2022 IGCSE Additional Mathematics May/June 0606/13 Q6
2022 IGCSE Additional Mathematics May/June 0606/13 Q6
10 marks
- It is given that\(\mathrm{f}:x\to2x^2\quad\text{for}\hspace{0.5em}x\ge0,\)\(\mathrm{g}:x\to2x+1\quad\text{for}\hspace{0.5em}x\ge0.\)Each of the expressions in the table can be written as one of the following.\(\mathrm{f}'\qquad\mathrm{f}''\qquad\mathrm{g}'\qquad\mathrm{g}''\qquad\mathrm{fg}\qquad\mathrm{gf}\qquad\mathrm{f}^2\qquad\mathrm{g}^2\qquad\mathrm{f}^{-1}\qquad\mathrm{g}^{-1}\)Complete the table. The first row has been completed for you.[5]
Expression Function notation \(2\) \(\mathrm{g}'\) \(0\) \(4x\) \(8x^2+8x+2\) \(4x+3\) \(\dfrac{x-1}{2}\) - It is given that \(\mathrm{h}(x)=(x-1)^2+3\) for \(x\ge a\). The value of \(a\) is as small as possible such that \(\mathrm{h}^{-1}\) exists.
- Write down the value of \(a\).[1]
- Write down the range of \(\mathrm{h}\).[1]
- Find \(\mathrm{h}^{-1}(x)\) and state its domain.[3]
Solution:
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Answer:(a) \(0\): \(\mathrm{g}''\); \(4x\): \(\mathrm{f}'\); \(8x^2+8x+2\): \(\mathrm{fg}\); \(4x+3\): \(\mathrm{g}^2\); \(\dfrac{x-1}{2}\): \(\mathrm{g}^{-1}\) (b)(i) \(a=1\) (ii) \(\mathrm{h}(x)\ge3\) (iii) \(\mathrm{h}^{-1}(x)=1+\sqrt{x-3}\), domain \(x\ge3\)
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