| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2005 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete the square |
| Difficulty | Moderate -0.8 This is a straightforward completing-the-square exercise with a follow-up question requiring only recognition that the line of symmetry occurs at x = -a. Both parts involve standard C1 techniques with no problem-solving required—just mechanical application of the completing-the-square algorithm and direct reading of the vertex form. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(3(x^2 + 4x) + 7\) | ||
| \(3(x+2)^2 - 12 + 7\) | ||
| \(3(x+2)^2 - 5\) | ||
| OR: \(3(x^2 + 2ax + a^2) + b\) | ||
| \(3x^2 + 6ax + 3a^2 + b\) | ||
| \(6a = 12\) | M1 | \(a = \frac{12}{6 \text{ or } 2}\) |
| \(a = 2\) | A1 | |
| \(3a^2 + b = 7\) | M1 | \(7 - a^2\) or \(7 - 3a^2\) or \(\frac{7}{3} - a^2\) (their \(a\)) |
| \(b = -5\) | A1 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(x = -2\) | B1ft 1 |
## Question 2(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER:** $3(x^2 + 4x) + 7$ | | |
| $3(x+2)^2 - 12 + 7$ | | |
| $3(x+2)^2 - 5$ | | |
| **OR:** $3(x^2 + 2ax + a^2) + b$ | | |
| $3x^2 + 6ax + 3a^2 + b$ | | |
| $6a = 12$ | M1 | $a = \frac{12}{6 \text{ or } 2}$ |
| $a = 2$ | A1 | |
| $3a^2 + b = 7$ | M1 | $7 - a^2$ or $7 - 3a^2$ or $\frac{7}{3} - a^2$ (their $a$) |
| $b = -5$ | A1 **4** | |
## Question 2(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $x = -2$ | B1ft **1** | |
---2 (i) Express $3 x ^ { 2 } + 12 x + 7$ in the form $3 ( x + a ) ^ { 2 } + b$.\\
(ii) Hence write down the equation of the line of symmetry of the curve $y = 3 x ^ { 2 } + 12 x + 7$.
\hfill \mbox{\textit{OCR C1 2005 Q2 [5]}}