HSC 2013 MX2 Marathon (archive) (4 Viewers)

Sy123 · Jun 25, 2013

Re: HSC 2013 4U Marathon

Here is a neat question I made, its a little hand-holdy I guess.

$$i) Consider the polynomial$ \ \ P(z) = z^m -1$

$$The roots of the polynomial is$ \ \ z_1, z_2, z_3, \dots , z_m$

$$Find these roots$ \ \ \ \fbox{1}$

$$ii) Let$ \ \ S_k = z_1^k + z_2^k + z_3^k + \dots + z_m^k$

$Prove that$

$S_k = \begin{cases} 0 \ \ \ $or$ \\ m \end{cases}$

$$Find for which positive integers$ \ \ k \ \ $results in either case$ \ \ \ \ \fbox{2}$

$iii) The Binomial theorem is given by$

$(1+x)^a = \binom{a}{0} + \binom{a}{1}x + \binom{a}{2}x^2 + \dots + \binom{a}{a} x^a$

$Prove that$

$\sum_{k=0}^n \binom{mn}{mk} = 2^{nm} \sum_{k=1}^{m} \cos^{nm} \left(\frac{\pi}{m}k \right) \cos(\pi nk) \ \ \ \ \ \fbox{4}$

$iv) Find$

$\sum_{k=0}^{n} \binom{3n}{3k} \ \ \ \fbox{1}$

Sy123 · Jun 26, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Consider for some positive real values$ \ \ a, \ \ b, \ \ k$

$$Define the sequence$ \ \ \{S_n \}$

$S_n = \frac{a}{kn+b} + \frac{a}{kn+2b} + \frac{a}{kn+3b} + \dots + \frac{a}{n(k+b)}$

$i) Prove that$

$0 < S_n < \frac{a}{k} \ \ \ \fbox{2}$

$ii) Prove that$

$\frac{b}{t+b} < \int_t^{t+b} \frac{dx}{x} < \frac{b}{t} \ \ \ \fbox{1}$

$iii) Hence prove that$

$\frac{a}{b} \ln \left(\frac{n(k+b)+b}{kn+b} \right) < S_n < \frac{a}{b} \ln \left(\frac{b+k}{k} \right)$

$$Hence find$ \ \ \lim_{n \to \infty}S_n \ \ \ \ \fbox{4}$

i) All variables within S_k is positive, all operations are positive, hence S_k > 0

$S_k = \frac{a}{k} \left(\frac{1}{n+ b/k} + \frac{1}{n+ 2b/k} + \frac{1}{3b/k} + \dots+ \frac{1}{n+ nb/k} \right) < \frac{a}{k} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n} \right) = \frac{a}{k}$

ii) Consider the upper and lower rectangle of the curve y=1/x, from x=t and x=t+b

iii) Consider the lower bound and the middle term for (ii)
Substitute, t=kn, t=kn+b, t=kn+2b, ... , t=kn+(n-1)b, add them all side by side:

$\ln\frac{b+k}{k} > \frac{b}{a}S_n$

Take the upper bound and middle term for (ii)
Substitute, t=kn+b, t=kn+2n, ... , t=kn+nb, and add them side by side, note that the logarithms telescope and cancel out

$\frac{b}{a} S_n > \ln \frac{n(k+b) + b}{kn+b}$

Then make the inequality into one whole one with b/a S_n in the middle, multiply everything by a/b to yield what is needed. Then taking the limit to infinity, makes the lower bound converge to the upper bound, by the squeeze theorem the middle term then becomes the upper and lower bound's limit.

