| Exam Board | Edexcel |
|---|---|
| Module | C12 (Core Mathematics 1 & 2) |
| Year | 2017 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Tangent or Normal Bounded Area |
| Difficulty | Standard +0.3 This is a straightforward multi-part integration question requiring basic skills: finding intercepts by substitution, solving a quadratic equation, and computing area between a curve and horizontal line. All steps are routine C2 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(18\) | B1 | \(P(0,18)\) or \(P=18\) fine; do not allow \(P(18,0)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3\) | B1 | \(R(3,0)\) or \(R=3\) fine; do not allow \(R(0,3)\) or if they state \(3\) and \(-4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(f(x) =\) their '18'; implied by \(\frac{(x-3)^2(x+4)}{2} = 18\) | M1 | Can be implied by sight of \((x-3)^2(x+4) = 2 \times \text{their } 18\) |
| Attempts to multiply out \((x-3)^2(x+4)\) correctly; expect \((x^2 \pm 6x \pm 9)(x+4)\) multiplied to a cubic | dM1 | Accept working from elsewhere in question |
| \(\Rightarrow x^3 - 2x^2 - 15x + 36 = 36 \Rightarrow x^3 - 2x^2 - 15x = 0 \Rightarrow x^2 - 2x - 15 = 0\) | A1* | Reaches given answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States \(x = 5\) | B1 | |
| \(y = \frac{(5-3)^2(5+4)}{2} \Rightarrow (5, 18)\) | M1A1 | Substitutes their \(5\) into \(f(x)\); allow \(x=5, y=18\); must be seen in part (c) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\!\left(\frac{1}{2}x^3 - x^2 - \frac{15}{2}x + 18\right)dx = \frac{1}{8}x^4 - \frac{1}{3}x^3 - \frac{15}{4}x^2 + 18x\) | M1A1 | Integrating cubic; all powers raised by one |
| Uses \(5\) as upper limit (subtracts \(0\)) to obtain area | M1 | May appear as two integrals \(0\) to \(3\) then \(3\) to \(5\) |
| Area of rectangle \(= 90\) | B1 | i.e. \(18 \times 5\) |
| Area \(=\) Area of rectangle \(-\) area beneath curve \(= 90 - 32\frac{17}{24} = 57\frac{7}{24}\) \(\left(\frac{1375}{24}\right)\) | dM1 A1cso | Dependent on both previous M's; note \(-57\frac{7}{24}\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\!\left(-\frac{1}{2}x^3 + x^2 + \frac{15}{2}x\right)dx = -\frac{1}{8}x^4 + \frac{1}{3}x^3 + \frac{15}{4}x^2\) | M1A1 | Integrating \(\pm(18 - f(x))\); all powers raised by one |
| Uses \(5\) as upper limit (subtracts \(0\)) | M1 | |
| Implied by correct answer \(57\frac{7}{24}\) | B1 | No need to use calculator on incorrect functions (score B0) |
| Implied by subtraction in integration \(= 57\frac{7}{24}\) \(\left(\frac{1375}{24}\right)\) | dM1 A1cso | Dependent on both previous M's; \(-57\frac{7}{24}\) is A0 |
## Question 15:
### Part (a)(i):
| Answer | Mark | Guidance |
|---|---|---|
| $18$ | B1 | $P(0,18)$ or $P=18$ fine; do not allow $P(18,0)$ |
### Part (a)(ii):
| Answer | Mark | Guidance |
|---|---|---|
| $3$ | B1 | $R(3,0)$ or $R=3$ fine; do not allow $R(0,3)$ or if they state $3$ and $-4$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $f(x) =$ their '18'; implied by $\frac{(x-3)^2(x+4)}{2} = 18$ | M1 | Can be implied by sight of $(x-3)^2(x+4) = 2 \times \text{their } 18$ |
| Attempts to multiply out $(x-3)^2(x+4)$ correctly; expect $(x^2 \pm 6x \pm 9)(x+4)$ multiplied to a cubic | dM1 | Accept working from elsewhere in question |
| $\Rightarrow x^3 - 2x^2 - 15x + 36 = 36 \Rightarrow x^3 - 2x^2 - 15x = 0 \Rightarrow x^2 - 2x - 15 = 0$ | A1* | Reaches given answer with no errors |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States $x = 5$ | B1 | |
| $y = \frac{(5-3)^2(5+4)}{2} \Rightarrow (5, 18)$ | M1A1 | Substitutes their $5$ into $f(x)$; allow $x=5, y=18$; must be seen in part (c) |
### Part (d) — Method 1 (Curve and line separate):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\!\left(\frac{1}{2}x^3 - x^2 - \frac{15}{2}x + 18\right)dx = \frac{1}{8}x^4 - \frac{1}{3}x^3 - \frac{15}{4}x^2 + 18x$ | M1A1 | Integrating cubic; all powers raised by one |
| Uses $5$ as upper limit (subtracts $0$) to obtain area | M1 | May appear as two integrals $0$ to $3$ then $3$ to $5$ |
| Area of rectangle $= 90$ | B1 | i.e. $18 \times 5$ |
| Area $=$ Area of rectangle $-$ area beneath curve $= 90 - 32\frac{17}{24} = 57\frac{7}{24}$ $\left(\frac{1375}{24}\right)$ | dM1 A1cso | Dependent on both previous M's; note $-57\frac{7}{24}$ is A0 |
### Part (d) — Method 2 (Curve $-$ line or line $-$ curve):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\!\left(-\frac{1}{2}x^3 + x^2 + \frac{15}{2}x\right)dx = -\frac{1}{8}x^4 + \frac{1}{3}x^3 + \frac{15}{4}x^2$ | M1A1 | Integrating $\pm(18 - f(x))$; all powers raised by one |
| Uses $5$ as upper limit (subtracts $0$) | M1 | |
| Implied by correct answer $57\frac{7}{24}$ | B1 | No need to use calculator on incorrect functions (score B0) |
| Implied by subtraction in integration $= 57\frac{7}{24}$ $\left(\frac{1375}{24}\right)$ | dM1 A1cso | Dependent on both previous M's; $-57\frac{7}{24}$ is A0 |15.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb1becd5-96c1-426d-9b85-4bbc4a61af27-42_695_1450_251_246}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of part of the graph $y = \mathrm { f } ( x )$, where
$$f ( x ) = \frac { ( x - 3 ) ^ { 2 } ( x + 4 ) } { 2 } , \quad x \in \mathbb { R }$$
The graph cuts the $y$-axis at the point $P$ and meets the positive $x$-axis at the point $R$, as shown in Figure 5.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item State the $y$ coordinate of $P$.
\item State the $x$ coordinate of $R$.
The line segment $P Q$ is parallel to the $x$-axis. Point $Q$ lies on $y = \mathrm { f } ( x ) , x > 0$
\end{enumerate}\item Use algebra to show that the $x$ coordinate of $Q$ satisfies the equation
$$x ^ { 2 } - 2 x - 15 = 0$$
\item Use part (b) to find the coordinates of $Q$.
The region $S$, shown shaded in Figure 5, is bounded by the curve $y = \mathrm { f } ( x )$ and the line segment $P Q$.
\item Use calculus to find the exact area of $S$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel C12 2017 Q15 [14]}}