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    Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
    Please give me some tips on solving em. Thanks in advance
    Given that f(x) = [2x^(3/2) - 3x^(-3/2)2 +5, x>0,
    (a) find, to 3 s.f., the value of x for which f(x) =5.
    (b) Show that f(x) may be written in the form Ax3 + B/x3 + C, where A, B and C are constants to be found.
    (c) Hence evaluate ∫f(x),1,2 dx
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    So can you show some working?
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    (Original post by Pherix)
    Hey guys, I'm stuck on a question while doing C2 A3 Practice Paper.
    Please give me some tips on solving em. Thanks in advance
    I assume that you can do the first part

    What did you try for the second part
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    I'm actually stuck on part A.
    f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
    [2x^(3/2) - 3x^(-3/2)]2 + 5 = 5
    4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
    4x^3 + 9/x^(3) -12 = 0
    A=4, B=9, C=-12

    As for part C, I manage to get 11.375 after integrating it.
    Still stuck on part A
    :confused:
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    (Original post by Pherix)
    I'm actually stuck on part A.
    f(x) = = [2x^(3/2) - 3x^(-3/2)]2 +5
    [2x^(3/2) - 3x^(-3/2)]2 + 5 = 5



    How did you go from there to

    4x^3 + 9x^(-3) -12 = 0 <-------(I'm not sure if this is related to part A or not but I'm stuck here for A)
    4x^3 + 9/x^(3) -12 = 0
    A=4, B=9, C=-12

    As for part C, I manage to get 11.375 after integrating it.
    Still stuck on part A
    :confused:
    • Thread Starter
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    I expanded [2x^(3/2) - 3x^(-3/2)]2 and brought the 5 from the LHS to the RHS.
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    (Original post by Pherix)
    I expanded [2x^(3/2) - 3x^(-3/2)]2 and brought the 5 from the LHS to the RHS.
    Aaaaaaah

    That was what you missed in the OP
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    I missed? I thought it's okay to expand them mentally. Been taught to do so :confused:
    Anyway, how do I get to solve for part A?
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    (Original post by Pherix)
    I missed? I thought it's okay to expand them mentally. Been taught to do so :confused:
    Anyway, how do I get to solve for part A?
    :confused:

    No I was referring to the missing bracket in your OP which changed the question A lot
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    (Original post by Pherix)
    I missed? I thought it's okay to expand them mentally. Been taught to do so :confused:
    Anyway, how do I get to solve for part A?
    Multiply by x^3 and solve the quadratic
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    Ahhhhhh. I just notice that I forgot to add that particular bracket. My bad
    Anyway, multiplying x^3 would give me 4x^6 - 12x^3 = -9
    I just can't solve it now. I'm confuse now.
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    (Original post by Pherix)
    Ahhhhhh. I just notice that I forgot to add that particular bracket. My bad
    Anyway, multiplying x^3 would give me 4x^6 - 12x^3 = -9
    I just can't solve it now. I'm confuse now.
    It is a quadratic
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    I still don't get it. My brain's not really processing things well at the moment.
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    (Original post by Pherix)
    I still don't get it. My brain's not really processing things well at the moment.
    Make a substitution t=x^3. Can you see the quadratic now?
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    AHHHHHHHHHHHHHHHHH. Thank you so much!
    Finally manage to solve it. I feel like an idiot now.
    Thanks again!
 
 
 
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