Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    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
    • Community Assistant
    Offline

    19
    ReputationRep:
    So can you show some working?
    Offline

    16
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    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:
    Offline

    16
    ReputationRep:
    (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
    Offline

    0
    ReputationRep:
    I expanded [2x^(3/2) - 3x^(-3/2)]2 and brought the 5 from the LHS to the RHS.
    Offline

    16
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    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?
    Offline

    16
    ReputationRep:
    (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
    Offline

    16
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    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.
    Offline

    16
    ReputationRep:
    (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
    • Thread Starter
    Offline

    0
    ReputationRep:
    I still don't get it. My brain's not really processing things well at the moment.
    • Community Assistant
    Offline

    19
    ReputationRep:
    (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?
    • Thread Starter
    Offline

    0
    ReputationRep:
    AHHHHHHHHHHHHHHHHH. Thank you so much!
    Finally manage to solve it. I feel like an idiot now.
    Thanks again!
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Will you be richer or poorer than your parents?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.