2026 IGCSE Additional Mathematics May/June 0606/11 Q6

2026 IGCSE Additional Mathematics May/June 0606/11 Q6

12 marks

The function \(\mathrm{f}\) is such that \(\mathrm{f}(x)=3x^2+8x\) for \(x\ge k\), where \(k\) is a constant.

It is given that \(\mathrm{f}^{-1}\) exists.

\(k\) has the least possible value for which \(\mathrm{f}^{-1}\) exists.

  1. Find the value of \(k\).[2]
  2. Find the range of \(\mathrm{f}\).[2]
  3. Find an expression for \(\mathrm{f}^{-1}\).[3]
  4. The function \(\mathrm{g}\) is such that \(\mathrm{g}(x)=3^x\) for \(x>-6\).
    1. Write down the domain of \(\mathrm{gf}\).[1]
    2. Solve the equation \(\mathrm{gf}(x)=27\).[4]

Solution:

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Answer:(a) \(k=-\dfrac43\) (b) \(\mathrm{f}(x)\ge-\dfrac{16}{3}\) (c) \(\mathrm{f}^{-1}(x)=\dfrac{-4+\sqrt{3x+16}}3\) (d)(i) \(x\ge-\dfrac43\) (ii) \(x=\dfrac13\)

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