The Student Room Group

Arsey's S15 FP1 Model Answers (attached to 1st post)

Here you go FP1 fans

I thought that was quite tricky in places.

I doubt I have used the most efficient methods as mine seemed too long on a couple.

Boundaries, no idea, probably a bit lower than normal.

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Reply 1
Original post by Arsey
Here you go FP1 fans

I thought that was quite tricky in places.

I doubt I have used the most efficient methods as mine seemed too long on a couple.

Boundaries, no idea, probably a bit lower than normal.


cheers m8.

how many marks would i drop if i got to the final quadratic on q and got 12+6i/18 which is right but then simplified it wrong?
(edited 8 years ago)
Thanks for doing this, Arsey. Couple of questions I have about marks I've lost - greatly appreciate it if you can give me your guesses as to how many marks I'd gain for each of these questions (know you've had a long day, sorry):


6) a) Went straight from (-1/4(5^k - 1) - 5^k) to (-1/4(5^(k+1) - 1)). Lose any marks for not showing rearranging it to be -5/4 etc?


6) b) Second part of induction - did basis, assumption and started n=k+1 correctly but couldn't get (k+1) out as a factor. Finished with just writing down Sum(K+1) = what they asked for. Assuming 2 marks for basis and starting n = k+1. Gain another method mark for conclusion or no?


7) Worked out B^-1 correctly (with correct determinant) but then (god know's why :/) calculated it a second time incorrectly. I did 30-15 instead of 30--15, resulting in 2c = 9 and c= 9/2. Lose all the marks, or gain one for method?


8) a) Did
SP = root((3p^2-3)^2 + (0-6p)^2)
SP = root(9p^4 - 18p^2 + 9 + 36p^2)
SP = 3p^2 + 18p + 3


and left it like that. Any marks out of the 3?


8) b) Worked out the equation of the tangent at p correctly (py - x = 3p^2), then said equation of tangent at q would be (qy - x = 3q^2). Made x the subjects and equalled them to each other, then attempted to solve (ended up with incorrect answer). Any idea how many marks out of 8?


Thanks so much!
Yay the King has arrived. All hail King Arsey. :biggrin:
Reply 4
Cheers Arsey
Reply 5
I have other work to do, so I will not be able to answer any questions tonight but if you are asking how many marks would i... look on past papers and MS, you'll find what you need.
Reply 6
Thanks mate
My son has got all of them right
Cheers
Reply 7
Isn't arg of z supposed the be -pi/3?
Or was I wrong?


Posted from TSR Mobile
I'm going to post this again as I didn't get a solid reply on the other thread:

In the first Induction question [Matrices]
I showed 5k(14)=20k\dfrac{-5^{k}}{(-\dfrac{1}{4})} = 20^{k} which I then quite stupidly went on to saying =4(5k)= 4(5^{k})
My teacher said the mark scheme allows one algebraic mistake, is this true?
because I then done 14[5k+4(5k)1] -\dfrac{1}{4}[5^{k}+4(5^{k})-1] \rightarrow 14[5(5k)1]-\dfrac{1}{4}[5(5^{k})-1] which then went on to being 14[5k+11]-\dfrac{1}{4}[5^{k+1}-1]
(edited 8 years ago)
Thank you big time arsey
For Q7 i left the coordinates in a matrix form, would i lose a mark?
Also for 4c) i wrote z3 not z1+z2 but drew it correctly, would i lose a mark on that?
Finally for the induction i did n=1 and found n=k but couldnt do the rest, how many marks would i get? 2 or 3?

Thanks a lot
Original post by Jelly150
Isn't arg of z supposed the be -pi/3?
Or was I wrong?


Posted from TSR Mobile


He put that, small writing with a * next to it :smile:
Reply 12
I put 3i on x axis instead of the y axis, will I lose just 1 out of the 2 marks?


Posted from TSR Mobile
Awesome, I reckon I dropped one mark :smile: will I still get 100 UMS, do you think?
Reply 14
Having seen the paper and solutions I feel as I mentioned earlier that A will be at around 60.
This is also backed up by the thread with the poll as to how people found it.
Arsey thank you sooo much 75/75 so happy. :biggrin:
Reply 16
Thank you!! I think I did alright:smile:
Original post by edothero
I'm going to post this again as I didn't get a solid reply on the other thread:

In the first Induction question [Matrices]
I showed 5k(14)=20k\dfrac{-5^{k}}{(-\dfrac{1}{4})} = 20^{k} which I then quite stupidly went on to saying =4(5k)= 4(5^{k})
My teacher said the mark scheme allows one algebraic mistake, is this true?
because I then done 14[5k+4(5k)1] -\dfrac{1}{4}[5^{k}+4(5^{k})-1] \rightarrow 14[5(5k)1]-\dfrac{1}{4}[5(5^{k})-1] which then went on to being 14[5k+11]-\dfrac{1}{4}[5^{k+1}-1]


Original post by TeeEm
Having seen the paper and solutions I feel as I mentioned earlier that A will be at around 60.
This is also backed up by the thread with the poll as to how people found it.


^^ I'm dying for an answer lmao, help :afraid:
Reply 18
Original post by TeeEm
Having seen the paper and solutions I feel as I mentioned earlier that A will be at around 60.
This is also backed up by the thread with the poll as to how people found it.

I agree. My bet would be 60 for an A
Original post by Jelly150
Isn't arg of z supposed the be -pi/3?
Or was I wrong?

Posted from TSR Mobile



Yes it is that, He's starred it and written it above

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