| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: product rule differentiation, stationary points, algebraic manipulation, and integration by parts. While it has many parts, each individual step is routine and well-signposted, making it slightly easier than average despite the length. |
| Spec | 1.05k Further identities: sec^2=1+tan^2 and cosec^2=1+cot^21.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08i Integration by parts1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = x \times 2\sin 2x + \cos 2x\) | M1, A1 | Product rule \(kx\sin 2x + \cos 2x\); no further incorrect working |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2α\sin 2α + \cos 2α = 0\); \(2α \sin 2α = \cos 2α\) or \(2α \tan 2α = 1\) or \(2α \tan 2α - 1 = 0\) | M1, A1 | Replacing \(x = α\) and writing equation equal to zero (at any line); AG; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(0.4) = 0.2\); \(f(0.5) = -0.6\); Change of sign ∴ \(0.4 < α < 0.5\) | M1, A1 | (0.9's unsubstantiated scores M0); AWRT o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x\tan 2x = 1\); \(\tan 2x = \frac{1}{2x}\); \(2x = \tan^{-1}\left(\frac{1}{2x}\right)\) or \(x = \frac{1}{2}\tan^{-1}\left(\frac{1}{2x}\right)\) | B1 | AG; CSO |
| Answer | Marks | Guidance |
|---|---|---|
| \(x_1 = 0.4\); \(x_2 = 0.4480...\); \(x_3 = 0.4200...\); \(= 0.42\) | M1, A1 | \(x_2 = 25.7\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x\cos 2x\); \(u = x\), \(du = 1\); \(dv = \cos 2x\), \(v = \frac{\sin 2x}{2}\); \(\int \frac{x\sin 2x}{2} - \int \frac{\sin 2x}{2}(dx) = \left[\frac{x\sin 2x}{2} + \frac{\cos 2x}{4}\right]_{0}^{(0.5)} = \left(\frac{\sin 1}{4} + \frac{\cos 1}{4}\right) - \left(\frac{\cos 0}{4}\right) = 0.0954\) | M1, m1, A1, m1, A1, A1 | Differentiate one term; integrate one term; must be \(k\sin 2x\); correct substitution of their values into parts formula using \(u = x\); correctly substituting values from previous 2 method marks; AWRT |
**3(a)**
| $\frac{dy}{dx} = x \times 2\sin 2x + \cos 2x$ | M1, A1 | Product rule $kx\sin 2x + \cos 2x$; no further incorrect working |
**Total: 2 marks**
**3(b)(i)**
| $-2α\sin 2α + \cos 2α = 0$; $2α \sin 2α = \cos 2α$ or $2α \tan 2α = 1$ or $2α \tan 2α - 1 = 0$ | M1, A1 | Replacing $x = α$ and writing equation equal to zero (at any line); AG; CSO |
**3(b)(ii)**
| $f(0.4) = 0.2$; $f(0.5) = -0.6$; Change of sign ∴ $0.4 < α < 0.5$ | M1, A1 | (0.9's unsubstantiated scores M0); AWRT o.e. |
**3(b)(iii)**
| $2x\tan 2x = 1$; $\tan 2x = \frac{1}{2x}$; $2x = \tan^{-1}\left(\frac{1}{2x}\right)$ or $x = \frac{1}{2}\tan^{-1}\left(\frac{1}{2x}\right)$ | B1 | AG; CSO |
**Total: 1 mark**
**3(b)(iv)**
| $x_1 = 0.4$; $x_2 = 0.4480...$; $x_3 = 0.4200...$; $= 0.42$ | M1, A1 | $x_2 = 25.7$ |
**3(c)**
| $y = x\cos 2x$; $u = x$, $du = 1$; $dv = \cos 2x$, $v = \frac{\sin 2x}{2}$; $\int \frac{x\sin 2x}{2} - \int \frac{\sin 2x}{2}(dx) = \left[\frac{x\sin 2x}{2} + \frac{\cos 2x}{4}\right]_{0}^{(0.5)} = \left(\frac{\sin 1}{4} + \frac{\cos 1}{4}\right) - \left(\frac{\cos 0}{4}\right) = 0.0954$ | M1, m1, A1, m1, A1, A1 | Differentiate one term; integrate one term; must be $k\sin 2x$; correct substitution of their values into parts formula using $u = x$; correctly substituting values from previous 2 method marks; AWRT |
**Total: 14 marks**
---3 A curve is defined for $0 \leqslant x \leqslant \frac { \pi } { 4 }$ by the equation $y = x \cos 2 x$, and is sketched below.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ce5aa0d-0a73-4bc4-aabc-314c0434e4f5-3_757_878_402_559}
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item The point $A$, where $x = \alpha$, on the curve is a stationary point.
\begin{enumerate}[label=(\roman*)]
\item Show that $1 - 2 \alpha \tan 2 \alpha = 0$.
\item Show that $0.4 < \alpha < 0.5$.
\item Show that the equation $1 - 2 x \tan 2 x = 0$ can be rearranged to become $x = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x } \right)$.
\item Use the iteration $x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x _ { n } } \right)$ with $x _ { 1 } = 0.4$ to find $x _ { 3 }$, giving your answer to two significant figures.
\end{enumerate}\item Use integration by parts to find $\int _ { 0 } ^ { 0.5 } x \cos 2 x \mathrm {~d} x$, giving your answer to three significant figures.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q3 [14]}}