HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

Sy123 · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

aDimitri said:
$\text{Testing for n=3, we get a triangle, and we know that the angle sum of a triangle is } \pi \\ \text{Therefore, statement is true for n=3} \\ \text{Assuming to be true for n=k} \\ \text{i.e. Angle sum of k sided polygon is given by } \pi(k-2) \\ \text{Construct a k sided polygon...} \\ \text{In order to transform it into a k+1 sided polygon, add a new vertex next to the kth side} \\ \text{It is quite clear that the only change that has been made is the addition of a new triangle adjacent to the pre-existing side} \\ \text{Hence, angle sum of k+1 sided polygon } = \pi(k-2) + \pi \\ = \pi((k+1) - 2) \\ \text{Therefore, blah blah blah true by mathematical induction}$

Yea well done

Nukeboy · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$i) For the polynomial$ \ \ P(x) = (x-a)^2 Q(x)$

$$Show that$ \ \ x=a \ \ $is a root of$ \ P'(a)$

$$ii) The polynomial$ \ \ S(y) = ay^3 + by + c \ \ $has a double zero at$ \ y= 1$

$$and has a remainder of$ \ 4 \ $when divided by$ \ \ y+1$

$$Find$ \ a,b,c$

Difficulty: Q13

(i) - By inspection
a is a double root of the polynomial P(x) as there is no remainder. This means that there is a turning point at x=a. Therefore, P'(a)=0 at x=a.

Would this be enough to answer the question. I am not sure how to do it mathematically, so would this explanation suffice?

aDimitri · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

Nukeboy said:
(i) - By inspection
a is a double root of the polynomial P(x) as there is no remainder. This means that there is a turning point at x=a. Therefore, P'(a)=0 at x=a.

Would this be enough to answer the question. I am not sure how to do it mathematically, so would this explanation suffice?

mathematically, you can just differentiate it.

aDimitri · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

For part ii)
$\\ S(1) = 0, S'(1) = 0, S(-1) = 4 \\ a+b+c = 0 \quad (1) \\ c-a-b = 4 \quad (2) \\ 3a + b = 0 \quad (3) \\ (1)+(2) \\ c = 2 \\ a+b = -2 \\ 2a - 2 = 0 \\ a = 1 \\ b = -3 \\ S(y) = y^3 -3y + 2$

Nukeboy · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

When the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?

hi im trash · Aug 14, 2014

Re: HSC 2013 3U Marathon Thread

Nukeboy said:
When the polynomial P(x) is divided by x^2-1 the remainder is 3x-1. What is the remainder when P(x) is divided by x-1?

2

Sy123 · Aug 16, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
Here is question I made:

$In this question take$

$\phi = \frac{1+\sqrt{5}}{2} , \ \tau = \frac{1-\sqrt{5}}{2}$

$Consider the identity$

$(x-\phi)(x-\tau) = x^2-x-1$

$Prove that$

$\sum_{k=0}^{2n} \binom{2n}{k}^2 \phi^{2n-k} \tau^k = \sum_{k=0}^n \binom{2n}{k} \binom{2n-k}{k}(-1)^k$

.

dunjaaa · Aug 16, 2014

Re: HSC 2013 3U Marathon Thread

A particle moves on the x-axis according to the acceleration equation of motion a=x. Initially the particle is at the origin with velocity v=2. Find the displacement x as a function of t.

integral95 · Aug 16, 2014

Re: HSC 2013 3U Marathon Thread

dunjaaa said:
A particle moves on the x-axis according to the acceleration equation of motion a=x. Initially the particle is at the origin with velocity v=2. Find the displacement x as a function of t.

