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    (Original post by chanc04)
    I know, they should do it like those music exams where you have pass, merit and distinuction. I completely agree with you!

    How annoying if someone was just one mark off a merit, but that other person completely failed that paper, and they both got U. Not fair, isn't it!
    Yeah it's not fair!!! Sorry I acciently negged you D: will pos later!!
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    (Original post by nick_the_big)
    got the same . hopefully is correct
    me too XD

    the only question i did quite well was the circle one with the square

    wat did u guys gat as the centre and the radius and was was ur max area and were u able to show that its a max???
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    (Original post by implus)
    This is how I did question 1:

    another way:

    tan[ (x-9) - 44 ] * tan[ (x-9) + 44 } = 1

    use tan double angle, you get difference of two square on top and bottom of the fractions and you can tidy up to:
    tan²(x-9) . [1+tan²44] = [1+tan²44]
    tan(x-9) = +/- 1
    so in the range 0<x<180
    tan(x-9) = 1
    and tan(x-9) = -1

    giving 54 and 144

    (of course, i forgot to deal with the +/- and so didnt mention the fact that tan(135)=-1 :rolleyes: )
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    (Original post by RK92)
    even simpler:

    tan[ (x-9) - 44 ] * tan[ (x-9) + 44 } = 1

    use tan double angle, you get difference of two square on top and bottom of the fractions and you can tidy up to:
    tan²(x-9) . [1+tan²44] = [1+tan²44]
    tan(x-9) = +/- 1
    so in the range 0<x<180
    tan(x-9) = 1
    and tan(x-9) = -1

    giving 54 and 144

    (of course, i forgot to check deal with the +/- and forgot that tan(135)=-1 :rolleyes: )
    Far simpler: as \cot(x) = \tan(90^{\circ}-x), we have \cot(\theta-53^{\circ}) = \tan(143^{\circ}-\theta); hence what we were asked to solve is  \tan(\theta+35^{\circ}) = \tan~(143^{\circ}-\theta) from which \theta = 54^{\circ}+90^{\circ}n. Obviously n = 0, and n = 1 give values of \theta in the desired range!
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    I could not get the first part of the vector question, which is annoying because I could have done the rest of it, |I just couldn't because I couldn't do part i.
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    is the answer for part C of Q3 be 70?
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    (Original post by ali25)
    I could not get the first part of the vector question, which is annoying because I could have done the rest of it, |I just couldn't because I couldn't do part i.
    Me too. I don't know how to do the reflection of the 3D vector
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    (Original post by Agent Fiasco)
    me too XD

    the only question i did quite well was the circle one with the square

    wat did u guys gat as the centre and the radius and was was ur max area and were u able to show that its a max???
    (0,1/2) mate, n i think i got the radius to be 1/2, I hope i got that right as it was the only question which seemed to go well

    As well, the value of alpha i got to be pi/6, can't remember what i got for area.

    I decided to leave determining whether it was max or not, i thought i wud leave it till the end as I already spent too much time (on Q1 mainly), and only had about an hour for Q5,6,7 so yeah, couldn't look at Q7 at all, only parts of Q6.
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    (Original post by lawwainam)
    is the answer for part C of Q3 be 70?
    yeah I got 70,

    how do yo show that a^2 etc... is what they gave, I could'nt get it :mad:
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    So who else thinks they got a U?
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    (Original post by OneMomentPlease)
    yeah I got 70,

    how do yo show that a^2 etc... is what they gave, I could'nt get it :mad:
    Do you mean Q5 (c) (i)?
    I couldn't get it too., sorry
    By the way, is q=6, p=3 for Q.7?
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    (Original post by lawwainam)
    Do you mean Q5 (c) (i)?
    I couldn't get it too., sorry
    By the way, is q=6, p=3 for Q.7?
    Ah ok, I really don't know why I couldnt get it, seemed simple enough...

    And yeah I think so, well thats what I remember putting
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    (Original post by Webbykun)
    Oh dude stop this! xD Okay I'm just gonna assume now I haven't got a distinction and prepare for my insurance uni .

