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# Official Edexcel C3 Discussion Thread watch

1. (Original post by kosy91)
52 is tooo low...its never gone below 57, i think
I think the June 06 paper was 56/75 for A. Cant find any lower.
2. sorry i dont agree with the last two solutions for question 8, you cannot have cot2x=0 as that implies that tan2x=0 and therefore the equation must be 1/0 = 0 which is plain wrong as that is infinity which is why the math error occurs on calculators??
3. (Original post by taffeta)
I think the June 06 paper was 56/75 for A. Cant find any lower.
yeah thats right....so this paper should be more or less??
4. (Original post by dan9)
sorry i dont agree with the last two solutions for question 8, you cannot have cot2x=0 as that implies that tan2x=0 and therefore the equation must be 1/0 = 0 which is plain wrong as that is infinity which is why the math error occurs on calculators??
dont use the calculator. tan2x will give you 1/0. but inverse of that is 0/1 which is 0, so the answer works!!
5. (Original post by kosy91)
dont use the calculator. tan2x will give you 1/0. but inverse of that is 0/1 which is 0, so the answer works!!
(Original post by Arsey)
Updated solutions.
You're wrong, it's impossible to divide by 0, so the cot2x=0 is impossible and not a valid answer.
6. (Original post by kosy91)
yeah thats right....so this paper should be more or less??
Well I thought June 06 was a tough one too. IMO boundaries should be similar for this - what we need is lots of people with bad marks then we'll be fine! Although too many seemed to cope ok on this thread
7. For questions 9) i) b) i used log instead of ln (the c2 method) and solved it and everything, but i got x=log(...) instead of ln(...) and i didnt cancel out e , how many mark will i lose, even though i got the answer as an exact but not in the way they wanted it. THEY DIDNT SPECIFY THAT THEY WANTED IT IN LN :P , anyway its my bad. xD
8. (Original post by kosy91)
fg(x)>3 as domain of g(x)>1 so we cant use x=1
Fg(x)>= 3

its equal to aswell cos (x+1)^2 + 3

draw that graph it touches three...simple quadratic (completing the square) from c1!!!

remember ((ln) and (e)) canceled to become the equation above ^^
9. (Original post by ArchedEdge)
You're wrong, it's impossible to divide by 0, so the cot2x=0 is impossible and not a valid answer.
yes true but cot2x is aswell as being 1/tan2x.....is also cos2x/sin2x
and from this you can put and get the values of 45 and 135!!
10. for Q8 i got x= 0, 22.5, 90, 112.5
11. i was under the impression that (x+1)^2 touches 1, but doesn't cross, but is then moved up 3 because of the + 3
12. what will the grade boundaries be..... I lost about 7-10 marks ;(
13. (Original post by taherq)
for Q8 i got x= 0, 22.5, 90, 112.5
thats exactly what I got, but I think we used tan2X=0 instead of cot2X=0
14. (Original post by eowl)
thats exactly what I got, but I think we used tan2X=0 instead of cot2X=0
ya thats wat i have done
15. I was just wondering then, does it become fg(x)>=3

as oppose to fg(x)>3
16. still anyone
can someoen please say if question 8)
45 and 135 cannot be solutions i put them back into the original equations!!!

someone correct that mark scheme or prove me wrong please!

(Original post by kosy91)
fg(x)>3 as domain of g(x)>1 so we cant use x=1
dammit
i lost that too
6 marks gone

oh wait -->

Fg(x)>= 3

its equal to aswell cos (x+1)^2 + 3

draw that graph it touches three...simple quadratic (completing the square) from c1!!!

remember ((ln) and (e)) canceled to become the equation above ^^
i did that and put fg>= 3
hope thats correct

unles with the composite function you have to take into account that g(x) is x>1

lost 6 marks (+ more if i had to include those two with the cotx=0) !!
17. What do you guys think the grade boundaries will be ? So far everyone I have spoken too found it fairly hard. My guess is 55 above for an A...
18. i think A will be in this range 56-58
19. can anyone tell me if this is right....for question 8, what i did was dis:

1/sin^(2)(2x) - cos2x/sin2x = 1

then crossed multiplied, crossed common terms and got this:

1 - cos(2x).sin(2x) = sin^(2) (2x)

1- sin^(2) (2x)= cos(2x).sin(2x)

cos^2 (2x) = cos(2x).sin(2x)

cos^2 (2x) - cos(2x).sin(2x)=0

cos(2x) (cos(2x) - sin(2x)) =0

cos2x=0 and cos2x=sine2x

2x= 90 and 270 & 2x=25 and 225
x= 45, 135 & x= 22.5, 112.5

is that right??????
20. (Original post by .:excel4100%:.)
Someone told me that aswell, I didn't understand what he meant but it was something to do with from the function f but I didn't know how it would work so I just put
So is it confirmed that the answer IS NOT

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