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OCR (Not MEI) C2 January 2013 Discussion

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Reply 120
Original post by Lard
IMG_20130118_182105[1].jpg
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Not sure if this has already been uploaded but I brought my exam paper out with me. You're welcome :wink:


ah brilliant, thanks!!

Original post by Mr M
As I said I don't have a paper as my school closed due to snow today. If someone sends me a scan or photos from their photo I will sort it out.
Reply 121
Original post by Lard
x


Ahh thanks a lot dude! :smile:
Reply 123
Original post by stirkee
Ah yeah of course. Can you remember which Q this was?

Something is telling me it was Q6 but I'm not sure.


Pretty sure it was Question 8 :smile:
Reply 124
Original post by Colroyd
Pretty sure it was Question 8 :smile:



well we know for sure now xD
Hard paper I think, I got 95% in the mock but I think I screwed this up badly.. may be doing a second retake. Most notably:

Question 1 ii) I somehow got a quadratic, used the formula, and got two answers: 16.7 and 3.06?
Q 8 iv) I think I knew how to do, but I accidentally expressed my answer as log(c)=2.7 or similar
9 ii) it says "other possible value" not "values", but I got two: (1/3)+(1/3)sqrt(7), (1/3)-(1/3)sqrt(7)

Other than that I can't think of any problems, but I keep thinking I made silly calculator mistakes and stuff.
Reply 126
Mr M, where u @ man.
Reply 127
Should have brought my answers out with me, I'm fairly good at maths so my answers are hopefully quite good, if you spot any mistakes tell me:
log2(x)-log2(x-3)=4=log2(2^4)=log(16)
therefore x/(x-3)=16 so 15x=48 and x=3.2
Coef y^3=80(3y+3y^2)^2 + 40(3y+3y^2)^2
=80(6y^3)+40(27y^3)=1560
got (1+sqrt7)/3 as A>0 but can't remember the question
For the circle q I got theta as0.5411 think it was give then
area/2=2thetaR^2/2 - sin(2theta)R^2/2
therefore area=R^2(2theta-sin(2theta)
R=7 theta=0.5411 so area=9.76
can anyone remember rest of qs?
(edited 11 years ago)
Reply 128
Original post by Lstigant
Should have brought my answers out with me, I'm fairly good at maths so my answers are hopefully quite good, if you spot any mistakes tell me:
log2(x)-log2(x-3)=4=log2(2^4)=log(16)
therefore x/(x-3)=16 so 15x=48 and x=3.2
Coef y^3=80(3y+3y^2)^2 + 40(3y+3y^2)^2
=80(6y^3)+40(27y^3)=1560
got (1+sqrt7)/3 as A>0 but can't remember the question
For the circle q I got theta as0.5411 think it was give then
area/2=2thetaR^2/2 - sin(2theta)R^2/2
therefore area=R^2(2theta-sin(2theta)
R=7 theta=0.5411 so area=9.76
can anyone remember rest of qs?


Just been posted above :smile:
Reply 129
Original post by Lstigant
Should have brought my answers out with me, I'm fairly good at maths so my answers are hopefully quite good, if you spot any mistakes tell me:
log2(x)-log2(x-3)=4=log2(2^4)=log(16)
therefore x/(x-3)=16 so 15x=48 and x=3.2
Coef y^3=80(3y+3y^2)^2 + 40(3y+3y^2)^2
=80(6y^3)+40(27y^3)=1560
got (1+sqrt7)/3 as A>0 but can't remember the question
For the circle q I got theta as0.5411 think it was give then
area/2=2thetaR^2/2 - sin(2theta)R^2/2
therefore area=R^2(2theta-sin(2theta)
R=7 theta=0.5411 so area=9.76
can anyone remember rest of qs?


Can't remember the circle q but the rest is correct going by what I got.
Reply 130
Can someone please explian how on Q9 i) the 4/x^2 integrated and put the 2a and a into becomes 2? Thanks.
Man, that was a nasty paper. Nice little poem my teacher told me...No point reflecting on what you did today, not going to help in any way!

But yeye Mr.M..the suspense is killing me :eek:
My process for 8)iv)

log2x = log2(x-3) + 4
log2x = log2(x-3) + log216
log2x = log2[16(x-3)] [get rid of the logs]
x = 16x - 48
48 = 15x
16/5 = x

Can anyone verify this? I'm awful with logs usually, but I surprised myself with this
Reply 133
Original post by DomRusky
My process for 8)iv)

log2x = log2(x-3) + 4
log2x = log2(x-3) + log216
log2x = log2[16(x-3)] [get rid of the logs]
x = 16x - 48
48 = 15x
16/5 = x

Can anyone verify this? I'm awful with logs usually, but I surprised myself with this


Yep, that is correct :smile:
I thought it was a really nice paper...
Reply 135
Original post by DomRusky
My process for 8)iv)

log2x = log2(x-3) + 4
log2x = log2(x-3) + log216
log2x = log2[16(x-3)] [get rid of the logs]
x = 16x - 48
48 = 15x
16/5 = x

Can anyone verify this? I'm awful with logs usually, but I surprised myself with this


Yeah the majority have argued that it is 3.2

Well done! :smile:
Original post by azns
mr m, where u @ man.


everything hangs on mr m
Original post by Lstigant
Should have brought my answers out with me, I'm fairly good at maths so my answers are hopefully quite good, if you spot any mistakes tell me:
log2(x)-log2(x-3)=4=log2(2^4)=log(16)
therefore x/(x-3)=16 so 15x=48 and x=3.2
Coef y^3=80(3y+3y^2)^2 + 40(3y+3y^2)^2
=80(6y^3)+40(27y^3)=1560
got (1+sqrt7)/3 as A>0 but can't remember the question
For the circle q I got theta as0.5411 think it was give then
area/2=2thetaR^2/2 - sin(2theta)R^2/2
therefore area=R^2(2theta-sin(2theta)
R=7 theta=0.5411 so area=9.76
can anyone remember rest of qs?


OMG that's so relieving, I did get x=3.2, the other numbers I got was when I was checking on the calculator.
Got the same on the area question, but I used different working. Forgot the formula for a segment, so I worked it out for the sector and tookaway the triangle. Area for each triangle was 6*sqrt(13)
Original post by olivers16
Just been posted above :smile:


I'd hate to argue this and look stupid,
but wouldn't " 80(3y+3y^2)^2 " make 80(9y^2 + 9y^4)
there's no y^3 in that part is there? or am I being stupid

I just got 40(3y+3y^2)^3 = 40x27y^3 + 40x27y^6
So co-efficient is just 40x27 = 1080

Am I being dumb here?
Reply 139
Question 1 is sine rule, cba to aanswer question 2 show they both equal 2650 or use algebra to show sum50-sum36=sum35
k2(4-3)=9=2x
x=9/2
32+80x+80x^2+40x^3+10x^4+x^5
tan(x)=sin(x)/cos(x) so 2sin^2(x)=4cos^2(x)-cos(x)
2-2sin^2(x)=4cos^2(x)-cos(x)
Therefore 6cos^2(x)-cos(x)-2

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