Edexcel FP3 - 27th June, 2016

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?

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.

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?

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.. ****.!

Am I the only one who liked the integrating questions ? You can literally check ur answers to see if they are correct

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

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?

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.. ****.!

How did you guys prove they were perpendicular? I did the dot product and got theta is 90

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.

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

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

How did you guys prove they were perpendicular? I did the dot product and got theta is 90