Thought this was one of the most challenging papers I've seen... on par with June 2012. First time I haven't been able to answer a question which was disappointing. Reckon about 67/75.
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C2 Maths AS aqa 2016 (unofficial mark scheme new)
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 101
 25052016 16:04

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 102
 25052016 16:06
(Original post by jordanwu)
Hmm for 9a I got y=1/2m6n... I'm pretty sure you had a negative indice on the bottom so it would become positive when you bring it to the top??? 
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 103
 25052016 16:06
[QUOTE=Eisobdxhsonw;65161255]
(Original post by Bosssman)
I'm still confused as to why it was 5+3cosx and not 3+5cosx also as to why the min value was 2, cheers tho
Here's the method:
(16 + 9(1  cos^2x))/(5  3cosx)
(25  9cos^2x)/(5  3cosx) then difference of two squares
(5 + 3cosx)(5  3cosx)/(5  3cosx)
The (5  3cosx) cancel to give 5 + 3 cosx
Minimum value of cosx is 1 so minimum value is 5 + 3(1) = 5  3 = 2 when x = pi 
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 104
 25052016 16:08
(Original post by Myachii)
Gotta love AQA
Thanks for the clarification though, that was annoying me. Awful lot of work for 2 marks (never seen anything like it before either) 
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 25052016 16:09
[QUOTE=Bosssman;65162331]
(Original post by Eisobdxhsonw)
The question was show that (16 + 9sin^2x)/(53cosx) can be written in the form p + qcosx and state the minim value, and the exact value of x for this point.
Here's the method:
(16 + 9(1  cos^2x))/(5  3cosx)
(25  9cos^2x)/(5  3cosx) then difference of two squares
(5 + 3cosx)(5  3cosx)/(5  3cosx)
The (5  3cosx) cancel to give 5 + 3 cosx
Minimum value of cosx is 1 so minimum value is 5 + 3(1) = 5  3 = 2 when x = pi 
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 106
 25052016 16:10
(Original post by jordanwu)
Hmm for 9a I got y=1/2m6n... I'm pretty sure you had a negative indice on the bottom so it would become positive when you bring it to the top???
Question was:
log3(c) = m
log27(d) = n
log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)
It wanted sqrt(c)/d^2 which is:
log3(sqrt(c))  log3(d^2) = log3(3^1/2m)  log3(3^6n)
sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m6n
...right? I've already lost enough marks on this paper
(Original post by Parhomus)
I find it hilarious that the value for x was pi, it felt like they were trying to trick us and it ended with a simple answer. :PLast edited by Myachii; 25052016 at 16:13. 
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 107
 25052016 16:12
(Original post by collegeboy66)
And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?
The solution to this question was to recognise that 0.2 = 1/5 = 5^1 and so resulted in 5^x = 0.2^x and hence the answer being a transformation of reflection in the y axis 
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 108
 25052016 16:13
(Original post by Myachii)
+1 OP pls fix
Another thing  6c is confusing the hell out of me
y=\sqrt{x^2+9}
onto
y=3\sqrt{x^2+1}
Let f(x) = \sqrt{x^2+9}
f(x) > 3f(x) = Stretch in y direction SF 3
However, your mark scheme says "Stretch in x direction SF 1/3"
For that to be the case, wouldn't the equation have to be the following:
f(x) > f(3x) = Your answer
Which would lead to the "original equation" having to be:
y=\sqrt{3x^2+1}
My logic was that the 3 is multiplying the entirety of the sqrt function.
For example:
5^x > 3 * 5^x = Stretch in y direction SF 3
5^x > 5^3x = Stretch in x direction SF 1/3
This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)
You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.
Forgive me if I'm talking out my ass too
So we started with y=sqrt(x^2+9)
and went to y=3*sqrt(x^2+1)
Ignore the y for now.
Square both equations to get x^2+9 > 9*(x^2+1)
Then we get x^2+9 > 9x^2+9
ignore the 9 and look what happened to x
We went from x^2>9x^2
We want to see what happened to x, not x^2
so next sqrt both, and we end up with:
x> 3x, therefore a stretch by a third in x direction! 
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 109
 25052016 16:14
(Original post by Myachii)
Got the same.
Question was:
log3(c) = m
log27(d) = n
log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)
It wanted sqrt(c)/d^2 which is:
log3(sqrt(c))  log3(d^2) = log3(3^1/2m)  log3(3^6n)
sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m6n
...right? I've already lost enough marks on this paper
Better than 1648x^10 making you doubt yourself for the remainder of the paper :P 
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 110
 25052016 16:14
(Original post by Myachii)
Got the same.
Question was:
log3(c) = m
log27(d) = n
log3(sqrt(c)) = log3(3^1/2m)
log3(d^2) = log3(3^6n)
It wanted sqrt(c)/d^2 which is:
log3(sqrt(c))  log3(d^2) = log3(3^1/2m)  log3(3^6n)
sqrt(c)/d^2 = (3^1/2m)/(3^6n)
= 3^1/2m6n
...right? I've already lost enough marks on this paper
Better than 1648x^10 making you doubt yourself for the remainder of the paper :P 
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 111
 25052016 16:15
(Original post by jake4198)
6c) Is that not a stretch along the yaxis SF 3? 
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 112
 25052016 16:15
(Original post by Bosssman)
Yes this is correct
OP pls fix:
9a) 3^1/2m6n (or if you like to factorise, 3^1/2(m12n) 
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 113
 25052016 16:16
Could you have done sum of 3 terms?

