| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting quadratic curve |
| Difficulty | Moderate -0.3 This is a structured multi-part question that guides students through standard techniques: substitution to form a quadratic, calculating discriminant, interpreting geometric meaning, and finding a normal. While it covers several concepts, each step is routine and heavily scaffolded, making it slightly easier than average for A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02d Quadratic functions: graphs and discriminant conditions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(y = x^2 - 5x + 15\); \(y = 5x - 10\); \(x^2 - 5x + 15 = 5x - 10\); \(x^2 - 10x + 25 = 0\) | M1, A1, 2 | Attempt to eliminate y; \(x^2 - 10x + 25 = 0\) AG (Obtained with no wrong working seen) |
| (ii) \(b^2 - 4ac = 100 - 100 = 0\) | B1, 1 | \(0\) (Do not allow \(\sqrt{(b^2 - 4ac)}\)) |
| (iii) Line is a tangent to the curve | B1√, 1 | Tangent or 'touches' N.B. Strict ft from their discriminant |
| (iv) \(x^2 - 10x + 25 = 0\); \((x-5)^2 = 0\); \(x = 5\) \(y = 15\) | M1, A1, A1, 3 | Correct method to solve 3 term quadratic; \(x = 5\); \(y = 15\) |
| (v) Gradient of tangent \(= 5\); Gradient of normal \(= -\frac{1}{5}\); \(y - 15 = -\frac{1}{5}(x-5)\); \(x + 5y = 80\) | B1, B1√, M1, A1, 4 | Gradient of tangent \(= 5\); Gradient of normal \(= -\frac{1}{5}\); Correct equation of straight line, any gradient, passing through (5, 15); \(x + 5y = 80\) |
**(i)** $y = x^2 - 5x + 15$; $y = 5x - 10$; $x^2 - 5x + 15 = 5x - 10$; $x^2 - 10x + 25 = 0$ | M1, A1, 2 | Attempt to eliminate y; $x^2 - 10x + 25 = 0$ **AG** (Obtained with no wrong working seen)
**(ii)** $b^2 - 4ac = 100 - 100 = 0$ | B1, 1 | $0$ (Do not allow $\sqrt{(b^2 - 4ac)}$)
**(iii)** Line is a tangent to the curve | B1√, 1 | Tangent or 'touches' **N.B. Strict ft from their discriminant**
**(iv)** $x^2 - 10x + 25 = 0$; $(x-5)^2 = 0$; $x = 5$ $y = 15$ | M1, A1, A1, 3 | Correct method to solve 3 term quadratic; $x = 5$; $y = 15$
**(v)** Gradient of tangent $= 5$; Gradient of normal $= -\frac{1}{5}$; $y - 15 = -\frac{1}{5}(x-5)$; $x + 5y = 80$ | B1, B1√, M1, A1, 4 | Gradient of tangent $= 5$; Gradient of normal $= -\frac{1}{5}$; Correct equation of straight line, any gradient, passing through (5, 15); $x + 5y = 80$8 (i) Given that $y = x ^ { 2 } - 5 x + 15$ and $5 x - y = 10$, show that $x ^ { 2 } - 10 x + 25 = 0$.\\
(ii) Find the discriminant of $x ^ { 2 } - 10 x + 25$.\\
(iii) What can you deduce from the answer to part (ii) about the line $5 x - y = 10$ and the curve $y = x ^ { 2 } - 5 x + 15$ ?\\
(iv) Solve the simultaneous equations
$$y = x ^ { 2 } - 5 x + 15 \text { and } 5 x - y = 10$$
(v) Hence, or otherwise, find the equation of the normal to the curve $y = x ^ { 2 } - 5 x + 15$ at the point $( 5,15 )$, giving your answer in the form $a x + b y = c$, where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{OCR C1 2006 Q8 [11]}}