| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Moderate -0.3 This is a standard C3 multi-part question testing quotient rule differentiation, finding turning points, and integration by substitution. Part (ii) is routine quotient rule application with algebraic simplification, part (iii) involves solving a quadratic and evaluating coordinates, and part (iv) is a guided substitution. All techniques are standard textbook exercises with clear scaffolding, making it slightly easier than average for C3. |
| Spec | 1.07j Differentiate exponentials: e^(kx) and a^(kx)1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08h Integration by substitution |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = 1/3\) | B1 | [1] or 0.33 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(3x-1)2x - x^2 \cdot 3}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2} = \frac{x(3x-2)}{(3x-1)^2}\) * | M1, A1 | Quotient rule www – must show both steps; penalise missing brackets. |
| E1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(dy/dx = 0\) when \(x(3x - 2) = 0\) \(\Rightarrow\) \(x = 0\) or \(x = 2/3\), so at \(P, x = 2/3\) | M1, A1 | if denom = 0 also then M0 o.e e.g. 0.6, but must be exact |
| when \(x = \frac{2}{3}\), \(y = \frac{(2/3)^2}{3x(2/3)-1} = \frac{4}{9x(2/3)-1}\) | M1, A1cao | o.e e.g. 0.4, but must be exact |
| when \(x = 0.6\), \(dy/dx = -0.1875\) | B1 | \(−3/16\) or \(−0.19\) or better |
| when \(x = 0.8\), \(dy/dx = 0.1633\) | B1 | \(8/49\) or \(0.16\) or better |
| Gradient increasing \(\Rightarrow\) minimum | E1 | [7] o.e. e.g. 'from negative to positive'. Allow ft on their gradients, provided –ve and +ve respectively. Accept table with indications of signs of gradient. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int \frac{x^2}{3x-1} dx\) \(u = 3x - 1 \Rightarrow du = 3dx\) | B1 | \((u+1)^2 / 9\) o.e. \(u\) |
| \(= \int \frac{(u+1)^2}{9u} \cdot \frac{1}{3} du\) | M1 | \(× 1/3 (du)\) |
| \(= \frac{1}{27} \int \frac{(u+1)^2}{u} du = \frac{1}{27} \int \frac{u^2 + 2u + 1}{u} du\) | M1 | Expanding |
| \(= \frac{1}{27} \int (u + 2 + \frac{1}{u}) du\) * | E1 | Condone missing \(du\)'s |
| Area = \(\int_{1/3}^1 \frac{x^2}{3x-1} dx\) | ||
| When \(x = 2/3\), \(u = 1\), when \(x = 1\), \(u = 2\) | ||
| \(= \frac{1}{27} \int_1^2 (u + 2 + 1/u) du\) | B1 | \([\frac{1}{2}u^2 + 2u + \ln u]\) |
| \(= \frac{1}{27} [\frac{1}{2}u^2 + 2u + \ln u]_1^2\) | M1 | Substituting correct limits, dep integration |
| \(= \frac{1}{27} [(2 + 4 + \ln 2) - (\frac{1}{2} + 2 + \ln 1)]\) | A1cao | o.e., but must evaluate \(\ln 1 = 0\) and collect terms. |
| \(= \frac{1}{27} (3 + \ln 2)\) [= \(\frac{7 + 2\ln 2}{54}\)] | [7] |
**Part (i)**
| $a = 1/3$ | B1 | [1] or 0.33 or better |
**Part (ii)**
| $\frac{dy}{dx} = \frac{(3x-1)2x - x^2 \cdot 3}{(3x-1)^2} = \frac{6x^2 - 2x - 3x^2}{(3x-1)^2} = \frac{3x^2 - 2x}{(3x-1)^2} = \frac{x(3x-2)}{(3x-1)^2}$ * | M1, A1 | Quotient rule www – must show both steps; penalise missing brackets. |
| | E1 | [3] |
**Part (iii)**
| $dy/dx = 0$ when $x(3x - 2) = 0$ $\Rightarrow$ $x = 0$ or $x = 2/3$, so at $P, x = 2/3$ | M1, A1 | if denom = 0 also then M0 o.e e.g. 0.6, but must be exact |
| when $x = \frac{2}{3}$, $y = \frac{(2/3)^2}{3x(2/3)-1} = \frac{4}{9x(2/3)-1}$ | M1, A1cao | o.e e.g. 0.4, but must be exact |
| when $x = 0.6$, $dy/dx = -0.1875$ | B1 | $−3/16$ or $−0.19$ or better |
| when $x = 0.8$, $dy/dx = 0.1633$ | B1 | $8/49$ or $0.16$ or better |
| Gradient increasing $\Rightarrow$ minimum | E1 | [7] o.e. e.g. 'from negative to positive'. Allow ft on their gradients, provided –ve and +ve respectively. Accept table with indications of signs of gradient. |
**Part (iv)**
| $\int \frac{x^2}{3x-1} dx$ $u = 3x - 1 \Rightarrow du = 3dx$ | B1 | $(u+1)^2 / 9$ o.e. $u$ |
| $= \int \frac{(u+1)^2}{9u} \cdot \frac{1}{3} du$ | M1 | $× 1/3 (du)$ |
| $= \frac{1}{27} \int \frac{(u+1)^2}{u} du = \frac{1}{27} \int \frac{u^2 + 2u + 1}{u} du$ | M1 | Expanding |
| $= \frac{1}{27} \int (u + 2 + \frac{1}{u}) du$ * | E1 | Condone missing $du$'s |
| | | |
| Area = $\int_{1/3}^1 \frac{x^2}{3x-1} dx$ | | |
| When $x = 2/3$, $u = 1$, when $x = 1$, $u = 2$ | | |
| $= \frac{1}{27} \int_1^2 (u + 2 + 1/u) du$ | B1 | $[\frac{1}{2}u^2 + 2u + \ln u]$ |
| $= \frac{1}{27} [\frac{1}{2}u^2 + 2u + \ln u]_1^2$ | M1 | Substituting correct limits, dep integration |
| $= \frac{1}{27} [(2 + 4 + \ln 2) - (\frac{1}{2} + 2 + \ln 1)]$ | A1cao | o.e., but must evaluate $\ln 1 = 0$ and collect terms. |
| $= \frac{1}{27} (3 + \ln 2)$ [= $\frac{7 + 2\ln 2}{54}$] | [7] | |
---9 Fig. 9 shows the curve $y = \frac { x ^ { 2 } } { 3 x - 1 }$.\\
P is a turning point, and the curve has a vertical asymptote $x = a$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1167a0e5-48c8-48e0-b2d1-76a50bad03ad-4_844_1486_447_331}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
(i) Write down the value of $a$.\\
(ii) Show that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x ( 3 x - 2 ) } { ( 3 x - 1 ) ^ { 2 } }$.\\
(iii) Find the exact coordinates of the turning point P .
Calculate the gradient of the curve when $x = 0.6$ and $x = 0.8$, and hence verify that P is a minimum point.\\
(iv) Using the substitution $u = 3 x - 1$, show that $\int \frac { x ^ { 2 } } { 3 x - 1 } \mathrm {~d} x = \frac { 1 } { 27 } \int \left( u + 2 + \frac { 1 } { u } \right) \mathrm { d } u$.
Hence find the exact area of the region enclosed by the curve, the $x$-axis and the lines $x = \frac { 2 } { 3 }$ and $x = 1$.
\hfill \mbox{\textit{OCR MEI C3 2009 Q9 [18]}}