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Class 12 Applications Of Derivatives · Long Answer3 Marksextrema_parameters

Class 12 Maths · Applications Of Derivatives · Long Answer #96

For real a, f(x)=ax²+6x−15 is strictly increasing before 3/4 and strictly decreasing after 3/4. Find the local maximum point of g(x)=ax²−6x+15.

✅ Step-by-Step Solution

Final Answer: (−3/4,69/4)
  1. The derivative of f is 2ax+6. Its sign change at 3/4 requires 2a(3/4)+6=0, giving a=−4.
  2. Then g(x)=−4x²−6x+15 and g′=−8x−6.
  3. g′=0 gives x=−3/4. Since g″=−8<0 it is a local (indeed global) maximum.
  4. g(−3/4)=−9/4+9/2+15=69/4, so the maximum point is (−3/4,69/4).

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] The derivative of f is 2ax+6. Its sign change at 3/4 requires 2a(3/4)+6=0, giving a=−4.
  • [1 mark] Then g(x)=−4x²−6x+15 and g′=−8x−6.
  • [1 mark] g′=0 gives x=−3/4. Since g″=−8<0 it is a local (indeed global) maximum.
  • [0.5 marks] g(−3/4)=−9/4+9/2+15=69/4, so the maximum point is (−3/4,69/4).

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