| 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 (with quotient rule), finding stationary points by solving a cubic numerator, integration by substitution with fractional powers, and calculating a definite integral. While each technique is C3 standard, the combination of steps, algebraic manipulation, and the need to work with the implicit form throughout makes this moderately harder than average. |
| Spec | 1.07s Parametric and implicit differentiation1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(a = \frac{1}{2}\) | B1 | allow \(x = \frac{1}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y^3 = \frac{x^3}{2x-1}\) | ||
| \(3y^2\frac{dy}{dx} = \frac{(2x-1)3x^2 - x^3 \cdot 2}{(2x-1)^2}\) | B1 | \(3y^2\frac{dy}{dx}\) |
| Quotient (or product) rule consistent with their derivatives | M1 | \((v\,du + u\,dv)/v^2\); M0 if wrong |
| \(= \frac{6x^3-3x^2-2x^3}{(2x-1)^2} = \frac{4x^3-3x^2}{(2x-1)^2}\) | A1 | correct RHS expression — condone missing bracket |
| \(\Rightarrow \frac{dy}{dx} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}\) | A1 | NB AG; penalise omission of bracket in QR at this stage |
| \(\frac{dy}{dx}=0\) when \(4x^3-3x^2=0\) | M1 | |
| \(\Rightarrow x^2(4x-3)=0\), \(x=0\) or \(\frac{3}{4}\) | A1 | if in addition \(2x-1=0\) giving \(x=\frac{1}{2}\), A0 |
| \(y^3 = \frac{(3/4)^3}{1/2} = \frac{27}{32}\) | M1 | must use \(x=\frac{3}{4}\); if \((0,0)\) given as additional TP, then A0 |
| \(y = 0.945\) (3sf) | A1 | can infer M1 from answer in range 0.94 to 0.95 inclusive |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | 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 |
| \(\times \frac{1}{2}\,du\) | M1 | 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 |
| \(\text{area} = \int_1^{4.5}\frac{x}{\sqrt[3]{2x-1}}\,dx\); when \(x=1\), \(u=1\); when \(x=4.5\), \(u=8\) | M1, A1 | correct integral and limits; \(u=1,8\) (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 | o.e. e.g. \(\left[u^{5/3}/(5/3)+u^{2/3}/(2/3)\right]\) |
| \(= \frac{1}{4}\left[\frac{96}{5}+6-\frac{3}{5}-\frac{3}{2}\right]\) | A1 | o.e. correct expression (may be inferred from correct final answer) |
| \(= 5\frac{31}{40} = 5.775\) or \(\frac{231}{40}\) | A1 | cao, must be exact; mark final answer |
# Question 2:
## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $a = \frac{1}{2}$ | B1 | allow $x = \frac{1}{2}$ |
**[1]**
## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $y^3 = \frac{x^3}{2x-1}$ | | |
| $3y^2\frac{dy}{dx} = \frac{(2x-1)3x^2 - x^3 \cdot 2}{(2x-1)^2}$ | B1 | $3y^2\frac{dy}{dx}$ |
| Quotient (or product) rule consistent with their derivatives | M1 | $(v\,du + u\,dv)/v^2$; M0 if wrong |
| $= \frac{6x^3-3x^2-2x^3}{(2x-1)^2} = \frac{4x^3-3x^2}{(2x-1)^2}$ | A1 | correct RHS expression — condone missing bracket |
| $\Rightarrow \frac{dy}{dx} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}$ | A1 | NB AG; penalise omission of bracket in QR at this stage |
| $\frac{dy}{dx}=0$ when $4x^3-3x^2=0$ | M1 | |
| $\Rightarrow x^2(4x-3)=0$, $x=0$ or $\frac{3}{4}$ | A1 | if in addition $2x-1=0$ giving $x=\frac{1}{2}$, A0 |
| $y^3 = \frac{(3/4)^3}{1/2} = \frac{27}{32}$ | M1 | must use $x=\frac{3}{4}$; if $(0,0)$ given as additional TP, then A0 |
| $y = 0.945$ (3sf) | A1 | can infer M1 from answer in range 0.94 to 0.95 inclusive |
**[9]**
## Part (iii)
| Answer | Mark | 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 |
| $\times \frac{1}{2}\,du$ | M1 | 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 |
| $\text{area} = \int_1^{4.5}\frac{x}{\sqrt[3]{2x-1}}\,dx$; when $x=1$, $u=1$; when $x=4.5$, $u=8$ | M1, A1 | correct integral and limits; $u=1,8$ (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 | o.e. e.g. $\left[u^{5/3}/(5/3)+u^{2/3}/(2/3)\right]$ |
| $= \frac{1}{4}\left[\frac{96}{5}+6-\frac{3}{5}-\frac{3}{2}\right]$ | A1 | o.e. correct expression (may be inferred from correct final answer) |
| $= 5\frac{31}{40} = 5.775$ or $\frac{231}{40}$ | A1 | cao, must be exact; mark final answer |
**[8]**
---2 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]{75eebbfb-7bfa-4382-a6d7-1c5a7f3f419a-2_754_870_478_609}
\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 Q2 [18]}}