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# C2 May 2013 WJEC Watch

1. (Original post by mabli)
Attachment 216704Attachment 216705Attachment 216706

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out

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weird right, everything did cancel leaving you with something totally unlike what it asked you to prove
2. (Original post by mabli)
Attachment 216704Attachment 216705Attachment 216706

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out

Posted from TSR Mobile
Last log question answers coming right up . . .
I'll do some more on request.
3. (Original post by mabli)
How did everyone find the exam? I thought it was okay about from the last logarithm question and the eighth question (circles)

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I twisted my knickers on question 2)b) a lot although it was only worth 3 marks. I believe I got one of the possible answers, but that was the one that I'm confident I lost marks on.

The other parts weren't so bad.
4. (Original post by Naomi32)
I did this too! I thought I went the long way around because the second part of the question asked for the co-ordinates of intersection which I had already found
Really hoping we get most of the marks for doing that, we still proved it wasn't the diameter?
I would give you two full marks if I were an examiner as the question was only worth 2 or 3 marks, and also for the exra effort you put in (I did it the shorter way, by the way)
5. 7)i)

So now we must show that:

ii)

iii)
6. Hi! I felt the exam was alright, except 2aii; I got confused because my values were less than one [(3x-2)(2x+3) x could not equal -3/2?]. Q3 was a mess, and same for last logs question and circles question.
I don't know how to feel about the exam, I know I answered all the questions and attempted the answers, but then got in a muddle in the middle of those odd questions..
7. (Original post by joostan)
Last log question answers coming right up . . .
I'll do some more on request.
Can you please do the area under the curve question? 6b I think. Thanks
8. I got 16.7cm^2 for that

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9. Does anyone have a solved paper from their teacher? Would like to compare my marks
10. For Proving Triangle: Cos A = x^2+(x-2)^2 - (x+2)^2 / 2(x)(x-2)

This gives: x^2+x^2-4x+4-x^2-4x-4 / 2x^2 - 4x

Simplifies to x^2 - 8x / 2x^2 - 4x

Divide by x = x-8 / 2x - 4
11. Part b) Angle BAC = 120

Formula for the angle is given from part a:

There Cos 120 = x+8 / 2x-4

-0.5(2x-4) = x+8

2-x = x+8

2x = -6

x=-3

Not sure if this is correct, as x is a negative.
12. Im going to try and use Latex for 6b)

Area under the curve between x=1 and x=3

This works out to be 44/3

Area under the line between x=0 and x=1

This works out to be 2

Therefore shaded region = 44/3 + 2 = 50/3
13. (Original post by mabli)
Attachment 216704Attachment 216705Attachment 216706

Taken a picture of the paper. I couldn't prove the first triangle question either, my answers kept cancelling each other out

Posted from TSR Mobile
Tbh the C2 May 2013 is one of the easiest question paper I have seen. Many of u will agree with me that question 2 (a) & 3(a) were not as expected but the rest of the question should have been really good if you would have done more past papers practice. e.g: the question that was about integration 6(b) that was taken from the C2 May 2007 paper...I'm very sure that I did very well in this exam :L
14. (Original post by Tynos)
Im going to try and use Latex for 6b)

Area under the curve between x=1 and x=3

This works out to be 44/3

Area under the line between x=0 and x=1

This works out to be 2

Therefore shaded region = 44/3 + 2 = 50/3
Thanks!! I got that too! Have you tried 8 or 9?
15. (Original post by Han96)
Thanks!! I got that too! Have you tried 8 or 9?
I jsut took a break lol. The latex is harder than the actual question xD hahah jokes

Ill try it now.
16. 8 Part a and b)

Therefore A(-1,3) Radius = Root 25 = 5

I did part ii first dont know if its allowed:

Intersects at: (3,6) and (2,7)

bi) If y=-x+9 is a diameter then length of this divided by 2 must equal 5.
Square root = Root 2

Root 2/2 is not equal to the radius of the circle therefore is not a diameter.
17. For the last part of part 8)

Centres of circles: (-1,3) and (11,8)

Using pythag:

= Root 169 = 13

Therefore shortest distance = 13-6-5 (Their radius's)=2
18. Do you think I'd get marks for changing the fraction into a whole number for 6.b)?

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19. (Original post by Han96)
Can you please do the area under the curve question? 6b I think. Thanks
Sorry, I was out yesterday evening, I've been beaten to it - I'll check the working
EDIT: I got the same answer - congrats
20. (Original post by mabli)
Do you think I'd get marks for changing the fraction into a whole number for 6.b)?

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Did you round to 3sf?

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