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Graphs and transformations: CJC 2024 Promo Q4

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Catholic Junior College2024 PromoPromo Q4●●● Challenging6 marks

What this question tests

Graphs (Rational Function).

Question

A curve $C$ has equation $y = \dfrac{x^2 - 4x + 4}{x - 1}$ .

Sketch $C$ , stating clearly the equations of any asymptotes, coordinates of turning points and points of intersection with the axes. [3]
By drawing a suitable graph on the same diagram in part (a), find the range of values of $b$ , where $b > 0$ , such that the equation $(x - 2)^2 + \frac{1}{b^2}\left(\frac{x^2 - 4x + 4}{x - 1} + 1\right)^{\!2} = 4$ has no real roots. [3]

Show full worked solution▾

(a)

Using Long Division: $\begin{aligned} y &= \frac{x^2 - 4x + 4}{x - 1} = x - 3 + \frac{1}{x - 1}. \end{aligned}$ Asymptotes: $x = 1$ , $y = x - 3$

$Catholic Junior College 2024 Promo Q4: solution diagram$

(b)

$\begin{aligned} (x - 2)^2 + \frac{1}{b^2}\left(\frac{x^2 - 4x + 4}{x - 1} + 1\right)^{\!2} &= 4\\ \frac{(x - 2)^2}{2^2} + \frac{1}{(2b)^2}\left(\frac{x^2 - 4x + 4}{x - 1} + 1\right)^{\!2} &= 1, \end{aligned}$ Ellipse centre $(2,-1)$ $\begin{aligned} 0 < 2b &< 1\\ 0 < b &< \frac{1}{2}. \end{aligned}$

Answers

(a)asymptotes $x = 1$ , $y = x - 3$ (b) $0 < b < \dfrac{1}{2}$

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