| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2008 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Find gradient at specific point |
| Difficulty | Standard +0.2 This is a straightforward multi-part C3 question testing standard techniques: (a) routine chain rule differentiation and substitution, (b) mid-ordinate rule application with calculator work, (c) standard volume of revolution formula. All parts follow textbook procedures with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.09f Trapezium rule: numerical integration4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = (e^{3x}+1)^{\frac{1}{2}}\); \(\frac{dy}{dx} = \frac{1}{2}(e^{3x}+1)^{-\frac{1}{2}} \times 3e^{3x}\); At \(x = \ln 2\): \(\frac{dy}{dx} = \frac{3}{2}(e^{\ln 8}+1)^{-\frac{1}{2}} \times e^{\ln 8} = \frac{3}{2} \times \frac{1}{3} \times 8 = 4\) | M1, A1, A1, M1, A1 | \(\frac{1}{2}(e^{3x}+1)^{-\frac{1}{2}}\); \(e^{3x}\); \(\frac{3}{2}\) (allow \(\frac{1}{2} \times 3\)) w.n.f.e; correct substitution into their \(\frac{dy}{dx}\) (must use ln8 or ln2³); CSO |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | Correct \(x\) values | |
| \(\int = 0.5 \times \sum y = 12.7\) | M1, A1 | 3 or 4 correct \(y\) values 4 s.f. or better; sc 12.7 with no working |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \pi \int y^2 dx = (\pi)\int(e^{3x}+1)(dx) = (\pi)\left[\frac{1}{3}e^{3x}+x\right]_{0}^{(2)} = (\pi)\left[\left(\frac{1}{3}e^6+2\right) - \left(\frac{1}{3}e^0+0\right)\right] = \pi\left[\frac{1}{3}e^6+\frac{5}{3}\right] = \frac{\pi}{3}(e^6+5)\) | M1, A1, m1, A1 | \(ke^{3x}+x\); correct substitution into \(f(e^{3x})\); CSO |
**6(a)**
| $y = (e^{3x}+1)^{\frac{1}{2}}$; $\frac{dy}{dx} = \frac{1}{2}(e^{3x}+1)^{-\frac{1}{2}} \times 3e^{3x}$; At $x = \ln 2$: $\frac{dy}{dx} = \frac{3}{2}(e^{\ln 8}+1)^{-\frac{1}{2}} \times e^{\ln 8} = \frac{3}{2} \times \frac{1}{3} \times 8 = 4$ | M1, A1, A1, M1, A1 | $\frac{1}{2}(e^{3x}+1)^{-\frac{1}{2}}$; $e^{3x}$; $\frac{3}{2}$ (allow $\frac{1}{2} \times 3$) w.n.f.e; correct substitution into their $\frac{dy}{dx}$ (must use ln8 or ln2³); CSO |
**Total: 5 marks**
**6(b)**
| | B1 | Correct $x$ values |
| $\int = 0.5 \times \sum y = 12.7$ | M1, A1 | 3 or 4 correct $y$ values 4 s.f. or better; sc 12.7 with no working |
**Total: 4 marks**
**6(c)**
| $v = \pi \int y^2 dx = (\pi)\int(e^{3x}+1)(dx) = (\pi)\left[\frac{1}{3}e^{3x}+x\right]_{0}^{(2)} = (\pi)\left[\left(\frac{1}{3}e^6+2\right) - \left(\frac{1}{3}e^0+0\right)\right] = \pi\left[\frac{1}{3}e^6+\frac{5}{3}\right] = \frac{\pi}{3}(e^6+5)$ | M1, A1, m1, A1 | $ke^{3x}+x$; correct substitution into $f(e^{3x})$; CSO |
**Total: 13 marks**
---6 The diagram shows the curve with equation $y = \left( \mathrm { e } ^ { 3 x } + 1 \right) ^ { \frac { 1 } { 2 } }$ for $x \geqslant 0$.\\
\includegraphics[max width=\textwidth, alt={}, center]{6ce5aa0d-0a73-4bc4-aabc-314c0434e4f5-5_483_611_402_717}
\begin{enumerate}[label=(\alph*)]
\item Find the gradient of the curve $y = \left( \mathrm { e } ^ { 3 x } + 1 \right) ^ { \frac { 1 } { 2 } }$ at the point where $x = \ln 2$.
\item Use the mid-ordinate rule with four strips to find an estimate for $\int _ { 0 } ^ { 2 } \left( \mathrm { e } ^ { 3 x } + 1 \right) ^ { \frac { 1 } { 2 } } \mathrm {~d} x$, giving your answer to three significant figures.
\item The shaded region $R$ is bounded by the curve, the lines $x = 0 , x = 2$ and the $x$-axis.
Find the exact value of the volume of the solid generated when the region $R$ is rotated through $360 ^ { \circ }$ about the $x$-axis.
\end{enumerate}
\hfill \mbox{\textit{AQA C3 2008 Q6 [13]}}