Join TSR now and get all your revision questions answeredSign up now

did anyone do ccea maths c1 today Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    i got everthing right apart from maybe question 8, i got
    - 4 - (2 root 2) < X < - 4 + (2 root 2). is that right
    Offline

    0
    ReputationRep:
    Yea i did it today! cant remember what i got but it was something like that!
    Offline

    0
    ReputationRep:
    What answers did use get in each question? What did use think of the paper overall?
    Offline

    0
    ReputationRep:
    Hi yeah I did it,
    I got something similar to that answer for question 8 as well. What did you get for question 7?
    Offline

    2
    ReputationRep:
    (Original post by Pyro2000x)
    i got everthing right apart from maybe question 8, i got
    - 4 - (2 root 2) < X < - 4 + (2 root 2). is that right
    This is right.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Alfie1993)
    Hi yeah I did it,
    I got something similar to that answer for question 8 as well. What did you get for question 7?
    what was it about, we had to give up our question book
    Offline

    2
    ReputationRep:
    (Original post by Pyro2000x)
    what was it about, we had to give up our question book
    The cost function which you had to say was increasing when t>2
    Offline

    0
    ReputationRep:
    Here are my answers can anyone give any feedback?
    1. Simplifying an equation - 5
    b) graph functions, f(x-1) moved one to the right
    f(-x) reflected in the y axis
    2. x=20 y=45 z=30
    3a)i) gradient of AB = 2
    ii) equation of line perpendic to AB going through (-1,1)
    2y=-x-1
    b) differentiating y=4x^3 - x^-1/3
    i got - 12x^2 + 1/3x^-4/3
    4. curve and line meeting at points C and D
    i) find co ordinates - C (2,-1) D (4,3)
    ii) exact length - sq root of 20/ 2root 5
    5a) factor theorem ii) factories fully - (2x-1)(x+2)(x+1)
    iii)solve equation so x=1/2, x=-2, x=-1
    b) triangle q
    6root 3 + 4
    6. horrible question about the road race
    couldn't do i)
    ii) 17/2
    7a)i) dy/dx of x^4 + 32x
    4x^3 + 32
    ii) find x co ord of turning point and its nature
    x= -2, minimum
    b) cost
    t > cube root of 6 ( i know thats wrong)
    8. curves that didn't meet
    i got the discriminant to be m^2 + 8m + 8<0 and instead of using the quad formula i did m^2 + 8m < - 8
    and ended up with like -8<m<-16 or something stupid

    anyone get any of the same answers?
    Offline

    2
    ReputationRep:
    (Original post by B18)
    Here are my answers can anyone give any feedback?
    1. Simplifying an equation - 5
    b) graph functions, f(x-1) moved one to the right
    f(-x) reflected in the y axis
    2. x=20 y=45 z=30
    3a)i) gradient of AB = 2
    ii) equation of line perpendic to AB going through (-1,1)
    2y=-x-1
    b) differentiating y=4x^3 - x^-1/3
    i got - 12x^2 + 1/3x^-4/3
    4. curve and line meeting at points C and D
    i) find co ordinates - C (2,-1) D (4,3)
    ii) exact length - sq root of 20/ 2root 5
    5a) factor theorem ii) factories fully - (2x-1)(x+2)(x+1)
    iii)solve equation so x=1/2, x=-2, x=-1
    b) triangle q
    6root 3 + 4
    6. horrible question about the road race
    couldn't do i)
    ii) 17/2
    7a)i) dy/dx of x^4 + 32x
    4x^3 + 32
    ii) find x co ord of turning point and its nature
    x= -2, minimum
    b) cost
    t > cube root of 6 ( i know thats wrong)
    8. curves that didn't meet
    i got the discriminant to be m^2 + 8m + 8<0 and instead of using the quad formula i did m^2 + 8m < - 8
    and ended up with like -8<m<-16 or something stupid

    anyone get any of the same answers?

    The cube root of 6 was wrong, it was supposed to be t>2. Number 8, you did wrong, but you would have gained some marks for getting as far as the quadratic inequality. I think your factorising in the long division may be wrong. Also, the equation of the perpendicular line; I think you may have got a sign or two wrong. Apart from that, all good
    • Thread Starter
    Offline

    0
    ReputationRep:
    unreal i am almost certain i got 100%
    Offline

    0
    ReputationRep:
    How did you think you had 100% when you didnt even attempt 6)i.
    Sorry to sound harsh!
    Offline

    2
    ReputationRep:
    It was a decent paper. I messed up in a few spots but I think I'll get a high B/ an A.
    Offline

    2
    ReputationRep:
    (Original post by acrawford957)
    How did you think you had 100% when you didnt even attempt 6)i.
    Sorry to sound harsh!
    It wasn't him who said that.

    (Original post by xDanny 117)
    It was a decent paper. I messed up in a few spots but I think I'll get a high B/ an A.
    What did you mess up on?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by acrawford957)
    How did you think you had 100% when you didnt even attempt 6)i.
    Sorry to sound harsh!
    that is not me
    Offline

    0
    ReputationRep:
    (Original post by acrawford957)
    How did you think you had 100% when you didnt even attempt 6)i.
    Sorry to sound harsh!
    lol yeah... i never said i thought i got 100% that was someone else
    Offline

    0
    ReputationRep:
    (Original post by CD315)
    The cube root of 6 was wrong, it was supposed to be t>2. Number 8, you did wrong, but you would have gained some marks for getting as far as the quadratic inequality. I think your factorising in the long division may be wrong. Also, the equation of the perpendicular line; I think you may have got a sign or two wrong. Apart from that, all good
    would you talk me through 6i) 7b) and 8? also what was the perpendicular equation meant to be?
    Offline

    2
    ReputationRep:
    (Original post by CD315)



    What did you mess up on?
    Used the wrong 'length of a line' formula in a question, couldn't get a final value for when the cost is increasing in Q7, only got to the quadratic inequality in Q8.
    Offline

    2
    ReputationRep:
    (Original post by B18)
    would you talk me through 6i) 7b) and 8? also what was the perpendicular equation meant to be?
    You had a negative 1 which I think was supposed to be a positive.

    6i; wasn't this the speed one? You just had something along the lines of 17/x + 13/(x-2) if I remember correctly? Then just add them and set equal to 4. (They're the expressions for time taken.)

    7b; you differentiated the function and said it was greater than 0. You ended up with it being 2.

    8; set the curves equal to each other and you'll form a quadratic. Since there is no solutions to this, due to them not meeting, you take the discriminant and say it's less than 0. I ended up with a quadratic inequality which I then drew a graph of.
 
 
 
Poll
How are you feeling about Results Day?

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.