Question 11 EITHER - A-Level Maths

CAIE FP1 2014 June — Question 11 EITHER

Exam Board	CAIE
Module	FP1 (Further Pure Mathematics 1)
Year	2014
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Partial Fractions
Type	Rational curve analysis with turning points and range restrictions
Difficulty	Standard +0.8 This is a multi-part Further Maths question requiring partial fractions with improper fractions, differentiation of a rational function (quotient rule), analysis of stationary points, and a comprehensive curve sketch including vertical and horizontal asymptotes. While each technique is standard, the combination and the requirement to analyze intersections with asymptotes elevates this beyond typical A-level questions to solid Further Maths territory.
Spec	1.02n Sketch curves: simple equations including polynomials 1.02y Partial fractions: decompose rational functions 1.07b Gradient as rate of change: dy/dx notation

Express \(\frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }\) in the form \(2 + \frac { A } { x - 1 } + \frac { B } { x + 1 }\), where \(A\) and \(B\) are integers to be found. The curve \(C\) has equation \(y = \frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }\). Show that there are two distinct values of \(x\) for which \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 0\). Sketch \(C\), stating the equations of the asymptotes and giving the coordinates of any points of intersection with the coordinate axes and with the asymptotes. You do not need to find the coordinates of the turning points.

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Question 11:

EITHER Path

Partial Fractions Setup:

Answer	Marks
\(2 + \frac{A}{x-1} + \frac{B}{x+1} = \frac{2x^2-1+A(x+1)+B(x-1)}{x^2-1} = \frac{2x^2+(A+B)x+A-B-2}{x^2-1} = \frac{2x^2-x+5}{x^2-1}\)	M1
\(\Rightarrow A = 3\), \(B = -4\)	A1A1 [3]

Turning Points:

Answer	Marks
\(y' = 0 \Rightarrow (x^2-1)(4x-1)-(2x^2-x+5).2x = 0 \Rightarrow x^2-14x+1=0\)	M1A1
\(B^2-4AC = (-14)^2 - 4 \times 1 \times 1 = 192 > 0 \Rightarrow\) two distinct turning points	M1A1 [4]

Asymptotes and Intersection:

Answer	Marks	Guidance
Asymptotes are \(x=1\), \(x=-1\), \(y=2\)	B1B1
\(y=2 \Rightarrow 2x^2-x+5=2x^2-2 \Rightarrow x=7 \Rightarrow (7,2)\)	M1A1	Accept if labelled on graph

Sketch:

Answer	Marks
Middle branch crossing \(y\)-axis at \((0,-5)\) and left branch	B1
Right branch	B1
Working to show no intersections with \(x\)-axis	B1 [7]

OR Path

(i):

Answer	Marks
\(\mathbf{r} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\lambda(\mathbf{i}+7\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-\mathbf{k}) \Rightarrow A\) is in \(\Pi_1\)	B1 [1]

(ii):

Answer	Marks
\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 7 & 1 \\ 3 & 1 & -1 \end{vmatrix} = \begin{pmatrix}-8\\4\\-20\end{pmatrix} \sim \begin{pmatrix}2\\-1\\5\end{pmatrix}\)	M1A1 [2]

(iii):

Answer	Marks
\(L\) is \((12,-6,6)\)	B1
\(2x-y+5z = 24+6+30=60\)	B1
\(\mathbf{n} = t(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) \Rightarrow 4t+t+25t=60 \Rightarrow t=2\)	M1A1\(\checkmark\)
\(\mathbf{n} = 4\mathbf{i}-2\mathbf{j}+10\mathbf{k}\)	A1 [5]

(iv):

Answer	Marks	Guidance
\(\mathbf{m} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\frac{3}{4}(8\mathbf{i}-4\mathbf{j}+4\mathbf{k}) = 10\mathbf{i}-5\mathbf{j}+5\mathbf{k}\) (AG)	B1
\(M\) is \((10,-5,5) \Rightarrow \overrightarrow{NM} = 6\mathbf{i}-3\mathbf{j}-5\mathbf{k}\)	B1\(\checkmark\)
\((6\mathbf{i}-3\mathbf{j}-5\mathbf{k})\times(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) = -\mathbf{i}+2\mathbf{j}\)	M1A1
Perpendicular distance is \(\dfrac{\	20(-\mathbf{i}+2\mathbf{j})\	}{\sqrt{30}} = \dfrac{20}{\sqrt{6}} = 8.16\)

