HSC 2013 Maths Marathon (archive) (6 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) (6 Viewers)

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