| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | March |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Area between curve and line |
| Difficulty | Standard +0.3 This is a straightforward P1 calculus question requiring differentiation to find a minimum (using fractional powers), then integration to find an area. Both are standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differentiate to obtain \(-\frac{4}{3}x^{-\frac{1}{3}} + x^{-\frac{4}{3}}\), or rewrite as quadratic in \(x^{-\frac{1}{3}}\) or \(x^{\frac{1}{3}}\) | B1 | Expect quadratic \(2\left(x^{-\frac{1}{3}}\right)^2 - 3x^{-\frac{1}{3}}+1\) OE. Allow \(2x^2-3x+1\) |
| Equate first derivative to zero and reach solution for \(x^{-\frac{1}{3}}\) or \(x^{\frac{1}{3}}\) with no error in indices, or complete the square to find minimum point \(2\left(a-\frac{3}{4}\right)^2 - \frac{1}{8}\) where \(a=x^{-\frac{1}{3}}\) | M1 | Substitution SOI if dealt with correctly later |
| Obtain \(x = \frac{64}{27}\) | A1 | Or exact equivalent. SC B1 if no working shown. Ignore extra solution \(x=0\) |
| \(y = -\frac{1}{8}\) seen | B1 | Or exact equivalent. Allow \(-0.125\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Recognise equation as quadratic in \(x^{-\frac{1}{3}}\) and attempt solution | M1 | \(2a^2-3a+1[=0]\) where \(a=x^{-\frac{1}{3}}\) |
| Obtain \(x^{-\frac{1}{3}}=1\) and \(x^{-\frac{1}{3}}=\frac{1}{2}\) | A1 | OE. SC B1 if no M mark awarded |
| Obtain 1 and 8 | A1 | SC B1 if no M mark awarded |
| Integrate to obtain form \(k_1x^{\frac{1}{3}} + k_2x^{\frac{2}{3}} + x\) or 2 out of 3 correct terms | *M1 | Expect \(6x^{\frac{1}{3}} - \frac{9}{2}x^{\frac{2}{3}} + x\) |
| Obtain correct \(6x^{\frac{1}{3}} - \frac{9}{2}x^{\frac{2}{3}} + x\) | A1 | No other terms from a second integral |
| Apply their limits correctly | DM1 | Their limits must be from their working |
| [Obtain \(-0.5\) and conclude area is] \(0.5\) | A1 |
## Question 11(a):
| Answer | Mark | Guidance |
|--------|------|----------|
| Differentiate to obtain $-\frac{4}{3}x^{-\frac{1}{3}} + x^{-\frac{4}{3}}$, or rewrite as quadratic in $x^{-\frac{1}{3}}$ or $x^{\frac{1}{3}}$ | B1 | Expect quadratic $2\left(x^{-\frac{1}{3}}\right)^2 - 3x^{-\frac{1}{3}}+1$ OE. Allow $2x^2-3x+1$ |
| Equate first derivative to zero and reach solution for $x^{-\frac{1}{3}}$ or $x^{\frac{1}{3}}$ with no error in indices, or complete the square to find minimum point $2\left(a-\frac{3}{4}\right)^2 - \frac{1}{8}$ where $a=x^{-\frac{1}{3}}$ | M1 | Substitution SOI if dealt with correctly later |
| Obtain $x = \frac{64}{27}$ | A1 | Or exact equivalent. SC B1 if no working shown. Ignore extra solution $x=0$ |
| $y = -\frac{1}{8}$ seen | B1 | Or exact equivalent. Allow $-0.125$ |
## Question 11(b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Recognise equation as quadratic in $x^{-\frac{1}{3}}$ and attempt solution | M1 | $2a^2-3a+1[=0]$ where $a=x^{-\frac{1}{3}}$ |
| Obtain $x^{-\frac{1}{3}}=1$ and $x^{-\frac{1}{3}}=\frac{1}{2}$ | A1 | OE. SC B1 if no M mark awarded |
| Obtain 1 and 8 | A1 | SC B1 if no M mark awarded |
| Integrate to obtain form $k_1x^{\frac{1}{3}} + k_2x^{\frac{2}{3}} + x$ or 2 out of 3 correct terms | *M1 | Expect $6x^{\frac{1}{3}} - \frac{9}{2}x^{\frac{2}{3}} + x$ |
| Obtain correct $6x^{\frac{1}{3}} - \frac{9}{2}x^{\frac{2}{3}} + x$ | A1 | No other terms from a second integral |
| Apply their limits correctly | DM1 | Their limits must be from their working |
| [Obtain $-0.5$ and conclude area is] $0.5$ | A1 | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{b5eb378d-a9cb-40e0-9203-374b58f1dcf9-14_467_757_262_653}
The diagram shows the curve with equation $\mathrm { y } = 2 \mathrm { x } ^ { - \frac { 2 } { 3 } } - 3 \mathrm { x } ^ { - \frac { 1 } { 3 } } + 1$ for $x > 0$. The curve crosses the $x$-axis at points $A$ and $B$ and has a minimum point $M$.
\begin{enumerate}[label=(\alph*)]
\item Find the exact coordinates of $M$.
\item Find the area of the region bounded by the curve and the line segment $A B$.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q11 [11]}}