4MA0 Higher June 2016 Paper 3 Q17
17 The diagram shows a trapezium.
Diagram NOT accurately drawn
All measurements on the diagram are in centimetres.
The area of the trapezium is 119 cm²
(i) Show that \(2x^2 - x - 120 = 0\)
(ii) Find the value of \(x\).
Show your working clearly.
Show your working clearly.
(6)
| Scheme | Marks |
|---|---|
(i) e.g. \(\dfrac{1}{2} \times (x + 6 + 3x - 4) \times (x - 1)\) or \((x + 6)(x - 1)\) or \((x - 1)(3x - 4)\) or \(\dfrac{1}{2} \times (x - 1)(3x - 4 - (x + 6))\) | M1 |
| eg. \(\dfrac{1}{2} \times (4x^2 - 2x - 2) = 119\) | M1 |
| shown | A1 |
(ii) \((2x \pm 15)(x \pm 8)\) (=0) or \(\dfrac{--1 \pm \sqrt{(-1)^2 - 4 \times 2 \times -120}}{2 \times 2}\) or \(\left(x - \dfrac{1}{4}\right)^2 - \left(\dfrac{1}{4}\right)^2 - 60 = 0\) | M1 |
\((2x + 15)(x - 8)\) (=0) or \(\dfrac{1 \pm \sqrt{1 + 960}}{4}\) or \(x = \dfrac{1}{4} \pm \sqrt{\left(\dfrac{1}{4}\right)^2 + 60}\) or −7.5 and 8 given as solutions | M1 |
| 8 | A1 |
| (6) |
Notes
M1: correct algebraic expression for any relevant area
M1: for correct equation with at least one pair of brackets expanded correctly
A1: for completion to given equation
M1: Start to solve quadratic condone one sign error in substitution if quadratic formula used; allow \(-1^2\) or \(1^2\) or 1 in place of \((-1)^2\)
ft from an incorrect 3 term quadratic equation
M1: dep
ft method from an incorrect 3 term quadratic equation
A1: Award all 3 marks if first M1 awarded and 8 alone given as final answer