Standard +0.3 This requires solving a modulus inequality by considering critical points (x = 3/5 and x = 7/3) and testing intervals, then solving linear inequalities in each region. It's slightly above average difficulty as it involves systematic case analysis and combining solution sets, but follows a standard algorithmic approach taught in P3 with no novel insight required.
State or imply non-modular inequality \((5x-3)^2 < 2^2(3x-7)^2\), or corresponding quadratic equation, or pair of linear equations \((5x-3) = \pm 2(3x-7)\)
B1
\(11x^2 - 138x + 187 > 0\)
Solve a 3-term quadratic, or solve two linear equations for \(x\)
M1
If no working is shown, the M1 is implied by the correct roots for an incorrect quadratic.
Obtain critical values \(x = \dfrac{17}{11}\) and \(x = 11\)
A1
Accept 1.55 or better.
State final answer \(x < \dfrac{17}{11},\ x > 11\)
A1
Strict inequality required. In set notation, allow notation for open sets but not closed sets e.g. accept \(\left(-\infty, \dfrac{17}{11}\right) \cup (11, \infty)\) or \(\left(-\infty, \dfrac{17}{11}\right[ \cup\ ]11, \infty)\) but not \(\left(-\infty, \dfrac{17}{11}\right] \cup [11, \infty)\). Allow 'or' but not 'and'. Accept \(\cup\). Final A0 for \(\dfrac{17}{11} > x > 11\). Exact values expected but ISW if exact inequalities seen followed by decimal approx.
Alternative Method for Question 1:
Answer
Marks
Guidance
Answer
Mark
Guidance
Obtain critical value \(x = 11\) from a graphical method, or by inspection, or by solving a linear equation or an inequality
B1
Obtain critical value \(x = \dfrac{17}{11}\) similarly
B2
Accept decimal value.
State final answer \(x < \dfrac{17}{11},\ x > 11\)
B1
Strict inequality required. See notes above.
4
**Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| State or imply non-modular inequality $(5x-3)^2 < 2^2(3x-7)^2$, or corresponding quadratic equation, or pair of linear equations $(5x-3) = \pm 2(3x-7)$ | **B1** | $11x^2 - 138x + 187 > 0$ |
| Solve a 3-term quadratic, or solve **two** linear equations for $x$ | **M1** | If no working is shown, the M1 is implied by the correct roots for an incorrect quadratic. |
| Obtain critical values $x = \dfrac{17}{11}$ and $x = 11$ | **A1** | Accept 1.55 or better. |
| State **final** answer $x < \dfrac{17}{11},\ x > 11$ | **A1** | Strict inequality required. In set notation, allow notation for open sets but not closed sets e.g. accept $\left(-\infty, \dfrac{17}{11}\right) \cup (11, \infty)$ or $\left(-\infty, \dfrac{17}{11}\right[ \cup\ ]11, \infty)$ but not $\left(-\infty, \dfrac{17}{11}\right] \cup [11, \infty)$. Allow 'or' but not 'and'. Accept $\cup$. Final A0 for $\dfrac{17}{11} > x > 11$. Exact values expected but ISW if exact inequalities seen followed by decimal approx. |
**Alternative Method for Question 1:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Obtain critical value $x = 11$ from a graphical method, or by inspection, or by solving a linear equation or an inequality | **B1** | |
| Obtain critical value $x = \dfrac{17}{11}$ similarly | **B2** | Accept decimal value. |
| State final answer $x < \dfrac{17}{11},\ x > 11$ | **B1** | Strict inequality required. See notes above. |
| | **4** | |