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AQA Core 3 - June 11th 2010

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Reply 20
This exam is on the morning of my 18th birthday! Thanks AQA for this lovely present :smile:

But seriously, I've done a fair few papers and am getting upwards of 90% each time, I think timing's going to be the only issue for me.
Maybe I should do some timed papers today?

Also thank you to Master Jomi for uploading the Jan 2010 papers :smile:
Reply 21
Only just noticed this thread :smile:
I'm finding C3 pretty good, there are some quite nice easy things on this paper. Integration I've found is my weakest point.
Just wanted to check if I'm missing out any trig identities. I've got:

Cosec^2x = 1+cot^2x
Sec^2x = 1+tan^2x
Sin^2x + cos^2x = 1
Tanx = sinx/cosx
MasterJomi
Here's the Jan 2010 paper that some of us did with the mark scheme!

Enjoy. :biggrin:

thank you very much :eek3:
Reply 23
kathx1
Only just noticed this thread :smile:
I'm finding C3 pretty good, there are some quite nice easy things on this paper. Integration I've found is my weakest point.
Just wanted to check if I'm missing out any trig identities. I've got:

Cosec^2x = 1+cot^2x
Sec^2x = 1+tan^2x
Sin^2x + cos^2x = 1
Tanx = sinx/cosx

I am pretty sure they're all the identities we need to know :smile:
Reply 24
kathx1
Only just noticed this thread :smile:
I'm finding C3 pretty good, there are some quite nice easy things on this paper. Integration I've found is my weakest point.
Just wanted to check if I'm missing out any trig identities. I've got:

Cosec^2x = 1+cot^2x
Sec^2x = 1+tan^2x
Sin^2x + cos^2x = 1
Tanx = sinx/cosx

S'it.
You only need to remember the sin^2x+cos^2x=1
And then divide by either sin^2x or cos^2x.
Jeester
I need some help!

I was just marking June '07 and I got to the final part of the last question:
http://store.aqa.org.uk/qual/gceasa/qp-ms/AQA-MPC3-W-MS-JUN07.PDF
You are supposed to use radian mode when doing this question but I did it in degrees, how do I know to use radions instead?

Thanks guys.


You have to use radians because the limits are 1 and 0.5 - it's not going to ask you to integrate between and 0.5º. And I think you can use those identities with degrees and radians - just look at the limits to know which one to use; it's usually really obvious when you think about it.
4cotx^2 + 12cosecx + 1=0

Got to the last part after putting identities inn. Cant seem to get final answer can anyhelp?
Va_Va_ Voom
4cotx^2 + 12cosecx + 1=0

Got to the last part after putting identities inn. Cant seem to get final answer can anyhelp?


You need to make sure you only have ONE trig identity in there - so either have the equation in terms of cosecx OR cotx; you can't have both because it relies on factorising or using the quadratic formula so you have the 2 possible values for x.
Va_Va_ Voom
4cotx^2 + 12cosecx + 1=0

Got to the last part after putting identities inn. Cant seem to get final answer can anyhelp?


What paper and question number are we looking at? (assuming it's from a past paper)

EDIT: decided to try and solve it, to see if I can help. If there's any mistakes in there, I'm sure someone will point them out. :smile:

sin^2(x) + cos^2(x) = 1

divide by sin x
1 + cot^2(x) = cosec^2(x)

rearrange for cot^2(x)
cot^2(x) = cosec^2(x) - 1

substitute into the question
4(cosec^2(x) - 1) + 12 cosec(x) + 1 = 0

Multiply out the bracket
4cosec^2(x) - 4 + 12cosec(x) + 1 = 0

Simplify
4cosec^2(x) + 12cosec(x) - 3 = 0

Doesn't factorise so put in quadratic formula
a = 4, b = 12, c = -3
cosec(x) = -(12)+-root((12)^2-4(4)(-3))/2(4) = (-3+-2root3)/2 (did on calc)

Do reciprocal to get in terms of sin(x)
sin(x) = 2/(-3+-2root3) = (6+-4root3)/3 (did on calc)

