SilentWaters
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Re: HSC 2015 3U Marathon
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HSC 2015 MX1 Marathon (archive) (4 Viewers)
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Drsoccerball
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Re: HSC 2015 3U Marathon
Same problem both from my phone and different computers
Same problem both from my phone and different computers
superstar12
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Re: HSC 2015 3U Marathon
Find the zeroes of the polynomial P(x) = 2x^3 - 13x^2 + 22x - 8, given that one zero is the product of the other two.
Find the zeroes of the polynomial P(x) = 2x^3 - 13x^2 + 22x - 8, given that one zero is the product of the other two.
Ekman
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Re: HSC 2015 3U Marathon
Product of roots: (ab)^2 = 4
Therefore ab = +-2
Test P(2) and P(-2), and P(-2) doesn't work. Factorize P(x) with the knowledge that one root is 2.
P(x) = (x-2)(2x^2-9x+4) = (x-2)(2x-1)(x-4)
Therefore the roots are 2,1/2 and 4
Let a and b be the roots of P(x). Therefore roots are ab, a, b.
Product of roots: (ab)^2 = 4
Therefore ab = +-2
Test P(2) and P(-2), and P(-2) doesn't work. Factorize P(x) with the knowledge that one root is 2.
P(x) = (x-2)(2x^2-9x+4) = (x-2)(2x-1)(x-4)
Therefore the roots are 2,1/2 and 4
FrankXie
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Re: HSC 2015 3U Marathon
new question-repost my last one
new question-repost my last one
Sy123
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Re: HSC 2015 3U Marathon
 Using a graph, show that$ \\ \int_{n}^{2n}\frac{dx}{x} - \frac{1}{2n} < \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} < \int_{n}^{2n} \frac{dx}{x} )
 Hence explain why$ \ \lim_{n \to \infty} \left(\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} \right) = \ln 2 )
 Hence, show that$ \ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots + \dots + \frac{(-1)^n}{n} + \dots = \ln 2 )
(If you allow me to build off of this)new question-repost my last one
Carrotsticks
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Re: HSC 2015 3U Marathon
^ limit as n->infinity.
^ limit as n->infinity.
Ekman
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Re: HSC 2015 3U Marathon
In a simple game of chance, two players must each select 3 numbers from a group of 7 numbers, namely 1,2,3,4,5,6 and 7. The only rule of the game is that the 3 numbers each player selects, must be consecutive. What is the probability that the number 2 is the smallest number both players have in common?
In a simple game of chance, two players must each select 3 numbers from a group of 7 numbers, namely 1,2,3,4,5,6 and 7. The only rule of the game is that the 3 numbers each player selects, must be consecutive. What is the probability that the number 2 is the smallest number both players have in common?
Ekman
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Re: HSC 2015 3U Marathon
I got 1/25 as well, but apparently the answers say 3/25. I just wanted to confirm whether the answers were wrong or not. Do some past hsc questions for Perms. Usually the questions between 2010 to 2000 tend to be a bit harder, but it is good practice.
Re: HSC 2015 3U Marathon
The 3 outcomes which satisfy the criteria are:
A: (1,2,3); B: (2,3,4)
A: (2,3,4); B: (1,2,3)
A: (2,3,4); B: (2,3,4)
The answer IS 3/25.I got 1/25 as well, but apparently the answers say 3/25. I just wanted to confirm whether the answers were wrong or not. Do some past hsc questions for Perms. Usually the questions between 2010 to 2000 tend to be a bit harder, but it is good practice.
The 3 outcomes which satisfy the criteria are:
A: (1,2,3); B: (2,3,4)
A: (2,3,4); B: (1,2,3)
A: (2,3,4); B: (2,3,4)
integral95
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Re: HSC 2015 3U Marathon
 Explain why$\ \ cosy\geq0 \\ \\ \\ ii) \ \ \ $Hence show that$ \ \ \ cosy = \sqrt{1-x^2}\\ \\ \\ iii)$Show that$ \ \ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} )
Kaido
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Re: HSC 2015 3U Marathon
Ahh geez, i read it as if "2 is the smallest number both players" have. Forgot the 'common' =.=
Kaido
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Re: HSC 2015 3U Marathon
ii) x=siny on re-arranging. From double angles, cosy=sqrt (1-sin^2y)=sqrt (1-x^2)
iii) dx/dy = cosy = sqrt (1-x^2)
-> dy/dx = 1/ sqrt (1-x^2)
i) take cos of both sides. sin-1x has a range of -pi/2<=y<=pi/2 and we know that cos(-pi/2<=y<=pi/2) is always >=0
ii) x=siny on re-arranging. From double angles, cosy=sqrt (1-sin^2y)=sqrt (1-x^2)
iii) dx/dy = cosy = sqrt (1-x^2)
-> dy/dx = 1/ sqrt (1-x^2)
Kaido
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Re: HSC 2015 3U Marathon
5 people are to be selected for a committee out of 9 people, two of which is Michael Jackson and Homer Simpson
a) what's the chance of both MJ and HS being selected
b) one of them being selected
c) they both die and not being selected? (for the luls)
5 people are to be selected for a committee out of 9 people, two of which is Michael Jackson and Homer Simpson
a) what's the chance of both MJ and HS being selected
b) one of them being selected
c) they both die and not being selected? (for the luls)
Re: HSC 2015 3U Marathon
Yes, people always say they have problems with probability. But a huge fraction of the time, the issue is first reading the question correctly, and then interpreting correctly what is being asked. And I am just as guilty, especially when I read the question quickly and assume it is exactly the same as some other question I have done previously, when in fact there is a subtle difference.
Re: HSC 2015 3U Marathon
(... and you do realise MJ is already dead? ...)
You do realise this is a 2 unit probability question?
(... and you do realise MJ is already dead? ...)
Last edited:
Kaido
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Re: HSC 2015 3U Marathon
yeah im not sure about the answer, would also like an explanation
Re: HSC 2015 3U Marathon
(b) Not clear whether you mean exactly one or at least one. Assuming the first:
(MJ in / HS out) OR (MJ out / HS in): 5/9 times 4/8 + 4/9 times 4/8 = 1/2
(c) I'm stumped ...
(a) MJ's chance of being selected is 5/9. HS's chance of being selected IF MJ is in is 4/8. Multiply to give 5/18.
(b) Not clear whether you mean exactly one or at least one. Assuming the first:
(MJ in / HS out) OR (MJ out / HS in): 5/9 times 4/8 + 4/9 times 4/8 = 1/2
(c) I'm stumped ...
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