| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Standard +0.3 This is a standard two-part integration question requiring finding intersection points by solving a quadratic equation, then computing area between curves using definite integration. While it involves multiple steps (solving equations, setting up and evaluating integrals), all techniques are routine for P1 level with no novel problem-solving required. Slightly above average difficulty due to the fractional power requiring careful algebraic manipulation. |
| Spec | 1.02q Use intersection points: of graphs to solve equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((3x-2)^{\frac{1}{2}} = \frac{1}{2}x+1 \Rightarrow 3x-2 = \left(\frac{1}{2}x+1\right)^2 = \frac{1}{4}x^2+x+1\) | M1 | Equating curve and line, attempt to square; \(\frac{1}{4}x^2+1\) M0 |
| \(\Rightarrow \frac{1}{4}x^2 - 2x + 3 [=0] \Rightarrow [x^2-8x+12=0] \Rightarrow (x-6)(x-2)[=0]\) | M1 | Forming and solving a 3TQ by factorisation, formula or completing the square |
| \((2, 2)\) and \((6, 4)\) | A1 A1 | A1 for each point, or A1 A0 for two correct \(x\)-values. If M0 for solving, SC B2 possible: B1 for each point or B1 B0 for two correct \(x\)-values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\text{Area} = \pm\int_{[2]}^{[6]}\left((3x-2)^{\frac{1}{2}} - \left(\frac{1}{2}x+1\right)\right)[dx]\) | *M1 | For intention to integrate and subtract (M0 if squared) |
| \(\pm\left[\frac{2}{9}(3x-2)^{\frac{3}{2}} - \left(\frac{1}{4}x^2+x\right)\right]_2^6\) | B1 B1 | B1 for each bracket integrated correctly (in any form) |
| \(\pm\left(\left[\frac{2}{9}(16)^{\frac{3}{2}} - \left(\frac{1}{4}\times36+6\right)\right] - \left[\frac{2}{9}(4)^{\frac{3}{2}} - \left(\frac{1}{4}\times4+2\right)\right]\right)\) | DM1 | \(\pm(F(\textit{their } 6) - F(\textit{their } 2))\) with *their* integral. Allow 1 sign error |
| \(\frac{4}{9}\) | A1 | AWRT 0.444. SC1 B1 for \(\frac{4}{9}\) if *M1 B1 B1 DM0*. SC2 B1 for \(\frac{4}{9}\) if *M1 B0 B0 DM0*, provided limits stated |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3x-2)^{\frac{1}{2}} = \frac{1}{2}x+1 \Rightarrow 3x-2 = \left(\frac{1}{2}x+1\right)^2 = \frac{1}{4}x^2+x+1$ | M1 | Equating curve and line, attempt to square; $\frac{1}{4}x^2+1$ M0 |
| $\Rightarrow \frac{1}{4}x^2 - 2x + 3 [=0] \Rightarrow [x^2-8x+12=0] \Rightarrow (x-6)(x-2)[=0]$ | M1 | Forming and solving a 3TQ by factorisation, formula or completing the square |
| $(2, 2)$ and $(6, 4)$ | A1 A1 | A1 for each point, or A1 A0 for two correct $x$-values. If M0 for solving, **SC B2** possible: B1 for each point or B1 B0 for two correct $x$-values |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\text{Area} = \pm\int_{[2]}^{[6]}\left((3x-2)^{\frac{1}{2}} - \left(\frac{1}{2}x+1\right)\right)[dx]$ | *M1 | For intention to integrate and subtract (M0 if squared) |
| $\pm\left[\frac{2}{9}(3x-2)^{\frac{3}{2}} - \left(\frac{1}{4}x^2+x\right)\right]_2^6$ | B1 B1 | B1 for each bracket integrated correctly (in any form) |
| $\pm\left(\left[\frac{2}{9}(16)^{\frac{3}{2}} - \left(\frac{1}{4}\times36+6\right)\right] - \left[\frac{2}{9}(4)^{\frac{3}{2}} - \left(\frac{1}{4}\times4+2\right)\right]\right)$ | DM1 | $\pm(F(\textit{their } 6) - F(\textit{their } 2))$ with *their* integral. Allow 1 sign error |
| $\frac{4}{9}$ | A1 | AWRT 0.444. **SC1 B1** for $\frac{4}{9}$ if *M1 B1 B1 DM0*. **SC2 B1** for $\frac{4}{9}$ if *M1 B0 B0 DM0*, provided limits stated |
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\includegraphics[max width=\textwidth, alt={}, center]{574b96b2-62f2-41b3-a178-8e68e16429ff-12_631_1031_267_534}
The diagram shows the curve with equation $y = ( 3 x - 2 ) ^ { \frac { 1 } { 2 } }$ and the line $y = \frac { 1 } { 2 } x + 1$. The curve and the line intersect at points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of $A$ and $B$.
\item Hence find the area of the region enclosed between the curve and the line.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q7 [9]}}