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

    1
    ReputationRep:
    I just got 69/75 in a very easy paper. I made a silly mistake when I was solving trig. equations. I always neglect the ranges they give you!
    Offline

    3
    ReputationRep:
    Aaaaah, I'm psyched for this exam!

    I need 90% to get halfway to that A*...
    Offline

    0
    ReputationRep:
    (Original post by jj.)
    I went through the book first, picked up on anything that was my weakness and then I started past papers. It's best to understand C3 first before doing the papers, if you understand it, the papers should be a breeze.
    (Original post by .:excel4100%:.)
    Yeah, go through the text book and start to understand the topics! Then start practice papers so you can get used to the format of the papers and timing also!

    If you do like all the Solomons and then the edexcel past papers, I can almost guarantee you will get at least an A!!
    Ok. So textbook then past papers it is...Thanks guys. I have 4 days left...is it possible? :confused: I spent Xmas revising for resits...hence, the lack of time. :cry:
    Offline

    16
    ReputationRep:
    Ok quick question, why is the range of the curve  y = e^{x^2} greater or equal to 1, and not just greater than 0?
    Offline

    2
    ReputationRep:
    (Original post by samir12)
    Ok quick question, why is the range of the curve  y = e^{x^2} greater or equal to 1, and not just greater than 0?
    Well for the graph of y=e^{x}, x would have to negative for the y to be less than one. However since it's x^{2} it cannot be negative hence the minimum value of y is 1 (when x=0)
    Offline

    0
    ReputationRep:
    General C3 question: Will I be forced to use the remainder theorem in the exam for a question? (I find the long division method easier)
    Offline

    0
    ReputationRep:
    I've been slamming my head into the wall trying to mark my answers to question papers. For the trig proofs, I always get there in the end but a different way to what's given in the mark scheme. Will I still get full marks for these questions, as long as I don't make any mistakes?
    Offline

    16
    ReputationRep:
    (Original post by milliondollarcorpse)
    Well for the graph of y=^{x}, x would have to negative for the y to be less than one. However since it's x^{2} it cannot be negative hence the minimum value of y is 1 (when x=0)
    Thanks mate, now I understand.
    Offline

    16
    ReputationRep:
    (Original post by gottastudy)
    General C3 question: Will I be forced to use the remainder theorem in the exam for a question? (I find the long division method easier)
    Yes, it only takes like 10 seconds to learn
    Offline

    0
    ReputationRep:
    (Original post by xSkyFire)
    Yes, it only takes like 10 seconds to learn


    I get the comparing coefficients part, just not really setting up the equation in the form.

    Will it always be (Ax^2 + Bx + C) (divisor)....

    I don't understand how to write the remainder part. Please help someone? Thank you.
    Offline

    16
    ReputationRep:
    (Original post by gottastudy)


    I get the comparing coefficients part, just not really setting up the equation in the form.

    Will it always be (Ax^2 + Bx + C) (divisor)....

    I don't understand how to write the remainder part. Please help someone? Thank you.
    Quotient x divisor + remainder, really straightforward. Just do the mixed exercises in chapter 1, C3 and it'll become clear :p:
    Offline

    0
    ReputationRep:
    (Original post by xSkyFire)
    Quotient x divisor + remainder, really straightforward. Just do the mixed exercises in chapter 1, C3 and it'll become clear :p:
    I already have done! Except I used long division...It didn't say in particular to use the remainder theorem!

    I don't get why in example 13, it's Dx + E and in example 12 why it's just + D (remainder part)? Is this to do with the powers?

    No worries. I figured it out.
    Offline

    0
    ReputationRep:
    (Original post by typewriter)
    Will I still get full marks for these questions, as long as I don't make any mistakes?
    yes.
    Offline

    0
    ReputationRep:
    need help with this

    Given that x = 4 sin (2y + 6), find dy/dx in terms of x.

    my answer to the Q was dy/dx= 1/8cos(2y+6)
    but i didnt know how to write it in term of x can someone help me plz
    Offline

    12
    ReputationRep:
    how would you do this

    if f(x)= 4-x^2

    solve ff(0)
    so 4-(4-x^2)^2
    x^4-8x^2-12=0

    i dont know how i would do it.
    xxx
    Offline

    2
    ReputationRep:
    (Original post by taherq)
    need help with this

    Given that x = 4 sin (2y + 6), find dy/dx in terms of x.

    my answer to the Q was dy/dx= 1/8cos(2y+6)
    but i didnt know how to write it in term of x can someone help me plz
    Rearrange x =4sin (2y + 6) in terms of y and substitute it into dy/dx
    (Original post by indie_couture)
    how would you do this

    if f(x)= 4-x^2

    solve ff(0)
    so 4-(4-x^2)^2
    x^4-8x^2-12=0

    i dont know how i would do it.
    xxx
    Let x^2=y

    Solve y^2-8y-12=0
    Offline

    0
    ReputationRep:
    AAAAAAAh c3! Does anyone have any tips? I can do all the questions I think, it's just carelessness that lets me down. I'm aiming for an A* as well Any advice?
    Offline

    0
    ReputationRep:
    can someone tell me why... cos5x-cosx = -2sin3xsin2x
    • PS Helper
    Offline

    0
    ReputationRep:
    PS Helper
    (Original post by MIKE ESSIEN IS QUITE SICK)
    can someone tell me why... cos5x-cosx = -2sin3xsin2x
    check your c3 text book. this is an identity that you have to learn.
    =-2sin(5x+x/2)sin(5x-x/2)
    =-2sin3xsin2x
    Offline

    0
    ReputationRep:
    (Original post by MIKE ESSIEN IS QUITE SICK)
    can someone tell me why... cos5x-cosx = -2sin3xsin2x
    The trig identity for cos(A) - cos(B) is in the formula booklet: -2sin(\frac{A+B}{2})sin(\frac{A-B}{2}).


    If you meant you wanted the proof:

    Use the identities
    cos(A + B) = cos(A)cos(B) - sin(A)sin(B)
    and
    cos(A - B) = cos(A)cos(B) + sin(A)sin(B)

    Subtract one from the other
    cos(A + B) - cos(A - B) = -2sin(A)sin(B)

    Let A + B = P and A - B = Q
    so, A = \frac{P + Q}{2} and B = \frac{P - Q}{2}

    Therefore, cos(P) + cos(Q) = -2sin(\frac{P + Q}{2})sin(\frac{P - Q}{2})

    And in your case, you'd just sub in the 5x and x as P and Q
 
 
 
  • 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.

  • Poll
    Would you rather give up salt or pepper?
  • 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

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