| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete the square |
| Difficulty | Moderate -0.8 Part (i) is a straightforward completing the square exercise with simple coefficients. Part (ii) applies this to find circle properties, which is a standard C1 technique requiring recognition that the equation needs rearranging into (x-a)²+(y-b)²=r² form. Both parts are routine applications of well-practiced methods with no problem-solving insight required, making this easier than average. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(x - \dfrac{5}{2}\right)^2 - \left(\dfrac{5}{2}\right)^2 + \dfrac{1}{4}\) | B1 | \(a = \dfrac{5}{2}\) |
| \(= \left(x - \dfrac{5}{2}\right)^2 - 6\) | M1 | \(\dfrac{1}{4} - a^2\) |
| A1 [3] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(x - \dfrac{5}{2}\right)^2 - 6 + y^2 = 0\) | ||
| Centre \(\left(\dfrac{5}{2}, 0\right)\) | B1, B1 | Correct \(x\) coordinate; Correct \(y\) coordinate |
| Radius \(= \sqrt{6}\) | B1 [3] |
## Question 7:
### Part (i):
$\left(x - \dfrac{5}{2}\right)^2 - \left(\dfrac{5}{2}\right)^2 + \dfrac{1}{4}$ | B1 | $a = \dfrac{5}{2}$
$= \left(x - \dfrac{5}{2}\right)^2 - 6$ | M1 | $\dfrac{1}{4} - a^2$
| A1 [3] | cao
### Part (ii):
$\left(x - \dfrac{5}{2}\right)^2 - 6 + y^2 = 0$ | |
Centre $\left(\dfrac{5}{2}, 0\right)$ | B1, B1 | Correct $x$ coordinate; Correct $y$ coordinate
Radius $= \sqrt{6}$ | B1 [3] |
---7 (i) Express $x ^ { 2 } - 5 x + \frac { 1 } { 4 }$ in the form $( x - a ) ^ { 2 } - b$.\\
(ii) Find the centre and radius of the circle with equation $x ^ { 2 } + y ^ { 2 } - 5 x + \frac { 1 } { 4 } = 0$.
\hfill \mbox{\textit{OCR C1 2009 Q7 [6]}}