2023 YTSS P2 Q10

2023 YTSS P2 Q10

Secondary 4
9 marks

The curve \(y=x\left(\dfrac12x-k\right)^3\) where \(k<4\), has a gradient of 0 at \(x=2\).

  1. Show that \(k=1\).[3]
  2. Explain why \(x=2\) is a point of inflexion on the curve and determine the nature of the other turning point.[6]

Solution:

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Answer:(a) \(k=1\). (b) Stationary inflexion at \((2,0)\); minimum at \((\dfrac12,-\dfrac{27}{128})\).

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