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Solving an inequality

  1. Jan 24, 2013 #1
    1. The problem statement, all variables and given/known data
    The range of k for which the inequality ##k\cos^2x-k\cos x+1≥0## for all x, is
    a. k<-1/2
    b. k>4
    c. -1/2≤k≤4
    d. -1/2≤k≤2


    2. Relevant equations



    3. The attempt at a solution
    I am not sure about how to begin with this. This seems to me a quadratic in cos(x) and here, the discriminant should be less than zero.
    ##k^2-4k<0##
    This gives me ##0<k<4## but this is not given in any of the options.

    Any help is appreciated. Thanks!
     
  2. jcsd
  3. Jan 24, 2013 #2

    I like Serena

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    Hey Pranav!

    It seems you are not done yet.
    Your solution would be right if cos x could take on any value.
    However, cos x is limited in range.
    This means that the actual solution will have to contain your solution and perhaps have more solution values.

    Is there any answer that matches that?

    Alternatively, you could try and find the minima and maxima of ##\cos^2x - \cos x##.
    Fill those in and solve for k.
    But of course that is more work...
     
  4. Jan 24, 2013 #3

    SammyS

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    For one thing, the discriminant should be non-negative.

    Then as I like Serena pointed out, you need be sure that k is such that -1 ≤ cos(x) ≤ 1 .
     
  5. Jan 24, 2013 #4

    haruspex

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    Do you mean non-positive?
     
  6. Jan 24, 2013 #5

    I like Serena

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    Shall we stick to less than zero? ;p
     
  7. Jan 24, 2013 #6

    haruspex

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    No, I think SammyS' point is that it's <= 0.
    Pranav, I suggest solving this by rewriting the inequality in the form <square involving cos x> <comparator> <function of k only>, but you'll have to treat different cases according to the sign of k.
     
  8. Jan 24, 2013 #7

    I like Serena

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    Sorry, my mistake.
     
  9. Jan 24, 2013 #8

    SammyS

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    Yes, I like Serena, you are right !

    I misread the problem.

    Of course, the discriminant must be less than or equal to zero so that [itex]\ \ k\cos^2(x)-k\cos(x)+1=0\ \ [/itex] has [STRIKE]no[/STRIKE] at most one real root .

    Edited above, per haruspex (next post) .
     
    Last edited: Jan 24, 2013
  10. Jan 24, 2013 #9

    haruspex

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    No, it can equal zero. One real root is ok; it just must not have two distinct real roots.
     
  11. Jan 24, 2013 #10

    ehild

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    The restriction that -1≤cosx≤1 can make the range of k wider than [0,4]. There can be roots of the equation ky^2-ky+1=0, but outside the interval (-1,1). The range should contain the interval [0,4] anyway. This is true for which of the given intervals?

    ehild
     
    Last edited: Jan 24, 2013
  12. Jan 24, 2013 #11

    SammyS

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    As I like Serena pointed out, -1 ≤ cos(x) ≤ 1 .

    Therefore, there are likely more permissible values of k than just those given by considering the discriminant.

    The problem at hand is equivalent to finding the values of k such that [itex]\ \ k\,t^2-k\,t+1\ge0\ \ [/itex] for -1 ≤ t ≤ 1 .
     
  13. Jan 25, 2013 #12
    Thanks everyone for the replies!

    Yes, discriminant should be less equal to zero. Sorry about that.
    I am not sure if I have understood all the replies properly. Following haruspex' post, I can rewrite the given inequality as
    ##(\cos x-\frac{1}{2})^2≥\frac{k-4}{4k}##
    How should I proceed from here?

    @ehild and ILS: The option C includes [0,4] so it should be the answer.
     
  14. Jan 25, 2013 #13

    SammyS

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    See the graph of [itex]\ \displaystyle kt^2-kt+1\,,\ [/itex] with k = -1/4, restricted to -1 ≤ t ≤ 1 .

    Because the domain of this function is restricted to [-1, 1], the function's range does not include any negative values.

    attachment.php?attachmentid=54993&stc=1&d=1359129189.gif from WolframAlpha

    Added in Edit:

    Of course, the graph of [itex]\ \displaystyle k\cos^2(x)-k\cos(x)+1\,,\ [/itex] looks nothing like the above graph, but both graphs have the same range for k = -1/4 .
     

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    Last edited: Jan 25, 2013
  15. Jan 25, 2013 #14

    ehild

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    That is not quite true. Depends on the sign of k.

    ehild
     
  16. Jan 25, 2013 #15

    haruspex

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    As I warned, and as ehild has pointed out, you have to split it according to the sign of k. (And k=0 as a third case.) The above is for k > 0.
    What is the range of values of the LHS?
     
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