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# AQA AS Core 2 maths unofficial MARK SCHEME (25/05/16) Watch

1. (Original post by Kayleigh216)
What did everyone get for the 16+9sin^2(x)/5-3cosx
Personally I used the identity that Sin^2(x) + Cos^2(x) = 1, rearranged to sub in 1-Cos^2(x) for Sin^2(x) then split the fraction to get (16+9)/5 + -9Cos^2(x)/-3Cosx, which worked out as 5 +3Cosx. Not entirely sure if thats right or not though
2. (Original post by voltz)
So noone else got 10.7 as the perimeter? I got 4.12 as the value of r??
Im pretty sure the perimeter was like 13.6?? I cant remember and r was either 3.something i think
I think b) was (0.2)^x = 4. Use logs to find out x to 3 sign fig
4. (Original post by Chisom99)
wasn't the question (2-x)^7
If it was (2-x)^7 then the coefficient of x^10 wont be -1648, and most people got -1648.
5. (Original post by Rager6amer)
2) a) sketch y = (0.2)^x [2] (negative gradient where the y value decreases exponentially as x increases, with y-intercept at 1 and x-axis being the curve's asymptote)

b) something about solving a log question e.g. (0.2)^x = 8 ? so log that to get x and provide answer to 3sf [3?]

c) transformation that maps the graph y =(0.2)^x onto y = (5)^x [1] answer: y = (1/5)^x there fore replace x with -x to get y = (5)^x so therefore transformation is reflection in y axis.
b was (0.2)^x = 4 !
6. (Original post by Rager6amer)
Im pretty sure the perimeter was like 13.6?? I cant remember and r was either 3.something i think
Yeah I think im wrong a lot of people got 13...hopefully I will get atleast 1 mark for some sort of working out
7. anyone remember what they got for the tan graphs ? i got 129 degress and 309 degrees
8. (Original post by voltz)
Yeah I think im wrong a lot of people got 13...hopefully I will get atleast 1 mark for some sort of working out
6 marks will carry at least 2 to 3 method marks so as long as you wrote down the steps you should be ok
9. (Original post by Rager6amer)
6 marks will carry at least 2 to 3 method marks so as long as you wrote down the steps you should be ok
I think I equaled the formula for area of a sector to half the area of the triangle or something... does this sound along the right lines?
10. (Original post by voltz)
I think I equaled the formula for area of a sector to half the area of the triangle or something... does this sound along the right lines?
Yeah that sounds right
11. (Original post by Rager6amer)
Yeah that sounds right
Thats okay then, and what did you get for the question asking to find the sum from 4 to 12?
12. Urrrmm, i somehow got a=10 for question one, the integration question, and it seems i was way off. Really not sure what i did. Does anyone remember the actually question and how many marks it was worth? Thank you
13. (Original post by LordCommander)
Personally I used the identity that Sin^2(x) + Cos^2(x) = 1, rearranged to sub in 1-Cos^2(x) for Sin^2(x) then split the fraction to get (16+9)/5 + -9Cos^2(x)/-3Cosx, which worked out as 5 +3Cosx. Not entirely sure if thats right or not though
That's what I got but then I couldn't get an answer for cos^-1= 5/3
14. (Original post by voltz)
Thats okay then, and what did you get for the question asking to find the sum from 4 to 12?
Yes I'd like to know this too because i got 72 but am beginning to think that was wrong. I'm so confused, i found this paper rather difficult.

Wait, was it from 4 to 12?? I'm sure i remember from 4-21??
15. (Original post by BlackthornRose)
Yes I'd like to know this too because i got 72 but am beginning to think that was wrong. I'm so confused, i found this paper rather difficult.

Wait, was it from 4 to 12?? I'm sure i remember from 4-21??
I got 90
16. (Original post by Kayleigh216)
I got 90
Yeah it seems everyone did.....I don't know what i did wrong, it looks like i forgot to add on 28 or something?
17. (Original post by BlackthornRose)
Yeah it seems everyone did.....I don't know what i did wrong, it looks like i forgot to add on 28 or something?
I got 88 - looks like I messed up
18. (Original post by voltz)
I got 88 - looks like I messed up
You and I both then, you and I both. Except, I also messed up on tons of other questions.....I don't even wanna think about it.
19. (Original post by timothygrealish)
I have contracted AIDS from that paper
I think everyone did 😂
20. (Original post by rosieivison)
what did people get for the transformation onto y=3√x^2+1? I'm not sure if that was the question asked correct me if I'm wrong??

Posted from TSR Mobile
Convert 3 to sqrt(9) and multiply by sqrt((x^2)+1) (surd rules)

= sqrt(9((x^2)+1))
= sqrt(9(x^2) + 9)
= sqrt((3x)^2 + 9)

therefore going from f(x) to f(3x) => stretch s.f. 1/3 parallell to x axis.

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