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

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3 years ago
#41
(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
0
#42
(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
0
#43
I think b) was (0.2)^x = 4. Use logs to find out x to 3 sign fig
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3 years ago
#44
(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.
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3 years ago
#45
(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 !
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3 years ago
#46
(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
0
3 years ago
#47
anyone remember what they got for the tan graphs ? i got 129 degress and 309 degrees
1
#48
(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
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3 years ago
#49
(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?
0
#50
(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
0
3 years ago
#51
(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?
0
3 years ago
#52
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
0
3 years ago
#53
(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
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3 years ago
#54
(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??
0
3 years ago
#55
(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
0
3 years ago
#56
(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?
0
3 years ago
#57
(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
1
3 years ago
#58
(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.
0
3 years ago
#59
(Original post by timothygrealish)
I have contracted AIDS from that paper
I think everyone did 😂
0
3 years ago
#60
(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.
0
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