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Edexcel FP3 June 2015 - Official Thread

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Reply 780
I got 2 as a repeated root for all of the eigenvalues... Did anybody else get that? xD
Reply 781
Hope the grade boundary low!!!
What did people do for area?


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Reply 783
any more? Need a decent polling sample (min 100)
http://www.thestudentroom.co.uk/showthread.php?t=3417321
Original post by physicsmaths
What did people do for area?


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0.5?
That went terribly for me... :frown: Also made stupid mistakes on the 'easy' questions, so things are definitely not looking up for me.
Never had an exam that's gone worse than that in my life.
Spet 1hr thinking 3 was wrong lol. Pissed up a few parts of later questions because of that. Thank **** I didn't need that for an a*
Reply 787
Original post by pengomar
I did. It wasn't that hard, just a bit long.


Sp you think the IAL paper is a bit easier?
Original post by nanairo
Sp you think the IAL paper is a bit easier?


I think so
Reply 789
Original post by Gome44
0.5?


fml I think I forgot to sub in the bottom limits. I had e and stuff over e. All over the place
**** i think i forgot to times by 1/2


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Reply 791
Original post by physicsmaths
**** i think i forgot to times by 1/2


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I got 1 for area of the triangle too. i had (1/2)sqrt2.e^t*sqrt2.e^-t=1
Original post by physicsmaths
**** i think i forgot to times by 1/2


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On what question?!
Original post by raff97
I got 1 for area of the triangle too. i had (1/2)sqrt2.e^t*sqrt2.e^-t=1


Yup I had area of OQR = 1 also, same as you're lengths also
:smile:
Original post by physicsmaths
**** i think i forgot to times by 1/2


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Most people are saying 1 so I'm probably wrong :/
Original post by raff97
i got 1 for area of the triangle too. I had (1/2)sqrt2.e^t*sqrt2.e^-t=1


me toooo
Original post by Gome44
Most people are saying 1 so I'm probably wrong :/


I got 1.


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Normalised eigenvectors if anyone's curious

For 2-√2, it was (0.5, -1/√2, 0.5)
For 2, it was (1/√2, 0, -1/√2)
For 2+√2, it was (0.5, 1/√2, 0.5)

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Original post by trooken
For the last question, I substituted y=mx+c into the equation of the ellipse, obtained values for x from a quadratic, worked out the corresponding values for y, then found the midpoints of X and Y in terms of c, then equated the two 'c's together to get a locus. The answer I got is not the same as what others before me posted, so is this method invalid, or did I most likely make a mistake with the algebra?

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I did this as well, I got y=-4mx , what did you get? :smile:


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