2u Mathematics Marathon v1.0 (1 Viewer)

[SoulSearcher]

ill take a shot even thou i havent learnt it yet...

Spoiler

2x/(x-2) >1

(multiply both sides by (x-2)^2)
= 2x(x-2) > (x-2)^2
= 2x^2 - 4x >x^2-4x+4
= x^2 - 4 >0
= (x+2)(x-2)>0
*test point x=0
= (2)(-2)>0 doesnt hold true
therefore
= x > 2

pft maybe i got it wrong...i dunno

Spoiler

In the bold lines, the first 2 lines are correct, but you forgot to account for the 2 values of x. Therefore the correct answer is x < -2 or x > 2

 

[SoulSearcher]

Next question:



Solve ==> solve simultaneously



x+2y-z = -5

2x-3y+4z = 28

4x+5y-3z = -10

Spoiler

some substitution is involved here

x + 2y - z = -5 ... (1), thus z = x + 2y + 5
2x-3y+4z = 28 ... (2)
4x+5y-3z = -10 ... (3)

sub in the value of z into (2) & (3)

2x - 3y + 4(x + 2y + 5) = 28, thus
6x + 5y = 8 ... (4)
4x + 5y - 3(x + 2y + 5) = -10, thus
x - y = 5 ... (5)
{ (5) x 5 } 5x - 5y = 25 ... (6)

add (6) to (4)

11x = 33
x = 3

sub value of x into (5)

3 - y = 5
y = -2

sub values of x and y into equation for z

z = 3 - 4 + 5
z = 4

therefore x = 3, y = -2 and z = 4

Next Question:

A bowl is formed by rotating the part of the curve y = x⁴ / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.

 

[Mountain.Dew]

Next Question:

A bowl is formed by rotating the part of the curve y = x⁴ / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.

Spoiler

y = x⁴ / 4, so 4y = x⁴, x = (4y)^1/4

when x = 0, y = 0.
when x= 2, y = 4.

so ur volume is thus:

V = pi * (Integral from 0 to 4 of (4y)^1/2)dy

V = pi * [(4y)<sup>3/2 / 6], 0 --> 4

V = pi * 64/6 = 32/3pi units3

hopefully that should be right.</sup>

<sup>Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.</sup>

 

[drynxz]

Next Question:

A bowl is formed by rotating the part of the curve y = x⁴ / 4 between x = 0 and x = 2 about the y axis. Find the volume of the bowl.

can someone tell me what i did wrong for soulsearchers question :S

Spoiler

V= *intergral B over A pi y ₂ dx

=*intergral 2 over 0 pi (x₄/4)₂ dx

=*intergral 2 over 0 pi x₈/16 dx

=pi [x^9/144x]2 over 0

=pi [512/288]

=(16/9) pi

=5.585 units ₃

yeh it seem wrong..lol maybe i used the wrong formula or some crap :S

 

[Mountain.Dew]

can someone tell me what i did wrong for soulsearchers question :S

Spoiler

V= *intergral B over A pi y ₂ dx

=*intergral 2 over 0 pi (x₄/4)₂ dx

=*intergral 2 over 0 pi x₈/16 dx

=pi [x^9/144x]2 over 0

=pi [512/288]

=(16/9) pi

=5.585 units ₃

yeh it seem wrong..lol maybe i used the wrong formula or some crap :S

mmmmmm theres something wrong here...

remember the volume is made by rotating the curve around the Y-AXIS not the x-axis, so you need to get y the subject from the original curve. you also needed to change the limits as well.

please see my solution (previous post).

oh, and with powers, it is (suP)...(/suP), not (suB)...(/suB) (replace the ( ) with [ ] though)

 

[drynxz]

yeh i realised i read the question wrong...and i guess its pretty obvious im new to this thread -_- lol

 

[sando]

has anyone got a question for me? preferably integration, cos i hav a test on that soon

 

[pLuvia]

Ok try this

∫ sec³x tanx dx

∫ x/(1+x²)² dx

 

[PLooB]

Ok try this

∫ sec³x tanx dx

Spoiler

∫ sec³x tanx dx = ∫ sec²x dx = tanx

I'm just guessing that works, not 100% sure.

 

[pLuvia]

I'm not sure how you got that PLooB but here's the solution

Spoiler

∫ sec³x tanx dx

Let u = secx
du = secxtanx dx

.: ∫ sec³x tanx dx
= ∫ sec²x (secxtanx dx)
= ∫ u² du
= u³ / 3 + C

 

[Mountain.Dew]

I'm not sure how you got that PLooB but here's the solution

Spoiler

∫ sec³x tanx dx

Let u = secx
du = secxtanx dx

.: ∫ sec³x tanx dx
= ∫ sec²x (secxtanx dx)
= ∫ u² du
= u³ / 3 + C

make sure u sub u = secx back into ur primtive function to get the correct answer

heres another question (which i posted earlier, lost in the sea of posts)

Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.

