HSC 2013 MX2 Marathon (archive) (5 Viewers)

gustavo28 · Oct 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Wait yea nevermind my method was a little different to yours, it involved a proof by contradiction (kinda).
Well done
---

$A function is defined for$ \ \ x,y > 0 \ \ $such that$

$f(xy) = f(x) + f(y)$

$$If$ \ \ f'(1) = 3 \ \ $find$ \ \ f(x)$

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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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:

$yf'(yx) = f'(x)$

Then set x=1 to get

$f'(y) = 3/y$

then solve

Carrotsticks · Oct 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice method, alternatively we can differentiate with respect to x (we aren't considering an x-y plane anyway)
To get:

$yf'(yx) = f'(x)$

Then set x=1 to get

$f'(y) = 3/y$

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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
For I, I'm getting:

$I(m,k,n) = \frac{k^n n!}{(m+1)(m+k+1)(m+2k+1) \dots (m+(n-1)k + 1)} \times \frac{1}{m + nk + 1}$

Which is a little different to yours

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

$\left( \frac{k}{m+(n-1)k + 1} \right)^n < \frac{m+nk+1}{n!} \int_0^1 x^m (1-x^k)^n \ dx < \left( \frac{k}{m+1} \right)^n$

oh yhy, i wrote the wrong thing online

ColdLipstick · Oct 16, 2013

Re: HSC 2013 4U Marathon

EDIT: Was an error in the image.

HeroicPandas · Oct 16, 2013

Re: HSC 2013 4U Marathon

ColdLipstick said:
EDIT: Was an error in the image.

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

Sy123 · Oct 16, 2013

Re: HSC 2013 4U Marathon

ColdLipstick said:
EDIT: Was an error in the image.

let $a$ be the double root

$P'(a) = na^{n-1} + k = 0 \Rightarrow a^{n-1} = \frac{-k}{n} \ \ \fbox{1}$

$P(a) = a^n + ka - s = 0$

$a(a^{n-1} + k) - s = a(k - \frac{k}{n}) - s = 0$

$a \frac{k}{n} = \frac{s}{n-1}$

$a^{n-1} \left( \frac{k}{n} \right)^{n-1} = \left( \frac{s}{n-1} \right)^{n-1}$

$$From$ \ \ \fbox{1}$

$a^{n-1} = -k/n \Rightarrow \ \ \left(\frac{s}{n-1}\right)^{n-1} + \left(\frac{k}{n}\right)^n = 0$

gustavo28 · Oct 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$0< x, y< 1$

$Prove that$

$\frac{x}{1+y} + \frac{y}{1+x} \leq 1$

$y \leq 1$

$x+y \leq 1+x$
$\frac{y}{1+x} \leq \frac{y}{y+x}$
Similarly,
$\frac{x}{1+y} \leq \frac{x}{y+x}$
Adding these two produce the required inequality.

gustavo28 · Oct 16, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Nice method, alternatively we can differentiate with respect to x (we aren't considering an x-y plane anyway)
To get:

$yf'(yx) = f'(x)$

Then set x=1 to get

$f'(y) = 3/y$

then solve

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

Sy123 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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 · Oct 16, 2013

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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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

$\sum_{m \ \ even} \binom{n}{m} = \sum_{m \ \ odd} \binom{n}{m}$

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

$0 = (1-1)^n = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \dots +(-1)^{n} \binom{n}{n}$

Take negatives to other sides to yield the desired equality.

----

gustavo28 · Oct 16, 2013

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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

Jesus..

Carrotsticks · Oct 16, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
0, Alice is a girl.

lol christ

ColdLipstick · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 · Oct 16, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
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 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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 ?

---

$\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$

seanieg89 · Oct 16, 2013

Re: HSC 2013 4U Marathon

gustavo28 said:
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?

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