FP2 June 2014 Q2
2. Use algebra to find the set of values of \(x\) for which \[\left|3x^2 - 19x + 20\right| < 2x + 2\] (6)
| Scheme | Marks |
|---|---|
| \(\left|3x^2 - 19x + 20\right| < 2x + 2\) | |
| \(3x^2 - 19x + 20 = 2x + 2\) \(\Rightarrow 3x^2 - 21x + 18 = 0 \Rightarrow x = \ldots\) \(3x^2 - 19x + 20 = 2x + 2\) and attempt to solve correctly May be solved as an inequality | M1 |
| \(x = 1,\quad x = 6\) Both (ie critical values seen) | A1 |
| \(-\left(3x^2 - 19x + 20\right) = 2x + 2\) \(\Rightarrow 3x^2 - 17x + 22 = 0 \Rightarrow x = \ldots\) \(-\left(3x^2 - 19x + 20\right) = 2x + 2\) and attempt to solve correctly May be solved as an inequality | M1 |
| \(x = \tfrac{11}{3},\quad x = 2\) Both (critical values seen) Accept awrt 3.67 | A1 |
| \(1 < x < 2,\quad \tfrac{11}{3} < x < 6\) Must be strict inequalities. Accept awrt 3.67 A1 either correct, A1 both correct. But give A1A0 if both correct apart from \(\leqslant\) seen somewhere in the final answers. Give A1A0 if both correct and extra intervals seen | A1, A1 |
| (6) | |
| (6 marks) |
Notes
If no algebra seen (implies a calculator solution) no marks.
With algebra:
M1 Squaring and reaching a quartic = 0
M1 Attempt to factorise and obtain at least one solution for \(x\). Coefficient of \(x^4\) and constant term correct for their quartic.
A1 Any 2 correct values
A1 All 4 correct values
Final 2 A marks as above
Accept set notation for the final 2 A marks. \(x \in (1, 2)\), \(x \in \left(\tfrac{11}{3}, 6\right)\) not \([1, 2]\)