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# C3 Past Paper Question watch

1. I'm a little confused with this one question, it's relatively simple as it's only 2 marks but I don't get how to work it out?

PART 4 of this question:

Thanks.
2. Well you should know that if , so you have to work out the values of for which .

Hint
if
3. (Original post by Josh-H)
I'm a little confused with this one question, it's relatively simple as it's only 2 marks but I don't get how to work it out?

State in terms of a, the set of values of x for which |ln(1/2 x - a)| = -ln(1/2 x - a)

Thanks.

P.s (1/2 x as in x/2)
Is this what you mean?

4. (Original post by steve2005)
Is this what you mean?

Yes that's what I meant but on the actual paper it's written as half x so I tried to demonstrate that then clarify what I meant but ended up failing :P
5. (Original post by nuodai)
Well you should know that if , so you have to work out the values of for which .

Hint
if
Ah I get it now, thank you!
7. this works when ln(1/2 x - a) < or = 0

assume y = x/2 - a
when y<e, ln(y) < 0

ln(y) = 0, where y =1, so y = 1 or y<e

so, x/2 -a < e
x/2 < a+e
x<2(a+e)

or x/2 -a = 1
x = 2a + 2
yes?
8. (Original post by titsmcgee)
this works when ln(1/2 x - a) < or = 0

assume y = x/2 - a
when y<e, ln(y) < 0

ln(y) = 0, where y =1, so y = 1 or y<e

so, x/2 -a < e
x/2 < a+e
x<2(a+e)

or x/2 -a = 1
x = 2a + 2
yes?
The answer I have is 2a < x ≤ 2a+2

It's from the OCR January 2007 C3 Paper.

Here is the question, sorry for any confusion before, it's part 4.

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