| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch rational with linear numerator |
| Difficulty | Standard +0.3 This is a straightforward FP1 question combining routine curve sketching of a simple rational function with standard interval bisection. Part (a) requires identifying asymptotes and sketching a basic reciprocal square function—both standard techniques. Part (b) involves verifying a root lies in an interval by sign change and applying the mechanical interval bisection algorithm twice. While it spans multiple techniques, each component is routine and well-practiced at this level, making it slightly easier than average overall. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.09a Sign change methods: locate roots |
| Answer | Marks | Guidance |
|---|---|---|
| 7(a)(i) | Asymptotes \(x = 2, y = 0\) | B1B1 |
| 7(a)(ii) | One correct branch | B1 |
| Both branches correct | B1 | 2 marks |
| 7(b)(i) | \(f(3) = -1, f(4) = 3\) | B1 |
| Sign change, so \(\alpha\) between 3 and 4 | E1 | 2 marks |
| 7(b)(ii) | \(f(3.5)\) considered first | M1 |
| \(f(3.5) > 0\) so \(3 < \alpha < 3.5\) | A1 | Some numerical value(s) needed |
| \(f(3.25) < 0\) so \(3.25 < \alpha < 3.5\) | A1 | 3 marks |
| Total for Q7 | 9 marks |
7(a)(i) | Asymptotes $x = 2, y = 0$ | B1B1 | 2 marks |
7(a)(ii) | One correct branch | B1 | |
| Both branches correct | B1 | 2 marks | no extra branches; $x = 2$ shown |
7(b)(i) | $f(3) = -1, f(4) = 3$ | B1 | where $f(x) = (x-3)(x - 2)^2 - 1$; OE |
| Sign change, so $\alpha$ between 3 and 4 | E1 | 2 marks |
7(b)(ii) | $f(3.5)$ considered first | M1 | OE but must consider $x = 3.5$ |
| $f(3.5) > 0$ so $3 < \alpha < 3.5$ | A1 | Some numerical value(s) needed |
| $f(3.25) < 0$ so $3.25 < \alpha < 3.5$ | A1 | 3 marks | Condone absence of values here |
| **Total for Q7** | **9 marks** |
---7 A curve $C$ has equation $y = \frac { 1 } { ( x - 2 ) ^ { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Write down the equations of the asymptotes of the curve $C$.
\item Sketch the curve $C$.
\end{enumerate}\item The line $y = x - 3$ intersects the curve $C$ at a point which has $x$-coordinate $\alpha$.
\begin{enumerate}[label=(\roman*)]
\item Show that $\alpha$ lies within the interval $3 < x < 4$.
\item Starting from the interval $3 < x < 4$, use interval bisection twice to obtain an interval of width 0.25 within which $\alpha$ must lie.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2010 Q7 [9]}}