| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points and nature |
| Difficulty | Standard +0.3 This is a structured multi-part question requiring product rule differentiation, finding stationary points, integration, and algebraic manipulation. While it covers several techniques, each part follows standard procedures with clear guidance. The substitution in part (iii) is routine algebra, and part (v) tests understanding of transformations rather than requiring novel insight. Slightly easier than average due to its scaffolded nature. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{dy}{dx} = 5x \cdot 6(2x-1)^2 + 5(2x-1)^3\) | M1A1 | |
| \(= 5(2x-1)^2(8x-1)\) | M1 | |
| Where the gradient is zero, \(x = \frac{1}{2}\) (double root, where the curve touches the \(x\)-axis) or \(x = \frac{1}{8}\) (at S) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Area has magnitude \(\left\ | \int_0^{\frac{1}{2}} 5x(2x-1)^3\,dx\right\ | \) |
| \(= \left\ | \int_0^{\frac{1}{2}}(40x^4 - 60x^3 + 30x^2 - 5x)\,dx\right\ | \) |
| \(= \left\ | \left[8x^5 - 15x^4 + 10x^3 - \frac{5x^2}{2}\right]_0^{\frac{1}{2}}\right\ | \) |
| \(= \left\ | \frac{1}{4} - \frac{15}{16} + \frac{10}{8} - \frac{5}{8}\right\ | = \left\ |
| So area is \(\frac{1}{16}\) units\(^2\) | A1 | |
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(f(x+0.5) = 5(x+0.5)(2[x+0.5]-1)^2\) | M1 | |
| \(= 40x^3(x+0.5)\) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int_{-\frac{1}{2}}^{0} 40x^3(x+0.5)\,dx\) | M1 | Multiply out |
| \(= \int_{-\frac{1}{2}}^{0}(40x^4 + 20x^3)\,dx\) | A1 | Terms |
| \(= \left[8x^5 + 5x^4\right]_{-\frac{1}{2}}^{0} = 0 - \left(-\frac{8}{32}+\frac{5}{16}\right) = -\frac{1}{16}\) | A1, A1 | Both terms |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Sketch showing \(y=f(x)\) and \(y=f(x+0.5)\) as translation | B1 | |
| The area representing the one integral is a translation of that representing the other, so their values are equal | E1 | |
| Total: 2 |
## Question 8:
### Part (i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{dy}{dx} = 5x \cdot 6(2x-1)^2 + 5(2x-1)^3$ | M1A1 | |
| $= 5(2x-1)^2(8x-1)$ | M1 | |
| Where the gradient is zero, $x = \frac{1}{2}$ (double root, where the curve touches the $x$-axis) or $x = \frac{1}{8}$ (at S) | A1 | |
| | **Total: 4** | |
### Part (ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Area has magnitude $\left\|\int_0^{\frac{1}{2}} 5x(2x-1)^3\,dx\right\|$ | B1 | (or at end if $-$ve sign is explained) |
| $= \left\|\int_0^{\frac{1}{2}}(40x^4 - 60x^3 + 30x^2 - 5x)\,dx\right\|$ | M1A1 | Expand or integrate by parts |
| $= \left\|\left[8x^5 - 15x^4 + 10x^3 - \frac{5x^2}{2}\right]_0^{\frac{1}{2}}\right\|$ | A2 | A1 for some terms right |
| $= \left\|\frac{1}{4} - \frac{15}{16} + \frac{10}{8} - \frac{5}{8}\right\| = \left\|-\frac{1}{16}\right\|$ | | |
| So area is $\frac{1}{16}$ units$^2$ | A1 | |
| | **Total: 6** | |
### Part (iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $f(x+0.5) = 5(x+0.5)(2[x+0.5]-1)^2$ | M1 | |
| $= 40x^3(x+0.5)$ | A1 | |
| | **Total: 2** | |
### Part (iv):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int_{-\frac{1}{2}}^{0} 40x^3(x+0.5)\,dx$ | M1 | Multiply out |
| $= \int_{-\frac{1}{2}}^{0}(40x^4 + 20x^3)\,dx$ | A1 | Terms |
| $= \left[8x^5 + 5x^4\right]_{-\frac{1}{2}}^{0} = 0 - \left(-\frac{8}{32}+\frac{5}{16}\right) = -\frac{1}{16}$ | A1, A1 | Both terms |
| | **Total: 4** | |
### Part (v):
| Answer | Mark | Guidance |
|--------|------|----------|
| Sketch showing $y=f(x)$ and $y=f(x+0.5)$ as translation | B1 | |
| The area representing the one integral is a translation of that representing the other, so their values are equal | E1 | |
| | **Total: 2** | |
---8 Fig. 8 shows part of the graph of the function $y = 5 x ( 2 x - 1 ) ^ { 3 }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3853d1e7-ae1f-4eca-93c7-96f03b6d31c3-4_508_803_450_703}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$ and hence find the $x$-coordinate of S , the turning point of the curve.\\
(ii) Find the area of the shaded region enclosed between the curve and the $x$-axis.\\
(iii) Given that $\mathrm { f } ( x ) = 5 x ( 2 x - 1 ) ^ { 3 }$, show that $\mathrm { f } ( x + 0.5 ) = 40 x ^ { 3 } ( x + 0.5 )$.\\
(iv) Find $\int _ { - \frac { 1 } { 2 } } ^ { 0 } 40 x ^ { 3 } ( x + 0.5 ) \mathrm { d } x$.\\
(v) Explain, with the aid of a sketch, the connection between your answer to parts (ii) and (iv).
\hfill \mbox{\textit{OCR MEI C3 Q8 [18]}}