# AQA MM1B and MPC3 exams January 23 2013 Watch

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#3

__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!

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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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#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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#5

(Original post by

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!

**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!

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#6

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

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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#7

(Original post by

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

**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

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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 :/

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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#10

(Original post by

------------------------------------------------------------------------

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.....

**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.....

Then it is (U^2 + 1) root U dU then from there multiply out brackets and integrate

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Dammit, just realised i made a stupid mistake on the 8 mark subsitution Q

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#12

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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#13

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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(Original post by

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!

**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!

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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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#16

(Original post by

did anyone do M1?

**primo2gd4u**)did anyone do M1?

what did you get for the vector question

the last part, finding the distance?

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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?

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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#18

(Original post by

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?

**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?

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#19

(Original post by

i did

what did you get for the vector question

the last part, finding the distance?

**ravxo**)i did

what did you get for the vector question

the last part, finding the distance?

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#20

(Original post by

45.6m which is the magnitude of the position vector [28 36] found from a time of 12 seconds.

**SherlockHolmes**)45.6m which is the magnitude of the position vector [28 36] found from a time of 12 seconds.

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