| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2007 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration by Substitution |
| Type | Finding maximum/minimum on curve |
| Difficulty | Standard +0.3 This is a straightforward multi-part C3 question requiring standard techniques: finding asymptotes by solving a simple equation, applying quotient rule differentiation (with clear guidance via 'show that'), solving a factorizable equation for turning points, and integration by substitution with the substitution essentially given by the denominator structure. All parts are routine applications of core techniques with no novel problem-solving required. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.07n Stationary points: find maxima, minima using derivatives1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(1 + 2x^3 = 0 \Rightarrow x^3 = -\frac{1}{2}\) | M1 | Setting denominator to zero |
| \(x = -\frac{1}{\sqrt[3]{2}} \approx -0.794\) | A1 | |
| \(a = -0.794\) | A1 | [3] Correct to 3 s.f. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{2x(1+2x^3) - x^2 \cdot 6x^2}{(1+2x^3)^2}\) | M1 | Quotient rule |
| \(= \frac{2x + 4x^4 - 6x^4}{(1+2x^3)^2} = \frac{2x - 2x^4}{(1+2x^3)^2}\) | A1 | Correct simplification |
| \(= \frac{2x(1-x^3)}{(1+2x^3)^2}\) | A1 | Shown |
| \(\frac{dy}{dx}=0 \Rightarrow x=0\) or \(x=1\) | M1 | Solving numerator = 0 |
| At \(x=0\): \(y=0\); at \(x=1\): \(y=\frac{1}{3}\) | A1 | Both \(y\)-values |
| Turning points: \((0,0)\) and \((1, \frac{1}{3})\) | A1A1 | [8] Both coordinates |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^1 \frac{x^2}{1+2x^3}\,dx\) | M1 | Correct integral set up |
| \(= \left[\frac{1}{6}\ln(1+2x^3)\right]_0^1\) | M1A1 | Integration using substitution/recognition |
| \(= \frac{1}{6}\ln 3 - \frac{1}{6}\ln 1 = \frac{1}{6}\ln 3\) | A1A1 | [5] Correct evaluation shown |
# Question 7:
**Part (i):**
| $1 + 2x^3 = 0 \Rightarrow x^3 = -\frac{1}{2}$ | M1 | Setting denominator to zero |
| $x = -\frac{1}{\sqrt[3]{2}} \approx -0.794$ | A1 | |
| $a = -0.794$ | A1 | [3] Correct to 3 s.f. |
**Part (ii):**
| $\frac{dy}{dx} = \frac{2x(1+2x^3) - x^2 \cdot 6x^2}{(1+2x^3)^2}$ | M1 | Quotient rule |
| $= \frac{2x + 4x^4 - 6x^4}{(1+2x^3)^2} = \frac{2x - 2x^4}{(1+2x^3)^2}$ | A1 | Correct simplification |
| $= \frac{2x(1-x^3)}{(1+2x^3)^2}$ | A1 | Shown |
| $\frac{dy}{dx}=0 \Rightarrow x=0$ or $x=1$ | M1 | Solving numerator = 0 |
| At $x=0$: $y=0$; at $x=1$: $y=\frac{1}{3}$ | A1 | Both $y$-values |
| Turning points: $(0,0)$ and $(1, \frac{1}{3})$ | A1A1 | [8] Both coordinates |
**Part (iii):**
| $\int_0^1 \frac{x^2}{1+2x^3}\,dx$ | M1 | Correct integral set up |
| $= \left[\frac{1}{6}\ln(1+2x^3)\right]_0^1$ | M1A1 | Integration using substitution/recognition |
| $= \frac{1}{6}\ln 3 - \frac{1}{6}\ln 1 = \frac{1}{6}\ln 3$ | A1A1 | [5] Correct evaluation shown |
---7 Fig. 7 shows the curve $y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }$. It is undefined at $x = a$; the line $x = a$ is a vertical asymptote.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0ee3d87a-0d9e-4fa5-b8f5-8b28489e65b5-3_654_1034_1505_497}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Calculate the value of $a$, giving your answer correct to 3 significant figures.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }$. Hence determine the coordinates of the turning points of the curve.\\
(iii) Show that the area of the region between the curve and the $x$-axis from $x = 0$ to $x = 1$ is $\frac { 1 } { 6 } \ln 3$.
\hfill \mbox{\textit{OCR MEI C3 2007 Q7 [16]}}