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# AQA maths core 1, 13th May 2015 watch

1. (Original post by Lizzie33)
Was the original translation equation:

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

(Before you completed the square originally)

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2. Believe it was 48π not 49
3. It was the first one, -3x
4. How many marks was finding out the values of k?
And also how many marks was the graph of the sin curve
5. For the cylinder question, I took Pie out and got an answer of -12
as there was 2 pie so I took it out
6. How many marks was finding out the values of k?
And also how many marks was the graph of the sin curve
7. (Original post by B82)
How many marks was finding out the values of k?
And also how many marks was the graph of the sin curve
Which question for fining out k? There was the inequality one and circle one.

No idea why you're on about a sin curve - trig graphs are C2.
8. (Original post by Why me)
For the cylinder question, I took Pie out and got an answer of -12
as there was 2 pie so I took it out
Did you eat the pie?
9. (Original post by aoxa)
Which question for fining out k? There was the inequality one and circle one.

No idea why you're on about a sin curve - trig graphs are C2.
The one about the circle
The sin graph where it cut the axis at 0 and 3 but one of them was repeated
10. 24pie r- pie/2r^3
differentiate
24- 3r^2/2

Differentiating Pie gives 1
11. http://www.thestudentroom.co.uk/show....php?t=3322885 can people vote on the difficulty of the C1 exam poll? interesting to know how everyone found it
12. (Original post by B82)
The one about the circle
The sin graph where it cut the axis at 0 and 3 but one of them was repeated
I think the answer was k=-2, k=8. I think it was 3-4 marks? I can't really remember.

And it wasn't a sin graph, it was a cubic graph, and the sketch I want to say was 2 marks, but don't take my word for that.
13. These are the answers I remember:
q1: -3/5 for gradient
AB: -5x+3y-1=0
(9,-4)

q2: 7+ root15

q3: circle: (x+1)^2 + (y-3)^2 =50
C(-1,3)
radius: root 50, which is 5 root 2
k=8 k=-2 or something like that, I think i've remembered that wrong, so correct that
minimum length: 7

q4: dy/dx: 4x^3 - 6x or something..
the line was something like y=-10x-4
intergration: 108/5
final answer: (54-108/5) 162/5

q5: polynomials sketch curve
remainder: 36
proof p(x)=0
(x+2)((x^2-5x+10)
x^2-5x+10 has no roots (b^2-4ac < 0) so x=-2 is the only root

q6: cylinder
h=(48pi - pir^2)/2pir
prove v= that equation thing by substituting h in
dy/dx= 24pi-3pir^2/2
r=4
d^2y/dx^2=6pir
so -12pi therefore maximum

q7: (x+(3/2))^2 - 1/4
Vertex: (-3/2, -1/4)
Los: x=-3/2
translated curve: x^2-x+4

q8: rearrange to get that equation
use b^2-4ac<0 to get the second equation
-4/9<k<4
thats most of it, my memory failed at some parts so correct me if I'm wrong..
14. How many marks was the second part of the integration question??
15. (Original post by aoxa)
I think the answer was k=-2, k=8. I think it was 3-4 marks? I can't really remember.And it wasn't a sin graph, it was a cubic graph, and the sketch I want to say was 2 marks, but don't take my word for that.
yeah sorry I meant a cubic graph!
but thanks!
16. Could you guys put the number of marks next to each answer please? Would greatly appreciate this!
17. (Original post by Sharkindustries)
Are you sure you are right??? Most of your answers match mine apart from 1c,3bi and bii
I'm not certain that i'm right, just the answers I got.
18. I'm in a2 so this is a retake for me! I revised everything apart from stuff about cylinders because I completely forgot about them!! D: so annoying lol. Apart from that I thought it was great!
19. I messed up on the cylinder questions?

6

ENTIre question

how can you find a max or min if you got nothing to substitute in to dy2
20. These are most of the answers I think (the question references may be wrong):

1. (a) gradient = -3/5
(b) 5x - 3y = -1
(c) A(9, -4)

2. 7 + √15

3. (a) y - 6 = -10(x + 1)
(b) 21.6
(c) 32.4

4. (a)(i) (x+1)^2 + (y-3)^2= 50
(a)(ii) C (-1,3)
(a)(iii) Radius = 5√2
(b) k = -2, k = 8
(c) Shortest distance = 7

5. (a)(i) (x+1.5)^2 - 0.25
(a)(ii) vertex (-1.5, -0.25)
(a)(iii) line of symmetry: x = -1.5
(b) translated curve = x^2-x+4

6. (a) h = 24/r - r/2
(b)(i) V = hπr^2 = πr^2(24/r - r/2) = 24πr - πr^3/2
(b)(ii) r = 4, d^2V/dr^2 = -12π which is less than 0 therefore it's a maximum point

7. (a) polynomial sketch, repeated root at the origin, another root at x = 3
(b)(i) remainder = 36
(b)(ii) p(-2)=0 therefore x+2 is a factor
(b)(iii) (x+2)(x^2-5x+10)
(b)(iv) discriminant (b^2-4ac) of x^2-5x+10 is less than 0 therefore x=-2 is the only real root

8. (a) show x^2 + 3(k-2)x + 13 - k = 0
(b) the line and the curve don't intersect therefore the discriminant is less than 0, so use that to show 9k^2-32k-16<0
(c) 9k^2-32k-16<0 = (9k+4)(k-4)<0
therefore -4/9<k<4

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