$a = x \\ \ \frac{d}{dx} (\frac{1}{2} v^2) = x \\ \\ \ \frac{1}{2} v^2 = \int x dx = x^2/2 + C \\ \\ v^2 = x^2 +4 \\ v = \sqrt{x^2+4} \\ \\ \ \frac{dx}{dt} = \sqrt{x^2+4} \\ \\ \ \frac{dt}{dx} = \frac{1}{\sqrt{x^2+4}}$ and yeah ceebs

wiliaamnguuyen · Aug 18, 2014

Re: HSC 2013 3U Marathon Thread

Hey guys. What's the answer to this question?
Find the exact value:
cos330?
Is it cos330 = cos(360-330)
= cos30
= root3/2?

photastic · Aug 18, 2014

Re: HSC 2013 3U Marathon Thread

wiliaamnguuyen said:
Hey guys. What's the answer to this question?
Find the exact value:
cos330?
Is it cos330 = cos(360-330)
= cos30
= root3/2?

Yep, all good.

Sy123 · Aug 21, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $i) Show that$ \ (1+x) + x(1+x) + x(1+x)^2 + \dots + x(1+x)^{n-1} = (1+x)^n$

$\\ $ii) Hence show that$ \ \sum_{k=m-1}^{n-1} \binom{k}{m-1} = \binom{n}{m} \ \ $for some$ \ 1 \leq m \leq n$

hi im trash · Aug 21, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$\\ $i) Show that$ \ (1+x) + x(1+x) + x(1+x)^2 + \dots + x(1+x)^{n-1} = (1+x)^n$

$\\ $ii) Hence show that$ \ \sum_{k=m-1}^{n-1} \binom{k}{m-1} = \binom{n}{m} \ \ $for some$ \ 1 \leq m \leq n$

i don't want to use latex to type up everything again ): took way too long last time.
i) (1+x) +x(1+x)+x(1+x)^2...+x(1+x)^n-1 <-- if we ignore the first term, second to last is a GP series, so a = x(1+x) r = (1+x). sub in formula, cancel you would end up with (1+x) + ((1+x)^n-1 -1) = (1+x)(1+(1+x)^n-1 -1) = (1+x)(1+x)^n-1 = (1+x)^n.

ii) equating coefficients of x^m. RHS = 1 + (nC1).(x) ... + (nCm).(x^m). thus, RHS = nCm.
LHS => if we expand everything, we would end up with (1+x) + x(1+x) ... + x(1+x)^m-2 + x(1+x)^m-1 ... + x(1+x)^n-1.
Since we're finding coefficient of x^m, and we have an additional "x" outside the bracket, we can assume the smallest power possible in binomial expansion is x^m-1 (since if we times that by x, we get x^m)
therefore we can ignore everything before x(1+x)^m-2 since the highest power of this expansion would be x^m-1
x(1+x)^m-1 + x(1+x)^m + x(1+x)^m+1 ... +x(1+x)^n+1. since we have (n+1) terms and the first term is m-1. thus

$\ \binom{m-1}{m-1} + \binom{m}{m-1} + \binom{m+1}{m-1} ... + \binom{n-1}{m-1}$

Therefore
$\ \sum_{k=m-1}^{n-1} \binom{k}{m-1} = \binom{n}{m}$

dunjaaa · Aug 22, 2014

Re: HSC 2013 3U Marathon Thread

Question:

Screen shot 2014-08-22 at 12.16.36 AM.png

hi im trash · Aug 22, 2014

Re: HSC 2013 3U Marathon Thread

Sy123 said:
.

god this Q <.<. im going to call the magnetic flux symbol "A" and the Telsa "B"
[(x-A)(x-B)]^2n = [(x(x-1)-1)]^2n

RHS = (x-A)^2n . (x-B)^2n
expanding both brackets of (x-A)^2n . (-B+x)^2n
(2nC2n-2.(-A)^2n-2.(x^2) + 2nC2n-1.(-A)^2n-1.(x) + 2nC2n(-A)^2n) . (2nC2n-2.(-B)^2.(x)^2n-2 + 2nC2n-1.(-B).(x)^2n-1 + 2nC2n.(x)^2n)
By equating coefficients of x^2n
(2nC2n-2).(2nC2n-2).(-A)^2n-2.(-B)^2 + (2nC2n-1).(2nC2n-1).(-A)^2n-1.(-B) ... (2nC2n-r)^2.(-A)^2n-r.(-B)^r
but 2nC2n-r = 2nCr (symetry of pascal)
therefore RHS = 2n[Sigma sign] r=o (2nCr)^2 (A)^2n-r.(B)^r (the minuses cancel each other out)

