| Exam Board | Edexcel |
|---|---|
| Module | C2 (Core Mathematics 2) |
| Year | 2005 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Standard +0.3 This is a standard C2 integration question requiring finding intersection points, setting up an area integral between a curve and line, and a straightforward differentiation to show increasing behavior. The integration involves basic polynomial and reciprocal terms with no complex algebraic manipulation, making it slightly easier than average. |
| Spec | 1.07o Increasing/decreasing: functions using sign of dy/dx1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int(2x + 8x^{-2} - 5)\,dx = x^2 + \frac{8x^{-1}}{-1} - 5x\) | M1, A1, A1 | One term wrong M1 A1 A0; two terms wrong M1 A0 A0 |
| \(\left[x^2 + \frac{8x^{-1}}{-1} - 5x\right]_1^4 = (16-2-20)-(1-8-5)\) \((= 6)\) | M1 | Substituting limits 4 and 1, subtracting correct way |
| \(x=1\): \(y=5\) and \(x=4\): \(y=3.5\) | B1 | |
| Area of trapezium \(= \frac{1}{2}(5+3.5)(4-1)\) \((= 12.75)\) | M1 | |
| Shaded area \(= 12.75 - 6 = 6.75\) | M1, A1 | Subtract either way round; (8 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = 2 - 16x^{-3}\) | M1, A1 | |
| Increasing where \(\frac{dy}{dx} > 0\); for \(x > 2\), \(\frac{16}{x^3} < 2\), \(\therefore \frac{dy}{dx} > 0\) (Allow \(\geq\)) | dM1, A1 | (4 marks) Alternative: M1 show \(x=2\) is minimum using e.g. 2nd derivative; A1 conclusion showing understanding of "increasing" with accurate working |
## Question 10:
**(a)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int(2x + 8x^{-2} - 5)\,dx = x^2 + \frac{8x^{-1}}{-1} - 5x$ | M1, A1, A1 | One term wrong M1 A1 A0; two terms wrong M1 A0 A0 |
| $\left[x^2 + \frac{8x^{-1}}{-1} - 5x\right]_1^4 = (16-2-20)-(1-8-5)$ $(= 6)$ | M1 | Substituting limits 4 and 1, subtracting correct way |
| $x=1$: $y=5$ and $x=4$: $y=3.5$ | B1 | |
| Area of trapezium $= \frac{1}{2}(5+3.5)(4-1)$ $(= 12.75)$ | M1 | |
| Shaded area $= 12.75 - 6 = 6.75$ | M1, A1 | Subtract either way round; (8 marks) |
**(b)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = 2 - 16x^{-3}$ | M1, A1 | |
| Increasing where $\frac{dy}{dx} > 0$; for $x > 2$, $\frac{16}{x^3} < 2$, $\therefore \frac{dy}{dx} > 0$ (Allow $\geq$) | dM1, A1 | (4 marks) Alternative: M1 show $x=2$ is minimum using e.g. 2nd derivative; A1 conclusion showing understanding of "increasing" with accurate working |10.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{135bc546-9274-4862-b2e7-c11e9c8e2c4f-13_1018_1029_287_445}
\end{center}
\end{figure}
Figure 1 shows part of the curve $C$ with equation $y = 2 x + \frac { 8 } { x ^ { 2 } } - 5 , x > 0$.\\
The points $P$ and $Q$ lie on $C$ and have $x$-coordinates 1 and 4 respectively. The region $R$, shaded in Figure 1, is bounded by $C$ and the straight line joining $P$ and $Q$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact area of $R$.
\item Use calculus to show that $y$ is increasing for $x > 2$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C2 2005 Q10 [12]}}