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Original post by Johann von Gauss
Not everyone can be handed out a pass... you're not entitled to a pass


Ha, i'm not asking for a pass, but when i do past papers all effing year with mostly all A* grades, well i'm understandably pissed off. They dole out this sh**e to us, and then expect to us to do it when past papers are fricking easy. So no i wasn't expecting an easy paper, but i don't think anyone expected this paper....
Original post by xKay
I got 1 over square root of p squared + 1


tan = sin / cos
therefor cos = sin/tan so sin/p
Reply 2322
Original post by Neilg99
The range was -4 and 2.5


Wasn't the range for g supposed to have an e value in it?
WHAT THE **** WAS THAT PROOF


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Original post by xKay
I got 1 over square root of p squared + 1


Correct
can someone make a difficulty poll on this? :frown:
Reply 2326
Original post by moment of truth
Was harder than June 2013 (imo) and those grade boundaries were ridiculously low, so these should be quite similar (hopefully!) Still annoyed at the paper, but there is nothing we can do now. Look forward to the next paper instead :redface:

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Original post by big fat poo
Ha, i'm not asking for a pass, but when i do past papers all effing year with mostly all A* grades, well i'm understandably pissed off. They dole out this sh**e to us, and then expect to us to do it when past papers are fricking easy. So no i wasn't expecting an easy paper, but i don't think anyone expected this paper....


Brace yourselves for C4....
Original post by sj97
Wasn't the range for g supposed to have an e value in it?


Youre right. It was 1/8e and something else e.
Original post by studentwiz
What was the values of k


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k < -2root5 k > 2root5
Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz
What did you do for the trig proof and is an unofficial mark scheme out yet?


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Original post by Chazley123
tan = sin / cos
therefor cos = sin/tan so sin/p


Yeah, then you have to use sin^2x + cos^2x = 1, to get rid of the sin
****
Was it an increasing or decreasing function?
Original post by Chazley123
how did everyone find it?


I found it alright actually :smile: at first the trig proof and the range question had me like WHAT but i got there in the end. I think i probably made one or two stupid mistakes though, like i realised that the gradient on the last q was positive but said that it was a decreasing function for some reason :frown: how about you?
My Answers:
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.
Reply 2336
Original post by jf1994
Did anyone else get that there was no asymptote for the second transformation graph? plz plz plz


No, I got y=-5 for the first one and y=5 for the second one? Messed up on the modulus questions though. Got something like x = 5/2.
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..?
Original post by Chazley123
k < -2root5 k > 2root5

Yes i got that wooop

The paper wasn't too bad.
It was just tricky and long. Like 9 questions is a lot...

I couldn't answer 1b) cosx in terms of tan :frown:
And 8a) The proving ahhh.
I got up to 1 + Sin 2A divided by Cos 2A and didn't know that you had to equate the 1 to sin^a+cos^2a apparently.

Hopefully I can get an A* still.
Original post by taichingkan
Was it an increasing or decreasing function?


increasing you mug

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