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Finding the function from given data

  1. Feb 9, 2014 #1

    utkarshakash

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    1. The problem statement, all variables and given/known data
    If f(x) is differentiable function satisfying [itex] 6x \displaystyle \int_0^1 f(tx)dt = 2x^3-3x^2+6x+5 [/itex] then select correct options(more than one may be correct)

    a)f(x) is symmetric about x=1
    b)f(x)=0 has no real roots
    c)f(-x)=f(x+1) for all x in R
    d) f(x)=1/2 has 2 real and equal roots


    3. The attempt at a solution
    From the options it is clear that the function f(x) is to be determined first. Now, if I divide both sides by 6,
    [itex] x \displaystyle \int_0^1 f(tx)dt = \int (x^2 - x +1) dx + \frac{5}{6} [/itex]

    I have no idea how to take it ahead.
     
  2. jcsd
  3. Feb 9, 2014 #2

    Curious3141

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    Homework Helper

    Hint: In [itex] 6x \displaystyle \int_0^1 f(tx)dt [/itex], substitute ##y = tx## and use Fundamental Theorem of Calculus to differentiate it after transforming the bounds.

    EDIT: I'm not at all certain that the constant term (5) belongs in the RHS of that equation. It makes no sense to me. If it is disregarded, the question is easily solvable. Maybe someone else will have an insight into this.
     
    Last edited: Feb 9, 2014
  4. Feb 9, 2014 #3

    utkarshakash

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    A clever approach. Thank You!
     
  5. Feb 9, 2014 #4
    I suggest you to make a note of this approach, its going to be helpful a lot of times. Test papers often include problems on FTOC.
     
  6. Feb 9, 2014 #5

    Curious3141

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    The consensus in the homework help forum is that the question is wrong. The presence of the constant term (5) on the RHS is inconsistent. Putting x = 0 makes the LHS vanish but not the RHS.

    *Another* wrong question?
     
  7. Feb 10, 2014 #6

    utkarshakash

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    Thanks for your suggestion. I'm going to make it right now.
     
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