| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 This is a standard multi-part C3 question testing routine application of quotient rule, product rule, tangent equations, substitution integration, and integration by parts. Part (a) is straightforward quotient rule application with algebraic simplification. Parts (b) and (c) follow textbook procedures with no novel insights required. The 'show that' format in (a) provides the target form, making it easier than open-ended questions. Slightly above average difficulty due to the length and variety of techniques, but all are standard C3 procedures. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08h Integration by substitution1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx}=\frac{(4x-3).4-4x(4)}{(4x-3)^2}\) | M1 | Must use quotient rule; condone one slip |
| \(=\frac{-12}{(4x-3)^2}\) | A1 | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{dy}{dx}=\frac{x.4}{4x-3}+\ln(4x-3)\) | M1 | \(\frac{f(x)}{4x-3}+g(x)\); \(f(x)\) may be constant |
| \(\frac{kx}{4x-3}+\ln(4x-3)\) | m1 | |
| Full correct answer | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(x=1\ y=0\) | B1 | |
| \(\frac{dy}{dx}=4\) | M1 | Sub \(x=1\) into their \(\frac{dy}{dx}\) |
| \(\therefore y=4(x-1)\) any correct form | A1 | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(u=4x-3\), \(du=4\,dx\) | M1 | |
| \(\int\frac{4x}{4x-3}\,dx=\int\frac{u+3}{u}\cdot\frac{du}{4}\) | A1 | Or \(\int\frac{4x}{4x-3}\,dx=\int\left(1+\frac{3}{4x-3}\right)dx\) |
| \(=\left(\frac{1}{4}\right)\int\left(1+\frac{3}{u}\right)(du)\) | m1 | \(=\int\left(1+\frac{3}{u}\right)du\) etc |
| \(=\frac{1}{4}\left[(4x-3)+3\ln(4x-3)\right](+c)\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(u=\ln(4x-3)\), \(\frac{dv}{dx}=1\) | M1 | In correct direction |
| \(\frac{du}{dx}=\frac{4}{4x-3}\), \(v=x\) | ||
| \(\int = x\ln(4x-3)-\int\frac{4x}{4x-3}\,dx\) | A1 | |
| \(=x\ln(4x-3)-\frac{1}{4}\left[(4x-3)+3\ln(4x-3)\right](+c)\) | m1 A1 | 4 |
# Question 9:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx}=\frac{(4x-3).4-4x(4)}{(4x-3)^2}$ | M1 | Must use quotient rule; condone one slip |
| $=\frac{-12}{(4x-3)^2}$ | A1 | 2 | $k=-12$ |
## Part (b)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{dy}{dx}=\frac{x.4}{4x-3}+\ln(4x-3)$ | M1 | $\frac{f(x)}{4x-3}+g(x)$; $f(x)$ may be constant |
| $\frac{kx}{4x-3}+\ln(4x-3)$ | m1 | |
| Full correct answer | A1 | 3 | |
## Part (b)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $x=1\ y=0$ | B1 | |
| $\frac{dy}{dx}=4$ | M1 | Sub $x=1$ into their $\frac{dy}{dx}$ |
| $\therefore y=4(x-1)$ any correct form | A1 | 3 | CSO Must have full marks in (b)(i) |
## Part (c)(i):
| Working | Mark | Guidance |
|---------|------|----------|
| $u=4x-3$, $du=4\,dx$ | M1 | |
| $\int\frac{4x}{4x-3}\,dx=\int\frac{u+3}{u}\cdot\frac{du}{4}$ | A1 | Or $\int\frac{4x}{4x-3}\,dx=\int\left(1+\frac{3}{4x-3}\right)dx$ |
| $=\left(\frac{1}{4}\right)\int\left(1+\frac{3}{u}\right)(du)$ | m1 | $=\int\left(1+\frac{3}{u}\right)du$ etc |
| $=\frac{1}{4}\left[(4x-3)+3\ln(4x-3)\right](+c)$ | A1 | 4 | CSO Condone missing $du$ |
## Part (c)(ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $u=\ln(4x-3)$, $\frac{dv}{dx}=1$ | M1 | In correct direction |
| $\frac{du}{dx}=\frac{4}{4x-3}$, $v=x$ | | |
| $\int = x\ln(4x-3)-\int\frac{4x}{4x-3}\,dx$ | A1 | |
| $=x\ln(4x-3)-\frac{1}{4}\left[(4x-3)+3\ln(4x-3)\right](+c)$ | m1 A1 | 4 | $x\ln(4x-3)-\text{their (c)(i)}$ |9
\begin{enumerate}[label=(\alph*)]
\item Given that $y = \frac { 4 x } { 4 x - 3 }$, use the quotient rule to show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { k } { ( 4 x - 3 ) ^ { 2 } }$, where $k$ is an integer.
\item \begin{enumerate}[label=(\roman*)]
\item Given that $y = x \ln ( 4 x - 3 )$, find $\frac { \mathrm { d } y } { \mathrm {~d} x }$.
\item Find an equation of the tangent to the curve $y = x \ln ( 4 x - 3 )$ at the point where $x = 1$.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Use the substitution $u = 4 x - 3$ to find $\int \frac { 4 x } { 4 x - 3 } \mathrm {~d} x$, giving your answer in terms of $x$.
\item By using integration by parts, or otherwise, find $\int \ln ( 4 x - 3 ) \mathrm { d } x$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2009 Q9 [16]}}