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

In the tables book, there are two formulae for binomial expansion. When do you know which method to use?

You know when you have 'cosOBA' and you need to find the angle?

Which vectors do you use?

Is it OB.BA/|OB|.|BA|?

SOMEONE PLEASE HELP IM STRESSING SO MUCH

3/2Ln3 - 2Ln2 becomes what? Im not sure what you do with the numbers infront of the Ln?

You need to use the log laws from core 2. Remember Ln is just a special type of log. So the numbers in front are dealt with using the rule aln(b) = ln(b^a). So for example 3/2ln(3) = ln(3^(3/2)).

If anyone likes a challenge:



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I've always used BO and BA, or OB and AB, never OB and BA.

For example in Jan 2013 Q6a(ii) you're asked to find the angle ACB. The mark scheme uses the vectors BC and AC. So it looks like you use the two vectors going towards the point at which the angle is. Using CB and CA should also work but I'm not sure that AC and CB would, but I could be wrong.

So just make sure your vectors are both pointing inwards OR both outwards from the angle (do not mix them)

CB and CA will work. If the two vectors are pointing away from the angle or inwards to the angle then you will obtain the same result. This is since vertically opposite angles are equal.

AC and CB will not work since now you are calculating adjacent angles. In the diagram If you are looking at the yellow angles as ACB then in this case you are calculating the red angles which are different.

So just make sure your vectors are both pointing inwards OR both outwards from the angle (do not mix them)

I've always used BO and BA, or OB and AB, never OB and BA.

For example in Jan 2013 Q6a(ii) you're asked to find the angle ACB. The mark scheme uses the vectors BC and AC. So it looks like you use the two vectors going towards the point at which the angle is. Using CB and CA should also work but I'm not sure that AC and CB would, but I could be wrong.

So what do you use, vectors going away from the angle of towards it???

So what do you use, vectors going away from the angle of towards it???

My teacher emailed me this vector reflection question and I can't do it at all!:

"The point P(1, 1, 2) is reflected in the line r = [1, 0, 2] + U[-1, 4, 3]. Find the co-ordinates of R, the reflect point."

Step by step anyone?

Anyone done question 12c in the jan 10 pilot paper? Or can help me integrate [10cos(x-0.64)]^2 thanks!!!


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I have no idea if there's a concrete method for doing a vector reflection, but this is how I would do it:

The reflected point, R, is of equidistance from the line r as P is, only in the opposite direction and the shortest distance from P to the line occurs when a line through P that intersects with the line is perpendicular to the line.

Let the point of intersection be A.

Therefore, PA = OA - OP. OA is equal to r for some value of U.
=> PA = [-U, 4U-1, 3U]

Doing the scalar product with PA . [-1, 4, 3] = 0 will allow you to solve to find the value for U at which A occurs. Substituting in the value for U will give you the position vector of A, from which you can find PA by doing OA - OP. However, as you already know PA = [-U, 4U-1, 3U] you can just substitute U into that vector.

Next, form an equation of a line(let said line be L₂ that passes through P and A with position vector [-U, 4U-1, 3U].
=> L₂: r = [-1, 4, 3] + N[-U, 4U-1, 3U]. (remember you know what U is at this point).


EDIT: in progress; note that I'm really tried at the moment so if I've done something stupid please just say .

any ideas as to what is likely to come up?

Try subbing theta for (x-0.64), squaring whole thing to 100cos^2(theta) and then using double angle formula.
As in, cos2theta= 2cos^2(theta) - 1
so cos^2(theta)=0.5(cos2(theta)+1)

Then integrate that and sub theta back in?
I think.