| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Improper integral to infinity |
| Difficulty | Standard +0.8 This question combines an improper integral requiring reverse chain rule with a related rates problem involving implicit differentiation. Part (a) is moderately challenging as improper integrals are beyond basic C1/C2 content. Part (b) requires recognizing that dy/dt = (dy/dx)(dx/dt) and solving for the point where dy/dx = -1, which demands conceptual understanding beyond routine integration practice. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates4.08c Improper integrals: infinite limits or discontinuous integrands |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left\{\frac{(4x+2)^{-1}}{-1}\right\}\{\div 4\}\) or e.g. \(\left\{\frac{1}{16}\right\}\left\{-(x+0.5)^{-1}\right\}\) or \(\frac{-1}{(16x+8)}\) | B1 B1 | OE. If more than one function of \(x\) present then B0 B0. |
| \(0-(-1/24)\) | M1 | Apply limits to an integral, \(\infty\) must be used correctly. |
| \(\frac{1}{24}\) | A1 | Allow 0.0417 AWRT. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = \left\{-2(4x+2)^{-3}\right\}\ \{\times 4\}\) | B1 B1 | Allow unsimplified forms. |
| Recognise \(\frac{dy}{dx} = -1\) | B1 | SOI |
| \(their\ \frac{-8}{(4x+2)^3} = their -1\) | M1 | Must be numerical. Must be some attempt to solve *their* equation and \(\frac{dy}{dx} \neq 0\). |
| \((0,\ \frac{1}{4})\) | A1 A1 | Accept \(x=0\), \(y=\frac{1}{4}\). \(y=\frac{1}{4}\) must be from \(x=0\) not \(x=-1\). |
## Question 10(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left\{\frac{(4x+2)^{-1}}{-1}\right\}\{\div 4\}$ or e.g. $\left\{\frac{1}{16}\right\}\left\{-(x+0.5)^{-1}\right\}$ or $\frac{-1}{(16x+8)}$ | B1 B1 | OE. If more than one function of $x$ present then B0 B0. |
| $0-(-1/24)$ | M1 | Apply limits to an integral, $\infty$ must be used correctly. |
| $\frac{1}{24}$ | A1 | Allow 0.0417 AWRT. |
---
## Question 10(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \left\{-2(4x+2)^{-3}\right\}\ \{\times 4\}$ | B1 B1 | Allow unsimplified forms. |
| Recognise $\frac{dy}{dx} = -1$ | B1 | SOI |
| $their\ \frac{-8}{(4x+2)^3} = their -1$ | M1 | Must be numerical. Must be some attempt to solve *their* equation and $\frac{dy}{dx} \neq 0$. |
| $(0,\ \frac{1}{4})$ | A1 A1 | Accept $x=0$, $y=\frac{1}{4}$. $y=\frac{1}{4}$ must be from $x=0$ not $x=-1$. |
---10 The function f is defined by $\mathrm { f } ( x ) = ( 4 x + 2 ) ^ { - 2 }$ for $x > - \frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find $\int _ { 1 } ^ { \infty } \mathrm { f } ( x ) \mathrm { d } x$.\\
A point is moving along the curve $y = \mathrm { f } ( x )$ in such a way that, as it passes through the point $A$, its $y$-coordinate is decreasing at the rate of $k$ units per second and its $x$-coordinate is increasing at the rate of $k$ units per second.
\item Find the coordinates of $A$.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2022 Q10 [10]}}