FP2 June 2013 (R) Q2
2. Use algebra to find the set of values of \(x\) for which \[\frac{6x}{3 - x} > \frac{1}{x + 1}\] (7)
| Scheme | Marks |
|---|---|
| NB Allow the first 5 marks with = instead of inequality \(\dfrac{6x}{3 - x} > \dfrac{1}{x + 1}\) | |
| \(6x(3 - x)(x + 1)^2 - (3 - x)^2(x + 1) > 0\) | M1 |
| \((3 - x)(x + 1)\left(6x^2 + 6x - 3 + x\right) > 0\) | |
| \((3 - x)(x + 1)(3x - 1)(2x + 3) > 0\) | M1dep |
| Critical values \(3,\ -1\) | B1 |
| and \(-\dfrac{3}{2},\ \dfrac{1}{3}\) | A1, A1 |
| Use critical values to obtain both of \(-\dfrac{3}{2} < x < -1\qquad \dfrac{1}{3} < x < 3\) | M1A1cso |
| (7 marks) |
Notes
M1 for multiplying through by \((x + 1)^2(3 - x)^2\)
OR: for collecting one side of the inequality and attempting to form a single fraction (see alternative in mark scheme)
M1dep for collecting on one side of the inequality and factorising the result of the above (usual rules for factorising the quadratic)
OR: for factorising the numerator – must be a three term quadratic – usual rules for factorising a quadratic (see alternative in mark scheme)
Dependent on the first M mark
B1 for the critical values \(3,\ -1\)
A1 for either \(-\dfrac{3}{2}\) or \(\dfrac{1}{3}\)
A1 for the second of these
NB: the critical values need not be shown explicitly – they may be shown on a sketch or just appear in the ranges or in the working for the ranges.
M1 using their 4 critical values to obtain appropriate ranges e.g. use a sketch graph of a quartic, (which must be the correct shape and cross the \(x\)-axis at the cvs) or a table or number line
A1cso for both of \(-\dfrac{3}{2} < x < -1,\quad \dfrac{1}{3} < x < 3\)
Set notation acceptable i.e. \(\left(-\dfrac{3}{2}, -1\right) \cup \left(\dfrac{1}{3}, 3\right)\) All brackets must be round; if square brackets appear anywhere then A0.
If both ranges correct, no working is needed for the last 2 marks, but any working shown must be correct.
Purely graphical methods are unacceptable as the question specifies “Use algebra…”.
Q2 – Alternative 1
| Scheme | Marks |
|---|---|
| \(\dfrac{6x}{3 - x} - \dfrac{1}{x + 1} > 0\) | |
| \(\dfrac{6x(x + 1) - (3 - x)}{(3 - x)(x + 1)} > 0\) | M1 |
| \(\dfrac{(3x - 1)(2x + 3)}{(3 - x)(x + 1)} > 0\) | M1dep |
| Critical values \(3,\ -1\) | B1 |
| and \(-\dfrac{3}{2},\ \dfrac{1}{3}\) | A1A1 |
| Use critical values to obtain both of \(-\dfrac{3}{2} < x < -1\qquad \dfrac{1}{3} < x < 3\) | M1A1cso |
| 7 Marks |