| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| 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-up parts. Part (i) requires straightforward application of the product rule and setting the derivative to zero. Parts (ii) and (iii) involve finding tangent equations and integration by parts, all using well-practiced techniques with no novel insights required. Slightly above average due to the multi-step nature and integration by parts, but still a typical exam question. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx}=\sin 2x + 2x\cos 2x\) | M1 | \(d/dx(\sin 2x)=2\cos 2x\) soi; can be inferred from \(dy/dx=2x\cos 2x\) |
| A1 | cao, mark final answer; e.g. \(dy/dx=\tan 2x+2x\) is A0 | |
| \(dy/dx=0\) when \(\sin 2x+2x\cos 2x=0\) | M1 | equating their derivative to zero, provided it has two terms |
| \(\Rightarrow \frac{\sin 2x+2x\cos 2x}{\cos 2x}=0\) | ||
| \(\Rightarrow \tan 2x+2x=0\)* | A1 [4] | must show evidence of division by \(\cos 2x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At P, \(x\sin 2x=0\) | M1 | Finding \(x=\pi/2\) using the given line equation is M0 |
| \(\Rightarrow \sin 2x=0\), \(2x=(0,)\pi \Rightarrow x=\pi/2\) | A1 | \(x=\pi/2\) |
| At P, \(dy/dx=\sin\pi+2(\pi/2)\cos\pi=-\pi\) | B1ft | ft their \(\pi/2\) and their derivative |
| Eqn of tangent: \(y-0=-\pi(x-\pi/2)\) | M1 | substituting \(0\), their \(\pi/2\) and their \(-\pi\) into \(y-y_1=m(x-x_1)\); or their \(-\pi\) into \(y=mx+c\) then evaluating \(c\): \(y=(-\pi)x+c\), \(0=(-\pi)(\pi/2)+c\) M1 \(\Rightarrow c=\pi^2/2\) |
| \(\Rightarrow y=-\pi x+\pi^2/2\) | ||
| \(\Rightarrow 2\pi x+2y=\pi^2\)* | A1 NB AG | |
| When \(x=0\), \(y=\pi^2/2\), so Q is \((0,\,\pi^2/2)\) | M1A1 [7] | can isw inexact answers from \(\pi^2/2\); \(\Rightarrow y=-\pi x+\pi^2/2\Rightarrow 2\pi x+2y=\pi^2\)* A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(=\) triangle \(OPQ\) – area under curve | M1 | soi (or area under \(PQ\) – area under curve); area under line may be expressed in integral form |
| Triangle \(OPQ=\tfrac{1}{2}\times\frac{\pi}{2}\times\frac{\pi^2}{2}\left[=\frac{\pi^3}{8}\right]\) | B1cao | allow art 3.9; or using integral: \(\int_0^{\pi/2}\!\left(\tfrac{1}{2}\pi^2-\pi x\right)dx=\left[\tfrac{1}{2}\pi^2 x-\tfrac{1}{2}\pi x^2\right]_0^{\pi/2}=\frac{\pi^3}{4}-\frac{\pi^3}{8}\left[=\frac{\pi^3}{8}\right]\) |
| Parts: \(u=x\), \(dv/dx=\sin 2x\); \(du/dx=1\), \(v=-\tfrac{1}{2}\cos 2x\) | M1 | condone \(v=k\cos 2x\) soi; \(v\) can be inferred from their '\(uv\)' |
| \(\int_0^{\pi/2}x\sin 2x\,dx=\left[-\tfrac{1}{2}x\cos 2x\right]_0^{\pi/2}-\int_0^{\pi/2}-\tfrac{1}{2}\cos 2x\,dx\) | A1ft | ft their \(v=-\tfrac{1}{2}\cos 2x\), ignore limits |
| \(=\left[-\tfrac{1}{2}x\cos 2x+\tfrac{1}{4}\sin 2x\right]_0^{\pi/2}\) | A1 | \([-\tfrac{1}{2}x\cos 2x+\tfrac{1}{4}\sin 2x]\) o.e., must be correct at this stage, ignore limits |
| \(=-\tfrac{1}{4}\pi\cos\pi+\tfrac{1}{4}\sin\pi-(-0\cos 0+\tfrac{1}{4}\sin 0)=\tfrac{1}{4}\pi[-0]\) | A1cao | (so dep previous A1) |
