AQA A2 Mathematics MPC4 Core 4 - 9th June 2015 [Discussion & unofficial markscheme]

Thank you i really appreciate your help are there any questions you'd like help with as I'd like to return the favour

Hopefully this helps



And yeah, I'm struggling with 7 b) i) on Jan 13! But I think I'll be able to sort it when I look at the mark scheme.

guys what shapes could they ask?~~~[vectors]
Rhombus
isosceles triangle
trapeziums
kites
rectangles
squares
perpendicular a.b=0
parallel=a.b=-1
cut put circles in there
maybe use cosine rule/sine rule also given that we find the magnitudes out???

June 2013 Question 7 anyone?3 marks but I just do not get it

It tells you the largest value that the range of change can be is 1.3

The largest number cos(ANYTHING) can be is 1.
Therefore to get 1.3, A must equal 1.3.

Then you have to find a point where cos() will equal 1 at every 12 hour occurrence. (i.e. when t=0, 12, 24, etc...)

If you draw a cos graph, you will see that this is at Pi/6.
Therefore k = Pi/6

June 2013 Question 7 anyone?

3 marks but I just do not get it

Does anyone know if there have been any recent spec changes at all, because I have been using past papers as my only revision source so if something comes up that isn't on there than I shall not be prepared!


Posted from TSR Mobile

Basically it's modelling the change in height of the tide over a period of time (i.e. rate of change of the height of the tide) as a cosine graph, whereby the maximum and minimum values for this are 1.3 and -1.3 (dh/dt = +/- 1.3).

From the question you know: dh/dt = acos(kt).

From a cosine graph you know that the maximum and minimum points occur at 1 and -1 respectively (i.e. cos = 1 or -1). Therefore a = 1.3 or -1.3. You can consider this as a stretch parallel to the y-axis of SF 1.3 if you like.

k is adjacent to the t in the model and t is being measured on the x-axis. You know the period of a cosine graph is 2pi and the tides occur every 12 hours (i.e. t = 12). This means that k becomes 2pi/12 in order to cancel with the 12 introduced through t so that you end up with cos(2pi) not cos(12).

June 2014, Q4ci - http://prntscr.com/7e7il0
they started in the mark scheme with the opening line

$5000p^{T-10} = 2500q^T$

would this:

$5000p^{T} = 2500q^{T+10}$

still give the right answer, and if so can someone post a solution using this to start with?
Thanks

June 2014, Q4ci - http://prntscr.com/7e7il0
they started in the mark scheme with the opening line

$5000p^{T-10} = 2500q^T$

would this:

$5000p^{T} = 2500q^{T+10}$

still give the right answer, and if so can someone post a solution using this to start with?
Thanks

Thank you so much!!!! It was too tricky for me

Can someone write out a step by step on June 2014, Q4)C)I) whilst it's been brought up.
Can't do it at all.

Sure if you give me about 5 minutes . I'll edit this post with working.

June 2014, Q4ci - http://prntscr.com/7e7il0
they started in the mark scheme with the opening line

$5000p^{T-10} = 2500q^T$

would this:

$5000p^{T} = 2500q^{T+10}$

still give the right answer, and if so can someone post a solution using this to start with?
Thanks