znx
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Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined
Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.
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RDKGames
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(Original post by znx)
Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined
Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.
Note that |x-a| &lt; b \Rightarrow -b &lt; x-a &lt; b
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znx
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(Original post by RDKGames)
Note that |x-a| &lt; b \Rightarrow -b &lt; x-a &lt; b
How would you go about finding a and b?
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RDKGames
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(Original post by znx)
How would you go about finding a and b?
Simultaneous equations.
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znx
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(Original post by RDKGames)
Simultaneous equations.
so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?
EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?
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mupsman2312
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(Original post by znx)
Express 1 < x < 3 in the form |x - a| < b, where a and b are to be determined
Any help would be appreciated on how to solve this question as I am fully stuck on how to even start it.
You could try using the fact that (x-1)(x-3) = {(x-2)^2} - 1. If you apply the right steps after this, then you should end-up with the correct answer.
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mupsman2312
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(Original post by znx)
so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?
EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?
Incredibly, I've never actually come across a question quite like this before, so I don't know what it's called (all that I can say is that it involves the modulus function). Is this in the new spec AS?
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znx
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(Original post by mupsman2312)
Incredibly, I've never actually come across a question quite like this before, so I don't know what it's called. Is this in the new spec AS?
Nope its in the old spec for C3, Its a question from the OCR MEI C3 Jan 2013 paper
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YouMadBro!
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-b + a = 1
b + a = 3
therefore, b=1 and a=2.

(Original post by RDKGames)
Note that |x-a| &lt; b \Rightarrow -b &lt; x-a &lt; b
Can you elaborate how you did that expansion please?

Edit: I just looked at it and realised I was being idiotic and forgetting the definition of what a modulus is, haha XD.
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(Original post by mupsman2312)
You could try using the fact that (x-1)(x-3) = {(x-2)^2} - 1. If you apply the right steps after this, then you should end-up with the correct answer.
This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.

(Original post by znx)
so am I on the right lines by doing x - 3 > 1 and x - 1 < 3?
EDIT: I realize that would not help solve anything. Do you know of the names of these sort of questions so I can go do some revision on them?
Following from my first post, adding a to everything gives a-b &lt; x &lt; a+b. This means a-b=1 and a+b=3

Get it?
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(Original post by RDKGames)
This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.
That's interesting, because I managed to get to the same answer in this way. It might be fairly long-winded, but I'm pretty sure that it still works. Though, yes, the other method is probably more efficient.

Spoiler:
Show



1 < x < 3, so (x-1)(x-3) < 0
=> x^2 - 4x + 3 < 0
=> (x^2 - 4x + 4) - 1 < 0
=> (x-2)^2 - 1 < 0
=> (x-2)^2 < 1
=> -1 < x-2 < 1, i.e. |x-2| < 1
Therefore, a = 2 and b = 1


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znx
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(Original post by RDKGames)
This is not correct - he is not being asked to 'solve' an inequality, simply rewrite one differently.



Following from my first post, adding a to everything gives a-b &lt; x &lt; a+b. This means a-b=1 and a+b=3

Get it?
Yes that makes sense now, so is it just another type of modulus question?
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(Original post by znx)
Yes that makes sense now, so is it just another type of modulus question?
Pretty much.

There is also an alternative (and faster) way to do this question if you understand how modulus inequalities work. When you begin with 1&lt;x&lt;3 your interval here is centered at \frac{1+3}{2}=2. You want to center it at 0. You can subtract this value from everything and get left with -1 &lt; x-2 &lt; 1 which is then saying that the the difference between x and 2 is between -1 and 1, which in other words says that the distance between x and 2 is less than 1, hence |x-2|&lt;1
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(Original post by RDKGames)
Pretty much.

There is also an alternative (and faster) way to do this question if you understand how modulus inequalities work. When you begin with 1&lt;x&lt;3 your interval here is centered at \frac{1+3}{2}=2. You want to center it at 0. You can subtract this value from everything and get left with -1 &lt; x-2 &lt; 1 which is then saying that the the difference between x and 2 is between -1 and 1, which in other words says that the distance between x and 2 is less than 1, hence |x-2|&lt;1
Yeah that sort of rings a bell, I'll go do some modulus revision now. Thank you and everyone else who spared some time to help me.
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