| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Implicit equations and differentiation |
| Type | Find stationary points |
| Difficulty | Standard +0.3 This is a standard C3 implicit differentiation question with routine steps: identifying an asymptote (trivial), applying implicit differentiation using quotient rule (standard technique), solving dy/dx=0 for stationary points, and performing a guided substitution for integration. All steps are clearly signposted with no novel insight required, making it slightly easier than average. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.07s Parametric and implicit differentiation1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \frac{x}{(2x-1)^{1/3}}\) | ||
| \(\Rightarrow \frac{dy}{dx} = \frac{(2x-1)^{1/3}\cdot 1 - x\cdot\frac{1}{3}(2x-1)^{-2/3}\cdot 2}{(2x-1)^{2/3}}\) | M1 | quotient rule or product rule on \(y\) – allow one slip |
| A1 | correct expression for the derivative | |
| \(= \frac{6x-3-2x}{3(2x-1)^{4/3}} = \frac{4x-3}{3(2x-1)^{4/3}}\) | M1 | factorising or multiplying top and bottom by \((2x-1)^{2/3}\) |
| A1 | ||
| \(= \frac{(4x-3)x^2}{3y^2(2x-1)^{2/3}(2x-1)^{4/3}} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}\) | A1 | establishing equivalence with given answer NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \left(\frac{x^3}{2x-1}\right)^{1/3}\) | ||
| \(\Rightarrow \frac{dy}{dx} = \frac{1}{3}\!\left(\frac{x^3}{2x-1}\right)^{-2/3} \cdot \frac{(2x-1)\cdot 3x^2 - x^3\cdot 2}{(2x-1)^2}\) | B1 | \(\frac{1}{3}\!\left(\frac{x^3}{2x-1}\right)^{-2/3}\times\ldots\) |
| M1A1 | \(\ldots\times\frac{(2x-1)\cdot 3x^2 - x^3\cdot 2}{(2x-1)^2}\) | |
| \(= \frac{1}{3}\cdot\frac{4x^3-3x^2}{x^2(2x-1)^{4/3}} = \frac{4x-3}{3(2x-1)^{4/3}}\) | A1 | |
| \(= \frac{(4x-3)x^2}{3y^2(2x-1)^{2/3}(2x-1)^{4/3}} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}\) | A1 | establishing equivalence with given answer NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y^3(2x-1) = x^3\) | ||
| \(3y^2\frac{dy}{dx}(2x-1) + y^3\cdot 2 = 3x^2\) | B1 | \(d/dx(y^3) = 3y^2(dy/dx)\) |
| M1 | product rule on \(y^3(2x-1)\) or \(2xy^3\) | |
| A1 | correct equation | |
| \(\frac{dy}{dx} = \frac{3x^2 - 2y^3}{3y^2(2x-1)}\) | ||
| \(= \frac{3x^2 - 2\cdot\frac{x^3}{(2x-1)}}{3y^2(2x-1)}\) | M1 | subbing for \(2y^3\) |
| \(= \frac{3x^2(2x-1)-2x^3}{3y^2(2x-1)^2} = \frac{6x^3-3x^2-2x^3}{3y^2(2x-1)^2} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}\) | A1 | NB AG |
## Question 9(ii)(1) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \frac{x}{(2x-1)^{1/3}}$ | | |
| $\Rightarrow \frac{dy}{dx} = \frac{(2x-1)^{1/3}\cdot 1 - x\cdot\frac{1}{3}(2x-1)^{-2/3}\cdot 2}{(2x-1)^{2/3}}$ | M1 | quotient rule or product rule on $y$ – allow one slip |
| | A1 | correct expression for the derivative |
| $= \frac{6x-3-2x}{3(2x-1)^{4/3}} = \frac{4x-3}{3(2x-1)^{4/3}}$ | M1 | factorising or multiplying top and bottom by $(2x-1)^{2/3}$ |
| | A1 | |
| $= \frac{(4x-3)x^2}{3y^2(2x-1)^{2/3}(2x-1)^{4/3}} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}$ | A1 | establishing equivalence with given answer **NB AG** |
## Question 9(ii)(2) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \left(\frac{x^3}{2x-1}\right)^{1/3}$ | | |
| $\Rightarrow \frac{dy}{dx} = \frac{1}{3}\!\left(\frac{x^3}{2x-1}\right)^{-2/3} \cdot \frac{(2x-1)\cdot 3x^2 - x^3\cdot 2}{(2x-1)^2}$ | B1 | $\frac{1}{3}\!\left(\frac{x^3}{2x-1}\right)^{-2/3}\times\ldots$ |
| | M1A1 | $\ldots\times\frac{(2x-1)\cdot 3x^2 - x^3\cdot 2}{(2x-1)^2}$ |
| $= \frac{1}{3}\cdot\frac{4x^3-3x^2}{x^2(2x-1)^{4/3}} = \frac{4x-3}{3(2x-1)^{4/3}}$ | A1 | |
| $= \frac{(4x-3)x^2}{3y^2(2x-1)^{2/3}(2x-1)^{4/3}} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}$ | A1 | establishing equivalence with given answer **NB AG** |
## Question 9(ii)(3) – Additional Solutions:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y^3(2x-1) = x^3$ | | |
| $3y^2\frac{dy}{dx}(2x-1) + y^3\cdot 2 = 3x^2$ | B1 | $d/dx(y^3) = 3y^2(dy/dx)$ |
| | M1 | product rule on $y^3(2x-1)$ or $2xy^3$ |
| | A1 | correct equation |
| $\frac{dy}{dx} = \frac{3x^2 - 2y^3}{3y^2(2x-1)}$ | | |
| $= \frac{3x^2 - 2\cdot\frac{x^3}{(2x-1)}}{3y^2(2x-1)}$ | M1 | subbing for $2y^3$ |
| $= \frac{3x^2(2x-1)-2x^3}{3y^2(2x-1)^2} = \frac{6x^3-3x^2-2x^3}{3y^2(2x-1)^2} = \frac{4x^3-3x^2}{3y^2(2x-1)^2}$ | A1 | **NB AG** |
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To extract mark scheme content, please share the actual mark scheme pages (typically containing tables with question numbers, answers, marks, and examiner guidance). Could you upload those pages instead?9 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]{28ce1bcc-e9d5-4ae6-98c0-67b5b8c50bc6-6_752_867_356_584}
\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 2013 Q9 [18]}}