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WJEC AS Maths C2 May 20th 2015 watch

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    So how was the paper
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    (Original post by matthewdjones)
    what was wrong with 8c? it was just pythagoras ?
    To find ST? I couldn't do it
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    Here's an unofficial mark scheme type thing that I've made. Some answers may not be in their correct places, and some could be wrong. I also probably left out a few questions, so let me know which ones. I'm pretty sure most of them are correct though, since I was able to check most of them.

    1. 0.4689

    2. You should end up with a quadratic and then get two solutions that are greater IN MAGNITUDE (modulus) than 1, and so cannot be solutions since the sine of angle cannot have a greater magnitude than 1.

    I got x=22 and x=143.

    I got phi=0, 180, 124.85 (I think).

    3. You had to use the sine rule to get the two values of the angle, 138 and 42 I think.

    Acute angle would therefore be 17.

    So area=33.33.

    4. nth term is a+(n-1)d, sub a and d in.

    It's your typical proof, just with numbers in, like Jan 2013 if I recall correctly.

    I think I got a=-7 and d=3.

    k=70

    5. r=-1/3, a=216.

    Sum to infinity clearly 162.

    Show that r satisfies the equation. Modulus of r less than 1, then pick the right solution.

    I don't know whether I've got the order of this question right.

    6. Integral is something like 3x^1/2 - (18x^7/3)/7 + c.

    50/3. But everyone got different answers .

    7. Basic log proof.

    1/2 and 2/3 I think.

    8. (3,-1)

    Just work out 1/2 the diameter PQ to find radius r = sqrt29.

    Got an angle of 66.8 I think.

    6 for length of ST.

    9. Use 1/2r^2(theta-sintheta) to find red segment. Find white segment by doing area of whole circle - area of red segment. Should get 1.04 or something to 2.09 something for ratio, so white is 2 times red (angle was 2.6 radians).

    Hope that's helpful, and again, any mistakes or questions left out please let me know.

    P.S. I'll give explanations to any of my answers if you don't understand how I got them.
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    (Original post by PrimeLime)
    Here's an unofficial mark scheme type thing that I've made. Some answers may not be in their correct places, and some could be wrong. I also probably left out a few questions, so let me
    know which ones. I'm pretty sure most of them are correct though, since I was able to check most of them.

    1. 0.4986 or something similar.

    2. You should end up with a quadratic and then get two solutions that are greater IN MAGNITUDE (modulus) than 1, and so cannot be solutions since the sine of angle cannot have a greater magnitude than 1.

    I got x=22 and x=143.

    I got phi=0, 180, 124.85 (I think).

    3. You had to use the sine rule to get the two values of the angle, 138 and 42 I think.

    Acute angle would therefore be 17.

    So area=33.33.

    4. nth term is a+(n-1)d, sub a and d in.

    It's your typical proof, just with numbers in, like Jan 2013 if I recall correctly.

    I think I got a=-7 and d=3.

    5. r=-1/3, a=216.

    Sum to infinity clearly 162.

    Show that r satisfies the equation. Modulus of r less than 1, then pick the right solution.

    I don't know whether I've got the order of this question right.

    6. Integral is something like 3x^1/2 - (18x^7/3)/7 + c.

    50/3. But everyone got different answers .

    7. Basic log proof.

    1/2 and 2/3 I think.

    8. (3,-1)

    Just work out 1/2 the diameter PQ to find radius r = sqrt29.

    Got an angle of 66.8 I think.

    6 for length of ST.

    9. Use 1/2r^2(theta-sintheta) to find red segment. Find white segment by doing area of whole circle - area of red segment. Should get 1.04 or something to 2.1 something for ratio, so white is 2 times red (angle was 2.6 radians).

    Hope that's helpful, and again, any mistakes or questions left out please let me know.

