MedVision ad

Mathematics Marathon (5 Viewers)

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
回复: Re: 回复: Re: Mathematics Marathon

Mark576 said:
A3 + B3 = (A + B)(A2 - AB + B2)
OMG. Silly me.

A^3+B^3 = (A+B)(A^2-AB+B^2) = (A+B)[(A+B)^2-2AB-AB] = (A+B)[(A+B)^2 -3AB]
A+B = 5/3, AB = -4/3

Thus A^3+B^3 = 85/9
 

zzdfa

Member
Joined
Mar 22, 2008
Messages
50
Location
vic
Gender
Male
HSC
2008
lyounamu said:
The answer for the infinity series is 2.

The sum -> 2 as x (numerator) -> infinity and y (denominator) -> infinity

prove it =p

bonus points for general soln for the sum to infinity of i/x^i
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
zzdfa said:
prove it =p

bonus points for general soln for the sum to infinity of i/x^i
S = a/(1-r) but you cannot use that formula here because the r is not constant.

I will just say in the question that as x/y -> 0, the S -> 2.
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
An easy one:

Find all real numbers x which satisfy the equation

4x4 = 4x2 + 3
 

Js^-1

No tengo pantelonès
Joined
Oct 14, 2007
Messages
318
Gender
Male
HSC
2008
4x<sup>4</sup> = 4x<sup>2</sup> + 3
4x<sup>4</sup> - 4x<sup>2</sup> - 3 = 0
Let x<sup>2</sup> = m
4m<sup>2</sup> - 4m - 3 = 0
4m<sup>2</sup> - 6m + 2m - 3 = 0
2m(2m - 3) + (2m - 3) = 0
(2m + 1)(2m - 3) = 0

So
2m + 1 = 0
m = -1/2
x<sup>2</sup> = -1/2
No real solutions.

2m - 3 = 0
m = 2/3
x<sup>2</sup> = 2/3
x = ± √(2/3)

.: x = ± √(2/3)

Think thats right...

If f (t) = mt<sup>2</sup> - 3e<sup>t</sup> + 7
Find f ' (t) and F (t) where F (t) = ∫ f (t) dt.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
回复: Re: Mathematics Marathon

f'(t) = 2mt - 3et
F(t) = ∫ mt2 - 3et + 7 = mt3/3 - 3et + 7t
 

bored of sc

Active Member
Joined
Nov 10, 2007
Messages
2,314
Gender
Male
HSC
2009
State the domain and range of the function:

y = 2 root (25-x2)


<SUP><SUB>[FONT=JFOEB D+ Times,Times]</SUP></SUB>[/FONT]
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
bored of sc said:
State the domain and range of the function:

y = 2 root (25-x2)


<SUP><SUB>[FONT=JFOEB D+ Times,Times]</SUP></SUB>[/FONT]
-5<= x <= 5

0 < = y <= 10

Integrate -sin 2x
 

Js^-1

No tengo pantelonès
Joined
Oct 14, 2007
Messages
318
Gender
Male
HSC
2008
1/2 cos 2x + C

Find the Co-ordinates of the point of inflection for
f (p) = p<sup>3</sup> - 3p<sup>2</sup> + 6p
 

lyounamu

Reborn
Joined
Oct 28, 2007
Messages
9,998
Gender
Male
HSC
N/A
Js^-1 said:
1/2 cos 2x + C

Find the Co-ordinates of the point of inflection for
f (p) = p<SUP>3</SUP> - 3p<SUP>2</SUP> + 6p
f'(p) = 3p^2 - 6p + 6
f''(p) = 6p - 6

Point of inflection is at f''(p) = 0
i.e. when 6p-6 = 0
p = 1

When p<1, f''(p) < 0
When p >1, f''(p) > 0 Therefore, change in concavity.

Therefore, point of inflection is at p=1

I am running out of question..by the way...I don't have any textbooks here to refer to...sorry guys...
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
There's not many questions that are difficult =\, only 3D trig and circle geo is what I struggle with but those require diagrams.
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Forbidden. said:
whoa theyre the same topics i struggled in too ... plus permuations and combinations back then
goddammit i hate permutations and combinations. i always miss a detail here and there :mad:
 

Ali92l

New Member
Joined
Jun 11, 2007
Messages
25
Gender
Male
HSC
2009
A line passes through the origin touches the curve with equation y=4x^3 +1 . Find the equation of this line.



meh
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Err wouldn't there be like a finite amount of solutions to that?
 

xMrRand0m

Member
Joined
Jan 3, 2008
Messages
88
Gender
Male
HSC
2009
can't you use the 'family of lines' technique?...and wouldn't the line cut through the curve?
 

lolokay

Active Member
Joined
Mar 21, 2008
Messages
1,015
Gender
Undisclosed
HSC
2009
Ali92l said:
A line passes through the origin touches the curve with equation y=4x^3 +1 . Find the equation of this line.
the gradient of the line is
y' = 12x2

hence it's equation is
y-0 = 12x2x - 0
y = 12x3
so, solving simultaneously
y = 4x3 + 1
8x3 = 1
x = 1/2 is where it touches the curve

so, solving for the equation of the line (passing through 1/2, y(1/2)=3/2 with gradient 3)
y - 3/2 = 3x - 3/2
y = 3x
 

ratcher0071

Member
Joined
Feb 17, 2008
Messages
617
Location
In Space
Gender
Male
HSC
2009
lolokay said:
the gradient of the line is
y' = 12x2

hence it's equation is
y-0 = 12x2x - 0
y = 12x3
so, solving simultaneously
y = 4x3 + 1
8x3 = 1
x = 1/2 is where it touches the curve

so, solving for the equation of the line (passing through 1/2, y(1/2)=3/2 with gradient 3)
y - 3/2 = 3x - 3/2
y = 3x
that's pretty neat :D
 

tommykins

i am number -e^i*pi
Joined
Feb 18, 2007
Messages
5,730
Gender
Male
HSC
2008
Still confused about 'touches' and intersects.
 

Users Who Are Viewing This Thread (Users: 0, Guests: 5)

Top