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# CCEA C4 Mathematics - 22nd May 2014 watch

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1. (Original post by stephen_mcgarry)
Hmm, isn't dy/dx = -4sint/(2-2cos2t)

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sorry just a typo error, but it works out as root2
2. Who got log 4/log 81?
I know its wrong...
3. Wrong topic, sorry.
4. (Original post by 1MacG)
sorry just a typo error, but it works out as root2
I cannot get it to work out as sqrt2 when I sup 5pi/4 in :/ it always works out as -2+2sqrt2
What am I doing wrong!?

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5. (Original post by stephen_mcgarry)
I cannot get it to work out as sqrt2 when I sup 5pi/4 in :/ it always works out as -2+2sqrt2
What am I doing wrong!?

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You're possibly not multiplying 5pi/4 x2?
6. (Original post by 1MacG)
You're possibly not multiplying 5pi/4 x2?
Ahh that was it! Thanks

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7. what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong
8. (Original post by jackanator)
what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong
I said the bottom of the bowl wasn't flat lol.
9. (Original post by jackanator)
what did you guys say for the modeling assumption? I had no idea, i said that if you put fruit in the bowl, it could hold a greater volume as it is over the top hahaha i assume this is wrong
I said you must assume the fishbowl is completely spherical, but in the real world the fishbowl must have a flat bottom and top.

I messed up the differential equation once. I started with ln(sqrtV), instead of 2V^1/2. Do you think I was lose all 10 marks? Or will I get some for working through, but using the wrong equation?
10. Did anyone say anything about the thickness of the bowl being negligible?
11. (Original post by 1MacG)
Did anyone say anything about the thickness of the bowl being negligible?
me!

cant see it being wrong tbh....
12. Could someone put up an unofficial mark scheme or even just post your answers if you think they're right lol. Specifically for questions 3 and 5?
13. (Original post by Matt857)
Could someone put up an unofficial mark scheme or even just post your answers if you think they're right lol. Specifically for questions 3 and 5?
Q3) I got 1072/45 = 23.8 (3sf)
Q5) 7.28PM (19.28)

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14. (Original post by Mr Tall)
me!

cant see it being wrong tbh....
that is so much more logical. i was thinking that, but then i tought the fruit was a better idea haha such a stupid answer when i look back
15. could you create a mark scheme please?
16. Not a hard paper. Kicking myself for losing marks in the last question though
17. did anyone else get -1/2 ln(5-x) - 3/2 ln(x+1) for q1?
i'm aware that could be horribly wrong as i seem to find a way to mess them up every time
18. Sorry for the delay everyone; it takes quite a bit of time to type out the mark scheme :P

Mark scheme is in first post.

Let me know if you have any problems with the mark scheme or if you notice any typos etc.

By the way, I stated the assumption was that the curved surface does not contribute to the volume/negligible thickness.
1i: 1/(2(5-x)) - 3/(2(1+x))
1ii: -1/2 (ln(5-x)) - 3/2 ln(1+x)

2) AB: 4i +9j
2ii) r = 3 + lamdba 4
-2 9
2iii) proved lambda was 2 for both

3) 23.8

4i) 24.0
4ii) assumed that the bottom of the bowl is flat and the insides are flat

5) 19.46 pm

6i) 25(sinx - 1.29)
6ii) -625 cot(x-1.29)

7 i) f-1 : x -> tan-1 x range = -0.666< x < 0.666
7ii) gf : x -> |tanx| range : 0< x < pi/4
7iii) reflection in y axis of right hand side of tan onto left

8) found t to be 3pi/4 and coordinates to be (3pi/2 + 1) (-2sqrt2
20. 1)i) 1/2(5-x) - 3/2(1+x)
ii) (-1/2)ln|5-x| (-3/2)ln|1+x|
2) I'll do it shortly don't have answers here, edit post later
3)1072/45 = 23.8 (3sf)
4)i) 24.0 (3sf)
ii) maybe the bowl must be perfectly symmetrical, must be smooth, thickness should be ignored? I'm not 100% sure sorry
5)7.28PM (19.28)
6)i) R=25 alpha=1.29 (3sf)
ii) -(cot(x-1.29))/625 + c
7)i) f^-1:x > arctan(y) [tan^-1(y)] domain (this is a guess) -1<=x<=1 range - pi/4<=f^-1(x)<=pi/4
ii) gf:x > |tanx| range 0<=gf(x)<=1
iii) u shaped curve, Centre point at 0,0 and goes up to pi/4,1 and -pi/4,1 (I'll attach a picture in next post)
8) (2pi,-4) I think you are supposed to disprove this point as, somebody stated, when you sub t into dy/dx, it does not equal sqrt2. The other coordinates are ((5pi-2)/2,-2sqrt2)

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