OCR FP2 2013 January — Question 2 10 marks

Exam BoardOCR
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPolynomial Division & Manipulation
TypeRational Function Asymptotes
DifficultyStandard +0.8 This is a Further Maths FP2 question requiring multiple techniques: finding vertical and oblique asymptotes (via polynomial division), axis intercepts, differentiation using quotient rule, and synthesizing all information into a coherent sketch. While each individual step is standard, the combination and the oblique asymptote (rather than horizontal) elevates this above typical A-level questions to moderately challenging Further Maths territory.
Spec1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.07n Stationary points: find maxima, minima using derivatives

2 The equation of a curve is \(y = \frac { x ^ { 2 } - 3 } { x - 1 }\).
  1. Find the equations of the asymptotes of the curve.
  2. Write down the coordinates of the points where the curve cuts the axes.
  3. Show that the curve has no stationary points.
  4. Sketch the curve and the asymptotes.

Question 2:
Part (i):
AnswerMarks Guidance
AnswerMarks Guidance
\(x = 1\)B1
\(y = \frac{x^2-3}{x-1} = \frac{(x-1)(x+1)-2}{x-1} = x+1\left[-\frac{2}{x-1}\right]\)M1 Or long division with quotient \(x+\ldots\)
\(\Rightarrow y = x+1\)A1 Must be stated
[3]
Part (ii):
AnswerMarks Guidance
AnswerMarks Guidance
\((0,3)\), \((\sqrt{3},0)\) and \((-\sqrt{3},0)\)B1 All three; allow when \(x=0\), \(y=3\) etc but do NOT allow \(y=3\) etc
[1]
Part (iii):
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{dy}{dx} = \frac{2x(x-1)-(x^2-3)}{(x-1)^2} = \frac{x^2-2x+3}{(x-1)^2}\)M1, A1 Differentiate; Gradient function
\(= \frac{(x-1)^2+2}{(x-1)^2} > 0\) for all \(x\), so no turning pointsA1 Conclusion; Alternative: \(\frac{dy}{dx} = 1 + \frac{2}{(x-1)^2} > 1\) so no turning points. Or "\(b^2-4ac\)"\(= -8 < 0\) so no roots
[3]
Part (iv):
AnswerMarks Guidance
AnswerMarks Guidance
Correct shape going through axes at correct pointsB1 Allow omission of \((0,3)\) if not in (ii)
Correct asymptotes includedB1 Oblique asymptote can be \(y=x+c\) with \(c \neq 1\)
Approaches correct asymptotes correctlyB1
[3]
## Question 2:

### Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $x = 1$ | B1 | |
| $y = \frac{x^2-3}{x-1} = \frac{(x-1)(x+1)-2}{x-1} = x+1\left[-\frac{2}{x-1}\right]$ | M1 | Or long division with quotient $x+\ldots$ |
| $\Rightarrow y = x+1$ | A1 | Must be stated |
| **[3]** | | |

### Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $(0,3)$, $(\sqrt{3},0)$ and $(-\sqrt{3},0)$ | B1 | All three; allow when $x=0$, $y=3$ etc but do NOT allow $y=3$ etc |
| **[1]** | | |

### Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = \frac{2x(x-1)-(x^2-3)}{(x-1)^2} = \frac{x^2-2x+3}{(x-1)^2}$ | M1, A1 | Differentiate; Gradient function |
| $= \frac{(x-1)^2+2}{(x-1)^2} > 0$ for all $x$, so no turning points | A1 | Conclusion; Alternative: $\frac{dy}{dx} = 1 + \frac{2}{(x-1)^2} > 1$ so no turning points. Or "$b^2-4ac$"$= -8 < 0$ so no roots |
| **[3]** | | |

### Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape going through axes at correct points | B1 | Allow omission of $(0,3)$ if not in (ii) |
| Correct asymptotes included | B1 | Oblique asymptote can be $y=x+c$ with $c \neq 1$ |
| Approaches correct asymptotes correctly | B1 | |
| **[3]** | | |

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2 The equation of a curve is $y = \frac { x ^ { 2 } - 3 } { x - 1 }$.\\
(i) Find the equations of the asymptotes of the curve.\\
(ii) Write down the coordinates of the points where the curve cuts the axes.\\
(iii) Show that the curve has no stationary points.\\
(iv) Sketch the curve and the asymptotes.

\hfill \mbox{\textit{OCR FP2 2013 Q2 [10]}}