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

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    (Original post by AaronA330)
    In the tables book, there are two formulae for binomial expansion. When do you know which method to use?
    The first one is the one you learnt in core 2. You will be unlikely to be asked to use this (this is the nice case where what you are expanding has the form (a + b)^n where n is a positive integer.)

    The second one is the one you use in core 4. (The not so nice case where n is now a real number (not necessarily an integer), you need a = 1 and the expansion is valid for -1<b<1 (b is x in the booklet).)
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    (Original post by Jankie)
    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
    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.
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    (Original post by Jed-Singh)
    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)).
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    (Original post by problemq)
    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)).
    oh yh, omg I did that on part a to that question aswell haha. Thank you very much friend
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    What does everyone think the hardest paper has been so far?
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    (Original post by CD223)
    If anyone likes a challenge:
    Attachment 418305


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    do you have the answers. i got root2,2root2 and -root2,-2root2.
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    (Original post by FeelsToWaltz)
    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.
    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)

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    (Original post by problemq)
    So just make sure your vectors are both pointing inwards OR both outwards from the angle (do not mix them)
    Quoting this for visibility. It's one of the simplest things to easily easily get wrong!
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    (Original post by problemq)
    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)

    Okay that's great, I was also getting confused because people were saying you were supposed to use AC and CB which gives a different answer. Thanks for the clarification!
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    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?
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    (Original post by FeelsToWaltz)
    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???
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    (Original post by maattwileman)
    So what do you use, vectors going away from the angle of towards it???
    Either one of those two options.
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    (Original post by maattwileman)
    So what do you use, vectors going away from the angle of towards it???
    Either one works. You can't use one going towards with one going away.
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    (Original post by 2014_GCSE)
    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?
    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 into the equation of the line will give you the position vector of A and, as you know PA = [-U, 4U-1, 3U] you substitute U in there as well to find PA.

    You want to find R, and as R is equidistance from the line in the question you know that PA = AR.
    => PA = OR - OA
    => OR = PA + OA
    => OR = [-U, 4U-1, 3U] + OA (where OA = the equation of the line with the value of U you found substituted in]
    => OR = whatever position vector that came to
    => R is (x, y, z) form of the position vector you just found

    NOTE: I'm really tried at the moment so if I've done something stupid please just say .
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    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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    (Original post by EmilyC96)
    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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    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.
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    (Original post by datpr0)
    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 .
    Working through it myself now and we seem to have done the same thing! Except, shouldn't it be AP, not PA?

    As at the right angle, the [-1, 4, 3] would be going away from the angle, so you would want AP (also going away from the angle)
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    any ideas as to what is likely to come up?
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    (Original post by maattwileman)
    any ideas as to what is likely to come up?
    I bet you there'll be a ***** of a vectors question. And probably one of those "form a differential" questions that leave you wondering if AQA's head office is in hell.
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    (Original post by moinoi)
    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.
    Thanks I keep forgetting to use the double angle formulae 😞


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