HSC 2015 MX2 Marathon (archive) (1 Viewer)

Drsoccerball · Jun 18, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Prove that for$ \ x \ $sufficiently close to$ \ 0 \ $that$ \ \cos x < \frac{\sin x}{x} < 1$

$\\ $ii) Hence show that$ \ \lim_{x \to 0} \frac{\sin x}{x} = 1$

$i) xcosx \leq sinx \leq x $ for very small positive values$

$ii) \lim_{x \to 0} 1 \leq \lim_{x \to 0}\frac{sinx}{x} \leq 1$

$$ Via squeeze theorem $ \lim_{x \to 0}\frac{sinx}{x} =1$
My teachers way of explaining squeeze theorem:
"If eating a burger and theres cheese in it, you will always eat the cheese"

InteGrand · Jun 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
My teachers way of explaining squeeze theorem:
"If eating a burger and theres cheese in it, you will always eat the cheese"

Apparently it's also called the 'two policemen and a drunk' theorem.

'The story is that if two policemen are escorting a drunk prisoner between them, and both officers go to a cell, then (regardless of the path taken, and the fact that the prisoner may be wobbling about between the policemen) the prisoner must also end up in the cell.' ( https://en.wikipedia.org/wiki/Squeeze_theorem )

Silly Sausage · Jun 18, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Apparently it's also called the 'two policemen and a drunk' theorem.

'The story is that if two policemen are escorting a drunk prisoner between them, and both officers go to a cell, then (regardless of the path taken, and the fact that the prisoner may be wobbling about between the policemen) the prisoner must also end up in the cell.' ( https://en.wikipedia.org/wiki/Squeeze_theorem )

squeeze theorem, pinch theorem, sandwich theorem and now two policemen and a drunk?
Give a me a break!

RealiseNothing · Jun 18, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$i) xcosx \leq sinx \leq x $ for very small positive values$

$ii) \lim_{x \to 0} 1 \leq \lim_{x \to 0}\frac{sinx}{x} \leq 1$

$$ Via squeeze theorem $ \lim_{x \to 0}\frac{sinx}{x} =1$
My teachers way of explaining squeeze theorem:
"If eating a burger and theres cheese in it, you will always eat the cheese"

Your first part (i) is not correct.

InteGrand · Jun 18, 2015

Re: HSC 2015 4U Marathon

Silly Sausage said:
squeeze theorem, pinch theorem, sandwich theorem and now two policemen and a drunk?
Give a me a break!

From that same Wikipedia article:

'In Italy, China, Chile, Russia, Poland, Hungary and France, the squeeze theorem is also known as the two carabinieri theorem, two militsioner theorem, sandwich theorem, two gendarmes theorem, "Double sided theorem" or two policemen and a drunk theorem.'

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ S_n(\alpha) = \sum_{k=1}^n \frac{1}{k^{\alpha}} \ $where$ \ \alpha > 0$

$\\ $i) Show through a graph or otherwise$ \ \int_1^{n-1} \frac{1}{x^{\alpha}} \ dx < S_n(\alpha) < 1 + \int_1^n \frac{1}{x^\alpha} \ dx$

$\\ $ii) Hence show that$ \ \lim_{n \to \infty} S_n(\alpha) \rightarrow \infty \ $for$ \ \alpha \leq 1$

Sy123 · Jun 18, 2015

Re: HSC 2015 4U Marathon

$\\ $i) Show that$ \ x^2 + y^2 \geq 2xy \ $for positive real$ \ x,y$

$\\ $ii) Hence show that$ \ \frac{2ab}{a+b} + \frac{2ac}{a+c} + \frac{2bc}{b+c} \leq a + b + c \ $for positive real$ \ a,b,c$

(note, harder than the previous couple, but still HSC exam level)

psyc1011 · Jun 19, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $i) Show that$ \ x^2 + y^2 \geq 2xy \ $for positive real$ \ x,y$

$\\ $ii) Hence show that$ \ \frac{2ab}{a+b} + \frac{2ac}{a+c} + \frac{2bc}{b+c} \leq a + b + c \ $for positive real$ \ a,b,c$

(note, harder than the previous couple, but still HSC exam level)

