HSC 2017 MX1 Marathon (5 Viewers)

eyeseeyou · Oct 8, 2016

Post any questions within the scope and level of Mathematics Extension 1. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-14 difficulty in the HSC.

Some housekeeping rules:

1. Provide neat working out when possible

2. Don't flood the thread with multiple unanswered questions

3. Since HSC MX1 Mathematics can be tested up to 30% of the maths you did in preliminary, feel free to post the often odd and challenging questions

Source(s): http://community.boredofstudies.org/13/mathematics-extension-1/344684/hsc-2016-3u-marathon.html
http://community.boredofstudies.org/14/mathematics-extension-2/344755/hsc-2016-4u-marathon.html

I encourage all current HSC students in particular to participate in this marathon.

To start off:

The point P(6p, 3p^2) is a point on the parabola x^2 = 12y .
(i)Find the equation of the tangent of P
(ii)The Tangent at P buts the y-axis at B. The point A divides PB internally in the ratio 1:2
Find the locus of the point A as P varies

pikachu975 · Oct 8, 2016

MX1 only goes up to question 14 lolol

\text{hello}

How do you do latex?

$\text{hello}$

Edit: Thanks trecex

InteGrand · Oct 8, 2016

pikachu975 said:
MX1 only goes up to question 14 lolol

\text{hello}

How do you do latex?

You need to enclose your code in tex tags: [.tex] [./tex] (without the full stops).

You can practise it here for instance: http://community.boredofstudies.org/3/non-school/345040/latex-practice.html .

pikachu975 · Oct 8, 2016

eyeseeyou said:
Post any questions within the scope and level of Mathematics Extension 1. Once a question is posted, it needs to be answered before the next question is raised.

This thread is mainly targeting Q1-15 difficulty in the HSC. Any questions of Q16 difficulty or harder should be posted in the Advanced thread.

Some housekeeping rules:

1. Provide neat working out when possible

2. Don't flood the thread with multiple unanswered questions

3. Since HSC MX1 Mathematics can be tested up to 30% of the maths you did in preliminary, feel free to post the often odd and challenging questions

Source(s): http://community.boredofstudies.org/13/mathematics-extension-1/344684/hsc-2016-3u-marathon.html
http://community.boredofstudies.org/14/mathematics-extension-2/344755/hsc-2016-4u-marathon.html

I encourage all current HSC students in particular to participate in this marathon.

To start off:

The point P(6p, 3p^2) is a point on the parabola x^2 = 12y .
(i)Find the equation of the tangent of P
(ii)The Tangent at P buts the y-axis at B. The point A divides PB internally in the ratio 1:2
Find the locus of the point A as P varies

$\\ \text{i)} \\ \text{This is legit just differentiating to find gradient then using point gradient formula...} \\ \\ \frac{dy}{dx} = \frac{x}{6} \\ \\ \text{Subbing in x = 6p} \\ \\ \frac{dy}{dx} = p \\ \\ \text{Sub into point gradient formula} \\ y - 3p^2 = p(x-6p) \\ y - 3p^2 = px - 6p^2 \\ y = px - 3p^2 \\ \\ \text{ii)} \\ \text{Find where it hits the y axis by subbing in x = 0 to the tangent equation} \\ \text{When x} = 0, y = -3p^2 \\ \text{Then use the internal division formula to find point A} \\ (\frac{1 \times 6p + 2 \times 0}{1 + 2}, \frac{1 \times 3p^2 + 2 \times -3p^2}{1+2}) \\ \\ = (2p, - p^2) \\ \text{Find point A in terms of the parameters then do the usual simultaneous equation to solve it} \\ x = 2p \\ \\ p = \frac{2}{x} \\ \\ \text{Sub into y} \\ y = -(\frac{2}{x})^2 \\ \\ y = \frac{-x^2}{4}$

This took years to type and the answer is probably wrong lol

leehuan · Oct 8, 2016

Drsoccerball · Oct 8, 2016

leehuan said:
$\text{Use mathematical induction to prove that}\\ (1+2+3+\dots+n)^2 = 1^3+2^3+3^3+\dots+n^3$

$Let P(n) be the proposition that :$

$(1+2+3+\dots+n)^2 = 1^3+2^3+3^3+\dots+n^3$

$$ Consider $ P(1):$

$(1)^2 = 1^3 = 1 $ Which is true. $$

$Let k be an integer for which P(k) is true:$

$(1+2+3+ \dots + k)^2 = 1^3 + 2^3 + \dots + k^3$

$$ Consider P(k+1): $ \\ (1+2+\dots + k + (k+1))^2 = 1^3 + 2^3 + \dots + k^3 + (k+1)^3$

$$ Consider the expansion $ (a+b)^2 = a^2 + b^2 + 2ab$

$LHS = (1+2+3+ \dots + k)^2 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^2(1 +k)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^3$

