| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find derivative of product |
| Difficulty | Moderate -0.8 This is a straightforward application of standard differentiation rules: (a) uses product rule with basic trig, (b)(i) uses chain rule, and (b)(ii) is a direct reverse application. All parts are routine textbook exercises requiring only recall and mechanical application of rules, making this easier than average. |
| Spec | 1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = x\sin 2x\), \(\frac{dy}{dx} = x \cdot 2\cos 2x + \sin 2x\) | M1 | Product rule |
| Both terms correct | A1, A1 | Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(y = (x^2-6)^4\), \(\frac{dy}{dx} = 4(x^2-6)^3(2x)\) (or better) | M1A1 | M1 for \((x^2-6)^3\); Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\int 8x(x^2-6)^3\,dx = (x^2-6)^4\) | M1 | For \(c(x^2-6)^4\) if correct attempt |
| \(\int = \frac{1}{8}(x^2-6)^4 (+c)\) | A1 | For \(\frac{1}{k}(x^2-6)^4\) at 'by parts'; M1A0 |
| \(k=8\) confirmed | A1 | Total: 3 marks. Or via expansion: \((x^2-6)^3 = x^6-18x^4+108x^2-216\) (M1A1); \(\int x(x^2-6)^3 = \frac{x^8}{8}-3x^6+27x^4-108x^2\) (A1) |
## Question 1:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = x\sin 2x$, $\frac{dy}{dx} = x \cdot 2\cos 2x + \sin 2x$ | M1 | Product rule |
| Both terms correct | A1, A1 | Total: 3 marks |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|---------|
| $y = (x^2-6)^4$, $\frac{dy}{dx} = 4(x^2-6)^3(2x)$ (or better) | M1A1 | M1 for $(x^2-6)^3$; Total: 2 marks |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\int 8x(x^2-6)^3\,dx = (x^2-6)^4$ | M1 | For $c(x^2-6)^4$ if correct attempt |
| $\int = \frac{1}{8}(x^2-6)^4 (+c)$ | A1 | For $\frac{1}{k}(x^2-6)^4$ at 'by parts'; M1A0 |
| $k=8$ confirmed | A1 | Total: 3 marks. Or via expansion: $(x^2-6)^3 = x^6-18x^4+108x^2-216$ (M1A1); $\int x(x^2-6)^3 = \frac{x^8}{8}-3x^6+27x^4-108x^2$ (A1) |
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\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $y = x \sin 2 x$.
\item \begin{enumerate}[label=(\roman*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ when $y = \left( x ^ { 2 } - 6 \right) ^ { 4 }$.
\item Hence, or otherwise, find $\int x \left( x ^ { 2 } - 6 \right) ^ { 3 } \mathrm {~d} x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q1 [8]}}