| Exam Board | OCR MEI |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Rational function curve sketching |
| Difficulty | Standard +0.8 This FP1 question requires identifying asymptotes and intercepts from a rational function (routine), solving an inequality using sign analysis (standard), and most significantly, finding the range of k for which a rational equation has no solutions by analyzing turning points or discriminant conditions. The final part demands connecting algebraic conditions to graphical features, requiring deeper conceptual understanding beyond mechanical application. The multi-step reasoning and synthesis of algebra with graphical interpretation elevates this above average difficulty. |
| Spec | 1.02g Inequalities: linear and quadratic in single variable1.02i Represent inequalities: graphically on coordinate plane1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Asymptotes \(y=0\), \(x=5\), \(x=8\) | B1, B1 | both |
| Crosses axes at \((4,\ 0)\), \(\left(0, -\frac{1}{10}\right)\) | B1 B1 | |
| \(\frac{x-4}{(x-5)(x-8)} > 0 \Rightarrow x > 8\) or \(4 < x < 5\) | B1 B1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{x-4}{(x-5)(x-8)} = k \Rightarrow x - 4 = kx^2 - 13kx + 40k\) | M1 | Attempt to remove fraction and simplify |
| \(\Rightarrow kx^2 - (13k+1)x + 40k + 4 = 0\) | A1 | 3 term quadratic \((= 0)\) |
| \(b^2 - 4ac = (13k+1)^2 - 4k(40k+4)\) | M1 | Attempt to use discriminant |
| \(= 9k^2 + 10k + 1\) | A1 | Correct 3-term quadratic |
| Critical values \(-1, -\frac{1}{9}\) | A1 | Roots found or factors shown |
| For no solutions to exist, \(9k^2 + 10k + 1 < 0 \Rightarrow -1 < k < -\frac{1}{9}\) | E1 | |
| No point on the graph has a \(y\) coordinate in the range \(\Rightarrow -1 < y < -\frac{1}{9}\) | E1 | Accept equivalent statement |
| [7] |
## Question 7:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Asymptotes $y=0$, $x=5$, $x=8$ | B1, B1 | both |
| Crosses axes at $(4,\ 0)$, $\left(0, -\frac{1}{10}\right)$ | B1 B1 | |
| $\frac{x-4}{(x-5)(x-8)} > 0 \Rightarrow x > 8$ or $4 < x < 5$ | B1 B1 | |
| | **[6]** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{x-4}{(x-5)(x-8)} = k \Rightarrow x - 4 = kx^2 - 13kx + 40k$ | M1 | Attempt to remove fraction and simplify |
| $\Rightarrow kx^2 - (13k+1)x + 40k + 4 = 0$ | A1 | 3 term quadratic $(= 0)$ |
| $b^2 - 4ac = (13k+1)^2 - 4k(40k+4)$ | M1 | Attempt to use discriminant |
| $= 9k^2 + 10k + 1$ | A1 | Correct 3-term quadratic |
| Critical values $-1, -\frac{1}{9}$ | A1 | Roots found or factors shown |
| For no solutions to exist, $9k^2 + 10k + 1 < 0 \Rightarrow -1 < k < -\frac{1}{9}$ | E1 | |
| No point on the graph has a $y$ coordinate in the range $\Rightarrow -1 < y < -\frac{1}{9}$ | E1 | Accept equivalent statement |
| | **[7]** | |
---7 Fig. 7 shows a sketch of $y = \frac { x - 4 } { ( x - 5 ) ( x - 8 ) }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e47c6fb-574b-4eee-81c8-4031fee9e2ba-3_696_975_406_529}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Write down the equations of the three asymptotes and the coordinates of the points where the curve crosses the axes. Hence write down the solution of the inequality $\frac { x - 4 } { ( x - 5 ) ( x - 8 ) } > 0$.\\
(ii) The equation $\frac { x - 4 } { ( x - 5 ) ( x - 8 ) } = k$ has no real solutions. Show that $- 1 < k < - \frac { 1 } { 9 }$. Relate this result to the graph of $y = \frac { x - 4 } { ( x - 5 ) ( x - 8 ) }$.
\hfill \mbox{\textit{OCR MEI FP1 2013 Q7 [13]}}