| Exam Board | Edexcel |
|---|---|
| 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 C1 completing-the-square question with standard parts: completing the square (routine algebraic manipulation), reading off the maximum from completed square form (direct observation), solving a quadratic (standard formula application), and sketching (plotting key features already found). All techniques are basic and well-practiced at this level, making it easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(f(x) = 9 - [x^2 - 6x]\) | M1 | |
| \(= 9 - [(x-3)^2 - 9]\) | M1 | |
| \(= 18 - (x-3)^2\), so \(A = 18, B = -3\) | A2 | |
| (b) Maximum value is \(18\) | B1 | |
| (c) \(18 - (x-3)^2 = 0\) gives \(x - 3 = \pm\sqrt{18}\) | M1 | |
| \(x = 3 \pm 3\sqrt{2}\) | M1 A1 | |
| (d) Sketch showing inverted parabola with vertex on \(y\)-axis and roots either side | B2 | (10) |
**(a)** $f(x) = 9 - [x^2 - 6x]$ | M1 |
$= 9 - [(x-3)^2 - 9]$ | M1 |
$= 18 - (x-3)^2$, so $A = 18, B = -3$ | A2 |
**(b)** Maximum value is $18$ | B1 |
**(c)** $18 - (x-3)^2 = 0$ gives $x - 3 = \pm\sqrt{18}$ | M1 |
$x = 3 \pm 3\sqrt{2}$ | M1 A1 |
**(d)** Sketch showing inverted parabola with vertex on $y$-axis and roots either side | B2 | (10)6.
$$f ( x ) = 9 + 6 x - x ^ { 2 } .$$
\begin{enumerate}[label=(\alph*)]
\item Find the values of $A$ and $B$ such that
$$\mathrm { f } ( x ) = A - ( x + B ) ^ { 2 }$$
\item State the maximum value of $\mathrm { f } ( x )$.
\item Solve the equation $\mathrm { f } ( x ) = 0$, giving your answers in the form $a + b \sqrt { 2 }$ where $a$ and $b$ are integers.
\item Sketch the curve $y = \mathrm { f } ( x )$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 Q6 [10]}}