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# AQA A2 Mathematics Pure Core 4 MPC4 - 17th June 2016

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1. (Original post by koolgurl14)
The answer is y = x/1-x I wasted this whole day to figure it out ...
I wrote the question down wrong, nvm!
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2. (Original post by koolgurl14)
The answer is y = x/1-x I wasted this whole day to figure it out ...

That was tricky! Where did it come from?
3. (Original post by bartbarrow)

That was tricky! Where did it come from?
OMG thank you so much I am glad i am not the only who thought it was tricky. It's from old spec past papers booklet our teacher gave us. examiners who did this question are probably already retired, so hopefully nothing crazy like that shows up tmmrw
4. (Original post by bartbarrow)

That was tricky! Where did it come from?
(Original post by koolgurl14)
OMG thank you so much I am glad i am not the only who thought it was tricky. It's from old spec past papers booklet our teacher gave us. examiners who did this question are probably already retired, so hopefully nothing crazy like that shows up tmmrw

Another way of doing it - I always multiply by denom and factorise it until I can get t on its own with this kinda thing
5. (Original post by Franckenstar)
You could square x and y and add x^2 + y^2
thats sounds complicated but would it not still have x^ + y^2 equaling to more t's?
6. can someone please explain vectors to me because im still struggling with it even after all of those past papers, i still dont know what they might ask and still not used to the style of the questions , can someone like a list of things that i need to know or list of questions that they usually ask
7. xy=2+1/(t-1)

(x-2)=-1/t
therefore xy=2-y(x-2)
No need to justify range or anything like that as question doesn't say to do so.

(Original post by bartbarrow)
I wrote the question down wrong, nvm!
8. In case it comes up again how would you differentiate (ln(x))^2?
9. (Original post by Mowerharvey)
In case it comes up again how would you differentiate (ln(x))^2?
Chain rule: let u = lnx so du/dx = 1/x

y = u^2 so dy/du = 2u

therefore dy/dx = dy/du x du/dx = 1/x x 2u = 2/x(u) = 2/x (lnx)
10. Do the Jan 13 paper, that's good revision because it was very very difficult
11. (Original post by ljb665)
Do the Jan 13 paper, that's good revision because it was very very difficult
I've just done it. It was more difficult than others, but I wouldn't even think of comparing it to the core 3 they gave us this year lol
That was honestly on a whole new level, im still not over it
12. (Original post by mickel_w)
I've just done it. It was more difficult than others, but I wouldn't even think of comparing it to the core 3 they gave us this year lol
That was honestly on a whole new level, im still not over it
It was just a bit of a joke really. They're resits for me too. Got an A last year but need an A* which means I need 96 UMS in this exam, LOL
13. (Original post by CasioGamer98)
The tide is said to repeat every 12 hours in the same way y=cos(x) repeats every 2pi radians. For the tide to be consistent with y=cos(x) divide 2pi by 12 and you find the constant k must be pi/6. Hence cos(pi/6 t) is now proportional to the rate of change of the tide.
Whats the purpose of the constant k?
Why is is k needed and how are we supposed to know k is 12/2pi ? Like even if it was in the question i would have had no idea what k was. Could you please explain this to me?
14. (Original post by khuman97)
can someone please explain vectors to me because im still struggling with it even after all of those past papers, i still dont know what they might ask and still not used to the style of the questions , can someone like a list of things that i need to know or list of questions that they usually ask
It it says anything about a right angle, a.b =0,if it says prove that it's a rhombus/square, the lengths will all be the same, isosceles triangle two lengths will be the same etc.
Have you done the questions where the new triangle is three times the area? It's because the top length is three times that of the original, if you draw your triangle properly you'll understand what I mean.
Prove two points intersect, use your vector equations and equate for X,y,z eg 2+lamda=6+mu, and lamda and mu should work for each coordinate
To show lines are askew, lamda and mu will not be the same for every coordinate.
If lines are parallel, they will have the same gradient which is the vector component after lamda or mu if that makes sense.
15. I reckon I need 100 ums to stand any chance of getting an A overall
16. I find vectors very hard anyone got any tips
17. (Original post by walkine)
Not to inconvenience you, but could you help at all with the parallelograms questions? I'm not sure how you do those.
Find the coordinate of the point you're looking for using a vector equation, eg X=6+lamda, Y=2+2lamda, Z=4+3lamda. You know that there are two sets of parallel lines so the two gradients are the same. Eg the gradient from AB and CD are the same. Then you know that the vector AB is the same as the vector CD, calculate lamda and sub back in. However if it says two possible positions there could be a numerous amount of ways.
18. (Original post by ljb665)
Find the coordinate of the point you're looking for using a vector equation, eg X=6+lamda, Y=2+2lamda, Z=4+3lamda. You know that there are two sets of parallel lines so the two gradients are the same. Eg the gradient from AB and CD are the same. Then you know that the vector AB is the same as the vector CD, calculate lamda and sub back in. However if it says two possible positions there could be a numerous amount of ways.
The question last year, you could work out the first position of E by using this method. To work out position two, you had to calculate the midpoint of AC then add this onto D to obtain the second position.
19. How do you turn an obtuse angle into an acute angle, and would we be asked to do this?
20. if anyones looking for extra vector/parameter/decay questions try looking through some of the edexcel papers

http://www.physicsandmathstutor.com/...rs/c4-solomon/

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