2024 H2 Maths Paper 1 Q4 — An Inequality, Then the Modulus Version

Both sides, drawnthe inequality

Test a value of

x

3.00

The left side The right side Where the left side wins

Part (a)

4 marks

Without using a calculator, solve the inequality $\dfrac{4}{x+2}\geq\dfrac{x-3}{x}$ .

Approach

Bring everything to one side, combine into a single fraction, factorise, and read the sign off a table. The one move you must not make is multiplying both sides by $x(x+2)$ — that quantity changes sign, and multiplying an inequality by a negative number flips it. Keeping the fraction whole avoids the whole issue.

First note the values that are not allowed at all: $x\neq-2$ and $x\neq0$ , since each makes a denominator zero. Now subtract:

\frac{4}{x+2}-\frac{x-3}{x}\geq0

\frac{4x-(x-3)(x+2)}{x(x+2)}\geq0

Expanding the numerator:

4x-\left(x^{2}-x-6\right)=4x-x^{2}+x+6=-x^{2}+5x+6

\frac{-x^{2}+5x+6}{x(x+2)}\geq0

Multiplying both sides by $-1$ — which does flip the inequality, but is safe because $-1$ is a constant:

\frac{x^{2}-5x-6}{x(x+2)}\leq0\ \Longrightarrow\ \frac{(x-6)(x+1)}{x(x+2)}\leq0

The critical values are $x=-2,\,-1,\,0,\,6$ . Each of the four factors changes sign once, so the quotient alternates:

Interval	x < −2	−2 < x < −1	−1 < x < 0	0 < x < 6	x > 6
x − 6	−	−	−	−	+
x + 1	−	−	+	+	+
x	−	−	−	+	+
x + 2	−	+	+	+	+
quotient	+	−	+	−	+

The inequality asks for $\leq0$ , so the intervals wanted are $-2\lt x\lt-1$ and $0\lt x\lt6$ . The endpoints $x=-1$ and $x=6$ make the numerator zero and are included; $x=-2$ and $x=0$ make the denominator zero and are excluded.

To check your answer:

Test one point from a claimed interval and one from outside. At $x=1$ : left side $\frac{4}{3}\approx1.33$ , right side $\frac{-2}{1}=-2$ , and $1.33\geq-2$ ✓ — inside. At $x=-\frac{1}{2}$ : left side $\frac{4}{1.5}\approx2.67$ , right side $\frac{-3.5}{-0.5}=7$ , and $2.67\geq7$ is false ✓ — correctly excluded.

$\therefore\ -2\lt x\leq-1\quad\text{or}\quad 0\lt x\leq6$ .

Where the marks go

Cross-multiplying to $4x\geq(x-3)(x+2)$ is the standard way to lose this question: it is only valid where $x(x+2)\gt0$ , and it silently produces $-1\leq x\leq6$ , which wrongly includes $x=-2$ and $x=0$ and wrongly includes $(-1,0)$ . The other loss is at the endpoints — strict at the excluded values, non-strict at the zeros of the numerator.

Part (b)

2 marks

Hence solve the inequality $\dfrac{4}{\left|x\right|+2}\geq\dfrac{\left|x\right|-3}{\left|x\right|}$ .

Approach

Replacing $x$ by $\left|x\right|$ throughout part (a):

-2\lt\left|x\right|\leq-1\quad\text{or}\quad 0\lt\left|x\right|\leq6

The first of these is impossible: $\left|x\right|\geq0$ for every $x$ , so it can never be $\leq-1$ . That branch is discarded, leaving

0\lt\left|x\right|\leq6

Now unwrap. $\left|x\right|\leq6$ means $-6\leq x\leq6$ , and $\left|x\right|\gt0$ means $x\neq0$ . Together:

To check your answer:

At $x=-3$ : left side $\frac{4}{5}=0.8$ , right side $\frac{0}{3}=0$ , and $0.8\geq0$ ✓. At $x=-7$ : left side $\frac{4}{9}\approx0.44$ , right side $\frac{4}{7}\approx0.57$ , false ✓ — correctly outside. And the whole solution set is symmetric about $0$ , as it must be, since replacing $x$ by $-x$ changes nothing in the inequality.

$\therefore\ -6\leq x\lt0\quad\text{or}\quad 0\lt x\leq6$ .

Note — why the symmetry is the giveaway

Every appearance of $x$ in part (b) is inside a modulus, so the two sides are even functions: whatever happens at $x=5$ happens at $x=-5$ . Any answer to part (b) that is not symmetric about the origin is wrong before you check a single value — and any answer that keeps $x=0$ has ignored a denominator.

What this question was really testing

One fraction, then a sign diagram. Rational inequalities are solved by collecting to a single fraction and reading signs — never by multiplying up by something whose sign you do not know.
Excluded values first. $x\neq-2$ , $x\neq0$ is written down before any algebra, and it is what makes two of the four endpoints strict.
"Hence" is an instruction. Part (b) carries two marks because it is meant to take two lines. Re-solving from scratch is not wrong, but it is not what the question is paying for.
Unwrapping a modulus at the end. $0\lt\left|x\right|\leq6$ becomes $-6\leq x\leq6$ with $x=0$ removed. Doing the modulus first, by splitting into cases, turns a two-line part into a page.

Never cross-multiply an inequality — and what "hence" is worth

Part (a)

Part (b)

What this question was really testing

This is how we mark, and how we teach.