| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2018 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - given gradient condition |
| Difficulty | Standard +0.3 This is a straightforward chain rule differentiation of ln(1-cos 2x) requiring the double angle identity to simplify to k cot x, followed by solving a standard trigonometric equation. The 'show that' format and multi-step nature add slight complexity, but all techniques are routine C3/C4 material with no novel insight required. |
| 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^21.07l Derivative of ln(x): and related functions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ln(1-\cos 2x) \Rightarrow \frac{dy}{dx} = \frac{2\sin 2x}{1-\cos 2x}\) | M1A1 | M1: differentiating to form \(\frac{\pm A\sin 2x}{1-\cos 2x}\); A1: correct expression |
| \(\Rightarrow \frac{dy}{dx} = \frac{4\sin x\cos x}{1-(1-2\sin^2 x)} = \frac{4\sin x\cos x}{2\sin^2 x} = 2\cot x\) | M1, A1 | M1: uses \(\sin 2x = 2\sin x\cos x\) and \(\cos 2x = 1-2\sin^2 x\); A1: simplifies to show \(\frac{dy}{dx} = 2\cot x\) with at least one correct intermediate line |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ln(1-\cos 2x) \Rightarrow y = \ln(2\sin^2 x) \Rightarrow \frac{dy}{dx} = \frac{4\sin x\cos x}{2\sin^2 x}\) | M1A1 | |
| \(\Rightarrow \frac{dy}{dx} = 2\frac{\cos x}{\sin x} = 2\cot x\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ln(2\sin^2 x) \Rightarrow y = \ln 2 + 2\ln\sin x \Rightarrow \frac{dy}{dx} = 0 + \frac{2\cos x}{\sin x}\) | M1A1 | |
| \(\Rightarrow \frac{dy}{dx} = 2\frac{\cos x}{\sin x} = 2\cot x\) | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = \ln(1-\cos 2x) \Rightarrow e^y = 1-\cos 2x \Rightarrow e^y\frac{dy}{dx} = 2\sin 2x\) | M1A1 | |
| Then as main scheme | M1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(2\cot x = 2\sqrt{3} \Rightarrow \tan x = \frac{1}{\sqrt{3}}\) | — | Uses \(\cot x = \frac{1}{\tan x}\) and proceeds to find \(x\) |
| \(x = \arctan\!\left(\frac{1}{\sqrt{3}}\right) \Rightarrow x = \frac{\pi}{6}\) | M1A1 | A1: \(x=\frac{\pi}{6}\); ignore additional incorrect values such as \(x=\frac{5\pi}{6}\); do not accept 30° |
| \(y = \ln\!\left(1-\cos\!\left(\frac{2\pi}{6}\right)\right) = \ln\frac{1}{2}\) or \(-\ln 2\) | M1A1 | M1: substitutes their \(x\) into \(y=\ln(1-\cos 2x)\); A1: \(\ln\frac{1}{2}\) or \(-\ln 2\); withhold if another \((x,y)\) given in range |
## Question 7:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln(1-\cos 2x) \Rightarrow \frac{dy}{dx} = \frac{2\sin 2x}{1-\cos 2x}$ | M1A1 | M1: differentiating to form $\frac{\pm A\sin 2x}{1-\cos 2x}$; A1: correct expression |
| $\Rightarrow \frac{dy}{dx} = \frac{4\sin x\cos x}{1-(1-2\sin^2 x)} = \frac{4\sin x\cos x}{2\sin^2 x} = 2\cot x$ | M1, A1 | M1: uses $\sin 2x = 2\sin x\cos x$ and $\cos 2x = 1-2\sin^2 x$; A1: simplifies to show $\frac{dy}{dx} = 2\cot x$ with at least one correct intermediate line |
**Alternative I:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln(1-\cos 2x) \Rightarrow y = \ln(2\sin^2 x) \Rightarrow \frac{dy}{dx} = \frac{4\sin x\cos x}{2\sin^2 x}$ | M1A1 | |
| $\Rightarrow \frac{dy}{dx} = 2\frac{\cos x}{\sin x} = 2\cot x$ | M1, A1 | |
**Alternative II:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln(2\sin^2 x) \Rightarrow y = \ln 2 + 2\ln\sin x \Rightarrow \frac{dy}{dx} = 0 + \frac{2\cos x}{\sin x}$ | M1A1 | |
| $\Rightarrow \frac{dy}{dx} = 2\frac{\cos x}{\sin x} = 2\cot x$ | M1, A1 | |
**Alternative III:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = \ln(1-\cos 2x) \Rightarrow e^y = 1-\cos 2x \Rightarrow e^y\frac{dy}{dx} = 2\sin 2x$ | M1A1 | |
| Then as main scheme | M1, A1 | |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $2\cot x = 2\sqrt{3} \Rightarrow \tan x = \frac{1}{\sqrt{3}}$ | — | Uses $\cot x = \frac{1}{\tan x}$ and proceeds to find $x$ |
| $x = \arctan\!\left(\frac{1}{\sqrt{3}}\right) \Rightarrow x = \frac{\pi}{6}$ | M1A1 | A1: $x=\frac{\pi}{6}$; ignore additional incorrect values such as $x=\frac{5\pi}{6}$; do not accept 30° |
| $y = \ln\!\left(1-\cos\!\left(\frac{2\pi}{6}\right)\right) = \ln\frac{1}{2}$ or $-\ln 2$ | M1A1 | M1: substitutes their $x$ into $y=\ln(1-\cos 2x)$; A1: $\ln\frac{1}{2}$ or $-\ln 2$; withhold if another $(x,y)$ given in range |
---7. A curve has equation
$$y = \ln ( 1 - \cos 2 x ) , \quad x \in \mathbb { R } , 0 < x < \pi$$
Show that
\begin{enumerate}[label=(\alph*)]
\item $\frac { \mathrm { d } y } { \mathrm {~d} x } = k \cot x$, where $k$ is a constant to be found.
Hence find the exact coordinates of the point on the curve where
\item $\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 \sqrt { 3 }$
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2018 Q7 [8]}}