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I am a little puzzled by a past exam question. Part (a) is very easy and part (b) is easy too. However, according to the question, and the mark scheme, I should be able to deduce the answer to part (b) from my answer to part (a) without any further calculation. I am able to find the answer to part (b), but I have to draw some graphs and it is far from straight forward.

Part (a) [6 marks]

Find the set of values of $x$ for which $$x+4>\frac{2}{x+3}$$ As I said: this is easy. Multiply both sides by $(x+3)^2$, take everything over to one side and then factorise. This gives $(x+2)(x+3)(x+5)>0$ and so $-5<x<-3$ or $x>-2$.

Part (b) [1 mark]

This is where it gets a little odd. The question says: Deduce the values of $x$ for which $$x+4>\frac{2}{|x+3|}$$ According to the solutions, part (b) should be blindingly obvious. The answer is $x>-2$. However, I don't see how one can deduce this trivially. I drew the graph and got the right answer, but a 1 mark question means it is obvious and there is no need for calculation.

Adding a modulus usually makes a problem more difficult and adds solutions, e.g. $x-1>2 \iff x>3$, while $|x-1|>2 \iff x<-1 \ \ \text{or} \ \ x>3$.

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The right-hand-side of (b) is positive. This forces $x > -4$. On the other hand, if $-4 < x < -3$, the left-hand side is less than $1$ but the right-hand-side is greater than $1$. Thus $x > -3$.

This means $|x + 3| = x+3$ so the two inequalities (a) and (b) are the same. Since the solution to (a) is $-5 < x < -3$ or $x > -2$, you must have $x > -2$.

I don't know if it is blindingly obvious, but there you go.

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  • $\begingroup$ I was able to come up with a similar argument using the graph. I think that this is worth more than one mark, and isn't the the method that the solutions expects. I'm beginning to think that the exam setter does not understand inequalities. $\endgroup$ – Fly by Night Jan 16 '15 at 19:56
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I don't know what they expected you to do. But notice that $$x+4>\frac{2}{|x+3|}\geq\frac{2}{x+3}$$

This means that any solution of part (b) is a solution of part (a) and that discarding solutions is enough. It is not difficult to get rid of the interval $(-5,-3)$, but I think some argument has to be written in the exam.

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You can see the result from the calc out of a). but not out of yours.

I multiply first $(x+3)$. Then you have 2 cases.

First case $x+3 < 0$ then you've got $(x+4)(x+3) < 2$ (you have to change the sign, because you multipy by a negative figure) this case return the solution $−5<x<−3$

And the second case $x+3 \geq 0$ returns $x>−2.$

It IS quite obvious, that if you multiply by $|x+3|$ there is only the second case. So only the solution of the second case is correct.

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