HSC 2016 MX1 Marathon (archive) (1 Viewer)

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Damn just do rectangles lel... Someone do the second limit.

Do you still call them "rectangles"?

Drsoccerball · Sep 29, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Do you still call them "rectangles"?

"Riemann sums" *In a strong German accent*.

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
"Riemann sums" *In a strong German accent*.

Lmfao.

______________________________

Here's a question to bring in a something more down to earth.

$\\ \text{Prove by mathematical induction that for all natural numbers }n \\ \text{The }n\text{th derivative of }e^{x^2}\text{ is }P(x)e^{x^2}\\ \text{where }P(x)\text{ is some polynomial of degree }n$

Note: The 0th derivative of a function is defined as just itself.

trecex1 · Sep 29, 2016

Re: HSC 2016 3U Marathon

leehuan said:
Lmfao.

______________________________

Here's a question to bring in a something more down to earth.

$\\ \text{Prove by mathematical induction that for all natural numbers }n \\ \text{The }n\text{th derivative of }e^{x^2}\text{ is }P(x)e^{x^2}\\ \text{where }P(x)\text{ is some polynomial of degree }n$

Note: The 0th derivative of a function is defined as just itself.

$n=1, \frac{d(e^{x^2})}{dx}= 2x\cdot e^{x^2} \text{where }P(x)\text{ is 2}x \text{ with degree 1} \therefore \text{true for n=1} \\ \text{Assuming statement is true for n=k, then } \frac{d^k(e^{x^2})}{dx}=P_1(x)e^{x^2}\\ \text{R.T.P true for n=k+1 i.e } \frac{d^{k+1}(e^{x^2})}{dx} =P_2(x)e^{x^2} \\ \text{From assumption, LHS: } \frac{d(P_1(x)e^{x^2})}{dx} \\\\ =P{}'_1(x)e^{x^2}+P_1(x)\cdot 2x\cdot e^{x^2} \\\\ =e^{x^2}\left (P{}'_1(x) +P_1(x)\cdot 2x \right ) \text \\ \\\\ =P_2(x)\cdot e^{x^2} \\\\= \text{RHS} \left ( \text{ as }P{}'_1(x) +P_1(x)\cdot 2x \text{ is any polynomial with degree k+1 as }P_1(x) \text{ is of degree k}\right ) \\\\ \text{Statement is true for n=1}, \therefore \text{true for n=1+1 =2, 2+1=3 and so on}\\ \text{So it is true for all } n\geq 1 \text{ by principle of Mathematical Induction.} \\\\ \text{Also: do markers even give marks for writing the conclusion? I usually skip it or just write "True by M.I"}$

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$n=1, \frac{d(e^{x^2})}{dx}= 2x\cdot e^{x^2} \text{where }P(x)\text{ is 2}x \text{ with degree 1} \therefore \text{true for n=1} \\ \text{Assuming statement is true for n=k, then } \frac{d^k(e^{x^2})}{dx}=P_1(x)e^{x^2}\\ \text{R.T.P true for n=k+1 i.e } \frac{d^{k+1}(e^{x^2})}{dx} =P_2(x)e^{x^2} \\ \text{From assumption, LHS: } \frac{d(P_1(x)e^{x^2})}{dx} \\ =P{}'_1(x)e^{x^2}+P_1(x)\cdot 2x\cdot e^{x^2} \\ =e^{x^2}\left (P{}'_1(x) +P_1(x)\cdot 2x \right ) \text) \\ =P_2(x)\cdot e^{x^2} = \text{RHS} \left ( \text{ as }P{}'_1(x) +P_1(x)\cdot 2x \text{ is any polynomial with degree k+1 as }P_1(x) \text{ is of degree k}\right ) \\\\ \text{Statement is true for n=1}, \therefore \text{true for n=1+1 =2, 2+1=3 and so on}\\ \text{So it is true for all } n\geq 1 \text{ by principle of Mathematical Induction.} \\\\ \text{Also: do markers even give marks for writing the conclusion? I usually skip it or just write "True by M.I"}$

The natural numbers start from 0, not 1.

But of course the inductive step is correct. Well done. I won't ask for any further technicality in 3U

To answer the question: True by mathematical induction is enough, but try not to abbreviate it. Fussy markers deduct you for it, others don't. (What's stupid is when they deduct you marks for not writing the paragraph.)

