2016 H2 Maths Paper 1 Q1 — Worked Solution

Solution

5 marks

We combine the two terms over the common denominator $x - 4$ .

\frac{4x^{2} + 4x - 14}{x - 4} - (x + 3)

= \frac{4x^{2} + 4x - 14 - (x + 3)(x - 4)}{x - 4}

= \frac{4x^{2} + 4x - 14 - \left( x^{2} - x - 12 \right)}{x - 4}

= \frac{3x^{2} + 5x - 2}{x - 4}

Factorising the numerator,

3x^{2} + 5x - 2 = (3x - 1)(x + 2)

\therefore\ \frac{4x^{2} + 4x - 14}{x - 4} - (x + 3) = \frac{(3x - 1)(x + 2)}{x - 4}

Approach

The inequality $\frac{4x^{2} + 4x - 14}{x - 4} < x + 3$ is exactly the statement that the expression found in part (i) is negative. This avoids cross-multiplying by $x - 4$ , whose sign depends on $x$ .

Since

\frac{4x^{2} + 4x - 14}{x - 4} - (x + 3) = \frac{(3x - 1)(x + 2)}{x - 4},

the inequality $\frac{4x^{2} + 4x - 14}{x - 4} < x + 3$ is equivalent to

\frac{(3x - 1)(x + 2)}{x - 4} < 0.

The critical values are $x = - 2,\ \frac{1}{3},\ 4$ , where $x = 4$ is excluded from the domain. We consider the sign of the expression on each interval:

x < - 2:\ \ \left( \text{−} \right)\left( \text{−} \right)/\left( \text{−} \right) < 0

- 2 < x < \frac{1}{3}:\ \ \left( \text{−} \right)\left( \text{+} \right)/\left( \text{−} \right) > 0

\frac{1}{3} < x < 4:\ \ \left( \text{+} \right)\left( \text{+} \right)/\left( \text{−} \right) < 0

x > 4:\ \ \left( \text{+} \right)\left( \text{+} \right)/\left( \text{+} \right) > 0

To check your answer: at

x = 1

, the left side of the original inequality is

\frac{4 + 4 - 14}{1 - 4} = \frac{- 6}{- 3} = 2

and

x + 3 = 4

, and indeed

2 < 4

. ✓

$\therefore$ $x < - 2$ or $\frac{1}{3} < x < 4$ .

2016 H2 Maths Paper 1, Question 1

Solution

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