A Maths question

• quailmaster

  15 Jan 11, 15:18

  Find the largest positive value of k for which the equation  2^x - 4^(x - 4k) =0 has real roots. Ans: 1/8  

• wee_ws

  15 Jan 11, 19:50

  Hi,

    I feel that something is wrong with this question.

    If we solve the equation, we obtain the solution x = 8k.

    Thus there will be a real root regardless of k.

    Thanks.

  Cheers,
  Wen Shih

• J2011

  15 Jan 11, 21:13

  Hi,

  I suppose the equation should be 2x² - 4(x – 4k) = 0 ?

• Lee012lee

  15 Jan 11, 21:43

  Deleted.

• quailmaster

  15 Jan 11, 22:31

  yeah i had x=8k as well .. thanks

• Lee012lee

  16 Jan 11, 13:46

  Originally posted by wee_ws:

  Hi,

    I feel that something is wrong with this question.

    If we solve the equation, we obtain the solution x = 8k.

    Thus there will be a real root regardless of k.

    Thanks.

  Cheers,
  Wen Shih

  By Subtituting x = 1, 2, 3 .... into x = 8k, the values of k become larger and larger.

  The question requires us to find the largest positive value of k.

  For k to be positive, x must be positive ie x > 0,

  When x > 0, 8k > 0, k > 0.

  Hence, the largest possible value of k will be k = infinity for one real root case 

  (not k = 1/8 as given by TS).

   

• wee_ws

  18 Jan 11, 14:53

  Hi,

    The problem is ill-posed in the first place :P

  Cheers,
  Wen Shih

• Lee012lee

  18 Jan 11, 16:05

  Hi Mr Wee,

  I concur with you.

  Regards.

  PS : I saw the newly published 5-years GCE "N" level Additional Mathematics

         (NA) TYS by Dyna at Big Bookshop a few days ago to which Mikethm, you

         and me were discussing about the possibility of publication of the TYS last

         year as there are only a small number of "N" level students taking "N" level

         AMaths. I ponder whether Dyna can make profit from this TYS.

• wee_ws

  18 Jan 11, 17:04

  Hi,

    I believe Dyna will monopolise this small market, so that is good enough already :)

  Cheers,
  Wen Shih