2023 IGCSE Additional Mathematics October/November 0606/12 Q8
2023 IGCSE Additional Mathematics October/November 0606/12 Q8
9 marks
- It is given that \(\mathrm{f}:x\to(3x+1)^2-4\) for \(x\ge a\), and that \(\mathrm{f}^{-1}\) exists.
- Find the least possible value of \(a\).[1]
- Using this value of \(a\), write down the range of \(\mathrm{f}\).[1]
- Using this value of \(a\), sketch the graphs of \(y=\mathrm{f}(x)\) and \(y=\mathrm{f}^{-1}(x)\) on the axes, stating the intercepts with the coordinate axes.[4]
- It is given that\(\mathrm{g}(x)=\ln\left(2x^2+5\right)\quad\text{for}\hspace{0.5em}x\ge0,\)\(\mathrm{h}(x)=3x-2\quad\text{for}\hspace{0.5em}x\ge0.\)Solve the equation \(\mathrm{hg}(x)=4\) giving your answer in exact form.[3]
Solution:
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Answer:(a)(i) \(a=-\dfrac13\) (ii) \(\mathrm{f}(x)\ge-4\) (iii) Sketch shown; \(y=\mathrm{f}(x)\): intercepts \(\left(\dfrac13,0\right)\) and \((0,-3)\); \(y=\mathrm{f}^{-1}(x)\): reflection in \(y=x\), intercepts \((-3,0)\) and \(\left(0,\dfrac13\right)\) (b) \(x=\sqrt{\dfrac{\mathrm{e}^2-5}{2}}\)
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