2024 IGCSE Additional Mathematics May/June 0606/13 Q6

2024 IGCSE Additional Mathematics May/June 0606/13 Q6

7 marks

It is given that \(y=\dfrac{\ln(2x^2+1)}{x+2}\), \(x\ne-2\).

  1. Find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\).[3]
  2. Given that \(x\) increases from \(1\) to \(1+h\), where \(h\) is small, find the approximate corresponding change in \(y\).[2]
  3. When \(x=1\), the rate of change in \(y\) is \(3\) units per second. Find the corresponding rate of change in \(x\).[2]

Solution:

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Answer:(a) \(\dfrac{(x+2)\dfrac{4x}{2x^2+1}-\ln(2x^2+1)}{(x+2)^2}\). (b) \(\Delta y\approx\dfrac{h(4-\ln3)}9\). (c) \(9.31\) units per second.

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