Edexcel C1/C2 June 2014
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I got that
Hopefully it's right
Hopefully it's right
Original post by randlemcmurphy
Sinx=0
x=0,pi,-pi also
Guess what?
I did all that and forgot to put 0!!
x=0,pi,-pi also
Guess what?
I did all that and forgot to put 0!!
I don't think it equalled Pi? Because it said greater than or equal to -Pi and less than Pi, not less that or equal to?
Original post by YGurung
2tanx -3sinx=0
2sinx/cosx -3sinx=0
sinx(2/cosx-3)=0
sinxx=0 and cosx=2/3
Therefore, x = 0, -0.84 or 0.84
2sinx/cosx -3sinx=0
sinx(2/cosx-3)=0
sinxx=0 and cosx=2/3
Therefore, x = 0, -0.84 or 0.84
I got this
Original post by YGurung
Oui Because the answer I replied to is wrong since you can't work out the cos-1 of anything that is greater than 1.
Because the answer I replied to is wrong since you can't work out the cos-1 of anything that is greater than 1.Sweet. Thanks
I think I got around 55 for this paper, what do you guys think that would be? In my school everyone found it hard
Original post by Uberphuq
I don't think it equalled Pi? Because it said greater than or equal to -Pi and less than Pi, not less that or equal to?
I put 0 and -pi as well
Original post by Jkizer
What topic was that question again?
It was the 3tanx-2sinx one.
since theres been much debate, the answers (which i'm nearly 100% sure are correct) are
38.9cm^2 for the area of ABDCDEA or whatever messed up shape that was
(sinx)(2-3cosx) giving you x=-pi,0,0.841,-0.841 (idk if it was 2sf, pi is NOT in the range)
2880 minimum surface area
k was 5root3
n is 44 not 43, you switch the inequality sign when dividing by the log because it's value is negative, so it was n>43.2 implying that n is 44. Subbing this back into the equation worked, with 44 it was <0.5 with 43 it was greater.
for the integration a was 2/3 b was -1/9
anything else?
38.9cm^2 for the area of ABDCDEA or whatever messed up shape that was
(sinx)(2-3cosx) giving you x=-pi,0,0.841,-0.841 (idk if it was 2sf, pi is NOT in the range)
2880 minimum surface area
k was 5root3
n is 44 not 43, you switch the inequality sign when dividing by the log because it's value is negative, so it was n>43.2 implying that n is 44. Subbing this back into the equation worked, with 44 it was <0.5 with 43 it was greater.
for the integration a was 2/3 b was -1/9
anything else?
(edited 9 years ago)
Original post by gyshmey nbofster
I thought the paper was fairly generous. My one concern, is that when finding the second derivative in part d of the last question, I acciedently wrote dy^2/dx^2. When it's actually d^2y/dx^2... Will I have lost an accuracy mark, as technically the notation of my second derivative was incorrect?
You would probably lose all the marks for that actually.
Original post by YGurung
A lot of people in my year( A2 students retaking C2) got 43 for n.
I got 43 as well.
Can some please show me how they got 44?
I got 43 as well.
Can some please show me how they got 44?
it came out to n>43.2, so n couldn't be 43 , and the answer has to be <0.5. so since n was the power of a fraction, the higher the power, the smaller the answer. so it was the next biggest integer
Original post by Miladrises
I think I got around 55 for this paper, what do you guys think that would be? In my school everyone found it hard
We all found it hard too!
Original post by YGurung
A lot of people in my year( A2 students retaking C2) got 43 for n.
I got 43 as well.
Can some please show me how they got 44?
I got 43 as well.
Can some please show me how they got 44?
You ended up with something like N>43.1 (You had to flip the inequality sign somewhere in the process as one of the logs was negative).
N has to be an integer, and 43 is too low, therefore it has to be 44. I checked this on my calculator as well just to confirm my answer, and 43 doesn't work.
Original post by Sumz.96
Apparently its 44,
Hmmmm...I checked using 43 as well and it was greater than 0.5, but I don't know where I went wrong so I will just put my working out here and hopefully someone can tell me where I have gone wrong:
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
Original post by YGurung
Hmmmm...I checked using 43 as well and it was greater than 0.5, but I don't know where I went wrong so I will just put my working out here and hopefully someone can tell me where I have gone wrong:
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
Log(⅞) is a negative and so is log(0.5/160) so when you divided through by a negative you change the inequality sign, which you didn't do. Hope this helps
Original post by YGurung
Hmmmm...I checked using 43 as well and it was greater than 0.5, but I don't know where I went wrong so I will just put my working out here and hopefully someone can tell me where I have gone wrong:
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
160-160(1-(7/8)n) < 0.5
I divided both sides by 160
1-(1-(7/8)n) <0.5/160
Opened up the brackets and the two give:
1-1+(7/8)n < 0.5/160
the ones cancelled each other out
(7/8)n<0.5/160
log both sides
n log(7/8) < log(0.5/160)
divide both sides by log(7/8)
then i got n < 43.19..
which is why got n= 43, if someone can point out my mistake please do so
you have to round up at the end because 43, as you have pointed out, will produce an incorrect answer but n must be an integer.
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