2026-08-31T11:35:12.659292 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 2 shows the curve with equation \(y = \dfr...

The whole difficulty in an inequality containing an algebraic fraction is that the denominator changes sign. Multiplying by \(\left(x - 2\right)\) would reverse the inequality when \(x\) is less than \(2\) and preserve it when \(x\) is greater, so the direction would be unknown. The question therefore instructs you to multiply by \(\left(x - 2\right)^{2}\), and understanding why is worth a mark in itself.

(a) Showing that the inequality may be written as \(\left(x - 2\right)\left(x - 5\right) \lt 0\) [3 marks]

Since \(x \ne 2\), the square \(\left(x - 2\right)^{2}\) is strictly positive for every permitted \(x\). Multiplying an inequality by a strictly positive quantity leaves its direction unchanged, which is exactly why the square is used:

\[\frac{x + 4}{x - 2} \gt 3 \quad \Longrightarrow \quad \left(x + 4\right)\left(x - 2\right) \gt 3\left(x - 2\right)^{2} \qquad \textbf{M1}\]

On the left one factor of \(\left(x - 2\right)\) cancels with the denominator, leaving the product shown. The M1 is for this multiplication with the inequality sign correctly preserved; a candidate who reverses it here cannot recover. Expand both sides:

\[x^{2} + 2x - 8 \gt 3\left(x^{2} - 4x + 4\right) = 3x^{2} - 12x + 12 \qquad \textbf{M1}\]

Collect everything on the right so that the leading coefficient stays positive, which avoids a sign reversal:

\[0 \gt 3x^{2} - 12x + 12 - x^{2} - 2x + 8 = 2x^{2} - 14x + 20\] \[0 \gt 2\left(x^{2} - 7x + 10\right) = 2\left(x - 2\right)\left(x - 5\right), \quad \text{that is} \quad \left(x - 2\right)\left(x - 5\right) \lt 0 \qquad \textbf{A1}\]

as required, the factor \(2\) being positive and so removable without affecting the direction. Since the target is printed, all three marks are for the derivation. The two traps are expanding \(3\left(x - 2\right)^{2}\) as \(3x^{2} + 12\), which drops the middle term, and dividing by \(-2\) at the end without reversing the inequality. Note the appearance of \(x = 2\) as a critical value even though the original expression is undefined there; that is a consequence of multiplying by \(\left(x - 2\right)^{2}\), and it turns out to be exactly the boundary the graph shows.

(b) The set of values of \(x\) [2 marks]

The critical values are where the product vanishes:

\[x = 2 \quad \text{and} \quad x = 5 \qquad \textbf{M1}\]

The M1 is for identifying both, and it is a follow-through from part (a). The expression \(\left(x - 2\right)\left(x - 5\right)\) is an upward parabola, so it is negative strictly between its roots and positive outside them:

\[2 \lt x \lt 5 \qquad \textbf{A1}\]

Both inequalities are strict. At \(x = 5\) the original fraction equals \(\dfrac{9}{3} = 3\), which is not greater than \(3\), and at \(x = 2\) the fraction is undefined, so neither endpoint can be included. The wrong turn this part is built to catch is choosing the outside regions, \(x \lt 2\) or \(x \gt 5\), which solve the inequality with the sign reversed.

Two independent checks are available. Numerically, take \(x = 3\), inside the claimed set: \(\dfrac{3 + 4}{3 - 2} = 7\), which is greater than \(3\). Take \(x = 6\), outside it: \(\dfrac{10}{4} = 2.5\), which is not. Take \(x = 0\), also outside: \(\dfrac{4}{-2} = -2\), which is not. Graphically, the figure shows the curve \(y = \dfrac{x + 4}{x - 2}\) together with the line \(y = 3\) and the asymptote \(x = 2\); the curve lies above the line exactly on the branch to the right of the asymptote and to the left of the crossing point, and that is the interval \(2 \lt x \lt 5\). Reading the answer off the figure like this is the fastest way to confirm which side of the critical values is wanted.

It is worth seeing why \(x \lt 2\) fails despite the algebra producing \(x = 2\) as a critical value. For \(x \lt 2\) the denominator \(x - 2\) is negative while the numerator \(x + 4\) is positive for \(x \gt -4\), so the fraction is negative there and cannot exceed \(3\). This is the branch on the far side of the asymptote, and no amount of algebra will place it in the solution set.

Examination takeaway. To clear a denominator in an inequality, multiply by its square, which is always positive, rather than by the denominator itself, whose sign is unknown. Then find the critical values, decide the region from the shape of the parabola, and test one value inside your answer against the original inequality, remembering that a value making the denominator zero is never a solution.