LHS - Set (x-a) = k
(kx-1)^2n = 2n[sigma]r=0 2nCr.(-1)^r.(kx-1)^2n-r
expand everything = ................(-1)^2n-1.[x(x-1)]^2n-2n+1.(2nC2n-1) + (-1)^2n.[x(x-1)]^2n-2n.(2nC2n)
However, the first term of this sequence must be r = n, because if we take into consideration [x(x-1)]^2n-r, if we have r = n-1,
then (x)^n-1 times (x-1)^n-1, in order to get x^2n, a term in (x-1) must equal to x^n+1, which is a higher power than n-1, therefore impossible.
so our equation now is (-1)^n.[x(x-1)]^2n-n.(2nC2n-n) .... + (-1)^2n-1.[x(x-1)]^2n-2n+1.(2nC2n-1) + (-1)^2n.[x(x-1)]^2n-2n.(2nC2n)
n[sigma]r=0 2nCr.(-1)^r.[x(x-1)]^2n-r
expanding -> [x(x-1)]^2n-r <- to find the coefficient of x^2n
x^2n-r.(x-1)^2n-r. by equating coefficients, x^2n-r.(... + 2n-rCr(-1)^2(n-r) x^r ...+ ...)
therefore, the coefficient for x^2n is 2n-rCr (we ignore -1 because since the power will always be even because its times by 2)
therefore: n[sigma]r=0 2nCr.(-1)^r.(2n-rCr)

and since (x-A)(x-B) = (x^2-x-1), RHS = LHS
therefore proven

sepseminar · Aug 25, 2014

Re: HSC 2013 3U Marathon Thread

This is a nice little problem about binomial coefficients.
$\\$a) Prove that $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\\$b) Prove that$ \binom{2^n-1}{k} $ is odd for all integers $ k $ such that $ 0 \leq k \leq 2^n-1 \\$c) Prove that$ \binom{2^n}{k} $ is even for all integers $ k $ such that $ 1 \leq k \leq 2^n-1$

dunjaaa · Sep 6, 2014

Re: HSC 2013 3U Marathon Thread

Screen shot 2014-09-06 at 9.06.19 PM.png

Harder 2u

Axio · Sep 6, 2014

Re: HSC 2013 3U Marathon Thread

sepseminar said:
This is a nice little problem about binomial coefficients.
$\\$a) Prove that $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}\\$b) Prove that$ \binom{2^n-1}{k} $ is odd for all integers $ k $ such that $ 0 \leq k \leq 2^n-1 \\$c) Prove that$ \binom{2^n}{k} $ is even for all integers $ k $ such that $ 1 \leq k \leq 2^n-1$

a)

$RHS=\binom{n-1}{k}+\binom{n-1}{k-1}\\=\frac{(n-1)!}{(n-k-1)!k!}+\frac{(n-1)!}{(n-k)!(k-1)!}\\=\frac{(n-k)(n-1)!+k(n-1)!}{(n-k)!k!}\\=\frac{(n-k+k)(n-1)!}{(n-k)!k!}\\=\frac{n!}{(n-k)!k!}=\binom{n}{k}=LHS$

Nukeboy · Sep 20, 2014

Re: HSC 2013 3U Marathon Thread

Harder 2unit question

Sy123 · Nov 22, 2014

Re: HSC 2013 3U Marathon Thread

$\\ $A sequence is defined as$ \\\\ a_0 = 1 \\ a_{n+1} = \frac{a_n}{a_n+1} \\ \\ $Prove by mathematical induction that$ \ \ a_n = \frac{1}{n+1} \ $for$ \ n \geq 0$

$\\ $EXTENSION (still within 3U knowledge), can you find another non-inductive method of finding$ \ a_n \ $(that is derivable without knowing the result as well (i.e. not ad-hoc))?$$

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