Applications Of Derivatives
The set of all values of 'a' for which f(x)=(a+2)x3−3ax2+9ax−1 decreases for all real x is:
Select the correct option:
Solution
(−∞,−3]
For f(x) to decrease for all x, f′(x) must be ≤0 for all x. f′(x)=3(a+2)x2−6ax+9a. Condition 1: Coeff of x2<0⟹3(a+2)<0⟹a<−2. Condition 2: Discriminant D≤0. D=(−6a)2−4∗3(a+2)∗9a =36a2−108a(a+2) =36a2−108a2−216a =−72a2−216a =−72a(a+3).
We need D≤0: −72a(a+3)≤0⟹a(a+3)≥0. This implies a is in (−∞,−3]∪[0,∞).
Combine with a<−2: Intersection of (−∞,−2) AND ((−∞,−3]∪[0,∞)) is (−∞,−3]. Thus, a belongs to (−∞,−3].
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About This Question
- Subject
- mathematics
- Chapter
- limit, continuity and differentiability
- Topic
- applications of derivatives
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
(−∞,−3]
For f(x) to decrease for all x, f′(x) must be ≤0 for all x. f′(x)=3(a+2)x2−6ax+9a. Condition 1: Coeff of x2<0⟹3(a+2)<0⟹a<−2. Condition 2: Discriminant D≤0. D=(−6a)2−4∗3(a+2)∗9a =36a2−108a(a+2) =36a2−108a2−216a =−72a2−216a =−72a(a+3).
We need D≤0: −72a(a+3)≤0⟹a(a+3)≥0. This implies a is in (−∞,−3]∪[0,∞).
Combine with a<−2: Intersection of (−∞,−2) AND ((−∞,−3]∪[0,∞)) is (−∞,−3]. Thus, a belongs to (−∞,−3].
This hard difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of applications of derivatives. It appeared in the 2025 exam.
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