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Math Help - an IIT-JEE(indian institue of technology joint entrance exam) 2007 question

  1. #1
    Member grgrsanjay's Avatar
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    an IIT-JEE(indian institue of technology joint entrance exam) 2007 question

    \displaystyle\lim_{x\to \pi/4} \left(\frac{\int_{2}^{sec^2 x} \left(f(t) dt) \right}{x^2 - \frac{\pi^2}{16}}\right) equals

    (A) \frac{8}{\pi} f(2)

    (B) \frac{2}{\pi} f(2)

    (C) \frac{8}{\pi} f(\frac{1}{2})

    (D) 4f(2)
    Last edited by grgrsanjay; August 31st 2010 at 07:10 AM.
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  2. #2
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    Quote Originally Posted by grgrsanjay View Post
    \lim_{x\to \pi/4} \left(\frac{\int_{2}^{secx^2} \left(f(t) dt) \right}{x^2 - \frac{\pi^2}{16}}\right) equals

    (A) \frac{8}{\pi} f(2)

    (B) \frac{2}{\pi} f(2)

    (C) \frac{8}{\pi} f(/frac{1}{2})

    (D) 4f(2)
    Do you mean sec^2(x) or sec(x^2)?

    Edit: If the first, I think using L'hopital's rule along with the chain rule and the fundamental theorem should work.
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    Member grgrsanjay's Avatar
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    its the first
    can you show me the methods i did it but i failed to get it
    i edited all my mistakes
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  4. #4
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    When you use L'hopital's rule, you get

    \frac{f(sec^2(x))}{2x}

    Substitute the limit x = π/4 and find the result.
    Last edited by sa-ri-ga-ma; August 31st 2010 at 08:16 AM.
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    Member grgrsanjay's Avatar
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    so is the answer coming out to be (A)??
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    I think it is (B)
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    but the answer is (A) given in all my books(4 books in total)
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  8. #8
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    Quote Originally Posted by sa-ri-ga-ma View Post
    When you use L'hopital's rule, you get

    \frac{f(sec^2(x))}{2x}

    Substitute the limit x = π/4 and find the result.
    This reply is mistaken.

    It should be  \displaystyle \frac{f(\sec^2(x))(2\sec^2(x)\tan(x))}{2x}

    So the answer is a.
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    OK. I will try again.
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  10. #10
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    yea,yea got it thanks sa-ri-ga-ma and plato
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