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2u Mathematics Marathon v1.0 (1 Viewer)

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pLuvia

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You sure you didn't mean e2x?
If so
e4x-4e2x+4=0
(e2x-2)(e2x-2)=0
e2x=2
x=(ln2)/2
 

angmor

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state the domain and range of the function y= 2( sqrt.[25-x<sup>2</sup>] )

(the sqrt. covers the whole 25-x<sup>2</sup> in case you dont get me). thanks. i would really like to know how you go about working this out, as my book only shows me the solution.
 

Riviet

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Whenever you have a function that involves a sqrt{F(x)}, the domain can usually be found by considering values of F(x) for which sqrt{F(x)} is defined, ie sqrt{F(x)} is only defined for F(x) > 0 since you can't take the square root of negative numbers in the real number system.

So 25-x2> 0
x2 - 25 > 0

Using a graph or by inspection,

Domain: x > 5 or x < -5

Let g(x) = sqrt(25-x2) which is the upper half of the circle (semicircle) with radius 5 units.

.'. range of g(x) is 0 < g(x) < 5


.'. 0 < 2.g(x) < 10

0 < 2.sqrt(25-x2) < 10


Range: 0 < y < 10
 
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P

pLuvia

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2x-3y-1=0
y=2x/3-1/3
y'=2/3
MP=[(1-1)/2,(3+9)/2]
=(0,6)
Hence
y-6=2x/3
2x-3y+18=0

New Question
Find values for x for 1+sqrt{3}tanx+3tan2x+3sqrt{3}tan3x+... to have a limiting sum, and hence find this the limiting sum
 
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mica

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wasnt the question

Show that 1+2cosx+4cos^2x+... has a limiting sum and then find it

or am i crazy.
anyway can someone tell me how to go about proving limiting sums. Like what is involved ?

thanks
Michael
 

SoulSearcher

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Yes, but that series did not have a limiting sum, as the range of 2cosx could be larger than 1 at certain times, therefore not having a limiting sum. To prove that a geometric series has a limiting sum, you have to show that the ratio lies between -1 and 1, i.e. |r| < 1
 

mica

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oh ok thanks SoulSearcher

So with 1+sqrt{3}tanx+3tan<sup>2</sup>x+3sqrt{3}tan<sup>3</sup>x+...

with a = 1

and r = t2 / t1
= sqrt{3}tanx

= t3 / t2
= 3tan^2x / sqrt{3}tanx
= 3tanx / sqrt{3}
= 3tanx / sqrt{3} * sqrt{3} / sqrt{3}
= sqrt{3}tanx

therefore to find values of x we place r = sqrt{3}tanx within -1 and 1 as such

-1 < sqrt{3}tanx < 1

= -1 / sqrt{3} < tanx < 1 / sqrt{3}

and 1 / sqrt{3} = pie / 6 using the trig triangles

therefore x = pie + pie / 6 to get -1 / sqrt{3}
and x = pie / 6 to get 1 / sqrt{3}

therefore x is any value between 30 to 210 degrees not including.
30 < x < 210

i wasnt quite sure so someone tell me what i have done wrong :)
 

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pLuvia said:
2x-3y-1=0
y=2x/3-1/3
y'=2/3
MP=[(1-1)/2,(3+9)/2]
=(0,6)
Hence
y-6=2x/3
2x-3y+18=0
......
So, it seems just about all Year 12 students doing Mathematics and/or its Extension courses drink Calculus when solving graphical equations .... I should have stated not to use calculus..

The answer I expected would look like this ...

2x-3y-1=0
3y=2x-1
y=2/3x-1/3
m1=2/3x

Midpoint=[(1+(-1))/2,(3+9)/2]
=(0,6)

y-y1=m(x-x1)
y-6=2/3(x-0)
y-6=2/3x

3y-18=2x

2x-3y+18=0
 

mica

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mica said:
oh ok thanks SoulSearcher

So with 1+sqrt{3}tanx+3tan<sup>2</sup>x+3sqrt{3}tan<sup>3</sup>x+...

with a = 1

and r = t2 / t1
= sqrt{3}tanx

= t3 / t2
= 3tan^2x / sqrt{3}tanx
= 3tanx / sqrt{3}
= 3tanx / sqrt{3} * sqrt{3} / sqrt{3}
= sqrt{3}tanx

therefore to find values of x we place r = sqrt{3}tanx within -1 and 1 as such

-1 < sqrt{3}tanx < 1

= -1 / sqrt{3} < tanx < 1 / sqrt{3}

and 1 / sqrt{3} = pie / 6 using the trig triangles

therefore x = pie + pie / 6 to get -1 / sqrt{3}
and x = pie / 6 to get 1 / sqrt{3}

therefore x is any value between 30 to 210 degrees not including.
30 < x < 210

i wasnt quite sure so someone tell me what i have done wrong :)

Was this correct ?
 

B35tY

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i think that's round the wrong way, i went from the line

-1 < sqrt(3)tanx < 1

= -1 / sqrt(3) < tanx < 1 / sqrt(3)

= tan ^ -1 (-1 / sqrt(3)) < x < tan ^ -1 (1 / sqrt(3))

= -30 < x < 30

but i might be wrong too...

EDIT: Just adding the = signs
 
P

pLuvia

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f3nr15 said:
So, it seems just about all Year 12 students doing Mathematics and/or its Extension courses drink Calculus when solving graphical equations .... I should have stated not to use calculus..
I didn't use calculus at all:confused:
 

mica

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still no clarification of the correct answer of values of x to satisfy

So with 1+sqrt{3}tanx+3tan<sup>2</sup>x+3sqrt{3}tan<sup>3</sup>x+...
 

Riviet

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-pi/6 < x < pi/6 is only one of the many solutions, because if you do a quick sketch of y=tanx, the curve repeats itself every pi radians. If you refer to the graph, you should observe that two other possibe solutions are -7pi/6 < x < -2pi/3 and 2pi/3 < x < 7pi/6.

So Sinfinity = a/(1-r)
=1/(1-sqrt.3tanx)
 
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