| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2016 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Tangent meets curve/axis — further geometry |
| Difficulty | Standard +0.3 This is a standard multi-part differentiation and integration question requiring routine techniques: expand and differentiate a cubic, find tangent equation using point-gradient form, then integrate to find area. All steps are textbook procedures with no novel insight required, making it slightly easier than average for A-level. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(y = x(x-1)(x-2) = x^3 - 3x^2 + 2x \Rightarrow \frac{dy}{dx} = 3x^2 - 6x + 2\) | M1, A1 | Attempts to multiply out \(x(x-1)(x-2)\) and differentiate each term; look for cubic being differentiated to quadratic; accept exact equivalents such as \(\frac{dy}{dx} = 3x^2 - 4x - 2x + 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| When \(x = \frac{1}{2}\), \(y = \frac{3}{8}\) | B1 | oe |
| Sub \(x = \frac{1}{2}\) into \(\frac{dy}{dx}\bigg | _{x=\frac{1}{2}} = \frac{3}{4} - 3 + 2 = -\frac{1}{4}\) | M1 |
| Uses gradient and \(\left(\frac{1}{2}, \frac{3}{8}\right) \Rightarrow y - \frac{3}{8} = -\frac{1}{4}\left(x - \frac{1}{2}\right) \Rightarrow 4y + x = 2\) | M1, A1 | Uses their gradient, \(x = \frac{1}{2}\) and \(y = \frac{3}{8}\) to find equation of tangent; \(k(4y + x = 2)\) where \(k\) is an integer. Accept \(a=1, b=4, c=2\) or multiples thereof. Finding normal is M0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int y\,dx = \frac{x^4}{4} - 3\frac{x^3}{3} + 2\frac{x^2}{2}(+c)\) | M1 A1 | Attempt to multiply out and integrate to quartic form; correct and unsimplified |
| Area under curve between \(x=\frac{1}{2}\) and \(x=1\): \(\left[\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right]_{\frac{1}{2}}^{1} = \left(\frac{7}{64}\right)\) | M1 | Uses limits \(x=\frac{1}{2}\) and \(x=1\) with integrated answer |
| Area under line between \(x=\frac{1}{2}\) and \(x=2\): \(= \frac{1}{2}\times\left(2-\frac{1}{2}\right)\times\frac{3}{8} = \left(\frac{9}{32}\right)\) | M1 | Correct method for area under line; triangle formula or correct integration |
| Area of \(R = \frac{1}{2}\times\left(2-\frac{1}{2}\right)\times\frac{3}{8} - \left[\left(\frac{1}{4}-1+1\right)-\left(\frac{1}{64}-\frac{1}{8}+\frac{1}{4}\right)\right] = \frac{9}{32}-\frac{7}{64} = \frac{11}{64}\) | dM1 A1 | Correct combination of areas with correct limits; dependent on all 3 M marks; answer \(\frac{11}{64}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int\left(\frac{1}{2}-\frac{1}{4}x\right)-(x(x-1)(x-2))\,dx = \left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)\) | M1 A1 | Attempts difference, multiplies out and integrates to quartic; correct unsimplified or simplified form \(\left(\frac{1}{2}x - \frac{9}{8}x^2 + x^3 - \frac{1}{4}x^4\right)\); do not follow through on tangent line |
| Area between line and curve from \(x=\frac{1}{2}\) to \(x=1\): \(= \left(\frac{3}{64}\right)\) | M1 | Uses limits \(x=\frac{1}{2}\) and \(x=1\) |
| Area under line between \(x=1\) and \(x=2\): \(= \frac{1}{2}\times(2-1)\times\frac{1}{4} = \frac{1}{8}\) | M1 | Correct triangle or correct integration with correct limits |
| Area of \(R = \frac{1}{8}+\left[\left(\frac{1}{8}\right)-\left(\frac{5}{64}\right)\right] = \frac{1}{8}+\frac{3}{64} = \frac{11}{64}\) | dM1 A1 | Correct combination, areas added; dependent on all 3 M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\int\left(\frac{1}{2}-\frac{1}{4}x\right)-(x(x-1)(x-2))\,dx = \left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)\) | M1 A1 | As ALT 1 |
| Total area between line and curve \(= \left[\left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)\right]_{\frac{1}{2}}^{2} = \left(\frac{27}{64}\right)\) | M1 | Uses limits \(x=\frac{1}{2}\) and \(x=2\) |
| Area under curve \(= \left[\frac{x^4}{4}-3\frac{x^3}{3}+x^2\right]_{1}^{2} = \left(-\frac{1}{4}\right)\) | M1 | Evaluates \(\pm\int_{1}^{2}x(x-1)(x-2)\,dx = \pm\left[\frac{x^4}{4}-x^3+x^2\right]_{1}^{2}\); limits and integration both correct |
| Area of \(R = \frac{27}{64}-\frac{1}{4} = \frac{11}{64}\) | dM1 A1 | Correct combination; equivalent to \(\left(\frac{27}{64}\right)+-\frac{1}{4}\) or \(\left(\frac{27}{64}\right)-\left |
## Question 16:
### Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $y = x(x-1)(x-2) = x^3 - 3x^2 + 2x \Rightarrow \frac{dy}{dx} = 3x^2 - 6x + 2$ | M1, A1 | Attempts to multiply out $x(x-1)(x-2)$ and differentiate each term; look for cubic being differentiated to quadratic; accept exact equivalents such as $\frac{dy}{dx} = 3x^2 - 4x - 2x + 2$ |
### Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| When $x = \frac{1}{2}$, $y = \frac{3}{8}$ | B1 | oe |
| Sub $x = \frac{1}{2}$ into $\frac{dy}{dx}\bigg|_{x=\frac{1}{2}} = \frac{3}{4} - 3 + 2 = -\frac{1}{4}$ | M1 | Substitute $x = \frac{1}{2}$ into their $\frac{dy}{dx}$ and either states/uses this as gradient, not the $y$-coordinate |
| Uses gradient and $\left(\frac{1}{2}, \frac{3}{8}\right) \Rightarrow y - \frac{3}{8} = -\frac{1}{4}\left(x - \frac{1}{2}\right) \Rightarrow 4y + x = 2$ | M1, A1 | Uses their gradient, $x = \frac{1}{2}$ and $y = \frac{3}{8}$ to find equation of tangent; $k(4y + x = 2)$ where $k$ is an integer. Accept $a=1, b=4, c=2$ or multiples thereof. Finding normal is M0 |
## Question (c):
---
**Main Scheme** (Area under curve from $x = \frac{1}{2}$ to $x = 1$):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int y\,dx = \frac{x^4}{4} - 3\frac{x^3}{3} + 2\frac{x^2}{2}(+c)$ | M1 A1 | Attempt to multiply out and integrate to quartic form; correct and unsimplified |
| Area under curve between $x=\frac{1}{2}$ and $x=1$: $\left[\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right]_{\frac{1}{2}}^{1} = \left(\frac{7}{64}\right)$ | M1 | Uses limits $x=\frac{1}{2}$ and $x=1$ with integrated answer |
| Area under line between $x=\frac{1}{2}$ and $x=2$: $= \frac{1}{2}\times\left(2-\frac{1}{2}\right)\times\frac{3}{8} = \left(\frac{9}{32}\right)$ | M1 | Correct method for area under line; triangle formula or correct integration |
| Area of $R = \frac{1}{2}\times\left(2-\frac{1}{2}\right)\times\frac{3}{8} - \left[\left(\frac{1}{4}-1+1\right)-\left(\frac{1}{64}-\frac{1}{8}+\frac{1}{4}\right)\right] = \frac{9}{32}-\frac{7}{64} = \frac{11}{64}$ | dM1 A1 | Correct combination of areas with correct limits; dependent on all 3 M marks; answer $\frac{11}{64}$ |
---
**ALT 1** (Area between line and curve, $x=\frac{1}{2}$ to $x=1$):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int\left(\frac{1}{2}-\frac{1}{4}x\right)-(x(x-1)(x-2))\,dx = \left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)$ | M1 A1 | Attempts difference, multiplies out and integrates to quartic; correct unsimplified or simplified form $\left(\frac{1}{2}x - \frac{9}{8}x^2 + x^3 - \frac{1}{4}x^4\right)$; do not follow through on tangent line |
| Area between line and curve from $x=\frac{1}{2}$ to $x=1$: $= \left(\frac{3}{64}\right)$ | M1 | Uses limits $x=\frac{1}{2}$ and $x=1$ |
| Area under line between $x=1$ and $x=2$: $= \frac{1}{2}\times(2-1)\times\frac{1}{4} = \frac{1}{8}$ | M1 | Correct triangle or correct integration with correct limits |
| Area of $R = \frac{1}{8}+\left[\left(\frac{1}{8}\right)-\left(\frac{5}{64}\right)\right] = \frac{1}{8}+\frac{3}{64} = \frac{11}{64}$ | dM1 A1 | Correct combination, areas added; dependent on all 3 M marks |
---
**ALT 2** (Area between line and curve, $x=\frac{1}{2}$ to $x=2$):
| Working | Mark | Guidance |
|---------|------|----------|
| $\int\left(\frac{1}{2}-\frac{1}{4}x\right)-(x(x-1)(x-2))\,dx = \left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)$ | M1 A1 | As ALT 1 |
| Total area between line and curve $= \left[\left(\frac{1}{2}x-\frac{1}{8}x^2\right)-\left(\frac{x^4}{4}-3\frac{x^3}{3}+2\frac{x^2}{2}\right)\right]_{\frac{1}{2}}^{2} = \left(\frac{27}{64}\right)$ | M1 | Uses limits $x=\frac{1}{2}$ and $x=2$ |
| Area under curve $= \left[\frac{x^4}{4}-3\frac{x^3}{3}+x^2\right]_{1}^{2} = \left(-\frac{1}{4}\right)$ | M1 | Evaluates $\pm\int_{1}^{2}x(x-1)(x-2)\,dx = \pm\left[\frac{x^4}{4}-x^3+x^2\right]_{1}^{2}$; limits and integration both correct |
| Area of $R = \frac{27}{64}-\frac{1}{4} = \frac{11}{64}$ | dM1 A1 | Correct combination; equivalent to $\left(\frac{27}{64}\right)+-\frac{1}{4}$ or $\left(\frac{27}{64}\right)-\left|-\frac{1}{4}\right|$; dependent on all 3 M marks |16.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{aa75f1c1-ee97-4fee-af98-957e6a3fbba1-25_739_1308_278_328}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
Figure 6 shows a sketch of part of the curve $C$ with equation
$$y = x ( x - 1 ) ( x - 2 )$$
The point $P$ lies on $C$ and has $x$ coordinate $\frac { 1 } { 2 }$\\
The line $l$, as shown on Figure 6, is the tangent to $C$ at $P$.
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$
\item Use part (a) to find an equation for $l$ in the form $a x + b y = c$, where $a$, $b$ and $c$ are integers.
The finite region $R$, shown shaded in Figure 6, is bounded by the line $l$, the curve $C$ and the $x$-axis.
The line $l$ meets the curve again at the point $( 2,0 )$
\item Use integration to find the exact area of the shaded region $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2016 Q16 [12]}}