| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 10 |
| 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 question with standard follow-up parts. Part (i) is routine algebraic manipulation, parts (ii) and (iii) follow directly from the completed square form, and part (iv) is a basic sketch. All techniques are standard C1 content with no problem-solving insight required, making it easier than average but not trivial due to the multi-part structure. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(f(x) = 9 - [x^2 - 6x] = 9 - [(x-3)^2 - 9] = 18 - (x-3)^2\), \(A = 18, B = -3\) | M1, M1, A2 | |
| (ii) \(18\) | B1 | |
| (iii) \(18 - (x-3)^2 = 0\); \(x - 3 = \pm\sqrt{18}\); \(x = 3 \pm 3\sqrt{2}\) | M1; M1, A1 | |
| (iv) Graph showing parabola with vertex above origin, crossing x-axis at two points, symmetric about \(x = 3\) | B2 | (10) |
**(i)** $f(x) = 9 - [x^2 - 6x] = 9 - [(x-3)^2 - 9] = 18 - (x-3)^2$, $A = 18, B = -3$ | M1, M1, A2 |
**(ii)** $18$ | B1 |
**(iii)** $18 - (x-3)^2 = 0$; $x - 3 = \pm\sqrt{18}$; $x = 3 \pm 3\sqrt{2}$ | M1; M1, A1 |
**(iv)** Graph showing parabola with vertex above origin, crossing x-axis at two points, symmetric about $x = 3$ | B2 | (10)8.
$$f ( x ) = 9 + 6 x - x ^ { 2 } .$$
(i) Find the values of $A$ and $B$ such that
$$\mathrm { f } ( x ) = A - ( x + B ) ^ { 2 }$$
(ii) State the maximum value of $\mathrm { f } ( x )$.\\
(iii) Solve the equation $\mathrm { f } ( x ) = 0$, giving your answers in the form $a + b \sqrt { 2 }$ where $a$ and $b$ are integers.\\
(iv) Sketch the curve $y = \mathrm { f } ( x )$.\\
\hfill \mbox{\textit{OCR C1 Q8 [10]}}