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AQA maths core 1, 13th May 2015

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for the cylinder question did anyone else get R as 4?
Original post by Mentalmirz
for the cylinder question did anyone else get R as 4?

Yes. which with d^2V/dr^2 implies max.
Original post by Lizzie33
Ah thank you! But wait sorry to be a pain but was the original qn

X^2 - 3x +2
Or
X^2 + 3x +2


I hope it was x^2 + 3x + 2
Original post by tanyapotter
yeah, what i did was the graph starts at the bottom left hand side quadrant, touches the origin, goes back down, has a minimum at the bottom right quadrant, then goes back up crossing the x axis at +3


Yeah same:smile:) woopa
Original post by Mentalmirz
for the cylinder question did anyone else get R as 4?


Yes, you had to say that it was +4 and -4 and therefore say that it was +4 as they wanted the positive value.
Reply 225
Avatar for rehmh05
rehmh05
Trapezium - parallel side sum was (30+6) x 1/2 x height

The height was 2 -(-1)

Therefore I got 48 -108/5
Anyone else?
Anyone think that other then that damn cylinder question, it wasn't that bad? Plus this being a bit harder means that Core 2 should be easier.. Should
Reply 227
Avatar for r-star
r-star
2
for the cubic graph i showed a positive x^3 graph but went through 0 and touched at 3 how many marks would this be?
Original post by student0042
Yes. which with d^2V/dr^2 implies max.
I put that it was a max :smile: awesome thanks!
Reply 229
Avatar for Jupers
Jupers
5
Original post by rehmh05
Trapezium - parallel side sum was (30+6) x 1/2 x height

The height was 2 -(-1)

Therefore I got 48 -108/5
Anyone else?


h was 3
0.5(3)(30+6)=54
Somehow I ended up with 4pi, and 12pi^2 for the last part of Q6...any idea how I did that?
Sniffed a naughty line of lem before the paper, so it was sweet as
Reply 232
Avatar for rehmh05
rehmh05
Original post by Jupers
h was 3
0.5(3)(30+6)=54
damn !
Are you sure you are right??? Most of your answers match mine apart from 1c,3bi and bii

Original post by student0042
1bi. 4x-3y+1=0 (I thought it was this form)
1c. A(9,-4)
3a. y-6=-10(x+1)
3bi. 108/5
3bii. 162/5
4a. (x+1)^2 + (y-3)^2 = 50
4bi. C(-1,3)
4bii. 2√5
4c. k=-8, 2
4d. minimum distance =7
5a. p = 3/2, q = -¼
5bi. (-3/2, -1/4)
5bii. 5c. y = x^2 - x + 4
6bii. =-12π therefore max (r=+4)
7a. draw x^2(x-2)
7bi. R = 36
7bii. R= 0 therefore root
7biii.(x-2)(x^2-5x+10)
7biv. x=2
8a. Show that x^2 + 3(k-2)x -13-k=0
8b. 9k^2 -32k -16 < 0
8c. -4/9 < k <4
Original post by rehmh05
Trapezium - parallel side sum was (30+6) x 1/2 x height

The height was 2 -(-1)

Therefore I got 48 -108/5
Anyone else?
I got 54 I found it as a triangle and a rectangle though my final answer came out as 32.4...
Can someone tell me if my working is right here:
(4,k)
(x+1)^2 + (y-3)^2 = 50
(4+1)^2 + (k-3)^2 = 50
25+k^2-6k+9=50
k^2-6k-16=0
(k-8)(k+2)
k=8
k=-2
is this how everyone else worked it out? if not, how did you guys do it?
Reply 236
Avatar for tadit
tadit
2
Original post by 2 4 MD 40
Sniffed a naughty line of lem before the paper, so it was sweet as

the cylinder question must have made you psychotic
The cylinder one:

48pi = piXr^2 + 2xpiXrxh (pi cancels)
48 = r^2 +2rh
48-r^2 = 2rh
h = (48-r^2)/2r
Was the original translation equation:

X^2 - 3x +2
Or
X^2 + 3x +2

(Before you completed the square originally)


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Reply 239
Avatar for tadit
tadit
2
Original post by Heffalump .
Can someone tell me if my working is right here:
(4,k)
(x+1)^2 + (y-3)^2 = 50
(4+1)^2 + (k-3)^2 = 50
25+k^2-6k+9=50
k^2-6k-16=0
(k-8)(k+2)
k=8
k=-2
is this how everyone else worked it out? if not, how did you guys do it?


Yep thats how I did it (but my quadratic was wrong hahah)

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