OCR MEI C3 2005 June — Question 8 17 marks

Exam BoardOCR MEI
ModuleC3 (Core Mathematics 3)
Year2005
SessionJune
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIntegration by Parts
TypeNormal/tangent then area with parts
DifficultyStandard +0.3 This is a structured multi-part question requiring standard techniques: solving sin equation, substitution, product rule differentiation, and integration by parts. While it involves several steps, each part is routine for C3 level with clear guidance through the problem structure. The integration by parts is straightforward with u=x, and the question explicitly tells students what to show, removing problem-solving challenge.
Spec1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts
8 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-3_421_789_1748_610} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the \(x\)-coordinate of P .
  2. Show that Q lies on the line \(y = x\).
  3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
  4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).
AnswerMarks Guidance
(i) At P, \(x \sin 3x = 0 \Rightarrow \sin 3x = 0 \Rightarrow 3x = \pi \Rightarrow x = \pi/3\)M1, A1, A1cao [3] \(x \sin 3x = 0\); \(3x = \pi\) or 180; \(x = \pi/3\) or 1.05 or better
(ii) When \(x = \pi/6, x \sin 3x = \frac{\pi}{6} \sin \frac{\pi}{2} = \frac{\pi}{6}\)
AnswerMarks Guidance
\(\Rightarrow Q(\pi/6, \pi/6)\) lies on line \(y = x\)E1 [1] \(y = \frac{\pi}{6}\) or \(x \sin 3x = x \Rightarrow \sin 3x = 1\) etc; Must conclude in radians, and be exact
(iii) \(y = x \sin 3x \Rightarrow \frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x\)
At Q: \(\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 1\)
\(=\) gradient of \(y = x\) (www)
AnswerMarks Guidance
So line touches curve at this pointB1, M1, A1cao, M1, E1 [6] d/dx\((sin 3x) = 3\cos 3x\); Product rule consistent with their derivs; \(3\cos 3x + \sin 3x\); substituting \(x = \pi/6\) into their derivative \(= 1\) ft dep 1st M1; \(=\) gradient of \(y = x\) (www)
(iv) Area under curve \(= \int_0^{\pi/6} x \sin 3x \, dx\)
Integrating by parts: \(u = x, dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\)
\(\int_0^{\pi/6} x \sin 3x \, dx = [-\frac{1}{3}x \cos 3x]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{3}\cos 3x \, dx = -\frac{1}{3} \cdot \frac{\pi}{6} \cos \frac{\pi}{2} + \frac{1}{3} \cdot 0 \cos 0 + [\frac{1}{9}\sin 3x]_0^{\pi/6}\)
\(= \frac{1}{9}\)
Area under line \(= \frac{1}{2} \cdot \frac{\pi}{6} \cdot \frac{\pi}{6} = \frac{\pi^2}{72}\)
AnswerMarks Guidance
So area required \(= \frac{\pi^2}{72} - \frac{1}{9} = \frac{\pi^2 - 8*}{72}\)M1, A1cao, A1ft, M1, A1, B1, E1 [7] Parts with \(u = x dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x\) [condone no negative]; \(...+ [\frac{1}{9}\sin 3x]_0^{\pi/6}\); substituting (correct) limits; \(\frac{1}{9}\) www; \(\frac{\pi^2}{72}\); www 
**(i)** At P, $x \sin 3x = 0 \Rightarrow \sin 3x = 0 \Rightarrow 3x = \pi \Rightarrow x = \pi/3$ | M1, A1, A1cao [3] | $x \sin 3x = 0$; $3x = \pi$ or 180; $x = \pi/3$ or 1.05 or better

**(ii)** When $x = \pi/6, x \sin 3x = \frac{\pi}{6} \sin \frac{\pi}{2} = \frac{\pi}{6}$

$\Rightarrow Q(\pi/6, \pi/6)$ lies on line $y = x$ | E1 [1] | $y = \frac{\pi}{6}$ or $x \sin 3x = x \Rightarrow \sin 3x = 1$ etc; Must conclude in radians, and be exact

**(iii)** $y = x \sin 3x \Rightarrow \frac{dy}{dx} = x \cdot 3\cos 3x + \sin 3x$

At Q: $\frac{dy}{dx} = \frac{\pi}{6} \cdot 3\cos \frac{\pi}{2} + \sin \frac{\pi}{2} = 1$

$=$ gradient of $y = x$ (www)

So line touches curve at this point | B1, M1, A1cao, M1, E1 [6] | d/dx$(sin 3x) = 3\cos 3x$; Product rule consistent with their derivs; $3\cos 3x + \sin 3x$; substituting $x = \pi/6$ into their derivative $= 1$ ft dep 1st M1; $=$ gradient of $y = x$ (www)

**(iv)** Area under curve $= \int_0^{\pi/6} x \sin 3x \, dx$

Integrating by parts: $u = x, dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$

$\int_0^{\pi/6} x \sin 3x \, dx = [-\frac{1}{3}x \cos 3x]_0^{\pi/6} + \int_0^{\pi/6} \frac{1}{3}\cos 3x \, dx = -\frac{1}{3} \cdot \frac{\pi}{6} \cos \frac{\pi}{2} + \frac{1}{3} \cdot 0 \cos 0 + [\frac{1}{9}\sin 3x]_0^{\pi/6}$

$= \frac{1}{9}$

Area under line $= \frac{1}{2} \cdot \frac{\pi}{6} \cdot \frac{\pi}{6} = \frac{\pi^2}{72}$

So area required $= \frac{\pi^2}{72} - \frac{1}{9} = \frac{\pi^2 - 8*}{72}$ | M1, A1cao, A1ft, M1, A1, B1, E1 [7] | Parts with $u = x dv/dx = \sin 3x \Rightarrow v = -\frac{1}{3}\cos 3x$ [condone no negative]; $...+ [\frac{1}{9}\sin 3x]_0^{\pi/6}$; substituting (correct) limits; $\frac{1}{9}$ www; $\frac{\pi^2}{72}$; www
8 Fig. 8 shows part of the curve $y = x \sin 3 x$. It crosses the $x$-axis at P . The point on the curve with $x$-coordinate $\frac { 1 } { 6 } \pi$ is Q .

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3efea8db-9fa1-47a8-89b8-e4888f87a313-3_421_789_1748_610}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}

(i) Find the $x$-coordinate of P .\\
(ii) Show that Q lies on the line $y = x$.\\
(iii) Differentiate $x \sin 3 x$. Hence prove that the line $y = x$ touches the curve at Q .\\
(iv) Show that the area of the region bounded by the curve and the line $y = x$ is $\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)$.

\hfill \mbox{\textit{OCR MEI C3 2005 Q8 [17]}}
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