The function $\mathrm{f}$ is defined by \(\mathrm{f}:x\mapsto \frac{1}{1-x},\ x\in\mathbb{R},\ x\n

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x\mapsto \frac{1}{1-x},\ x\in\mathbb{R},\ x\ne 1$. Explain why $\mathrm{f}^2$ does not exist.[2] The function $\mathrm{g}$ is defined by $\mathrm{g}:x\mapsto \frac{1}{1-x},\ x\in\mathbb{R},\ x\ne 0,1$. Given that $\mathrm{g}^2$ exists, find $\mathrm{g}^2(x)$. State its domain and range.[3] Show that $\mathrm{g}^3(x)=x$ and hence solve the equation $\mathrm{g}^{2025}(x)+\mathrm{g}(x)=0$.[3]

Accepted Answer

(a) $\mathrm{f}^2$ does not exist (b) $\mathrm{g}^2(x)=1-\dfrac1x$, domain and range $\mathbb R\setminus\{0,1\}$ (c) $x=\dfrac{1\pm\sqrt5}{2}$

2025 EJC Promo Q8

2025 EJC Promo Q8

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