2025 BVSS PRELIMS P1 Q12

2025 BVSS PRELIMS P1 Q12

Secondary 4
8 marks
2025 Bedok View Secondary School Prelims A Math Paper 1

The gradient at any point on a curve \(y\) is given by \(6x - \frac{1}{kx^3}\). The line \(4x - 4y = 3\) is a normal to the curve at the point where \(x = \frac{1}{2}\).

  1. Show that \(k = 2\).[2]
  2. Hence find the \(x\)-coordinates of the stationary points.[2]
  3. Find \(\frac{d^2y}{dx^2}\) and explain whether the gradient \(6x - \frac{1}{kx^3}\) has a turning point.[4]

Solution:

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Answer:\((a)\ k=2\) \((b)\ x=\pm\frac{1}{\sqrt[4]{12}}\) \((c)\ \frac{d^2y}{dx^2}=6+\frac{3}{2x^4}\); no turning point.

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