4MA0 Higher January 2015 Paper 4 Q17
17
Diagram NOT accurately drawn
The diagram shows a trapezium.
The trapezium has an area of 17 cm2
(a) Show that \(2x^2 + 7x - 17 = 0\) (3)
(b) Work out the value of \(x\).
Give your answer correct to 3 significant figures.
Show your working clearly. (3)
Give your answer correct to 3 significant figures.
Show your working clearly. (3)
| Scheme | Marks |
|---|---|
\(\dfrac{1}{2} \times 2x \times (x + x + 7)\) or \(2x \times x\) or \(\dfrac{1}{2} \times 7 \times 2x\) or \(2x \times (x + 7)\) | M1 |
\(\dfrac{1}{2} \times 2x \times (x + x + 7) = 17\) or \(2x \times x + \dfrac{1}{2} \times 7 \times 2x = 17\) or \(2x \times (x + 7) - \dfrac{1}{2} \times 7 \times 2x = 17\) | M1 |
| show | A1 |
| (3) |
Notes
M1: for area of trapezium or any relevant area
Allow \(2x\) in place of \(x + x\)
M1: Allow \(2x\) in place of \(x + x\)
A1: for deriving the given answer \(2x^2 + 7x - 17 = 0\) correctly
| Scheme | Marks |
|---|---|
| \(\dfrac{-7 \pm \sqrt{7^2 - 4 \times 2 \times -17}}{2 \times 2}\) | M1 |
| \(\sqrt{185}\) or \(\sqrt{49 + 136}\) or 13.6.... | M1 |
| 1.65 | A1 |
| (3) | |
| (6 marks) |
Notes
M1: for correct substitution; condone one sign error in substitution
Accept + in place of ±
NB. Terms may be simplified eg. accept 4 in place of 2×2 in denominator
M1: (independent) for correct simplification of discriminant (if evaluated at least 3sf rounded or truncated)
A1: dep on 1st M mark scored
for value rounding to 1.65 given as final answer
Award 3 marks if first M1 scored and answer correct