HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

RealiseNothing · Feb 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $The definitions of the sequences are$ \ a_{n} + a_{n+1} = b_n \ (1) \ $and$ \ a_n a_{n+1} = 2^n \ (2) \\ a_1 = 1 \ $and$ \ b_1 = 3$

$\\ \frac{2^{n+1}}{2^n} = \frac{a_{n+2} a_{n+1}}{a_{n+1}a_n} \Rightarrow \ a_{n+2} = 2a_n$

$\\ $So, given$ \ m=1,2,3,\dots \\ \ x_m = a_{2m} \\ y_m = a_{2m-1}$

$\\ $So,$ \ x_{m+1} = 2x_m \ $and$ \ y_{m+1} = 2y_m \ $and$ \ x_1 = 2, y_1 = 1$

$\\ S_n = b_1 + b_2 + \dots + b_n = (a_1+ a_2) + (a_2+a_3) + \dots + (a_n + a_{n+1}) \\ = (a_{n+1} + a_1) + 2(a_2 + a_3 + \dots + a_{n-1} + a_n) \\ \\ $If$ \ n = 2m \Rightarrow \ S_{2m} = y_{m+1} + y_1 + 2((x_1 + x_2 + \dots + x_m) + (y_2 + \dots + y_m))$

$\\ $If$ \ n = 2m -1 \Rightarrow \ S_{2m-1} = x_m + y_1 + 2((x_1 + \dots + x_{m-1}) + (y_2 + \dots + y_m))$

$\\ \therefore \ S_{2m} = 7\cdot 2^m - 7$

$\\ \therefore \ S_{2m-1} = 7\cdot 2^{m} - 7 \\\\ \therefore \ S_n = 7 \cdot 2^{\lceil \frac{n}{2} \rceil } - 7$

I get this answer but only when n is even.

When n is odd I get:

$10(2^{\frac{n-1}{2}})-7$

RealiseNothing · Feb 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

My way was a bit different I think:

Note that from the product of roots like Sy did, we get:

$a_{n+2}=2a_n$

But:

$b_{n+1}-b_n = a_{n+2}-a_n = 2a_n - a_n = a_n$

$b_{n+1}-b_n=a_n$

This allows us to use a telescoping sum:

$S_n=2(a_1+a_2+a_3+...+a_{n+1})-(a_1+a_{n+1})$

$S_n = 2((b_2-b_1)-(b_3-b_2)-...-(b_{n+2}-b_{n+1}))-(a_1+a_{n+1})$

$S_n=2(b_{n+2}-b_1)-(a_1+a_{n+1})$

But $a_1=1$ and $b_1=3$ and $a_{n+1}=b_{n+2}-b_{n+1}$ which gives:

$S_n=b_{n+2}+b_{n+1}-7$

Now we just split it into the "n is even/odd" cases to find what $b_{n+2}+b_{n+1}$ is and we are done.

This isn't hard knowing that $b_{n+2}=2b_n$ and so $b_{n+2}+b_{n+1}=2(b_n+b_{n-1})$ with $b_1+b_2=7$ amd $b_2+b_3=10$

InteGrand · Feb 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\text{Consider a square pyramid with base length } s \text{ and height } H.$

$\text{ A line is drawn on one of the triangular faces parallel to the triangle's base edge and at a height } h \text{ above the ground } (0\leq h < H). \text{ Another such line is drawn on the opposite triangular face, at a height } k\in [h,H) \text{ above the ground.}$

$\text{ The pyramid is sliced through the plane containing the two drawn lines to form two ``pieces''}.$

$\text{What is the volume of the lower piece (i.e. the one containing the base)?}$

Sy123 · Mar 18, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let$ \ T_N = \sum_{k=1}^N \frac{1}{\sqrt{k+\frac{1}{2}}} \\ S_N = \sum_{k=1}^N \frac{1}{\sqrt{k}} \\ \\ $Find$ \ \lim_{N \to \infty} \frac{T_N}{S_N}$

