FP2 June 2014 (R) Q2
2. Using algebra, find the set of values of \(x\) for which \[3x - 5 \lt \frac{2}{x}\] (5)
| Scheme | Marks |
|---|---|
| \(3x - 5 - \dfrac{2}{x} = 0\) (or \(\lt\)) or mult through by \(x^2\) | |
| \(\dfrac{3x^2 - 5x - 2}{x} = 0\) (or \(\lt\)) | |
| \(\dfrac{(3x + 1)(x - 2)}{x} = 0\) or \(x(3x + 1)(x - 2) = 0\) | M1 |
| CVs \(x = -\dfrac{1}{3},\ 2\) | A1 |
| \(x = 0\) | B1 |
| \(x \lt -\dfrac{1}{3},\ 0 \lt x \lt 2\) or in set language (with curved brackets for A1) | M1A1 |
| Special case If \(\leqslant\) used deduct final mark only. | |
| (5) | |
| (5 marks) |
Notes
M1 obtaining two non-zero cvs by any valid method (not calculator)
A1 non-zero cvs correct
B1 \(x = 0\)
M1 deducing one appropriate range from their cvs
A1 both ranges correct
First 3 marks – award with inequalities or =
M1A0 if strict inequality not used
ALT: If multiplied through by \(x\):
| Scheme | Marks |
|---|---|
| \(x \gt 0\): | |
| \(3x^2 - 5x \lt 2 \quad (3x + 1)(x - 2) \lt 0\) | |
| cvs \(x = -\dfrac{1}{3},\ x = 2\) | M1(solve quad) |
| \(\therefore\ 0 \lt x \lt 2\) | B1,A1 |
| \(x \lt 0\) | |
| \(3x^2 - 5x - 2 \gt 0\) | M1 |
| cvs \(x = -\dfrac{1}{3},\ x = 2\) | |
| \(\therefore\ x \lt -\dfrac{1}{3}\) | A1 |