| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Sketch two trig curves and count intersections/solutions |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question requiring standard graph sketching of basic trig functions, simple verification by substitution using exact values, reading intersection points from the graph, and interpreting inequalities graphically. While it requires multiple skills, each component is routine for P1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05f Trigonometric function graphs: symmetries and periodicities1.05g Exact trigonometric values: for standard angles1.05o Trigonometric equations: solve in given intervals |
**(i)**
- B1: $y = \sin x$ $(0,0)$, $(\pi,0)$ + curve
- B1: $y = \cos 2x$ One full cycle
- B1: $y = \cos 2x$ starts and finishes at $(0, 1)$ and oscillates between $-1$ and $+1$. Do not penalise graphs from $0$ to $360$.
**(ii)**
- B1: Evidence of $\sin 30° = \cos 60° = 0.5$
- B1: Other root is $150°$
**(iii)**
- B1: $0 \leq x < 30$ and $150 < x \leq 180$ ($(x < 30$ or $x > 150$ ok). Condone $<$ or $\leq$ throughout
- B1: co
---5 (i) Sketch, on the same diagram, the graphs of $y = \sin x$ and $y = \cos 2 x$ for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
(ii) Verify that $x = 30 ^ { \circ }$ is a root of the equation $\sin x = \cos 2 x$, and state the other root of this equation for which $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$.\\
(iii) Hence state the set of values of $x$, for $0 ^ { \circ } \leqslant x \leqslant 180 ^ { \circ }$, for which $\sin x < \cos 2 x$.
\hfill \mbox{\textit{CAIE P1 2011 Q5 [7]}}