| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Solve equation involving composites |
| Difficulty | Moderate -0.3 This question involves straightforward composition of functions followed by solving standard trigonometric equations. Part (i) requires substituting f(x) into g and solving cos(expression)=1, which is direct recall. Part (ii) requires solving a linear equation involving cos(x), also routine. Both parts use basic A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\cos\left(\frac{1}{2}x + \frac{\pi}{6}\right) = (-1)\); \(x = -\frac{\pi}{3}\) | B1, B1 | cao [2] |
| (ii) \(\frac{1}{2}\cos x + \frac{\pi}{6} = 1\); \(\cos x = 0.9528\); \(x = \pm 0.31\) | B1, B1, B1/\(\checkmark\) | \(\checkmark\) for negative of first answer [4] |
(i) $\cos\left(\frac{1}{2}x + \frac{\pi}{6}\right) = (-1)$; $x = -\frac{\pi}{3}$ | B1, B1 | cao [2]
(ii) $\frac{1}{2}\cos x + \frac{\pi}{6} = 1$; $\cos x = 0.9528$; $x = \pm 0.31$ | B1, B1, B1/$\checkmark$ | $\checkmark$ for negative of first answer [4]6 The functions f and g are defined for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$ by
$$\begin{aligned}
& f ( x ) = \frac { 1 } { 2 } x + \frac { 1 } { 6 } \pi \\
& g ( x ) = \cos x
\end{aligned}$$
Solve the following equations for $- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi$.\\
(i) $\operatorname { gf } ( x ) = 1$, giving your answer in terms of $\pi$.\\
(ii) $\operatorname { fg } ( x ) = 1$, giving your answers correct to 2 decimal places.
\hfill \mbox{\textit{CAIE P1 2012 Q6 [6]}}