HSC 2013 Maths Marathon (archive) (2 Viewers)

Badsiii · Oct 19, 2013

Re: HSC 2013 2U Marathon

Show that
sec(x) - sec(x)cos^4(x) = sin(x)tan(x)
1 + cos^2(x)

Sy123 · Oct 19, 2013

Re: HSC 2013 2U Marathon

mahmoudali said:
y=ln(x) mad bro?

Yes I'm quite furious that you managed to answer that

----

$i) Show that the values of$ \ \ a \ \ $that satisfy$

$\int_{-a}^a e^{x} - 1 \ dx$

$Satisfy the equation$

$e^{a} - e^{-a} = 2a$

$ii) Show that$ \ \ a=0 \ \ $is the only solution$

Menomaths · Oct 19, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
Yes I'm quite furious that you managed to answer that

----

$i) Show that the values of$ \ \ a \ \ $that satisfy$

$\int_{-a}^a e^{x} - 1 \ dx$

$Satisfy the equation$

$e^{a} - e^{-a} = 2a$

$ii) Show that$ \ \ a=0 \ \ $is the only solution$

integral = (e^x - x) bounded by a and -a
(e^a - a)-(e^-a + a)
e^a -a -e^-a - a = 0
e^a - e^-a = 2a

logify both sides

a+a = log_e(2a)

2a = log_e(2a)

e^2a = 2a

therefore a has to = 0?

HeroicPandas · Oct 19, 2013

Re: HSC 2013 2U Marathon

Menomaths said:
integral = (e^x - x) bounded by a and -a
(e^a - a)-(e^-a + a)
e^a -a -e^-a - a = 0
e^a - e^-a = 2a

logify both sides

a+a = log_e(2a) <---- from here on, its wrong, because you invented your own log law

2a = log_e(2a)

e^2a = 2a

therefore a has to = 0?

(in red)

and if u sub in a =0,
e^2a = 2a
e^0 = 0
1 = 0
trollface.jpg

Badsiii · Oct 19, 2013

Re: HSC 2013 2U Marathon

Equation x^2 - 6x -8y= 23
i) Find the coordinates of the focus

ii) what is the equation of directrix

Menomaths · Oct 19, 2013

Re: HSC 2013 2U Marathon

HeroicPandas said:
(in red)

and if u sub in a =0,
e^2a = 2a
e^0 = 0
1 = 0
trollface.jpg

ahahhahahah sorry for inventing new things in maths, i have a habit of doing that :<

Shazer2 · Oct 19, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
Equation x^2 - 6x -8y= 23
i) Find the coordinates of the focus

ii) what is the equation of directrix

$x^{2} - 6x -8y = 23 \\ \\ \indent (x-3)^{2} = 8y + 32 $ (Completing the square)$ \\ \\ \indent (x-3)^2 = 8(y+4) \\ \\ \indent \therefore $ Vertex at $(3, -4) \\ \\ \\ \indent 4a = 8 \\ \\ \indent a = 2 \\ \\ \indent \therefore $ Focus at $(3, -2). \\ \\ \indent \therefore $ Hence, directrix is $ y = -6.$

Badsiii · Oct 19, 2013

Re: HSC 2013 2U Marathon

Simplfy:

sin(2pi-x)
Sin(pi/2 -x)

SpiralFlex · Oct 19, 2013

Re: HSC 2013 2U Marathon

Shazer2 said:
$x^{2} - 6x -8y = 23 \\ \\ \indent (x-3)^{2} = 8y + 32 $ (Completing the square)$ \\ \\ \indent (x-3)^2 = 8(y+4) \\ \\ \indent \therefore $ Vertex at $(3, -4) \\ \\ \\ \indent 4a = 8 \\ \\ \indent a = 2 \\ \\ \indent \therefore $ Focus at $(3, -2). \\ \\ \indent \therefore $ Hence, directrix is $ y = -6.$

Therefore hence?

anthony_wheels · Oct 19, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
Simplfy:

sin(2pi-x)
Sin(pi/2 -x)

π = pi
sin(2π-x) = -sin(x)
sin(π/2 - x) = cosx

∴ sin(2π-x)/sin(π/2 - x) = -sin(x)/cosx = -tanx

Shazer2 · Oct 19, 2013

SpiralFlex said:
Therefore hence?

