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.

I got < or equal to and > or equal to. Is that not right?

Reply 2367

10ict019

3

What did people get for the range of g??

Reply 2368

Chazley123

2

Original post by emu_neutrino

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?

as k was a negative integer it would have meant that the function was increasing, also i think for another question was that it was a one to many function?

Reply 2369

BFP=big fat poo

2

guys for the range question, I got that the range was between 0.3 and -0.7? Please someone agree with me :s-smilie:

Reply 2370

blueynuey1

2

Please vote on poll:

http://strawpoll.me/4609429

Reply 2371

lilminster

8

Guess that A* has gone out of the window for me :frown:

Reply 2372

piercy42

1

so...seems ill be redoing some maths exams next year, no way am i settling for my insurance choice just bc a paper killed me

Reply 2373

emu_neutrino

10

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..?

didnt it say after that, why does g(x) not have an inverse? and how did you manage to work out an inverse for that, it was a very messy function. since it said hence in part b, i put g'(x) =0 and found the x coordinates of the turning points then used these to find the y values which had e in them, the graph had two turning points which is why i did that. also the hint from the previous q to use g'(x).

Reply 2374

xyzmaster

8

Original post by lilminster

Guess that A* has gone out of the window for me :frown:

C4?

Reply 2375

mahatma ghandi

1

Original post by big fat poo

guys for the range question, I got that the range was between 0.3 and -0.7? Please someone agree with me :s-smilie:

no. you're wrong lol

Reply 2376

Chazley123

2

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..?

6 marks for that I couldnt believe it, should have been 6 marks for a in my opinion as there were two product rules, however I didnt realise you could equal dy/dx to zero find the coord of stationary points thus leading to the answer of the range

Reply 2377

Devran22

11

Original post by edexcelbastards

F*** you, your mother is a bear

I am sorry you don't know that it is impossible for a bear to have a human child. I guess you worse in biology then you are in maths?

Reply 2378

HappyHumanist

2

I did a reverse proof thing for the identity question (I multiplied by the denominator, and then again) and managed to get the LHS. Do you think I'd be credited for this?

Reply 2379

Half Crazy

Original post by phoenixsilver

But how?
g^-1(x)=(cubic function)e^-2x
was what I got, you then put that with the range,

Nooo it wasn't G^-1(x) = f(x)e^-2x, it was G'(x) = f(x)e^-2x.

G'(x) is differential of G(x),
G^-1(x) is inverse function of G(x)

Edexcel A2 C3 Mathematics 12th June 2015

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