| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2008 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Line intersecting quadratic curve |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing basic skills: finding gradient from linear equation, writing equation of parallel line, and solving simultaneous equations with one linear and one quadratic. All parts are routine procedures with no problem-solving required, making it easier than average, though the multiple parts and algebraic manipulation prevent it from being trivial. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks |
|---|---|
| Gradient \(= -\frac{1}{2}\) | B1 1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[y - 5 = -\frac{1}{2}(x - 6)\] | M1 | Equation of straight line through (6, 5) with any non-zero numerical gradient. Uses gradient found in (i) in their equation of line |
| \[2y - 10 = -x + 6\] | ||
| \[x + 2y - 16 = 0\] | A1 3 | Correct answer in correct form (integer coefficients) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\frac{4-x}{2} = x^2 + x + 1\] | *M1 | Substitute to find an equation in x (or y) |
| \[4 - x = 2x^2 + 2x + 2\] | ||
| \[2x^2 + 3x - 2 = 0\] | DM1 | Correct method to solve quadratic |
| \[(2x - 1)(x + 2) = 0\] | ||
| \[x = \frac{1}{2}, x = -2\] | A1 | |
| \[y = \frac{7}{4}, y = 3\] | A1 4 | |
| SR one correct (x,y) pair www | B1 |
| Answer | Marks |
|---|---|
| \[y = (4-2y)^2 + (4-2y) + 1\] | *M |
| \[y = 16 - 16y + 4y^2 + 4 - 2y + 1\] | |
| \[0 = 21 - 19y + 4y^2\] | |
| \[0 = (4y - 7)(y - 3)\] | DM |
| \[y = \frac{7}{4}, y = 3\] | A1 |
| \[x = \frac{1}{2}, x = -2\] | A1 |
## 7(i)
Gradient $= -\frac{1}{2}$ | B1 1 |
## 7(ii)
$$y - 5 = -\frac{1}{2}(x - 6)$$ | M1 | Equation of straight line through (6, 5) with any non-zero numerical gradient. Uses gradient found in (i) in their equation of line |
$$2y - 10 = -x + 6$$ | |
$$x + 2y - 16 = 0$$ | A1 3 | Correct answer in correct form (integer coefficients) |
## 7(iii)
$$\frac{4-x}{2} = x^2 + x + 1$$ | *M1 | Substitute to find an equation in x (or y) |
$$4 - x = 2x^2 + 2x + 2$$ | | |
$$2x^2 + 3x - 2 = 0$$ | DM1 | Correct method to solve quadratic |
$$(2x - 1)(x + 2) = 0$$ | | |
$$x = \frac{1}{2}, x = -2$$ | A1 | |
$$y = \frac{7}{4}, y = 3$$ | A1 4 | |
**SR** one correct (x,y) pair **www** | B1 |
OR
$$y = (4-2y)^2 + (4-2y) + 1$$ | *M | |
$$y = 16 - 16y + 4y^2 + 4 - 2y + 1$$ | | |
$$0 = 21 - 19y + 4y^2$$ | | |
$$0 = (4y - 7)(y - 3)$$ | DM | |
$$y = \frac{7}{4}, y = 3$$ | A1 |
$$x = \frac{1}{2}, x = -2$$ | A1 |
---7 (i) Find the gradient of the line $l$ which has equation $x + 2 y = 4$.\\
(ii) Find the equation of the line parallel to $l$ which passes through the point ( 6,5 ), giving your answer in the form $a x + b y + c = 0$, where $a , b$ and $c$ are integers.\\
(iii) Solve the simultaneous equations
$$y = x ^ { 2 } + x + 1 \quad \text { and } \quad x + 2 y = 4$$
\hfill \mbox{\textit{OCR C1 2008 Q7 [8]}}