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HSC 2013 MX2 Marathon (archive) (3 Viewers)

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RealiseNothing

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Re: HSC 2013 4U Marathon

Consider the quadratic:



If one of the roots of the quadratic is where is not real.

Find the range of values for

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Consider the quadratic:



If one of the roots of the quadratic is where is not real.

Find the range of values for

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.
Like I know what the answer should be, but I did it a different way and got something different. Trying to work out why that is.
 

seanieg89

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Re: HSC 2013 4U Marathon

Consider the quadratic:



If one of the roots of the quadratic is where is not real.

Find the range of values for

I got my answer which I found interesting but I'm not sure if it's 100% correct. I want to see if anyone gets the same as me.
 

Sy123

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Re: HSC 2013 4U Marathon

























Really just some random results, might have done a few parts before.
Great question there, here is my attempt:

iii)

Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j.
The number of arrangements of k after j and k before j are equal.

Hence the number of arrangements are total/2


iv)

We must arrange n letters into n slots.

If k and j have no spaces between them, there are n-1 ways they can be arranged ( i.e. _ _ _ k j _ _ .. _ _ )
If k and j hve 1 space between them, there are n-2 ways to arrange them (i.e. _ _ k _ j _ _ .. _ _ )

and so on.

So if we consider the number of ways we can arrange k and j, such that k is before j, when they are both together, 1 space apart, 2 spaces ... n-2 spaces apart the number of arrangements is 1 + 2 + 3 + ... + n-1 = T_(n-1). Arrange other letters in (n-2)! ways
Hence total arrangemetns is: T_(n-1) (n-2)!

v)

They both are the number of arrangements of same scenario hence they are equal:



vi)



vii) Using the result from part ii



viii) From part v



From part ii





Hence when we sub it in, we get recurrence.

 
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seanieg89

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Re: HSC 2013 4U Marathon

What was your method that produced an incorrect answer?
 

seanieg89

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Re: HSC 2013 4U Marathon

Oh right you said not real, yeah, then just take the complement.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Great question there, here is my attempt:

iii)

Two letters k, and j, can only be arranged in a set of n letters, either that k is BEFORE j, or k is AFTER j.
The number of arrangements of k after j and k before j are equal.

Hence the number of arrangements are total/2


iv)

We must arrange n letters into n slots.

If k and j have no spaces between them, there are n-1 ways they can be arranged ( i.e. _ _ _ k j _ _ .. _ _ )
If k and j hve 1 space between them, there are n-2 ways to arrange them (i.e. _ _ k _ j _ _ .. _ _ )

and so on.

So if we consider the number of ways we can arrange k and j, such that k is before j, when they are both together, 1 space apart, 2 spaces ... n-2 spaces apart the number of arrangements is 1 + 2 + 3 + ... + n-1 = T_(n-1). Arrange other letters in (n-2)! ways
Hence total arrangemetns is: T_(n-1) (n-2)!

v)

They both are the number of arrangements of same scenario hence they are equal:



vi)



vii) Using the result from part ii



viii) From part v



From part ii





Hence when we sub it in, we get recurrence.

That was quick lol, very nice.
 

GoldyOrNugget

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Re: HSC 2013 4U Marathon

























Really just some random results, might have done a few parts before.
iii) Let the two letters be a and b. In exactly half the permutations, a will come before b. There are n! total permutations, therefore a can come before b in n!/2 ways. (this seems very obvious to me, so I'm not sure how much more mathematically it must be phrased. If you imagine that a comes before b in more than half the permutations, then since a and b can refer to any pair of letters, we can swap their assignments so that a now refers to the b letter and vice versa, but now you'd have a coming before b in less than half the permutations, which is a contradiction. Similarly if you start off by assuming that a comes before b in less than half the permutations.)

iv) Choose the 2 spots where we will place a and b in that order, then permute the others in the remaining spots: nC2 * (n-2)!. Now nC2 can be visualised by picking one item from n, then picking the second one from the remaning (n-1) items, then dividing by 2 to eliminate duplicates -- i.e. n(n-1)/2, which we know is Tn-1 (from arithmetic series, or i and ii), so an equivalent expression for iii) is Tn-1 * nC2 as required.

v) The two methods solve the same problem, so the expressions can be equated. Then multiply both sides by 2.







EDIT: aww Sy already answered it :( it takes so long to type this shit up...
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Oh right you said not real, yeah, then just take the complement.
Here's what I did which ended up wrong (I think?)

Since one root is , the other must be

So we get:





Using:

and adding the two equation together we obtain:



Using the identity:



We get a quadratic in which when solved using the quadratic formula gives:

or

Now the first solution works for all , but the second solution doesn't as for certain values of A within the domain of

Hence solving for A:





Which is a different result to

(note I forgot to add the fact that )

What's wrong with this solution?
 
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Sy123

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Re: HSC 2013 4U Marathon

Here's what I did which ended up wrong (I think?)

Since one root is , the other must be

So we get:





Using:

and adding the two equation together we obtain:

2cos2\theta + 2Acos\theta + 1 = 0

Using the identity:



We get a quadratic in which when solved using the quadratic formula gives:

or

Now the first solution works for all , but the second solution doesn't as for certain values of A within the domain of

Hence solving for A:





Which is a different result to

(note I forgot to add the fact that )

What's wrong with this solution?
Bolded what was wrong, when you add the two equations, you get:



when you use that formula, you get:



Intuitive method though
 

seanieg89

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Re: HSC 2013 4U Marathon



? Assuming theta is meant to be the argument of z, this is not true.
 

RealiseNothing

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Re: HSC 2013 4U Marathon

Bolded what was wrong, when you add the two equations, you get:



when you use that formula, you get:



Intuitive method though
Woops that was a typo, hit 1 instead of 2. In my working I did have +2, and it worked out to be what I had.
 

seanieg89

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Re: HSC 2013 4U Marathon

Shortest way I can think of is:

1. Whatever we choose A to be the poly has two complex roots counting multiplicity.

2. One of them being non-real is the opposite of both of them being real.

3. Both of them are real <=> A=-(r+1/r) for some real r.

4. The range of -(r+1/r) is |A| >= 2, A real. (By AM-GM or calculus or w/e).

5. Take complements.

6. Profit.
 

seanieg89

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Re: HSC 2013 4U Marathon

How do we know the roots are of unit modulus?
 
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