2017 H2 Maths Paper 1 Q2 — Worked Solution

Part (i)

2 marks

The graph of $y = \frac{1}{x - a}$ is a rectangular hyperbola with vertical asymptote $x = a$ and horizontal asymptote $y = 0$ : it lies above the $x$ -axis for $x > a$ (decreasing towards $y = 0$ as $x \rightarrow \infty$ , and $y \rightarrow \infty$ as $x \rightarrow a^{\text{+}}$ ), and below the $x$ -axis for $x < a$ (increasing towards $y = 0$ as $x \rightarrow - \infty$ , and $y \rightarrow - \infty$ as $x \rightarrow a^{-}$ ).

The graph of $y = b|x - a|$ is a V-shape with vertex $(a,\ 0)$ , gradient $- b$ for $x < a$ and gradient $b$ for $x > a$ .

Part (ii)

4 marks

Approach

For $x < a$ , $\frac{1}{x - a}$ is negative while $b|x - a|$ is non-negative, so the inequality holds for every $x < a$ with no further work. For $x > a$ , both expressions are positive; the sketch in (i) shows $\frac{1}{x - a}$ starting above $b|x - a|$ near the asymptote and ending below it as $x \rightarrow \infty$ , so the two graphs cross exactly once, and the inequality holds beyond that crossing. We find the crossing point algebraically.

For $x > a$ , $|x - a| = x - a$ , so we solve the equation

\frac{1}{x - a} = b(x - a)

\Rightarrow 1 = b(x - a)^{2}

\Rightarrow (x - a)^{2} = \frac{1}{b}

\Rightarrow x - a = \frac{1}{\sqrt{b}}\ \left( \text{since x} > a,\ \text{so x} - a > 0 \right)

\Rightarrow x = a + \frac{1}{\sqrt{b}}

Near the asymptote ( $x \rightarrow a^{\text{+}}$ ), $\frac{1}{x - a} \rightarrow \infty$ while $b(x - a) \rightarrow 0$ , so the curve lies above the line there and the inequality fails; beyond the crossing point the line lies above the curve, so the inequality holds for $x > a + \frac{1}{\sqrt{b}}$ .

Combining both regions:

x < a\ \ \text{or}\ \ x > a + \frac{1}{\sqrt{b}}

To check your answer: at

x = a - 1

(in the region

x < a

\frac{1}{x - a} = - 1

while

b|x - a| = b > 0

, so the inequality holds. ✓ At

x = a + \frac{2}{\sqrt{b}}

(beyond the crossing),

\frac{1}{x - a} = \frac{\sqrt{b}}{2}

while

b|x - a| = 2\sqrt{b}

, and

\frac{\sqrt{b}}{2} < 2\sqrt{b}

since

b > 0

. ✓

$\therefore$ the solution is $x < a$ or $x > a + \frac{1}{\sqrt{b}}$ .

2017 H2 Maths Paper 1, Question 2

Part (i)

Part (ii)

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