question on inequality involving absolute value

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bluenotewitt
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My question concerns the following question from the 2015 june edexcel FP2 exam:
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I know how to go about part a), but (even though I can figure what the correct answer is; x>1) I don't quite understand the answer to b) (marked).

What is the idea behind it? ( please non-algebraic)

Thank you for any help
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nerak99
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Well the modulus signs put the left hand part of the graph as positive and so the place where x+2 is greater shift to only that region to he right of x=1. See sketch to show curve including modulus and line x+2

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bluenotewitt
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(Original post by nerak99)
Well the modulus signs put the left hand part of the graph as positive and so the place where x+2 is greater shift to only that region to he right of x=3. See sketch to show curve including modulus and line x+2

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OK, I see. Is that how you would have answered it in the exam? It seems the examiners want me to use my answer to part (a)? But you appear to be answering it independent of (a)...?

(EDIT: "Hence, or otherwise,..." indicates that one can use a) to answer b))
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nerak99
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In part 1 you find the critical values and interpret them.

If you realised what you were actually dealing with then part b becomes a hence and you realise that the x=1 value was the only one remaining in force. Fro your answer I thought you wanted an explanation rather than a solution.
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bluenotewitt
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(Original post by nerak99)
In part 1 you find the critical values and interpret them.

If you realised what you were actually dealing with then part b becomes a hence and your realise that the x=1 value was the only one remaining in force. Fro your answer I thought you wanted an explanation rather than a solution.
Thank you for your explanation so far. Yes, I was more asking for a solution, but your answers have been helpful. Could you clarify your last answer? i.e. How would you interpret the c.v.s ( roots of a cubic?) ? What are we dealing with? Why is x=1 the only remaining c.v.?
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nerak99
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Mmmm. The original problem had no modulus and so you were dealing with a curve of the form y=1/x and a line y=mx+c. You should be able to "see" that the solutions are going to involve three critical values. The left hand end between where the line crosses the curve and the asymptote and then to the right of x=1 where the line is above he RH part of the curve.

The mechanism is to multiply up and factories the cubic to find the three values of x concerned. This could be accomplished by a person who does not "see" the curves as I describe above.

When we come to (b) the answer is obvious if you can see the curves because the left hand curve is now clearly going to be above any line of the form y=mx+c so long as c<0 but if you can't see, the mechanism to get a solution out of (b) is possible but a bit messy. It is possible-likely that a weaker student will not even try.

In other words the 'hence' is a quick few marks for the more able and is s distinguishing part Q for higher grade students.

You might do well to read the examiners report on that Q
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bluenotewitt
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(Original post by nerak99)
Mmmm. The original problem had no modulus and so you were dealing with a curve of the form y=1/x and a line y=mx+c. You should be able to "see" that the solutions are going to involve three critical values. The left hand end between where the line crosses the curve and the asymptote and then to the right of x=1 where the line is above he RH part of the curve.

The mechanism is to multiply up and factories the cubic to find the three values of x concerned. This could be accomplished by a person who does not "see" the curves as I describe above.

When we come to (b) the answer is obvious if you can see the curves because the left hand curve is now clearly going to be above any line of the form y=mx+c so long as c<0 but if you can't see, the mechanism to get a solution out of (b) is possible but a bit messy. It is possible-likely that a weaker student will not even try.

In other words the 'hence' is a quick few marks for the more able and is s distinguishing part Q for higher grade students.

You might do well to read the examiners report on that Q
Ah OK! Thank you very much! I don't think I would have a problem with imagining the curve. I just wanted to make sure I wasn't missing any trick, as the question was only worth 1 mark. A'levels seem to me very "mechanical", so I don't naturally expect such an approach. Thanks again!
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nerak99
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(Original post by bluenotewitt)
Ah OK! Thank you very much! I don't think I would have a problem with imagining the curve. I just wanted to make sure I wasn't missing any trick, as the question was only worth 1 mark. A'levels seem to me very "mechanical", so I don't naturally expect such an approach. Thanks again!
Part of the examiner's report on this Q (Quoting this is "fair use")

Some students still interpret “using algebra” to mean only using algebra. An algebraic rearrangement to identify all 3 critical values is all that is required. Many students wasted time by then considering sets of values to work out their final inequality. Those who realised a graphical approach is acceptable here, once critical values had been identified, were usually able to identify the required inequalities efficiently. Students who graphed both sides of the original inequality were in a better position to answer part (b) as they could quickly identify which parts of the graph would now be positive. Those who had drawn resulting cubics and other expressions either proceeded to get part (b) wrong or had to draw a second graph, wasting time. Many students felt the need to justify their answer to part (b), when for a single B mark, x > 1 was all that was required. Many responses just restated the inequalities found in (a).
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Chittesh14
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Long way: When x+2 > \frac{12}{x+3}

(x+3)(x+6)(x+1) > 0

Clearly, the solutions are: - 6 < x < -3 or x > 1.

Other one would be x+2 > \frac{12}{-(x+3)} because of the modulus sign.
Multiplying by (x+3)^2 would give:

(x+2)(x+3)^2 &gt; \frac{12(x+3)^2}{-(x+3)}



(x+2)(x+3)^2 &gt; 12(x+3)^2/-(x+3)

(x+2)(x+3)^2 &gt; 12(x+3)/-1



Multiplying both sides by -1 gives:



-(x+2)(x+3)^2 &lt; 12(x+3)

So, 0 &lt; (x+2)(x+3)^2 + 12(x+3)

Therefore, (x+2)(x+3)^2 + 12(x+3) &gt; 0 

(x+3)[(x+2)(x+3) + 12] &gt; 0

(x+3)(x^2+5x+18) &gt; 0

x > -3.

The only solution that satisfies both inequalities is x > 1.

You can draw both inequalities out separately:
One of when x+3 > 0 and one for when x + 3 < 0.

So:

(x+2)(x+3) > 12 (when x+3 > 0) Solution: x < -6 or x > 1. But, as x +3 > 0, x > -3, so solution: x > -3 and x > 1 --> x > 1.
or (x+2)(x+3) < 12 (when x + 3 < 0) hence why inequality sign changes as you are multiplying by (x+3) which is negative. Solution: -6 < x < 1 but as (x+3) < 0, x < -3. So, - 6 < x < -3.
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