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    (Original post by danielhx)
    this may seem like a silly question but is it highly doable to get full marks for the MCQs? my section B rly isn't great.
    Personally it was very doable in the older papers but the questions seem to have gotten considerably harder. I know I won't be able to especially under exam setting and being mega stressed but that's just me...
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    How would you do this?

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    (Original post by theaverage)
    How would you do this?

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    Is the answer a? I tried subbing in values such as pi/2 and 3pi/2 and through elimination..
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    (Original post by danielhx)
    Is the answer a? I tried subbing in values such as pi/2 and 3pi/2 and through elimination..
    Yupp, ahh I didn't even think of that! Thanks

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    (Original post by theaverage)
    Yupp, ahh I didn't even think of that! Thanks

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    no worries are they past MAT questions?
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     2 = sin x + sin^2 x + sin^3 x + sin^4 x + . . .
    how many solutions?
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    (Original post by Mystery.)
     2 = sin x + sin^2 x + sin^3 x + sin^4 x + . . .
    how many solutions?
    Find what sin x has to be, then find how many times sin x equals that value in the range under consideration (which you haven't provided, which makes this impossible to answer concretely).
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    (Original post by danielhx)
    no worries are they past MAT questions?
    Yupp 2002 I think

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    (Original post by AGFilsell)
    (this is probably wrong because other people got a different answer, oh well)
    I extended AP and BQ out of the circle until they touched (lets say at point 'S' and formed a kite shape. From there i said that angle RPS= pi - x (because angles on a straight line = pi) and from that i said that angle RQS= pi - (pi - x)= x (opposite angles in a kite add to pi). Finally, since RQB and RQS are on a straight line, they equal pi too, Therefore my answer is 'x + y = pi'.

    Can someone please check if this is wrong or right please? I feel like ive definitely made a mistake here
    The reason why this is wrong is that PQRS is never a kite. For a kite to be a kite, it must have two pair of adjacent sides equal. Moreover, since there is no tangent in the question, why would you arrive to anything such as pi-x? btw Mystery are you asking the same question lol
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    (Original post by Mystery.)
     2 = sin x + sin^2 x + sin^3 x + sin^4 x + . . .
    how many solutions?
    there is another way.One way is that you can quickly get (2/sinx)-1=2 by some easy algebra, or since the right hand side is a geometric sequence you can just find sum of terms to infinity then you can get sin x= 2/3. However, where is your 0<x<pi or 0<x<2pi sort of thing? Have you forgotten that one?
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    (Original post by theaverage)
    How would you do this?

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    1J was a really .... I basically used compound, triple and double angle formulas to get the final answer. I think I used 2 minutes on this, which is quite time-consuming.
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    (Original post by LaserRanger)
    1J was a really .... I basically used compound, triple and double angle formulas to get the final answer. I think I used 2 minutes on this, which is quite time-consuming.
    I used de moivre's theorem.
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    De moivre's? not quite get how the complex becomes real
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    Hey guys I'm applying for straight computer science and was wondering if someone could help me out with the grading for 6 and 7. I technically have the right solutions for the problems but the way I word them makes it seem that I would get them wrong on the exam. I was wondering if they have lots of leeway when it comes to long explanations and if they will support any mentality of solving a problem if it is marginally right.
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    (Original post by LaserRanger)
    De moivre's? not quite get how the complex becomes real
    cos 5x + i sin 5x = (cos x + i sin x)^5

    So sin 5x = Im (cos x + i sin x)^5 = 5c^4 s - 10 c^2 s^3 + s^5 (writing c for cos x and s for sin x).

    Replacing c^4 by (1-s^2)^2 and c^2 by (1-s^2) we get

    5 (1 - 2s^2 + s^4) s - 10(1-s^2) s^3 + s^5

    = 16 s^5 - 20 s^3 + 5s

    FWIW, to answer the actual question, I'd definitely do it by plugging in values for x and elimination. (Actually, from experience I'd know it would be all odd powers of sin, therefore (c), (d) ruled out immediately).
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    (Original post by AmanGottu)
    Hey guys I'm applying for straight computer science and was wondering if someone could help me out with the grading for 6 and 7. I technically have the right solutions for the problems but the way I word them makes it seem that I would get them wrong on the exam. I was wondering if they have lots of leeway when it comes to long explanations and if they will support any mentality of solving a problem if it is marginally right.
    I don't think it's possible to answer this without actually seeing a solution from you.

    That is, I don't think they're going to be needlessly strict, but it's possible to get the right answer (by luck) with a totally flawed argument, (or more likely, make an educated guess but have no real argument to justify your answer). You won't get many marks in those cases.
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    (Original post by physicsmaths)
    A very good way is sometimes process of elimination which speeds stuff up sometimes although it isn't intended I think. Not always best way though, you would have tl judge it on question by question.
    Especially with the graph questions sometimes this is the only (reasonable) way of doing things. The mark schemes often do it this way. But this year there's 5 choices rather than the 4 in most past papers, so I guess that discourages process of elimination for questions where that's not the method they want you to use.
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    help with this q?

    I have tried to understand the marking scheme so many times but I just don't get it.

    seems like it won't let me load the attachment but its 2011 1G?
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    (Original post by Mystery.)
    help with this q?

    I have tried to understand the marking scheme so many times but I just don't get it.

    seems like it won't let me load the attachment but its 2011 1G?
    What i did was just deduce that the equation of the graph -|x|+1 and then integrate though this may not be the best method
 
 
 
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