| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas by integration |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward multi-part question requiring standard techniques: completing the square or differentiation to find the maximum point A, then using the gradient condition to find B, followed by routine integration and area calculation. While it has multiple steps (7+4=11 marks), each step uses basic AS-level methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = -x^2 + 8x - 10\) | ||
| (i) \(\frac{dy}{dx} = -2x + 8\) | B1 | co |
| \(= 0\) when \(x = 4\), \(A\) is \((4, 6)\) | M1 A1 | Sets to 0 and attempt to solve for x. co. Correct form of equation. |
| Equation of AB is \(y - 6 = 2(x - 4)\) | M1 | Eliminates x or y completely |
| Sim eqns with eqn of curve | M1 | Method for quadratic eqn = 0. co (Must not be guessed from diagram) |
| \(\to x^2 - 6x + 8 = 0\) or \(y^2 - 8y + 12 = 0\) | A1 | |
| \(\to B(2, 2)\) | A1 | [7] |
| (ii) \(\int -x^2 + 8x - 10 dx = -\frac{x^3}{3} + 4x^2 - 10x\) | B2,1 | 3 terms, loses 1 for each error |
| Uses his x limits 2 to 4 | M1 | Uses x limits correctly – allow ± co – allow ± (2 must have been correctly found, not guessed) |
| \(\to 9\frac{1}{3}\) | A1 | [4] |
$y = -x^2 + 8x - 10$ | |
(i) $\frac{dy}{dx} = -2x + 8$ | B1 | co
$= 0$ when $x = 4$, $A$ is $(4, 6)$ | M1 A1 | Sets to 0 and attempt to solve for x. co. Correct form of equation.
Equation of AB is $y - 6 = 2(x - 4)$ | M1 | Eliminates x or y completely
Sim eqns with eqn of curve | M1 | Method for quadratic eqn = 0. co (Must not be guessed from diagram)
$\to x^2 - 6x + 8 = 0$ or $y^2 - 8y + 12 = 0$ | A1 |
$\to B(2, 2)$ | A1 | [7]
(ii) $\int -x^2 + 8x - 10 dx = -\frac{x^3}{3} + 4x^2 - 10x$ | B2,1 | 3 terms, loses 1 for each error
Uses his x limits 2 to 4 | M1 | Uses x limits correctly – allow ± co – allow ± (2 must have been correctly found, not guessed)
$\to 9\frac{1}{3}$ | A1 | [4]\includegraphics{figure_9}
The diagram shows part of the curve $y = -x^2 + 8x - 10$ which passes through the points $A$ and $B$. The curve has a maximum point at $A$ and the gradient of the line $BA$ is $2$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of $A$ and $B$. [7]
\item Find $\int y \, dx$ and hence evaluate the area of the shaded region. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2012 Q9 [11]}}