| Exam Board | AQA |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2005 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Volumes of Revolution |
| Type | Rotation about x-axis: polynomial or root function |
| Difficulty | Moderate -0.3 This is a straightforward C3 volumes of revolution question with standard techniques. Part (a) involves sketching a simple parabola and applying the volume formula V=π∫y²dx with no algebraic complications. Part (b) requires sketching an absolute value function and solving basic modulus equations/inequalities using standard methods. All components are routine textbook exercises requiring recall and direct application rather than problem-solving insight, making it slightly easier than average. |
| Spec | 1.02l Modulus function: notation, relations, equations and inequalities1.02n Sketch curves: simple equations including polynomials1.02t Solve modulus equations: graphically with modulus function4.08d Volumes of revolution: about x and y axes |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Sketch: arch shape, max at \((0,4)\), crossing \(x\)-axis at \(\pm 2\) | M1 | Shape symmetrical about \(y\)-axis |
| All correct | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(V = (\pi)\int(4-x^2)^2\,dx\) | M1 | |
| \(= (\pi)\int 16-8x^2+x^4\,dx\) | B1 | Expanding bracket |
| \(= (\pi)\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]\) | M1 | Correctly integrating 2 of their terms |
| \(= \pi\frac{256}{15}\) | A1 | Total: 4 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Sketch: modulus graph (V-shapes pointing up, max at \((0,4)\)) | M1 | Modulus graph |
| Correct shape | A1 | Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\lvert 4-x^2\rvert = 3\); \(4-x^2 = 3 \Rightarrow x = +1,-1\) | M1, A1 | Attempt at solving correct equation; 2 correct |
| \(4-x^2 = -3 \Rightarrow x = \pm\sqrt{7}\) (or exact equivalent) | A1 | 2 correct; Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(-\sqrt{7} < x < -1\) | B1F | Condone \(\sqrt{7} = 2.6\) (or better) |
| \(1 < x < \sqrt{7}\) | B1F | Total: 2 marks |
## Question 6:
### Part (a)(i)
| Working | Marks | Guidance |
|---------|-------|---------|
| Sketch: arch shape, max at $(0,4)$, crossing $x$-axis at $\pm 2$ | M1 | Shape symmetrical about $y$-axis |
| All correct | A1 | Total: 2 marks |
### Part (a)(ii)
| Working | Marks | Guidance |
|---------|-------|---------|
| $V = (\pi)\int(4-x^2)^2\,dx$ | M1 | |
| $= (\pi)\int 16-8x^2+x^4\,dx$ | B1 | Expanding bracket |
| $= (\pi)\left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]$ | M1 | Correctly integrating 2 of their terms |
| $= \pi\frac{256}{15}$ | A1 | Total: 4 marks |
### Part (b)(i)
| Working | Marks | Guidance |
|---------|-------|---------|
| Sketch: modulus graph (V-shapes pointing up, max at $(0,4)$) | M1 | Modulus graph |
| Correct shape | A1 | Total: 2 marks |
### Part (b)(ii)
| Working | Marks | Guidance |
|---------|-------|---------|
| $\lvert 4-x^2\rvert = 3$; $4-x^2 = 3 \Rightarrow x = +1,-1$ | M1, A1 | Attempt at solving correct equation; 2 correct |
| $4-x^2 = -3 \Rightarrow x = \pm\sqrt{7}$ (or exact equivalent) | A1 | 2 correct; Total: 3 marks |
### Part (b)(iii)
| Working | Marks | Guidance |
|---------|-------|---------|
| $-\sqrt{7} < x < -1$ | B1F | Condone $\sqrt{7} = 2.6$ (or better) |
| $1 < x < \sqrt{7}$ | B1F | Total: 2 marks |6
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = 4 - x ^ { 2 }$, indicating the coordinates of the points where the graph crosses the coordinate axes.
\item The region between the graph and the $x$-axis from $x = 0$ to $x = 2$ is rotated through $360 ^ { \circ }$ about the $x$-axis. Find the exact value of the volume of the solid generated.
\end{enumerate}\item \begin{enumerate}[label=(\roman*)]
\item Sketch the graph of $y = \left| 4 - x ^ { 2 } \right|$.
\item Solve $\left| 4 - x ^ { 2 } \right| = 3$.
\item Hence, or otherwise, solve the inequality $\left| 4 - x ^ { 2 } \right| < 3$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C3 2005 Q6 [13]}}