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# AQA A2 Mathematics MPC3 Core 3 - Wednesday 15th June 2016 [Official Thread] Watch

1. Was meant to be getting A* but now I think this has ****ed it all.
2. (Original post by ashwinderk)
Nah, it was 177/224.
swear i said that

3. This is like the 3rd hitler reacts video to this exam but this is the best xD
4. (Original post by Parhomus)
All i remember was when I put that 7 marker for integration into my calculator it took 20 years to churn out the answer.
thought mine was going to set on fire at one point.. I had to do it twice as well as I thought I messed it up oh dear oh dear

Here's my (half complete) unofficial mark scheme with worked solutions

Think they're all right, have checked with desmos/wolfram

Couldn't remember some questions or the questions numbers!

The answer to Q9 is wrong somewhere, -21.3 degrees is the only actual right answer but idk why
6. (Original post by vilifyme)
at the maxium point of the graph, the gradient is 0. So differentiate 16x-e^(2x) and equate it to 0. 16-2e^2x=0 so 16=2e^2x so 8=e^2x so sqrt(8)=e^x. ln(sqrt(8))=x which is the same as 0.5ln(8). so f(x) is less than or equal to 0.5ln(8). I got 0/5 in that question probably but as soon as I got out the of exam, I realised the gradient is 0 and solved it in my head! So annoyed with myself.
You had to also substitute 0.5ln8 into f(x), as the range is f(x)<= the Y coordinate. I got 8-8ln(8) or something like that.
8. (Original post by lollypop101)
loooooooool man got -123 uno looool ffs
the entire exam robbed me of the ability to think straight
9. EDIT: Nevermind!
10. (Original post by WurfWurf)
Please just **** off. You made this account today just to annoy people. http://prntscr.com/bgqh6h . Busted
yeah i made it today, but to be honest that paper wasnt bad at all
11. (Original post by smartsy)
the entire exam robbed me of the ability to think straight
don't man i know, the whole thing was basically a fking mess its acc disgusting when u think about what aqa made us answer
12. That paper was fine until question 3...😂😂
i) We know secx +tan x = -5 and sec^2x -tan^2x = 1 (as an identity) factorising the identity gives (secx+tanx)(secx-tanx)=1 and by substituing in the first equality we find secx-tanx=-0.2
ii) add secx+tanx = -5 and secx-tanx=-.02 to get 2secx = -5.2
secx=-2.6
cosx=-5/13
iii) let theta equal the old x
2x-70=theta
2x-70=+-arcos(5/13) + 360n
Rearrange for x and you get two answers (-23.somethinng and almost -90 can't remember exactly)
You substitute them back in to the original equation to check, the -23.something one works and the other doesn't, leaving the -23.something as the only solution
14. (Original post by abfb)
not sure if this has been posted and i didn't get part a in the exam but realized it at at home
question 9
a)
Spoiler:
Show
secx-tanx=-5
secx=tanx-5
sec2x=(tanx-5)2 ==>sec2x=tan2x-10tanx+25
since sec2x=tan2x+1
tan2x+1=tan2x-10tanx+25
tan2x's cancel so 10tanx=24 ==> tanx=2.4
secx=tanx-5 ==> secx=-2.6
therefore secx+tanx=-0.2
b)
Spoiler:
Show
secx-tanx=-5 and secx+tanx=-0.2
add the 2 equations together ==> 2secx=-5.2 ==> cosx=1/-2.6
c)
Spoiler:
Show
use cos(2x+70)=1/-2.6 to solve
if there are any mistakes i've made feel free to point them out
thanks and do you know how to the the wholoe of the integral by parts question
15. does any1 know where u can find the specimen papers for core 4, i did have a website but idno where its gone help pls
16. (Original post by matty9)

This is like the 3rd hitler reacts video to this exam but this is the best xD
its the fourth one I've seen so far LOL
and it's gold how they've collectively covered all the questions in the paper
17. Where can i find the unofficial mark scheme?
18. (Original post by 09noblei)
Where can i find the unofficial mark scheme?

and

http://www.thestudentroom.co.uk/show....php?t=4166871
19. (Original post by jojoj)
thanks and do you know how to the the wholoe of the integral by parts question
integration by parts question
Spoiler:
Show
∫(ln(3x))2/x2 dx

let u= (ln(3x))2 dv/dx=1/x2
du/dx=2ln(3x)/x v=-1/x

∫(ln(3x))2/x2 dx = -(ln(3x))2 /x - ∫(2(ln(3x))2)/x)*(-1/x)dx

∫(ln(3x))2/x2 dx = -(ln(3x))2 /x - ∫-(2ln(3x)/x2 dx

∫(ln(3x))2/x2 dx = -(ln(3x))2 /x +2∫(ln(3x)/x2 dx

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