| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Integration with Partial Fractions |
| Type | Rational function curve sketching |
| Difficulty | Challenging +1.2 This is a Further Maths question requiring curve sketching of a rational function, involving asymptote identification, differentiation using quotient rule, and finding intersection points. While it requires multiple techniques and careful analysis across four parts, each individual step is relatively standard for FP2 level—asymptotes from denominator zeros, stationary points via differentiation, and intersection solving. The multi-part nature and Further Maths context place it above average difficulty, but it doesn't require novel insight or particularly complex algebraic manipulation. |
| Spec | 1.02n Sketch curves: simple equations including polynomials1.02y Partial fractions: decompose rational functions1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x = -1\) | B1 | B1 for each |
| \(x = 7\) | B1 | |
| \(y = 1\) | B1 | \(-1\) for any extras |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = \frac{(x^2-6x-7)2x-(x^2+1)(2x-6)}{(x+1)^2(x-7)^2}\) | M1, A1 | Differentiation using quotient rule. Or expand as partial fractions and use fn of fn rule |
| \(= 0\) when \((x^2-6x-7)2x-(x^2+1)(2x-6)=0\) | ||
| \(3x^2+8x-3=0\) | A1 | Quadratic |
| \(\Rightarrow x=-3, \frac{1}{3}\); \(\quad y=\frac{1}{2}, -\frac{1}{8}\) | A1 | Both \(x\) values. Or: A1 one pair |
| i.e. \(\left(-3,\frac{1}{2}\right), \left(\frac{1}{3},-\frac{1}{8}\right)\) | A1 | Both \(y\) values. A1 other pair |
| Answer | Marks | Guidance |
|---|---|---|
| When \(y=1\), \(x^2-6x-7=x^2+1\) | M1, A1 | |
| \(\Rightarrow 6x=-8 \Rightarrow x=-\frac{4}{3} \Rightarrow \left(-\frac{4}{3},1\right)\) | A1 | Coordinate pair needs to be seen |
| Answer | Marks | Guidance |
|---|---|---|
| [Sketch of curve] | B1 | Left section, cutting asymptote and approaching \(y=1\) from below |
| B1 | Right hand section | |
| B1 | Middle section all below \(x\)-axis, labelling intercept on graph or by a statement |
# Question 7(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = -1$ | B1 | B1 for each |
| $x = 7$ | B1 | |
| $y = 1$ | B1 | $-1$ for any extras |
| **[3]** | | |
# Question 7(ii):
$\frac{dy}{dx} = \frac{(x^2-6x-7)2x-(x^2+1)(2x-6)}{(x+1)^2(x-7)^2}$ | M1, A1 | Differentiation using quotient rule. Or expand as partial fractions and use fn of fn rule |
$= 0$ when $(x^2-6x-7)2x-(x^2+1)(2x-6)=0$ | |
$3x^2+8x-3=0$ | A1 | Quadratic |
$\Rightarrow x=-3, \frac{1}{3}$; $\quad y=\frac{1}{2}, -\frac{1}{8}$ | A1 | Both $x$ values. Or: A1 one pair |
i.e. $\left(-3,\frac{1}{2}\right), \left(\frac{1}{3},-\frac{1}{8}\right)$ | A1 | Both $y$ values. A1 other pair |
**[5]**
---
# Question 7(iii):
When $y=1$, $x^2-6x-7=x^2+1$ | M1, A1 |
$\Rightarrow 6x=-8 \Rightarrow x=-\frac{4}{3} \Rightarrow \left(-\frac{4}{3},1\right)$ | A1 | Coordinate pair needs to be seen |
**[3]**
---
# Question 7(iv):
[Sketch of curve] | B1 | Left section, cutting asymptote and approaching $y=1$ from below |
| B1 | Right hand section |
| B1 | Middle section all below $x$-axis, labelling intercept on graph or by a statement |
**[3]**
---7 The equation of a curve is $y = \frac { x ^ { 2 } + 1 } { ( x + 1 ) ( x - 7 ) }$.\\
(i) Write down the equations of the asymptotes.\\
(ii) Find the coordinates of the stationary points on the curve.\\
(iii) Find the coordinates of the point where the curve meets one of its asymptotes.\\
(iv) Sketch the curve.
\hfill \mbox{\textit{OCR FP2 2013 Q7 [14]}}