| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Partial Fractions |
| Type | Rational curve sketching with asymptotes and inequalities |
| Difficulty | Standard +0.3 This is a straightforward FP1 question combining asymptote identification (routine), algebraic manipulation to form a quadratic (standard), and using the tangency condition b²-4ac=0 (well-practiced technique). While it has multiple parts, each step follows standard procedures without requiring novel insight or complex problem-solving, making it slightly easier than average. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.02q Use intersection points: of graphs to solve equations |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Mark |
| (a) | Asymptotes \(x = 1\); \(y = 1\) | B1 B1 |
| (b) | \(-4x + c = \frac{x}{x-1}\) | M1 |
| \((-4x + c)(x - 1) = x\) | A1 | OE (denominators cleared) |
| \(-4x^2 + cx + 4x - c = x\) | ||
| \(-4x^2 + cx + 3x - c = 0\) | ||
| \(4x^2 - (c + 3)x + c = 0\) | A1 | 3 marks; CSO AG No incorrect algebraic expressions etc |
| (c)(i) | Discriminant is \((c + 3)^2 - 4(4c)\) | B1 |
| For tangency \(c^2 - 10c + 9 = 0\) | M1 | Forming a quadratic eqn in c after equating discriminant to zero |
| \((c - 9)(c - 1) = 0 \Rightarrow c = 1\), \(c = 9\) | A1 | 3 marks; Correct values 1, 9 for c |
| (c)(ii) | \(c = 1\): \(4x^2 - 4x + 1 = 0\) \(c = 9\): \(4x^2 - 12x + 9 = 0\) | M1 |
| \(4x^2 - 4x + 1 = 0 \Rightarrow x = 1/2\) (= 0.5) \(4x^2 - 12x + 9 = 0 \Rightarrow x = 3/2\) (= 1.5) | A1 A1 | No other root from quadratic |
| When \(x = 1/2\), \(y = -1\); when \(x = 3/2\), \(y = 3\) \(\left(\frac{1}{2}, -1\right)\) \(\left(\frac{3}{2}, 3\right)\) | A1 | 4 marks; Accept in either format |
| Part | Answer/Working | Mark | Guidance |
|------|---|---|---|
| (a) | Asymptotes $x = 1$; $y = 1$ | B1 B1 | 2 marks; $x = 1$ OE; $y = 1$ OE |
| (b) | $-4x + c = \frac{x}{x-1}$ | M1 | Elimination of y PI by next line |
| | $(-4x + c)(x - 1) = x$ | A1 | OE (denominators cleared) |
| | $-4x^2 + cx + 4x - c = x$ | | |
| | $-4x^2 + cx + 3x - c = 0$ | | |
| | $4x^2 - (c + 3)x + c = 0$ | A1 | 3 marks; CSO AG No incorrect algebraic expressions etc |
| (c)(i) | Discriminant is $(c + 3)^2 - 4(4c)$ | B1 | OE |
| | For tangency $c^2 - 10c + 9 = 0$ | M1 | Forming a quadratic eqn in c after equating discriminant to zero |
| | $(c - 9)(c - 1) = 0 \Rightarrow c = 1$, $c = 9$ | A1 | 3 marks; Correct values 1, 9 for c |
| (c)(ii) | $c = 1$: $4x^2 - 4x + 1 = 0$ $c = 9$: $4x^2 - 12x + 9 = 0$ | M1 | Substitutes at least one of c's values for c from (c)(i) either into the given quadratic in (b) OE or into $\frac{c+3}{8}$ |
| | $4x^2 - 4x + 1 = 0 \Rightarrow x = 1/2$ (= 0.5) $4x^2 - 12x + 9 = 0 \Rightarrow x = 3/2$ (= 1.5) | A1 A1 | No other root from quadratic |
| | When $x = 1/2$, $y = -1$; when $x = 3/2$, $y = 3$ $\left(\frac{1}{2}, -1\right)$ $\left(\frac{3}{2}, 3\right)$ | A1 | 4 marks; Accept in either format |9 A curve has equation
$$y = \frac { x } { x - 1 }$$
\begin{enumerate}[label=(\alph*)]
\item Find the equations of the asymptotes of this curve.
\item Given that the line $y = - 4 x + c$ intersects the curve, show that the $x$-coordinates of the points of intersection must satisfy the equation
$$4 x ^ { 2 } - ( c + 3 ) x + c = 0$$
\item It is given that the line $y = - 4 x + c$ is a tangent to the curve.
\begin{enumerate}[label=(\roman*)]
\item Find the two possible values of $c$.\\
(No credit will be given for methods involving differentiation.)
\item For each of the two values found in part (c)(i), find the coordinates of the point where the line touches the curve.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2012 Q9 [12]}}