For the function \( f(x)=2x^{3}-21x^{2}+36x-100 \);
Match the LIST-I with LIST-II
| LIST-I | LIST-II | ||
| A. | Point of local maxima is | I. | -208 |
| B. | Point of local minima is | II. | -83 |
| C. | Local maximum value is | III. | 6 |
| D. | Local minimum value is | IV. | 1 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → A-IV, B-III, C-II, D-I
The critical points come from \( f'(x)=0 \):
\( \displaystyle f'(x)=6x^{2}-42x+36 \)
\( \displaystyle =6(x-1)(x-6) \)
So the critical points are x = 1 and x = 6. By the second derivative test, with \( f''(x)=12x-42 \),
\( \displaystyle f''(1)=-30\lt0 \)
\( \displaystyle f''(6)=30\gt0 \)
(A) Point of local maxima
Since \( f''(1)\lt0 \), the point of local maxima is x = 1. This matches (IV) 1.
(B) Point of local minima
Since \( f''(6)\gt0 \), the point of local minima is x = 6. This matches (III) 6.
(C) Local maximum value
\( \displaystyle f(1)=2-21+36-100 \)
\( \displaystyle =-83 \)
This matches (II) −83.
(D) Local minimum value
\( \displaystyle f(6)=432-756+216-100 \)
\( \displaystyle =-208 \)
This matches (I) −208.
So the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I).