Poll: After these 2 exams, how would you rate them?
Mechanics: Easy (4)
8%
Mechanics: Okay (10)
20%
Mechanics: Could have been better (7)
14%
Mechanics: I've failed (2)
4%
Core 3: Easy (12)
24%
Core 3: Okay (4)
8%
Core 3: Could have been bettwe (8)
16%
Core 3: I've failed (3)
6%
The H
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#1
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#1
Mechanics and Core 3, how do we feel about these exams?
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Kanda
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#2
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Has anyone got the AQA M1 Jun 2012 Paper
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KING2011BOB
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#3
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To sum up the core3

Prove a (sum) = cosec^2(x)

Then use that

To say cosec^2(x) - cosecx - 3 - 0

Doesn't factorise,so then use the formula

Get cosecx values from formula

Do 1/answers to get sinx

Then do sin^-1(1/answers) to get x values,then do 180-(answers) +-360

In other words a whole world of hell in some questions!

------------------------------------------------------------------------
The last question literally I have no idea how to finish:

From integral between x and y of sec^4 (x) root(tanx)

u = tanx so du/dx = sec^2(x)

So therefore integral of 4 and 0 between sec^2 (x) root (u)

WTF do you do from here.....
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2013mercs
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#4
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#4
cosec^2(x) - cosecx - 3 = 0, this didnt factorise as you pointed out, what i done was to plug the equation into my graphics calculator as a quadratic equation (x^2-x-3=0), this forced out two horrible answers of x=2.3027 and x=-1.302, then using these values as cosecx i worked my way through this question, seems wrong though!
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kitkat19
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#5
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(Original post by 2013mercs)
cosec^2(x) - cosecx - 3 = 0, this didnt factorise as you pointed out, what i done was to plug the equation into my graphics calculator as a quadratic equation (x^2-x-3=0), this forced out two horrible answers of x=2.3027 and x=-1.302, then using these values as cosecx i worked my way through this question, seems wrong though!
everyone got this...its right
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2013mercs
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#6
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Thats good, a mate of mine had me worried because he made cosec^2(x)-cosecx=3, factorised LHS to
cosecx(cosecx-1)=3, then either cosecx=3 or cosecx=4,

The very last question, i ended up with a fraction
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bugsuper
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#7
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#7
(Original post by 2013mercs)
Thats good, a mate of mine had me worried because he made cosec^2(x)-cosecx=3, factorised LHS to
cosecx(cosecx-1)=3, then either cosecx=3 or cosecx=4,

The very last question, i ended up with a fraction
Your friend's logic doesn't work. If cosecx = 3 then his equation gives 3*2=6; if cosecx=4 then he should get 12. it's inconsistent. the correct method was quadratic formula to get, like, {1 +/- rt(13)}/2
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primo2gd4u
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#8
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did anyone do M1?
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primo2gd4u
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#9
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#9
the last question on projectiles....
finding the horizontal component of velocity:
was it - displacement x time
(or) distance (divided by) time :/

finding the initial speed,
u find the vertical and horizontal components of velocity... and then find the magnitude
however i just realised that gives u the FINAL velecity and not the INITIAL velocity, so i'm wondering did we need to use a SUVAT equation to find U? it was 5 marks afterall :/
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OliT95
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#10
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#10
(Original post by KING2011BOB)
To sum up the core3

------------------------------------------------------------------------
The last question literally I have no idea how to finish:

From integral between x and y of sec^4 (x) root(tanx)

u = tanx so du/dx = sec^2(x)

So therefore integral of 4 and 0 between sec^2 (x) root (u)

WTF do you do from here.....
The integral was pie/4 and 0 btw. But then you make U in terms of sec^2x, so you square both sides so U^2 = tan^2x, then add 1 so U^2 + 1 = tan^2x + 1. With the identity it means U^2 + 1 = sec^2 so then you just subsitute in U^2 + 1.

Then it is (U^2 + 1) root U dU then from there multiply out brackets and integrate
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The H
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#11
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#11
Dammit, just realised i made a stupid mistake on the 8 mark subsitution Q
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AvaSofia
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#12
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Didn't think it went brilliantly, knew I could have done so much better, however I did get the substitution and trig question right, so I'll take that as a positive!
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con31773
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#13
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Nailed the substitution question, but that trig question was awful. When I saw the question didn't factorize I used the quadratic formula to punch out two awful values making me think my calculation was wrong and moved on. The paper overall was pretty tricky, some pretty awkward questions.
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The H
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#14
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(Original post by AvaSofia)
Didn't think it went brilliantly, knew I could have done so much better, however I did get the substitution and trig question right, so I'll take that as a positive!
They happen to be the 2 questions I've just found out I made terrible mistakes on
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Lucas96
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#15
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#15
In M1, question 7c how many marks will I lose for using r=ut+1/2at^2 to find the time instead of v=u+at?
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ravxo
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#16
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#16
(Original post by primo2gd4u)
did anyone do M1?
i did

what did you get for the vector question
the last part, finding the distance?
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ravxo
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#17
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#17
i thought M1 was alright
last question was confusing, i had to find alpha first then find V
ended up with 26.2 ms-1 and 39.1 degrees. anybody get the same?
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Lucas96
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#18
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#18
(Original post by ravxo)
i thought M1 was alright
last question was confusing, i had to find alpha first then find V
ended up with 26.2 ms-1 and 39.1 degrees. anybody get the same?
I got the same angle but I think I got a different velocity, because my horizontal component was 16
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SherlockHolmes
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#19
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#19
(Original post by ravxo)
i did

what did you get for the vector question
the last part, finding the distance?
45.6m which is the magnitude of the position vector [28 36] found from a time of 12 seconds.
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ravxo
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#20
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(Original post by SherlockHolmes)
45.6m which is the magnitude of the position vector [28 36] found from a time of 12 seconds.
i got a different answer
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