$\lim_{n \to \infty}S_n = \frac{a}{n} \ln \left( \frac{b+k}{k} \right )$

Sy123 said:
If you don't mind me expanding on this

$ii) Using part (i), prove that$

$e = \sum_{k=0}^{n} \frac{1}{k!} + \frac{(-1)^{n+1}}{n!}e \int_0^1 t^n e^{-t} \ dt \ \ \ \fbox{3}$

$Let$

$J_n = \int_0^{1} t^n e^{-t} \ dt$

$iii) Show that$

$0 < J_n < \frac{1}{n+1} \ \ \ \fbox{1}$

$iv) Hence prove that$

$e = \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \ \ \ \fbox{1}$

$v) Prove by mathematical induction$

$\frac{1}{(n+1)!} < e - \sum_{k=0}^n \frac{1}{k!} < \frac{1}{n \cdot n!} \ \ \ \fbox{2}$

$Assume that there exists some integers$ \ \ p \ \ $and$ \ \ q \ \ $ with no common factor such that$

$e= \frac{p}{q}$

$vi) Prove that$

$\frac{1}{n+1} < \frac{p}{q} n! - N < \frac{1}{n} \ \ \ \fbox{1}$

$$For some integer$ \ \ N$

$$vii) Hence prove that$ \ \ e \ \ $is irrational$ \ \ \ \ \fbox{2}$

ii) Make f(t) = e^t. Then simplify out, once integral is simplified, sub in x=1.

iii) $t^n e^{-t} < t^n$

Integrate both sides from 0 to 1. Both are positive over this interval.

iv) (iii), proves that the error integral in (ii) converges to zero as n approaches infinity (by squeeze theorem in part (iii). Hence as n approaches infinity, for part (ii) the integral converges to zero and the sum remains.

v) general inductive proof, we do not refer to the lower bound in this proof, because due to the infinite series in iv)

$e -\sum_{k=0}^{n} \frac{1}{k!} = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \dots > \frac{1}{(n+1)!}$ by default since all terms are positive.

The upper bound is proven by induction, by using the infinite series in the assumed step, then subtracting 1/(m+1)! to both sides of the upper bound inequality.

vi) From what is proven in part (v), multiply everything by n!. The sum in the middle becomes a series of integers, hence is N.

viii) Let e=p/q, p and q are integers, since p and q are integers and n! = n(n-1)(n-2)(n-3)....(q)....(3)(2)(1). n! is divisible by q and hence p/q n! is an integer.
But since N is an integer therefore p/qn! - N is an integer, BUT the lower bound is 1/(n+1) < 1 and the upper bound is 1/n < 1. Therefore the middle CANNOT be an integer.
Hence a contradiction has arisen due to letting e=p/q

Therefore by proof by contradiction, e is irrational.

Sy123 · Jun 26, 2013

Re: HSC 2013 4U Marathon

$Prove that$

$\left(1- \frac{1}{4} \right) \left(1- \frac{1}{9} \right ) \left( 1 - \frac{1}{16} \right ) \dots \left( 1 - \frac{1}{n^2} \right ) > \frac{1}{2} \ \ \ \ \fbox{2}$

(added in mark count as an estimated length of soln)

asianese · Jun 26, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\left(1- \frac{1}{4} \right) \left(1- \frac{1}{9} \right ) \left( 1 - \frac{1}{16} \right ) \dots \left( 1 - \frac{1}{n^2} \right ) > \frac{1}{2}$

I'm guessing you're not looking for a proof by induction?

Sy123 · Jun 26, 2013

Re: HSC 2013 4U Marathon

asianese said:
I'm guessing you're not looking for a proof by induction?

Yep.
If I want an inductive solution, I will specify that in the question

RealiseNothing · Jun 26, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\left(1- \frac{1}{4} \right) \left(1- \frac{1}{9} \right ) \left( 1 - \frac{1}{16} \right ) \dots \left( 1 - \frac{1}{n^2} \right ) > \frac{1}{2}$

If you simplify the LHS, it becomes:

$\prod_{k=2}^{n} \frac{k^2-1}{k^2} > \frac{1}{2}$

My first thought is to construct an infinite product that actually equals to $\frac{1}{2}$ . The most simple form would be:

$\frac{1}{2^{a_2}} \times \frac{1}{2^{a_3}} \times \frac{1}{2^{a_4}} \times ...$

Multiplying all the 2's together will add the powers to get:

$\frac{1}{2^{a_2 + a_3 + a_4 + ...}}$

Now for this to equal $\frac{1}{2}$ , we want:

$a_2 + a_3 + a_4 + ... = 1$

The easiest way to do this is to let:

$a_j = \frac{1}{2^{j-1}}$ such that:

