A sequence of negative real numbers $u_1, u_2, u_3, \dots$ is given by \[u_1 = -1 \quad \text{and}

Question

A sequence of negative real numbers $u_1, u_2, u_3, \dots$ is given by $$u_1 = -1 \quad \text{and} \quad u_{n+1} = \frac{-2u_n + 6}{u_n - 1}, \text{ for}\hspace{0.5em} n \ge 1.$$ Given that $u_n \to l$ as $n \to \infty$, find the value of $l$.[3] Sketch the graph of $y = \frac{-2x + 6}{x - 1} - x$, stating the equations of the asymptotes and the coordinates of the points where it crosses the axes.[3] Hence show that $u_{n+1} > u_n$ if $u_n < l$.[2]

Accepted Answer

(a) $l=-3$ (b) Asymptotes $x=1$, $y=-x-2$; intercepts $(-3,0)$, $(2,0)$, $(0,-6)$ (c) Shown

2025 CJC P1 Q5

2025 CJC P1 Q5

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