(6-4root3)/3 = -0.309, so can be this
(6+4root3)/3 = 4.309, so cant be this

sin(x) = (6-4root3)/3
Then find the solutions in the range given. :smile:
bluespacetiger
What paper and question number are we looking at? (assuming it's from a past paper)


Got it from college has no date on it just speciman paper A
Va_Va_ Voom
Got it from college has no date on it just speciman paper A


Had a look at it and the method I used in my edit is seemingly correct. Hope that helps. :smile:
Reply 31
Does anyone know if AQA want the answers to be top heavey fractions or the other one where you put a number in front of the fraction?
After revising for FP2 and FP3... C3 seems... rather okay! I need to go through a few papers tonight and have a textbook flick though.
Reply 33
I'm relatively confident. More confident than I am in Core 4, anyway! There's one (REALLY SIMPLE) thing I can't seem to do, though: finding the range/domain for a function? Some of them it's obvious, but I don't have a method, I just randomly plug numbers in and see if it tends towards anything. Is there a menthod?

That being said, I'm getting 85-95% on the past papers, usually. Lets hope this happens in the exam, I often fall apart during exams. I think timing will be an issue.
Reply 34
steph_v
I'm relatively confident. More confident than I am in Core 4, anyway! There's one (REALLY SIMPLE) thing I can't seem to do, though: finding the range/domain for a function? Some of them it's obvious, but I don't have a method, I just randomly plug numbers in and see if it tends towards anything. Is there a menthod?

That being said, I'm getting 85-95% on the past papers, usually. Lets hope this happens in the exam, I often fall apart during exams. I think timing will be an issue.

I draw the graph on the calculator then you can see what it tends to.
S'what I do.

Or if you're working out the inverse then it's just the previous domain.
bluespacetiger
What paper and question number are we looking at? (assuming it's from a past paper)

EDIT: decided to try and solve it, to see if I can help. If there's any mistakes in there, I'm sure someone will point them out. :smile:

sin^2(x) + cos^2(x) = 1

divide by sin x
1 + cot^2(x) = cosec^2(x)

rearrange for cot^2(x)
cot^2(x) = cosec^2(x) - 1

substitute into the question
4(cosec^2(x) - 1) + 12 cosec(x) + 1 = 0

Multiply out the bracket
4cosec^2(x) - 4 + 12cosec(x) + 1 = 0

Simplify
4cosec^2(x) + 12cosec(x) - 3 = 0

Doesn't factorise so put in quadratic formula
a = 4, b = 12, c = -3
cosec(x) = -(12)+-root((12)^2-4(4)(-3))/2(4) = (-3+-2root3)/2 (did on calc)

Do reciprocal to get in terms of sin(x)
sin(x) = 2/(-3+-2root3) = (6+-4root3)/3 (did on calc)

(6-4root3)/3 = -0.309, so can be this
(6+4root3)/3 = 4.309, so cant be this

sin(x) = (6-4root3)/3
Then find the solutions in the range given. :smile:


thanks for the help bruv, my calculater was on radians.
Reply 36
pricel18
On January 2010 paper, does anyone know if there is a quicker way to do question 7b? It is the d2y/dx^2 one? It's taking me about a page to do it all.

Chain rule?
Reply 37
Do we need double angle formulas or are they C4? C3 and C4 have just blended together.
steph_v
I'm relatively confident. More confident than I am in Core 4, anyway! There's one (REALLY SIMPLE) thing I can't seem to do, though: finding the range/domain for a function? Some of them it's obvious, but I don't have a method, I just randomly plug numbers in and see if it tends towards anything. Is there a method?

That being said, I'm getting 85-95% on the past papers, usually. Lets hope this happens in the exam, I often fall apart during exams. I think timing will be an issue.


Excuse me, are you me?
Because everything you wrote is exactly the same with me!
I always get the range/domain bit wrong - I've had it explained to me a thousand times, but I still get it wrong!
Reply 39
KitCat93
Do we need double angle formulas or are they C4? C3 and C4 have just blended together.

Core 4.

(But you do need to know how to differentiate Cos2X etc.

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