| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting quadratic curve |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question testing standard C1 techniques: solving simultaneous equations by substitution (leading to a quadratic with repeated root), interpreting this geometrically as tangency, and finding a normal using the perpendicular gradient rule. While it requires connecting multiple concepts, each step follows routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 - 3x + 2 = 3x - 7 \Rightarrow x^2 - 6x + 9 = 0\). Hence \((x-3)^2 = 0\). So \(x = 3\) and \(y = 2\) | M1 A1 M1 A1 A1 | For equating two expressions for \(y\); For correct 3-term quadratic in \(x\); For factorising, or other solution method; For correct value of \(x\); For correct value of \(y\) |
| 5 | ||
| (ii) The line \(y = 3x - 7\) is tangent to the curve \(y = x^2 - 3x + 2\) at the point \((3, 2)\) | B1 B1 | For stating tangency; For identifying \(x = 3, y = 2\) as coordinates |
| 2 | ||
| (iii) Gradient of curve is \(2x - 3 = 3\) at \(x = 3\). Hence gradient of normal is \(-\frac{1}{3}\). Equation of normal is \(y - 2 = -\frac{1}{3}(x-3)\), i.e. \(x + 3y - 9 = 0\) | B1 B1 √ M1 A1 | For stating correct gradient of given line; For stating corresponding perpendicular grad; For appropriate use of straight line equation; For correct equation in required form |
| 4 |
(i) $x^2 - 3x + 2 = 3x - 7 \Rightarrow x^2 - 6x + 9 = 0$. Hence $(x-3)^2 = 0$. So $x = 3$ and $y = 2$ | M1 A1 M1 A1 A1 | For equating two expressions for $y$; For correct 3-term quadratic in $x$; For factorising, or other solution method; For correct value of $x$; For correct value of $y$
| | 5 |
(ii) The line $y = 3x - 7$ is tangent to the curve $y = x^2 - 3x + 2$ at the point $(3, 2)$ | B1 B1 | For stating tangency; For identifying $x = 3, y = 2$ as coordinates
| | 2 |
(iii) Gradient of curve is $2x - 3 = 3$ at $x = 3$. Hence gradient of normal is $-\frac{1}{3}$. Equation of normal is $y - 2 = -\frac{1}{3}(x-3)$, i.e. $x + 3y - 9 = 0$ | B1 B1 √ M1 A1 | For stating correct gradient of given line; For stating corresponding perpendicular grad; For appropriate use of straight line equation; For correct equation in required form
| | 4 |
---5 (i) Solve the simultaneous equations
$$y = x ^ { 2 } - 3 x + 2 , \quad y = 3 x - 7$$
(ii) What can you deduce from the solution to part (i) about the graphs of $y = x ^ { 2 } - 3 x + 2$ and $y = 3 x - 7$ ?\\
(iii) Hence, or otherwise, find the equation of the normal to the curve $y = x ^ { 2 } - 3 x + 2$ at the point ( 3,2 ), giving your answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C1 Q5 [11]}}