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In a style I've never seen before. I'm not really sure where to begin with this one. I had a go, but by the time I got to part (v) I was just scratching my head really:
They don't seem to write questions like that anymore! Does anybody know what they're looking for here (no answer provided).
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Let's first look at A. What that notation means is we are looking at the set of all points P for which the line segment PQ has length 1. Do you know what that might be?
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Sure, unless I really have forgotten everything! If we were to show it on a graph we would have a circle centre (1,1) (i.e. Q) with radius 1. The set A will therefore be infinitely large, but some members of that set will be the point (1,2), the point (1,0), the point (2,1) and the point (0, 1).
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hi Au101
Does anybody know what they're looking for here ?
Well it tests a lot of maths:
Do you understand set builder notation ?
Can you draw each locus correctly ?
Can you work out intersections and unions correctly ?
Your first answer is correct.
Post the others if you want a second opinion on your answers.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Oh, sure, okay Thanks
Well, if we were to show B on a graph it would obviously be a straight line with equation x = y. So, the gradient is 1 and the y-intercept is zero. The set B will also be infinitely large, but some members of that set will be the point (0,0), the point (1,1) (i.e. Q), the point (2,2), etc.
Now, C doesn't seem that hard either. On a graph it will be a straight vertical line with x = 2. So it is also infinitely large, but includes the point (2,0), the point (2,1) and the point (2,2).
Now, D is less generous, but I believe, on a graph, the locus would be two lines parallel to x = y. One of the lines will be at a distance of 1 unit from x = y to the "left" (x < y) and the other will be at a distance of 1 unit from x = y to the "right" (x > y). Obviously another infinitely large set. Not 100% sure but I think a couple of points in this set would be:
So, it seems to me that: A and B will have two members in common, since the straight line that is the locus of B passes through the centre of the circle that is the locus of A.
B and C will have one member in common: the point (2,2).
A and C will have one member in common: the point (2,1).
There is no element common to all of A, B and C.
The two lines which form the locus of D will cross the line x = 2 at one point each, so the C and D will have two members in common.
B and D can have no members in common.
A and D will, I think, have four members in common - two on each of the lines which form the locus of D.
I don't think there will be an element common to all of A, C and D.
Now, how to draw that? Assuming I'm right!
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hi Au101,
Here's a diagram for all of those:
I don't think your points are right yet.
I'll just check.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Ooooh that's lovely, and now I see it all laid out nicely I can see how I was wrong about D in that A and D will only have two members in common - one on each of the lines, because of course, they must be tangents to the circle that is the locus of A. The points I suggest as elements of D came from choosing the point (0,0) (which is on the line x = y).
I proceeded as follows: The perpendicular distance between the point (0,0) and the line D1 is 1. Thus we can draw a right angle triangle - can we not? - with hypotenuse 1 and the other two sides of length a:
By inspection it is clear, then, that the point on D1 which is 1 unit from (0,0) is (-1/√2, 1/√2).
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Thanks for that calculation. That makes it very clear.
Now for E.
(1) Get A intersect B
(2) Get A intersect D
(3) Then form the union of those two answers.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Well indeed, now comes the tricky bit - I assumed I was supposed to convert all of this lovely (hard-earned ) information into a Venn diagram. And this is where Au101's brain goes: well forget that, I'm off for a cup of tea!
Er, this is what I have, assuming I've done this uploading an image thing right
On second thoughts, how does one upload an image
Let's try:
Last edited by Au101 (2015-01-05 07:34:19)
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For this question, you don't need a Venn diagram. The answer should be a set containing members. So just identify the members and list them like this
{answer 1, answer 2, .....}
Bob
Got to log off until later........
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hmmmm, well, that's what I took "illustrate the following sets on the same diagram" to mean, but if you think we just need the elements of the set:
Then it should be easy enough to work them out, I get:
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I think the diagram is the graph on post 6.
Just checking your co-ords.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Co-ords good too.
B
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thank you both very very much
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