A curve $C$ has equation $y=\mathrm{f}\left( x \right)$, where \(\mathrm{f}\left( x \right)=\fra

Question

A curve $C$ has equation $y=\mathrm{f}\left( x \right)$, where $\mathrm{f}\left( x \right)=\frac{x-1}{a}+\frac{a}{x-1}$, $x\in \mathbb{R}$, $x\ne 1$, with $a$ being a constant such that $-1 < a < 0$. Find algebraically, the set of values that $\mathrm{f}\left( x \right)$ can take.[3] Sketch $C$, showing clearly the equations of any asymptotes and coordinates of any axial intercepts and stationary points in terms of $a$.[4] On separate diagrams, sketch the graph of $y=\frac{1}{\mathrm{f}\left( x \right)}$,[3] $y=\mathrm{f}'\left( x \right)$,[2] showing clearly in each case, the equations of any asymptotes and the coordinates of any axial intercepts and stationary points in terms of $a$. Describe a sequence of transformations that transform the graph of $y=\mathrm{f}\left( x \right)$ to $y=\frac{x}{2a}+\frac{a}{2x}$.[2]

Accepted Answer

(i) $\left\{ y:y\in \mathbb{R},\,y\le -2 \right.$ or $\left. y\ge 2 \right\}$ (ii) Asymptotes: $x=1$, $y=\frac{x}{a}-\frac1a$; y-intercept $\left(0,-\frac{1+a^2}{a}\right)$; stationary points $(1+a,2)$, $(1-a,-2)$; no x-intercept. (iii)(a) Horizontal asymptote $y=0$, excluded hole $(1,0)$, y-intercept $\left(0,-\frac{a}{1+a^2}\right)$, stationary points $(1+a,\frac12)$, $(1-a,-\frac12)$. (iii)(b) Asymptotes $x=1$, $y=\frac1a$; x-intercepts $(1+a,0)$, $(1-a,0)$; y-intercept $\left(0,\frac1a-a\right)$; no stationary point. (iv) Translate one unit left, then scale by factor $\frac12$ parallel to the y-axis.

2017 HCI Promo Q9

2017 HCI Promo Q9

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