InteGrand · Mar 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ T_N = \sum_{k=1}^N \frac{1}{\sqrt{k+\frac{1}{2}}} \\ S_N = \sum_{k=1}^N \frac{1}{\sqrt{k}} \\ \\ $Find$ \ \lim_{N \to \infty} \frac{T_N}{S_N}$

By sketching graphs of $y= \frac{1}{\sqrt{x}}$ and $y=\frac{1}{\sqrt{x+\frac{1}{2}}}$ and comparing relevant areas, we come up with the following:

$\int_{1}^{N+1}\frac{\text{d}x}{\sqrt{x+\frac{1}{2}}}<\sum_{k=1}^{N}\frac{1}{\sqrt{k+\frac{1}{2}}}<\frac{1}{\sqrt{1+\frac{1}{2}}}+\int_{1}^{N}\frac{ \text{d}x}{\sqrt{x+\frac{1}{2}}} \Leftrightarrow \int_{1}^{N+1}\frac{\text{d}x}{\sqrt{x+\frac{1}{2}}}<T_N<\sqrt{\frac{2}{3}}+\int_{1}^{N} \frac{\text{d}x}{\sqrt{x+\frac{1}{2}}}$ $(1)$

and

$\int_{1}^{N+1}\frac{\text{d}x}{\sqrt{x}}<\sum_{k=1}^{N}\frac{1}{\sqrt{k}}<1+\int_{1}^{N}\frac{\text{d}x}{\sqrt{x}} \Leftrightarrow \int_{1}^{N+1}\frac{\text{d}x}{\sqrt{x}}<S_N<1+\int_{1}^{N}\frac{\text{d}x}{\sqrt{x}}$ $(2)$ .

Evaluation of the integrals and simplification yields

$2\sqrt{N+\frac{3}{2}}-\sqrt{6}<T_N<2\sqrt{N+\frac{1}{2}}-\frac{4}{\sqrt{6}}$ $(1^\prime)$

and

$2\sqrt{N+1}-2<S_N<2\sqrt{N}-1$ $(2^\prime)$ .

Division of $(1^\prime)$ by $(2^\prime)$ yields

$\frac{2\sqrt{N+\frac{3}{2}}-\sqrt{6}}{2\sqrt{N+1}-2}<\frac{T_N}{S_N}<\frac{2\sqrt{N+\frac{1}{2}}- \frac{4}{\sqrt{6}}}{2\sqrt{N}-1} $ $(*)$ .

Taking limits in $(*)$ and using the squeeze theorem gives $\lim_{N\to\infty}\frac{T_N}{S_N}=1$ (and the quotient approaches 1 from below, as clearly $T_N < S_N$ ).

Sy123 · Mar 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

My method:

$\\ \sum_{k=2}^{n+1} \frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} < \frac{1}{2\sqrt{2}}t_n < \sum_{k=1}^{n}\frac{1}{\sqrt{2k-1} + \sqrt{2k+1}}$

$\frac{1}{\sqrt{2k-1} + \sqrt{2k+1}} = \frac{1}{2} (\sqrt{2k+1} - \sqrt{2k-1})$

$\Rightarrow \ \sqrt{2}(\sqrt{2n+3} - \sqrt{3}) < t_n < \sqrt{2}(\sqrt{2n+1} - 1)$

$\Rightarrow \ \lim_{n \to \infty} \frac{t_n}{\sqrt{n}} = 2 \ \ (*)$

$\\ $Now,$ \ \sum_{k=1}^n \frac{1}{\sqrt{k} + \sqrt{k+1}} < \frac{s_n}{2} < \sum_{k=0}^{n-1} \frac{1}{\sqrt{k} + \sqrt{k+1}}$

