2021 IGCSE Additional Mathematics February/March 0606/22 Q10
2021 IGCSE Additional Mathematics February/March 0606/22 Q10
8 marks
The function \(\mathrm{f}\) is defined by \(\mathrm{f}(x)=\dfrac{\sqrt{4x^2-1}}{2x}\) for \(0.5\le x\le1.5\).
The diagram shows a sketch of \(y=\mathrm{f}(x)\).
- It is given that \(\mathrm{f}^{-1}\) exists. Find the domain and range of \(\mathrm{f}^{-1}\).[3]
- Find an expression for \(\mathrm{f}^{-1}(x)\).[3]
- The function \(\mathrm{g}\) is defined by \(\mathrm{g}(x)=\mathrm{e}^{x^2}\) for all real \(x\). Show that \(\mathrm{gf}(x)=\mathrm{e}^{\left(1-\frac{a}{bx^2}\right)}\), where \(a\) and \(b\) are integers.[2]
Solution:
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Answer:(a)(i) Domain \(0\le x\le\dfrac{2\sqrt{2}}{3}\), range \(0.5\le\mathrm{f}^{-1}(x)\le1.5\) (ii) \(\mathrm{f}^{-1}(x)=\sqrt{\dfrac{1}{4-4x^2}}\) (b) \(\mathrm{gf}(x)=\mathrm{e}^{\left(1-\frac{1}{4x^2}\right)}\), so \(a=1\), \(b=4\)
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