Yeah I didn't mean that

was my solution correct though?

Sent from my iPhone using Tapatalk 2

ForOneEon · Oct 19, 2013

Re: HSC 2013 2U Marathon

Badsiii said:
Simplfy:

sin(2pi-x)
Sin(pi/2 -x)

sin (-x)
cos (x)

= -tan(x)

EDIT: Beaten to the punch by Lord Wheels! His solution is more in-depth too! >_<

Sy123 · Oct 19, 2013

Re: HSC 2013 2U Marathon

$i) Show that$

$1 + (\cos \theta + \sin \theta) + (\cos^2 \theta + \sin^2 \theta) + (\cos^3 \theta + \sin^3 \theta) + \dots = \frac{1-\sin \theta \cos \theta}{(1-\cos \theta)(1-\sin \theta)}$

$ii) Show that this sum is always positive$

girlworld_club · Oct 19, 2013

Re: HSC 2013 2U Marathon

Differentiate:

x Times (49-x^2)^1/2

Shazer2 · Oct 19, 2013

Re: HSC 2013 2U Marathon

girlworld_club said:
Differentiate:

x Times (49-x^2)^1/2

$x(49-x^2)^{1/2} \\ \\ \indent \frac {d}{dx} \quad x(49-x^{2})^{1/2} \\ \\ \indent = x \times -x(49-x^{2})^{-\frac {1}{2}} + (49-x^{2})^{1/2} \\ \\ \indent = \frac {-x^{2}}{\sqrt {49-x^{2}}} + \sqrt {49-x^{2}}$

RealiseNothing · Oct 19, 2013

Re: HSC 2013 2U Marathon

Show that:

$1 + e^{-x} + e^{2x} + e^{-3x} + e^{4x} + ...$

goes to infinity for all real x.

asianese · Oct 19, 2013

Re: HSC 2013 2U Marathon

Sy123 said:
$i) Show that$

$1 + (\cos \theta + \sin \theta) + (\cos^2 \theta + \sin^2 \theta) + (\cos^3 \theta + \sin^3 \theta) + \dots = \frac{1-\sin \theta \cos \theta}{(1-\cos \theta)(1-\sin \theta)}$

$ii) Show that this sum is always positive$

so trixxy

asianese · Oct 19, 2013

Re: HSC 2013 2U Marathon

Prove the cosine rule for triangles.

spanky125 · Oct 19, 2013

Re: HSC 2013 2U Marathon

I posted this in the general thread but it must have been to complex for their simple minds as I did not get a reply.

Sample Question:

John has the choice of either taking Stacey or Kristie to the prom.
The chance of Stacey putting it out is equivalent to x in the equation:
4x-3=8x+1
The chance of Kristie putting out is equivalent to y in the equation:
y+3=2-5y

Which girl should John take to the prom in order to have the greatest chance of getting laid?

RealiseNothing · Oct 19, 2013

Re: HSC 2013 2U Marathon

spanky125 said:
I posted this in the general thread but it must have been to complex for there simple minds as I did not get a reply.

Sample Question:

John has the choice of either taking Stacey or Kristie to the prom.
The chance of Stacey putting it out is equivalent to x in the equation:
4x-3=8x+1
The chance of Kristie putting out is equivalent to y in the equation:
y+3=2-5y

Which girl should John take to the prom in order to have the greatest chance of getting laid?

Sick negative probability.

HSC 2013 Maths Marathon (archive) (2 Viewers)

Member

This too shall pass

Exaı̸̸̸̸̸̸̸̸lted Member

Heroic!

Member

Exaı̸̸̸̸̸̸̸̸lted Member

Member

Member

Well-Known Member

Member

Member

Member

This too shall pass

Member

Member

what is that?It is Cowpea

Σ

Σ

Just waitin' for a mate

what is that?It is Cowpea

Users Who Are Viewing This Thread (Users: 0, Guests: 2)