2026 IGCSE Additional Mathematics May/June 0606/22 Q8

12 marks

The function \(\mathrm f\) is defined, for \(x\ge-1\), by \(\mathrm f(x)=1+2^x\).

Find the range of \(\mathrm f\).[1]
1. Write down the domain of \(\mathrm f^{-1}\).[1]
2. Find an expression for \(\mathrm f^{-1}(x)\).[2]
3. On the axes, sketch the graph of \(y=\mathrm f(x)\) and hence the graph of \(y=\mathrm f^{-1}(x)\).
  State any intercepts with the coordinate axes.[4]
The function \(\mathrm g\) is defined, for \(x\ge-1\), by \(\mathrm g(x)=x(x+k)\) where \(k\) is a constant.
1. Explain why \(\mathrm{gf}\) exists.[1]
2. Find an expression for \(\mathrm{gf}(x)\).[1]
3. It is given that \(\mathrm{gf}(3)=126\).
  Find the value of \(k\).[2]

Solution:

Solution locked

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Answer:(a) \(\left[\dfrac32,\infty\right)\) (b)(i) \(x\ge\dfrac32\) (ii) \(\log_2(x-1)\) (iii) Graphs shown; intercepts \((0,2)\), \((2,0)\). (c)(i) Range of \(\mathrm f\) lies in domain of \(\mathrm g\). (ii) \((1+2^x)(1+2^x+k)\) (iii) \(k=5\).

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