Official Edexcel C2 May 2006 Thread! Watch

Rebecki
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#41
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#41
(Original post by insparato)
I got the answer 14 lol

[8 + 12 - 2] - [1 + 5 - 4]

argh what was the limit?
I did the same and got 14 too. Is this answer right?

As for the minimum value of n I got 2 which is completely wrong lol God knows how I got to that answer if its supposed to be 15 :p:
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insparato
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#42
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#42
My friend did this one on the calculator afterwards it can do integration and he got 14 and so did i so i guess we are right .
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ali p
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#43
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I got 15 as well. I think its right as 14 gave just under 24 and 15 just over.
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Rebecki
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(Original post by insparato)
My friend did this one on the calculator afterwards it can do integration and he got 14 and so did i so i guess we are right .
Yay at least I know I've got something right
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futureaussiecto
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#45
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#45
as far as im aware the answer for the radians (second angle) was as follows

the angle between the two outer lines was obviously 90 degrees

180 degrees is pie radians

therefore 90 degrees is pie/2 radians, which is something like 1.57 radians

The angle from the first part was (if i remember) 0.65 radians

therefore the angle of the second Q. is 1.57-0.65= 0.92 radians

so i dont see how you got 1.5 radians for the triangle.....
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futureaussiecto
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#46
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(Original post by Rebecki)
Yay at least I know I've got something right
yeh i got 14 because it was 16-2


also got 14 for the minimum value of n for geometric.
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Deeksha
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#47
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#47
Hey guys do u think the grade boundaries will be quite high for this paper?...It was 64/75 for an A in Jan.
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vector
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#48
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#48
Hi all,

Not Edexcel, but a report on WJEC.

To be honest, i felt it a lot harder than any of the past papers. I must say that without Insparato's help last night (many thanks )I would have been in a bit of a mess.

One of the G Series q's was VERY difficult, and required simaltaneous equations with a qaudratic and a polynomial. Also, the final circle question was relatively difficult, until you got your head round it (had to prove rather than calculate part of the area of a sector).

Think I did OK anyway!
Michael
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Lykabelle
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#49
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#49
Can someone verify whether that tan question was
tan theta=5
or
tan theta=1/5 ?
Thanks
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theoneandonly
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#50
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it was tan theta= 5
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Mohit_C
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#51
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(Original post by insparato)
My friend did this one on the calculator afterwards it can do integration and he got 14 and so did i so i guess we are right .
The question said the minimum value to exceed 24. And i used my teacher's advice to check whether it's correct and 14 was 23.9 and 15 was just a little over 24. So what is right?
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CalculusMan
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#52
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#52
I got 6, so I must have messed up the signs somewhere as i got it to 16 - 10.
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abc_123_kes
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#53
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i put 14 as well but sum of my frinds that i talked to said they out 15..
but i'm pretty sure it was 14 cuz n<15 as the signs changed when we divided by a negative...
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theoneandonly
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#54
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#54
was the question not smallest value to exeed 24

so whatever we got as 14.25 or whatever, we must go above that

(i dont know it it said whole number, so i guess anything reasonable above 14.25)
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jasperstory
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#55
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#55
I followed the same logic as above and put 15.
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CalculusMan
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#56
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#56
(Original post by MrTrig)
I got 6, so I must have messed up the signs somewhere as i got it to 16 - 10.
This is for 2), the int question. Can anyone confirm the answer they got?

Also, on 10), how many people forgot the +c on one of the integrations? I made sure I didn't as thast what I did in C1.
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abc_123_kes
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#57
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oh well its only 1 mark cuz i did the working right - i got 14.2..

anyway, everything else was prettty good - i think i did really well on this one.. waaay better than on C1 cuz i know i lost at least 3marks on C1 - it kinda harder than C2...
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Desi
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#58
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#58
did u hve to put c???????/
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CalculusMan
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#59
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#59
On the INT question with no limits, yes you did.
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Chimpyang
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#60
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#60
Ok, here is my reasoning (+ slightly convoluted solution i wrote in the exam) as to why n = 15...

Step 1
Sn=25(1-r^n)
Sn>24

Step 2
24/25<(1-r^n)

Step 3
1-r^n>0.96
r^n<1-0.96
r^n<0.04

Step 4
r=4/5

therefore (4/5)^n<0.04
but pretend that (4/5)^n=0.04
and change it to
log4/5(0.04)=n

Step 5

log10(0.04)/log10(4/5)=n
n = 14.42513488

But this value make Sn = 24 and being the number of terms, it has to be an integer and therefore must be rounded up to 15.
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