$\therefore P(k+1) $ is true. $$

$Hence by mathematical induction the proposition P(n) is true.$

leehuan · Oct 8, 2016

Drsoccerball said:
$Let P(n) be the proposition that :$

$(1+2+3+\dots+n)^2 = 1^3+2^3+3^3+\dots+n^3$

$$ Consider $ P(1):$

$(1)^2 = 1^3 = 1 $ Which is true. $$

$Let k be an integer for which P(k) is true:$

$(1+2+3+ \dots + k)^2 = 1^3 + 2^3 + \dots + k^3$

$$ Consider P(k+1): $ \\ (1+2+\dots + k + (k+1))^2 = 1^3 + 2^3 + \dots + k^3 + (k+1)^3$

$$ Consider the expansion $ (a+b)^2 = a^2 + b^2 + 2ab$

$LHS = (1+2+3+ \dots + k)^2 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^2(1 +k)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^3$

$\therefore P(k+1) $ is true. $$

$Hence by mathematical induction the proposition P(n) is true.$

Ok already mr first year

fancy shmancy layout

InteGrand · Oct 8, 2016

leehuan · Oct 8, 2016

That 4u induction in the 3u marathon though

trecex1 · Oct 8, 2016

InteGrand said:
$\noindent Consider the sequence of numbers $F_{1},F_{2},F_{3},\ldots$, defined recursively by $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$, and $F_{1} = F_{2} = 1$. (This is the \textit{Fibonacci sequence}.)$

$\noindent (i) Use induction to show that for any real number $x$ satisfying $x^2 - x -1 = 0$, we have $x^n = xF_{n} + F_{n-1}$ for all integers $n \geq 2$.$

$\noindent (ii) Hence, show that $F_{n} = \frac{1}{\sqrt{5}} \left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$ for all positive integers $n$. (This is known as \emph{Binet's formula} for the Fibonacci numbers.)$

Here's the induction, part, I'll try the next part later
$\text{n=2} \\ LHS: x^2 \\ RHS: xF_2+F_1 \\ =x+1 = x+x^2-x \\ =x^2 = LHS \therefore \text{ the statement is true for n=2} \\\\ \text{Assume statement is true for all n up to and including }n=k \\ x^k = xF_k + F_{k-1} \\\\ \text{R.T.P the statement is true for }n=k+1 \\\\ x^{k+1}=xF_{k+1}+F_k\\ \text{From our assumption}\\ LHS: x\cdot x^k\\ =x\left ( xF_k+F_{k-1} \right )\\ =x^2F_k + xF_{k-1}\\ =\left ( 1+x \right )F_k + xF_{k-1} \\ =F_k + x\left ( F_k + F_{k-1} \right ) \\ =xF_{k+1}+F_k \\ =RHS \therefore \text{ true for all }n\geq 2 \text{ by principle of mathematical induction}$

Drsoccerball · Oct 8, 2016

InteGrand said:
$\noindent Consider the sequence of numbers $F_{1},F_{2},F_{3},\ldots$, defined recursively by $F_{n} = F_{n-1} + F_{n-2}$ for $n \geq 3$, and $F_{1} = F_{2} = 1$. (This is the \textit{Fibonacci sequence}.)$

$\noindent (i) Use induction to show that for any real number $x$ satisfying $x^2 - x -1 = 0$, we have $x^n = xF_{n} + F_{n-1}$ for all integers $n \geq 2$.$

$\noindent (ii) Hence, show that $F_{n} = \frac{1}{\sqrt{5}} \left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$ for all positive integers $n$. (This is known as \emph{Binet's formula} for the Fibonacci numbers.)$

$i) Let P(n) be the proposition that:$

$x^n - x F_n - F_{n-1} = 0 $ for some x which satisfies $ x^2 - x - 1 = 0.$

$Consider P(2) :$

$x^2 - x F_1 - F_ 0 = 0 = x^2 - x - 1 $ and thus P(2) is true. $$

$Let k be an integer for which P(k) is true:$

$x^n - x F_k - F_{k-1} = 0 $ for some x which satisfies $ x^2 - x - 1 = 0.$

$Now consider P(k+1) :$

$x \cdot x^k -x F_{k+1} - F_k = 0$

$x(x^k - x F_k - F_{k - 1}) + F_k(x^2 - 1) - x (F_{k+1} - F_{k-1}) = 0$

$F_k(x^2 - 1) -x(F_k) = 0 $ From assumption and definition of a fib sequence. $$

$F_k(x^2-x-1) = 0$

$\therefore P(k+1) $ is true. $$

$Hence by Mathematical Induction the proposition is true.$

$ii) Since x is a solution to the equation we find solve the characteristic equation to find x.$

