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

    0
    ReputationRep:
    (Original post by mabli)
    Attachment 216704Attachment 216705Attachment 216706

    Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


    Posted from TSR Mobile
    weird right, everything did cancel leaving you with something totally unlike what it asked you to prove
    Offline

    11
    ReputationRep:
    (Original post by mabli)
    Attachment 216704Attachment 216705Attachment 216706

    Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


    Posted from TSR Mobile
    Last log question answers coming right up . . .
    I'll do some more on request.
    Offline

    0
    ReputationRep:
    (Original post by mabli)
    How did everyone find the exam? I thought it was okay about from the last logarithm question and the eighth question (circles)


    Posted from TSR Mobile
    I twisted my knickers on question 2)b) a lot although it was only worth 3 marks. I believe I got one of the possible answers, but that was the one that I'm confident I lost marks on.:mad:

    The other parts weren't so bad.
    Offline

    0
    ReputationRep:
    (Original post by Naomi32)
    I did this too! I thought I went the long way around because the second part of the question asked for the co-ordinates of intersection which I had already found
    Really hoping we get most of the marks for doing that, we still proved it wasn't the diameter?
    I would give you two full marks if I were an examiner as the question was only worth 2 or 3 marks, and also for the exra effort you put in (I did it the shorter way, by the way)
    Offline

    11
    ReputationRep:
    7)i)
    \log_a(xy)=\log_a(x) + \log_a(y)
    \Leftrightarrow xy = a^{\log_a(x) + \log_a(y)}
    So now we must show that:
    xy = a^{\log_a(x) + \log_a(y)}
    a^{\log_a(x) + \log_a(y)} = a^{\log_a(x)} \times a^{\log_a(y)} = xy

    ii)
    5^{2-3x} = 8 \Rightarrow (3x-2)\log_2(5) = 3

\Rightarrow 3x\log_2(5) = 3 + 2\log_2(5)

\Rightarrow x = \dfrac{3+2\log_2(5)}{3\log_2(5)} \apprx 0.236

    iii)
    \log_a(90x^2) - \log_a(\frac{5}{x}) = \frac{1}{2}\log_a(144x^8)

\Rightarrow \log_a(90) + 2\log_a(x)-\log_a(5) = \log_a(12) + 4\log_a(x)

\Rightarrow \log_a(\frac{3}{2}) = \log_a(x)

\Rightarrow x = \dfrac{3}{2}
    Offline

    0
    ReputationRep:
    Hi! I felt the exam was alright, except 2aii; I got confused because my values were less than one [(3x-2)(2x+3) x could not equal -3/2?]. Q3 was a mess, and same for last logs question and circles question.
    I don't know how to feel about the exam, I know I answered all the questions and attempted the answers, but then got in a muddle in the middle of those odd questions..
    :erm:
    Offline

    0
    ReputationRep:
    (Original post by joostan)
    Last log question answers coming right up . . .
    I'll do some more on request.
    Can you please do the area under the curve question? 6b I think. Thanks
    • Thread Starter
    Offline

    2
    ReputationRep:
    I got 16.7cm^2 for that


    Posted from TSR Mobile
    Offline

    0
    ReputationRep:
    Does anyone have a solved paper from their teacher? Would like to compare my marks
    Offline

    2
    ReputationRep:
    For Proving Triangle: Cos A = x^2+(x-2)^2 - (x+2)^2 / 2(x)(x-2)

    This gives: x^2+x^2-4x+4-x^2-4x-4 / 2x^2 - 4x

    Simplifies to x^2 - 8x / 2x^2 - 4x

    Divide by x = x-8 / 2x - 4
    Offline

    2
    ReputationRep:
    Part b) Angle BAC = 120

    Formula for the angle is given from part a:

