| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Graphical equation solving with auxiliary line |
| Difficulty | Standard +0.3 This question requires students to recognize that solving x² - 3x + 1 = 0 is equivalent to finding where x² - 4x + 3 = mx + c by rearranging to match the given curve. The algebraic manipulation is straightforward (m = -1, c = -2), and reading approximate roots from a graph is a standard skill. The inequality shading in part (c) is routine A-level work. Overall, this is slightly easier than average due to its methodical, step-by-step nature with no conceptual surprises. |
| Spec | 1.02i Represent inequalities: graphically on coordinate plane1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x^2 - 3x + 1 \Rightarrow x^2 - 4x + 3 = -x + 2\) | M1 | Attempt form equation of form \(x^2 - 4x + 3 = mx + c\). NB \(x^2 - 3x + 1 = x^2 - 4x + 3\): M0 unless this leads to \(y = mx + c\) seen |
| \(m = -1\), \(c = 2\) or \(y = -x + 2\) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Line \(y = -x + 2\) drawn | M1 | Good attempt at draw their line from (a). Ignore \(y\)-coords |
| \(x = 0.4\ (\pm0.1)\), \(x = 2.6\ (\pm0.1)\) | A1 | cao NB, correct answers do NOT score marks unless they clearly come from the correct line seen, except: SC: correct answers from graph of \(y = x^2 - 3x + 1\) B0B1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| At least one region indicated wholly above line \(y = -x+2\), ft their line, no omission. Follow only correct line or their line from (a) | B1ft | |
| Wholly below the curve \(y = x^2 - 4x + 3\), no omissions. Follow their line as drawn with its shading | B1ft | |
| All correct cao | B1 | Accept any correct indication, e.g. shading in, shading out, arrows, letters etc |
| [3] |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x^2 - 3x + 1 \Rightarrow x^2 - 4x + 3 = -x + 2$ | M1 | Attempt form equation of form $x^2 - 4x + 3 = mx + c$. NB $x^2 - 3x + 1 = x^2 - 4x + 3$: M0 unless this leads to $y = mx + c$ seen |
| $m = -1$, $c = 2$ or $y = -x + 2$ | A1 | |
| **[2]** | | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Line $y = -x + 2$ drawn | M1 | Good attempt at draw their line from (a). Ignore $y$-coords |
| $x = 0.4\ (\pm0.1)$, $x = 2.6\ (\pm0.1)$ | A1 | cao NB, correct answers do NOT score marks unless they clearly come from the correct line seen, except: SC: correct answers from graph of $y = x^2 - 3x + 1$ B0B1 |
| **[2]** | | |
---
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| At least one region indicated wholly above line $y = -x+2$, ft their line, no omission. Follow only correct line or their line from (a) | B1ft | |
| Wholly below the curve $y = x^2 - 4x + 3$, no omissions. Follow their line as drawn with its shading | B1ft | |
| All correct cao | B1 | Accept any correct indication, e.g. shading in, shading out, arrows, letters etc |
| **[3]** | | |
---3 The diagram in the Printed Answer Booklet shows part of the graph of $y = x ^ { 2 } - 4 x + 3$.
\begin{enumerate}[label=(\alph*)]
\item It is required to solve the equation $x ^ { 2 } - 3 x + 1 = 0$ graphically by drawing a straight line with equation $y = m x + c$ on the diagram, where $m$ and $c$ are constants.
Find the values of $m$ and $c$.
\item Use the graph to find approximate values of the roots of the equation $x ^ { 2 } - 3 x + 1 = 0$.
\item By shading, or otherwise, indicate clearly the regions where all of the following inequalities are satisfied. You should use the values of $m$ and $c$ found in part (a).\\
$x \geqslant 0$\\
$x \leqslant 4$\\
$y \leqslant x ^ { 2 } - 4 x + 3$\\
$y \geqslant m x + c$
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q3 [7]}}