Edexcel FP2 2005 June — Question 6 12 marks

Exam BoardEdexcel
ModuleFP2 (Further Pure Mathematics 2)
Year2005
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicModulus function
TypeSketch modulus functions involving quadratic or other non-linear
DifficultyStandard +0.3 This is a standard Further Maths modulus question requiring sketching two modulus graphs (one quadratic, one linear), solving the equation algebraically by considering cases, and using the graphs to find an inequality solution. While it involves multiple parts and case analysis, the techniques are routine for FP2 students with no novel insight required, making it slightly easier than average.
Spec1.02l Modulus function: notation, relations, equations and inequalities1.02p Interpret algebraic solutions: graphically1.02q Use intersection points: of graphs to solve equations1.02s Modulus graphs: sketch graph of |ax+b|1.02t Solve modulus equations: graphically with modulus function
6. (a) On the same diagram, sketch the graphs of \(y = \left| x ^ { 2 } - 4 \right|\) and \(y = | 2 x - 1 |\), showing the coordinates of the points where the graphs meet the axes.
(b) Solve \(\left| x ^ { 2 } - 4 \right| = | 2 x - 1 |\), giving your answers in surd form where appropriate.
(c) Hence, or otherwise, find the set of values of \(x\) for which \(\left| x ^ { 2 } - 4 \right| > | 2 x - 1 |\).
(3)(Total 12 marks)
Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
Shape – Symmetric about \(y\)-axisB1
Shape – Vertex on positive \(x\)-axisB1
\(-z, z\) markedB1
\(\frac{1}{2}\) markedB1 4 marks total
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x^2 - 4 = 2x - 1\)M1
\(x^2 - 2x - 3 = 0 \Rightarrow x = 3, -1\)A1
\(x^2 - 4 = -(2x-1)\)M1
\(x^2 + 2x - 5 = 0 \Rightarrow x = \frac{-2 \pm \sqrt{4+20}}{2}\)A1
\(x = -1 \pm \sqrt{6}\)A1 5 marks total; correct 3-term quadratic = 0
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(x < -1-\sqrt{6}\ ;\ -1 < x < \sqrt{6} - 1,\ x > 3\)B1ft; B1ft; B1 Accept 3sf; 3 marks total 
# Question 6:

## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Shape – Symmetric about $y$-axis | B1 | |
| Shape – Vertex on positive $x$-axis | B1 | |
| $-z, z$ marked | B1 | |
| $\frac{1}{2}$ marked | B1 | 4 marks total |

## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x^2 - 4 = 2x - 1$ | M1 | |
| $x^2 - 2x - 3 = 0 \Rightarrow x = 3, -1$ | A1 | |
| $x^2 - 4 = -(2x-1)$ | M1 | |
| $x^2 + 2x - 5 = 0 \Rightarrow x = \frac{-2 \pm \sqrt{4+20}}{2}$ | A1 | |
| $x = -1 \pm \sqrt{6}$ | A1 | 5 marks total; correct 3-term quadratic = 0 |

## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x < -1-\sqrt{6}\ ;\ -1 < x < \sqrt{6} - 1,\ x > 3$ | B1ft; B1ft; B1 | Accept 3sf; 3 marks total |

---
6. (a) On the same diagram, sketch the graphs of $y = \left| x ^ { 2 } - 4 \right|$ and $y = | 2 x - 1 |$, showing the coordinates of the points where the graphs meet the axes.\\
(b) Solve $\left| x ^ { 2 } - 4 \right| = | 2 x - 1 |$, giving your answers in surd form where appropriate.\\
(c) Hence, or otherwise, find the set of values of $x$ for which $\left| x ^ { 2 } - 4 \right| > | 2 x - 1 |$.\\
(3)(Total 12 marks)\\

\hfill \mbox{\textit{Edexcel FP2 2005 Q6 [12]}}
This paper (11 questions)
View full paper
Q1 5 Q2 7 Q3 12 Q4 13 Q5 7 Q6 12 Q7 14 Q8 13 Q9 11 Q10 12 Q11 14