Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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

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

hope this makes sense. anyone correct me if im wrong pls

Please could someone explain exercise F question 16b on here https://8fd9eafbb84fdb32c73d8e44d980d7008581d86e.googledrive.com/host/0B1ZiqBksUHNYTnpyeF8xQlZweHc/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).

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.

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

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

Well, not always the case, but yeah, if you just draw some triangles and use trig you usually get the answer.

Can anyone explain to me how to work out the area of polar coordinates : example 10c on page 138 of edexcel fp2 book.

This is my first time using latex by the way .

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