AQA AS Further Maths Paper 1 Pure 2019 (13th May) unofficial markscheme Watch

Poll: What you think you got? (Marks)
10-20 (12)
6.52%
20-30 (21)
11.41%
30-40 (39)
21.2%
40-50 (41)
22.28%
50-60 (33)
17.93%
60-70 (16)
8.7%
70-80 (22)
11.96%
asifmahmoud
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#1
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#1
How did everyone find it? It was hard to finish everything in time. Can we start an unofficial markscheme for today's paper? Thank you!!
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√-1
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#2
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#2
really hard
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asifmahmoud
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Specially that vector question was urggghhh
(Original post by √-1)
really hard
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lemmens
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I thought it was very difficult, but could’ve been a lot worse. I find the further pure exams so pushed for time and so always end up leaving some questions blank!! hoping that the grade boundaries are as low as last year🙈
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Max360
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#5
Asif I see you.
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Aksagar
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#6
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volume of revolution was 2.2 radians??
(Original post by asifmahmoud)
How did everyone find it? It was hard to finish everything in time. Can we start an unofficial markscheme for today's paper? Thank you!!
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Max360
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Nah I got something like 13/12 pie a^ 3 which simplifies to 3.40 as m
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george_0liver29
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(Original post by asifmahmoud)
How did everyone find it? It was hard to finish everything in time. Can we start an unofficial markscheme for today's paper? Thank you!!
Pretty Hard, hopefully they'll have lowered the boundaries, though I doubt that'll happen after last year
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√-1
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(Original post by george_0liver29)
Pretty Hard, hopefully they'll have lowered the boundaries, though I doubt that'll happen after last year
grade boundaries are based on how everyone did and it looked like everyone found it hard so hopefully they will be lowered
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Max360
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#10
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What did u guys get for the last question?
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123an456
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(Original post by Aksagar)
volume of revolution was 2.2 radians??
i got 4.4?
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lemmens
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I think the maclaurin series question was cosh(x) = 1 + x^2/2 + x^4/24 and so cosh(ix) is the same as cos(x)

for the graph of x = cosh (y+b) I got a parabola on the positive x-axis that had its minimum point at (1,0) translated down by the constant b. hence the minimum distance to the y-axis would be 1 unit

r = k / sin(theta) can be rearranged to give rsin(theta)=k which is just y=k, so a horizontal line

the proof by induction question involving the matrices could be done by multiplying A^k by A again

these are the only ones I remember off the top of my head!!
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lemmens
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(Original post by Max360)
What did u guys get for the last question?
alpha + beta + gamma = 0 as there’s no x^2 term, therefore alpha + beta = -gamma

I think p>2 because the stationary point would have to be above the x-axis for there to be no real roots

and I found the new equation by rearranging the new roots and substituting them back into the equation, giving me the value where there is no x term as ‘p+2’
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√-1
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this is AS? did u do the A level cus the last question was about monkeys
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Max360
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what was the final line for the hyperbolic equation?
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alicestudies
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this sounds promising!!! For that first one I started off accidentally using sinhx then realising it was a plus sign not a minus so I had to start again but I got that in the end!
(Original post by lemmens)
I think the maclaurin series question was cosh(x) = 1 + x^2/2 + x^4/24 and so cosh(ix) is the same as cos(x)

for the graph of x = cosh (y+b) I got a parabola on the positive x-axis that had its minimum point at (1,0) translated down by the constant b. hence the minimum distance to the y-axis would be 1 unit

r = k / sin(theta) can be rearranged to give rsin(theta)=k which is just y=k, so a horizontal line

the proof by induction question involving the matrices could be done by multiplying A^k by A again

these are the only ones I remember off the top of my head!!
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lemmens
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(Original post by √-1)
this is AS? did u do the A level cus the last question was about monkeys
this is AQA as
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lemmens
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(Original post by Max360)
what was the final line for the hyperbolic equation?
I think it was x=1/2 cosh^-1 (1+root3)
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lemmens
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(Original post by alicestudies)
this sounds promising!!! For that first one I started off accidentally using sinhx then realising it was a plus sign not a minus so I had to start again but I got that in the end!
ahaha hooray!! glad you got the same
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eclipseboi
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(Original post by lemmens)
I think the maclaurin series question was cosh(x) = 1 + x^2/2 + x^4/24 and so cosh(ix) is the same as cos(x)

for the graph of x = cosh (y+b) I got a parabola on the positive x-axis that had its minimum point at (1,0) translated down by the constant b. hence the minimum distance to the y-axis would be 1 unit

r = k / sin(theta) can be rearranged to give rsin(theta)=k which is just y=k, so a horizontal line

the proof by induction question involving the matrices could be done by multiplying A^k by A again

these are the only ones I remember off the top of my head!!
I agree with the last three, couldn't do the Maclaurin though so I have no idea

Matrices almost ruined me as I initially tried A x A^k, but it worked once I used A^k x A

I think I got the vectors proof right, I ended up with something similar-looking to the required answer, multiplied top and bottom by 17 and got it.

For feasible values of y I got y< or equal to (7-root6)/2 and y> than or equal to (7+root6)/2

Can't remember any others
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