question - help is appreciated (1 Viewer)

lyounamu

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Q. Elections are held for the positions of Chairperson and Secretary in a committee of ten people seated around this table. What is the probability that the two people elected are sitting directly opposite each other? Give brief reason for your answer .
 

munch0r

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is the answer 1/9? i'm currently doubting it but meh
 

lyounamu

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munch0r said:
is the answer 1/9?
Yep. But I was a bit confused...about the wording of the question.

Can you go a little depth of explanation?
 

u-borat

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yeah 1/9

you place either CP or secretary at a seat, there are 9 seats left, and there's a 1/9 chance that the person opposite is the CP/secretary.
 

munch0r

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i just did 8!/9!
number of perms with them sitting opposite / number of total perms
 

lyounamu

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Q. Divide the polynomial f(x) = 2x^4 - 10x^3 + 12x^2 + 2x -3 by g(x) = x^2 - 3x + 1

Q. Hence write f(x) = g(x) . q(x) + r(x) where q(x) and r(x) are polynomials and r(x) has degree less than 2.

Q. Hence show that f(x) and g(x) have no zeros in common.

I did all questions that are not highlighted. But I don't really understand how I should approach the last question.

Only solution that I can think of (and I did) was that I showed that g(x) has 0 at certain points and f(x) doesn't have 0 at that point. Would I be correct if I approached that way?

Thanks.
 

12o9

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lyounamu said:
Q. Divide the polynomial f(x) = 2x^4 - 10x^3 + 12x^2 + 2x -3 by g(x) = x^2 - 3x + 1

Q. Hence write f(x) = g(x) . q(x) + r(x) where q(x) and r(x) are polynomials and r(x) has degree less than 2.

Q. Hence show that f(x) and g(x) have no zeros in common.

I did all questions that are not highlighted. But I don't really understand how I should approach the last question.

Only solution that I can think of (and I did) was that I showed that g(x) has 0 at certain points and f(x) doesn't have 0 at that point. Would I be correct if I approached that way?

Thanks.
Since it's a hence question you would have to relate it to part b).

I would have assumed that there was a common root.
So if a root of g(x) is a root of f(x) [let's call the root 'a' from now on], then g(x).q(x) would have to equal zero
Since we have assumed that 'a' is a root of f(x), then r(x) would also have to equal zero.

so a = 5/4

I would have then subbed it back into g(x) to prove that it wasn't a root. Hence proving the assumption wrong

Edit:
 
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lyounamu

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12o9 said:
Since it's a hence question you would have to relate it to part b).

I would have assumed that there was a common root.
So if a root of g(x) is a root of f(x) [let's call the root 'a' from now on], then g(x).q(x) would have to equal zero
Since we have assumed that 'a' is a root of f(x), then r(x) would also have to equal zero.

so a = 5/4

I would have then subbed it back into g(x) to prove that it wasn't a root. Hence proving the assumption wrong

Edit:
Yep, yep. That's great. Thanks for your help.
 

lyounamu

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Here is a question. I got all the answers corrects except the last section of this.

Help is appreciated:

Q:

Consider the function f(x) = 1/4 ((x-1)^2 + 7)

i) Sketch the function y = f(x), showing any intercepts with the axes, and the coordinates of tis vertex.

ii) What is the largest domain containing the vluae x=3, for which the function has an inverse funciton f-1(x)?

iii) Sketch the graph of y=f-1(x) on the same set of axes as your graph in part (i). Label two graphs clearly.

iv) What is the doamin of the inverse function?

v) Let A be the a real number not in the domain found in part (ii). Find f-1(f(a))

Thanks in advance.
 

QuLiT

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is the answer 1 - sqrt(4a-7) if not i don't get the question:p
 

QuLiT

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ok well, you know that at a point a which is a<1 a has the same value y value of another point, an equal length from the axis of the parabola but in the other direction. so the distance from a to the axis of the parabola is 1-a and therefore the point 1-a away from x= 1 is 1+ 1-a = 2-a and since f-1(f(b)) for a value in the domain is = b therefore if f(a) = f(2-a) and f(2-a) is the domain then f-1(f(2-a)) = 2 - a
 
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lyounamu

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QuLiT said:
ok well, you know that at a point a which is a<1 a has the same value y value of another point, an equal length from the axis of the parabola but in the other direction. so the distance from a to the axis of the parabola is 1-a and therefore the point 1-a away from x= 1 is 1+ 1-a = 2-a
I am sorry, I am a bit confused. Do you mind writing a solution right? I am sorry for bothering you. :(
 

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