Class 12 Applications Of Derivatives · Long Answer3 Marksmonotonicity
Class 12 Maths · Applications Of Derivatives · Long Answer #97
Find the maximal intervals of strict increase and decrease for f(x)=3x⁴−4x³−12x²+5.
✅ Step-by-Step Solution
Final Answer: Increasing [−1,0] and [2,∞); decreasing (−∞,−1] and [0,2].
- f′=12x³−12x²−24x=12x(x+1)(x−2).
- The stationary inputs are −1,0,2.
- Its derivative signs on successive intervals are −,+,−,+. Include finite endpoints by continuity and strict monotonicity on each interior.
- Thus increasing intervals are [−1,0] and [2,∞); decreasing intervals are (−∞,−1] and [0,2].
📋 CBSE Stepwise Marking Scheme
- [0.5 marks] f′=12x³−12x²−24x=12x(x+1)(x−2).
- [1 mark] The stationary inputs are −1,0,2.
- [1 mark] Its derivative signs on successive intervals are −,+,−,+. Include finite endpoints by continuity and strict monotonicity on each interior.
- [0.5 marks] Thus increasing intervals are [−1,0] and [2,∞); decreasing intervals are (−∞,−1] and [0,2].
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