N2012 P1 Q8

N2012 P1 Q8

9 marks
Past Year National Exams
Free

The curve \(C\) has equation \(x-y={{\left( x+y \right)}^{2}}\). It is given that \(C\) has only one turning point.

  1. Show that \(1+\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{2x+2y+1}\).[4]
  2. Hence, or otherwise, show that \(\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=-{{\left( 1+\frac{\mathrm{d}y}{\mathrm{d}x} \right)}^{3}}\).[3]
  3. Hence state, with one reason, whether the turning point is a maximum or a minimum.[2]

Video Solution:

Default solution

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Answer:\(1 + \frac{dy}{dx} = \frac{2}{2x+2y+1}\); \(\frac{d^2y}{dx^2} = -\left(1+\frac{dy}{dx}\right)^3\); maximum

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