2025 TJC P1 Q2

2025 TJC P1 Q2

4 marks

The curve \(C\) has equation : \(xy = 1 + (x - y)^3\)

Find the gradient of the tangent to \(C\) at the point \(A\) where \(y = 0\).[4]

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Answer:\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac32\) at \(A=(-1,0)\).

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