| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Shaded region area with quadratic |
| Difficulty | Standard +0.3 This is a standard C1 question combining completing the square, finding intersections, and basic integration to find an area. All techniques are routine textbook exercises with clear scaffolding through multiple parts. The area calculation requires subtracting integrals, which is slightly above pure recall, but this is a well-practiced procedure at this level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.02e Complete the square: quadratic polynomials and turning points1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(x - \frac{3}{2}\right)^2\) | M1 | or \(p = 1.5\) stated |
| \(\left(x - \frac{3}{2}\right)^2 + \frac{11}{4}\) | A1 (2 marks) | \((x-1.5)^2 + 2.75\); Mark their final line as their answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = \frac{3}{2}\) | B1\(\checkmark\) (1 mark) | Correct or FT their "\(x = p\)" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x^2 - 3x + 5 = x + 5 \Rightarrow x^2 = 4x\) | M1 | Eliminating \(x\) or \(y\) and collecting like terms (condone one slip); or \((y-5)^2 - 3(y-5) + 5 = y \Rightarrow y^2 - 14y + 45 = 0\) |
| \((x \neq 0) \Rightarrow x = 4\) | A1 | |
| \(y = 9\) | A1 (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{x^3}{3} - \frac{3x^2}{2} + 5x\ (+c)\) | M1, A1, A1 (3 marks) | One of these terms correct; another term correct; all correct (need not have \(+c\)) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left[\ \right]_0^4 = \frac{4^3}{3} - 3 \times \frac{4^2}{2} + 5 \times 4\) | M1 | Must have earned M1 in part (b)(ii); \(F(\text{their } x_B)\{-F(0)\}\) "correctly sub'd" |
| \(= 17\frac{1}{3}\) | A1 | \(\left(\frac{64}{3} - 24 + 20 =\right)\ \frac{52}{3}\) or \(\frac{104}{6}\) etc; condone 17.3 but not \(16\frac{4}{3}\) etc |
| Area of trapezium \(= \frac{1}{2}(x_B)(5 + y_B)\) | B1\(\checkmark\) | FT their numerical values of \(x_B, y_B\); Area \(= \frac{1}{2} \times 4 \times 14\ (= 28)\) |
| Area of shaded region \(= 28 - 17\frac{1}{3}\) | ||
| \(= 10\frac{2}{3}\) | A1 (4 marks) | CSO; \(\frac{32}{3}\), accept 10.7 or better |
## Question 5:
**Part 5(a)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(x - \frac{3}{2}\right)^2$ | M1 | or $p = 1.5$ stated |
| $\left(x - \frac{3}{2}\right)^2 + \frac{11}{4}$ | A1 (2 marks) | $(x-1.5)^2 + 2.75$; Mark their final line as their answer |
**Part 5(a)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = \frac{3}{2}$ | B1$\checkmark$ (1 mark) | Correct or FT their "$x = p$" |
**Part 5(b)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x^2 - 3x + 5 = x + 5 \Rightarrow x^2 = 4x$ | M1 | Eliminating $x$ or $y$ and collecting like terms (condone one slip); **or** $(y-5)^2 - 3(y-5) + 5 = y \Rightarrow y^2 - 14y + 45 = 0$ |
| $(x \neq 0) \Rightarrow x = 4$ | A1 | |
| $y = 9$ | A1 (3 marks) | |
**Part 5(b)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{x^3}{3} - \frac{3x^2}{2} + 5x\ (+c)$ | M1, A1, A1 (3 marks) | One of these terms correct; another term correct; all correct (need not have $+c$) |
**Part 5(b)(iii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left[\ \right]_0^4 = \frac{4^3}{3} - 3 \times \frac{4^2}{2} + 5 \times 4$ | M1 | Must have earned M1 in part (b)(ii); $F(\text{their } x_B)\{-F(0)\}$ "correctly sub'd" |
| $= 17\frac{1}{3}$ | A1 | $\left(\frac{64}{3} - 24 + 20 =\right)\ \frac{52}{3}$ or $\frac{104}{6}$ etc; condone 17.3 but not $16\frac{4}{3}$ etc |
| Area of trapezium $= \frac{1}{2}(x_B)(5 + y_B)$ | B1$\checkmark$ | FT their numerical values of $x_B, y_B$; Area $= \frac{1}{2} \times 4 \times 14\ (= 28)$ |
| Area of shaded region $= 28 - 17\frac{1}{3}$ | | |
| $= 10\frac{2}{3}$ | A1 (4 marks) | CSO; $\frac{32}{3}$, accept 10.7 or better |
---5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $x ^ { 2 } - 3 x + 5$ in the form $( x - p ) ^ { 2 } + q$.
\item Hence write down the equation of the line of symmetry of the curve with equation $y = x ^ { 2 } - 3 x + 5$.
\end{enumerate}\item The curve $C$ with equation $y = x ^ { 2 } - 3 x + 5$ and the straight line $y = x + 5$ intersect at the point $A ( 0,5 )$ and at the point $B$, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{dbc25177-4a28-480f-93d5-41acb2a2d28c-4_471_707_653_676}
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the point $B$.
\item Find $\int \left( x ^ { 2 } - 3 x + 5 \right) \mathrm { d } x$.
\item Find the area of the shaded region $R$ bounded by the curve $C$ and the line segment $A B$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2012 Q5 [13]}}