AS June 2022 Q1
1. Use algebra to find the set of values of \(x\) for which
\[x \geqslant \frac{2x + 15}{2x + 3}\](6)
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{2x + 15}{2x + 3} \Rightarrow 2x^2 + 3x = 2x + 15 \Rightarrow 2x^2 + x - 15 = 0 \Rightarrow x = \ldots\) Alternative 1: \((2x + 3)^2 x \geqslant (2x + 3)(2x + 15) \Rightarrow (2x + 3)(2x^2 + 3x - 2x - 15) \geqslant 0\) \((2x + 3)(x + 3)(2x - 5) \geqslant 0\) Alternative 2; \(x - \dfrac{2x + 15}{2x + 3} \geqslant 0 \Rightarrow \dfrac{x(2x + 3) - 2x - 15}{2x + 3} \geqslant 0 \Rightarrow \dfrac{(x + 3)(2x - 5)}{2x + 3} \geqslant 0\) | M1 | 1.1b |
| \(\Rightarrow (x + 3)(2x - 5) = 0 \Rightarrow \text{CVs are } -3, \dfrac{5}{2}\) | A1 | 1.1b |
| Also \(2x + 3 = 0 \Rightarrow x = -\dfrac{3}{2}\) a CV | B1 | 2.3 |
| Hence from graph (oe) the solution set is \(\left\{x \in \mathbb{R} : -3 \leqslant x \lt -\dfrac{3}{2}, x \geqslant \dfrac{5}{2}\right\} \quad \left\{x : -3 \leqslant x \lt -\dfrac{3}{2}, x \geqslant \dfrac{5}{2}\right\}\) | M1 A1 A1 | 1.1b 2.2a 2.5 |
| (6) | ||
| (6 marks) |
Notes
M1: For a complete method to find the critical values other than \(-\dfrac{3}{2}\).
Alternative 1: Multiplies by \((2x + 3)^2\), collects terms onto one side and factorises into three brackets.
Alternative 2: Collects terms onto one side and combines into single fraction using a common denominator and factorises the numerator
A1: Correct critical values \(-3\) and \(\dfrac{5}{2}\)
B1: For the critical value \(-\dfrac{3}{2}\)
M1: Selects the correct regions for their three CV’s. Should include the right hand side open ended and another bounded region. CV’s of \(a \lt b \lt c\) then must be of the form \(a \leqslant x \leqslant b,\ x \geqslant c\) or \(a \lt x \lt b,\ x \gt c\) the direction of the inequalities must be correct with or without strict inequalities.
A1: At least one correct interval identified. Alternatively allow for both intervals with correct end points but incorrect strict or inclusive inequalities
A1: Fully correct solution as a set – accept alternative set notations e.g. \(\left[-3, -\dfrac{3}{2}\right) \cup \left[\dfrac{5}{2}, \infty\right)\), but not just inequalities. Minimum use of set notation \(-3 \leqslant x \lt -\dfrac{3}{2} \cup x \geqslant \dfrac{5}{2}\)
Note: Correct answer with no working scores M0 A0 but can score B1 M1 A1 A1
No working shown to factorise a cubic equation e.g.
\(4x^3 + 8x^2 - 27x - 45 = (x + 3)(2x + 3)(2x - 5)\) is M0 A0 but can still score B1 M1 A1 A1
A0 for \(-3 \leqslant x \lt -\dfrac{3}{2} \cap x \geqslant \dfrac{5}{2}\) or \(-3 \leqslant x \lt -\dfrac{3}{2}\) and \(x \geqslant \dfrac{5}{2}\)
Special case: If they have a repeated root final 3 marks M1 A1 A0 is possible e.g.