| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Harmonic Form |
| Type | Express and solve equation |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 transformations and trigonometric identities. Part (a) requires routine graph sketching with transformations; part (b) is a standard R-cos(x+α) conversion using the harmonic form; part (c) uses given stationary points to find A; part (d) solves a trigonometric equation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.05n Harmonic form: a sin(x)+b cos(x) = R sin(x+alpha) etc1.05o Trigonometric equations: solve in given intervals |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Sketch showing: maximum at origin, crossing x-axis at \(x = \pm 135\), minimum value of \(-1\) at \((-135, -1)\) and \((135, -1)\) | B3 | |
| (a)(ii) Sketch showing: maximum at \((-45, 15)\), minimum at \((135, -1)\), crossing x-axis appropriately | M1 A2 | |
| (b) \(2\sqrt{2}\cos x - 2\sqrt{2}\sin x = R\cos x \cos\alpha - R\sin x \sin\alpha\) | M1 A1 | |
| \(R\cos\alpha = 2\sqrt{2}\), \(R\sin\alpha = 2\sqrt{2}\), \(\therefore R = \sqrt{8+8} = 4\) | B1 | |
| \(\tan\alpha = 1\), \(\alpha = 45°\) | B1 | |
| \(\therefore f(x) = A + 4\cos(x+45)°\) | ||
| (c) \(3\) | B1 | |
| (d) \(3 + 4\cos(x+45) = 0\), \(\cos(x+45) = -\frac{3}{4}\) | M1 | |
| \(x + 45 = 180 - 41.4, 180 + 41.4 = 138.6, 221.4\) | M1 | |
| \(x = 93.6, 176.4\) (1dp) | A2 | (14) |
(a)(i) Sketch showing: maximum at origin, crossing x-axis at $x = \pm 135$, minimum value of $-1$ at $(-135, -1)$ and $(135, -1)$ | B3 |
(a)(ii) Sketch showing: maximum at $(-45, 15)$, minimum at $(135, -1)$, crossing x-axis appropriately | M1 A2 |
(b) $2\sqrt{2}\cos x - 2\sqrt{2}\sin x = R\cos x \cos\alpha - R\sin x \sin\alpha$ | M1 A1 |
$R\cos\alpha = 2\sqrt{2}$, $R\sin\alpha = 2\sqrt{2}$, $\therefore R = \sqrt{8+8} = 4$ | B1 |
$\tan\alpha = 1$, $\alpha = 45°$ | B1 |
$\therefore f(x) = A + 4\cos(x+45)°$ |
(c) $3$ | B1 |
(d) $3 + 4\cos(x+45) = 0$, $\cos(x+45) = -\frac{3}{4}$ | M1 |
$x + 45 = 180 - 41.4, 180 + 41.4 = 138.6, 221.4$ | M1 |
$x = 93.6, 176.4$ (1dp) | A2 | **(14)**
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**Total: (75)**\includegraphics{figure_1}
Figure 1 shows the curve $y = f(x)$ which has a maximum point at $(-45, 7)$ and a minimum point at $(135, -1)$.
\begin{enumerate}[label=(\alph*)]
\item Showing the coordinates of any stationary points, sketch on separate diagrams the graphs of
\begin{enumerate}[label=(\roman*)]
\item $y = f(|x|)$,
\item $y = 1 + 2f(x)$. [6]
\end{enumerate}
\end{enumerate}
Given that
$$f(x) = A + 2\sqrt{2} \cos x^{\circ} - 2\sqrt{2} \sin x^{\circ}, \quad x \in \mathbb{R}, \quad -180 \leq x \leq 180,$$
where $A$ is a constant,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item show that f(x) can be expressed in the form
$$f(x) = A + R \cos (x + \alpha)^{\circ},$$
where $R > 0$ and $0 < \alpha < 90$, [3]
\item state the value of $A$, [1]
\item find, to $1$ decimal place, the $x$-coordinates of the points where the curve $y = f(x)$ crosses the $x$-axis. [4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel C3 Q7 [14]}}