2024 IGCSE Additional Mathematics May/June 0606/21 Q9

2024 IGCSE Additional Mathematics May/June 0606/21 Q9

7 marks

The functions f and g are defined by

\(\mathrm{f}(x)=\dfrac{3x^2}{4x-1}\quad\text{for}\hspace{0.5em}x<0\)

\(\mathrm{g}(x)=\dfrac1{x^2}\quad\text{for}\hspace{0.5em}x<0.\)

  1. Explain why the function \(\mathrm{fg}\) does not exist.[1]
  2. Given that the function \(\mathrm{gf}\) does exist, find and simplify an expression for \(\mathrm{gf}(x)\).[2]
  3. Show that \(\mathrm{f}^{-1}(x)\) can be written as \(\dfrac{px-\sqrt{x(qx+r)}}3\), where \(p\), \(q\) and \(r\) are integers.[4]

Solution:

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Answer:(a) The positive range of \(\mathrm{g}\) is outside the domain of \(\mathrm{f}\). (b) \(\dfrac{(4x-1)^2}{9x^4}\). (c) \(\mathrm{f}^{-1}(x)=\dfrac{2x-\sqrt{x(4x-3)}}3\), \(p=2,q=4,r=-3\).

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