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Edexcel FP2 Official 2016 Exam Thread - 8th June 2016

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Avatar for edothero
edothero
OP
15
Original post by Zacken
sqrt((x1 - x2)^2 + (y1-y2)^2) give the distance between (x1, y1) and (x2, y2).


Ah of course, thanks
In the equalities topic how would you know how to sketch this graph? would you literally just have to sub in loads of values?
FP2 graph.png
Original post by Music With Rocks
In the equalities topic how would you know how to sketch this graph? would you literally just have to sub in loads of values?
FP2 graph.png


It's in the form y = |f(x)|. Think about how you'd sketch C3 mod functions.
Original post by Ayman!
It's in the form y = |f(x)|. Think about how you'd sketch C3 mod functions.


Well the modulus causes it to be reflected in the x-axis doesn't it? Still not sure how to sketch a graph with an x term on both the denominator and numerator
Reply 504
Avatar for Rkai01
Rkai01
6
https://8fd9eafbb84fdb32c73d8e44d980d7008581d86e.googledrive.com/host/0B1ZiqBksUHNYTnpyeF8xQlZweHc/CH3.pdf

Ex I q 10 B why isn't the range x>2 and y> -4 since that point onwards is the locus but it states x>0 and y>0??
Please assistance, thanks.
Original post by Music With Rocks
Well the modulus causes it to be reflected in the x-axis doesn't it? Still not sure how to sketch a graph with an x term on both the denominator and numerator


Perhaps this might help

xx+2=x+2−2x+2=1−2x+2\displaystyle \frac{x}{x+2}= \frac{x+2 -2}{x+2}=1 -\frac{2}{x+2}.

Using the above, it's just C1 transformations.
Original post by Ayman!
Perhaps this might help

xx+2=x+2−2x+2=1−2x+2\displaystyle \frac{x}{x+2}= \frac{x+2 -2}{x+2}=1 -\frac{2}{x+2}.

Using the above, it's just C1 transformations.


That is great, it certainly helps. Thank you very much! :smile:
Is cover-up a valid method for part a's (finding partial fractions generally) in Method of Difference questions? (just writing down 'using coverup, A = 1/2, B = -1/2')
I think this might have been answered before, but how would I go about answering 9 part b)?

Screenshot_67.png
I understood that I needed to find rsinθrsin \theta and rcosθrcos \theta, and then differentiate each with respect to x, but after that, I don't know what to do :/
Reply 509
Avatar for edothero
edothero
OP
15
Original post by tripleseven
I think this might have been answered before, but how would I go about answering 9 part b)?

Screenshot_67.png
I understood that I needed to find rsinθrsin \theta and rcosθrcos \theta, and then differentiate each with respect to x, but after that, I don't know what to do :/


Find the points of where the line tangent to the pole meets the curve and the points where the line parallel to the pole meets the curve. See where it takes you from there
Original post by Student403
Is cover-up a valid method for part a's (finding partial fractions generally) in Method of Difference questions? (just writing down 'using coverup, A = 1/2, B = -1/2':wink:


Hoping so, they award like 1-2 marks just for finding the answer. The mark scheme usually has nothing but accuracy marks. :s
Original post by Ayman!
Hoping so, they award like 1-2 marks just for finding the answer. The mark scheme usually has nothing but accuracy marks. :s


Yeah I was wondering because on J09 it was 1 mark, J10 it was 2 marks, but both times it was just A marks

Thanks!
Original post by 1asdfghjkl1
for this question (i have attached the mark scheme) can someone explain where the very first line of working came from? thanks :smile:


The 6r^2+2 comes from multiplying out (r+1)^3-(r-1)^3. You can the use this identity to do the method of differences.
Original post by tripleseven
I think this might have been answered before, but how would I go about answering 9 part b)?

Screenshot_67.png
I understood that I needed to find rsinθrsin \theta and rcosθrcos \theta, and then differentiate each with respect to x, but after that, I don't know what to do :/


First, you need to find the y co-ordinate where the tangent to the curve is parallel to the initial line by differentiating y= rsinθrsin \theta. Once you've got this you can find out the total width PS by doubling your y value. To find the length PQ of the rectangle you can work it out without having to differentiate rcosθrcos \theta. When θ\theta is 0 r=3 so the length PQ is twice this which is 6. Now you've got the area of the rectangle. All you need to do now is subtract the area enclosed by the curve from part a. This is my first time using latex so I'm not quite sure what I'm doing. I just copied and pasted what you put in your comment so hopefully it works :smile:
(edited 7 years ago)
Original post by Music With Rocks
Well the modulus causes it to be reflected in the x-axis doesn't it? Still not sure how to sketch a graph with an x term on both the denominator and numerator


You can think about what the value of the function is at x=0, what the function does as x tends to +- infinity and the asymptote at x=-2
(edited 7 years ago)
Original post by Music With Rocks
In the equalities topic how would you know how to sketch this graph? would you literally just have to sub in loads of values?
FP2 graph.png

I think this is a good point, and it's the reason why in more recent papers you are prompted to 'use algebra to find the set of values of x for which...', as a fully graphical approach may give those with graphics calculators an advantage over those who take longer to work out the shape of the curve manually. Working out the shape of this curve however shouldn't take too much work, but I think it's worth pointing out that there are other approaches to modular inequalities. One that relies almost entirely on algebra is considering individually the cases that may arise according to the value of x. In this example, it would be wise to consider the sign of x/(x+2) for x > 0, x = 0, -2 < x < 0, x = -2, x < -2. Obviously, in the case x = -2 the function is not defined so we can discard this immediately. Working in this way should give the same result as the graphical approach offered by the solutionbank.
(edited 7 years ago)
I don't understand the difference between parametric curves and polar curved? Are they different or the same? Thanks


Posted from TSR Mobile
Original post by jkhan9625
I don't understand the difference between parametric curves and polar curved? Are they different or the same? Thanks


Posted from TSR Mobile

They are the same. Some curves that you meet in FP2 like the cardioid can be very difficult to describe in Cartesian coordinates, so to make life easier we introduce parameters like r and theta.
Original post by Music With Rocks
In the equalities topic how would you know how to sketch this graph? would you literally just have to sub in loads of values?
FP2 graph.png


The way I like to make sure the graph is right is using extreme cases.
So for x/(x+2) I already know it's a hyperbole and know the asymptote x = -2. Then I substitute in extreme cases, such as x = -1.999999 and x = -2.000001. Then I can sub in extremely large/small values of x, such as x = 5000 or x =-5000 and using these cases you can sketch the graph of x/(x+2). For the modulus of this you just reflect off the graph that is beneath the x axis as you would normally in C3. It's kinda tedious and you can lose more time doing this but it makes it so much easier when you're solving inequalities, as you can see exactly how many points of intersection and can roughly see where these intersections occur.
(edited 7 years ago)
In this question why is the angle not marked as 3π/43\pi/4?

fp2 arg diagram.png(apologies for rubbish quality)

I am going to be keeping you guys busy with questions as I am basically self teaching FP2 having started on Friday :tongue: you have been a great help so far

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