AQA Core 3 - Thursday 6th June 2013 (AM) - Official Thread

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Very last question, bummed me over because I was being stupid. I know I've lost a few here and there, but hopefully I've done enough to get close to an A which would help. Do you think grade boundaries will be similar to Jan 13 core 3, higher or maybe ever lower?

I think higher tbh

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Reply 61

beccac94

Original post by beautywithbrains

Same! Well I thought they would be high cos people who I talked to said they liked that exam

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Everyone seems to be saying different things, I just really hope I get an A seen as I got 1 mark off in Jan

Reply 62

sophonax1

People at my school thought it was harder so hopefully lower GB's. I got 2ln(3/2) for last q

Reply 63

Relf

This paper was very easy, but I messed up a few questions. I need and A* for uni. Think I got around 66-68 marks. However I think it will be 70 or maybe even 71 for 90ums. Core 4 has got to go really well for me or I'm retaking the year for 2 stupid mistakes. Aaahhhhhhhhhh

Reply 64

nothepreacher

Had a question f anyone could answer it. On the cos inverse question I reflected the graph in y axis instead of x but moved pi units in y axis to give 3 over 2 pi as the x intersection. Would I get marks for moving it pi unts in y direction?

Reply 65

edina101morris

i got 2ln(3/2) on the last one as well :biggrin:

for the geometric translation question does it matter which order? would it make any difference if it was reflect in the y-axis or stretch in x-direction S.F. 2 first?

Reply 66

beccac94

Original post by sophonax1

People at my school thought it was harder so hopefully lower GB's. I got 2ln(3/2) for last q

I got that and when I put the equation into my graphical equation it said it was the right answer, I'm only hoping that you didn't have to write it as ln(9/4) for the full marks

Reply 67

Relf

What were the graphs meant to look like. -f(3x), f(|x|) and y=cos^-1x , y=pi-cos^-1x and I think there was an other one...

(edited 10 years ago)

Reply 68

GeneralOJB

Original post by beccac94

I got that and when I put the equation into my graphical equation it said it was the right answer, I'm only hoping that you didn't have to write it as ln(9/4) for the full marks

No... 2ln(3/2) is correct and they could not take marks for it unless they asked for it in the form ln(a/b) where a and b are integers.

(edited 10 years ago)

Reply 69

stuart_aitken

Original post by stuart_aitken

I think for the first one it was
(-1,pi) (1,0)
And the second one was
(-1,0) (1,pi)

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Original post by Danatlive

I also did this but it didnt make sense as the graph had been translated up the y axis by pi units so I added pi to my coords hmmm

[

(edited 10 years ago)

Reply 70

beccac94

Original post by GeneralOJB

No... 2ln(2/3) is correct and they could not take marks for it unless they asked for it in the form ln(a/b) where a and b are integers.

Surely that would mean that it would be negative something when my calculator said it was apositive??? I suppose we'll just have to see what the mark scheme is

Reply 71

edina101morris

Original post by Relf

What were the graphs meant to look like. -f(3x), f(|x|) and y=cos^-1x , y=pi-cos^-1x and I think there was an other one...

urmm there was a ln one as well i think it was y= ln(3x+2) and then you had to do the inverse of it?

the f(|x|) one was being reflected it in the y-axis i think cus the values of y would be the same on the negative side as the positive

(edited 10 years ago)

Reply 72

GeneralOJB

Original post by beccac94

Surely that would mean that it would be negative something when my calculator said it was apositive??? I suppose we'll just have to see what the mark scheme is

huh? What are you talking about?

edit: oh sorry, I meant 2ln(3/2)

Reply 73

edina101morris

Original post by GeneralOJB

huh? What are you talking about? The answer was 2ln(2/3) lol

wasn't the answer 2ln(3/2)???

edit: ignore just seen you're other post

do you know which order we had to write the transformations in? was it reflect first the stretch or does it not matter?

(edited 10 years ago)

Reply 74

nothepreacher

Original post by stuart_aitken

[

the graph you have drawn is wrong. It is supposed to be reflected in x axis and not y

Reply 75

GeneralOJB

Original post by edina101morris

wasn't the answer 2ln(3/2)???

Yeah, sorry, people above confused by posting the wrong answer. Cattle!

Reply 76

AunShah

For the transformation question f=(x) to f=(-2x) did it matter, which order you did the transformation in??
I put (reflection in the y axis) and then (stretch scale factor 0.5 in x axis), I think. Would that be right?

Reply 77

stuart_aitken

For the -f(3x) graph, I reflected it in the x-axis, with the x-intercept at (2,0) because of the 1/3 stretch (original x-intercept was (6,0), and the y-intercept remained at (0,4)

For f|x| I just drew the original graph but bounced it off the x-axis when it hit, so everything stayed in the upper two quadrants.

And opinions on that?

Reply 78

Icycandle

Any willing to, or confident enough to outline their steps in working out the last question? Where you have to integrate 1/x+x^0.5 between the limits 1 and 4 using the substitution of u = x^0.5? (7marks)