OCR MEI C3 — Question 6 18 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProduct & Quotient Rules
TypeShow derivative satisfies condition
DifficultyStandard +0.3 This is a structured multi-part question covering standard C3 techniques: odd/even function properties, quotient rule differentiation, substitution integration, and function transformations. While it requires multiple skills (differentiation, integration, sketching), each part follows routine procedures with clear guidance. The substitution is explicitly suggested, and the final part simply requires recognizing the relationship between transformations and definite integrals. Slightly easier than average due to the scaffolding and standard methods throughout.
Spec1.02w Graph transformations: simple transformations of f(x)1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.08h Integration by substitution

The function \(\text{f}(x) = \frac{\sin x}{2 - \cos x}\) has domain \(-\pi \leqslant x \leqslant \pi\). Fig. 8 shows the graph of \(y = \text{f}(x)\) for \(0 \leqslant x \leqslant \pi\). \includegraphics{figure_6}
  1. Find \(\text{f}(-x)\) in terms of \(\text{f}(x)\). Hence sketch the graph of \(y = \text{f}(x)\) for the complete domain \(-\pi \leqslant x \leqslant \pi\). [3]
  2. Show that \(\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}\). Hence find the exact coordinates of the turning point P. State the range of the function \(\text{f}(x)\), giving your answer exactly. [8]
  3. Using the substitution \(u = 2 - \cos x\) or otherwise, find the exact value of \(\int_0^\pi \frac{\sin x}{2 - \cos x} dx\). [4]
  4. Sketch the graph of \(y = \text{f}(2x)\). [1]
  5. Using your answers to parts (iii) and (iv), write down the exact value of \(\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx\). [2]

The function $\text{f}(x) = \frac{\sin x}{2 - \cos x}$ has domain $-\pi \leqslant x \leqslant \pi$.

Fig. 8 shows the graph of $y = \text{f}(x)$ for $0 \leqslant x \leqslant \pi$.

\includegraphics{figure_6}

\begin{enumerate}[label=(\roman*)]
\item Find $\text{f}(-x)$ in terms of $\text{f}(x)$. Hence sketch the graph of $y = \text{f}(x)$ for the complete domain $-\pi \leqslant x \leqslant \pi$. [3]

\item Show that $\text{f}'(x) = \frac{2\cos x - 1}{(2 - \cos x)^2}$. Hence find the exact coordinates of the turning point P.

State the range of the function $\text{f}(x)$, giving your answer exactly. [8]

\item Using the substitution $u = 2 - \cos x$ or otherwise, find the exact value of $\int_0^\pi \frac{\sin x}{2 - \cos x} dx$. [4]

\item Sketch the graph of $y = \text{f}(2x)$. [1]

\item Using your answers to parts (iii) and (iv), write down the exact value of $\int_0^{\frac{\pi}{2}} \frac{\sin 2x}{2 - \cos 2x} dx$. [2]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C3  Q6 [18]}}