Higher November 2020 Paper 2 Q17
17 The diagram shows a rectangle.
Diagram NOT accurately drawn
The area of the rectangle is \(A\) cm2
Given that \(A \lt 3x + 27\)
find the range of possible values for \(x\).
(5)
| Scheme | Marks |
|---|---|
| \((3x + 2)(2x - 4) \lt 3x + 27\) oe eg \(6x^2 - 8x - 8 \lt 3x + 27\) | M1 |
| eg \(6x^2 - 11x - 35 \lt 0\) | M1 |
| \((2x - 7)(3x + 5)\;(= 0)\) or \(\dfrac{11 \pm \sqrt{(-11)^2 - 4 \times 6 \times (-35)}}{2 \times 6}\) | M1 |
| \(-\dfrac{5}{3}, \dfrac{7}{2}\) | A1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2 \lt x \lt \dfrac{7}{2}\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: condone incorrect symbol
M1: expanding and rearranging to get a correct 3 term quadratic, condone incorrect symbol
M1: first step to find the critical values dep on M1 for solving their 3 term quadratic using any correct method (allow one sign error and some simplification – allow as far as the equivalent of \(\dfrac{11 \pm \sqrt{121 + 840}}{12}\)) or if factorising, allow brackets which expanded give 2 out of 3 terms correct)
A1: oe the positive critical value only or both critical values (if both they must be correct)
A1: accept \(2 \leqslant x \lt \dfrac{7}{2}\) may be seen as two separate inequalities \(x \gt 2\) (\(x \geqslant 2\)) and \(x \lt \dfrac{7}{2}\)
(corrected from the printed mark scheme: “(\(x \leqslant 2\))”)