    P.S. I'll give explanations to any of my answers if you don't understand how I got them.
    Thanks very much this has made my day! Only got 9 and a little 8 wrong and on 5 I couldn't show the equation was 3n^2 eft, how many marks would this have been out of 8. I did the rest
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    (Original post by Mr rump)
    Thanks very much this has made my day! Only got 9 and a little 8 wrong and on 5 I couldn't show the equation was 3n^2 eft, how many marks would this have been out of 8. I did the rest
    You're welcome, but I'm surprised I remembered that much of the paper!
    I don't know about the mark allocations, but it seems like you did well .
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    (Original post by PrimeLime)
    You're welcome, but I'm surprised I remembered that much of the paper!
    I don't know about the mark allocations, but it seems like you did well .
    Haha I couldn't 😂 and yeah I think i did quite well, although not as high as c1 but still an a, that question for number 5 was 8 marks but it wasn't split up, I hope showing the equation was only 2 and the harder bit was 6
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    (Original post by Mr rump)
    To find ST? I couldn't do it
    ST involved drawing a tangent (which was ST) at that point S on the circle (you could assume that S was any point on the circle, so it doesn't matter where you put S). Radius will be perpendicular to tangent and so we can form a right-angled triangle AST. Calculate length of AT, and then use Pythagoras with the radius and the hypotenuse (AT) to find ST, which I got to be 6.
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    (Original post by Mr rump)
    Haha I couldn't 😂 and yeah I think i did quite well, although not as high as c1 but still an a, that question for number 5 was 8 marks but it wasn't split up, I hope showing the equation was only 2 and the harder bit was 6
    Ah, showing the equation was probably not too much. It's a good thing they gave you the equation though, right?
    Hoping for lower grade boundaries.
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    (Original post by Niarhys)
    Helo, thank you very much for the answers it the first one deffos right?
    It's actually 0.4689. I'll change that.
    And yep, my calculator does integration and gave me 0.4685..., which is close enough seen as this is the Trapezium Rule, i.e. and approximation to the integral.
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    (Original post by PrimeLime)
    It was a pretty standard paper I think. The only exception was the last question which caught a lot of people out. It was tricky, but it was still the same stuff as usual, just sort of in disguise.
    Looking back, I'd have to say that grade boundaries won't be too low, since there wasn't too much else that was trickier than usual. Integration for areas was a bit awkward and possibly some of the trig questions. So grade boundaries will probably be slight lower than usual, say, 4 marks lower for an A or something.
    I did pretty badly tbh but it wasn't an impossible paper. I'd say it was slightly more challenging than June 2014 but again I wouldn't expect a massive grade boundaries drop so I'd expect something similar to last summer, maybe a tad higher in some areas
    A - 57
    B - 49
    C - 42
    D - 36
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    Honestly I just dont see the point of the formula booklet, it does not help at all seeing as they make the formulas look like another language.

    I'm going to have to ace the stats exam do get a decent grade now haha.....
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    (Original post by Jack1066)
    Honestly I just dont see the point of the formula booklet, it does not help at all seeing as they make the formulas look like another language.

    I'm going to have to ace the stats exam do get a decent grade now haha.....
    same with stats i never look at the formula booklet, its just offputting
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    Very happy to see this, although I can see some mistakes I've made. Got Q 9 wrong and only managed to get as close to 3 radians. So I'm hoping I can at least get 2 or 3 method marks. I think I lost an overall of 15 marks totals (considering that I get no method marks). If I recall correctly, I got 50/3 as well for area of shaded region.

    For the equation that involved tan. I got one of the solutions. But I made a silly mistake which I took sin over and divided tan by sin to give 1/cos and then solved for cos, which gave me one solution. But because Sin was eliminated, I didn't get the other 2 solutions. I should've taken taken Sin out as a common factor. Then that would've equated Sin to 0 to give me 0 & 180

    Didn't get that radius one, didn't think it would've been that easy lmaoooo, I was overthinking it and just wrote any rubbish.

    Didn't get the angle PQR, had no clue what I was doing and it didn't help that my diagram was really bad so I couldn't see properly as I rushed it. I ended up with like 70.0 degrees.

