AS October 2020 Q2
2. Use algebra to determine the values of \(x\) for which
\[\frac{x + 1}{2x^2 + 5x - 3} \gt \frac{x}{4x^2 - 1}\](5)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{x + 1}{2x^2 + 5x - 3} \gt \dfrac{x}{4x^2 - 1}\) | ||
| \(\dfrac{2x^2 + 3x + 1 - x^2 - 3x}{(2x - 1)(2x + 1)(x + 3)} \gt 0\) or \((x + 1)(2x - 1)(2x + 1)^2(x + 3) - x(2x - 1)(2x + 1)(x + 3)^2 \gt 0\) | M1 | 2.1 |
| \(\dfrac{x^2 + 1}{(2x - 1)(2x + 1)(x + 3)} \gt 0\) or \((x + 3)(2x - 1)(2x + 1)(x^2 + 1) \gt 0\) | dM1 | 1.1b |
| All three critical values \(-3, -\dfrac{1}{2}, \dfrac{1}{2}\) | A1 | 1.1b |
| \(\left\{x \in \mathbb{R} : -3 \lt x \lt -\dfrac{1}{2}\right\} \cup \left\{x \in \mathbb{R} : x \gt \dfrac{1}{2}\right\}\) | dM1 A1 | 2.2a 2.5 |
| (5) | ||
| (5 marks) |
Notes
M1: Gathers terms on one side and puts over a common denominator, or multiplies by \((2x + 1)^2(2x - 1)^2(x + 3)^2\) and gathers terms on one side
dM1: Expands and simplifies numerator or factorises into 4 factors. Depends on the previous method mark.
A1: Correct critical values and no “extras” but ignore any attempts to solve \(x^2 + 1 = 0\) (correct or otherwise)
dM1: Deduces that 1 “inside” inequality and 1 “outside” inequality is required with critical values in ascending order. Depends on the previous method mark.
A1: Exactly 2 correct intervals, accepting equivalent notation
Special Case: Allow M1M0A0M0A0
\[\frac{x + 1}{2x^2 + 5x - 3} \gt \frac{x}{4x^2 - 1} \Rightarrow \frac{x + 1}{(2x - 1)(x + 3)} \gt \frac{x}{(2x - 1)(2x + 1)} \Rightarrow \frac{x + 1}{(x + 3)} \gt \frac{x}{(2x + 1)}\]\[\Rightarrow (x + 1)(x + 3)(2x + 1)^2 \gt x(x + 3)^2(2x + 1) \text{ etc.}\]