HSC 2016 MX1 Marathon (archive) (3 Viewers)

Drsoccerball · Nov 5, 2015

Re: HSC 2016 3U Marathon

leehuan said:
NEXT QUESTION:

$\\ $Use mathematical induction to prove the general arithmetic mean-geometric mean inequality: $ \\ \frac { { a }_{ 1 }+{ a }_{ 2 }+\dots +{ a }_{ k } }{ k } \ge \sqrt [ k ]{ { a }_{ 1 }{ a }_{ 2 }\dots { a }_{ k } }$

I'll tone the difficulty back down after this one.

LOL only 5 people in the state (apparently) in extension 2 got this right and there were steps LOL... I also don't think reverse induction is 3U or even 4U difficulty...

calamebe · Nov 6, 2015

Re: HSC 2016 3U Marathon

Well I don't have an answer to the latest one (yet

) but I finally did part g) of the other question. You just let d = 0 and the area is root(s(s-a)(s-b)(s-c)), and the property of being a cyclic triangle is not required because every triangle is a cyclic triangle.

leehuan · Nov 6, 2015

Re: HSC 2016 3U Marathon

Do you really need reverse induction to prove it?

Though, I was thinking about whether or not that one was too harsh lol, I'll post a new question anyway if even 2 days later nobody has come up with a solution.

Paradoxica · Nov 6, 2015

Re: HSC 2016 3U Marathon

Drsoccerball said:
LOL only 5 people in the state (apparently) in extension 2 got this right and there were steps LOL... I also don't think reverse induction is 3U or even 4U difficulty...

There is a more direct method of induction that doesn't involve forwards-backwards induction. The proof is on Wikipedia.

leehuan · Nov 6, 2015

Re: HSC 2016 3U Marathon

Paradoxica said:
There is a more direct method of induction that doesn't involve forwards-backwards induction. The proof is on Wikipedia.

Except, y u do dis...
---------------------------------------
NEXT QUESTION:

$\\ $The points P$(2ap,a{p}^{2})$ and Q$(2aq,a{q}^{2})$ lie on the parabola ${x}^{2}=4ay$ such that their tangents are always perpendicular to each other. \\ (a) Deduce that PQ is a focal chord. \\ (b) Find the locus of the intersection of the tangents.$

kawaiipotato · Nov 6, 2015

Re: HSC 2016 3U Marathon

Paradoxica said:
There is a more direct method of induction that doesn't involve forwards-backwards induction. The proof is on Wikipedia.

Is there a limk to the backwards induction proof? Or is it on Wikipedia?

Paradoxica · Nov 6, 2015

Re: HSC 2016 3U Marathon

kawaiipotato said:
Is there a limk to the backwards induction proof? Or is it on Wikipedia?

Also on wikipedia. Although I'm not quite certain we're talking about the same proof here...

InteGrand · Nov 6, 2015

Re: HSC 2016 3U Marathon

kawaiipotato said:
Is there a limk to the backwards induction proof? Or is it on Wikipedia?

It's this one (Cauchy's forward-backward induction proof): https://en.wikipedia.org/wiki/Inequ...uchy_using_forward.E2.80.93backward_induction

This proof was also an Extension Q in a 3U Pender (Cambridge) textbook exercise.

relativity1 · Nov 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
Except, y u do dis...
---------------------------------------
NEXT QUESTION:

$\\ $The points P$(2ap,a{p}^{2})$ and Q$(2aq,a{q}^{2})$ lie on the parabola ${x}^{2}=4ay$ such that their tangents are always perpendicular to each other. \\ (a) Deduce that PQ is a focal chord. \\ (b) Find the locus of the intersection of the tangents.$

equation of pq = y=((p+q)/2)x - apq
For this to be a focal chord must pass through o,a
sub o,a --> a=-apq therefore pq=-1 (not too sure if this would count as proof that its a focal chord but if i recall the textbook states this is general form)

b)d(1/4a.x^2)/dx =x/2a
subbing 2ap into x/2a = p
tangent P =
y-ap^2=p(x-2ap) ---> y=xp-ap^2
similarly for q ---> y=xq-aq^2
solving simultaneously
xq-aq^2=xp-ap^2
x(q-p)=a(q-p)(q+p)
x=a(p+q)
therefore since y=xq-aq^2
y=aq(p+q)-aq^2
=aqp+aq^2-aq^2
=aqp
therefore point of intersection is (a(p+q),apq)
since y=apq and from a) we know pq=-1
therefore locus is y=-a

leehuan · Nov 6, 2015

Re: HSC 2016 3U Marathon

Yep
----------------------------
NEXT QUESTION:
Troll time
$$How many different arrangements are possible using the letters of 'SUPERCALIFRAGILISTICEXPIALIDOCIOUS'?$

VBN2470 · Nov 6, 2015

Re: HSC 2016 3U Marathon

ANSWER: $1.41246953 \times 10^{30}$

NEW QUESTION:

Prove that $\dfrac{\text{d}}{\text{d}x}(\tan{x})=\sec^{2}x$ from first principles.

Paradoxica · Nov 6, 2015

Re: HSC 2016 3U Marathon

VBN2470 said:
ANSWER: $1.41246953 \times 10^{30}$

NEW QUESTION:

Prove that $\dfrac{\text{d}}{\text{d}x}(\tan{x})=\sec^{2}x$ from first principles.

