| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Area under curve using integration |
| Difficulty | Standard +0.3 This is a straightforward multi-part calculus question requiring standard techniques: differentiation to find stationary points, equation of tangent, and integration using power rule. All steps are routine for P1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}x^{-1/2} - \frac{1}{2}k^2x^{-3/2}\) | B1 B1 | Allow any correct unsimplified form |
| \(\frac{1}{2}x^{-1/2} - \frac{1}{2}k^2x^{-3/2} = 0\) leading to \(\frac{1}{2}x^{-1/2} = \frac{1}{2}k^2x^{-3/2}\) | M1 | OE. Set to zero and one correct algebraic step towards the solutions. \(\frac{\mathrm{d}y}{\mathrm{d}x}\) must only have 2 terms |
| \((k^2, 2k)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(x = 4k^2\), \(\frac{\mathrm{d}y}{\mathrm{d}x} = \left[\frac{1}{4k} - \frac{1}{16k}\right] = \frac{3}{16k}\) | B1 | OE |
| \(y = \left[2k + k^2 \times \frac{1}{2k}\right] = \frac{5k}{2}\) | B1 | OE. Accept \(2k + \frac{k}{2}\) |
| Equation of tangent is \(y - \frac{5k}{2} = \frac{3}{16k}(x - 4k^2)\) or \(y = mx + c \rightarrow \frac{5k}{2} = \frac{3}{16k}(4k^2) + c\) | M1 | Use of line equation with *their* gradient and \((4k^2, \text{their } y)\) |
| When \(x = 0\), \(y = \left[\frac{5k}{2} - \frac{3k}{4}\right] = \frac{7k}{4}\) or from \(y = mx + c\), \(c = \frac{7k}{4}\) | A1 | OE |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int\left(x^{\frac{1}{2}}+k^2x^{-\frac{1}{2}}\right)dx = \dfrac{2x^{\frac{3}{2}}}{3}+2k^2x^{\frac{1}{2}}\) | B1 | Any unsimplified form |
| \(\left(\dfrac{16k^3}{3}+4k^3\right)-\left(\dfrac{9k^3}{4}+3k^3\right)\) | M1 | Apply limits \(\dfrac{9}{4}k^2 \to 4k^2\) to an integration of \(y\). M0 if volume attempted. |
| \(\dfrac{49k^3}{12}\) | A1 | OE. Accept \(4.08\,k^3\) |
| 3 |
## Question 11(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}x^{-1/2} - \frac{1}{2}k^2x^{-3/2}$ | B1 B1 | Allow any correct unsimplified form |
| $\frac{1}{2}x^{-1/2} - \frac{1}{2}k^2x^{-3/2} = 0$ leading to $\frac{1}{2}x^{-1/2} = \frac{1}{2}k^2x^{-3/2}$ | M1 | OE. Set to zero and one correct algebraic step towards the solutions. $\frac{\mathrm{d}y}{\mathrm{d}x}$ must only have 2 terms |
| $(k^2, 2k)$ | A1 | |
---
## Question 11(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $x = 4k^2$, $\frac{\mathrm{d}y}{\mathrm{d}x} = \left[\frac{1}{4k} - \frac{1}{16k}\right] = \frac{3}{16k}$ | B1 | OE |
| $y = \left[2k + k^2 \times \frac{1}{2k}\right] = \frac{5k}{2}$ | B1 | OE. Accept $2k + \frac{k}{2}$ |
| Equation of tangent is $y - \frac{5k}{2} = \frac{3}{16k}(x - 4k^2)$ or $y = mx + c \rightarrow \frac{5k}{2} = \frac{3}{16k}(4k^2) + c$ | M1 | Use of line equation with *their* gradient and $(4k^2, \text{their } y)$ |
| When $x = 0$, $y = \left[\frac{5k}{2} - \frac{3k}{4}\right] = \frac{7k}{4}$ or from $y = mx + c$, $c = \frac{7k}{4}$ | A1 | OE |
## Question 11(c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int\left(x^{\frac{1}{2}}+k^2x^{-\frac{1}{2}}\right)dx = \dfrac{2x^{\frac{3}{2}}}{3}+2k^2x^{\frac{1}{2}}$ | **B1** | Any unsimplified form |
| $\left(\dfrac{16k^3}{3}+4k^3\right)-\left(\dfrac{9k^3}{4}+3k^3\right)$ | **M1** | Apply limits $\dfrac{9}{4}k^2 \to 4k^2$ to an integration of $y$. M0 if volume attempted. |
| $\dfrac{49k^3}{12}$ | **A1** | OE. Accept $4.08\,k^3$ |
| | **3** | |11\\
\includegraphics[max width=\textwidth, alt={}, center]{aaba3158-b5be-464e-bea3-1a4c460f9637-16_622_1091_260_525}
The diagram shows part of the curve with equation $y = x ^ { \frac { 1 } { 2 } } + k ^ { 2 } x ^ { - \frac { 1 } { 2 } }$, where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the minimum point of the curve, giving your answer in terms of $k$.\\
The tangent at the point on the curve where $x = 4 k ^ { 2 }$ intersects the $y$-axis at $P$.
\item Find the $y$-coordinate of $P$ in terms of $k$.\\
The shaded region is bounded by the curve, the $x$-axis and the lines $x = \frac { 9 } { 4 } k ^ { 2 }$ and $x = 4 k ^ { 2 }$.
\item Find the area of the shaded region in terms of $k$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2021 Q11 [11]}}