Challenging +1.2 This requires systematic case analysis of modulus inequalities with a parameter, squaring both sides to eliminate moduli, and expressing the solution in terms of a. While methodical, it demands careful algebraic manipulation across multiple cases and is more challenging than routine single-variable modulus problems, but remains a standard P3 technique without requiring novel insight.
State or imply non-modular inequality \(2^2(3x+a)^2 < (2x+3a)^2\), or corresponding quadratic equation, or pair of linear equations
B1
e.g. \((6x+2a)^2 = (2x+3a)^2\) or \(32x^2+12xa-5a^2=0\); \(2(3x+a)=(2x+3a)\) and \(-2(3x+a)=(2x+3a)\)
Solve 3-term quadratic, or solve two linear equations for \(x\)
M1
Apply general rules for solving quadratic by formula or factors. Instead of \(x=\{\text{formula}\}\), have \(\{\text{formula}\}=0\) and try to solve for \(a\) then M0
Obtain critical values \(x=\frac{1}{4}a\) and \(x=-\frac{5}{8}a\)
A1
State final answer \(-\frac{5}{8}a < x < \frac{1}{4}a\) or \(-0.625a < x < 0.25a\); or \(x > -\frac{5}{8}a\) and \(x < \frac{1}{4}a\); or \(x > -\frac{5}{8}a \cap x < \frac{1}{4}a\)
A1
Do not condone \(\leqslant\) for \(<\) in final answer. Do not ISW. SC Set \(a\) to value (say \(a=1\)), after initial B1 gained, then \(-\frac{5}{8} < x < \frac{1}{4}\): B1 maximum 2 out of 4
Alternative Method for Question 1:
Answer
Marks
Guidance
Answer
Mark
Guidance
Obtain critical value \(x=\frac{1}{4}a\) from graphical method, or by solving a linear equation or linear inequality
B1
Obtain critical value \(x=-\frac{5}{8}a\) similarly
B2
State final answer \(-\frac{5}{8}a < x < \frac{1}{4}a\) or \(-0.625a < x < 0.25a\); or \(x > -\frac{5}{8}a\) and \(x < \frac{1}{4}a\); or \(x > -\frac{5}{8}a \cap x < \frac{1}{4}a\)
B1
Do not condone \(\leqslant\) for \(<\) in final answer. Do not ISW. SC Set \(a\) to value (say \(a=1\)), after initial B1 gained, then \(-\frac{5}{8} < x < \frac{1}{4}\): B1 maximum 2 out of 4
Total
4
## Question 1:
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality $2^2(3x+a)^2 < (2x+3a)^2$, or corresponding quadratic equation, or pair of linear equations | **B1** | e.g. $(6x+2a)^2 = (2x+3a)^2$ or $32x^2+12xa-5a^2=0$; $2(3x+a)=(2x+3a)$ and $-2(3x+a)=(2x+3a)$ |
| Solve 3-term quadratic, or solve two linear equations for $x$ | **M1** | Apply general rules for solving quadratic by formula or factors. Instead of $x=\{\text{formula}\}$, have $\{\text{formula}\}=0$ and try to solve for $a$ then M0 |
| Obtain critical values $x=\frac{1}{4}a$ and $x=-\frac{5}{8}a$ | **A1** | |
| State final answer $-\frac{5}{8}a < x < \frac{1}{4}a$ or $-0.625a < x < 0.25a$; or $x > -\frac{5}{8}a$ **and** $x < \frac{1}{4}a$; or $x > -\frac{5}{8}a \cap x < \frac{1}{4}a$ | **A1** | Do not condone $\leqslant$ for $<$ in final answer. Do not ISW. **SC** Set $a$ to value (say $a=1$), after initial B1 gained, then $-\frac{5}{8} < x < \frac{1}{4}$: **B1** maximum 2 out of 4 |
**Alternative Method for Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain critical value $x=\frac{1}{4}a$ from graphical method, or by solving a linear equation or linear inequality | **B1** | |
| Obtain critical value $x=-\frac{5}{8}a$ similarly | **B2** | |
| State final answer $-\frac{5}{8}a < x < \frac{1}{4}a$ or $-0.625a < x < 0.25a$; or $x > -\frac{5}{8}a$ **and** $x < \frac{1}{4}a$; or $x > -\frac{5}{8}a \cap x < \frac{1}{4}a$ | **B1** | Do not condone $\leqslant$ for $<$ in final answer. Do not ISW. **SC** Set $a$ to value (say $a=1$), after initial B1 gained, then $-\frac{5}{8} < x < \frac{1}{4}$: **B1** maximum 2 out of 4 |
| **Total** | **4** | |
1 Find, in terms of $a$, the set of values of $x$ satisfying the inequality
$$2 | 3 x + a | < | 2 x + 3 a |$$
where $a$ is a positive constant.\\
\hfill \mbox{\textit{CAIE P3 2022 Q1 [4]}}