 

[insert-username]

Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.

Annuities is 2U, but this one is particularly hard, since it's not straightforward and requires fiddling with the resulting equation. You might want to try an easier one to keep it 2U.


I_F

 

[Mountain.Dew]

Annuities is 2U, but this one is particularly hard, since it's not straightforward and requires fiddling with the resulting equation. You might want to try an easier one to keep it 2U.


I_F

oh dear. sorry about that guys. i will post up a new question that is 2U.

in the meantime, the 3U question is still up there for your interest and viewing pleasure, as the saying goes.

Next Question:

a particle travels in a straight line. its position x cm from the centre at time t is given by the equation: x = 6 - 5t - 2t^2

a) find the initial velocity
b) find when the particle will come to rest
c) find time when the particle passes through the centre.

 

[SoulSearcher]

Next Question:

A sum of $24000 is borrowed now at the rate of 15% p.a. reducible interest. Payment is made by n equal annual installments of $3840 beginning at the end of a year.

Show that 1 – (1.15)^-n = 0.9375 and hence find n.

note: if necessary, please tell me if this question is 3U only, so that I can repost up a new question.

This question is just an extension of the 2 unit course, but it isn't only for 3 unit students, it just requires some index law fiddling.

Spoiler

Now
A_n = 24000(1.15)ⁿ - { 3840 ( 1.15ⁿ - 1) } / { 1.15 - 1 }

but at the end of the loan, A_n = 0, therefore

0 = 24000(1.15)ⁿ - { 3840 ( 1.15ⁿ - 1) } / { 1.15 - 1 }
0 = 24000(1.15)ⁿ - 25600( 1.15ⁿ - 1)
25600( 1.15ⁿ - 1) = 24000(1.15)ⁿ
25600( 1.15ⁿ ) - 24000( 1.15ⁿ ) = 25600
1600( 1.15ⁿ ) = 25600
1.15ⁿ = 16

now remembering 1.15ⁿ = 1 / 1.15^-n
1 / 1.15^-n = 16
1 = 16(1.15^-n)
0.0625 = 1.15^-n

add 1 to both sides
1 + 0.0625 = 1.15^-n + 1

take the values that are not equal to 1 over
1 - 1.15^-n = 0.9375

thus showing what was needed.

to find n , refer back to the working out
1.15ⁿ = 16
n ln 1.15 = ln 16
n = ln 16 / ln 1.15
n = 19.83
n=20 rounded off to nearest integer (edited)

Ill do the other question later

 

[Mountain.Dew]

SoulSearcher, remember n is an integer, because u it involves the element of reality.

therefore, n = 20 installments.

 

[SoulSearcher]

Whoops, edited now

Next Question:

a particle travels in a straight line. its position x cm from the centre at time t is given by the equation: x = 6 - 5t - 2t^2

a) find the initial velocity
b) find when the particle will come to rest
c) find time when the particle passes through the centre.

Spoiler

(a) x = 6 - 5t - 2t²
therefore dx/dt = -5 - 4t which is velocity
when t = 0,
v = -5 - 0
therefore initial velocity is -5cms^-1

(b) to find the time when the point rests, v = 0
0 = -5 - 4t
4t = -5
t = -5/4 (notice here that time cannot be negative and therefore there is no resting point)

(c)this occurs when x = 0
0 = 6 - 5t - 2t²
2t² + 5t - 6 = 0
since that cannot be factorised easily, use quadratic formula
t = { -5 ± sqrt(25 + 48) } / 4
t = { -5 ± sqrt(73) } / 4
therefore t = ( -5 - sqrt(73) } / 4 or t = { -5 + sqrt(73) }/4
but as time cannot be negative, t = { -5 + sqrt(73) }/4

 

[Mountain.Dew]

time to revive an old thread...

Question:
y = log_e(tanx) + x^2 + 25x - 12

find y'

 

[Sober]

time to revive an old thread...

Question:
y = log_e(tanx) + x^2 + 25x - 12

find y'

y' = sec²(x) / tan(x) + 2x + 25

= sec(x).cosec(x) + 2x + 25

Early bird gets to post the next question.

 

[Riviet]

Question continued:
Hence find ∫ 3.sec(x).cosec(x) + sin(5-2x) dx

 

[darkliight]

Spoiler

3ln(tan(x)) + cos(5-2x)/2

y = sqrt(x*tan(x)), find dy/dx

Edit: Fixed -- thanks Riviet