FP2 June 2007 Q2
2.
The diagram above shows a sketch of the curve with equation \[y = \frac{x^2 - 1}{|x + 2|}, \qquad x \neq -2.\]
The curve crosses the \(x\)-axis at \(x = 1\) and \(x = -1\) and the line \(x = -2\) is an asymptote of the curve.
(a) Use algebra to solve the equation \(\dfrac{x^2 - 1}{|x + 2|} = 3(1 - x)\). (6)
(b) Hence, or otherwise, find the set of values of \(x\) for which \[\frac{x^2 - 1}{|x + 2|} \lt 3(1 - x).\] (3)
| Scheme | Marks |
|---|---|
| \([(x \gt -2)]\): Attempt to solve \(x^2 - 1 = 3(1 - x)(x + 2)\) \([4x^2 + 3x - 7 = 0]\) | M1 |
| \(x = 1\), or \(-\dfrac{7}{4}\) | B1, A1 |
| \([(x \lt -2)]\): Attempt to solve \(x^2 - 1 = -3(1 - x)(x + 2)\) | M1 |
| Solving \(x + 1 = 3x + 6 \qquad (2x^2 + 3x - 5 = 0)\) | M1dep |
| \(x = -\dfrac{5}{2}\) | A1 |
| (6) |
Notes
“Squaring”
| Scheme | Marks |
|---|---|
| If candidates do not notice the factor of \((x - 1)^2\) they have quartic to solve; Squaring and finding quartic \(= 0\ [8x^4 + 18x^3 - 25x^2 - 36x + 35 = 0]\) | |
| Finding one factor and factorising \((x - 1)(8x^3 + 26x^2 + x - 35) = 0\) | M1 |
| Finding one other factor and reducing other factor to quadratic, likely to be \((x - 1)^2(8x^2 + 34x + 35) = 0\) | M1 |
| Complete factorisation \((x - 1)^2(2x + 5)(4x + 7) = 0\) | M1 |
| [Second M1 implies the first, if candidate starts there or cancels \((x - 1)^2\)] | |
| \(x = 1\) B1, \(x = -7/4\) A1, \(x = -5/2\) | A1 |
\(x = 1\) allowed anywhere, no penalty in (b)
| Scheme | Marks |
|---|---|
| \(-\dfrac{7}{4} \lt x \lt 1\) One part | M1 |
| Both correct and enclosed | A1 |
| \(x \lt -\dfrac{5}{2}\) {Must be for \(x \lt -2\) and only one value} | B1ft |
| (3) | |
| (9 marks) |
Notes
Correct answers seen with no working is independent of (a) (graphical calculator) mark as scheme.
Only allow the accuracy mark if no other interval, in both parts
\(\leqslant\) used penalise first time used