Pre-U Pre-U 9795/1 2011 June — Question 7 11 marks

Exam BoardPre-U
ModulePre-U 9795/1 (Pre-U Further Mathematics Paper 1)
Year2011
SessionJune
Marks11
TopicCurve Sketching
TypeSketch rational with quadratic numerator
DifficultyChallenging +1.2 This is a comprehensive curve sketching question requiring identification of asymptotes (vertical at x=1/2, oblique via polynomial division), intercepts, and behavior analysis. While it involves multiple techniques (polynomial division, limits, asymptotic behavior), these are standard Further Maths skills applied systematically rather than requiring novel insight. The 11-mark allocation reflects thoroughness of justification rather than exceptional conceptual difficulty.
Spec1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02y Partial fractions: decompose rational functions
Sketch the curve with equation \(y = \frac{x^2 + 4x}{2x - 1}\), justifying all significant features. [11]
AnswerMarks Guidance
\(y = \frac{x^2+4x}{2x-1} = \frac{1}{2}x(2x-1) + \frac{1}{2}(2x-1) + \frac{1}{2} = \frac{1}{2}x + \frac{9}{4} + \frac{4}{2x-1}\)M1
For \(y = \frac{1}{2}x + c\)A1, A1
For \(c = \frac{9}{4}\) (ignore rend'. term)B1
Vertical asymptote \(x = \frac{1}{2}\) noted or clear from graphB1
\(\frac{dy}{dx} = \frac{(2x-1)(2x+4)-(x^2+4x) \cdot 2}{(2x-1)^2}\)M1 Diff'. and setting num'. = 0
Solving a quadratic in \(x\) (\(x^2 - x - 2 = 0\))M1
TPs at \((-1, 1)\) and \((2, 4)\) One each; or one for both \(x\)'s ✓ but \(y\)'s missingA1, A1
Crossing-points on the axes at \((0, 0)\) and \((-4, 0)\)B1
[Graph showing correct shape with vertical asymptote at \(x = \frac{1}{2}\), turning points, and axis crossings]B1, B1
[11] 
$y = \frac{x^2+4x}{2x-1} = \frac{1}{2}x(2x-1) + \frac{1}{2}(2x-1) + \frac{1}{2} = \frac{1}{2}x + \frac{9}{4} + \frac{4}{2x-1}$ | M1 |
For $y = \frac{1}{2}x + c$ | A1, A1 |
For $c = \frac{9}{4}$ (ignore rend'. term) | B1 |
Vertical asymptote $x = \frac{1}{2}$ noted or clear from graph | B1 |
$\frac{dy}{dx} = \frac{(2x-1)(2x+4)-(x^2+4x) \cdot 2}{(2x-1)^2}$ | M1 | Diff'. and setting num'. = 0
Solving a quadratic in $x$ ($x^2 - x - 2 = 0$) | M1 |
TPs at $(-1, 1)$ and $(2, 4)$ One each; or one for both $x$'s ✓ but $y$'s missing | A1, A1 |
Crossing-points on the axes at $(0, 0)$ and $(-4, 0)$ | B1 |
[Graph showing correct shape with vertical asymptote at $x = \frac{1}{2}$, turning points, and axis crossings] | B1, B1 |
| [11] |

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Sketch the curve with equation $y = \frac{x^2 + 4x}{2x - 1}$, justifying all significant features. [11]

\hfill \mbox{\textit{Pre-U Pre-U 9795/1 2011 Q7 [11]}}
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