| Exam Board | OCR |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Conic sections |
| Type | Conic translation and transformation |
| Difficulty | Challenging +1.2 This is a Further Maths FP2 question on rational functions and asymptotes. Part (i) requires polynomial division to find oblique asymptote (standard technique). Part (ii) requires rearranging to a quadratic in x and using discriminant ≥ 0, which is a well-established method for range problems. While this is Further Maths content, it follows predictable procedures without requiring novel insight, placing it moderately above average difficulty. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x = -1\) is an asymptote | B1 | For correct equation of vertical asymptote |
| \(y = 2x + 1 + \frac{2}{x+1}\) | M1 | For algebraic division, or equivalent |
| Hence \(y = 2x + 1\) is an asymptote | A1 | For correct equation of oblique asymptote |
| \(-\) | 3 | |
| (ii) EITHER: Quadratic \(2x^2 + (3-y)x + (3-y) = 0\) | M1 | For using discriminant of relevant quadratic |
| has no real roots if \((3-y)^2 < 8(3-y)\) | A1 | For correct inequality or equation in \(y\) |
| Hence \((3-y)(−5−y) < 0\) | M1 | For factoring, or equivalent |
| So required values are \(3\) and \(−5\) | A1 | For given answer correctly shown |
| OR: \(\frac{dy}{dx} = 2 - \frac{2}{(x+1)^2} = 0\) | M1 | For differentiating and equating to zero |
| Hence \((x+1)^2 = 1\) | A1 | For correct simplified quadratic in \(x\) |
| So \(x = -2\) and \(0 \Rightarrow y = -5\) and \(3\) | M1 | For solving for \(x\) and substituting to find \(y\) |
| \(-\) | A1 | For given answer correctly shown |
| \(-\) | 4 |
(i) $x = -1$ is an asymptote | B1 | For correct equation of vertical asymptote |
$y = 2x + 1 + \frac{2}{x+1}$ | M1 | For algebraic division, or equivalent |
Hence $y = 2x + 1$ is an asymptote | A1 | For correct equation of oblique asymptote |
$-$ | **3** | |
(ii) **EITHER:** Quadratic $2x^2 + (3-y)x + (3-y) = 0$ | M1 | For using discriminant of relevant quadratic |
has no real roots if $(3-y)^2 < 8(3-y)$ | A1 | For correct inequality or equation in $y$ |
Hence $(3-y)(−5−y) < 0$ | M1 | For factoring, or equivalent |
So required values are $3$ and $−5$ | A1 | For given answer correctly shown |
**OR:** $\frac{dy}{dx} = 2 - \frac{2}{(x+1)^2} = 0$ | M1 | For differentiating and equating to zero |
Hence $(x+1)^2 = 1$ | A1 | For correct simplified quadratic in $x$ |
So $x = -2$ and $0 \Rightarrow y = -5$ and $3$ | M1 | For solving for $x$ and substituting to find $y$ |
$-$ | A1 | For given answer correctly shown |
$-$ | **4** | |
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\includegraphics[max width=\textwidth, alt={}, center]{e4e1c424-8dd5-4d18-9950-e902de0301b0-2_728_951_486_534}
The diagram shows the graph of
$$y = \frac { 2 x ^ { 2 } + 3 x + 3 } { x + 1 }$$
(i) Find the equations of the asymptotes of the curve.\\
(ii) Prove that the values of $y$ between which there are no points on the curve are - 5 and 3 .
\hfill \mbox{\textit{OCR FP2 Q2 [7]}}