$\frac{1}{\sqrt{k} + \sqrt{k+1}} = \sqrt{k+1} - \sqrt{k}$

$\Rightarrow \ 2(\sqrt{n+1} - 1) < s_n < 2\sqrt{n}$

$\Rightarrow \ \lim_{n \to \infty} \frac{s_n}{\sqrt{n}} = 2 \ \ (**)$

$\\ \frac{(*)}{(**)}\Rightarrow \ \lim_{n \to \infty} \frac{t_n}{s_n} = 1$

Sy123 · Mar 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Define the sequence$ \ x_1 = \frac{1}{2} \ , \ x_n = \frac{x_{n-1}}{2nx_{n-1} + 1}$

$\\ $Find$ \ \sum_{k=1}^n x_k$

glittergal96 · Mar 20, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Define the sequence$ \ x_1 = \frac{1}{2} \ , \ x_n = \frac{x_{n-1}}{2nx_{n-1} + 1}$

$\\ $Find$ \ \sum_{k=1}^n x_k$

Let $y_n=x_n^{-1}.$

The recurrence then implies

$y_n=2n+y_{n-1},y_1=2$

which has unique solution $y_n=n(n+1).$

(The form of the recurrence made it clear to me that it would probably have a quadratic solution, so I just looked for something of the form an+b).

So $x_n=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$

The sum in question then telescopes to

$\frac{n}{n+1}.$

InteGrand · Mar 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

This is a question from the 1973 HSC 4U paper:

$\paragraph{A man inherits, on his 21st birthday (31 December, 1970), two houses - one in the city and one in the country. On the first day after this birthday, i.e., on 1 January, 1971, he moves into the city house. In this house there is a box containing one red and two white balls and in the country house there is a similar box containing one red and three white balls. Each day he draws at random a ball from the box in the house where he is, notes its colour, and then returns it to the box: if it is red he moves to spend the next day at the other house (otherwise he stays where he is and awaits the outcome of the next drawing).}$

$\paragraph{(I) Calculate the probability that he is in residence at the city house:}$

$\paragraph{(a) on 3 January, 1971;}$
$\paragraph{(b) on the \textit{n}-th day after his 21st birthday.}$
$\paragraph{(II) Assuming that he lives to a ripe old age estimate (approximately) the probability that he will die in the city house.}$

$\paragraph{You may use the following facts:}$

$\bullet \ln\left ( \frac{1+x}{1-x} \right )=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1} \text{ for }|x|<1$ .
$\bullet \ln 2 \approx 0.69, \text{ with error}<1\%$

FrankXie · Mar 26, 2015

Re: HSC 2015 4U Marathon - Advanced Level

InteGrand said:
This is a question from the 1973 HSC 4U paper:

$\paragraph{You may use the following facts:}$

$\bullet \ln\left ( \frac{1+x}{1-x} \right )=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1} \text{ for }|x|<1$ .
$\bullet \ln 2 \approx 0.69, \text{ with error}<1\%$

i got $\frac{19}{36}$ for part (i) and $\frac{3}{7}$ for part (ii). i am not sure this is true or not, because i didn't use those facts.

VBN2470 · Mar 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

NEW QUESTION:

$$ Given that $f$ is a continuous function on $\mathbb{R}$ such that $f(0)=0$ and $\lim_{x\to\infty} f(x)=0$, prove that $f$ attains a maximum on $[0, \infty)$.$

Don't know if this was the right place to post this q.

seanieg89 · Mar 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

VBN2470 said:
NEW QUESTION:

$$ Given that $f$ is a continuous function on $\mathbb{R}$ such that $f(0)=0$ and $\lim_{x\to\infty} f(x)=0$, prove that $f$ attains a maximum on $[0, \infty)$.$

Don't know if this was the right place to post this q.

I think any proper proof of this will need a precise definition of the reals, which is comfortably outside the scope of mx2. (Unless you allow people to assume the extreme value theorem, which implies the claimed result almost trivially).

InteGrand · Mar 27, 2015

Re: HSC 2015 4U Marathon - Advanced Level

VBN2470 said:
NEW QUESTION:

$$ Given that $f$ is a continuous function on $\mathbb{R}$ such that $f(0)=0$ and $\lim_{x\to\infty} f(x)=0$, prove that $f$ attains a maximum on $[0, \infty)$.$

Don't know if this was the right place to post this q.

