Standard +0.8 This requires systematic case analysis of the modulus inequality by considering critical points x = a/2 and x = -3a, solving multiple linear inequalities in each region, then combining solutions. While methodical, it demands careful algebraic manipulation and logical organization beyond routine modulus problems, placing it moderately above average difficulty.
22 ( 2x−a )2 <( x+3a )2, or corresponding quadratic equation, or pair of
linear equations
Answer
Marks
2(2x – a) = ± (x + 3a)
B1
Make reasonable attempt at solving a 3-term quadratic, or solve two linear
Answer
Marks
equations for x
M1
5 1
Obtain critical values x= a and x=− a
Answer
Marks
3 5
A1
1 5
State final answer − a<x< a
Answer
Marks
5 3
A1
5
OR: Obtain critical value x= a from a graphical method, or by inspection, or
3
Answer
Marks
by solving a linear equation or an inequality
B1
1
Obtain critical value x=− a similarly
Answer
Marks
5
B2
1 5
State final answer − a<x< a
5 3
Answer
Marks
[Do not condone ⩽ for < in the final answer.]
B1
4
Answer
Marks
Guidance
Question
Answer
Marks
Question 1:
1 | EITHER: State or imply non-modular inequality
22 ( 2x−a )2 <( x+3a )2, or corresponding quadratic equation, or pair of
linear equations
2(2x – a) = ± (x + 3a) | B1
Make reasonable attempt at solving a 3-term quadratic, or solve two linear
equations for x | M1
5 1
Obtain critical values x= a and x=− a
3 5 | A1
1 5
State final answer − a<x< a
5 3 | A1
5
OR: Obtain critical value x= a from a graphical method, or by inspection, or
3
by solving a linear equation or an inequality | B1
1
Obtain critical value x=− a similarly
5 | B2
1 5
State final answer − a<x< a
5 3
[Do not condone ⩽ for < in the final answer.] | B1
4
Question | Answer | Marks | Guidance
Find the set of values of $x$ satisfying the inequality $2|2x - a| < |x + 3a|$, where $a$ is a positive constant. [4]
\hfill \mbox{\textit{CAIE P3 2018 Q1 [4]}}