| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simultaneous equations |
| Type | Tangency condition for line and curve |
| Difficulty | Moderate -0.3 Part (i) is a standard quadratic-linear simultaneous equations problem requiring substitution and solving a quadratic. Part (ii) is trivial interpretation. Part (iii) requires using the discriminant condition b²-4ac=0 for tangency, which is a step beyond routine but still a well-practiced C1 technique. Overall slightly easier than average due to straightforward algebraic manipulation with no conceptual surprises. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02q Use intersection points: of graphs to solve equations1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = x^2 - 5x + 4\) | ||
| \(y = x - 1\) | ||
| \(x^2 - 5x + 4 = x - 1\) | M1 | Substitute to find an equation in x (or y) |
| \(x^2 - 6x + 5 = 0\) | M1 | |
| \((x-1)(x-5) = 0\) | M1 | Correct method to solve quadratic |
| \(x = 1, x = 5\) | A1 | |
| \(y = 0, y = 4\) | A1 | 4 marks; (N.B. This final A1 may be awarded in part (ii) if y coordinates only seen in part (ii)) |
| SR one correct (x,y) pair www | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 points of intersection | B1 | 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| EITHER | ||
| \(x^2 - 5x + 4 = x + c\) has 1 solution | M1 | \(x^2 - 5x + 4 = x + c\) has 1 soln seen or implied |
| \(x^2 - 6x + (4-c) = 0\) | M1 | |
| \(b^2 - 4ac = 0\) | M1 | Discriminant = 0 or \((x-a)^2 = 0\) soi |
| \(36 - 4(4-c) = 0\) | A1 | |
| \(c = -5\) | A1 | 4 marks; \(36 - 4(4-c) = 0\) or \(9 = 4 - c\) |
| OR | ||
| \(\frac{dy}{dx} = 1 = 2x - 5\) | M1 | Algebraic expression for gradient of curve = non-zero gradient of line used |
| \(x = 3, y = -2\) | A1 | \(2x - 5 = 1\) |
| \(-2 = 3 + c\) | A1 | |
| \(c = -5\) | A1 | 4 marks |
| SR \(c = -5\) without any working | B1 |
## (i)
$y = x^2 - 5x + 4$ | |
$y = x - 1$ | |
$x^2 - 5x + 4 = x - 1$ | M1 | Substitute to find an equation in x (or y)
$x^2 - 6x + 5 = 0$ | M1 |
$(x-1)(x-5) = 0$ | M1 | Correct method to solve quadratic
$x = 1, x = 5$ | A1 |
$y = 0, y = 4$ | A1 | 4 marks; (N.B. This final A1 may be awarded in part (ii) if y coordinates only seen in part (ii))
SR one correct (x,y) pair www | B1 |
## (ii)
2 points of intersection | B1 | 1 mark
## (iii)
EITHER | |
$x^2 - 5x + 4 = x + c$ has 1 solution | M1 | $x^2 - 5x + 4 = x + c$ has 1 soln seen or implied
$x^2 - 6x + (4-c) = 0$ | M1 |
$b^2 - 4ac = 0$ | M1 | Discriminant = 0 or $(x-a)^2 = 0$ soi
$36 - 4(4-c) = 0$ | A1 |
$c = -5$ | A1 | 4 marks; $36 - 4(4-c) = 0$ or $9 = 4 - c$
OR | |
$\frac{dy}{dx} = 1 = 2x - 5$ | M1 | Algebraic expression for gradient of curve = non-zero gradient of line used
$x = 3, y = -2$ | A1 | $2x - 5 = 1$
$-2 = 3 + c$ | A1 |
$c = -5$ | A1 | 4 marks
SR $c = -5$ without any working | B1 |
---\begin{enumerate}[label=(\roman*)]
\item Solve the simultaneous equations
$$y = x^2 - 5x + 4, \quad y = x - 1.$$ [4]
\item State the number of points of intersection of the curve $y = x^2 - 5x + 4$ and the line $y = x - 1$. [1]
\item Find the value of $c$ for which the line $y = x + c$ is a tangent to the curve $y = x^2 - 5x + 4$. [4]
\end{enumerate}
\hfill \mbox{\textit{OCR C1 2006 Q7 [9]}}