InteGrand · Sep 29, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$n=1, \frac{d(e^{x^2})}{dx}= 2x\cdot e^{x^2} \text{where }P(x)\text{ is 2}x \text{ with degree 1} \therefore \text{true for n=1} \\ \text{Assuming statement is true for n=k, then } \frac{d^k(e^{x^2})}{dx}=P_1(x)e^{x^2}\\ \text{R.T.P true for n=k+1 i.e } \frac{d^{k+1}(e^{x^2})}{dx} =P_2(x)e^{x^2} \\ \text{From assumption, LHS: } \frac{d(P_1(x)e^{x^2})}{dx} \\ =P{}'_1(x)e^{x^2}+P_1(x)\cdot 2x\cdot e^{x^2} \\ =e^{x^2}\left (P{}'_1(x) +P_1(x)\cdot 2x \right ) \text) \\ =P_2(x)\cdot e^{x^2} = \text{RHS} \left ( \text{ as }P{}'_1(x) +P_1(x)\cdot 2x \text{ is any polynomial with degree k+1 as }P_1(x) \text{ is of degree k}\right ) \\\\ \text{Statement is true for n=1}, \therefore \text{true for n=1+1 =2, 2+1=3 and so on}\\ \text{So it is true for all } n\geq 1 \text{ by principle of Mathematical Induction.} \\\\ \text{Also: do markers even give marks for writing the conclusion? I usually skip it or just write "True by M.I"}$

Don't need to waste time writing a long conclusion involving the 'if true for n = 1, then true for n = 2' etc. mantra (in fact, it's theoretically bad to). Just need to write something like 'hence the result is true by induction'.

Paradoxica · Sep 29, 2016

Re: HSC 2016 3U Marathon

leehuan said:
The natural numbers start from 0, not 1.

No they don't. You don't control the international standards which nobody has even agreed upon.

On that note though, I include 0 as a natural number.

pikachu975 · Sep 29, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
Don't need to waste time writing a long conclusion involving the 'if true for n = 1, then true for n = 2' etc. mantra (in fact, it's theoretically bad to). Just need to write something like 'hence the result is true by induction'.

Fitzpatrick does it like:

True for n=k+1 if n=k is true
True for n=1
By induction, it is true for all positive integers n

Would this be alright too

leehuan · Sep 29, 2016

Re: HSC 2016 3U Marathon

Paradoxica said:
No they don't. You don't control the international standards which nobody has even agreed upon.

On that note though, I include 0 as a natural number.

I stopped caring about authors who said otherwise after too much exposure to "positive integers" everywhere.

Paradoxica · Sep 29, 2016

Re: HSC 2016 3U Marathon

leehuan said:
I stopped caring about authors who said otherwise after too much exposure to "positive integers" everywhere.

The only ones who do nowadays are those who work in Real Analysis... But that's not my only priority so my definition will not align with theirs.

Paradoxica · Sep 30, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
Damn just do rectangles lel...

KingOfActing · Sep 30, 2016

Re: HSC 2016 3U Marathon

Drsoccerball said:
The induction is left uncompleted but heres a follow up :
$$ Prove that: $ \frac{2n}{3} \cdot \sqrt{n} \leq \sqrt{1} + \sqrt{2} + ... + \sqrt{n} \leq \frac{(4n+3)}{6} \cdot \sqrt{n}$

$$Consider the trapezoidal approximation of the same integral as before. As the function is monotonic increasing, we get:$\\\begin{align*}\sum_{k = 1}^n \frac{\sqrt{k-1} + \sqrt{k}}{2}\times 1 &\leq \frac{2}{3}n\sqrt{n} \\\frac{1}{2}\left(0 + \sqrt{1} + \sqrt{1} + \sqrt{2} + \dots + \sqrt{n-1} + \sqrt{n} \right ) &\leq\frac23n\sqrt{n}\\\frac12\left(2\left(\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}\right) - \sqrt{n}\right) &\leq \frac23n\sqrt{n}\\\sqrt{1} + \sqrt{2} + \dots + \sqrt{n} &\leq \left(\frac{2n}{3} + \frac12\right)\sqrt{n} = \frac{4n + 3}{6} \times \sqrt{n}\end{align*}$

Let me guess, next part is squeeze theorem?

Drsoccerball · Sep 30, 2016

Re: HSC 2016 3U Marathon

pikachu975 said:
Fitzpatrick does it like:

True for n=k+1 if n=k is true
True for n=1
By induction, it is true for all positive integers n

Would this be alright too

Just write : "The proposition P(n) is true for all positive integers n." *Assuming you said "let the proposition/statement be P(n)."*

Also as Integrand says it is not theoretically right as because when do you stop counting ? (As Prof. Peter Brown said)

Mathematical Induction is one of or the only thing in maths that proves something for all infinite cases from my knowledge.

KingOfActing · Sep 30, 2016

Re: HSC 2016 3U Marathon

trecex1 said:
$\\ \text{Prove that }x\leq e^{x-1}\\ \text{Hence or otherwise:} \\\\ i)\text{ For }a_1,a_2...a_n > 0 \text{ then } \frac{a_1+a_2+...+a_n}{n}\geq \sqrt[n]{a_1a_2..a_n}\\ ii)\text{For }a,b,c,d >0 \text{ then }(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 9 \\ iii) \text{For }a,b,c,> 0 \text{ then } \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geq \frac{3}{2} \\ \\ \text{Bonus: more methods to prove }i)$