$Using first principles, we have$

$\tan(x)' = \lim_{\delta x \rightarrow 0}\frac{\tan{(x+\delta x)} - \tan{x}}{\delta x}$

$Expand using compound angles for the tangent function.$

$= \lim_{\delta x \rightarrow 0}\frac{\frac{\tan{x}+ \tan{\delta x}}{1-\tan{x}\tan{\delta x}} - \tan{x}}{\delta x}= \lim_{\delta x \rightarrow 0}\frac{\tan{x}+ \tan{\delta x} - \tan{x} + \tan^2{x}\tan{\delta x}}{\delta x - \delta x \tan{x} \tan{\delta x}}$

$$Negate and divide everything by $\tan{\delta x}$

$\lim_{\delta x \rightarrow 0} \frac{1+\tan^2{x}}{\frac{\delta x}{\tan{\delta x}} - \delta x \tan{x}}$

$$Invoke the trigonometric limit for $\frac{\sin{x}}{x}$

$\lim_{ x \rightarrow 0}\frac{\sin{x}}{x} = 1 \Rightarrow \lim_{x \rightarrow 0}\frac{x}{\sin{x}} = 1$

$\lim_{x \rightarrow 0}\frac{x}{\tan{x}} = \lim_{x \rightarrow 0}\frac{x\cos{x}}{\sin{x}} = 1$

$\lim_{\delta x \rightarrow 0} \frac{1+\tan^2{x}}{\frac{\delta x}{\tan{\delta x}} - \delta x \tan{x}} = \frac{\sec^2{x}}{1 - 0}$

$\therefore \tan(x)' = \sec^2{x}$

leehuan · Nov 6, 2015

Re: HSC 2016 3U Marathon

NEXT QUESTION:

$\\ $Prove the tangent half-angle substitution results (aka. t-formulae) \\ You may choose to use a diagram.$$

KingOfActing · Nov 6, 2015

Re: HSC 2016 3U Marathon

leehuan said:
NEXT QUESTION:

$\\ $Prove the tangent half-angle substitution results (aka. t-formulae) \\ You may choose to use a diagram.$$

$\begin{align*}\sin{x}&=\sin{\left ( \frac{x}{2} + \frac{x}{2} \right )} \\&=2\sin{\frac{x}{2}}\cos{\frac{x}{2}} \text{ \fbox{Addition of angles Formula}} \\&=2\frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}}\cos^2{\frac{x}{2}} \\&= \frac{2\tan{\frac{x}{2}}}{\sec^2{\frac{x}{2}}} \\ &= \frac{2\tan{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}} \text{ \fbox{Pythagorean Identity} (*)} \end{align*}$

$\begin{align*}\cos{x}&=\cos{\left ( \frac{x}{2} + \frac{x}{2} \right )} \\&=\cos^2{\frac{x}{2}} - \sin^2{\frac{x}{2}} \text{ \fbox{Addition of angles Formula}} \\&=\cos^2{\frac{x}{2}} - \frac{\sin^{2}{\frac{x}{2}}}{\cos^{2}{\frac{x}{2}}}\cos^2{\frac{x}{2}}\\&=\cos^2{\frac{x}{2}}\left (1 - \tan^2{\frac{x}{2}} \right )\\&=\frac{1 - \tan^2{\frac{x}{2}}}{\sec^2{\frac{x}{2}}}\\&=\frac{1-\tan^2{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}} \text{ \fbox{Pythagorean Identity} (**)} \end{align*}$

$\begin{align*}\tan{x}&=\frac{\sin{x}}{\cos{x}}\\&=\frac{\left ( \frac{2\tan{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}}\right )}{\left ( \frac{1-\tan^2{\frac{x}{2}}}{1 + \tan^2{\frac{x}{2}}}\right )} \text{ \fbox{Using (*) and (**)}}\\&=\frac{2\tan{\frac{x}{2}}}{1-\tan^2{\frac{x}{2}}}\end{align*}$

leehuan · Nov 7, 2015

Re: HSC 2016 3U Marathon

NEXT QUESTION

$\int { x\sqrt { x+1 } dx } $ using the substitution $ \\ $(a) $u=x+1 \\ $(b) ${u}^{2}=x+1$

RachelGreen · Nov 7, 2015

Re: HSC 2016 3U Marathon

RachelGreen · Nov 7, 2015

Re: HSC 2016 3U Marathon

leehuan · Nov 7, 2015

Re: HSC 2016 3U Marathon

NEXT QUESTION:
$\\ $Prove the statement of the product rule: $ \\ \frac { d }{ dx } \left( f\left( x \right) g\left( x \right) \right) =g\left( x \right) f^{ \prime }\left( x \right) +f\left( x \right) g^{ \prime }\left( x \right)$

rand_althor · Nov 7, 2015

Re: HSC 2016 3U Marathon

leehuan said:
NEXT QUESTION:
$\\ $Prove the statement of the product rule: $ \\ \frac { d }{ dx } \left( f\left( x \right) g\left( x \right) \right) =g\left( x \right) f^{ \prime }\left( x \right) +f\left( x \right) g^{ \prime }\left( x \right)$

$\begin{align*}f'(x)&=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \textup{ and }g'(x)=\lim_{h \to 0} \frac{g(x+h)-g(x)}{h}. \\\frac{\mathrm{d}}{\mathrm{d}x}\left (f(x)g(x) \right )&=\lim_{h \to 0} \frac{f(x+h)g(x+h)-f(x)g(x)}{h} \\&=\lim_{h \to 0}\frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x)}{h} \\&=\lim_{h \to 0}f(x+h)\cdot\lim_{h \to 0}\frac{g(x+h)-g(x)}{h} + \lim_{h \to 0}g(x)\cdot\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \\&=f(x)g'(x)+g(x)f'(x)\end{align*}$

leehuan · Nov 7, 2015

Re: HSC 2016 3U Marathon

$\\ $leehuan guessed all 10 questions in his MX1 multiple choice. What is the probability that he got at least 9/10 right anyway?$$

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