2024 HCI Promo Q6

2024 HCI Promo Q6

8 marks
Promo

The curve \(C\) has equation \(\ln {{y}^{2}}={{x}^{2}}y\), where \(x\in \mathbb{R}\), \(-1\le y<0\).

It is given that \(C\) has only one turning point.

  1. Show that \(\frac{\mathrm{d}y}{\mathrm{d}x}\left( 2-{{x}^{2}}y \right)=2x{{y}^{2}}\).[3]
  2. Find the coordinates of the turning point.[2]
  3. Calculate the value of \(\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}\) at the turning point and determine whether the turning point is a maximum or a minimum.[3]

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Answer:(b) Since \(-1\le y<0\), the turning point is \(\left( 0,-1 \right)\). (c) \(\frac{{{\text{d}}^{2}}y}{\text{d}{{x}^{2}}}=1>0\) Therefore \(\left( 0,-1 \right)\) is a minimum point.

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