2025 IGCSE Additional Mathematics May/June 0606/11 Q6

2025 IGCSE Additional Mathematics May/June 0606/11 Q6

8 marks

A curve has equation \(y=\left(\dfrac{x^2-1}{x^2+1}\right)^4\).

  1. Show that \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) can be written as \(\dfrac{Ax(x^2-1)^3}{(x^2+1)^5}\), where \(A\) is a positive integer to be found.[5]
    1. Show that the curve has stationary points where \(x=-1\), \(x=0\) and \(x=1\).[1]
    2. Use the first derivative test to determine which two stationary points have the same nature and state whether they are maximum or minimum points.[2]

Solution:

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Answer:(a) \(\dfrac{16x(x^2-1)^3}{(x^2+1)^5}\), \(A=16\) (shown) (b)(i) \(x=-1,0,1\) (shown) (b)(ii) \(x=-1\) and \(x=1\) are minimum points.

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