Standard +0.3 This is a straightforward modulus inequality requiring consideration of cases based on sign changes at x=0 and x=-5/2. While it involves multiple cases and some algebraic manipulation, it's a standard textbook exercise testing routine application of modulus inequality techniques without requiring novel insight.
State or imply non-modular inequality \(x^2 < (5 + 2x)^2\), or corresponding equation, or pair of linear equations \(x = \pm(5 + 2x)\)
M1
Obtain critical values \(-5\) and \(-\frac{5}{3}\) only
A1
Obtain final answer \(x < -5, x > -\frac{5}{3}\)
A1
[3]
OR:
Answer
Marks
Guidance
State one critical value e.g. \(-5\), by solving a linear equation or inequality, or from a graphical method, or by inspection
B1
State the other critical value, e.g. \(-\frac{5}{3}\), and no other
B1
Obtain final answer \(x < -5, x > -\frac{5}{3}\)
B1
[3]
[Do not condone \(\le\) or \(\ge\)]
State or imply non-modular inequality $x^2 < (5 + 2x)^2$, or corresponding equation, or pair of linear equations $x = \pm(5 + 2x)$ | M1 |
Obtain critical values $-5$ and $-\frac{5}{3}$ only | A1 |
Obtain final answer $x < -5, x > -\frac{5}{3}$ | A1 | [3]
**OR:**
State one critical value e.g. $-5$, by solving a linear equation or inequality, or from a graphical method, or by inspection | B1 |
State the other critical value, e.g. $-\frac{5}{3}$, and no other | B1 |
Obtain final answer $x < -5, x > -\frac{5}{3}$ | B1 | [3]
[Do not condone $\le$ or $\ge$]