It shouldn't have been possible to get +/ as the factorised quadratic was (2x5)^2 = 0 which just gives x = 5/2 repeated root(Original post by alkaline.)
can I just say I did that for 9b at first and got plus/minus 5/2
but then I thoughts
you can't just "say" that you can't put logs at negative so +5/2 would be the answer so I crossed it out and used b^24ac= 0 on the quadratic that it formed  will I still get the marks???? FMLLLLLLL
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C2 Maths AS aqa 2016 (unofficial mark scheme new)
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Why bother with a post grad? Are they even worth it? Have your say!  26102016 

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 25052016 15:06

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 25052016 15:10
(Original post by tajtsracc)
Would do you guys predict an A will be? More than or less than 60?
it was a hard paper but I knew all the tricks 
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 25052016 15:10
(Original post by Bosssman)
It shouldn't have been possible to get +/ as the factorised quadratic was (2x5)^2 = 0 which just gives x = 5/2 repeated root
i got 2x+5 and 2x5 somehow......
I er factorised it and there were negatives in the quadratic so one has to be positive and one has to be negative right?
oh well I crossed it out though and used b24ac so that proves it right?
I dk if crossed out x=5/2 though so kengko;weo[fkdpknsgkm im panicking n stressing. 
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 84
 25052016 15:10
(Original post by Parhomus)
Mtangent was=1/2 therefore Mnormal=2. Therefore you just write y5=2(x4) or whatever. I think he just expanded and rearranged it which is fine too. 
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 25052016 15:13
(Original post by Bosssman)
Wasn't it 13.8 for perimeter? Some are saying its 13.6 but pretty sure it was 13.8 if you didn't round before the final answer 
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 25052016 15:15
(Original post by Hbassett26)
Thank you! I get how you would get that equation from there but where was the gradient of the tangent from 
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 25052016 15:23
(Original post by alkaline.)
****.
i got 2x+5 and 2x5 somehow......
I er factorised it and there were negatives in the quadratic so one has to be positive and one has to be negative right?
oh well I crossed it out though and used b24ac so that proves it right?
I dk if crossed out x=5/2 though so kengko;weo[fkdpknsgkm im panicking n stressing.
And yeah, proves the repeated root 
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 25052016 15:25
(Original post by Bosssman)
Using the first derivative, inputting the x value of P 
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 25052016 15:34
(Original post by Hbassett26)
Ah that's what I did! So I think the X coordinate in the mark scheme is wrong because X would have to be 4 instead of 5 in order for it to work! 
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 25052016 15:43
(Original post by Bosssman)
Yes, the coordinate of P was (4,5)
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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 91
 25052016 15:44
For 7b if I expanded the other bracket fully but didn't get the actual value of the coefficient how many marks would I get

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 25052016 15:45
[QUOTE=Bosssman;65156935]
(Original post by Eisobdxhsonw)
The answer to the log question was definitely 5/2
The question was
Show that log4(2x+3) + log4(2x+15) = 1 + log4(14x + 5)
Also the trig question, it resulted in 5 + 3cosx, and so when x = pi, 3cosx = 3 so minimum value is 2 
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 25052016 15:46
And for 2c divided 5 by 0.2 and got a stretch of sf 25 in the y axis. Why isnt this correct?

