1 Solve the inequality \(| 2 x - 5 | > x\).
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Question 1:
Answer Marks
Guidance
Answer Mark
Guidance
Solve \(2x - 5 = x\) to obtain \(x = 5\) B1 Attempt solution of linear equation where signs of \(2x\) and \(x\) are different M1 Obtain \(x = \frac{5}{3}\) A1 Conclude \(x < \frac{5}{3}\), \(x > 5\) A1 Must be 2 separate inequalities. Allow equivalents \(\left(-\infty, \frac{5}{3}\right) \cup (5, \infty)\)
Alternative method for Question 1:
Answer Marks
Guidance
Answer Mark
Guidance
State or imply non-modulus equation \((2x-5)^2 = x^2\) B1 Attempt solution of 3-term quadratic equation M1 Obtain \(\frac{5}{3}\) and \(5\) A1 Conclude \(x < \frac{5}{3}\), \(x > 5\) A1 Must be 2 separate inequalities. Allow equivalents \(\left(-\infty, \frac{5}{3}\right) \cup (5, \infty)\)
Total: 4 marks
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**Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Solve $2x - 5 = x$ to obtain $x = 5$ | **B1** | |
| Attempt solution of linear equation where signs of $2x$ and $x$ are different | **M1** | |
| Obtain $x = \frac{5}{3}$ | **A1** | |
| Conclude $x < \frac{5}{3}$, $x > 5$ | **A1** | Must be 2 separate inequalities. Allow equivalents $\left(-\infty, \frac{5}{3}\right) \cup (5, \infty)$ |
**Alternative method for Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modulus equation $(2x-5)^2 = x^2$ | **B1** | |
| Attempt solution of 3-term quadratic equation | **M1** | |
| Obtain $\frac{5}{3}$ and $5$ | **A1** | |
| Conclude $x < \frac{5}{3}$, $x > 5$ | **A1** | Must be 2 separate inequalities. Allow equivalents $\left(-\infty, \frac{5}{3}\right) \cup (5, \infty)$ |
**Total: 4 marks**
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1 Solve the inequality $| 2 x - 5 | > x$.\\
\hfill \mbox{\textit{CAIE P2 2022 Q1 [4]}}