Since is continuous on $[0,\infty)$ , it continuous on $[0,t]$ for any positive real number $t$ . If $f(x)$ is never positive, then clearly attains a maximum (0) at x = 0. So assume is positive somewhere in the interval $[0,a]$ , where $a$ is some positive real number.

By the Extreme Value Theorem, attains a maximum value $K>0$ on $[0,a]$ , say at $x=a^*$ . Since $\lim_{x\to \infty}f(x)=0$ , it means that for all $x$ greater than some positive number $M$ (which must be greater than $a^*$ ), we will have $-K<f(x)<K$ . Now consider the interval $[a, M]$ . By the Extreme Value Theorem, attains a maximum $K_2$ on here. If $K_2 \geq K$ , then attains the maximum value $K_2$ on the interval $[0,\infty)$ (since $f(x)\leq K \leq K_2$ for all $x \in [0,a]$ and $-K<f(x)<K\leq K_2$ for all $x\in [M,\infty)$ ). Otherwise (i.e. $K>K_2$ ), attains the maximum value $K$ on $[0,\infty)$ , since $f(x)\leq K$ for all $x \in [0,a]$ , $f(x)\leq K_2 < K$ for all $x\in [a,M]$ , and $f(x)<K$ for all $x\in [M,\infty)$ . In either case, indeed attains a maximum on $[0,\infty)$ (the maximum being $\max\{K,K_2\}$ ).

seanieg89 · Mar 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

As a sidenote for any HS students that want to know where the extreme value theorem comes from, here is what you need to know for one proof.

1. The least upper bound property of the reals. (Any set A of real numbers which has an upper bound M has a LEAST upper bound U. Being the least upper bound means that U =< M for any other upper bound M). In fact, the reals are can in a certain sense be defined as the smallest extension of the rational numbers that has this property.

2. The epsilon delta definition of a limit, and the definition of Cauchy sequences.

3. The sequential characterisation of continuity.

Then: Show that any sequence in the interval [a,b] has a convergent subsequence, and use this to prove that $f:[a,b]\rightarrow \mathbb{R}$ attains a maximum for any continuous function f.

Sy123 · Mar 28, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Let$ \ a_2, a_3, \dots , a_n \ $be among the positive reals so that$ \ a_1 a_2 a_3 \dots a_n = 1 \\ \\ $Prove that$ \ (1+a_2)^2(1+a_3)^3(1+a_4)^4\dots (1+a_n)^n > n^n$

Sy123 · Apr 2, 2015

Re: HSC 2015 4U Marathon - Advanced Level

$\\ $Find all pairs of non-negative integers$ \ (x,y) \ $that satisfy$ \\ x^3y + x + y = xy + 2xy^2$

simpleetal · Apr 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Let$ \ a_2, a_3, \dots , a_n \ $be among the positive reals so that$ \ a_1 a_2 a_3 \dots a_n = 1 \\ \\ $Prove that$ \ (1+a_2)^2(1+a_3)^3(1+a_4)^4\dots (1+a_n)^n > n^n$

Lol I remember this question from my exam 3 weeks ago lol. What I did during the exam was apply AM-GM (they made us prove the general form in the previous parts) for each factor on the LHS, i.e. (1+a3)^3>= 3^3 /2^2 a3 then I just multiplied all the factors and then afterwards I simplied showed equality can't be achieved

simpleetal · Apr 6, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
$\\ $Find all pairs of non-negative integers$ \ (x,y) \ $that satisfy$ \\ x^3y + x + y = xy + 2xy^2$

Okay, so the expression just becomes xy(x^2-2y)+1=(x-1)(y-1)
Case 1: Suppose both x,y are positive. If x^2>2y the LHS grows faster than the RHS (xy>(x-1)(y-1) for positive x,y) so equality cannot be achieved. However if x^2<2y, then the second factor on lhs is negative. but if x and y are positive then rhs is non negatuve. hence equality can be achieved if xy (x^2-2y) is -1 which can be achieved via the solution (1,1)
Case 2: if x,y both zero, it is a solution (0,0)
Case 3: If one of x,y is greater than zero while the other is zero, then RHs is always negative while the LHS is 1