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 25052016 15:49
[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 
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 95
 25052016 15:53
(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
To get stretch parallel to xaxis s.f. 1/3 I did this
Squared both sides of original equation to give y^2 = x^2 + 9
Then I squared the new equation to give y^2 = 9(x^2 + 1)
Expanded the bracket to give y^2 = 9x^2 + 9
Now the only difference between the two equations is there's a 9x^2 instead of x^2
So then you can have y^2 = (3x)^2 + 9 hence giving the transformation
Your idea of multiplying the entire function is a misread of the equation as y = 3root(x^2 + 9) when it is y = 3root(x^2 +1)Last edited by Bosssman; 25052016 at 15:55.Post rating:1 
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 25052016 15:58
(Original post by Bosssman)
The differentiation was for the normal through P translated to pass through maximum point M (different question)
To get stretch parallel to xaxis s.f. 1/3 I did this
Squared both sides of original equation to give y^2 = x^2 + 9
Then I squared the new equation to give y^2 = 9(x^2 + 1)
Expanded the bracket to give y^2 = 9x^2 + 9
Now the only difference between the two equations is there's a 9x^2 instead of x^2
So then you can have y^2 = (3x)^2 + 9 hence giving the transformation
Your idea of multiplying the entire function is a misread of the equation as y = 3root(x^2 + 9) when it is y = 3root(x^2 +1)
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 16:03
(Original post by IsThisReallyLife)
QUESTION 8B:
(Copied from first post):
8) find the value of tan(x) tan(x) = 5/4 [2]
b) rearrange 16+9sin^{2}x / 3  5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]
For part B, wouldnt x= 0, 2pi?
As it was 5  3cos(theta)
If theta = pi then its:
5  (3*1) = 8
hence theta should be 0 or 2pi as this gives:
5  (3*1) =2 = Minimum value?
Also i think the question had it as plural??
Or am i missing something here? 
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 25052016 16:04
(Original post by beanigger)
Unoffical Mark scheme for C2 AQA 2016
It would help if you could correct the amount of marks in each question if it is wrong please
Questions:
1)a) integrate something to get 36x^{1} + ax^{2 }/2 [3]
b) find the value of a when the answer to 1)a) is equal to 16 with limits of 3 to 1, a =  2 [2]
2)a) Draw the graph of (0.2)^{x }and state where itcrosses axis (0,1)[2]
b) use logarithms to solve (0.2)^{x} = 4 , x = 0.861 [2]
c) describe transformation of (0.2)^{x } onto (5)^{x} reflection in y axis [1]
3)a) differentiate something to get 3x^{1/2 }1 [2]
b) find y coordinate of the maximum point M M(9,6) [3]
c) equation of normal to curve at P i think P was (5,6) y=2x+13 [2]
d) the normal to curve at P is translated by (0,k) find value of k, k = 5.5 [3]
4)a) show that a+10d=8 [2]
b)sum to 2nd and 3rd term is 50, work out the 12th term of the series D=2 and A=28, u12 = 6 [3]
c) work out the sum of the 21st term minus sum of 3rd term = 90 [3]
5)a) show that thetha (in radians) was 0.568 [2]
b) work out area of triangle 19.9 [2]
c) work out perimeter of shaded area, first piece of info given > area of sector = area of shaded area
you had to realise that the area of sector = area of triangle  area of sector
rearrange to get 2(area of sector) = 19.9 and solve for r to get 5.36?
perimeter of shaded = (9r) + (8r) + 5 + (length of arc = r*theta)
perimeter of shaded = 13.8 [6]
6)a)trapezium rule to get 65.6 [4]
b) describe transformation of graph y=(x^{2} +9)^{1/2} onto y = 5 + (x^{2} +9)^{1/2}
Translation by (0, 5) [2]
c) describe transformation of graph y=(x^{2} +9)^{1/2} onto y=3(x^{2} +1)^{1/2} Stretch by scale factor 1/3 in x direction [2]
7)a) expand (2+x)^{5} work out p q r, i think p was 10 q was 80 and r was 80 (correct me if im wrong) [4]
b)find the coefficient of x^{10} in the expansion of (2+x)^{5} * (something else)^{7}
it was something like 1648 [5]
8)a)find the value of tan(x) tan(x) = 5/4 [2]
b) values of tanx in the range of 0 to 360 degrees 45, 129, 225, 309 [3]
b) rearrange 16+9sin^{2}x / 3  5cosx into the form p + qcosx and find the min value = 2, this is when x is equal to pi [4]
9)rearrange something into the form (c)^{1/2} / d^2 to get 3^{y }y = 1/2 ( m + 12n) [4]
b) find the one and only solution of something, x = 5/2 ( you should get two equal roots i think) [4] 
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 99
 25052016 16:04
(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? 
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 100
 25052016 16:04
(Original post by habibtii)
yup, I most definitely failed. with dignity x)
I took Edexcel C2. not this. no wonder the mark scheme looked like a different language to me.
tbh, so did the Edexcel paper. but it's okay. I'll get there.
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Updated: June 4, 2016
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