| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2014 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a standard multi-part integration question requiring finding intercepts, tangent equations, intersection points, and area between curve and line. All techniques are routine C1/C2 material (differentiation, solving cubics by factoring, definite integration) with no novel problem-solving required. Slightly easier than average due to straightforward algebraic manipulation and the 'show that' scaffolding in parts (a) and (b). |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.07i Differentiate x^n: for rational n and sums1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sub \(x=3\): \(y = 9 - 9 = 0\) | B1* | Must show at least one intermediate line. Can sub \(y=0\) to get \(x=3\) instead |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = x^2 - \frac{1}{3}x^3 \Rightarrow \dfrac{dy}{dx} = 2x - x^2\) | M1A1 | M1 for any term correct. A1 fully correct (unsimplified acceptable) |
| Sub \(x=3\): \(\dfrac{dy}{dx} = 6 - 9 = -3\) | dM1 | Dependent on previous M |
| Tangent: \(-3 = \dfrac{y-0}{x-3} \Rightarrow y = -3x + 9\) | ddM1, A1* | Dependent on both M marks. cso, given answer, all aspects must be correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x^2 - \frac{1}{3}x^3 = -3x + 9 \Rightarrow x^3 - 3x^2 - 9x + 27 = 0\) | M1, A1 | Setting equations equal. Must use algebra |
| \((x-3)^2(x+3) = 0 \Rightarrow x = -3\) | dM1, A1 | Solving by factorisation for another value. cso \(x=-3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\displaystyle\int x^2 - \frac{1}{3}x^3\,dx = \left[\frac{1}{3}x^3 - \frac{1}{12}x^4\right]\) | M1A1 | M1 for one term correct. A1 fully correct simplified expression |
| Area of triangle \(= \frac{1}{2}\times(3 - x_B)\times y_B = 54\) | M1 | Correct method for triangle area |
| Shaded area \(= {'}54{'} - \left[\left(\frac{1}{3}(3)^3 - \frac{1}{12}(3)^4\right) - \left(\frac{1}{3}(-3)^3 - \frac{1}{12}(-3)^4\right)\right] = 36\) | dM1, A1 | Subtract area under curve from triangle, correct limits. cso 36 |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sub $x=3$: $y = 9 - 9 = 0$ | B1* | Must show at least one intermediate line. Can sub $y=0$ to get $x=3$ instead |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = x^2 - \frac{1}{3}x^3 \Rightarrow \dfrac{dy}{dx} = 2x - x^2$ | M1A1 | M1 for any term correct. A1 fully correct (unsimplified acceptable) |
| Sub $x=3$: $\dfrac{dy}{dx} = 6 - 9 = -3$ | dM1 | Dependent on previous M |
| Tangent: $-3 = \dfrac{y-0}{x-3} \Rightarrow y = -3x + 9$ | ddM1, A1* | Dependent on both M marks. cso, given answer, all aspects must be correct |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - \frac{1}{3}x^3 = -3x + 9 \Rightarrow x^3 - 3x^2 - 9x + 27 = 0$ | M1, A1 | Setting equations equal. Must use algebra |
| $(x-3)^2(x+3) = 0 \Rightarrow x = -3$ | dM1, A1 | Solving by factorisation for another value. cso $x=-3$ |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\displaystyle\int x^2 - \frac{1}{3}x^3\,dx = \left[\frac{1}{3}x^3 - \frac{1}{12}x^4\right]$ | M1A1 | M1 for one term correct. A1 fully correct simplified expression |
| Area of triangle $= \frac{1}{2}\times(3 - x_B)\times y_B = 54$ | M1 | Correct method for triangle area |
| Shaded area $= {'}54{'} - \left[\left(\frac{1}{3}(3)^3 - \frac{1}{12}(3)^4\right) - \left(\frac{1}{3}(-3)^3 - \frac{1}{12}(-3)^4\right)\right] = 36$ | dM1, A1 | Subtract area under curve from triangle, correct limits. cso 36 |12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b85872d4-00b2-499b-9765-f7559d3de66a-19_1011_1349_237_310}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of part of the curve $C$ with equation $y = x ^ { 2 } - \frac { 1 } { 3 } x ^ { 3 } C$ touches the $x$-axis at the origin and cuts the $x$-axis at the point $A$.
\begin{enumerate}[label=(\alph*)]
\item Show that the coordinates of $A$ are $( 3,0 )$.
\item Show that the equation of the tangent to $C$ at the point $A$ is $y = - 3 x + 9$
The tangent to $C$ at $A$ meets $C$ again at the point $B$, as shown in Figure 5.
\item Use algebra to find the $x$ coordinate of $B$.
The region $R$, shown shaded in Figure 5, is bounded by the curve $C$ and the tangent to $C$ at $A$.
\item Find, by using calculus, the area of region $R$.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2014 Q12 [15]}}