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    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.
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    Well you should know that |x| = -x if x \le 0, so you have to work out the values of x for which \ln \left( \frac{x}{2} - a \right) \le 0.

    Hint
    \ln u \le 0 if 0 < u \le 1
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    (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?

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    (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
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    (Original post by nuodai)
    Well you should know that |x| = -x if x \le 0, so you have to work out the values of x for which \ln \left( \frac{x}{2} - a \right) \le 0.

    Hint
    \ln u \le 0 if 0 < u \le 1
    Ah I get it now, thank you!
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    what paper was this for?
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    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?
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    (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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