$a_2 + a_3 + a_4 + ... = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 1$

So we get the infinite product:

$\prod_{j=2}^{\infty} \frac{1}{2^{\frac{1}{2^{j-1}}}} = \frac{1}{2}$

Which in expanded terms is:

$\frac{1}{2^{\frac{1}{2}}} \times \frac{1}{2^{\frac{1}{4}}} \times \frac{1}{2^{\frac{1}{8}}} \times ... = \frac{1}{2}$

Now lets consider what we can do with this infinite product:

1) Since all terms are less than 1, then by multiplying by an extra term decreases the actual product each time. Hence the product will be a minimum when there are infinite terms.

2) Since the product is a minimum when there are infinite terms, then all we have to do is show that the infinite product of the result that needs to be proven is larger than the infinite product we constructed.

Now trying to figure out how to finish this off...

Sy123 · Jun 26, 2013

Re: HSC 2013 4U Marathon

RealiseNothing said:
If you simplify the LHS, it becomes:

$\prod_{k=2}^{n} \frac{k^2-1}{k^2} > \frac{1}{2}$

My first thought is to construct an infinite product that actually equals to $\frac{1}{2}$ . The most simple form would be:

$\frac{1}{2^{a_2}} \times \frac{1}{2^{a_3}} \times \frac{1}{2^{a_4}} \times ...$

Multiplying all the 2's together will add the powers to get:

$\frac{1}{2^{a_2 + a_3 + a_4 + ...}}$

Now for this to equal $\frac{1}{2}$ , we want:

$a_2 + a_3 + a_4 + ... = 1$

The easiest way to do this is to let:

$a_j = \frac{1}{2^{j-1}}$ such that:

$a_2 + a_3 + a_4 + ... = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = 1$

So we get the infinite product:

$\prod_{j=2}^{\infty} \frac{1}{2^{\frac{1}{2^{j-1}}}} = \frac{1}{2}$

Which in expanded terms is:

$\frac{1}{2^{\frac{1}{2}}} \times \frac{1}{2^{\frac{1}{4}}} \times \frac{1}{2^{\frac{1}{8}}} \times ... = \frac{1}{2}$

Now lets consider what we can do with this infinite product:

1) Since all terms are less than 1, then by multiplying by an extra term decreases the actual product each time. Hence the product will be a minimum when there are infinite terms.

2) Since the product is a minimum when there are infinite terms, then all we have to do is show that the infinite product of the result that needs to be proven is larger than the infinite product we constructed.

Now trying to figure out how to finish this off...

Quite clever, I'm unsure about how to prove that:

$1-\frac{1}{x^2} > \frac{1}{2^{2^{\frac{1}{j-1}}}}$

Maybe using calculus, taking f(x) = LHS - RHS, then differentiating etc etc.
But differentiating the RHS would be..... yeah lol
EDIT: I guess logarithmic differentiation would be good. But I'll let you continue.
Its an interesting solution though.

seanieg89 · Jun 27, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Prove that$

$\left(1- \frac{1}{4} \right) \left(1- \frac{1}{9} \right ) \left( 1 - \frac{1}{16} \right ) \dots \left( 1 - \frac{1}{n^2} \right ) > \frac{1}{2} \ \ \ \ \fbox{2}$

(added in mark count as an estimated length of soln)

$\displaystyle\prod_{k=2}^n \frac{k^2-1}{k^2}=\frac{\left(\displaystyle\prod_{k=2}^n (k-1)\right)\left(\displaystyle\prod_{k=2}^n (k+1)\right)}{(n!)^2}=\frac{n+1}{2n}>\frac{1}{2}.$

Sy123 · Jun 27, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
$\displaystyle\prod_{k=2}^n \frac{k^2-1}{k^2}=\frac{\left(\displaystyle\prod_{k=2}^n (k-1)\right)\left(\displaystyle\prod_{k=2}^n (k+1)\right)}{(n!)^2}=\frac{n+1}{2n}>\frac{1}{2}.$

Similar to what I had in mind:

$\prod \frac{k^2-1}{k^2} = \prod \frac{k-1}{k} \prod \frac{k+1}{k}$ we are allowed to do this since order in multiplication does not matter.

