| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - direct evaluation |
| Difficulty | Standard +0.8 This is a substantial multi-part C3 question requiring quotient rule differentiation, integration using a given result, inverse function derivation, and domain considerations. While each individual technique is standard, the combination of parts (especially proving the integration result and finding the inverse function algebraically) elevates this above average difficulty for A-level. |
| Spec | 1.02v Inverse and composite functions: graphs and conditions for existence1.05a Sine, cosine, tangent: definitions for all arguments1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1/(1+\cos\frac{\pi}{3}) = 2/3\) | B1 | or 0.67 or better |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(f'(x) = -1(1+\cos x)^{-2} \cdot -\sin x\) | M1 | chain rule or quotient rule |
| \(= \dfrac{\sin x}{(1+\cos x)^2}\) | B1 | \(\frac{d}{dx}(\cos x) = -\sin x\) soi |
| A1 | correct expression | |
| When \(x = \pi/3\), \(f'(x) = \dfrac{\sin(\pi/3)}{(1+\cos(\pi/3))^2}\) | M1 | substituting \(x = \pi/3\) |
| \(= \dfrac{\sqrt{3}/2}{(1\frac{1}{2})^2} = \dfrac{\sqrt{3}}{2}\times\dfrac{4}{9} = \dfrac{2\sqrt{3}}{9}\) | A1 | oe or 0.38 or better |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| deriv \(= \dfrac{(1+\cos x)\cos x - \sin x\cdot(-\sin x)}{(1+\cos x)^2}\) | M1 | Quotient or product rule — condone \(uv' - u'v\) for M1 |
| \(= \dfrac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}\) | A1 | correct expression |
| \(= \dfrac{\cos x + 1}{(1+\cos x)^2}\) | M1dep | \(\cos^2 x + \sin^2 x = 1\) used dep M1 |
| \(= \dfrac{1}{1+\cos x}\ *\) | E1 | www |
| Area \(= \displaystyle\int_0^{\pi/3} \frac{1}{1+\cos x}\,dx\) | ||
| \(= \left[\dfrac{\sin x}{1+\cos x}\right]_0^{\pi/3}\) | B1 | |
| \(= \dfrac{\sin\frac{\pi}{3}}{1+\cos\frac{\pi}{3}} - 0\) | M1 | substituting limits |
| \(= \dfrac{\sqrt{3}}{2}\times\dfrac{2}{3} = \dfrac{\sqrt{3}}{3}\) | A1cao | or \(1/\sqrt{3}\) — must be exact |
| [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(y = 1/(1+\cos x),\ x \leftrightarrow y\) | ||
| \(x = 1/(1+\cos y)\) | M1 | attempt to invert equation |
| \(\Rightarrow 1 + \cos y = 1/x\) | A1 | |
| \(\Rightarrow \cos y = 1/x - 1\) | ||
| \(\Rightarrow y = \arccos(1/x - 1)\ *\) | E1 | www |
| Domain is \(\frac{1}{2} \le x \le 1\) | B1 | |
| [graph: reasonable reflection in \(y=x\)] | B1 | reasonable reflection in \(y=x\) |
| [5] |
# Question 4:
## Part (i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1/(1+\cos\frac{\pi}{3}) = 2/3$ | B1 | or 0.67 or better |
| **[1]** | | |
---
## Part (ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $f'(x) = -1(1+\cos x)^{-2} \cdot -\sin x$ | M1 | chain rule or quotient rule |
| $= \dfrac{\sin x}{(1+\cos x)^2}$ | B1 | $\frac{d}{dx}(\cos x) = -\sin x$ soi |
| | A1 | correct expression |
| When $x = \pi/3$, $f'(x) = \dfrac{\sin(\pi/3)}{(1+\cos(\pi/3))^2}$ | M1 | substituting $x = \pi/3$ |
| $= \dfrac{\sqrt{3}/2}{(1\frac{1}{2})^2} = \dfrac{\sqrt{3}}{2}\times\dfrac{4}{9} = \dfrac{2\sqrt{3}}{9}$ | A1 | oe or 0.38 or better |
| **[5]** | | |
---
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| deriv $= \dfrac{(1+\cos x)\cos x - \sin x\cdot(-\sin x)}{(1+\cos x)^2}$ | M1 | Quotient or product rule — condone $uv' - u'v$ for M1 |
| $= \dfrac{\cos x + \cos^2 x + \sin^2 x}{(1+\cos x)^2}$ | A1 | correct expression |
| $= \dfrac{\cos x + 1}{(1+\cos x)^2}$ | M1dep | $\cos^2 x + \sin^2 x = 1$ used dep M1 |
| $= \dfrac{1}{1+\cos x}\ *$ | E1 | www |
| Area $= \displaystyle\int_0^{\pi/3} \frac{1}{1+\cos x}\,dx$ | | |
| $= \left[\dfrac{\sin x}{1+\cos x}\right]_0^{\pi/3}$ | B1 | |
| $= \dfrac{\sin\frac{\pi}{3}}{1+\cos\frac{\pi}{3}} - 0$ | M1 | substituting limits |
| $= \dfrac{\sqrt{3}}{2}\times\dfrac{2}{3} = \dfrac{\sqrt{3}}{3}$ | A1cao | or $1/\sqrt{3}$ — must be exact |
| **[7]** | | |
---
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $y = 1/(1+\cos x),\ x \leftrightarrow y$ | | |
| $x = 1/(1+\cos y)$ | M1 | attempt to invert equation |
| $\Rightarrow 1 + \cos y = 1/x$ | A1 | |
| $\Rightarrow \cos y = 1/x - 1$ | | |
| $\Rightarrow y = \arccos(1/x - 1)\ *$ | E1 | www |
| Domain is $\frac{1}{2} \le x \le 1$ | B1 | |
| [graph: reasonable reflection in $y=x$] | B1 | reasonable reflection in $y=x$ |
| **[5]** | | |4 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]{1d12cd0d-07b0-429c-ad3b-e3bccb0fae18-4_820_815_551_715}
\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 Q4 [18]}}