Hence the solution set is (1,1) and (0,0)

Sy123 · Apr 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

simpleetal said:
Okay, so the expression just becomes xy(x^2-2y)+1=(x-1)(y-1)
Case 1: Suppose both x,y are positive. If x^2>2y the LHS grows faster than the RHS (xy>(x-1)(y-1) for positive x,y) so equality cannot be achieved. However if x^2<2y, then the second factor on lhs is negative. but if x and y are positive then rhs is non negatuve. hence equality can be achieved if xy (x^2-2y) is -1 which can be achieved via the solution (1,1)
Case 2: if x,y both zero, it is a solution (0,0)
Case 3: If one of x,y is greater than zero while the other is zero, then RHs is always negative while the LHS is 1

Hence the solution set is (1,1) and (0,0)

You missed one solution (2,2)

$\\ $Manipulating the equation results in$ \ \frac{y}{x} =y + 2y^2 - 1 -x^2y \\ \\ $Since the RHS is an integer, then so is the LHS, and so$ \ y = kx \ $for some integer$ \ k \\ $Putting this back into the original equation results in a polynomial of degree$ \\ kx^4 - 2k^2x^3 - kx^2 + (k+1)x = 0 \\ \\ $So$ \ x = 0 \ $is a solution$ \\ kx^3 - 2k^2x^2 = kx + (k+1) = 0 \ $also must only have integer solutions for$ \ x \\ \\ $But then dividing the polynomial by$ \ k \ $we see that$ \ \frac{1}{k} = 2kx^2 - x^3 +x - 1 \\ $Since the RHS is an integer, so is the LHS, and since$ \ (x,y) \ $are a non-negative pair, the only solution for$ \ k \ $is$ \ k = 1 \\ \\ $So$ \ y =x$. Putting this back into the original polynomial we see$ \\ x^3 - 2x^2 - x + 2 = x(x^2-1) - 2(x^2-1) = (x-2)(x-1)(x+1) = 0 \\ $So$ \ x = 1,2 \ $are also solutions (discounting the negative solutions), and hence the total solutions are, given that$ \ y =x \\ \\ \therefore (2,2), (1,1), (0,0)$

simpleetal · Apr 7, 2015

Re: HSC 2015 4U Marathon - Advanced Level

Sy123 said:
You missed one solution (2,2)

$\\ $Manipulating the equation results in$ \ \frac{y}{x} =y + 2y^2 - 1 -x^2y \\ \\ $Since the RHS is an integer, then so is the LHS, and so$ \ y = kx \ $for some integer$ \ k \\ $Putting this back into the original equation results in a polynomial of degree$ \\ kx^4 - 2k^2x^3 - kx^2 + (k+1)x = 0 \\ \\ $So$ \ x = 0 \ $is a solution$ \\ kx^3 - 2k^2x^2 = kx + (k+1) = 0 \ $also must only have integer solutions for$ \ x \\ \\ $But then dividing the polynomial by$ \ k \ $we see that$ \ \frac{1}{k} = 2kx^2 - x^3 +x - 1 \\ $Since the RHS is an integer, so is the LHS, and since$ \ (x,y) \ $are a non-negative pair, the only solution for$ \ k \ $is$ \ k = 1 \\ \\ $So$ \ y =x$. Putting this back into the original polynomial we see$ \\ x^3 - 2x^2 - x + 2 = x(x^2-1) - 2(x^2-1) = (x-2)(x-1)(x+1) = 0 \\ $So$ \ x = 1,2 \ $are also solutions (discounting the negative solutions), and hence the total solutions are, given that$ \ y =x \\ \\ \therefore (2,2), (1,1), (0,0)$

whoops hahaha I see my mistake in case 1 lol mustve been too tired. but nice solution sir!

HSC 2015 MX2 Marathon ADVANCED (archive) (1 Viewer)

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