| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Composite transformation sketch |
| Difficulty | Moderate -0.3 This is a standard P1 function transformation question requiring sketching f(2x) and f(x)+3, finding intersection by equating expressions, and algebraic manipulation with surds. While it involves multiple parts and surd simplification, all techniques are routine for A-level: horizontal stretch (factor 1/2), vertical translation (+3), and solving √(2x) = √x + 3. The 'show that' in part (b) provides the target form, reducing problem-solving demand. Slightly easier than average due to straightforward transformations and guided algebraic steps. |
| Spec | 1.07d Second derivatives: d^2y/dx^2 notation1.07e Second derivative: as rate of change of gradient1.08a Fundamental theorem of calculus: integration as reverse of differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One correct sketch drawn and labelled correctly | M1 | For \(f(2x)\): curve should start at \(O\), be above and remain above \(f(x)\), not head back towards it significantly. For \(f(x)+3\): curve should start on positive \(y\)-axis, approximately same shape as \(f(x)\) |
| One correct sketch drawn and labelled with correct point for that curve | A1 | Point does not have to be in correct relative position — just look for values |
| Completely correct sketches with both points correct and at least one correctly labelled | A1 | Allow \(f(2x)\) to cross \(f(x)+3\) as long as it is beyond \((9,6)\) but with no other intersections for \(x>0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Sets \(\sqrt{x}+3=\sqrt{2x}\) | B1 | Correct equation |
| \(3=(\sqrt{2}-1)\sqrt{x}\) | M1 | Writes \(\sqrt{2x}\) as \(\sqrt{2}\sqrt{x}\) and collects terms in \(\sqrt{x}\) |
| \(\sqrt{x}=\dfrac{3}{(\sqrt{2}-1)}\times\dfrac{(\sqrt{2}+1)}{(\sqrt{2}+1)}=3(\sqrt{2}+1)\) | A1* | Proceeds to given answer showing at least the steps; dependent — must reach printed answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{x}=3(\sqrt{2}+1)\Rightarrow x=9(\sqrt{2}+1)^2=\ldots\) | M1 | Attempts to square the given expression to find \(x\); may condone slip on the 3 but must result in form \(\alpha+\beta\sqrt{2}\) |
| \(x=9(3+2\sqrt{2})\), i.e. \(x=27+18\sqrt{2}\) | A1 | Allow equivalent correct forms |
| \(y=3\sqrt{2}+6\) | B1 | Note \(y=\sqrt{18\sqrt{2}+27}+3\) is correct but not simplified so scores B0 |
## Question 9:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| One correct sketch drawn and labelled correctly | M1 | For $f(2x)$: curve should start at $O$, be above and remain above $f(x)$, not head back towards it significantly. For $f(x)+3$: curve should start on positive $y$-axis, approximately same shape as $f(x)$ |
| One correct sketch drawn and labelled with correct point for that curve | A1 | Point does **not** have to be in correct relative position — just look for values |
| Completely correct sketches with both points correct and at least one correctly labelled | A1 | Allow $f(2x)$ to cross $f(x)+3$ as long as it is beyond $(9,6)$ but with no other intersections for $x>0$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $\sqrt{x}+3=\sqrt{2x}$ | B1 | Correct equation |
| $3=(\sqrt{2}-1)\sqrt{x}$ | M1 | Writes $\sqrt{2x}$ as $\sqrt{2}\sqrt{x}$ and collects terms in $\sqrt{x}$ |
| $\sqrt{x}=\dfrac{3}{(\sqrt{2}-1)}\times\dfrac{(\sqrt{2}+1)}{(\sqrt{2}+1)}=3(\sqrt{2}+1)$ | A1* | Proceeds to given answer showing at least the steps; dependent — must reach printed answer |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{x}=3(\sqrt{2}+1)\Rightarrow x=9(\sqrt{2}+1)^2=\ldots$ | M1 | Attempts to square the given expression to find $x$; may condone slip on the 3 but must result in form $\alpha+\beta\sqrt{2}$ |
| $x=9(3+2\sqrt{2})$, i.e. $x=27+18\sqrt{2}$ | A1 | Allow equivalent correct forms |
| $y=3\sqrt{2}+6$ | B1 | Note $y=\sqrt{18\sqrt{2}+27}+3$ is correct but not simplified so scores B0 |
---9. In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-26_595_716_420_662}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Figure 5 shows a sketch of the curve with equation $y = \mathrm { f } ( x )$ where
$$\mathrm { f } ( x ) = \sqrt { x } \quad x > 0$$
The point $P ( 9,3 )$ lies on the curve and is shown in Figure 5.\\
On the next page there is a copy of Figure 5 called Diagram 1.
\begin{enumerate}[label=(\alph*)]
\item On Diagram 1, sketch and clearly label the graphs of
$$y = \mathrm { f } ( 2 x ) \text { and } y = \mathrm { f } ( x ) + 3$$
Show on each graph the coordinates of the point to which $P$ is transformed.
The graph of $y = \mathrm { f } ( 2 x )$ meets the graph of $y = \mathrm { f } ( x ) + 3$ at the point $Q$.
\item Show that the $x$ coordinate of $Q$ is the solution of
$$\sqrt { x } = 3 ( \sqrt { 2 } + 1 )$$
\item Hence find, in simplest form, the coordinates of $Q$.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-27_599_720_274_660}
\end{center}
\section*{Diagram 1}
Turn over for a copy of Diagram 1 if you need to redraw your graphs.
Only use this copy if you need to redraw your graphs.\\
\includegraphics[max width=\textwidth, alt={}, center]{f1e1d4f5-dd27-4839-a6f3-f6906666302c-29_600_718_1991_660}
Copy of Diagram 1
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q9 [9]}}