The seperate products all cancel out if you expand it out to yield n+1/2n > 1/2

Sy123 · Jun 27, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$For some positive real values$ \ \ \{x_k \}$

$Prove that$

$\frac{1}{n} \sum_{k=1}^{n} x_k \geq \left(\frac{1}{n} \sum_{k=1}^n x_k^{-1} \right)^{-1}$

$State the condition for equality$

(don't feel the obligation to answer the questions I'm posting in order, answer whichever ones you want to)

LHS - RHS, make it a common denominator, the result simplifies into proving:

$(x_1+x_2+ \dots + x_n) (\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}) \geq n^2$

This can be done by expanding the LHS, you will get, an n number of 1's
And $\binom{n}{2}$ pairs of reciprocals, i.e. x1/x2 + x2/x1, an nC2 number of these pairs

And we know that:

$a+\frac{1}{a} \geq 2$

That means in the above expression:

$LHS\geq \binom{n}{2} \times 2 + n = n^2$

proof is then complete.

seanieg89 · Jun 27, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
LHS - RHS, make it a common denominator, the result simplifies into proving:

$(x_1+x_2+ \dots + x_n) (\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n}) \geq n^2$

This can be done by expanding the LHS, you will get, an n number of 1's
And $\binom{n}{2}$ pairs of reciprocals, i.e. x1/x2 + x2/x1, an nC2 number of these pairs

And we know that:

$a+\frac{1}{a} \geq 2$

That means in the above expression:

$LHS\geq \binom{n}{2} \times 2 + n = n^2$

proof is then complete.

(and gives you the equality condition of all numbers being equal).

Sy123 · Jun 27, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
(and gives you the equality condition of all numbers being equal).

Oops yeah hehe.
Coming from x_k/x_(k+1) = x_(k+1)/x_k, and x_k is positive. Hence x_k=x_(k+1)

Sy123 · Jun 29, 2013

Re: HSC 2013 4U Marathon

Just some inequalities, they are not in order of difficulty, nor will the answer to one necessarily help with the others:

$Prove that$

$$a)$ \ \ (a+b)(a+c)(b+c) \geq 8abc$

$$b)$ \ \ a^4+b^4+c^4 \geq abc(a+b+c)$

$$c)$ \ \ \frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

asianese · Jun 29, 2013

Re: HSC 2013 4U Marathon

$$c)$ \left( \sqrt a-\sqrt b \right)^2 \ge 0 \Rightarrow a+b-2\sqrt{ab} \ge 0 \Rightarrow a+b\ge 2\sqrt{ab}\\\\ \begin{align*} \dfrac{a+b+c+d}{4} &\ge \dfrac{2\sqrt{ab}+2\sqrt{cd}}{4} \\&= \dfrac{\sqrt{ab}+\sqrt{cd}}{2} \\&\ge \dfrac{2\sqrt{\sqrt{ab}\sqrt{cd}}}{2} \\&= \sqrt[4]{abcd}.\end{align*}$

I didn't see this method until later in the HSC year. Always used to use another method (can't remember it).

Makematics · Jun 29, 2013

Re: HSC 2013 4U Marathon

soz to kill the party but does anyone have any harder Q's on topics other than Harder 3 unit? i mean it is the HSC 4U Marathon after all!

Sy123 · Jun 29, 2013

Re: HSC 2013 4U Marathon

Makematics said:
soz to kill the party but does anyone have any harder Q's on topics other than Harder 3 unit? i mean it is the HSC 4U Marathon after all!

