| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Sketch function and inverse graphs |
| Difficulty | Standard +0.3 This is a straightforward composite/inverse functions question requiring: (a) stating range from a restricted quadratic (routine), (b) sketching inverse by reflection in y=x (standard technique), and (c) solving f(x)=x algebraically (leads to a cubic but solvable by inspection/factoring). All parts are textbook-standard with no novel insight required, making it slightly easier than average. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(f \geqslant -5\) | B1 | Accept \(y \geqslant -5\), \(f(x) \geqslant -5\), \(f \in [-5,\infty)\) or correct formal set notation. Not just \(x \geqslant -5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Curve starting on negative \(x\)-axis and passing through positive \(y\)-axis, in quadrants 1 and 2 only | M1 | For curve starting on negative \(x\)-axis and passing through positive \(y\)-axis in quadrants 1 and 2 only |
| Shape and position correct | A1 | Correct shape (curvature) and position. Must be increasing with decreasing gradient |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2x^2 - 5 = x\) or \(2x^2-5 = \sqrt{\dfrac{x+5}{2}}\) or \(x = \sqrt{\dfrac{x+5}{2}}\) or \(2(2x^2-5)^2-5=x\) | B1 | Sets up correct equation. Allow "=0" implied if attempt to solve. Just \(2x^2-x-5\) is B0 with no further working |
| Full attempt to solve \(2x^2 - x - 5 = 0 \Rightarrow x = \ldots\) exact | M1 | Full attempt to solve correct equation leading to exact answers |
| \(x = \dfrac{1+\sqrt{41}}{4}\) | A1 | \(x = \dfrac{1+\sqrt{41}}{4}\) ONLY |
## Question 4:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f \geqslant -5$ | B1 | Accept $y \geqslant -5$, $f(x) \geqslant -5$, $f \in [-5,\infty)$ or correct formal set notation. Not just $x \geqslant -5$ |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Curve starting on negative $x$-axis and passing through positive $y$-axis, in quadrants 1 and 2 only | M1 | For curve starting on negative $x$-axis and passing through positive $y$-axis in quadrants 1 and 2 only |
| Shape and position correct | A1 | Correct shape (curvature) and position. Must be increasing with decreasing gradient |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2x^2 - 5 = x$ or $2x^2-5 = \sqrt{\dfrac{x+5}{2}}$ or $x = \sqrt{\dfrac{x+5}{2}}$ or $2(2x^2-5)^2-5=x$ | B1 | Sets up correct equation. Allow "=0" implied if attempt to solve. Just $2x^2-x-5$ is B0 with no further working |
| Full attempt to solve $2x^2 - x - 5 = 0 \Rightarrow x = \ldots$ exact | M1 | Full attempt to solve correct equation leading to exact answers |
| $x = \dfrac{1+\sqrt{41}}{4}$ | A1 | $x = \dfrac{1+\sqrt{41}}{4}$ ONLY |\begin{enumerate}
\item The function f is defined by
\end{enumerate}
$$\mathrm { f } ( x ) = 2 x ^ { 2 } - 5 \quad x \geqslant 0 \quad x \in \mathbb { R }$$
(a) State the range of f
On the following page there is a diagram, labelled Diagram 1, which shows a sketch of the curve with equation $y = \mathrm { f } ( x )$.\\
(b) On Diagram 1, sketch the curve with equation $y = \mathrm { f } ^ { - 1 } ( x )$.
The curve with equation $y = \mathrm { f } ( x )$ meets the curve with equation $y = \mathrm { f } ^ { - 1 } ( x )$ at the point $P$
Using algebra and showing your working,\\
(c) find the exact $x$ coordinate of $P$
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{bef290fb-fbac-4c9c-981e-5e323ac7182e-09_607_610_248_731}
\end{center}
\section*{Diagram 1}
\hfill \mbox{\textit{Edexcel P3 2023 Q4 [6]}}