AS June 2018 Q3
3. Use algebra to find the values of \(x\) for which
\[\frac{x}{x^2 - 2x - 3} \leqslant \frac{1}{x + 3}\](7)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x}{x^2 - 2x - 3} \leqslant \dfrac{1}{x + 3}\) | ||
| \(\dfrac{x(x + 3) - (x^2 - 2x - 3)}{(x^2 - 2x - 3)(x + 3)} \leqslant 0\) or \(x(x - 3)(x + 1)(x + 3)^2 - (x - 3)^2(x + 1)^2(x + 3) \leqslant 0\) or \(x(x^2 - 2x - 3)(x + 3)^2 - (x^2 - 2x - 3)^2(x + 3) \leqslant 0\) | M1 | 2.1 |
| \(\dfrac{5x + 3}{(x - 3)(x + 3)(x + 1)}\ \{\leqslant 0\}\) or \((x - 3)(x + 1)(x + 3)(5x + 3)\ \{\leqslant 0\}\) | M1 A1 | 1.1b 1.1b |
| All three critical values \(-3, 3, -1\) | B1 | 1.1b |
| Critical value \(-\dfrac{3}{5}\) | B1ft | 1.1b |
| \(\left\{x \in \mathbb{R} : -3 \lt x \lt -1\right\} \cup \left\{x \in \mathbb{R} : -\dfrac{3}{5} \leqslant x \lt 3\right\}\) | M1 A1 | 2.2a 2.5 |
| (7) | ||
| (7 marks) |
Notes
M1: Gathers terms on one side and puts over a common denominator,
or multiplies by \((x + 1)^2(x - 3)^2(x + 3)^2\) (or by the equivalent \((x^2 - 2x - 3)^2(x + 3)^2\)) and gathers terms onto one side
M1: Expands and simplifies fully the numerator or takes out a factor of \((x - 3)(x + 1)(x + 3)\) (or the equivalent \((x^2 - 2x - 3)(x + 3)\)) and then simplifies fully their remaining factor
A1: \(\dfrac{5x + 3}{(x - 3)(x + 3)(x + 1)}\) or \((x - 3)(x + 1)(x + 3)(5x + 3)\)
B1: Correct critical values of \(-3\), \(3\) and \(-1\) which can be implied, e.g. from their inequalities
B1ft: Correct critical value of \(-\dfrac{3}{5}\) which can be implied, e.g. from their inequalities
Note: B1ft: You can follow through their fourth factor which is in the form \((ax + b)\), \(a, b \neq 0\) to give C.V. \(= -\dfrac{b}{a}\), if their fourth factor is not any of either \((x - 3)\), \((x + 3)\) or \((x + 1)\)
M1: Deduces that 2 “inside” inequalities are required with critical values in ascending order
A1: Exactly 2 correct intervals, condoning omission of the union symbol
Note: Also accept, e.g.
- \(-3 \lt x \lt -1,\ -\dfrac{3}{5} \leqslant x \lt 3\)
- \((-3, -1),\ \left[-\dfrac{3}{5}, 3\right)\)
- \(-1 \gt x \gt -3,\ 3 \gt x \geqslant -\dfrac{3}{5}\)
Note: Give 1st A0 for \((x^2 - 2x - 3)(x + 3)(5x + 3)\ \{\leqslant 0\}\) with no other working seen
Note: Give 1st A1 (implied) for \((x^2 - 2x - 3)(x + 3)(5x + 3)\ \{\leqslant 0\}\) with \(x = 3,\ x = -1\) stated
Note: Give 1st A0 for \(\dfrac{5x + 3}{(x^2 - 2x - 3)(x + 3)}\ \{\leqslant 0\}\) with no other working seen
Note: Give 1st A1 (implied) for \(\dfrac{5x + 3}{(x^2 - 2x - 3)(x + 3)}\ \{\leqslant 0\}\) with \(x = 3,\ x = -1\) stated
Note: Give 1st A0 for \(\dfrac{5x + 3}{x^3 + x^2 - 9x - 9}\ \{\leqslant 0\}\) with no other working seen
Note: Give 1st A1 (implied) for \(\dfrac{5x + 3}{x^3 + x^2 - 9x - 9}\ \{\leqslant 0\}\) with \(x = 3,\ x = -1,\ x = -3\) stated
Note: Allow special case final M1 for any of
- \(-3 \lt x \lt -1\) (condoning closed inequalities or a mixture of open and closed inequalities)
- \(-\dfrac{3}{5} \leqslant x \lt 3\) (condoning closed inequalities or a mixture of open and closed inequalities)
but do not allow M1 for any of
- e.g. \(-3 \lt x \lt -1,\ -1 \lt x \leqslant -\dfrac{3}{5}\) (“continuing inequalities”)
- e.g. \(-3 \lt x \lt 1,\ -\dfrac{3}{5} \leqslant x \lt 3\) (“overlapping inequalities”)
Alternative Method
\(x(x - 3)(x + 1)(x + 3)^2 \leqslant (x - 3)^2(x + 1)^2(x + 3)\)
\(x^5 + 4x^4 - 6x^3 - 36x^2 - 27x \leqslant x^5 - x^4 - 14x^3 + 6x^2 + 45x + 27\)
\(5x^4 + 8x^3 - 42x^2 - 72x - 27 \leqslant 0\)
Note: \(5x^4 + 8x^3 - 42x^2 - 72x - 27 \leqslant 0\) without any other working is M1M0A0
Note: \(5x^4 + 8x^3 - 42x^2 - 72x - 27 \leqslant 0 \Rightarrow x = -3, -1, 3\) is M1M1A1B1
Note: \(5x^4 + 8x^3 - 42x^2 - 72x - 27 \leqslant 0 \Rightarrow x = -3, -1, 3, -\dfrac{3}{5}\) is M1M1A1B1B1