FP2 June 2017 Q2
2. Use algebra to find the set of values of \(x\) for which \[\frac{x - 2}{2(x + 2)} \leqslant \frac{12}{x(x + 2)}\] (9)
| Scheme | Marks |
|---|---|
| \(\dfrac{x - 2}{2(x + 2)} \leqslant \dfrac{12}{x(x + 2)}\) | |
| NB Question states "Use algebra..." so purely graphical solutions score max 1/9 (the B1). A sketch and some algebra to find CVs or intersection points can score according to the method used. | |
| Can use \(\leqslant, \lt\) or \(=\) for the first 6 marks in all methods | |
| \(\dfrac{x - 2}{2(x + 2)} - \dfrac{12}{x(x + 2)}\ (\leqslant 0)\) Collects expressions to one side. | M1 |
| \(\dfrac{x^2 - 2x - 24}{2x(x + 2)}\ (\leqslant 0)\) M1: Attempt common denominator A1: Correct single fraction | M1A1 |
| \(x = 0,\ -2\) Correct critical values | B1 |
| \(x^2 - 2x - 24 \Rightarrow (x + 4)(x - 6)\ (= 0) \Rightarrow x = \ldots\) Attempt to solve their quadratic as far as \(x = \ldots\) | M1 |
| \(x = -4,\ 6\) Correct critical values. May be seen on a sketch. | A1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) with \(\leqslant\) or \(\lt\) throughout M1: Attempt two inequalities using their 4 critical values in ascending order. (dependent on at least one previous M mark) A1: All 4 CVs in the inequalities correct | dM1A1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) \([-4, -2) \cup (0, 6]\) A1: Inequality signs correct Set notation may be used. \(\cup\) or "or" but not "and" | A1cao |
| (9 marks) |
Notes
Alternative 1: Multiplies both sides by \(x^2(x + 2)^2\)
| Scheme | Marks |
|---|---|
| \(x^2(x - 2)(x + 2) \leqslant 24x(x + 2)\) \(x^3(x + 2) - 2x^2(x + 2) \leqslant 24x(x + 2)\) Both sides \(\times x^2(x + 2)^2\) May multiply by more terms but must be a positive multiplier containing \(x^2(x + 2)^2\) | M1 |
| \(x^3(x + 2) - 2x^2(x + 2) - 24x(x + 2)\ (\leqslant 0)\) M1: Collects expressions to one side A1: Correct inequality | M1A1 |
| \(x = 0,\ -2\) Correct critical values | B1 |
| \(x^4 - 28x^2 - 48x\ (= 0)\) \(x(x + 2)(x - 6)(x + 4)\ (= 0) \Rightarrow x = \ldots\) Attempt to solve their quartic as far as \(x = \ldots\) to obtain the other critical values Can cancel \(x\) and solve a cubic or \(x\) and \((x + 2)\) and solve a quadratic. | M1 |
| \(x = -4,\ 6\) Correct critical values | A1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) with \(\leqslant\) or \(\lt\) throughout M1: Attempt two inequalities using their 4 critical values in ascending order. (dependent on at least one previous M mark) A1: All 4 CVs in the inequalities correct | dM1A1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) \([-4, -2) \cup (0, 6]\) A1: Inequality signs correct Set notation may be used. \(\cup\) or "or" but not "and" | A1cao |
| (9) |
Alternative 2: using a sketch graph
(probably from calculator)
| Scheme | Marks |
|---|---|
| Draw graphs of \(y = \dfrac{x - 2}{2(x + 2)}\) and \(y = \dfrac{12}{x(x + 2)}\) | |
| CVs \(x = 0,\ -2\) (Vertical asymptotes of graphs.) | B1 |
| \(\dfrac{x - 2}{2(x + 2)} = \dfrac{12}{x(x + 2)}\) Eliminate \(y\) | M1 |
| \(x(x - 2) = 24\) M1: Obtains a quadratic equation A1: Correct equation | M1A1 |
| \(x^2 - 2x - 24 \Rightarrow (x + 4)(x - 6) = 0 \Rightarrow x = \ldots\) Attempt to solve their quadratic as far as \(x = \ldots\) | M1 |
| CVs \(x = -4,\ 6\) Correct critical values | A1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) with \(\leqslant\) or \(\lt\) throughout M1: Attempt two inequalities using their 4 critical values in ascending order. (dependent on at least one previous M mark) | dM1 |
| \(-4 \leqslant x \lt -2,\ \ 0 \lt x \leqslant 6\) A1: All 4 CVs in the inequalities correct A1: All inequality signs correct | A1 A1cao |
| (9) |
NB As above, but with no sketch graph shown: CVs \(x = 0, -2\) must be stated somewhere. (B1)
Otherwise no marks available.