    There Cos 120 = x+8 / 2x-4

    -0.5(2x-4) = x+8

    2-x = x+8

    2x = -6

    x=-3

    Not sure if this is correct, as x is a negative.
    Offline

    2
    ReputationRep:
    Im going to try and use Latex for 6b)

    y=x^2+3 and y=4x



x^2+3=4x

x^2-4x+3=0

(x-1)(x-3)=0



Therefore intersects at x=1 and x=3

    Area under the curve between x=1 and x=3

    \displaystyle \int_1^3 \{x^2 + 3\} dx = \left[ \frac{x^3}{3} + 3x \right]_1^3

    This works out to be 44/3


    Area under the line between x=0 and x=1

    \displaystyle \int_0^1 \{4x\} dx = \left[2x^2\right]_0^1

    This works out to be 2

    Therefore shaded region = 44/3 + 2 = 50/3
    Offline

    1
    ReputationRep:
    (Original post by mabli)
    Attachment 216704Attachment 216705Attachment 216706

    Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out


    Posted from TSR Mobile
    Tbh the C2 May 2013 is one of the easiest question paper I have seen. Many of u will agree with me that question 2 (a) & 3(a) were not as expected but the rest of the question should have been really good if you would have done more past papers practice. e.g: the question that was about integration 6(b) that was taken from the C2 May 2007 paper...I'm very sure that I did very well in this exam :L
    Offline

    0
    ReputationRep:
    (Original post by Tynos)
    Im going to try and use Latex for 6b)

    y=x^2+3 and y=4x



x^2+3=4x

x^2-4x+3=0

(x-1)(x-3)=0



Therefore intersects at x=1 and x=3

    Area under the curve between x=1 and x=3

    \displaystyle \int_1^3 \{x^2 + 3\} dx = \left[ \frac{x^3}{3} + 3x \right]_1^3

    This works out to be 44/3


    Area under the line between x=0 and x=1

    \displaystyle \int_0^1 \{4x\} dx = \left[2x^2\right]_0^1

    This works out to be 2

    Therefore shaded region = 44/3 + 2 = 50/3
    Thanks!! I got that too! Have you tried 8 or 9?
    Offline

    2
    ReputationRep:
    (Original post by Han96)
    Thanks!! I got that too! Have you tried 8 or 9?
    I jsut took a break lol. The latex is harder than the actual question xD hahah jokes

    Ill try it now.
    Offline

    2
    ReputationRep:
    8 Part a and b)

    



x^2+y^2+2x-6y-15=0

(x+1)^2+(y-3)^2=25

    Therefore A(-1,3) Radius = Root 25 = 5

    I did part ii first dont know if its allowed:
    

y=-x+9

(x+1)^2+(-x+9-3)^2=25

Expand and simplify to:

2x^2-10x+12=0

x^2-5x+6=0

(x-3)(x-2)=0

    Intersects at: (3,6) and (2,7)

    bi) If y=-x+9 is a diameter then length of this divided by 2 must equal 5.
    (2-3)^2+(7-6)^2 Square root = Root 2

    Root 2/2 is not equal to the radius of the circle therefore is not a diameter.
    Offline

    2
    ReputationRep:
    For the last part of part 8)

    Centres of circles: (-1,3) and (11,8)

    Using pythag:

    (11+1)^2+(8-3)^2

    = Root 169 = 13

    Therefore shortest distance = 13-6-5 (Their radius's)=2
    • Thread Starter
    Offline

    2
    ReputationRep:
    Do you think I'd get marks for changing the fraction into a whole number for 6.b)?


    Posted from TSR Mobile
    Offline

    11
    ReputationRep:
    (Original post by Han96)
    Can you please do the area under the curve question? 6b I think. Thanks
    Sorry, I was out yesterday evening, I've been beaten to it - I'll check the working
    EDIT: I got the same answer - congrats
    Offline

    2
    ReputationRep:
    (Original post by mabli)
    Do you think I'd get marks for changing the fraction into a whole number for 6.b)?


    Posted from TSR Mobile
    Did you round to 3sf?
 
 
 
  • 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
    What's your favourite Christmas sweets?
  • 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.