Year 13 Maths Help Thread

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    Hi
    i need some help with this questions with the last part.
    i have the question and solution attached but i don't really understand whats going on with it?Attachment 582200582202Attachment 582200582202
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    (Original post by ARASH123g)
    Hi
    i need some help with this questions with the last part.
    i have the question and solution attached but i don't really understand whats going on with it?Attachment 582200582202Attachment 582200582202
    What dont u understand?
    Move this to the step thread for some proper help.


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    Exactly how far into the solution do you go before losing understanding?
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    (Original post by ARASH123g)
    Hi
    i need some help with this questions with the last part.
    i have the question and solution attached but i don't really understand whats going on with it?Attachment 582200582202Attachment 582200582202
    Can you get to  \cot \frac{1}{2}(q+p)=\cot \frac{1}{2} (s+r) ?
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    (Original post by coolsul98)
    Attachment 580752part iii

    you ****er are you re-sitting some units? Tell me bc
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    Help - ques 6.b) iii) where did i go wrong?

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    (Original post by kiiten)
    Help - ques 6.b) iii) where did i go wrong?

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    Your answer is correct, it just isn't in the 'simplest form' I guess. Multiply both the numerator and denominator by -1 and the rest should come quite easily.

    \dfrac{6-x}{-x+5} \Rightarrow \dfrac{x-6}{x-5} \Rightarrow \dfrac{x-5-1}{x-5} \Rightarrow \dfrac{x-5}{x-5} - \dfrac{1}{x-5} \Rightarrow 1 - \dfrac{1}{x-5}
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    (Original post by justinawe)
    Your answer is correct, it just isn't in the 'simplest form' I guess. Multiply both the numerator and denominator by -1 and the rest should come quite easily.

    \dfrac{6-x}{-x+5} \Rightarrow \dfrac{x-6}{x-5} \Rightarrow \dfrac{x-5-1}{x-5} \Rightarrow \dfrac{x-5}{x-5} - \dfrac{1}{x-5} \Rightarrow 1 - \dfrac{1}{x-5}
    Ohh i see thanks

    Would i still get the marks with my answer?
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    (Original post by kiiten)
    Ohh i see thanks

    Would i still get the marks with my answer?
    I'm not too sure about that I'm afraid, it's been a couple years since I did my A-levels, I don't quite remember how the markschemes go. The question doesn't say anything about what form you should leave your answer in, so my guess is you'd get the marks. At worst you'd be docked 1 mark.
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    (Original post by justinawe)
    I'm not too sure about that I'm afraid, it's been a couple years since I did my A-levels, I don't quite remember how the markschemes go. The question doesn't say anything about what form you should leave your answer in, so my guess is you'd get the marks. At worst you'd be docked 1 mark.
    Ahh ok . Anyone else know??
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    (Original post by kiiten)
    Ohh i see thanks

    Would i still get the marks with my answer?
    Your answer is fine. There's no need to split up the fraction.

    You could make it nicer if you multiply top and bottom by -1 but a mark scheme would allow either form.

    Why did you think your answer was wrong?
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    (Original post by notnek)
    Your answer is fine. There's no need to split up the fraction.

    You could make it nicer if you multiply top and bottom by -1 but a mark scheme would allow either form.

    Why did you think your answer was wrong?
    First i thought my answer was different to the right answer. Then I confused myself by trying to simplify my answer to match the mark scheme but ended up with 1 + 6/5 (x cancel out) - I must have completely forgot that you cant do that :3 - just a silly mistake
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    (Original post by kiiten)
    First i thought my answer was different to the right answer. Then I confused myself by trying to simplify my answer to match the mark scheme but ended up with 1 + 6/5 (x cancel out) - I must have completely forgot that you cant do that :3 - just a silly mistake
    Lack of understanding of cancelling fractions is very common at A Level and I've seen it cost many marks in exams.