    Didn't have time to go back to do proving the Sn equation by first principles, left it out.

    But unlike C1, the questions I left out are relatively low marks. A 2 marker here and a couple of 3 markers. I'm quite confident with the rest and even if I get them wrong, I'm hoping for a couple of method marks.

    Overall, great paper, very pleased with it.

    I'm not sure about you, but I noticed a trend. The people who managed to do very well on C1 had ended up completely flopping this C2 exam. Whereas I completely flopped C1 and found C2 a very nice paper? I was shocked to see many people unhappy with that exam. Hopefully that just means boundaries will be relatively lower to compensate for the questions I missed out lmao.
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    Actually, I just checked, turns out the paper was out of 75 not 70 (as I thought). And considering the questions I've left blank, that should be a big enough margin to keep me in an A, pray for me guys.
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    And you have the cross over of those like me who flopped on both C1 and C2 (Definitely worse on C2 though).


    I am ridiculously bad at learning the proofs but I reckon I got a mark for it even though I got the last bit wrong
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    How did people approach the curve/line intergration question?(6b)
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    (Original post by Faisalshamallakh)
    Very happy to see this, although I can see some mistakes I've made. Got Q 9 wrong and only managed to get as close to 3 radians. So I'm hoping I can at least get 2 or 3 method marks. I think I lost an overall of 15 marks totals (considering that I get no method marks). If I recall correctly, I got 50/3 as well for area of shaded region.

    For the equation that involved tan. I got one of the solutions. But I made a silly mistake which I took sin over and divided tan by sin to give 1/cos and then solved for cos, which gave me one solution. But because Sin was eliminated, I didn't get the other 2 solutions. I should've taken taken Sin out as a common factor. Then that would've equated Sin to 0 to give me 0 & 180

    Didn't get that radius one, didn't think it would've been that easy lmaoooo, I was overthinking it and just wrote any rubbish.

    Didn't get the angle PQR, had no clue what I was doing and it didn't help that my diagram was really bad so I couldn't see properly as I rushed it. I ended up with like 70.0 degrees.

    Didn't have time to go back to do proving the Sn equation by first principles, left it out.

    But unlike C1, the questions I left out are relatively low marks. A 2 marker here and a couple of 3 markers. I'm quite confident with the rest and even if I get them wrong, I'm hoping for a couple of method marks.

    Overall, great paper, very pleased with it.

    I'm not sure about you, but I noticed a trend. The people who managed to do very well on C1 had ended up completely flopping this C2 exam. Whereas I completely flopped C1 and found C2 a very nice paper? I was shocked to see many people unhappy with that exam. Hopefully that just means boundaries will be relatively lower to compensate for the questions I missed out lmao.
    I don't think I flopped the C2...
    Yep it was a nice paper in which I screwed up on a couple of questions but managed to correct last minute. I can't wake up so early and then do an exam at 9:00 completely switched on...
    Loads of people (probably everyone I talked to) lost the other two solutions on the trig question because they just divided through by tantheta and so were 'dividing by 0' for those two other solutions. I wouldn't worry too much. I did what you said and took sin out as a common factor XD.
    I would have thought the first principles arithmetic series proof would have been quite quick? It was quicker than doing it algebraically like other papers XD.
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    (Original post by NathBrook)
    How did people approach the curve/line intergration question?(6b)
    Area under curve between -1 and 3 minus that triangle under the line.
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    (Original post by PrimeLime)
    Area under curve between -1 and 3 minus that triangle under the line.
    yeah i thought the integration was a really nice question tbf
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    (Original post by PrimeLime)
    Area under curve between -1 and 3 minus that triangle under the line.
    The integration question was out of 9 yes? I did it wrong and did it between -1 and 6. Think the drawing confused me!

    The 9 marks was for both parts though? Just asking to see if I could gain any marks for correct integration, even though I used the wrong limits?
 
 
 
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