Higher June 2017 Paper 2 Q21
21.
Diagram not drawn to scale
(a) Show that \(7x^2 + 4x - 93{\cdot}4 = 0\). [3]
(b) Use the quadratic formula to solve \(7x^2 + 4x - 93{\cdot}4 = 0\).
Give both of your answers correct to 2 decimal places. [3]
Give both of your answers correct to 2 decimal places. [3]
(c) Find each of the lengths of the parallel sides of the trapezium.
You must justify any decisions that you make. [3]
You must justify any decisions that you make. [3]
| Scheme | Marks | Comments |
|---|---|---|
| \(\tfrac12 \times (x+1) \times (4x - 5 + 3x + 2)\) or \(\tfrac12 \times (x+1) \times (7x - 3)\) | M1 | Mark intention, i.e. brackets may be missing |
| \(45.2 = \tfrac12(7x^2 + 7x - 3x - 3)\) or equivalent | m1 | |
| \(90.4 = 7x^2 + 4x - 3\) with \(7x^2 + 4x = 93.4\) or \(7x^2 + 4x - 93.4 = 0\) | A1 |
| Scheme | Marks | Comments |
|---|---|---|
| (\(x =\)) \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 7 \times -93.4}}{2 \times 7}\) | M1 | Allow if seen in (a), provided not contradicted in (b) Allow 1 slip in substitution, but must be correct formula |
| \(= \dfrac{-4 \pm \sqrt{2631.2}}{14}\) | A1 | OR for sight of 3.38 or \(-3.95\) |
| 3.38 with \(-3.95\) | A1 | CAO. Both solutions to 2dp |
| Scheme | Marks | Comments |
|---|---|---|
| 8.5(129…cm) or 8.52 (cm) and 12.1(346… cm) or 12.14 (cm) | B2 | FT use of ‘their positive value’ for B1 only provided previous M1 in (b) awarded B1 for sight of 4×3.38 – 5 and 3×3.38 + 2, or for sight of either correct length |
| Decision and justification, e.g. ‘that the negative solution in (b) was not valid, as lengths can only be positive’ | E1 | Do not accept a decision alone, e.g. ‘did not use the negative answer’ |
| (9) |