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 114
 25052016 16:16
(Original post by Parhomus)
I got the coefficient for the power of ten wrong; just because I messed up the addition of the coefficients at the end; even though I expanded both of them correctly. :C 
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 115
 25052016 16:17
(Original post by Jjohnson919)
Could you have done sum of 3 terms? 
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 116
 25052016 16:18
(Original post by Myachii)
+1 OP pls fix
Another thing  6c is confusing the hell out of me
y=\sqrt{x^2+9}
onto
y=3\sqrt{x^2+1}
Let f(x) = \sqrt{x^2+9}
f(x) > 3f(x) = Stretch in y direction SF 3
However, your mark scheme says "Stretch in x direction SF 1/3"
For that to be the case, wouldn't the equation have to be the following:
f(x) > f(3x) = Your answer
Which would lead to the "original equation" having to be:
y=\sqrt{3x^2+1}
My logic was that the 3 is multiplying the entirety of the sqrt function.
For example:
5^x > 3 * 5^x = Stretch in y direction SF 3
5^x > 5^3x = Stretch in x direction SF 1/3
This would be the case because the number 3 is only multiplying the x by 3, not the entire function (which in this case is 5^x)
You said you proved it was x direction SF 1/3 using differentiation, could you show this please? Using methods and formulae that I'd been taught, I got to a y stretch SF 3.
Forgive me if I'm talking out my ass too 
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 117
 25052016 16:19
I don't think the test was that bad, pretty decent; but it was very hard, I know I lost a few marks. For the trapezium rule question I even checked it with my calculator and it said 65.3 or something so I was like phew I got the answer correct. (Answer was 65.6 but calculator can do integration to more precision).

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 118
 25052016 16:20
(Original post by Myachii)
Like using a sledgehammer to crack a nut :P 
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 119
 25052016 16:21
(Original post by Parhomus)
I don't think the test was that bad, pretty decent; but it was very hard, I know I lost a few marks. For the trapezium rule question I even checked it with my calculator and it said 65.3 or something so I was like phew I got the answer correct. (Answer was 65.6 but calculator can do integration to more precision). 
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 120
 25052016 16:25
(Original post by Jjohnson919)
Ummmm... Do you know what question I'm talking about?
You could do the sum of the first three numbers using the formula (which is what I presume you used), or you could just figure them out (26, 24, 22) because it was an arithmetic series so subtracting two wasn't that hard :P
I said it was like using a sledgehammer to crack a nut because if you used the formula to calculate the sum of the first three, you made yourself a hell of a lot more work and more room for error.
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Updated: June 4, 2016
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