| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 20 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.3 This is a multi-part question testing standard product rule differentiation, basic function properties (odd/even), and integration by parts. While it has many parts, each individual step is routine C3 material: applying product rule to x·cos(2x), showing a function is odd, finding turning points by setting derivative to zero, and standard integration by parts. The algebraic manipulation is straightforward and no novel insight is required. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| P is where \(x\cos 2x = 0\), \(x \neq 0 \Rightarrow \cos 2x = 0\) | M1 | Setting \(y=0\) for positive \(x\) |
| \(2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}\) | A1 | Correct \(x\) |
| \(P = \left(\frac{\pi}{4}, 0\right)\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(-x) = (-x)\cos(-2x) = -x\cos 2x = -f(x)\) | M1 | Correct algebraic manipulation |
| Hence odd function | A1 | Stated |
| Graphically: rotational symmetry of order 2 about the origin | B1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \cos 2x - 2x\sin 2x\) | M1 | Product rule |
| — | A1 | [2] Correct derivative |
| Answer | Marks | Guidance |
|---|---|---|
| At turning point: \(\cos 2x - 2x\sin 2x = 0\) | M1 | Setting derivative to zero |
| \(\cos 2x = 2x\sin 2x \Rightarrow x\tan 2x = \frac{1}{2}\) (dividing by \(\cos 2x\)) | A1 | [2] Shown |
| Answer | Marks | Guidance |
|---|---|---|
| At \(x=0\): \(\frac{dy}{dx} = \cos 0 = 1\), so gradient \(= 1\) | B1 | Correct gradient |
| \(\frac{d^2y}{dx^2} = -2\sin 2x - 2\sin 2x - 4x\cos 2x = -4\sin 2x - 4x\cos 2x\) | M1 | Differentiating again |
| At \(x=0\): \(-4\sin 0 - 0 = 0\) | A1 | Shown |
| — | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^{\frac{\pi}{4}} x\cos 2x\,dx\): integrate by parts, \(u=x\), \(\frac{dv}{dx}=\cos 2x\) | M1 | By parts |
| \(= \left[\frac{x\sin 2x}{2}\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\frac{\sin 2x}{2}\,dx\) | A1 | Correct first step |
| \(= \frac{\pi}{8} - \left[-\frac{\cos 2x}{4}\right]_0^{\frac{\pi}{4}}\) | A1 | Correct integration |
| \(= \frac{\pi}{8} - \left(\frac{1}{4} - \frac{1}{4}\cdot(-1)^{...}\right)\) | — | — |
| \(= \frac{\pi}{8} + \left[\frac{\cos 2x}{4}\right]_0^{\frac{\pi}{4}} = \frac{\pi}{8} + \frac{\cos\frac{\pi}{2}}{4} - \frac{1}{4}\) | A1 | Substituting limits |
| \(= \frac{\pi}{8} - \frac{1}{4}\) | A1 | Correct answer |
| Graphically: area between curve and \(x\)-axis from \(0\) to \(\frac{\pi}{4}\) | B1 | [6] Correct interpretation |
# Question 8:
**Part (i):**
| P is where $x\cos 2x = 0$, $x \neq 0 \Rightarrow \cos 2x = 0$ | M1 | Setting $y=0$ for positive $x$ |
| $2x = \frac{\pi}{2} \Rightarrow x = \frac{\pi}{4}$ | A1 | Correct $x$ |
| $P = \left(\frac{\pi}{4}, 0\right)$ | A1 | [3] |
**Part (ii):**
| $f(-x) = (-x)\cos(-2x) = -x\cos 2x = -f(x)$ | M1 | Correct algebraic manipulation |
| Hence odd function | A1 | Stated |
| Graphically: rotational symmetry of order 2 about the origin | B1 | [3] |
**Part (iii):**
| $\frac{dy}{dx} = \cos 2x - 2x\sin 2x$ | M1 | Product rule |
| — | A1 | [2] Correct derivative |
**Part (iv):**
| At turning point: $\cos 2x - 2x\sin 2x = 0$ | M1 | Setting derivative to zero |
| $\cos 2x = 2x\sin 2x \Rightarrow x\tan 2x = \frac{1}{2}$ (dividing by $\cos 2x$) | A1 | [2] Shown |
**Part (v):**
| At $x=0$: $\frac{dy}{dx} = \cos 0 = 1$, so gradient $= 1$ | B1 | Correct gradient |
| $\frac{d^2y}{dx^2} = -2\sin 2x - 2\sin 2x - 4x\cos 2x = -4\sin 2x - 4x\cos 2x$ | M1 | Differentiating again |
| At $x=0$: $-4\sin 0 - 0 = 0$ | A1 | Shown |
| — | A1 | [4] |
**Part (vi):**
| $\int_0^{\frac{\pi}{4}} x\cos 2x\,dx$: integrate by parts, $u=x$, $\frac{dv}{dx}=\cos 2x$ | M1 | By parts |
| $= \left[\frac{x\sin 2x}{2}\right]_0^{\frac{\pi}{4}} - \int_0^{\frac{\pi}{4}}\frac{\sin 2x}{2}\,dx$ | A1 | Correct first step |
| $= \frac{\pi}{8} - \left[-\frac{\cos 2x}{4}\right]_0^{\frac{\pi}{4}}$ | A1 | Correct integration |
| $= \frac{\pi}{8} - \left(\frac{1}{4} - \frac{1}{4}\cdot(-1)^{...}\right)$ | — | — |
| $= \frac{\pi}{8} + \left[\frac{\cos 2x}{4}\right]_0^{\frac{\pi}{4}} = \frac{\pi}{8} + \frac{\cos\frac{\pi}{2}}{4} - \frac{1}{4}$ | A1 | Substituting limits |
| $= \frac{\pi}{8} - \frac{1}{4}$ | A1 | Correct answer |
| Graphically: area between curve and $x$-axis from $0$ to $\frac{\pi}{4}$ | B1 | [6] Correct interpretation |8 Fig. 8 shows part of the curve $y = x \cos 2 x$, together with a point P at which the curve crosses the $x$-axis.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0ee3d87a-0d9e-4fa5-b8f5-8b28489e65b5-4_421_965_349_550}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the exact coordinates of P .\\
(ii) Show algebraically that $x \cos 2 x$ is an odd function, and interpret this result graphically.\\
(iii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(iv) Show that turning points occur on the curve for values of $x$ which satisfy the equation $x \tan 2 x = \frac { 1 } { 2 }$.\\
(v) Find the gradient of the curve at the origin.
Show that the second derivative of $x \cos 2 x$ is zero when $x = 0$.\\
(vi) Evaluate $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x$, giving your answer in terms of $\pi$. Interpret this result graphically.
\hfill \mbox{\textit{OCR MEI C3 2007 Q8 [20]}}