| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of composite quotient/product |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: evaluating a function, applying quotient rule, verifying a derivative for integration, and finding an inverse function. Each part is routine with clear guidance, requiring no novel insight—slightly easier than the typical average A-level question due to its scaffolded nature. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 1/(1+\cos\frac{\pi}{3}) = \frac{2}{3}\) | B1 | or 0.67 or better |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f'(x) = -1(1+\cos x)^{-2} \cdot -\sin x\) | M1 | chain rule or quotient rule |
| \(= \frac{\sin x}{(1+\cos x)^2}\) | B1 | \(\frac{d}{dx}(\cos x) = -\sin x\) soi |
| A1 | correct expression | |
| When \(x = \frac{\pi}{3}\), \(f'(x) = \frac{\sin(\pi/3)}{(1+\cos(\pi/3))^2}\) | M1 | substituting \(x = \frac{\pi}{3}\) |
| \(= \frac{\sqrt{3}/2}{(1\frac{1}{2})^2} = \frac{\sqrt{3}}{2} \times \frac{4}{9} = \frac{2\sqrt{3}}{9}\) | A1 | oe or 0.38 or better. (0.385, 0.3849) |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| deriv \(= \frac{(1+\cos x)\cos x - \sin x \cdot(-\sin x)}{(1+\cos x)^2}\) | M1 | Quotient or product rule – condone \(uv' - u'v\) for M1 |
| \(= \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}\) | A1 | correct expression |
| \(= \frac{\cos x + 1}{(1+\cos x)^2}\) | M1dep | \(\cos^2 x + \sin^2 x = 1\) used dep M1 |
| \(= \frac{1}{1+\cos x}\) * | E1 | www |
| Area \(= \int_0^{\pi/3} \frac{1}{1+\cos x}\,dx\) | ||
| \(= \left[\frac{\sin x}{1+\cos x}\right]_0^{\pi/3}\) | B1 | |
| \(= \frac{\sin\frac{\pi}{3}}{1+\cos\frac{\pi}{3}} - 0\) | M1 | substituting limits |
| \(= \frac{\sqrt{3}}{2} \cdot \frac{2}{3} = \frac{\sqrt{3}}{3}\) | A1 cao | or \(\frac{1}{\sqrt{3}}\) – must be exact |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = 1/(1+\cos x)\), \(x \leftrightarrow y\), \(x = 1/(1+\cos y)\) | M1 | attempt to invert equation |
| \(1 + \cos y = \frac{1}{x}\) | A1 | |
| \(\cos y = \frac{1}{x} - 1\) | E1 | www |
| \(y = \arccos\!\left(\frac{1}{x}-1\right)\) * | ||
| Domain is \(\frac{1}{2} \leq x \leq 1\) | B1 | |
| Reasonable reflection in \(y = x\) | B1 | |
| [5] |
# Question 8:
## Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 1/(1+\cos\frac{\pi}{3}) = \frac{2}{3}$ | B1 | or 0.67 or better |
| | [1] | |
## Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f'(x) = -1(1+\cos x)^{-2} \cdot -\sin x$ | M1 | chain rule or quotient rule |
| $= \frac{\sin x}{(1+\cos x)^2}$ | B1 | $\frac{d}{dx}(\cos x) = -\sin x$ soi |
| | A1 | correct expression |
| When $x = \frac{\pi}{3}$, $f'(x) = \frac{\sin(\pi/3)}{(1+\cos(\pi/3))^2}$ | M1 | substituting $x = \frac{\pi}{3}$ |
| $= \frac{\sqrt{3}/2}{(1\frac{1}{2})^2} = \frac{\sqrt{3}}{2} \times \frac{4}{9} = \frac{2\sqrt{3}}{9}$ | A1 | oe or 0.38 or better. (0.385, 0.3849) |
| | [5] | |
## Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| deriv $= \frac{(1+\cos x)\cos x - \sin x \cdot(-\sin x)}{(1+\cos x)^2}$ | M1 | Quotient or product rule – condone $uv' - u'v$ for M1 |
| $= \frac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}$ | A1 | correct expression |
| $= \frac{\cos x + 1}{(1+\cos x)^2}$ | M1dep | $\cos^2 x + \sin^2 x = 1$ used dep M1 |
| $= \frac{1}{1+\cos x}$ * | E1 | www |
| Area $= \int_0^{\pi/3} \frac{1}{1+\cos x}\,dx$ | | |
| $= \left[\frac{\sin x}{1+\cos x}\right]_0^{\pi/3}$ | B1 | |
| $= \frac{\sin\frac{\pi}{3}}{1+\cos\frac{\pi}{3}} - 0$ | M1 | substituting limits |
| $= \frac{\sqrt{3}}{2} \cdot \frac{2}{3} = \frac{\sqrt{3}}{3}$ | A1 cao | or $\frac{1}{\sqrt{3}}$ – must be exact |
| | [7] | |
## Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = 1/(1+\cos x)$, $x \leftrightarrow y$, $x = 1/(1+\cos y)$ | M1 | attempt to invert equation |
| $1 + \cos y = \frac{1}{x}$ | A1 | |
| $\cos y = \frac{1}{x} - 1$ | E1 | www |
| $y = \arccos\!\left(\frac{1}{x}-1\right)$ * | | |
| Domain is $\frac{1}{2} \leq x \leq 1$ | B1 | |
| Reasonable reflection in $y = x$ | B1 | |
| | [5] | |
---8 Fig. 8 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 1 } { 1 + \cos x }$, for $0 \leqslant x \leqslant \frac { 1 } { 2 } \pi$.\\
P is the point on the curve with $x$-coordinate $\frac { 1 } { 3 } \pi$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8feffafd-4eba-4968-b4d2-88fa364d6170-3_825_816_571_662}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the $y$-coordinate of P .\\
(ii) Find $\mathrm { f } ^ { \prime } ( x )$. Hence find the gradient of the curve at the point P .\\
(iii) Show that the derivative of $\frac { \sin x } { 1 + \cos x }$ is $\frac { 1 } { 1 + \cos x }$. Hence find the exact area of the region enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = \frac { 1 } { 3 } \pi$.\\
(iv) Show that $\mathrm { f } ^ { - 1 } ( x ) = \arccos \left( \frac { 1 } { x } - 1 \right)$. State the domain of this inverse function, and add a sketch of $y = \mathrm { f } ^ { - 1 } ( x )$ to a copy of Fig. 8.
\hfill \mbox{\textit{OCR MEI C3 2008 Q8 [18]}}