| So shaded area \(=\pi^3/8-\pi/4=\pi(\pi^2-2)/8\)* | A1 NB AG [7] | must be from fully correct work |
## Question 8(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx}=\sin 2x + 2x\cos 2x$ | M1 | $d/dx(\sin 2x)=2\cos 2x$ soi; can be inferred from $dy/dx=2x\cos 2x$ |
| | A1 | cao, mark final answer; e.g. $dy/dx=\tan 2x+2x$ is A0 |
| $dy/dx=0$ when $\sin 2x+2x\cos 2x=0$ | M1 | equating their derivative to zero, **provided it has two terms** |
| $\Rightarrow \frac{\sin 2x+2x\cos 2x}{\cos 2x}=0$ | | |
| $\Rightarrow \tan 2x+2x=0$* | A1 **[4]** | must show evidence of division by $\cos 2x$ |
---
## Question 8(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At P, $x\sin 2x=0$ | M1 | Finding $x=\pi/2$ using the given line equation is M0 |
| $\Rightarrow \sin 2x=0$, $2x=(0,)\pi \Rightarrow x=\pi/2$ | A1 | $x=\pi/2$ |
| At P, $dy/dx=\sin\pi+2(\pi/2)\cos\pi=-\pi$ | B1ft | ft their $\pi/2$ and their derivative |
| Eqn of tangent: $y-0=-\pi(x-\pi/2)$ | M1 | substituting $0$, their $\pi/2$ and their $-\pi$ into $y-y_1=m(x-x_1)$; or their $-\pi$ into $y=mx+c$ then evaluating $c$: $y=(-\pi)x+c$, $0=(-\pi)(\pi/2)+c$ M1 $\Rightarrow c=\pi^2/2$ |
| $\Rightarrow y=-\pi x+\pi^2/2$ | | |
| $\Rightarrow 2\pi x+2y=\pi^2$* | A1 **NB AG** | |
| When $x=0$, $y=\pi^2/2$, so Q is $(0,\,\pi^2/2)$ | M1A1 **[7]** | can isw inexact answers from $\pi^2/2$; $\Rightarrow y=-\pi x+\pi^2/2\Rightarrow 2\pi x+2y=\pi^2$* A1 |
---
## Question 8(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $=$ triangle $OPQ$ – area under curve | M1 | soi (or area under $PQ$ – area under curve); area under line may be expressed in integral form |
| Triangle $OPQ=\tfrac{1}{2}\times\frac{\pi}{2}\times\frac{\pi^2}{2}\left[=\frac{\pi^3}{8}\right]$ | B1cao | allow art 3.9; or using integral: $\int_0^{\pi/2}\!\left(\tfrac{1}{2}\pi^2-\pi x\right)dx=\left[\tfrac{1}{2}\pi^2 x-\tfrac{1}{2}\pi x^2\right]_0^{\pi/2}=\frac{\pi^3}{4}-\frac{\pi^3}{8}\left[=\frac{\pi^3}{8}\right]$ |
| Parts: $u=x$, $dv/dx=\sin 2x$; $du/dx=1$, $v=-\tfrac{1}{2}\cos 2x$ | M1 | condone $v=k\cos 2x$ soi; $v$ can be inferred from their '$uv$' |
| $\int_0^{\pi/2}x\sin 2x\,dx=\left[-\tfrac{1}{2}x\cos 2x\right]_0^{\pi/2}-\int_0^{\pi/2}-\tfrac{1}{2}\cos 2x\,dx$ | A1ft | ft their $v=-\tfrac{1}{2}\cos 2x$, ignore limits |
| $=\left[-\tfrac{1}{2}x\cos 2x+\tfrac{1}{4}\sin 2x\right]_0^{\pi/2}$ | A1 | $[-\tfrac{1}{2}x\cos 2x+\tfrac{1}{4}\sin 2x]$ o.e., must be correct at this stage, ignore limits |
| $=-\tfrac{1}{4}\pi\cos\pi+\tfrac{1}{4}\sin\pi-(-0\cos 0+\tfrac{1}{4}\sin 0)=\tfrac{1}{4}\pi[-0]$ | A1cao | (so dep previous A1) |
| So shaded area $=\pi^3/8-\pi/4=\pi(\pi^2-2)/8$* | A1 **NB AG** **[7]** | must be from fully correct work |
---8 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]{7b77c646-2bc5-4166-b22e-3c1229abd722-4_712_923_463_571}
\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 2012 Q8 [18]}}