The function $\mathrm{f}$ is defined by \[\mathrm{f} : x \mapsto \frac{-x - 2}{5x + 1}, \quad x \i

Question

The function $\mathrm{f}$ is defined by $$\mathrm{f} : x \mapsto \frac{-x - 2}{5x + 1}, \quad x \in \mathbb{R}, x \neq -\frac{1}{5}.$$ Explain why both the function $\mathrm{f}^{-1}$ and composite function $\mathrm{f}^2$ exist.[4] Find $\mathrm{f}^2(x)$ and state the range of $\mathrm{f}^2$.[3] Determine the solution of the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.[2] The function $g$ is defined by $$g : x \mapsto \frac{2x - 2}{5x + 4}, \quad x \in \mathbb{R}, x \neq -\frac{4}{5}, x \neq -\frac{1}{5}, x \neq \frac{2}{5}.$$ Verify that $g^2(x) = \mathrm{f}(x)$.[1] Given that $k$ is an integer, find $g^{49}(k)$, giving your answer in terms of $k$.[3]

Accepted Answer

(i) $f$ is one-one and $R_f=D_f=\mathbb R\setminus\left\{-\dfrac15\right\}$ (ii) $f^2(x)=x$, range $\mathbb R\setminus\left\{-\dfrac15\right\}$ (iii) $x\in\mathbb R\setminus\left\{-\dfrac15\right\}$ (iv) $g^2(x)=f(x)$ (v) $g^{49}(k)=\dfrac{2k-2}{5k+4}$

2014 VJC Promo Q8

2014 VJC Promo Q8

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