HSC 2015 MX2 Marathon (archive) (5 Viewers)

Drsoccerball · Mar 23, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
I understand what you're trying to say, but I don't understand how you intended to use the double derivative to acquire what you want to acquire.

Also, the turning points don't have to be under the x axis in order for the poly to have a maximal value less than 1. What if the function values were say 0.7, 0.5, 0.3 etc?

By proving that f"(x) has stationary points above or below the axis this shows thst there is only one root. If there is only one root in f"(x) then there is only one turning point in f'(x)

FrankXie · Mar 23, 2015

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ 8x^3 - 3x^2 - 3x - 1 = 0 \\$

lol. so it is -1 at the end but not +1, I was thinking if it was +1, there is no solution in that form.

seanieg89 · Mar 23, 2015

Re: HSC 2015 4U Marathon

Newton's method (as taught in HS) definitely isn't good enough to prove inequalities like the one Sy123 posted without additional work.

You are not taught any criteria by to check whether the resulting sequence does indeed converge to an exact root for given seed values $x_0$ , nor do you have any quantitative estimate for how large the error is between the n-th approximate root and the exact root.

You need these things to make such a proof attempt rigorous.

Carrotsticks · Mar 23, 2015

Re: HSC 2015 4U Marathon

seanieg89 said:
Newton's method (as taught in HS) definitely isn't good enough to prove inequalities like the one Sy123 posted without additional work.

You are not taught any criteria by to check whether the resulting sequence does indeed converge to an exact root for given seed values $x_0$ , nor do you have any quantitative estimate for how large the error is between the n-th approximate root and the exact root.

You need these things to make such a proof attempt rigorous.

Looking at DrSoccer's posts in here, I doubt this would have been understood.

I tried stepping down a level and countering their 'proof' conceptually, but even that seemed to be too much.

Drsoccerball · Mar 23, 2015

Re: HSC 2015 4U Marathon

Carrotsticks said:
Looking at DrSoccer's posts in here, I doubt this would have been understood.

I tried stepping down a level and countering their 'proof' conceptually, but even that seemed to be too much.

What does converge mean

Drsoccerball · Mar 23, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
What does converge mean

Besides that sounds pretty legit

seanieg89 · Mar 23, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
What does converge mean

A sequence converges to blah if the limiting value of the sequence is blah.

The formal definition involves topological spaces.

turntaker · Mar 24, 2015

Re: HSC 2015 4U Marathon

Integral of SecX DX using two methods

InteGrand · Mar 24, 2015

Re: HSC 2015 4U Marathon

turntaker said:
Integral of SecX DX using two methods

The "trick" method:

$\int \sec x \text{ d}x=\int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}\text{ d}x=\int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\text{ d}x=\int \frac{(\sec x + \tan x)^{\prime}}{\sec x + \tan x}\text{ d}x=\ln |\sec x + \tan x|+C$

turntaker · Mar 24, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
The "trick" method:

$\int \sec x \text{ d}x=\int \sec x \frac{\sec x + \tan x}{\sec x + \tan x}\text{ d}x=\int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x}\text{ d}x=\int \frac{(\sec x + \tan x)^{\prime}}{\sec x + \tan x}\text{ d}x=\ln |\sec x + \tan x|+C$

What's that

InteGrand · Mar 24, 2015

Re: HSC 2015 4U Marathon

turntaker said:
Integral of SecX DX using two methods

Partial fractions method:

$\int \sec x \text{ d}x=\int \frac{1}{\cos x}\text{ d}x$

$=\int \frac{\cos x}{\cos^2 x}\text{ d}x$

$=\int \frac{\text{d}(\sin x)}{1-\sin ^2 x}$

$=\int \frac{\text{d}(\sin x)}{(1+\sin x)(1-\sin x)}$

$=\int \left ( \frac{\frac{1}{2}}{1+\sin x}+\frac{\frac{1}{2}}{1-\sin x} \right )\text{d}(\sin x)$

$=\frac{1}{2}\ln \left | \frac{1+\sin x}{1-\sin x} \right |+C$ .

InteGrand · Mar 24, 2015

Re: HSC 2015 4U Marathon

turntaker said:
What's that

What's what?

Drsoccerball · Mar 24, 2015

Re: HSC 2015 4U Marathon

Can someone integrate $\frac{-1}{\sqrt{16 -x^2}}dx$

And explain why the answer isnt

$-sin^{-1}(x/4) + c$

swagswagyoloswag · Mar 24, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Can someone integrate $\frac{-1}{\sqrt{16 -x^2}}dx$

And explain why the answer isnt

$-sin^{-1}(x/4) + c$

It is correct. But notice there is a +c. Sin^(-1)(x/4) and cos^(-1)(x/4) differ by pi/2.
to prove this: d/dx sin^(-1)(x/4) + cos^(-1)(x/4)
= 0 (cbb typing the differentation step)
siince it equals zero and if two functions equal you can integrate.
Int d/dx sin^(-1)(x/4) + cos^(-1)(x/4) = int 0 dx
therefore, sin^(-1)(x/4) + cos^(-1)(x/4) = D where D is a constant
now you can sub any number as they are congruent. Sub x = 4 you get D = pi/2.
Therefore they differ by pi/2 and therefore your answer is still correct

turntaker · Mar 24, 2015

Re: HSC 2015 4U Marathon

InteGrand said:
What's what?

ttrick method.
how did you know to multiply by secx + tanx

what would you multiply by if the question was 1/cosec instead of sec

swagswagyoloswag · Mar 24, 2015

Re: HSC 2015 4U Marathon

turntaker said:
ttrick method.
how did you know to multiply by secx + tanx

what would you multiply by if the question was 1/cosec instead of sec

1/cosecx is just sinx

InteGrand · Mar 24, 2015

Re: HSC 2015 4U Marathon

turntaker said:
ttrick method.
how did you know to multiply by secx + tanx

what would you multiply by if the question was 1/cosec instead of sec

$\int \csc x \text{ d}x=\int \csc x \cdot \frac{\csc x + \cot x}{\csc x + \cot x}\text{ d}x=\int \frac{\csc^2 x + \csc x\cot x}{\csc x + \cot x}\text{ d}x=\int \frac{-(\csc x + \cot x)^{\prime}}{\csc x + \cot x}\text{ d}x=-\ln|\csc x + \cot x|+C$

turntaker · Mar 24, 2015

Re: HSC 2015 4U Marathon

swagswagyoloswag said:
1/cosecx is just sinx

oh I meant integral of cosecx using the trick method

Sy123 · Mar 28, 2015

Re: HSC 2015 4U Marathon

$\\ $Let there be a triangle$ \ ABC \ $and points$ \ P,Q,R \ $on$ \ BC, AC, AB \ $respectively, such that$ \ P,Q,R \ $are midpoints on the sides$ \ BC,AC,AB \ $respectively$ \\ \\ $Show that$ \\ 4(AP^2 + BQ^2 + CR^2) = 3(AB^2 + AC^2 + BC^2)$

Sy123 · Mar 29, 2015

Re: HSC 2015 4U Marathon

$\\ $Let$ \ P(x) \ $and$ \ Q(x) \ $be polynomials of degree$ \ n \ $and are such that$ \\ P(x) = Q(x) \ $for some$ \ n + 1 \ $distinct values of$ \ x \\\\ $Show that$ \ P(x) = Q(x) \ $for all$ \ x$

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