Sorry for grammar and i have tried my best to condense working

i) $(x-y)^2 > 0 \iff x^2 + y^2 \geq 2xy, \quad x,y >0 \tag{1}$
ii) From (1)
x^2 -> x, y^2 -> y, z^2 -> z and divide give
$\frac{1}{x+y} \leq 2 \sqrt{xy} \tag{2}$

Use (2) and change (x,y) -> (a,b), (a,c) and (b,c) (label these equations as (3)) and sub into q give
$\leq \frac{2ab}{2\sqrt{ab}} + \frac{2ac}{2\sqrt{ac}} + \frac{2bc}{2\sqrt{bc}} \leq \sqrt{ab} + \sqrt{ac} + \sqrt{bc}.$

Recall (3) and add them up
$a+b \geq 2 \sqrt{ab}$
$c+b \geq 2 \sqrt{bc}$
$a+c \geq 2 \sqrt{ac}$
after cancel 2 give
$a+b+c \geq \sqrt{ab} + \sqrt{bc }+ \sqrt{ac}$

which answers the question!

BreathingMaths · Jun 19, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
$$ Prove that $ \lim _{x\rightarrow 0}}xlnx =0$

Let $x = \frac{1}{u}$
$\lim_{u \rightarrow \infty} \frac{1}{u} \text{ln}\left ( \frac{1}{u} \right ) = \lim_{u \rightarrow \infty} 0 - \frac{\text{ln}(u)}{u} = 0$
As u approaches infinity, u increases more rapidly than ln(u), so the fraction ln(u)/u will tend to 0

Trebla said:
Can you explain why

$\displaystyle\lim_{x\rightarrow 0} x^x = 1$

Using the above proved limit, we have
$\lim_{x \rightarrow 0} x^x = \lim_{x \rightarrow 0} e^{\text{ln}(x^x)} = \lim_{x \rightarrow 0}e^{x\text{ln}(x)} = e^{\lim_{x \rightarrow 0} x\text{ln}(x) } = 1$

Drsoccerball · Jun 19, 2015

Re: HSC 2015 4U Marathon

BreathingMaths said:
Let $x = \frac{1}{u}$
$\lim_{u \rightarrow \infty} \frac{1}{u} \text{ln}\left ( \frac{1}{u} \right ) = \lim_{u \rightarrow \infty} 0 - \frac{\text{ln}(u)}{u} = 0$
As u approaches infinity, u increases more rapidly than ln(u), so the fraction ln(u)/u will tend to 0

Using the above proved limit, we have
$\lim_{x \rightarrow 0} x^x = \lim_{x \rightarrow 0} e^{\text{ln}(x^x)} = \lim_{x \rightarrow 0}e^{x\text{ln}(x)} = e^{\lim_{x \rightarrow 0} x\text{ln}(x) } = 1$

Its not proved by elementary methods

BreathingMaths · Jun 19, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Its not proved by elementary methods

Which one? How is that not elementary methods?

Sy123 · Jun 19, 2015

Re: HSC 2015 4U Marathon

psyc1011 said:
Sorry for grammar and i have tried my best to condense working

i) $(x-y)^2 > 0 \iff x^2 + y^2 \geq 2xy, \quad x,y >0 \tag{1}$
ii) From (1)
x^2 -> x, y^2 -> y, z^2 -> z and divide give
$\frac{1}{x+y} \leq 2 \sqrt{xy} \tag{2}$

Use (2) and change (x,y) -> (a,b), (a,c) and (b,c) (label these equations as (3)) and sub into q give
$\leq \frac{2ab}{2\sqrt{ab}} + \frac{2ac}{2\sqrt{ac}} + \frac{2bc}{2\sqrt{bc}} \leq \sqrt{ab} + \sqrt{ac} + \sqrt{bc}.$

Recall (3) and add them up
$a+b \geq 2 \sqrt{ab}$
$c+b \geq 2 \sqrt{bc}$
$a+c \geq 2 \sqrt{ac}$
after cancel 2 give
$a+b+c \geq \sqrt{ab} + \sqrt{bc }+ \sqrt{ac}$

which answers the question!

I think there are some typos there but I see what you're trying to say, what I did:

$x^2 + y^2 \geq 2xy \Rightarrow (x+y)^2 \geq 4xy \Rightarrow \ \frac{x+y}{2} \geq \frac{2xy}{x+y}$

$\\ \therefore \ \frac{2ab}{a+b} + \frac{2ac}{a+c} + \frac{2bc}{b+c} \leq \frac{a+b}{2} + \frac{a+c}{2} + \frac{b+c}{2} = a + b + c \ \square$

glittergal96 · Jun 20, 2015

Re: HSC 2015 4U Marathon

BreathingMaths said:
Which one? How is that not elementary methods?

