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प्रश्न
Read the following passage and answer the questions given below.
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- Is the function differentiable in the interval (0, 12)? Justify your answer.
- If 6 is the critical point of the function, then find the value of the constant m.
- Find the intervals in which the function is strictly increasing/strictly decreasing.
OR
Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute ‘maximum/absolute minimum values of the function.
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उत्तर
i. f(x) = –0.1x2 + mx + 98.6, being a polynomial function, is differentiable everywhere, hence, differentiable in (0, 12)
ii. f'(x) = –0.2x + m
Since, 6 is the critical point,
f'(6) = 0
⇒ m = 1.2
iii. f(x) = –0.1x2 + 1.2x + 98.6
f'(x) = –0.2x + 1.2 = –0.2(x – 6)
| In the Interval | f'(x) | Conclusion |
| (0, 6) | +ve | f is strictly increasing in [0, 6] |
| (6, 12) | -ve | f is strictly decreasing in [6, 12] |
OR
iii. f(x) = –0.1x2 + mx + 98.6,
f'(x) = –0.2x + 1.2, f'(6) = 0,
f"(x) = –0.2
f"(6) = –0.2 < 0
Hence, by second derivative test 6 is a point of local maximum. The local maximum value = f(6) = −0.1 × 62 + 1.2 × 6 + 98.6 = 102.2
We have f(0) = 98.6, f(6) = 102.2, f(12) = 98.6
6 is the point of absolute maximum and the absolute maximum value of the function = 102.2.
0 and 12 both are the points of absolute minimum and the absolute minimum value of the function = 98.6.
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