| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Completing the square, form and properties |
| Difficulty | Moderate -0.5 This is a standard C1 completing-the-square question with a straightforward application to distance minimization. Part (a) is routine algebraic manipulation, while part (b) connects it to coordinate geometry through a guided structure that explicitly tells students to use their previous result. The question requires multiple techniques but follows a predictable pattern with clear signposting, making it slightly easier than average. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(2(x + 1.5)^2\) | M1 | OE |
| \(2\left(x + \frac{3}{2}\right)^2 + 0.5\) | A1 | \(2\left(x + \frac{3}{2}\right)^2 + \frac{1}{2}\) OE; 2 marks total |
| (ii) (Minimum value is) \(0.5\) | B1 | ft their \(q\); 1 mark total |
| (b)(i) \(\left(AB^2 =\right) (x + 3)^2 + (3x + 9 - 5)^2\) | M1 | condone one sign error inside one bracket |
| \((3x + 4)^2 = 9x^2 + 24x + 16\) | B1 | OE |
| \(AB^2 = x^2 + 6x + 9 + 9x^2 + 24x + 16 = 10x^2 + 30x + 25\) | ||
| \(\Rightarrow AB^2 = 5(2x^2 + 6x + 5)\) | A1cso | AG; 3 marks total |
| (ii) Either \(\sqrt{5 \times \text{'their'(a)(ii)}}\) or \(5 \times \text{'their'(a)(ii)}\) | M1 | using their minimum value from (a)(ii) and 5; provided "their" (a)(ii) > 0 |
| \(\left(\text{Minimum length of } AB =\right) \frac{1}{2}\sqrt{10}\) | A1cso | 2 marks total |
**(a)(i)** $2(x + 1.5)^2$ | M1 | OE
$2\left(x + \frac{3}{2}\right)^2 + 0.5$ | A1 | $2\left(x + \frac{3}{2}\right)^2 + \frac{1}{2}$ OE; 2 marks total
**(ii)** (Minimum value is) $0.5$ | B1 | ft their $q$; 1 mark total
**(b)(i)** $\left(AB^2 =\right) (x + 3)^2 + (3x + 9 - 5)^2$ | M1 | condone one sign error inside one bracket
$(3x + 4)^2 = 9x^2 + 24x + 16$ | B1 | OE
$AB^2 = x^2 + 6x + 9 + 9x^2 + 24x + 16 = 10x^2 + 30x + 25$ | |
$\Rightarrow AB^2 = 5(2x^2 + 6x + 5)$ | A1cso | AG; 3 marks total
**(ii)** Either $\sqrt{5 \times \text{'their'(a)(ii)}}$ or $5 \times \text{'their'(a)(ii)}$ | M1 | using their minimum value from (a)(ii) and 5; provided "their" (a)(ii) > 0
$\left(\text{Minimum length of } AB =\right) \frac{1}{2}\sqrt{10}$ | A1cso | 2 marks total
**Total: 8 marks**
---5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $2 x ^ { 2 } + 6 x + 5$ in the form $2 ( x + p ) ^ { 2 } + q$, where $p$ and $q$ are rational numbers.
\item Hence write down the minimum value of $2 x ^ { 2 } + 6 x + 5$.
\end{enumerate}\item The point $A$ has coordinates $( - 3,5 )$ and the point $B$ has coordinates $( x , 3 x + 9 )$.
\begin{enumerate}[label=(\roman*)]
\item Show that $A B ^ { 2 } = 5 \left( 2 x ^ { 2 } + 6 x + 5 \right)$.
\item Use your result from part (a)(ii) to find the minimum value of the length $A B$ as $x$ varies, giving your answer in the form $\frac { 1 } { 2 } \sqrt { n }$, where $n$ is an integer.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q5 [8]}}