| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.3 This is a standard C3 product rule question with routine follow-through parts. Part (i) requires straightforward application of the product rule and setting the derivative to zero. Parts (ii) and (iii) involve standard techniques (finding tangent equations, integration by parts) but are multi-step. The question is slightly above average due to the integration by parts in (iii) and multiple connected parts, but all techniques are standard C3 material with no novel insights required. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.05o Trigonometric equations: solve in given intervals1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks |
|---|---|
| \(\frac{dy}{dx} = \sin 2x + 2x \cos 2x\) | M1 |
| \(\frac{dy}{dx} = 0\) when \(\sin 2x + 2x \cos 2x = 0\) | A1 |
| \(\frac{\sin 2x + 2x \cos 2x}{\cos 2x} = 0\) | M1 |
| \(\tan 2x + 2x = 0\) | A1 |
| Answer | Marks |
|---|---|
| At \(P\), \(x \sin 2x = 0\) | M1 |
| \(\sin 2x = 0\), \(2x = 0\) or \(\pi\), \(x = \frac{\pi}{2}\) | A1 |
| At \(P\), \(\frac{dy}{dx} = \sin \pi + 2(\frac{\pi}{2}) \cos \pi = -\pi\) | B1 ft |
| Eqn of tangent: \(y - 0 = -\pi(x - \frac{\pi}{2})\) | M1 |
| \(y = -\pi x + \frac{\pi^2}{2}\) | A1 |
| \(2x + 2y = \pi^2\) | M1 A1 |
| Answer | Marks |
|---|---|
| Area \(=\) triangle \(OPQ\) \(-\) area under curve | M1 |
| Triangle \(OPQ = \frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi^2}{2} = \frac{\pi^3}{8}\) | B1 cao |
| Parts: \(u = x\), \(\frac{dv}{dx} = \sin 2x\) | M1 |
| \(\frac{du}{dx} = 1\), \(v = -\frac{1}{2} \cos 2x\) | A1 ft |
| \(\int_0^{\pi/2} x \sin 2x \, dx = \left[-\frac{1}{2} x \cos 2x\right]_0^{\pi/2} + \int_0^{\pi/2} \frac{1}{2} \cos 2x \, dx\) | A1 |
| \(= \left[-\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x\right]_0^{\pi/2}\) | A1 |
| \(= -\frac{1}{4}\pi \cos \pi - \sin \pi - (0 - \cos 0 - \sin 0) = -\frac{1}{4}\pi[0]\) | A1 |
| So shaded area \(= \frac{\pi^3}{8} - \frac{\pi}{4} = \frac{(2\pi^2 - 2\pi)}{8}\) | A1 |
## (i)
$\frac{dy}{dx} = \sin 2x + 2x \cos 2x$ | M1
$\frac{dy}{dx} = 0$ when $\sin 2x + 2x \cos 2x = 0$ | A1
$\frac{\sin 2x + 2x \cos 2x}{\cos 2x} = 0$ | M1
$\tan 2x + 2x = 0$ | A1
**Guidance:**
- $\frac{d}{dx}(\sin 2x) = 2\cos 2x$ soi
- cao, mark final answer
- equating their derivative to zero, provided it has two terms
- must show evidence of division by $\cos 2x$ (can be inferred from $\frac{dy}{dx} = 2x \cos 2x$)
- e.g. $\frac{dy}{dx} = \tan 2x + 2x$ is A0
## (ii)
At $P$, $x \sin 2x = 0$ | M1
$\sin 2x = 0$, $2x = 0$ or $\pi$, $x = \frac{\pi}{2}$ | A1
At $P$, $\frac{dy}{dx} = \sin \pi + 2(\frac{\pi}{2}) \cos \pi = -\pi$ | B1 ft
Eqn of tangent: $y - 0 = -\pi(x - \frac{\pi}{2})$ | M1
$y = -\pi x + \frac{\pi^2}{2}$ | A1
$2x + 2y = \pi^2$ | M1 A1
**Guidance:**
- Finding $x = \frac{\pi}{2}$ using the given line equation is M0
- ft their $\frac{\pi}{2}$ and their derivative
- substituting $0$, their $\frac{\pi}{2}$ and their $-\pi$ into $y - y_1 = m(x - x_1)$ or their $-\pi$ into $y = mx + c$, and then evaluating $c$: $y = (-\pi)x + c$, $0 = (-\pi)(\frac{\pi}{2}) + c$ → $c = \frac{\pi^2}{2}$ → $y = -\pi x + \frac{\pi^2}{2}$ → $2x + 2y = \pi^2$
- NB AG
- can isw inexact answers from $\frac{\pi^2}{2}$
## (iii)
Area $=$ triangle $OPQ$ $-$ area under curve | M1
Triangle $OPQ = \frac{1}{2} \cdot \frac{\pi}{2} \cdot \frac{\pi^2}{2} = \frac{\pi^3}{8}$ | B1 cao
Parts: $u = x$, $\frac{dv}{dx} = \sin 2x$ | M1
$\frac{du}{dx} = 1$, $v = -\frac{1}{2} \cos 2x$ | A1 ft
$\int_0^{\pi/2} x \sin 2x \, dx = \left[-\frac{1}{2} x \cos 2x\right]_0^{\pi/2} + \int_0^{\pi/2} \frac{1}{2} \cos 2x \, dx$ | A1
$= \left[-\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x\right]_0^{\pi/2}$ | A1
$= -\frac{1}{4}\pi \cos \pi - \sin \pi - (0 - \cos 0 - \sin 0) = -\frac{1}{4}\pi[0]$ | A1
So shaded area $= \frac{\pi^3}{8} - \frac{\pi}{4} = \frac{(2\pi^2 - 2\pi)}{8}$ | A1
**Guidance:**
- soi (or area under $PQ$ – area under curve; allow art 3.9)
- condone $v = k \cos 2x$ soi
- ft their $v = -\frac{1}{2} \cos 2x$, ignore limits
- $\left[-\frac{1}{2} x \cos 2x + \frac{1}{4} \sin 2x\right]$ o.e., must be correct at this stage, ignore limits (so dep previous A1)
- NB AG must be from fully correct work
- area under line may be expressed in integral form or using integral: $\frac{1}{2}\int_0^{\pi/2} (2\pi - 2x) \, dx = \frac{1}{2}\left[\pi^2 - \pi x^2\right]_0^{\pi/2}$
---1 Fig. 8 shows a sketch of part of the curve $y = x \sin 2 x$, where $x$ is in radians.\\
The curve crosses the $x$-axis at the point P . The tangent to the curve at P crosses the $y$-axis at Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{35646966-3747-4f1d-bf94-60e9e3130afe-1_706_920_489_606}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$. Hence show that the $x$-coordinates of the turning points of the curve satisfy the equation $\tan 2 x + 2 x = 0$.\\
(ii) Find, in terms of $\pi$, the $x$-coordinate of the point P .
Show that the tangent PQ has equation $2 \pi x + 2 y = \pi ^ { 2 }$.\\
Find the exact coordinates of Q.\\
(iii) Show that the exact value of the area shaded in Fig. 8 is $\frac { 1 } { 8 } \pi \left( \pi ^ { 2 } - 2 \right)$.
\hfill \mbox{\textit{OCR MEI C3 Q1 [18]}}