| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2007 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Trig Graphs & Exact Values |
| Type | Find function constants from given conditions |
| Difficulty | Moderate -0.8 This is a straightforward question requiring substitution of given values to form simultaneous equations, solving for constants, then finding zeros and sketching a cosine graph. All steps are routine P1 techniques with no problem-solving insight needed, making it easier than average but not trivial due to the multiple parts. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f(x) = a + b\cos 2x\), \(\rightarrow a + b = -1\) | B1 | co |
| and \(a - b = 7\) | B1 | co |
| Solution \(\rightarrow a = 3\) and \(b = -4\) | B1 [3] | co |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(3 - 4\cos 2x = 0 \rightarrow \cos 2x = \frac{3}{4}\) | M1 | \(\cos 2x\) subject and finds \(\cos^{-1}\) before \(\div 2\) |
| \(\rightarrow x = 0.36\) and \(2.78\) | A1, A1\(\sqrt{}\) [3] | co; \(\sqrt{}\) for \(\pi - 1^{\text{st}}\) answer and no other answers in range (Degrees max \(\frac{1}{3}\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Graph: 1 oscillation, \(y\)-intercept at \(-1\), max at \(7\) | B1 | Ignore anything outside \(0\) to \(\pi\). Must be 1 oscillation only |
| Shape correct with curves, from \(-1\) to \(7\) | B1 [2] | Everything ok including curves, not blatant lines and from \(-1\) to \(7\) |
## Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f(x) = a + b\cos 2x$, $\rightarrow a + b = -1$ | B1 | co |
| and $a - b = 7$ | B1 | co |
| Solution $\rightarrow a = 3$ and $b = -4$ | B1 [3] | co |
## Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $3 - 4\cos 2x = 0 \rightarrow \cos 2x = \frac{3}{4}$ | M1 | $\cos 2x$ subject and finds $\cos^{-1}$ before $\div 2$ |
| $\rightarrow x = 0.36$ and $2.78$ | A1, A1$\sqrt{}$ [3] | co; $\sqrt{}$ for $\pi - 1^{\text{st}}$ answer and no other answers in range (Degrees max $\frac{1}{3}$) |
## Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Graph: 1 oscillation, $y$-intercept at $-1$, max at $7$ | B1 | Ignore anything outside $0$ to $\pi$. Must be 1 oscillation only |
| Shape correct with curves, from $-1$ to $7$ | B1 [2] | Everything ok including curves, not blatant lines and from $-1$ to $7$ |
---8 The function f is defined by $\mathrm { f } ( x ) = a + b \cos 2 x$, for $0 \leqslant x \leqslant \pi$. It is given that $\mathrm { f } ( 0 ) = - 1$ and $\mathrm { f } \left( \frac { 1 } { 2 } \pi \right) = 7$.\\
(i) Find the values of $a$ and $b$.\\
(ii) Find the $x$-coordinates of the points where the curve $y = \mathrm { f } ( x )$ intersects the $x$-axis.\\
(iii) Sketch the graph of $y = \mathrm { f } ( x )$.
\hfill \mbox{\textit{CAIE P1 2007 Q8 [8]}}