Moderate -0.8 This is a straightforward quadratic inequality requiring factorisation (or quadratic formula), identification of critical values, and sketching/testing to determine solution regions. It's a standard C1 exercise with routine steps and no complications, making it easier than average but not trivial since it requires correct application of the sign-change method.
Or \(3(x + \frac{1}{3})(x+3)\); or for \(-\frac{1}{3}\) and \(-3\) found as endpoints eg by use of formula
\(x < -3\)
A1
Mark final answers
[or] \(x > -\frac{1}{3}\) oe
A1
Allow only A1 for \(-3 > x > -\frac{1}{3}\) oe as final answer or for \(x \leq -3\) and \(x \geq -\frac{1}{3}\); A0 for combinations with only one part correct eg \(-3 > x < -\frac{1}{3}\), though this would earn M1 if not already awarded
If M0, allow SC1 for sketch of parabola the right way up with their solutions ft their endpoints
[3]
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(3x+1)(x+3)$ | M1 | Or $3(x + \frac{1}{3})(x+3)$; or for $-\frac{1}{3}$ and $-3$ found as endpoints eg by use of formula |
| $x < -3$ | A1 | Mark final answers |
| [or] $x > -\frac{1}{3}$ oe | A1 | Allow only A1 for $-3 > x > -\frac{1}{3}$ oe as final answer or for $x \leq -3$ and $x \geq -\frac{1}{3}$; A0 for combinations with only one part correct eg $-3 > x < -\frac{1}{3}$, though this would earn M1 if not already awarded |
| | If M0, allow SC1 for sketch of parabola the right way up with their solutions ft their endpoints | |
| | **[3]** | |
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