| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Area Using Substitution or Rearrangement |
| Difficulty | Standard +0.3 This is a straightforward area-between-curves problem with a helpful substitution provided. Part (a) requires solving a quartic equation that factors nicely (2u^4 = u^2 + 1 leads to a quadratic in u^2), and part (b) involves standard integration with the given substitution. The substitution is explicitly given, making this slightly easier than average but still requiring competent algebraic manipulation and integration technique. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u = 2x-3\) leading to \(2u^4 = u^2 + 1\) leading to \(2u^4 - u^2 - 1[=0]\) | B1 | |
| \((2u^2+1)(u^2-1)[=0]\) | M1 | Factors or formula or completing square must be shown |
| \(u = \pm 1\) leading to \(2x-3 = \pm 1\) leading to \(x = 1\) or \(2\) | A1 | |
| \((1, 2),\ (2, 2)\) | A1 | Special case: If B1 M0 scored then SC B2 for correct coordinates or SC B1 for correct \(x\) values only. Special case: \(2(2x-3)^4 = (2x-3)^2 + 1\), \(32x^4 - 192x^3 + 428x^2 - 420x + 152 = 0\), \(x = 1, 2\) finding both from correct quartic SC B1, \((1,2),(2,2)\) SC DB1. Trial and improvement without quartic: both \(x\) correct B1, both coordinates correct B2. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left\{\frac{(2x-3)^3}{3\times 2} + x\right\} [-] \left\{\frac{2(2x-3)^5}{5\times 2}\right\}\) | B1 B1 | Integrate the 2 functions |
| \(\left(\frac{1}{6}+2\right) - \left(-\frac{1}{6}+1\right) - \left\{\frac{1}{5} - \left(-\frac{1}{5}\right)\right\}\) | M1 | Apply limits \(1 \to 2\) to an integral. Some evidence of substitution. Minimum \(\left(\frac{13}{6} - \frac{5}{6}\right) - \left(\frac{1}{5} + \frac{1}{5}\right)\) or equivalent. Allow 1 sign error for 1st M1 |
| \(\frac{4}{3} - \frac{2}{5}\) | M1 | Subtract (at some point) the 2 areas. Must subtract areas not just integrals |
| \(\frac{14}{15}\) | A1 | Special case: If M0 for substitution of limits award SC B1 for correct answer. Condone \(-\frac{14}{15}\) if corrected. If subtraction wrong way round award B1 B1 M1 M1 A0. \(\int y^2\, dx\) or \(\int x\, dy\) scores 0/5. \(\pi\int y\, dx\) used: award B1 B1 M1 M1 A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(u = 2x-3\), \(\int(u^2+1-2u^4)\,du\) | B2,1,0 | |
| \(\left\{\frac{1}{2}\right\}\left(\left\{\frac{1}{3}u^3+u\right\}-\left\{\frac{2}{5}u^5\right\}\right)\) | ||
| \(\frac{1}{2}\left(\left(\frac{1}{3}+1-\frac{2}{5}\right)-\left(\frac{-1}{3}-1+\frac{2}{5}\right)\right)\) | M1 | Applies limits \(-1 \to 1\) |
| M1 | Subtract (at some point) the 2 areas | |
| \(\frac{1}{2}\left(\frac{14}{15}+\frac{14}{15}\right) = \frac{14}{15}\) | A1 | |
| Total | 5 |
## Question 8(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = 2x-3$ leading to $2u^4 = u^2 + 1$ leading to $2u^4 - u^2 - 1[=0]$ | B1 | |
| $(2u^2+1)(u^2-1)[=0]$ | M1 | Factors or formula or completing square must be shown |
| $u = \pm 1$ leading to $2x-3 = \pm 1$ leading to $x = 1$ or $2$ | A1 | |
| $(1, 2),\ (2, 2)$ | A1 | Special case: If B1 M0 scored then SC B2 for correct coordinates or SC B1 for correct $x$ values only. Special case: $2(2x-3)^4 = (2x-3)^2 + 1$, $32x^4 - 192x^3 + 428x^2 - 420x + 152 = 0$, $x = 1, 2$ finding both from correct quartic SC B1, $(1,2),(2,2)$ SC DB1. Trial and improvement without quartic: both $x$ correct B1, both coordinates correct B2. |
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## Question 8(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left\{\frac{(2x-3)^3}{3\times 2} + x\right\} [-] \left\{\frac{2(2x-3)^5}{5\times 2}\right\}$ | B1 B1 | Integrate the 2 functions |
| $\left(\frac{1}{6}+2\right) - \left(-\frac{1}{6}+1\right) - \left\{\frac{1}{5} - \left(-\frac{1}{5}\right)\right\}$ | M1 | Apply limits $1 \to 2$ to an integral. Some evidence of substitution. Minimum $\left(\frac{13}{6} - \frac{5}{6}\right) - \left(\frac{1}{5} + \frac{1}{5}\right)$ or equivalent. Allow 1 sign error for 1st M1 |
| $\frac{4}{3} - \frac{2}{5}$ | M1 | Subtract (at some point) the 2 areas. Must subtract **areas** not just integrals |
| $\frac{14}{15}$ | A1 | Special case: If M0 for substitution of limits award SC B1 for correct answer. Condone $-\frac{14}{15}$ if corrected. If subtraction wrong way round award B1 B1 M1 M1 A0. $\int y^2\, dx$ or $\int x\, dy$ scores 0/5. $\pi\int y\, dx$ used: award B1 B1 M1 M1 A0 |
## Question 8(b) (Alternative method):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $u = 2x-3$, $\int(u^2+1-2u^4)\,du$ | B2,1,0 | |
| $\left\{\frac{1}{2}\right\}\left(\left\{\frac{1}{3}u^3+u\right\}-\left\{\frac{2}{5}u^5\right\}\right)$ | | |
| $\frac{1}{2}\left(\left(\frac{1}{3}+1-\frac{2}{5}\right)-\left(\frac{-1}{3}-1+\frac{2}{5}\right)\right)$ | M1 | Applies limits $-1 \to 1$ |
| | M1 | Subtract (at some point) the 2 areas |
| $\frac{1}{2}\left(\frac{14}{15}+\frac{14}{15}\right) = \frac{14}{15}$ | A1 | |
| **Total** | **5** | |
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\includegraphics[max width=\textwidth, alt={}, center]{5e3e5418-7976-4232-8550-1da6420a3fcb-12_684_776_274_680}
The diagram shows the curves with equations $y = 2 ( 2 x - 3 ) ^ { 4 }$ and $y = ( 2 x - 3 ) ^ { 2 } + 1$ meeting at points $A$ and $B$.
\begin{enumerate}[label=(\alph*)]
\item By using the substitution $u = 2 x - 3$ find, by calculation, the coordinates of $A$ and $B$.
\item Find the exact area of the shaded region.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q8 [9]}}