Mei c3 2019 Watch

uauai
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#1
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#1
Anyone resit C3?
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Aaaaaaab
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#2
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Went pretty well apart from qu 7 I got tanx 3^1/2
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Heidilw2102
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Yes went well, hated the last parts of question 9 though
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uauai
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#4
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Yeah, I thought it was decent too and was question 7 the proving one?
(Original post by Aaaaaaab)
Went pretty well apart from qu 7 I got tanx 3^1/2
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uauai
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#5
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Me too! I got the denominator they were asking for but screwed up the top bit so I hope I got method marks 🤞
(Original post by Heidilw2102)
Yes went well, hated the last parts of question 9 though
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uauai
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#6
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What do we reckon the grade boundaries will be like? I think the exam was easier than the 2018 paper so can the boundaries be higher than last year’s?
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Heidilw2102
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Question 7 was the proving one. I used the quotient rule implicitly and then ended up with an equation in sin and cos = 0 and when it rearranged to equal dy/dx it got exactly the end point
(Original post by uauai)
Yeah, I thought it was decent too and was question 7 the proving one?
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Heidilw2102
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#8
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Last year was 58 for an 80 UMS (if you have 18 in the coursework). I reckon it will be exactly the same, maybe one mark slightly lower. I though last years paper was actually easier.
(Original post by uauai)
What do we reckon the grade boundaries will be like? I think the exam was easier than the 2018 paper so can the boundaries be higher than last year’s?
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uauai
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#9
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Yeah, that was a nice question imo
(Original post by Heidilw2102)
Question 7 was the proving one. I used the quotient rule implicitly and then ended up with an equation in sin and cos = 0 and when it rearranged to equal dy/dx it got exactly the end point
I probably found this year easier cos I was more prepared but hopefully the grade boundaries do stay the same, good luck!
(Original post by Heidilw2102)
Last year was 58 for an 80 UMS (if you have 18 in the coursework). I reckon it will be exactly the same, maybe one mark slightly lower. I though last years paper was actually easier.
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sophierob103
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I preferred this paper to last year, struggled abit with the last part of question 9, I couldn’t get the top part of the equation but I got the bottom
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uauai
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Same, I’m a bit scared my answer to the part before might’ve been wrong cos I was really struggling to form the numerator
(Original post by sophierob103)
I preferred this paper to last year, struggled abit with the last part of question 9, I couldn’t get the top part of the equation but I got the bottom
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Heidilw2102
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I also couldn't get the top part wtf! lol
I reckon there will be ECF marks and method marks. The final answer will only be one mark lost on each question. If the rest of the paper went well I wouldn't stress about it.
(Original post by uauai)
Same, I’m a bit scared my answer to the part before might’ve been wrong cos I was really struggling to form the numerator
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sophierob103
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#13
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Is anyone doing C4 on Thursday?
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iDeviceJ
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#14
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Let’s compile answers, this is what I remmeber getting:

1) 1/9(x^3+2)^0.5 ??? tbh can’t remember this but was integration
2) ln4/9
3) a= 24.5, k= 0.0623
4) (think this was the connected rated one) dy/dt = -2/15

5) can’t remmeber

6) can’t rmemeber

7) can’t rmemebe

8) f’’(x) = 0, so point of inflection

9) integration: 1/k^2 - 2e^-1k^2

and then the prove one
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BioChemDave
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#15
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Answers i remeber:

2ln(3/2)
a = 24.5 and k = 0.063
36.1 Years for 90% of max mass
dy/dt = 1/30
value of coordinate p was likee root 2
Area was 1/2
Last questions, you hard to work out the area of the triangle, which was like 1/2K^2e and subtract it from the area which was like 1/k^2(1-1/k^2e) .
Last edited by BioChemDave; 4 weeks ago
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iDeviceJ
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#16
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answer was ln(4/9), it said to express answer as a single logarithm. i agree with everything else apart from ur -3/8. how can it be a maximum, there was a 12x on the top of the f’’(x)
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BioChemDave
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ln(4/9) gives you a negative area. 2ln(3/2) is a single logarithm, with a positive value for the area. integrated the equation then on my calculator and it gave me the equavilvent value.
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iDeviceJ
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#18
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#18
2ln3/2 isn’t a single logarithm
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iDeviceJ
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#19
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#19
Sorry i meant ln(9/4)
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uauai
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#20
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I think I got 1/2 for the normal gradient and 2 for the inverse gradient idk if it was right
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