Easy -1.2 This is a straightforward quadratic inequality requiring rearrangement to standard form, factorisation of x²+2x-3=(x+3)(x-1), and identification of the solution region x∈(-3,1). It's a routine C1 exercise testing basic algebraic manipulation with no conceptual challenges, making it easier than average.
B3 for \(-3\) and \(1\) or M1 for \(x^2+2x-3\ [< 0]\) or \((x+1)^2 < / = 4\) and M1 for \((x+3)(x-1)\) or \(x = \frac{-2 \pm 4}{2}\) or for \((x+1)\) and \(\pm 2\) on opposite sides of equation or inequality; if 0, then SC1 for one of \(x < 1\), \(x > -3\)
## Question 13:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $-3 < x < 1$ [condone $x < 1, x > -3$] | 4 | B3 for $-3$ and $1$ or M1 for $x^2+2x-3\ [< 0]$ or $(x+1)^2 < / = 4$ and M1 for $(x+3)(x-1)$ or $x = \frac{-2 \pm 4}{2}$ or for $(x+1)$ and $\pm 2$ on opposite sides of equation or inequality; if 0, then SC1 for one of $x < 1$, $x > -3$ |
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