    What's yr firm? Are you majoring in math in future?

    (Original post by lawwainam)
    Do you mean Q5 (c) (i)?
    I couldn't get it too., sorry
    By the way, is q=6, p=3 for Q.7?
    Yeah that is correct.
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    (Original post by superkinetic)
    Yeah you're to use dot product, to prove \vec{AP} \cdot \vec{AB} = 0, as cos90=0. But then you got to find B first which depends on the P' from (a). That is the only method I can think of.
    I didn't use vectors at all.


    Since you were given 120, the opposite angle is 60 degrees.

    Thus the necessary angle is 180 - 60 - 30 = 90.


    The reason I had to do it this way is because I didn't know how to find P', or B, or anything.


    How did people find the inverse function of m at the end?
    I couldn't find 'q' either. That was annoying.

    I didn't do question 1 either.

    I got 60 for the sum to infinity...

    I couldn't show the last parts of the circle questions - I said that the radius of the circle was the normal to the curve, and that equals y - 1/2 - but I couldn't get the algebra out. FML.

    Definitely not got a distinction. Ugh. Might not be going to uni. D:
    *fingers crossed I got a 2 SOMEWHERE on STEP*
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    (Original post by AnonyMatt)
    I didn't use vectors at all.


    Since you were given 120, the opposite angle is 60 degrees.

    Thus the necessary angle is 180 - 60 - 30 = 90.


    The reason I had to do it this way is because I didn't know how to find P', or B, or anything.


    How did people find the inverse function of m at the end?
    I couldn't find 'q' either. That was annoying.

    I didn't do question 1 either.

    I got 60 for the sum to infinity...

    I couldn't show the last parts of the circle questions - I said that the radius of the circle was the normal to the curve, and that equals y - 1/2 - but I couldn't get the algebra out. FML.

    Definitely not got a distinction. Ugh. Might not be going to uni. D:
    *fingers crossed I got a 2 SOMEWHERE on STEP*
    I used the quadratic formula to find the inverse function.
    Am I right?
    And for Q4, alpha=pi/6, right?
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    Hey guys, a couple of things.

    Firstly, what did people get for 7 c) (10 marks)?

    Part i) was to find m^(-1), can't remember the rest of the question though

    Also what are your thoughts about what the grade boundaries are going to be this year?

    I'd say this is the hardest paper I've seen and 2006/7 had distinction grades at 67 and 68 respectively so maybe 65 ish?

    The only thing I'm clinging onto was the fact I got all of the vector question, struggled with 3 b, c, 5 c and 7 b & c.

    Thanks.
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    (Original post by SmileyGurl13)
    So who else thinks they got a U?
    ME!
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    (Original post by jumping_homeless)
    Hey guys, a couple of things.

    Firstly, what did people get for 7 c) (10 marks)?

    Part i) was to find m^(-1), can't remember the rest of the question though

    Also what are your thoughts about what the grade boundaries are going to be this year?

    I'd say this is the hardest paper I've seen and 2006/7 had distinction grades at 67 and 68 respectively so maybe 65 ish?

    The only thing I'm clinging onto was the fact I got all of the vector question, struggled with 3 b, c, 5 c and 7 b & c.

    Thanks.
    The grade boundaries of 2010 scared me.
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    (Original post by lawwainam)
    The grade boundaries of 2010 scared me.
    2010 was an aberration - very easy paper.
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    (Original post by AnonyMatt)
    I didn't use vectors at all.


    Since you were given 120, the opposite angle is 60 degrees.

    Thus the necessary angle is 180 - 60 - 30 = 90.


    The reason I had to do it this way is because I didn't know how to find P', or B, or anything.
    But that's not true that opposite angles in a kite add up to 180 deg; it's coincidental that both opposite angles add up to 180 deg, as one of the angle is 90 deg.

    Assume there are 2 similar angles of 100 deg in a kite. Do the remaining opposite angles add up to 180 deg?

    Anw, do u require AEA for yr offer?
 
 
 
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