| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Two Curves Intersection Area |
| Difficulty | Standard +0.3 This is a straightforward integration question requiring students to find the area between two curves with given intersection points and then find a normal line equation. Part (a) involves setting up and evaluating a single definite integral with simple fractional powers, while part (b) requires basic differentiation and straight-line geometry. Both parts are standard textbook exercises with no novel problem-solving required, making this slightly easier than average. |
| Spec | 1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int\left(\frac{5}{2} - x^{\frac{1}{2}} - x^{-\frac{1}{2}}\right)dx\) | M1 | OR as 2 separate integrals \(\int\left(\frac{5}{2} - x^{1/2}\right)dx - \int\left(x^{-1/2}\right)dx\) |
| \(\left\{\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}}\right\}\left\{-\right\}\left\{2x^{\frac{1}{2}}\right\}\) | A1 A1 A1 | If two separate integrals with no subtraction, SC B1 for each correct integral |
| \(\left(10 - \frac{16}{3} - 4\right) - \left(\frac{5}{8} - \frac{1}{12} - 1\right)\) | DM1 | Substitute limits \(\frac{1}{4} \to 4\) at least once, must be seen |
| \(\frac{9}{8}\) or \(1.125\) | A1 | WWW. Cannot be awarded if \(\pi\) appears in any integral |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[\frac{dy}{dx} =\right] -\frac{1}{2}x^{-\frac{3}{2}}\) | B1 | |
| When \(x = 1\), \(m = -\frac{1}{2}\) | M1 | Substitute \(x = 1\) into a differential |
| [Equation of normal is] \(y - 1 = 2(x-1)\) | M1 | Through \((1, 1)\) with gradient \(-\frac{1}{m}\) or \(\frac{1-p}{1} = 2\) |
| [When \(x = 0\),] \(p = -1\) | A1 | WWW |
## Question 8(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\left(\frac{5}{2} - x^{\frac{1}{2}} - x^{-\frac{1}{2}}\right)dx$ | M1 | OR as 2 separate integrals $\int\left(\frac{5}{2} - x^{1/2}\right)dx - \int\left(x^{-1/2}\right)dx$ |
| $\left\{\frac{5}{2}x - \frac{2}{3}x^{\frac{3}{2}}\right\}\left\{-\right\}\left\{2x^{\frac{1}{2}}\right\}$ | A1 A1 A1 | If two separate integrals with no subtraction, SC B1 for each correct integral |
| $\left(10 - \frac{16}{3} - 4\right) - \left(\frac{5}{8} - \frac{1}{12} - 1\right)$ | DM1 | Substitute limits $\frac{1}{4} \to 4$ at least once, must be seen |
| $\frac{9}{8}$ or $1.125$ | A1 | WWW. Cannot be awarded if $\pi$ appears in any integral |
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## Question 8(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[\frac{dy}{dx} =\right] -\frac{1}{2}x^{-\frac{3}{2}}$ | B1 | |
| When $x = 1$, $m = -\frac{1}{2}$ | M1 | Substitute $x = 1$ into a differential |
| [Equation of normal is] $y - 1 = 2(x-1)$ | M1 | Through $(1, 1)$ with gradient $-\frac{1}{m}$ or $\frac{1-p}{1} = 2$ |
| [When $x = 0$,] $p = -1$ | A1 | WWW |
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\includegraphics[max width=\textwidth, alt={}, center]{cba7391d-2cf7-483c-9a21-bc9fd49860ee-12_570_961_260_591}
The diagram shows the curves with equations $y = x ^ { - \frac { 1 } { 2 } }$ and $y = \frac { 5 } { 2 } - x ^ { \frac { 1 } { 2 } }$. The curves intersect at the points $A \left( \frac { 1 } { 4 } , 2 \right)$ and $B \left( 4 , \frac { 1 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item Find the area of the region between the two curves.
\item The normal to the curve $y = x ^ { - \frac { 1 } { 2 } }$ at the point $( 1,1 )$ intersects the $y$-axis at the point $( 0 , p )$. Find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q8 [10]}}