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Edexcel FP3 - 27th June, 2016 watch

  • View Poll Results: How did you find the Edexcel FP3 exam?
    Very hard
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    (Original post by oShahpo)
    I can't remember, I think I might have not substituted in the limits. Do you remember which question this was? Was it two parts or just one part?
    it was the integral of 1 over the square root of a quadratic, where it had -x^2 as a coefficient. you had to complete the square, etc
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    (Original post by Ewanclementson)
    Solutions - Let me know if any need changing.
    1)3 and -2
    2)5/8+ln(3/2)
    3)k=2
    4)pi/3 2/3arctan(e^x/3)
    5)25/3-32/15root2
    6)9 then p=7 and q=5, the eigenvector is (2,1,-2) and then there will be varying orders of the eigenvectors for P.
    7)1/12(pi+6root3-6)
    8)7x+5y-9z=8 and my a was (11/8,-13/8,0) and b was (11/8,-1/8,1) although others may differ.
    wasnt for the line of intersection a=(11/8, -13/40, 0) and b looks the same as mine, it seems like we both cancelled z and worked like that, i dont understand how u got 8 as denominator?? i may be wrong
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    (Original post by Anon-)
    it was the integral of 1 over the square root of a quadratic, where it had -x^2 as a coefficient. you had to complete the square, etc
    I can't remember if I subbed in the limits. How many marks was it?
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    (Original post by oShahpo)
    I can't remember if I subbed in the limits. How many marks was it?
    5 marks i think
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    (Original post by oShahpo)
    I agree with all but I don't remember this pi/3 at all. Do you remember the question?
    AHHH **** I forgot to sub the e^x in U.. ****.!
    It was the one where we had to integrate a polynomial but it was 1/the polynomial and the coefficient of x^2 was -1
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    (Original post by Major-fury)
    Am I the only one who liked the integrating questions ? You can literally check ur answers to see if they are correct
    to be honest my casio fx-991es saved my life in A2 maths, i can just check all my integrals, differentials, matrices on it
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    (Original post by Major-fury)
    It was the one where we had to integrate a polynomial but it was 1/the polynomial and the coefficient of x^2 was -1
    Was the polynomial root( 4 - (x-1)^2 )?
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    Im so glad the questions I couldn't do were worth 3 marks😂😂
    Quite a nice paper honestly I just couldn't figure out the reduction formula and the proof for arcoth.
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    (Original post by oShahpo)
    I can't remember, I think I might have not substituted in the limits. Do you remember which question this was? Was it two parts or just one part?
    it was part i, where part ii had part a and b within it, was with the substitution question, right above it
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    How did you guys prove they were perpendicular? I did the dot product and got theta is 90
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    (Original post by oShahpo)
    I agree with all but I don't remember this pi/3 at all. Do you remember the question?
    AHHH **** I forgot to sub the e^x in U.. ****.!
    The question was the integral between 3 and 5 of 1/root(15+2x-x^2) if I remember correctly which integrated to arcsin((x-1)/4) with those bounds which went to pi/2 - pi/6 which is pi/3.
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    (Original post by Major-fury)
    How did you guys prove they were perpendicular? I did the dot product and got theta is 90
    yhh i just worked out the normal vector for plane 2 and normal vector for plane one was pretty much given, then i just dotted the two together and got 0, same way as u got pheta=90 tbh
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    (Original post by Ewanclementson)
    The question was the integral between 3 and 5 of 1/root(15+2x-x^2) if I remember correctly which integrated to arcsin((x-1)/4) with those bounds which went to pi/2 - pi/6 which is pi/3.
    Oh yea now I remember. It was arcsin(5-1/4) - arcsin (2/4) = pi/3, thanks man I was scared I had left it without subbing.
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    (Original post by Ewanclementson)
    Solutions - Let me know if any need changing.
    1)3 and -2
    2)5/8+ln(3/2)
    3)k=2
    4)pi/3 2/3arctan(e^x/3)
    5)25/3-32/15root2
    6)9 then p=7 and q=5, the eigenvector is (2,1,-2) and then there will be varying orders of the eigenvectors for P.
    7)1/12(pi+6root3-6)
    8)7x+5y-9z=8 and my a was (11/8,-13/8,0) and b was (11/8,-1/8,1) although others may differ.
    ahh i got a different answer for 2. I got 19/12ln(6) i might be wrong
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    (Original post by MarocMan)
    wasnt for the line of intersection a=(11/8, -13/40, 0) and b looks the same as mine, it seems like we both cancelled z and worked like that, i dont understand how u got 8 as denominator?? i may be wrong
    I believe I recorded mine wrong as I remember also getting -13/40. I shall correct that and repost the answers.
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    Solutions - updated - Let me know if any need changing.
    1)3 and -2
    2)5/8+ln(3/2)
    3)k=2
    4)pi/3 2/3arctan(e^x/3)
    5)25/3-32/15root2
    6)9 then p=7 and q=5, the eigenvector is (2,1,-2) and then there will be varying orders of the eigenvectors for P.
    7)1/12(pi+6root3-6)
    8)7x+5y-9z=8 and my a was (11/8,-13/40,0) and b was (11/8,-1/8,1) although others may differ
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    How many marks would I lose for not putting u=e^x? I.e. not resubbing the e^x. I am not even sure I didn't do it, but I think so.
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    Need help with last question could someone upload a photo please of working out from memory If possible
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    (Original post by Ewanclementson)
    Solutions - updated - Let me know if any need changing.
    1)3 and -2
    2)5/8+ln(3/2)
    3)k=2
    4)pi/3 2/3arctan(e^x/3)
    5)25/3-32/15root2
    6)9 then p=7 and q=5, the eigenvector is (2,1,-2) and then there will be varying orders of the eigenvectors for P.
    7)1/12(pi+6root3-6)
    8)7x+5y-9z=8 and my a was (11/8,-13/40,0) and b was (11/8,-1/8,1) although others may differ
    I envy your memory my friend. All's correct, high five.
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    (Original post by Major-fury)
    How did you guys prove they were perpendicular? I did the dot product and got theta is 90
    i did the dot product of the two normal vectors, which is 0 (cos90)
 
 
 
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