| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.8 This is a substantial multi-part question requiring implicit differentiation (a C3 topic that students often find challenging), finding stationary points by solving a cubic equation, and integration by substitution with fractional powers. The implicit differentiation using quotient rule is non-trivial, and the final area calculation requires careful setup and exact arithmetic with surds. While each technique is standard C3 material, the combination and length elevate this above average difficulty. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(u = 2x-1 \Rightarrow du = 2\,dx\) | ||
| \(\int \frac{x}{\sqrt[3]{2x-1}}\,dx = \int \frac{\frac{1}{2}(u+1)}{u^{1/3}} \cdot \frac{1}{2}\,du\) | M1 | \(\frac{\frac{1}{2}(u+1)}{u^{1/3}}\) — if missing brackets, withhold A1 |
| M1 | \(\times \frac{1}{2}\,du\) — condone missing \(du\) here, but withhold A1 | |
| \(= \frac{1}{4}\int \frac{u+1}{u^{1/3}}\,du = \frac{1}{4}\int (u^{2/3} + u^{-1/3})\,du\) | A1 | NB AG |
| \(\text{area} = \int_1^{4.5} \frac{x}{\sqrt[3]{2x-1}}\,dx\) | M1 | Correct integral and limits — may be inferred from a change of limits and their attempt to integrate \(\frac{1}{4}(u^{2/3}+u^{-1/3})\) |
| When \(x=1,\,u=1\); when \(x=4.5,\,u=8\) | A1 | \(u=1,8\) (or substituting back to \(x\)'s and using 1 and 4.5) |
| \(= \frac{1}{4}\int_1^8 (u^{2/3}+u^{-1/3})\,du\) | ||
| \(= \frac{1}{4}\left[\frac{3}{5}u^{5/3}+\frac{3}{2}u^{2/3}\right]_1^8\) | B1 | \(\left[\frac{3}{5}u^{5/3}+\frac{3}{2}u^{2/3}\right]\) o.e. e.g. \([u^{5/3}/(5/3)+u^{2/3}/(2/3)]\) |
| \(= \frac{1}{4}\left[\frac{96}{5}+6-\frac{3}{5}-\frac{3}{2}\right]\) | A1 | o.e. correct expression (may be inferred from a correct final answer) |
| \(= 5\frac{31}{40} = 5.775\) or \(\frac{231}{40}\) | A1 | cao, must be exact; mark final answer |
| [8] |
# Question 1:
## Part (iii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $u = 2x-1 \Rightarrow du = 2\,dx$ | | |
| $\int \frac{x}{\sqrt[3]{2x-1}}\,dx = \int \frac{\frac{1}{2}(u+1)}{u^{1/3}} \cdot \frac{1}{2}\,du$ | M1 | $\frac{\frac{1}{2}(u+1)}{u^{1/3}}$ — if missing brackets, withhold A1 |
| | M1 | $\times \frac{1}{2}\,du$ — condone missing $du$ here, but withhold A1 |
| $= \frac{1}{4}\int \frac{u+1}{u^{1/3}}\,du = \frac{1}{4}\int (u^{2/3} + u^{-1/3})\,du$ | A1 | **NB AG** |
| $\text{area} = \int_1^{4.5} \frac{x}{\sqrt[3]{2x-1}}\,dx$ | M1 | Correct integral and limits — may be inferred from a change of limits and their attempt to integrate $\frac{1}{4}(u^{2/3}+u^{-1/3})$ |
| When $x=1,\,u=1$; when $x=4.5,\,u=8$ | A1 | $u=1,8$ (or substituting back to $x$'s and using 1 and 4.5) |
| $= \frac{1}{4}\int_1^8 (u^{2/3}+u^{-1/3})\,du$ | | |
| $= \frac{1}{4}\left[\frac{3}{5}u^{5/3}+\frac{3}{2}u^{2/3}\right]_1^8$ | B1 | $\left[\frac{3}{5}u^{5/3}+\frac{3}{2}u^{2/3}\right]$ o.e. e.g. $[u^{5/3}/(5/3)+u^{2/3}/(2/3)]$ |
| $= \frac{1}{4}\left[\frac{96}{5}+6-\frac{3}{5}-\frac{3}{2}\right]$ | A1 | o.e. correct expression (may be inferred from a correct final answer) |
| $= 5\frac{31}{40} = 5.775$ or $\frac{231}{40}$ | A1 | cao, must be exact; mark final answer |
| **[8]** | | |
---1 Fig. 9 shows the curve with equation $y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }$. It has an asymptote $x = a$ and turning point P .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{0b4c4935-998c-404f-8fed-9b39b849168e-1_752_864_443_617}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Write down the value of $a$.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }$.
Hence find the coordinates of the turning point P , giving the $y$-coordinate to 3 significant figures.\\
(iii) Show that the substitution $u = 2 x - 1$ transforms $\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x$ to $\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u$.
Hence find the exact area of the region enclosed by the curve $y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }$, the $x$-axis and the lines $x = 1$ and $x = 4.5$.
\hfill \mbox{\textit{OCR MEI C3 Q1 [18]}}