4MA0 Higher June 2017 Paper 4 Q15
15 Here is a trapezium.
Diagram NOT accurately drawn
All measurements are in centimetres.
The area of the trapezium is 60 cm²
(a) Show that \(3x^2 + 10x - 117 = 0\) (3)
(b) Work out the value of \(x\)
Show your working clearly.
Give your answer correct to 3 significant figures. (3)
Show your working clearly.
Give your answer correct to 3 significant figures. (3)
| Scheme | Marks |
|---|---|
\(\dfrac{1}{2} \times (x + 5 + 2x - 4) \times (x + 3)\) or \((3x + 1)(x + 3) = 120\) or \((2x - 4)(x + 3) + \tfrac{1}{2}(9 - x)(x + 3)\) or \((x + 5)(x + 3) - \tfrac{1}{2}(9 - x)(x + 3)\) | M1 |
| \(\dfrac{1}{2} \times (3x^2 + 9x + x + 3) = 60\) oe | M1 |
| \(3x^2 + 10x + 3 = 120\) or \(1.5x^2 + 5x + 1.5 = 60\) Answer: shown | A1 |
| (3) |
Notes
M1: correct expression for area (trapezium)
(rectangle + triangle)
(rectangle – triangle)
M1: correct expansion of (all pairs) brackets in a correct equation
A1: dep on fully correct working to get to \(3x^2 + 10x - 117 = 0\)
| Scheme | Marks |
|---|---|
\(\dfrac{-10 \pm \sqrt{1504}}{6}\) or \(\dfrac{-10 \pm \sqrt{10^2 - -1404}}{2 \times 3}\) oe or \(\dfrac{-10 \pm 4\sqrt{94}}{6}\) NB: denominator must be 2 × 3 or 6 and there must be evidence for correct order of operations in the numerator | M2 |
| 4.80 | A1 |
| (3) | |
| (6 marks) |
Notes
M2: If not M2 then M1 for
\(\dfrac{-10 \pm \sqrt{10^2 - 4 \times 3 \times -117}}{2 \times 3}\)
(may have just + rather than ±)
Condone one sign error in substitution; allow partial evaluation
A1: Award M2A1 for answers in range 4.796 – 4.8 (and no other answer) with sufficient correct working that would gain at least M1
[Award M2A0 for working sufficient for M1 with both the –ve and +ve answers (–8.13 & 4.80)]