2u Mathematics Marathon v1.0 (1 Viewer)

[pLuvia]

You sure you didn't mean e^2x?
If so

Spoiler

e^4x-4e^2x+4=0
(e^2x-2)(e^2x-2)=0
e^2x=2
x=(ln2)/2

 

[Riviet]

Yep yep, that looks about right too. Next!

 

[angmor]

state the domain and range of the function y= 2( sqrt.[25-x²] )

(the sqrt. covers the whole 25-x² 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]

Spoiler

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

 

[SoulSearcher]

Question?

 

[Forbidden.]

Question?

Find the equation of the straight line parallel to the line 2x-3y-1=0 and through the midpoint of (1,3) and (-1,9)

 

[pLuvia]

Spoiler

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

 

[mica]

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]

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]

oh ok thanks SoulSearcher

So with 1+sqrt{3}tanx+3tan²x+3sqrt{3}tan³x+...

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

 

[Forbidden.]

Spoiler

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 ...

Spoiler

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

 

[sinist4]

post another Q

 

[Forbidden.]

post another Q

Diffrentiate to the first principle 2x³-16x⁵ if x=6

 

[mica]

oh ok thanks SoulSearcher

So with 1+sqrt{3}tanx+3tan²x+3sqrt{3}tan³x+...

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]

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

 

[pLuvia]

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

 

[drynxz]

y'=2/3 <--- use of y'

 

[pLuvia]

y'=2/3 <--- use of y'

Too lazy to write m₁, means the same thing anyway Except that there was no calculus involved

 

[mica]

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

So with 1+sqrt{3}tanx+3tan²x+3sqrt{3}tan³x+...

 

[Riviet]

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