| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2024 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Standard Integrals and Reverse Chain Rule |
| Type | Curve transformation after integration |
| Difficulty | Standard +0.3 Part (a) is a straightforward reverse chain rule integration with constant determination from a given point. Part (b) requires applying two transformations sequentially, which is more conceptually demanding than typical P1 questions but still follows standard procedures. The integration itself is routine, and while the transformation combination requires care, it's methodical rather than requiring novel insight. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Integrate to obtain form \(k(5x-3)^{-1}\) | \*M1 | OE |
| \(4(5x-3)^{-1}\) | A1 | Or unsimplified equivalent. Condone absence of \(\ldots + c\) so far. |
| Substitute \(x = \frac{4}{5}\) and \(y = -3\) to attempt value of \(c\) | DM1 | DM0 for substituting \(\left(-3, \frac{4}{5}\right)\) |
| \(y = 4(5x-3)^{-1} - 7\); allow \(f(x)\) or \(f = 4(5x-3)^{-1} - 7\) | A1 | OE. Condone \(c = -7\) as the final answer providing \(y = \) or \(f(x) = \frac{4}{(5x-3)} + c\) OE is seen earlier. Attempts to write equation in \(y = mx + c\) form scores A0. Do not ISW. Gains max 3/4. |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Carry out stretch by replacing \(x\) by \(2x\) in their equation | M1 | Award if given as the second transformation. Do not ignore sign errors. |
| Carry out translation by replacing \(x\) by \(x - 2\) and \(y\) by \(y - 10\) | M1 | OE. Award if given as the first transformation. Do not ignore sign errors. |
| \(y = \frac{4}{10x - 23} + 3\) | A1 | Or similarly simplified equivalent, WWW. |
| 3 |
## Question 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Integrate to obtain form $k(5x-3)^{-1}$ | \*M1 | OE |
| $4(5x-3)^{-1}$ | A1 | Or unsimplified equivalent. Condone absence of $\ldots + c$ so far. |
| Substitute $x = \frac{4}{5}$ and $y = -3$ to attempt value of $c$ | DM1 | DM0 for substituting $\left(-3, \frac{4}{5}\right)$ |
| $y = 4(5x-3)^{-1} - 7$; allow $f(x)$ or $f = 4(5x-3)^{-1} - 7$ | A1 | OE. Condone $c = -7$ as the final answer providing $y = $ or $f(x) = \frac{4}{(5x-3)} + c$ OE is seen earlier. Attempts to write equation in $y = mx + c$ form scores A0. Do not ISW. Gains max 3/4. |
| | **4** | |
## Question 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Carry out stretch by replacing $x$ by $2x$ in their equation | M1 | Award if given as the second transformation. Do not ignore sign errors. |
| Carry out translation by replacing $x$ by $x - 2$ and $y$ by $y - 10$ | M1 | OE. Award if given as the first transformation. Do not ignore sign errors. |
| $y = \frac{4}{10x - 23} + 3$ | A1 | Or similarly simplified equivalent, WWW. |
| | **3** | |6 A curve passes through the point $\left( \frac { 4 } { 5 } , - 3 \right)$ and is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { - 20 } { ( 5 x - 3 ) ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the equation of the curve.
\item The curve is transformed by a stretch in the $x$-direction with scale factor $\frac { 1 } { 2 }$ followed by a translation of $\binom { 2 } { 10 }$.
Find the equation of the new curve.\\
\includegraphics[max width=\textwidth, alt={}, center]{d6976a4b-aecf-43f1-a3f2-bcad37d03585-08_2716_38_143_2009}
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2024 Q6 [7]}}