Saying that log(x)/x -> 0 as x -> inf because the latter "increases more rapidly" is vague and not rigorous. You would have to state what your mean by "increasing more rapidly", and what this means for the limits of quotients. (This amounts to the L'Hopital rule, which is outside syllabus).

A proof without using this technique is as follows:

Note the the LHS is positive for all x>1.

We also have $\log(x) < \sqrt{x-1}$ for $x > 1$ (*)$$ .

To prove (*), we note that equality holds at 1, and

$(RHS-LHS)'(x)=\frac{1}{2\sqrt{x-1}}-\frac{1}{x}=\frac{x-2\sqrt{x-1}}{2x\sqrt{x-1}}.$

Now the denominator of this quotient is positive, and the numerator is non-negative, as

$x^2-(2\sqrt{x-1})^2=(x-2)^2\geq 0.$

This tells us that (*) holds for all x>1.

So, returning to our original problem, we have for x > 1 that:

$0<\frac{\log(x)}{x}\leq \frac{\sqrt{x-1}}{x} \leq \frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\rightarrow 0.$

From here, the squeeze theorem completes the proof.

(Sorry about blending LaTeX and plaintext, I am a little lazy tonight.)

glittergal96 · Jun 20, 2015

Re: HSC 2015 4U Marathon

To justify a statement drsoccerball used earlier without proof, I am going to prove the following:

$\sin(x)< x < \tan(x)$

for

$x\in (0,\pi/2).$

We do it geometrically (it is tempting to fall into the trap of using calculus, but things like the formulae for differentiating trig functions COME from such inequalities, so we would need to be extremely careful to avoid circular logic).

Draw the unit circle and the point P on this circle with angle x radians from the positive x-axis in the anti-clockwise direction.

Drop a perpendicular from P to A on the x-axis. Let B be the point (1,0). Draw a tangent to the circle at B and let this meet the ray OP at C.

We note that the coordinates of these points are:

A=(cos(x),0), B=(1,0), P=(cos(x),sin(x)), C=(1,tan(x)).

Now:

{Area of triangle OPB} < {Area of circular sector OPB} < {Area of triangle OCB}

since these regions are nested.

Computing these, we get

$\frac{\sin x}{2}< \frac{x}{2}< \frac{\tan x}{2}$

as claimed.

glittergal96 · Jun 20, 2015

Re: HSC 2015 4U Marathon

A pretty easy but nice application of these ideas is to prove the differentiation formulae for the trig functions from first principles.

Note that the differentiation formulae trivially imply the limit in Sy's original question, since

$\lim_{x\rightarrow 0}\frac{\sin x}{x}=\lim_{x\rightarrow 0}\frac{\sin x-\sin 0}{x-0}=\frac{d}{dx}(\sin x)|_{x=0}=\cos(0)=1.$

Drsoccerball · Jun 20, 2015

Re: HSC 2015 4U Marathon

How do you rotate graphs around other graphs ?

braintic · Jun 20, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
How do you rotate graphs around other graphs ?

It is really only meaningful to rotate a graph about a point in 2 dimensions, or about a straight line in three dimensions.
Pointwise rotation of one curve about another curve will distort the shape of the first curve. Even then, the idea only carries meaning if every point on the first curve lies on one and only one normal to the second curve.

What motivates this question? Are you asking for a geometrical interpretation, or an algebraic method?

InteGrand · Jun 20, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
How do you rotate graphs around other graphs ?

Are you talking about for volumes?

Drsoccerball · Jun 20, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
Are you talking about for volumes?

yah

Drsoccerball · Jun 20, 2015

Re: HSC 2015 4U Marathon

braintic said:
It is really only meaningful to rotate a graph about a point in 2 dimensions, or about a straight line in three dimensions.
Pointwise rotation of one curve about another curve will distort the shape of the first curve. Even then, the idea only carries meaning if every point on the first curve lies on one and only one normal to the second curve.

What motivates this question? Are you asking for a geometrical interpretation, or an algebraic method?

I just want to go further than rotating around straight x= and y= lines...

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