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2026 NYGH P2 Q8

2026 NYGH P2 Q8

IP 4
8 marks

The tangent to the curve \(y=\ln\sqrt{\dfrac{1-2x}{2+3x}}\) at the point \(x=k\), is perpendicular to the line \(2x-7y+5=0\).

  1. Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac7{2(2x-1)(2+3x)}\).[4]
  2. Find the values of \(k\).[4]

Solution:

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Answer:(a) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac7{2(2x-1)(2+3x)}\) (b) \(k=-\dfrac12\) or \(k=\dfrac13\)

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