# Question 11:

## EITHER Path

**Partial Fractions Setup:**
| $2 + \frac{A}{x-1} + \frac{B}{x+1} = \frac{2x^2-1+A(x+1)+B(x-1)}{x^2-1} = \frac{2x^2+(A+B)x+A-B-2}{x^2-1} = \frac{2x^2-x+5}{x^2-1}$ | M1 | |
|---|---|---|
| $\Rightarrow A = 3$, $B = -4$ | A1A1 [3] | |

**Turning Points:**
| $y' = 0 \Rightarrow (x^2-1)(4x-1)-(2x^2-x+5).2x = 0 \Rightarrow x^2-14x+1=0$ | M1A1 | |
|---|---|---|
| $B^2-4AC = (-14)^2 - 4 \times 1 \times 1 = 192 > 0 \Rightarrow$ two distinct turning points | M1A1 [4] | |

**Asymptotes and Intersection:**
| Asymptotes are $x=1$, $x=-1$, $y=2$ | B1B1 | |
|---|---|---|
| $y=2 \Rightarrow 2x^2-x+5=2x^2-2 \Rightarrow x=7 \Rightarrow (7,2)$ | M1A1 | Accept if labelled on graph |

**Sketch:**
| Middle branch crossing $y$-axis at $(0,-5)$ and left branch | B1 | |
|---|---|---|
| Right branch | B1 | |
| Working to show no intersections with $x$-axis | B1 [7] | |

---

## OR Path

**(i):**
| $\mathbf{r} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\lambda(\mathbf{i}+7\mathbf{j}+\mathbf{k})+\mu(3\mathbf{i}+\mathbf{j}-\mathbf{k}) \Rightarrow A$ is in $\Pi_1$ | B1 [1] | |
|---|---|---|

**(ii):**
| $\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 7 & 1 \\ 3 & 1 & -1 \end{vmatrix} = \begin{pmatrix}-8\\4\\-20\end{pmatrix} \sim \begin{pmatrix}2\\-1\\5\end{pmatrix}$ | M1A1 [2] | |
|---|---|---|

**(iii):**
| $L$ is $(12,-6,6)$ | B1 | |
|---|---|---|
| $2x-y+5z = 24+6+30=60$ | B1 | |
| $\mathbf{n} = t(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) \Rightarrow 4t+t+25t=60 \Rightarrow t=2$ | M1A1$\checkmark$ | |
| $\mathbf{n} = 4\mathbf{i}-2\mathbf{j}+10\mathbf{k}$ | A1 [5] | |

**(iv):**
| $\mathbf{m} = 4\mathbf{i}-2\mathbf{j}+2\mathbf{k}+\frac{3}{4}(8\mathbf{i}-4\mathbf{j}+4\mathbf{k}) = 10\mathbf{i}-5\mathbf{j}+5\mathbf{k}$ (AG) | B1 | |
|---|---|---|
| $M$ is $(10,-5,5) \Rightarrow \overrightarrow{NM} = 6\mathbf{i}-3\mathbf{j}-5\mathbf{k}$ | B1$\checkmark$ | |
| $(6\mathbf{i}-3\mathbf{j}-5\mathbf{k})\times(2\mathbf{i}-\mathbf{j}+5\mathbf{k}) = -\mathbf{i}+2\mathbf{j}$ | M1A1 | |
| Perpendicular distance is $\dfrac{\|20(-\mathbf{i}+2\mathbf{j})\|}{\sqrt{30}} = \dfrac{20}{\sqrt{6}} = 8.16$ | M1A1 [6] | Mark various alternative methods similarly |

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Express $\frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }$ in the form $2 + \frac { A } { x - 1 } + \frac { B } { x + 1 }$, where $A$ and $B$ are integers to be found.

The curve $C$ has equation $y = \frac { 2 x ^ { 2 } - x + 5 } { x ^ { 2 } - 1 }$. Show that there are two distinct values of $x$ for which $\frac { \mathrm { d } y } { \mathrm {~d} x } = 0$.

Sketch $C$, stating the equations of the asymptotes and giving the coordinates of any points of intersection with the coordinate axes and with the asymptotes. You do not need to find the coordinates of the turning points.

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