Standard +0.3 This is a standard modulus inequality requiring case-by-case analysis based on critical points x = 3 and x = -1/3. Students must consider three intervals, square both sides or use sign analysis, then combine solutions. Slightly above average difficulty due to the algebraic manipulation required, but follows a well-practiced technique for A-level Pure Maths 3.
State or imply non-modular inequality \(2(x-3)^2 > (3x+1)^2\), or corresponding quadratic equation, or pair of linear equations \(2(x-3) = \pm(3x+1)\)
B1
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations
M1
Obtain critical values \(x = -7\) and \(x = 1\)
A1
State answer \(-7 < x < 1\)
A1
[4 marks total]
OR:
Answer
Marks
Obtain critical value \(x = -7\) or \(x = 1\) from a graphical method, by inspection, or by solving a linear equation or inequality
B1
Obtain critical values \(x = -7\) and \(x = 1\)
B2
State answer \(-7 < x < 1\)
B1
[Do not condone: < for <.]
[4 marks total]
State or imply non-modular inequality $2(x-3)^2 > (3x+1)^2$, or corresponding quadratic equation, or pair of linear equations $2(x-3) = \pm(3x+1)$ | B1 |
Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations | M1 |
Obtain critical values $x = -7$ and $x = 1$ | A1 |
State answer $-7 < x < 1$ | A1 |
[4 marks total]
OR:
Obtain critical value $x = -7$ or $x = 1$ from a graphical method, by inspection, or by solving a linear equation or inequality | B1 |
Obtain critical values $x = -7$ and $x = 1$ | B2 |
State answer $-7 < x < 1$ | B1 |
[Do not condone: < for <.] | [4 marks total]