HSC 2013 MX2 Marathon (archive) (7 Viewers)

[gustavo28]

Re: HSC 2013 4U Marathon

Wait yea nevermind my method was a little different to yours, it involved a proof by contradiction (kinda).
Well done
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So, substitute x=y=1 to get that f(1)=0
Set y=1+t/x into the identity to get f(x+t)=f(x)+f(1+t/x).
So, f(x+t)-f(x)=f(1+t/x)-f(1)
So, x(f(x+t)-f(x))/t=(f(1+t/x)-f(1))/(t/x)

So, as t goes to 0, the RHS is f'(1), while the LHS is the limit of x((f(x+t)-f(x))/t as t goes to 0. Thus, since this limit is equal to 3, then f is differentiable at all positive x. So, xf'(x)=3 for all positive x. So, f(x)=3ln(x)+c. Set x=1 to get that f(1)=c=0. So, f(x)=3ln(x) and then we can check this.

 

[Sy123]

Re: HSC 2013 4U Marathon

So, substitute x=y=1 to get that f(1)=0
Set y=1+t/x into the identity to get f(x+t)=f(x)+f(1+t/x).
So, f(x+t)-f(x)=f(1+t/x)-f(1)
So, x(f(x+t)-f(x))/t=(f(1+t/x)-f(1))/(t/x)

So, as t goes to 0, the RHS is f'(1), while the LHS is the limit of x((f(x+t)-f(x))/t as t goes to 0. Thus, since this limit is equal to 3, then f is differentiable at all positive x. So, xf'(x)=3 for all positive x. So, f(x)=3ln(x)+c. Set x=1 to get that f(1)=c=0. So, f(x)=3ln(x) and then we can check this.

Nice method, alternatively we can differentiate with respect to x (we aren't considering an x-y plane anyway)
To get:



Then set x=1 to get



then solve

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

Nice method, alternatively we can differentiate with respect to x (we aren't considering an x-y plane anyway)
To get:



Then set x=1 to get



then solve

Could be a slight issue because how do we know that f(xy) is differentiable at other places other than x=1?

Nvm, missed that you set x=1, so it's cool.

 

[rural juror]

Re: HSC 2013 4U Marathon

For I, I'm getting:



Which is a little different to yours

The inequality is still wrong though my bad, it should be:

oh yhy, i wrote the wrong thing online

 

[ColdLipstick]

Re: HSC 2013 4U Marathon

EDIT: Was an error in the image.

 

[HeroicPandas]

Re: HSC 2013 4U Marathon

EDIT: Was an error in the image.

i've seen this somewhere...Cambridge 4U?

 

[Sy123]

Re: HSC 2013 4U Marathon

EDIT: Was an error in the image.

let be the double root

 

[gustavo28]

Re: HSC 2013 4U Marathon

Similarly,

Adding these two produce the required inequality.

 

[gustavo28]

Re: HSC 2013 4U Marathon

Nice method, alternatively we can differentiate with respect to x (we aren't considering an x-y plane anyway)
To get:



Then set x=1 to get



then solve

Even though you set x=1, I think differentiability might have to be proven first to reach that conclusion.

 

[Sy123]

Re: HSC 2013 4U Marathon

Even though you set x=1, I think differentiability might have to be proven first to reach that conclusion.

I got the question from art of problem solving problem number 9.3.18, and it assumes differentiability, is there a way to oprove that it is differentiable using only the property?

(can someone post a question please)

 

[gustavo28]

Re: HSC 2013 4U Marathon

Prove that the number of subsets of {1,2,...,n } with even cardinality is equal to the number of subsets of odd cardinality of the same set.

 

[Sy123]

Re: HSC 2013 4U Marathon

Prove that the number of subsets of {1,2,...,n } with even cardinality is equal to the number of subsets of odd cardinality of the same set.

If I understood the definition of cardinality and subsets correctly, we simply need to prove that



Since C(n,m) is the number of ways to pick m elements from a set of n.



Take negatives to other sides to yield the desired equality.

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[gustavo28]

Re: HSC 2013 4U Marathon

If Alice is allowed to toss a coin 10 times and Bob 9 times, then what is the probability that Alice gets more heads than Bob?

 

[RealiseNothing]

Re: HSC 2013 4U Marathon

If Alice is allowed to toss a coin 10 times and Bob 9 times, then what is the probability that Alice gets more head than Bob?

0, Alice is a girl.

 

[asianese]

Re: HSC 2013 4U Marathon

Jesus..

 

[Carrotsticks]

Re: HSC 2013 4U Marathon

0, Alice is a girl.

lol christ

 

[ColdLipstick]

Re: HSC 2013 4U Marathon

 

[gustavo28]

Re: HSC 2013 4U Marathon

So close, but why "must" we have Q=pR? Why must any Q be divisible by p? This needs proof but otherwise good.

So, first of all suppose there is a monic rational quadratic with the root X+1/X with X as the cube root of a non-cube rational. From now on this root will be known as T.
Let this quadratic be x^2+ax+b. So, (x-a)(x^2+ax+b)=x^3+ax^2+bx-ax^2-a^2x-ab=x^3+x(b-a^2)-ab=Q.

So, using the P that has been referred to previously, we get P-Q=some linear rational polynomial which also must have T as a root. But this cant have the irrational root T.

Now, rational polynomials of degree 1 and 2 cant have T as a root then. Suppose some rational polynomial S(x) has T as a root. Note it must be of degree greater than or equal to 2. So, S(x)=M(x)P(x)+R(x) where deg R is 2 or less. Now, R has T as a root. But R can't be a constant, linear or quadratic rational polynomial. So R=0 and P has to divide S.

 

[Sy123]

Re: HSC 2013 4U Marathon

If Alice is allowed to toss a coin 10 times and Bob 9 times, then what is the probability that Alice gets more heads than Bob?

1/2 ?

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[seanieg89]

Re: HSC 2013 4U Marathon

So, first of all suppose there is a monic rational quadratic with the root X+1/X with X as the cube root of a non-cube rational. From now on this root will be known as T.
Let this quadratic be x^2+ax+b. So, (x-a)(x^2+ax+b)=x^3+ax^2+bx-ax^2-a^2x-ab=x^3+x(b-a^2)-ab=Q.

So, using the P that has been referred to previously, we get P-Q=some linear rational polynomial which also must have T as a root. But this cant have the irrational root T.

Now, rational polynomials of degree 1 and 2 cant have T as a root then. Suppose some rational polynomial S(x) has T as a root. Note it must be of degree greater than or equal to 2. So, S(x)=M(x)P(x)+R(x) where deg R is 2 or less. Now, R has T as a root. But R can't be a constant, linear or quadratic rational polynomial. So R=0 and P has to divide S.

Exactly, well done. Are you a student or a teacher?