# Question 1:

$$\frac{1}{x+2} > 2x+3$$

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**Step 1 – Algebraic manipulation to obtain a 3TQ or 4TC:**

$\frac{1-(x+2)(2x+3)}{x+2} > 0 \Rightarrow 2x^2+7x+5=0$

or $x+2>(2x+3)(x+2)^2$

or $\frac{1}{x+2}=2x+3 \Rightarrow (2x+3)(x+2)-1=2x^2+7x+5=0$

| M1 | Uses algebra to obtain a 3TQ, $(x+2)$ multiplied by a 3TQ or a 4TC. Allow slips and condone incorrect inequality signs but the first algebraic step should be otherwise appropriate. Do not accept e.g., $(2x+3)(x+2)=0$. Squaring first acceptable (obtaining a 5TQ). |

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**Step 2 – Critical values from quadratic:**

$x = -\frac{5}{2}, -1$

| A1 | Both $-1$ and $-\frac{5}{2}$ from appropriate work, no extra incorrect critical values. May only appear in solution set. Allow solving 3TQ by calculator. |

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**Step 3 – Critical value $x=-2$:**

$x=-2$

| B1 | Identifies $-2$ as a critical value. May only appear in solution set. This mark available even if no algebraic manipulation seen. Allow from e.g., $(2x+3)(x+2)=0$. |

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**Step 4 – Final solution set:**

$x < -\frac{5}{2},\ -2 < x < -1$

or e.g., $\left(-\infty, -2.5\right),\left(-2,-1\right)$

| M1 A1 | **M1**: Regions $x<a$, $-2<x<b$ with real cvs $a<-2$ and $b>-2$ (condone $b<x<-2$ as notational slip). Condone non-strict inequalities and poor notation. Not dependent but must follow algebraic manipulation attempt. **A1**: Correct solution set in any form. Do not isw if correct inequalities subsequently incorrectly amended. Allow all marks even if incorrect inequality sign seen earlier. |

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**Total: (5)**