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

$(\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n = F_n(\frac{1+\sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}) +F_{n-1} - F_{n-1}$

$F_{n} = \frac{1}{\sqrt{5}} \left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$

InteGrand · Oct 8, 2016

Drsoccerball said:
$i) Let P(n) be the proposition that:$

$x^n - x F_n - F_{n-1} = 0 $ for some x which satisfies $ x^2 - x - 1 = 0.$

$Consider P(2) :$

$x^2 - x F_1 - F_ 0 = 0 = x^2 - x - 1 $ and thus P(2) is true. $$

$Let k be an integer for which P(k) is true:$

$x^n - x F_k - F_{k-1} = 0 $ for some x which satisfies $ x^2 - x - 1 = 0.$

$Now consider P(k+1) :$

$x \cdot x^k -x F_{k+1} - F_k = 0$

$x(x^k - x F_k - F_{k - 1}) + F_k(x^2 - 1) - x (F_{k+1} - F_{k-1}) = 0$

$F_k(x^2 - 1) -x(F_k) = 0 $ From assumption and definition of a fib sequence. $$

$F_k(x^2-x-1) = 0$

$\therefore P(k+1) $ is true. $$

$Hence by Mathematical Induction the proposition is true.$

$ii) Since x is a solution to the equation we find solve the characteristic equation to find x.$

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

$(\frac{1+\sqrt{5}}{2})^n - (\frac{1-\sqrt{5}}{2})^n = F_n(\frac{1+\sqrt{5}}{2} - \frac{1-\sqrt{5}}{2}) +F_{n-1} - F_{n-1}$

$F_{n} = \frac{1}{\sqrt{5}} \left[\left(\frac{1+\sqrt{5}}{2}\right)^{n} - \left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]$

Well done!

Green Yoda · Oct 8, 2016

Drsoccerball said:
$Let P(n) be the proposition that :$

$(1+2+3+\dots+n)^2 = 1^3+2^3+3^3+\dots+n^3$

$$ Consider $ P(1):$

$(1)^2 = 1^3 = 1 $ Which is true. $$

$Let k be an integer for which P(k) is true:$

$(1+2+3+ \dots + k)^2 = 1^3 + 2^3 + \dots + k^3$

$$ Consider P(k+1): $ \\ (1+2+\dots + k + (k+1))^2 = 1^3 + 2^3 + \dots + k^3 + (k+1)^3$

$$ Consider the expansion $ (a+b)^2 = a^2 + b^2 + 2ab$

$LHS = (1+2+3+ \dots + k)^2 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 + (k+1)^2 +2(1+2+3+...+k)(k+1)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^2(1 +k)$

$LHS = 1^3 + 2^3 + \dots + k^3 +(k+1)^3$

$\therefore P(k+1) $ is true. $$

$Hence by mathematical induction the proposition P(n) is true.$

3rd last line: where did 2(1+2+3+..+k) go?

leehuan · Oct 8, 2016

Rathin said:
3rd last line: where did 2(1+2+3+..+k) go?

Summed the AP up.

Drsoccerball · Oct 8, 2016

In how many ways can you climb 15 steps by only taking 1 or 3 steps at a time?

Green Yoda · Oct 8, 2016

leehuan said:
Summed the AP up.

AP? sorry I just started Induction

jathu123 · Oct 8, 2016

Rathin said:
AP? sorry I just started Induction

arithmetic progression;

https://en.wikipedia.org/wiki/Arithmetic_progression

si2136 · Oct 8, 2016

Drsoccerball said:
In how many ways can you climb 15 steps by only taking 1 or 3 steps at a time?

Is it 18?

Paradoxica · Oct 8, 2016

Drsoccerball said:
In how many ways can you climb 15 steps by only taking 1 or 3 steps at a time?

95

si2136 · Oct 8, 2016

Paradoxica said:
95

Could you show your working out? I read the question wrong with 10 steps but with 15 steps, this is my working out:

15 Steps

3 Step // 1 Step

5 // 0 = 1 Way
4 // 3 = 35 Ways
3 // 6 = 84 Ways
2 // 9 = 55 Ways
1 // 12 = 13 Ways
0 // 15 = 1 Way

Therefore 189 ways.

HSC 2017 MX1 Marathon (5 Viewers)

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