| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring product rule differentiation, solving transcendental equations, and integration by parts. Part (ii) requires showing a non-trivial result involving tan, and part (iii) needs integration by parts of x cos 3x. More demanding than a standard C3 question due to the proof element and multiple techniques required. |
| Spec | 1.05o Trigonometric equations: solve in given intervals1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| At P, \(x\cos 3x = 0 \Rightarrow \cos 3x = 0\) | M1 | Or verification |
| \(3x = \pi/2,\; 3\pi/2 \Rightarrow x = \pi/6,\; \pi/2\) | M1, A1, A1 [4] | \(3x = \pi/2,\,(3\pi/2\ldots)\); dep both Ms; condone degrees |
| So P is \((\pi/6,\,0)\) and Q is \((\pi/2,\,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{dy}{dx} = -3x\sin 3x + \cos 3x\) | M1, B1, A1 | Product rule; \(\frac{d}{dx}(\cos 3x) = -3\sin 3x\); cao |
| At P, \(\dfrac{dy}{dx} = -\dfrac{\pi}{2}\sin\dfrac{\pi}{2} + \cos\dfrac{\pi}{2} = -\dfrac{\pi}{2}\) | M1, A1cao | Substituting their \(\pi/6\) (must be rads); \(-\pi/2\) |
| At TPs \(\dfrac{dy}{dx} = -3x\sin 3x + \cos 3x = 0 \Rightarrow \cos 3x = 3x\sin 3x\) | M1 | \(dy/dx = 0\) and \(\sin 3x / \cos 3x = \tan 3x\) used |
| \(\Rightarrow 1 = 3x\tan 3x \Rightarrow x\tan 3x = \frac{1}{3}\) | E1 [7] | www |
| Answer | Marks | Guidance |
|---|---|---|
| \(A = \displaystyle\int_0^{\pi/6} x\cos 3x\,dx\) | B1 | Correct integral and limits (soi); must be in radians |
| Parts with \(u=x\), \(dv/dx = \cos 3x\), \(du/dx=1\), \(v = \frac{1}{3}\sin 3x\) | M1 | |
| \(A = \left[\dfrac{1}{3}x\sin 3x\right]_0^{\pi/6} - \displaystyle\int_0^{\pi/6}\dfrac{1}{3}\sin 3x\,dx\) | A1 | Can be without limits |
| \(= \left[\dfrac{1}{3}x\sin 3x + \dfrac{1}{9}\cos 3x\right]_0^{\pi/6}\) | A1 | Dep previous A1 |
| \(= \dfrac{\pi}{18} - \dfrac{1}{9}\) | M1dep, A1cao [6] | Substituting correct limits; o.e. but must be exact |
## Question 6:
**(i)**
| At P, $x\cos 3x = 0 \Rightarrow \cos 3x = 0$ | M1 | Or verification |
|---|---|---|
| $3x = \pi/2,\; 3\pi/2 \Rightarrow x = \pi/6,\; \pi/2$ | M1, A1, A1 [4] | $3x = \pi/2,\,(3\pi/2\ldots)$; dep both Ms; condone degrees |
| So P is $(\pi/6,\,0)$ and Q is $(\pi/2,\,0)$ | | |
**(ii)**
| $\dfrac{dy}{dx} = -3x\sin 3x + \cos 3x$ | M1, B1, A1 | Product rule; $\frac{d}{dx}(\cos 3x) = -3\sin 3x$; cao |
|---|---|---|
| At P, $\dfrac{dy}{dx} = -\dfrac{\pi}{2}\sin\dfrac{\pi}{2} + \cos\dfrac{\pi}{2} = -\dfrac{\pi}{2}$ | M1, A1cao | Substituting their $\pi/6$ (must be rads); $-\pi/2$ |
| At TPs $\dfrac{dy}{dx} = -3x\sin 3x + \cos 3x = 0 \Rightarrow \cos 3x = 3x\sin 3x$ | M1 | $dy/dx = 0$ and $\sin 3x / \cos 3x = \tan 3x$ used |
| $\Rightarrow 1 = 3x\tan 3x \Rightarrow x\tan 3x = \frac{1}{3}$ | E1 [7] | www |
**(iii)**
| $A = \displaystyle\int_0^{\pi/6} x\cos 3x\,dx$ | B1 | Correct integral and limits (soi); must be in radians |
|---|---|---|
| Parts with $u=x$, $dv/dx = \cos 3x$, $du/dx=1$, $v = \frac{1}{3}\sin 3x$ | M1 | |
| $A = \left[\dfrac{1}{3}x\sin 3x\right]_0^{\pi/6} - \displaystyle\int_0^{\pi/6}\dfrac{1}{3}\sin 3x\,dx$ | A1 | Can be without limits |
| $= \left[\dfrac{1}{3}x\sin 3x + \dfrac{1}{9}\cos 3x\right]_0^{\pi/6}$ | A1 | Dep previous A1 |
| $= \dfrac{\pi}{18} - \dfrac{1}{9}$ | M1dep, A1cao [6] | Substituting correct limits; o.e. but must be exact |
---6 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]{11877196-83d9-4283-9eef-e617bea50c63-3_553_1178_622_529}
\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 Q6 [17]}}