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# Edexcel A2 C3 Mathematics 12th June 2015 watch

1. (Original post by jf1994)
Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz
Nah. I got y is 5 for the asymptote. It was just reflected on the y axis.

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2. (Original post by Chazley123)
tan = sin / cos
therefor cos = sin/tan so sin/p
I put this at first, but its not correct. Cause they wanted it with respect to p so you had to use the identity tan^2x =1+sec^2x (or whatever it is)

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3. (Original post by sj97)
Wasn't the range for g supposed to have an e value in it?
4. (Original post by PaulCJH)
1. 2p/(1-p^2), 1/((p^2+1)^1/2), (1+p)/(p-1)
2. x> or equal to ln5/2, Solutions: ln3/2, ln7/2.
3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
4. 20, ln2/40, 93
5. 4pi^2, 1/3pi
7. -4e^-4<g(x)<1/(8e)
8. 2.820, 5.962,
9. increasing, -3k/(x-2k)^2

Some might be wrong please bear that in mind.
Yup same
5. (Original post by sj97)
Wasn't the range for g supposed to have an e value in it?
4e and -2.5e
6. (Original post by taichingkan)
Was it an increasing or decreasing function?
Increasing, due to the positive gradient.
7. (Original post by jf1994)
Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz
Well, I think there is one. When x<ln(5/2), the asymptote would be y=5.
8. (Original post by phoenixsilver)
Why are people getting values with e in them for the range of g(x)?
The only hard questions in this paper was the trig prove one and (although easier) the range one.

I don't understand why people are getting e values for the range though.
The question gave the domain of g(x), so this must be equal to the range of the inverse function. I equated this to inverse g(x), getting a quadratic with x taken out of it.
How do you get e values..?
9. POLL ----> http://strawpoll.me/4609428
POLL ----> http://strawpoll.me/4609428
POLL ----> http://strawpoll.me/4609428
POLL ----> http://strawpoll.me/4609428
10. (Original post by PaulCJH)
1. 2p/(1-p^2), 1/((p^2+1)^1/2), (1+p)/(p-1)
2. x> or equal to ln5/2, Solutions: ln3/2, ln7/2.
3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
4. 20, ln2/40, 93
5. 4pi^2, 1/3pi
7. -4e^-4<g(x)<1/(8e)
8. 2.820, 5.962,
9. increasing, -3k/(x-2k)^2

Some might be wrong please bear that in mind.
Exactly the same as mine.
11. Asymptote for second graph was y=5, I think?
12. 8a)

1/sin2a + 2sinacosa/cos^2a - sin^2 a
1 = cos^2a + sin^2

therefore = cos^2a + sin^2a + 2sinacosa/ cos^2a - sin^2a

the top forms a quadratic and the bottom is a difference of two squares thus leading to the final answer

http://strawpoll.me/4609429
14. (Original post by HannahSan)
Well, I think there is one. When x<ln(5/2), the asymptote would be y=5.
*x=5 you mean
15. (Original post by MCMXCVIII)
Nah. I got y is 5 for the asymptote. It was just reflected on the y axis.

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But it was the mod of f(x). right? So anything under the x axis would be reflected up, with anything over the x axis staying where it is. So there can't have been an asymptote at y=5 because the portion of graph that remains where it is clearly cuts through y = 5
16. Can someone post an unofficial mark scheme?
17. (Original post by phoenixsilver)
Increasing, due to the positive gradient.
Good then. Just making sure. Thanks!
18. (Original post by RThornton)
Yeah, then you have to use sin^2x + cos^2x = 1, to get rid of the sin
oh did it have to be only in terms of tan then? thats shame
19. (Original post by Neilg99)
The paper was quite easy.. I think I'll get above 70/75.. Any solutions you need?

Do all
20. (Original post by PaulCJH)
3. 26.57, 2 root 5, 51.8 and -25.3, k<2 root 5 and k>2 root 5
I got k<-4 root 5, k>4 root 5 for that

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