I think this counts under Polynomials:

Sy123 said:
$Let$

$P(x) = x^6-x^5-x^3-x^2-x$

$Q(x) = x^4-x^3-x^2 -1$

$z_1, z_2, z_3, z_4 \ \ $are roots of the polynomial$ \ \ Q(x)$

$Prove that$

$P(z_1)+P(z_2)+P(z_3)+P(z_4) = 6$

Sy123 · Jun 29, 2013

Re: HSC 2013 4U Marathon

Here is a mechanics one:

$A particle is travelling in circular motion, \\ show that its acceleration is given by$

$a = r\sqrt{\phi^2 + \omega^4}$

$Where$ \ \ \omega \ \ $is the angular velocity$

$\phi \ \ $is the angular acceleration$$

$r \ \ $is the radius of motion$$

seanieg89 · Jul 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
I think this counts under Polynomials:

As no-one else seems to be answering it:

$P(z)=Q(z)(z^2+1)+(z^2-z+1)\\ \Rightarrow \sum_{n=1}^4 P(z_n)=\sum_{n=1}^4 z_n^2 -\sum_{n=1}^4 z_n +4\\=(\sum_{n=1}^4 z_n)^2-2\sum_{m<n}z_mz_n - \sum_{n=1}^4 z_n +4 = 6.$

Sy123 · Jul 3, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
Here is a mechanics one:

$A particle is travelling in circular motion, \\ show that its acceleration is given by$

$a = r\sqrt{\phi^2 + \omega^4}$

$Where$ \ \ \omega \ \ $is the angular velocity$

$\phi \ \ $is the angular acceleration$$

$r \ \ $is the radius of motion$$

First lets model the particle travelling in circular motion with parametric equations:

$x=r\cos \theta$

$y = r\sin \theta$

Where r is the radius, and theta is the angle the particle makes with the x-axis at some time t. And x and y are the x and y displacements with respect to the origin on a cartesian plane.

Now differentiate with respect to t

$\dot{x} = - r \frac{d\theta}{dt} \sin \theta$

$\dot{y} = r\frac{d\theta}{dt} \cos \theta$

$$But$ \ \ \omega = \frac{d\theta}{dt}$

$\therefore \ \ \dot{x} = -r\omega \sin \theta$

$\dot{y} = r \omega \cos \theta$

Then differentiate again, BUT:

$\phi = \frac{d^2 \theta}{dt^2} \neq 0$

Because I haven't specified non-uniform circular motion, merely circular motion, therefore, when we differentiate again, we cannot take omega as constant, and hence must use product rule

$\ddot{x} = -r(\phi \cos \theta + \omega \sin \theta \omega)$

$\ddot{y} = r(\phi \cos \theta - \omega \sin \theta \omega)$

Now, we know that in Physics, centripetal acceleration, which is (a), is towards the centre of the circular motion.

$\therefore \ a = \sqrt{\ddot{x}^2 + \ddot{y}^2}$

Substitute the $\ddot{x}$ and $\ddot{y}$ that was in terms of theta, phi and omega into this equation.
And you will in the end arrive at the above answer.

(Note that if we assume uniform circular motion, then phi = 0, and we arrive at the familiar formula a=rw^2 )

Sy123 said:
Just some inequalities, they are not in order of difficulty, nor will the answer to one necessarily help with the others:

$Prove that$

$$a)$ \ \ (a+b)(a+c)(b+c) \geq 8abc$

$$b)$ \ \ a^4+b^4+c^4 \geq abc(a+b+c)$

$$c)$ \ \ \frac{a+b+c+d}{4} \geq \sqrt[4]{abcd}$

a) (I should of specified that a > 0, b > 0, c > 0)

$a+b \geq 2\sqrt{ab}$

$a+c \geq 2\sqrt{ac}$

$b+c \geq 2\sqrt{bc}$

Multiply all inequalities together to get the answer.

b)

$x^2+y^2 \geq 2xy$

$x^2+z^2 \geq 2xz$

$y^2 + z^2 \geq 2yz$

Add all inequalities side by side:

$x^2+y^2+z^2 \geq xy+xz+yz$

Substitute:

$x=a^2 \ \ y=b^2 \ \ z= c^2$

$a^4+b^4+c^4 \geq a^2b^2+a^2c^2 + b^2c^2 \geq (ab)(ac) + (ab)(bc) + (ac)(bc) = abc(a+b+c)$

And c) asianese answered that.

Sy123 · Jul 3, 2013

Re: HSC 2013 4U Marathon

$Solve the equation$

$a^2+b^2+c^2+d^2 - ab - bc -cd - d + \frac{2}{5} = 0$

HSC 2013 MX2 Marathon (archive) (4 Viewers)

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