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    (Original post by tripleseven)
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    Those two parts are independent. The working out doesn't follow on. It stops at x = -\dfrac{1}{2} \pm \sqrt{2a^{2}+\dfrac{1}{4}}

    As you should know, when a modulus is involved you consider both the positive and the negative, so for this question you would do

    x^{2}-a^{2} = a^{2} - x

    and

    x^{2}-a^{2} = - (a^{2} - x)

    So what the person has done is work out the x values for when x^{2}-a^{2} = a^{2} - x and then straight underneath work out the x values for when  x^{2}-a^{2} = x - a^{2}
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    (Original post by edothero)
    Those two parts are independent. The working out doesn't follow on. It stops at x = -\dfrac{1}{2} \pm \sqrt{2a^{2}+\dfrac{1}{4}}

    As you should know, when a modulus is involved you consider both the positive and the negative, so for this question you would do

    x^{2}-a^{2} = a^{2} - x

    and

    x^{2}-a^{2} = - (a^{2} - x)

    So what the person has done is work out the x values for when x^{2}-a^{2} = a^{2} - x and then straight underneath work out the x values for when  x^{2}-a^{2} = x - a^{2}
    Ahh silly me, thank you for pointing that out! And it makes sense, because they intersect at 4 different points, hence the need for 4 solutions. Thank you very much!
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    (Original post by 1asdfghjkl1)
    can someone help me with this? i end up with the inequality x^2 -x +1 <0 but you can't factorise that so i'm stuck.. the mark scheme says the answer is -0.5<x<2/3
    Here you go, my working, then just draw your usual graph with critical values to show where it's greater (or less in your case) than zero.... Name:  IMAG5061.jpg
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    Make sure you bring it over to one side and factorise fully...
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    (Original post by ChuckNorriss)
    hope this makes sense. anyone correct me if im wrong pls
    Ahh so you change the given one to cartesian - got it. Thanks for your help
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    Please could someone explain exercise F question 16b on here https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH3.pdf how do we know the minimum value of arg z is pi/2 from the diagram? Thanks
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    (Original post by economicss)
    Please could someone explain exercise F question 16b on here https://8fd9eafbb84fdb32c73d8e44d980...lZweHc/CH3.pdf how do we know the minimum value of arg z is pi/2 from the diagram? Thanks
    From the diagram, you can see that the points on the circle have arguments from pi/2 to more than that, i.e: look at the blue lines below:



    The angle each one makes with the positive x-axis is greater than the straight vertical one that is tangent to the circle. That has te least argument of pi/2 since the circle is tangent to the Im-axis (because of its centre and radius).
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    (Original post by Zacken)
    From the diagram, you can see that the points on the circle have arguments from pi/2 to more than that, i.e: look at the blue lines below:



    The angle each one makes with the positive x-axis is greater than the straight vertical one that is tangent to the circle. That has te least argument of pi/2 since the circle is tangent to the Im-axis (because of its centre and radius).
    I see, thank you, so is it also because the circle is in the second quadrant that we can tell this, as all angles in the second quadrant are obtuse, or is that wrong? Haha, thanks
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    (Original post by economicss)
    I see, thank you, so is it also because the circle is in the second quadrant that we can tell this, as all angles in the second quadrant are obtuse, or is that wrong? Haha, thanks
    Yeah, I guess so. If that's how you want to look at it.
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    (Original post by Zacken)
    Yeah, I guess so. If that's how you want to look at it.
    Okay great, thank you, I think I'm still in FP1 mode thinking about quadrants!
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    (Original post by Zacken)
    From the diagram, you can see that the points on the circle have arguments from pi/2 to more than that, i.e: look at the blue lines below:



    The angle each one makes with the positive x-axis is greater than the straight vertical one that is tangent to the circle. That has te least argument of pi/2 since the circle is tangent to the Im-axis (because of its centre and radius).
    How would you work out the minimum arg z for part c please? Thanks
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    (Original post by economicss)
    How would you work out the minimum arg z for part c please? Thanks
    I presume you mean maximum? What don't you understand from the solutions bank?
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    (Original post by Zacken)
    I presume you mean maximum? What don't you understand from the solutions bank?
    Oops yes sorry, so I understand now the minimum as you explained so I get the pi/2, do we just have to add on the argument worked out by using the x and y coordinates of the centre of the circle, is this always the case for maximum arg questions? Thanks
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    (Original post by economicss)
    Oops yes sorry, so I understand now the minimum as you explained so I get the pi/2, do we just have to add on the argument worked out by using the x and y coordinates of the centre of the circle, is this always the case for maximum arg questions? Thanks
    Well, not always the case, but yeah, if you just draw some triangles and use trig you usually get the answer.
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    (Original post by Zacken)
    Well, not always the case, but yeah, if you just draw some triangles and use trig you usually get the answer.
    Okay great thanks for your help!
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    Can anyone explain to me how to work out the area of polar coordinates : example 10c on page 138 of edexcel fp2 book.
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    (Original post by fpmaniac)
    Can anyone explain to me how to work out the area of polar coordinates : example 10c on page 138 of edexcel fp2 book.
    They have used the formula for the area printed in the formula booklet.

    Area = \frac 12 \int_{\beta}^\alpha r^2 d\theta

    They have used this for each sector individually, and have split some of the sectors up so they do not cancel out. This is why the areas have been doubled, as, with polar coordinates, the graphs are very often symmetrical. The area for r= 2 + \cos\theta is calculated by squaring r and then dividing the total area by 2 (which cancels with the half from the formula). A similar thing is then done with r = 5\cos\theta.

    The angles for r = 2 + \cos\theta are \frac\pi3 and as the intersection at the top is at an angle of \frac\pi3 and this refers to the red curve in the diagram. The angles for r = 5\cos\theta are \frac\pi3 and \frac\pi2 as the curve goes from an angle of \frac\pi3 up to the angle of \frac\pi2. This is then doubled for the angle at the bottom

    If you want me to go into any more detail please ask, what follows on from here is relatively easy integration (the more you practice this style of integration, the more second nature it becomes), the double angle formula for \cos2\theta is often needed, where the 2 may be changed for another value.

    This is my first time using latex by the way .
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    (Original post by Craig1998)
    This is my first time using latex by the way .
    Lovely job!
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    (Original post by Alby1234)
    Very quick complex number question, when the locus is a half-line, how do you know which way the line 'goes'? i.e. above the starting point or below? For example: Arg(z+1-2i) = pi/3
    Treat the angle the same as you normally treat arguments.

    So like the domain is  \big( - \dfrac{ \pi}{2} , \dfrac{ \pi}{2} \big]
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    Could someone help me on how to solve part b, the markscheme doesnt make sense to me. Where does the -pi/9*2^2 part come from?

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