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# AQA Maths Core 1+2 May 2013 Watch

1. Looking forward to this exam!

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2. Best of luck to all!
3. Guys, remember, im gonna need everyones input for this!

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4. Thanks for the support!
5. Guys remember to get as many answers down!

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6. (Core 2) Here's what I can remember (question parts/numbers might be mixed up)

1 (b) sum to infinity = 160

3) (c) Integral = 39/2

For the question that was something like show that the x-coordinate of the stationary point can be written in the form 2p, where p is a rational number, I got p =6/5

For the long logarithm one where you had to show that the equation had only one solution, I ended up with (x + 3)^2 = 0, so x = -3 was the only solution.

For the last trig equation one, I got x = 16, x = 104, or x = 136.

That's all I can remember for now.
7. This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15

Final question:
20/7 <= k <= 4

Correct or not...?

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8. a. Write log(a)b = c in index form.
b. Show that only one value of x satisfies 2log(2)(x+7) - log(2)(x-5) = 3

a. a^c = b

b. log(2)x = 3
x = 2^3 = 8
log(2)(x+7)^2 - log(2)(x-5) = log(2)8
log(2)((x+7)^2)/(x-5) = log(2)8
((x-7)^2)/(x-5) = 8
(x+7)^2 = 8(x-5)
x^2 + 14x + 49 = 8x - 40
x^2 + 6x + 9 = 0
(x+3)^2 = 0
x = -3
9. (Original post by mike256)
Here's what I can remember (question parts/numbers might be mixed up)

1 (b) sum to infinity = 160

3) (c) Integral = 39/2

For the question that was something like show that the x-coordinate of the stationary point can be written in the form 2p, where p is a rational number, I got p =6/5

For the long logarithm one where you had to show that the equation had only one solution, I ended up with (x + 3)^2 = 0, so x = -3 was the only solution.

For the last trig equation one, I got x = 16, x = 104, or x = 136.

That's all I can remember for now.
I got all exactly the same as you!
In addition,for the trapezium rule one,I got 7.12.

For the sum of 12 terms(the last part of the first question),I got 159.96

For the obtuse angle of the sector,I got 1.87(to 3sf)

And for the transformation problem,I guess it is stretching the original graph
surd(8x^3 -1) in the x-direction of scale factor 2

For the g(x) and g(4) question,I got g(4) = 2.3

for the term u3(the one that right after asking you to show p = 2/3),I got 56

For the equation of normal one:

-4/9 = (y-11)/(x-4)
9y-99 = -4x + 16
4x + 9y -115 = 0

Anyone else can share their answers to us ?
10. The log question was:

log2(x+7)2 - log2(x+5) - 3 = 0

→log2(x2+14x+49) = 3 + log2(x+5)
→log2(x2+14x+49) = log223 + log2(x+5)
→log2(x2+14x+49) = log2(8x+40)
→x2+14x+49 = 8x+40
→x2+6x+9 = 0
→(x+3)2

So x = -3

(However, for some reason I thought 14-8=8 so messed up the last bit... )
11. (Original post by stuart_aitken)
This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15

Final question:
20/7 <= k <= 4

Correct or not...?

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Integration under curve was 63/2 for me and shaded area was 27?
12. (Original post by Andrreewwww)
Integration under curve was 63/2 for me and shaded area was 27?
Your answer is probably correct. My values do look a bit ridiculous. I checked it so thoroughly but couldn't see anything wrong. Must've made a daft tiny error somewhere. Boooo.

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13. (Original post by stuart_aitken)
Your answer is probably correct. My values do look a bit ridiculous. I checked it so thoroughly but couldn't see anything wrong. Must've made a daft tiny error somewhere. Boooo.

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Unfortunately for the last question i forgot to turn back to x values, and got two k vales of 22/7 and 20/7, not sure if i'll get full marks for that question
14. Solutions for tan x=-1 for me was 180+-45 = 135
360+-45 = 315

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15. (Original post by Andrreewwww)
Integration under curve was 63/2 for me and shaded area was 27?
I got those too
I think people didnt realise that it was a trapezium.
Area of that was (6+33)*3/2 ?
16. The transformation of [2 -0.7] was √((x-2)3 +1) -0.7

And the 2k question, I got k = 6/5
17. (Original post by Alifff)
I got those too
I think people didnt realise that it was a trapezium.
Area of that was (6+33)*3/2 ?
Yeah, I couldn't think what the trapezium rule was, so i simply used a square and a triangle, got 117/2 for the trapezium
18. (Original post by stuart_aitken)
This is all I want to know:

For core 1

Integration question:
Area under curve was 524/15

Final question:
20/7 <= k <= 4

Correct or not...?

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The inequalities are right I think but was the variable K? I think i left it as X
19. (Original post by Alifff)
The inequalities are right I think but was the variable K? I think i left it as X
It did ask for values in X
20. (Original post by Andrreewwww)
Yeah, I couldn't think what the trapezium rule was, so i simply used a square and a triangle, got 117/2 for the trapezium
It does the job haha but do you remember the transformation of the circle?
from (X-5)^2 + (Y-7)^2 = 49 to (X+1)^2 + Y^2 = 49?

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