Question 5 - A-Level Maths

Edexcel F2 2023 January — Question 5 6 marks

Exam Board	Edexcel
Module	F2 (Further Pure Mathematics 2)
Year	2023
Session	January
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Modulus function
Type	Solve \|quadratic\| compared to linear: algebraic inequality
Difficulty	Challenging +1.2 This is a Further Maths question requiring systematic case analysis of the modulus function, solving quadratic inequalities in each case, and combining solution sets. While methodical, it's a standard technique for FM students with multiple algebraic steps but no novel insight required.
Spec	1.02g Inequalities: linear and quadratic in single variable 1.02l Modulus function: notation, relations, equations and inequalities

In this question you must show all stages of your working.

Solutions relying entirely on calculator technology are not acceptable.
Use algebra to determine the set of values of $x$ for which $$\frac { x ^ { 2 } - 9 } { | x + 8 | } > 6 - 2 x$$

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
$(x=)-8$	B1	This critical value stated or used
For cv's $3, -\frac{19}{3}$ OR for cv's $3, -13$	M1	Any valid attempt to find at least one critical value other than $x=-8$. Condone use of $=$, $>$, $<$ etc. as part of working. NB leads to $3x^2+10x-57=0$ or $x^2+10x-39=0$
Any two of: $x=-13,\ -\frac{19}{3},\ 3$	A1	For any two of these cv's. May be seen embedded in inequalities. Depends on at least one previous M mark.
$-13 < x < -8,\quad -8 < x < -\frac{19}{3},\quad x > 3$	A1 A1	A1: Any 2 of these inequalities. A1: All correct and no other regions. Depends on all previous marks. Note $-13 < x < -\frac{19}{3},\ x\neq -8$ counts as 2 correct. Also condone $-13 < x < -\frac{19}{3},\ x>3$ as 2 correct inequalities.

Total Q5: 6 marks

## Question 5:

| $(x=)-8$ | B1 | This critical value stated or used |

| For cv's $3, -\frac{19}{3}$ **OR** for cv's $3, -13$ | M1 | Any valid attempt to find at least one critical value other than $x=-8$. Condone use of $=$, $>$, $<$ etc. as part of working. NB leads to $3x^2+10x-57=0$ or $x^2+10x-39=0$ |

| Any two of: $x=-13,\ -\frac{19}{3},\ 3$ | A1 | For any two of these cv's. May be seen embedded in inequalities. Depends on at least one previous M mark. |

| $-13 < x < -8,\quad -8 < x < -\frac{19}{3},\quad x > 3$ | A1 A1 | A1: Any 2 of these inequalities. A1: All correct and no other regions. Depends on all previous marks. Note $-13 < x < -\frac{19}{3},\ x\neq -8$ counts as 2 correct. Also condone $-13 < x < -\frac{19}{3},\ x>3$ as 2 correct inequalities. |

**Total Q5: 6 marks**

Show LaTeX source

\begin{enumerate}
  \item In this question you must show all stages of your working.
\end{enumerate}

Solutions relying entirely on calculator technology are not acceptable.\\
Use algebra to determine the set of values of $x$ for which

$$\frac { x ^ { 2 } - 9 } { | x + 8 | } > 6 - 2 x$$

\hfill \mbox{\textit{Edexcel F2 2023 Q5 [6]}}

This paper (9 questions)

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Q1 8 Q2 6 Q3 9 Q4 7 Q5 6 Q6 8 Q7 8 Q8 10 Q9 13

Answer	Marks	Guidance
\((x=)-8\)	B1	This critical value stated or used
For cv's \(3, -\frac{19}{3}\) OR for cv's \(3, -13\)	M1	Any valid attempt to find at least one critical value other than \(x=-8\). Condone use of \(=\), \(>\), \(<\) etc. as part of working. NB leads to \(3x^2+10x-57=0\) or \(x^2+10x-39=0\)
Any two of: \(x=-13,\ -\frac{19}{3},\ 3\)	A1	For any two of these cv's. May be seen embedded in inequalities. Depends on at least one previous M mark.
\(-13 < x < -8,\quad -8 < x < -\frac{19}{3},\quad x > 3\)	A1 A1	A1: Any 2 of these inequalities. A1: All correct and no other regions. Depends on all previous marks. Note \(-13 < x < -\frac{19}{3},\ x\neq -8\) counts as 2 correct. Also condone \(-13 < x < -\frac{19}{3},\ x>3\) as 2 correct inequalities.