$\frac{d^2}{dx^2}e^{x-1} = e^{x-1} > 0 \quad \forall x \in \mathbb R\\\\\therefore e^{x-1}$ is a convex function.$\\$Note that the tangent at $x = 1$ of $e^{x-1}$ is $ y = x$. Since $e^{x-1}$ is a convex function, it is always above its tangents, and hence for all real $x,\; e^{x-1} \geq x\\\\i)$ Let $\overline{a} = \frac{a_1+a_2+\dots+a_n}{n}.\\\frac{a_1}{\overline{a}}\leq e^{a_1/\overline{a}-1}\\\\\frac{a_1a_2}{\overline{a}^2}\leq e^{a_1/\overline{a} + a_2/\overline{a}-2}\\\\\dots\\\\\frac{a_1a_2a_3\dots a_n}{\overline{a}^n}\leq e^{a_1/\overline{a}+a_2/\overline{a}+a_3/\overline{a}+\dots+a_n/\overline{a} - n} = e^{n\overline{a}/\overline{a} - n} = e^0 = 1 \\\left(\frac{a_1+a_2+\dots+a_n}{n} \right )^n \geq a_1a_2\dots a_n\\\frac{a_1+a_2+\dots+a_n}{n} \geq \sqrt[n]{a_1a_2\dots a_n}\\ii) \left(a+b+c \right )\left(\frac1a+\frac1b+\frac1c \right )\geq 3 \times \sqrt[3]{abc} \times 3\times \sqrt[3]{\frac{1}{abc}} = 9$

I need sleep, maybe iii tomorrow

davidgoes4wce · Sep 30, 2016

Re: HSC 2016 3U Marathon

InteGrand said:
$\noindent There's no $O$ in the diagram (and it should be $S$ rather than $R$ to use that result). So what we could say is $PT^2 = PS \times PQ$.$

$\noindent Edit: oh, yeah, if we define $O$ as the foot of the wall, then that is true.$

Am I also right in saying with this question , is when P and T are at equal points ), theta is the maximum? Do they have to be on the tangent line?

davidgoes4wce · Sep 30, 2016

Re: HSC 2016 3U Marathon

davidgoes4wce said:
This was the last question in the 2009 HSC

they put the reasoning

$OT^2=OR \times OQ $ (The square of the length of the tangent from an external point is equal to the product of the intercepts of the secant passing through this point) $$

I didn't see it first time but the external point in this case is at the foot of the wall. Could someone confirm?

Am I also right in saying with this question , is when P and T are at equal points ), theta is the maximum? Do they have to be on the tangent line?

InteGrand · Sep 30, 2016

Re: HSC 2016 3U Marathon

Yeah, theta is maximised when P = T.

Paradoxica · Sep 30, 2016

Re: HSC 2016 3U Marathon

mini8658 said:
$$\noindent Solve $\sin^{-1}{x}-\cos^{-1}{y}=\frac{\pi}{12} $ and $ \cos^{-1}{x}+\sin^{-1}{y}=\frac{5\pi}{12}$

$$\noindent Add and subtract both equations to and from each other.$ \\\\ \frac{\pi}{2} + \sin^{-1}{y} - \cos^{-1}{y} = \frac{\pi}{2} \\\\ \cos^{-1}{x} - \sin^{-1}{x} + \frac{\pi}{2} = \frac{\pi}{3} \\\\ $Use the $\frac{\pi}{2}$ on both equations to complement the angles.$ \\\\ 2\sin^{-1}{y} = \frac{\pi}{2} \\\\ 2\cos^{-1}{x} = \frac{\pi}{3} \\\\ y = \frac{1}{\sqrt{2}}, x = \frac{\sqrt{3}}{2}$

KingOfActing · Oct 1, 2016

Re: HSC 2016 3U Marathon

$\begin{align*}iii)\; \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} &=(a+b+c)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&=\frac{1}{2}\left((a+b) + (b + c) + (c + a)\right)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&\geq\frac{1}{2}\times 9 - 3\\&=\frac{3}{2}\end{align*}$

I sincerely remember there being a method to do this one by showing that equality like in the AM-GM inequality was when a=b=c, but I just can't remember how.

Paradoxica · Oct 1, 2016

Re: HSC 2016 3U Marathon

KingOfActing said:
$\begin{align*}iii)\; \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} &=(a+b+c)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&=\frac{1}{2}\left((a+b) + (b + c) + (c + a)\right)\left(\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{a+c} \right ) - 3\\&\geq\frac{1}{2}\times 9 - 3\\&=\frac{3}{2}\end{align*}$

I sincerely remember there being a method to do this one by showing that equality like in the AM-GM inequality was when a=b=c, but I just can't remember how.

Magically complete some squares?

HSC 2016 MX1 Marathon (archive) (1 Viewer)

Well-Known Member

Well-Known Member

Well-Known Member

Active Member

Well-Known Member

Well-Known Member

-insert title here-

Premium Member

Well-Known Member

-insert title here-

-insert title here-

lukewarm mess

Well-Known Member

lukewarm mess

Well-Known Member

Well-Known Member

Well-Known Member

-insert title here-

lukewarm mess

-insert title here-

Users Who Are Viewing This Thread (Users: 0, Guests: 1)