| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2010 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.3 This is a straightforward multi-part C3 question requiring standard product rule differentiation, solving cos(3x)=0 for x-intercepts, and integration by parts. All techniques are routine for this level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At P, \(x\cos 3x = 0 \Rightarrow \cos 3x = 0\) | M1 | or verification |
| \(3x = \pi/2, 3\pi/2\) | M1 | \(3x = \pi/2, (3\pi/2\ldots)\) |
| \(x = \pi/6, \pi/2\) | A1, A1 | dep both Ms; condone degrees |
| So P is \((\pi/6, 0)\) and Q is \((\pi/2, 0)\) | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -3x\sin 3x + \cos 3x\) | M1, B1, A1 | Product rule; \(d/dx(\cos 3x) = -3\sin 3x\); cao |
| At P, \(\frac{dy}{dx} = -\frac{\pi}{2}\sin\frac{\pi}{2} + \cos\frac{\pi}{2} = -\frac{\pi}{2}\) | M1, A1cao | substituting their \(-\pi/6\) (must be rads); \(-\pi/2\) |
| At TPs \(\frac{dy}{dx} = -3x\sin 3x + \cos 3x = 0\) | M1 | \(dy/dx=0\) and \(\sin 3x/\cos 3x = \tan 3x\) used |
| \(\Rightarrow \cos 3x = 3x\sin 3x \Rightarrow 1 = 3x\tan 3x\) | ||
| \(\Rightarrow x\tan 3x = \frac{1}{3}\) * | E1 [7] | www |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A = \int_0^{\pi/6} x\cos 3x\,dx\) | B1 | Correct integral and limits (soi) – ft their P, must be in radians |
| Parts with \(u=x\), \(dv/dx = \cos 3x\), \(du/dx=1\), \(v = \frac{1}{3}\sin 3x\) | M1 | |
| \(A = \left[\frac{1}{3}x\sin 3x\right]_0^{\pi/6} - \int_0^{\pi/6}\frac{1}{3}\sin 3x\,dx\) | A1 | can be without limits |
| \(= \left[\frac{1}{3}x\sin 3x + \frac{1}{9}\cos 3x\right]_0^{\pi/6}\) | A1 | dep previous A1 |
| \(= \frac{\pi}{18} - \frac{1}{9}\) | M1dep, A1cao [6] | substituting correct limits, dep 1st M1: ft their P in radians; o.e. but must be exact |
# Question 8:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At P, $x\cos 3x = 0 \Rightarrow \cos 3x = 0$ | M1 | or verification |
| $3x = \pi/2, 3\pi/2$ | M1 | $3x = \pi/2, (3\pi/2\ldots)$ |
| $x = \pi/6, \pi/2$ | A1, A1 | dep both Ms; condone degrees |
| So P is $(\pi/6, 0)$ and Q is $(\pi/2, 0)$ | [4] | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -3x\sin 3x + \cos 3x$ | M1, B1, A1 | Product rule; $d/dx(\cos 3x) = -3\sin 3x$; cao |
| At P, $\frac{dy}{dx} = -\frac{\pi}{2}\sin\frac{\pi}{2} + \cos\frac{\pi}{2} = -\frac{\pi}{2}$ | M1, A1cao | substituting their $-\pi/6$ (must be rads); $-\pi/2$ |
| At TPs $\frac{dy}{dx} = -3x\sin 3x + \cos 3x = 0$ | M1 | $dy/dx=0$ and $\sin 3x/\cos 3x = \tan 3x$ used |
| $\Rightarrow \cos 3x = 3x\sin 3x \Rightarrow 1 = 3x\tan 3x$ | | |
| $\Rightarrow x\tan 3x = \frac{1}{3}$ * | E1 [7] | www |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A = \int_0^{\pi/6} x\cos 3x\,dx$ | B1 | Correct integral and limits (soi) – ft their P, must be in radians |
| Parts with $u=x$, $dv/dx = \cos 3x$, $du/dx=1$, $v = \frac{1}{3}\sin 3x$ | M1 | |
| $A = \left[\frac{1}{3}x\sin 3x\right]_0^{\pi/6} - \int_0^{\pi/6}\frac{1}{3}\sin 3x\,dx$ | A1 | can be without limits |
| $= \left[\frac{1}{3}x\sin 3x + \frac{1}{9}\cos 3x\right]_0^{\pi/6}$ | A1 | dep previous A1 |
| $= \frac{\pi}{18} - \frac{1}{9}$ | M1dep, A1cao [6] | substituting correct limits, dep 1st M1: ft their P in radians; o.e. but must be exact |
---8 Fig. 8 shows part of the curve $y = x \cos 3 x$.\\
The curve crosses the $x$-axis at $\mathrm { O } , \mathrm { P }$ and Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b3e20ee-457c-46be-b2e5-12573bee2fbf-3_551_1189_925_479}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the exact coordinates of P and Q .\\
(ii) Find the exact gradient of the curve at the point P .
Show also that the turning points of the curve occur when $x \tan 3 x = \frac { 1 } { 3 }$.\\
(iii) Find the area of the region enclosed by the curve and the $x$-axis between O and P , giving your answer in exact form.
\hfill \mbox{\textit{OCR MEI C3 2010 Q8 [17]}}