Turn on thread page Beta
    • Thread Starter
    Offline

    14
    ReputationRep:
    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.
    • PS Helper
    Offline

    14
    PS Helper
    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
    Offline

    14
    ReputationRep:
    (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?

    • Thread Starter
    Offline

    14
    ReputationRep:
    (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
    • Thread Starter
    Offline

    14
    ReputationRep:
    (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!
    Offline

    2
    ReputationRep:
    what paper was this for?
    Offline

    0
    ReputationRep:
    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?
    • Thread Starter
    Offline

    14
    ReputationRep:
    (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.

 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 17, 2010

University open days

  1. Norwich University of the Arts
    Postgraduate Open Days Postgraduate
    Thu, 19 Jul '18
  2. University of Sunderland
    Postgraduate Open Day Postgraduate
    Thu, 19 Jul '18
  3. Plymouth College of Art
    All MA Programmes Postgraduate
    Thu, 19 Jul '18
Poll
How are you feeling in the run-up to Results Day 2018?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.