OCR MEI C3 2014 June — Question 2 5 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeFind gradient at specific point
DifficultyStandard +0.3 This is a straightforward differentiation question requiring the chain rule applied twice (once for ln and once for the composite trigonometric function), followed by substitution of a standard angle. While it involves multiple steps, the techniques are standard C3 material with no problem-solving insight required, making it slightly easier than average.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
Find the exact gradient of the curve \(y = \ln(1 - \cos 2x)\) at the point with \(x\)-coordinate \(\frac{1}{4}\pi\). [5]
AnswerMarks Guidance
\(y = \ln(1 - \cos 2x)\), let \(u = 1 - \cos 2x\)M1 \(1/(1 - \cos 2x)\) soi
\(\Rightarrow \quad dy/dy = dy/du \cdot du/dx = (1/u) \cdot 2\sin 2x\)M1 \(d/dx(1 - \cos 2x) = \pm 2\sin 2x\)
\(= \frac{2\sin 2x}{1 - \cos 2x}\)A1 cao
When \(x = \pi/6\), \(\frac{dy}{dx} = \frac{2\sin(\pi/3)}{1 - \cos(\pi/3)}\)M1 substituting \(\pi/6\) or \(30°\) into their deriv; must be in at least two places
\(= 2\sqrt{3}\)A1 cao [5] isw after correct answer seen 
$y = \ln(1 - \cos 2x)$, let $u = 1 - \cos 2x$ | M1 | $1/(1 - \cos 2x)$ soi |
$\Rightarrow \quad dy/dy = dy/du \cdot du/dx = (1/u) \cdot 2\sin 2x$ | M1 | $d/dx(1 - \cos 2x) = \pm 2\sin 2x$ |
$= \frac{2\sin 2x}{1 - \cos 2x}$ | A1 cao | |
When $x = \pi/6$, $\frac{dy}{dx} = \frac{2\sin(\pi/3)}{1 - \cos(\pi/3)}$ | M1 | substituting $\pi/6$ or $30°$ into their deriv; must be in at least two places |
$= 2\sqrt{3}$ | A1 cao [5] | isw after correct answer seen |
Find the exact gradient of the curve $y = \ln(1 - \cos 2x)$ at the point with $x$-coordinate $\frac{1}{4}\pi$. [5]

\hfill \mbox{\textit{OCR MEI C3 2014 Q2 [5]}}
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