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Find least value of a

  1. Nov 1, 2012 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    Find all the values of the parameter 'a' for which the inequality [itex]4^x-a2^x-a+3 \leq 0[/itex] is satisfied by at least one real x.


    2. Relevant equations

    3. The attempt at a solution
    Let [itex]2^x=t[/itex]
    Then the inequality changes to
    [itex]t^2-at-(a-3)\leq0[/itex]
    For the required condition D must be greater than or equal to 0.

    [itex]a^2+4a-12\geq0[/itex]
    [itex](a-2)(a+6)\geq0[/itex]
    [itex]a\in (-∞,6]U[2,∞)[/itex]
     
  2. jcsd
  3. Nov 1, 2012 #2

    lurflurf

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    I do not understand what you have done.
    I would have written
    2<=(4^x+3)/(2^x+1)<=a
     
  4. Nov 1, 2012 #3

    utkarshakash

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    How can you write this?
     
  5. Nov 1, 2012 #4

    SammyS

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    I take it that 'D' refers to the discriminant .

    What you have so far is correct.

    Can 2x ever be negative?
     
  6. Nov 1, 2012 #5

    lurflurf

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    using your
    t=2^x
    (t^2+3)-a(t^2+1)<=0
    (t^2+3)<=a(t^2+1)<=0
    (t^2+3)/(t^2+1)<=a
    in fact s=(2^x+1)/2 is a nicer substitution

    (t^2+3)/(t^2+1)=2((2^x+1)/2)+1/((2^x+1)/2)-1)=2(s+1/s-1)
    since ((2^x+1)/2)(1/((2^x+1)/2))=s/s=1 the minimum occurs when
    ((2^x+1)/2)=(1/((2^x+1)/2)) or s=1/s

    or use calculus if you like
     
  7. Nov 1, 2012 #6

    utkarshakash

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    You are right. It can never be negative. But substituting 6 for a does not give a -ve t. So that must be included.
     
  8. Nov 1, 2012 #7

    SammyS

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    That should be [itex]\displaystyle a\in (-\infty,-6]\cup[2,\infty)[/itex] in your Original Post.
     
  9. Nov 2, 2012 #8

    utkarshakash

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    Yes it should be -6 instead of 6 (that was just a typing mistake) but the answer is still incorrect.
     
  10. Nov 2, 2012 #9

    SammyS

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    Check a = -6. See what that makes t and thus 2x.
     
  11. Nov 2, 2012 #10

    ehild

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    Remember t=2x must be positive. What values of a would yield positive solutions for t?

    ehild
     
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