    If you're confident that you won't make this mistake again then ignore it. But if you're even slightly unsure then I recommend you rethink about why what you did was wrong. Feel free to ask if you want an explanation.

    Some A Level students just don't like to admit to their teacher that they're unsure about cancelling
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    (Original post by notnek)
    Lack of understanding of cancelling fractions is very common at A Level and I've seen it cost many marks in exams.

    If you're confident that you won't make this mistake again then ignore it. But if you're even slightly unsure then I recommend you rethink about why what you did was wrong. Feel free to ask if you want an explanation.

    Some A Level students just don't like to admit to their teacher that they're unsure about cancelling
    I agree - i wouldnt ask my teacher this. Yeah i know where i went wrong, i know how to cancel fractions but sometimes forget because i dont have to do it often.

    The only way i remember is to factorise the top and bottom and see if anything can be simplified/cancelled out. I wasnt taught/didnt know that you can also split up the numbers to simplify the fraction. Thanks

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    Im confused about drawing modulus graphs.

    E.g. y = IxI + Ix-3I

    I did -x + (-x+3)
    Y = -2x +3

    Then x + (-x+3) etc.

    These are the lines you draw to make up the graph but how do you find the range/bounds for each of these. E.g. y=-2x+3 has the bound x <0 but how do you know that?

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    (Original post by kiiten)
    Im confused about drawing modulus graphs.

    E.g. y = IxI + Ix-3I

    I did -x + (-x+3)
    Y = -2x +3

    Then x + (-x+3) etc.

    These are the lines you draw to make up the graph but how do you find the range/bounds for each of these. E.g. y=-2x+3 has the bound x <0 but how do you know that?

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    You have to think about what values of x change the expressions inside the molulii from positive to negative.

    For |x|, that point is x = 0.
    For |x-3|, that points is x =3.

    So you have two critical points, x = 0 and x = 3 which gives you three regions that you need to consider:

    x &lt; 0

    0 \leq x &lt; 3

    x \geq 3.


    If we consider the first region x&lt;0:


    Here x&lt;0 and x-3 &lt; 0 so the equation becomes

    y = -x -(x-3) = -2x + 3

    and so you draw the line y=-2x + 3 in the region x&lt;0.
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    (Original post by notnek)
    You have to think about what values of x change the expressions inside the molulii from positive to negative.

    For |x|, that point is x = 0.
    For |x-3|, that points is x =3.

    So you have two critical points, x = 0 and x = 3 which gives you three regions that you need to consider:

    x &lt; 0

    0 \leq x &lt; 3

    x \geq 3.


    If we consider the first region x&lt;0:


    Here x&lt;0 and x-3 &lt; 0 so the equation becomes

    y = -x -(x-3) = -2x + 3

    and so you draw the line y=-2x + 3 in the region x&lt;0.
    Ok i understand where you got the critical points from but how do you know x<0 and not x>0 for example?
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    (Original post by kiiten)
    Ok i understand where you got the critical points from but how do you know x<0 and not x>0 for example?
    The critical points x = 0 and x = 3 are where things change (either side of them). If you draw these two points on a number line then you'll see that you have created 3 distinct regions.

    You have to consider these three regions to see what happens to the expression in each of the regions.

    x>0 is considered but this is part of the middle region.
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    (Original post by notnek)
    The critical points x = 0 and x = 3 are where things change (either side of them). If you draw these two points on a number line then you'll see that you have created 3 distinct regions.

    You have to consider these three regions to see what happens to the expression in each of the regions.

    x>0 is considered but this is part of the middle region.
    Oh I see. What if you had 3 critical values e.g. I(x-1)(x-2)(x-3)I - the values wold be 1, 2 and 3 but what about the bounds??
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    (Original post by kiiten)
    Oh I see. What if you had 3 critical values e.g. I(x-1)(x-2)(x-3)I - the values wold be 1, 2 and 3 but what about the bounds??
    Have you tried sketching the graph, both in the 2 and 3 cv case?
 
 
 
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