| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Position vectors and magnitudes |
| Difficulty | Moderate -0.8 This is a straightforward vectors question requiring basic dot product application (perpendicular vectors) and standard vector arithmetic. Part (i) uses the condition that perpendicular vectors have zero dot product to find p, while part (ii) involves routine vector subtraction and normalization. Both parts are direct applications of standard techniques with no problem-solving insight required, making this easier than average for A-level. |
| Spec | 1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{(1+c)^2 + s^2}{s(1+c)} = \frac{1+2c+c^2+s^2}{s(1+c)}\) | M1 | Correct use of fractions |
| \(= \frac{2+2c}{s(1+c)} = \frac{2(1+c)}{s(1+c)} \rightarrow \frac{2}{s}\) | M1 A1 | Use of trig identity, A1 needs evidence of cancelling |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{2}{s} = \frac{3}{c} \rightarrow t = \frac{2}{3}\) | M1 | Use part (i) and \(t = s \div c\), may restart from given equation |
| \(\rightarrow \theta = 33.7°\) or \(213.7°\) | A1 A1FT | FT for \(180° +\) 1st answer. 2nd A1 lost for extra solutions in range |
| Total: 3 |
## Question 3(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{(1+c)^2 + s^2}{s(1+c)} = \frac{1+2c+c^2+s^2}{s(1+c)}$ | **M1** | Correct use of fractions |
| $= \frac{2+2c}{s(1+c)} = \frac{2(1+c)}{s(1+c)} \rightarrow \frac{2}{s}$ | **M1 A1** | Use of trig identity, **A1** needs evidence of cancelling |
| **Total: 3** | | |
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## Question 3(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{2}{s} = \frac{3}{c} \rightarrow t = \frac{2}{3}$ | **M1** | Use part (i) and $t = s \div c$, may restart from given equation |
| $\rightarrow \theta = 33.7°$ or $213.7°$ | **A1 A1FT** | FT for $180° +$ 1st answer. 2nd **A1** lost for extra solutions in range |
| **Total: 3** | | |
---3 (i) Prove the identity $\frac { 1 + \cos \theta } { \sin \theta } + \frac { \sin \theta } { 1 + \cos \theta } \equiv \frac { 2 } { \sin \theta }$.\\
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(ii) Hence solve the equation $\frac { 1 + \cos \theta } { \sin \theta } + \frac { \sin \theta } { 1 + \cos \theta } = \frac { 3 } { \cos \theta }$ for $0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }$.\\
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\hfill \mbox{\textit{CAIE P1 2017 Q3 [6]}}