extreme values are found among the values at the critical points and the end points:
\(f(0) = 80\)
\(f(1) = 2 - 21 + 36 + 80 = 97\)
\(f(6) = 432 - 756 + 216 + 80 = -28\)
(A) one of its critical points is at \(x\) =
This matches (IV) 6, since f'(6) = 0.
(B) Its absolute maximum value is
This matches (III) 97, the largest of 80, 97 and −28.
(C) Its absolute minimum value is
This matches (I) −28, the smallest of 80, 97 and −28.
(D) Its second derivative at \(x = 0\) is
This matches (II) −42:
\(f''(x) = 12x - 42,\quad f''(0) = -42\)
So the matching is (A)-(IV), (B)-(III), (C)-(I), (D)-(II).