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Sequences and series

  1. Apr 21, 2007 #1
    A={xεR:X^11+2X^5<2} let a=supA By choosing a suitable sequence of elements of belonging to A and which tends to a as n->inf, or otherwise, show that a^11+2a^5=<2.Choose another sequence this time of all real numbers not belonging to A to show that a^11+2a^5>=2 and hence show that a^11+2a^5=2,so the equation x^11+2x^5=2 has a real solution

    any help would be really appreciated!how can i solve it
     
  2. jcsd
  3. Apr 23, 2007 #2

    Office_Shredder

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    For the first sequence, use a-1/n. a-1/n<a for all n, so is in A (since the function is increasing, all elements less than a are in A). When you plug a-1/n into the function, you get something like:

    a^11 + 2a^5 +1/n*(bunch of stuff) < 2

    Similiarly for taking a+1/n for the second part gives

    a^11 + 2a^5 + 1/n*(bunch of stuff) >=2

    Try to work it from there
     
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