Standard +0.3 This is a straightforward modulus inequality requiring consideration of critical points (x = 0 and x = -3/4) and testing regions, which is a standard technique taught in P3. While it requires systematic case analysis, it's a routine application of modulus inequality methods with no novel insight needed, making it slightly easier than average.
State or imply non-modular inequality \((4x + 3)^2 > x^2\), or corresponding equation or pair of equations \(4x + 3 = \pm x\)
M1
Obtain a critical value, e.g. \(-1\)
A1
Obtain a second critical value, e.g. \(-\frac{3}{5}\)
A1
State final answer \(x < -1, x > -\frac{3}{5}\)
A1
[4 marks total]
OR:
Answer
Marks
Guidance
Obtain critical value \(x = -1\), by solving a linear equation or inequality, or from a graphical method or by inspection
B1
Obtain the critical value \(-\frac{3}{5}\) similarly
B2
State final answer \(x < -1, x > -\frac{3}{5}\)
B1
[4 marks total]
Guidance: Do not condone \(\leq\) or \(\geq\)
State or imply non-modular inequality $(4x + 3)^2 > x^2$, or corresponding equation or pair of equations $4x + 3 = \pm x$ | M1 |
Obtain a critical value, e.g. $-1$ | A1 |
Obtain a second critical value, e.g. $-\frac{3}{5}$ | A1 |
State final answer $x < -1, x > -\frac{3}{5}$ | A1 | [4 marks total]
**OR:**
Obtain critical value $x = -1$, by solving a linear equation or inequality, or from a graphical method or by inspection | B1 |
Obtain the critical value $-\frac{3}{5}$ similarly | B2 |
State final answer $x < -1, x > -\frac{3}{5}$ | B1 | [4 marks total]
**Guidance:** Do not condone $\leq$ or $\geq$
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