C3: June 2005 Q6 Help! Watch

adellie
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#1
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I was doing the edexcel June 2005 and I'm a bit confused with Q6. It's regarding the modulus sign and graphs so I can't type the question up! Instead, it can be found here ...

http://www.mathspapers.co.uk/Papers/edex/C3Jun05Q.pdf

I'd be very grateful if someone could explain how to (and a quick reasoning behind each step) on how to get the answers for 6(c) and (d).

I do have the markscheme but it isn't very detailed.

Thank you!
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Plato123
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c) sub  x=1 into  f(x) to get a

0 = |x-1| - 2 to get b
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linus_k
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last time i gave an answer i got told off, so i shall miss bits:

(c) at y=b, x=0 \implies |x-1| = 1 \implies f(x) = \ldots
at  y=a, x=1 \implies |x-1| = 0 \implies f(x) = \ldots

(d) the best way is to this of the graph as two different strait line graphs, one for which x-1 > 0, one for x-1 <= 0, and consider the two separate equations
x-1-2=5x \quad(x-1&gt;0)
1-x-2=5x \quad (x-1\leq 0)
only one of which has a solution satisfying the inequality in brackets

(edited to remove modulus signs)
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adellie
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(Original post by linus_k)
last time i gave an answer i got told off, so i shall miss bits:

(c) at y=b, x=0 \implies |x-1| = 1 \implies f(x) = \ldots
at  y=a, x=1 \implies |x-1| = 0 \implies f(x) = \ldots

(d) the best way is to this of the graph as two different strait line graphs, one for which x-1 > 0, one for x-1 <= 0, and consider the two separate equations
x-1-2=5x \quad(x-1&gt;0)
1-x-2=5x \quad (x-1\leq 0)
only one of which has a solution satisfying the inequality in brackets

(edited to remove modulus signs)
Thanks for the hints!

So as far I see it, I just use the modulus sign for how it should be used in (c)!

However, I don't quite get how the graph would look like in (d) nor how inequlities would come into this...

Should I draw these line on a graph:
x-1-2=y
1-x-2=y
5x=y

?

... I attempted to just solve: 1-x-2=5x and x-1-2=5x for which I got -1/6 and -3/4 respectively. However, only -1/6 is specified as the answer! I'm somehow thinking that this would be solved by drawing graphs or something but as I've mentioned I'm certaion of what graphs to draw?

On a side note, how come i've reversed the signs for x and y in 1-x-2 ?

Thanks for the help so far, I've really appreciated it.
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adellie
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(Original post by Lou Reed)
for d) you could do it using graphs...i mean that, if you draw the line y=5x, and notice that it crosses y=f(x) on the first line segment. Using the values for a and b you can work out the equation of this line to be y = x - 1. So by setting 5x = x - 1 you get, again, x = -1/6.

The -3/4 is from presuming that the second line segment continues (but it doesnt in the graph!)

thats how im understanding it anyways

"On a side note, how come i've reversed the signs for x and y in 1-x-2"

|x - 1| = +(x - 1) AND -(x - 1) thats just the nature of the modulus (someone can probably give you a better reason tho).
Thanks! ... I think I was really confused becuase I thoght the graph for y=5x was the line x=5 ... my mistake :p: This does clear it up though!
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