4MA0 Higher June 2011 Paper 4 Q24
24
Diagram NOT accurately drawn
The diagram shows a trapezium \(PQRS\).
\(PS\) is parallel to \(QR\).
\(PS = 4QR\).
\(\overrightarrow{PQ} = \mathbf{a}\) \(\overrightarrow{QR} = \mathbf{b}\)
The point \(T\) lies on the line \(PR\) such that \(PT : TR = 4 : 1\)
| Scheme | Marks |
|---|---|
| (i) \(4\mathbf{b}\) | B1 |
| (ii) \(\mathbf{a} + \mathbf{b}\) | B1 |
| (iii) \(3\mathbf{b} - \mathbf{a}\) oe | B1 |
| (3) |
Notes
B1: 4 × b etc Do not accept upper case letters
B1: Do not accept upper case letters
B1: needs not be simplified (e.g \(-\mathbf{b} - \mathbf{a} + 4\mathbf{b}\)) No upper case
Apart from questions 5b, 8, 15d, 20b, 21b, 23, 24b (where the mark scheme states otherwise) the correct answer, unless clearly obtained by an incorrect method, should be taken to imply correct working.
| Scheme | Marks |
|---|---|
| \(TS = \frac{1}{5}(\mathbf{a} + \mathbf{b}) + 3\mathbf{b} - \mathbf{a}\) \(QT = -\mathbf{a} + \frac{4}{5}(\mathbf{a} + \mathbf{b})\) | M1 |
| \(TS = -\frac{4}{5}\mathbf{a} + \frac{16}{5}\mathbf{b}\) \(QT = -\frac{1}{5}\mathbf{a} + \frac{4}{5}\mathbf{b}\) \(TS = \frac{4}{5}(-\mathbf{a} + 4\mathbf{b})\) and \(QT = \frac{1}{5}(-\mathbf{a} + 4\mathbf{b})\) | M1 |
| \(k = 4\) | A1 |
| (3) | |
| (Total 6 marks) |
Notes
M1: for any correct route from T to S or from Q to T using capitals or lower case e.g. TS =TR + RS or QT = QP + PT
M1: for both correct simplified routes from T to S and Q to T (must be lower case vectors here)
A1: dep on B1 in aii) and aiii) and at least M1