2026 AHS P1 Q14

2026 AHS P1 Q14

Secondary 4
6 marks

The equation of a curve is \(y=(x-2)\sqrt{4x+1}\) where \(x>-\dfrac14\).

  1. Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{3(2x-1)}{\sqrt{4x+1}}\).[3]
  2. Hence find the range of values of \(x\) for which the function \(y=(x-2)\sqrt{4x+1}\) is decreasing.[3]

Solution:

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Answer:(a) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{3(2x-1)}{\sqrt{4x+1}}\). (b) \(-\dfrac14<x<\dfrac12\).

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