| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Parametric or Inverse Function Area |
| Difficulty | Standard +0.8 This question requires finding intersection points by solving a quadratic-linear system, then computing area between curves using integration. The parametric nature (y² = 2x - 1) requires careful handling of square roots and choosing correct branches. Students must set up ∫(x_line - x_curve)dy rather than the standard dx form, requiring conceptual understanding beyond routine integration. The algebraic manipulation and multi-step reasoning elevate this above average difficulty. |
| Spec | 1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y^2 = 3y \Rightarrow y(y-3) = 0 \Rightarrow y = 3\) (or 0) | M1 | OR form equation in \(x\) and attempt solution |
| \(x = \frac{1}{2}\) or 5 (\(\Rightarrow a = 5\)) | AG A1 | OR sub \(x=5\) each eq (M1) \(\to y = 3\) (twice) (A1). \((5,3)\) subst only once scores 0/2 |
| (ii) \(\Delta = \frac{(2x-1)^2}{3}\) \([+2]\), \([\frac{2}{3} \times \frac{x^2}{2} - \frac{4}{3}]\) | B1B1B1 | Or \(\frac{1}{2}(5-\frac{1}{2}) \times 3\) |
| \([\frac{27}{3} - 0]\), \([\frac{25}{3} - \frac{5}{12} - (\frac{1}{12} - \frac{1}{6})]\) | M1 | Apply limits \(\frac{1}{2}\) and 5 for, at least, curve |
| Subtract areas at some stage | M1 | Dependent on some integration |
| \(\frac{9}{4}\) oe | A1 | cao. 9/4 with no working scores 0/6, but 9 − 27/4 = 9/4 scores 1/6 (M1 subtraction) |
**(i)** $y^2 = 3y \Rightarrow y(y-3) = 0 \Rightarrow y = 3$ (or 0) | M1 | OR form equation in $x$ and attempt solution
$x = \frac{1}{2}$ or 5 ($\Rightarrow a = 5$) | AG A1 | OR sub $x=5$ each eq (M1) $\to y = 3$ (twice) (A1). $(5,3)$ subst only once scores 0/2 | [2]
**(ii)** $\Delta = \frac{(2x-1)^2}{3}$ $[+2]$, $[\frac{2}{3} \times \frac{x^2}{2} - \frac{4}{3}]$ | B1B1B1 | Or $\frac{1}{2}(5-\frac{1}{2}) \times 3$
$[\frac{27}{3} - 0]$, $[\frac{25}{3} - \frac{5}{12} - (\frac{1}{12} - \frac{1}{6})]$ | M1 | Apply limits $\frac{1}{2}$ and 5 for, at least, curve
Subtract areas at some stage | M1 | Dependent on some integration
$\frac{9}{4}$ oe | A1 | cao. 9/4 with no working scores 0/6, but 9 − 27/4 = 9/4 scores 1/6 (M1 subtraction) | [6]8\\
\includegraphics[max width=\textwidth, alt={}, center]{e69332d0-2e45-4a86-a1f9-5d83bca1ad9b-3_629_853_251_644}
The diagram shows the curve $y ^ { 2 } = 2 x - 1$ and the straight line $3 y = 2 x - 1$. The curve and straight line intersect at $x = \frac { 1 } { 2 }$ and $x = a$, where $a$ is a constant.\\
(i) Show that $a = 5$.\\
(ii) Find, showing all necessary working, the area of the shaded region.
\hfill \mbox{\textit{CAIE P1 2012 Q8 [8]}}