Mathematics Marathon HSC 09 (1 Viewer)

[00iCon]

i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t³+sec²t

with initial conditions x(0)=6 m and v(0)=4 ms^-1

using i), find x as a function of t

Do we have to know part (i)?

 

[luv2luvurmama]

ò <SUP>(Penis)3 Dp
</SUP>Find volume generated through rotating the area between Y=Penis<SUP> </SUP>and the axis around the y axis.
Tip: if having difficulty sketching graph, google penis.

 

[xFusion]

i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t³+sec²t

with initial conditions x(0)=6 m and v(0)=4 ms^-1

using i), find x as a function of t

i)∫tan(x)dx =∫sinx/cosx dx
let u=cosx
du/dx=-sinx
---> -∫du/u =-ln(u) =-ln(cosx) +C
ii)a(t)=t³+sec²t
v(t)=t^4/4 + tan(t) +C
whent=0, v=4, C=4
v(t)=(t^4)/4 + tan(t) +4
x(t)=(t^5)/20 -ln(cos(t)) +4t +C
when x=6, t=0, C=6
hence x=(t^5)/20 -ln(cos(t)) +4t +6

 

[max powers]

i) find ∫tan(x)dx by writing the integral in the form f'(x)/f(x)

ii) a(t)=t³+sec²t

with initial conditions x(0)=6 m and v(0)=4 ms^-1

using i), find x as a function of t

part i.
you let tanx = sinx/cosx
= - ln cosx ?

part ii.
a= t^3 + Sec^2 t
v= t^4/3 + tan t + C
let v=4 and t=0
therefore C=4
v= t^4/3 + tan t +4
x= t^5/15 - ln cosx + 4t +C
let x=6 and t=0
x= t^5/15 - ln cosx + 4t +6

...I'm no good at typing it out lol.

 

[tommykins]

39th Question (guesstimate.)

This is a higher-order question. Note that it requires English language rather than mathematical notation; which is not always necessary for the skilled mathematician.

Question: Explain why a circular function is equal to infinity. (And yes, there IS an answer...)



NOW no taking the piss!! Serious answer only pls .....

Wow....what a shit question.

 

[xFusion]

3 sin(^2)x cos x

Question:

Find the value of n for which the equation (n+2)x^2 + 3x - 5 = 0 has one root triple the other.

let roots be @, 3@
-b/a = -3/(n+2) =@ +3@ =4@
@= -3/[4(n+2)] -------1
c/a= -5/(n+2) =3@^2 --------2
sub 1 into 2
n= -2.3375
lol nice answer

Question: Jodi takes out a $50,000 loan. The interest is 6% P/A (calculated at the end of each quarter before she pays). She pays this back in quarterly installments. How much would she have to pay each quarter to pay it off in 5 years?

 

[jet]

I'm gonna make this question up and see how it goes.

To cut a plank of timber, builders cut a rectangular prism out of a perfectly circular log of timber. If a cross section were viewed, it would be a rectangle inscribed into a circle. Consider a log with cross sectional radius 'r'. What is the maximum area that the cross section of the plank can have? i.e. the maximum area of the rectangle.

 

[untouchablecuz]

let A be the area of the rectangle and [x, y] be the sides of this rectangle

A=xy => A²=x²y² [1]

x²+y²=(2r)²=4r² => y²=4r²-x² [2]

sub [2] in [1]

A²=x²y²=x²(4r²-x²)=4r²x²-x⁴

maximising A² is equivalent to maximising A, since dA²/dx=(dA²/dA)(dA/dx) = 2A(dA/dx)

dA²/dx=8r²x-4x³=4x(2r²-x²)=4x(√2r-x)(√2r+x)

let dA²/dx=2A(dA/dx)=0

x=√2r, since x>0

d²A²/dx²=8r²-12x²

when x=√2r, d²A²/dx²=8r²-24r²<0

.'. x=√2r is a maximum turning point

sub in [2], y=√2r

sub x=√2r and y=√2r into [1]

A_max=(√2r)²=2r²

 

[untouchablecuz]

you can also do it like this:

consider rectangle WXYZ, where XY=2r is a diameter

let angle WXZ=k, WZ=a and WX=b

A_rectangle=ab [1]

in ∆WXZ,

sin(k)=a/(2r) => a=2rsin(k) [2]

cos(k)=b/(2r) => b=2rcos(k) [3]

sub [2] and [3] in [1]

A_rectangle=(2r²)(2sinkcosk)=(2r²)sin(2k)

A_rectangle is maximum when sin(2k) is maximum

i.e. when sin(2k)=1

.'. A_max=2r²

 

[random-1005]

can som1 plzz integrate: [1-sqroot(x)]/[1-x] using a 2unit method....i kinda gave up on 3unit afta the trials

YOU CANT!

theres a reason why it is 3 unit content

 

[random-1005]

no, they meet at two points.

(0,0) and (1,1)

ohhh you got ripped oicon lol

 

[random-1005]

Q.It is estimated that 85% of students in Australia own a mobile phone.
(i) Two students are selected at random. What is the probability that neither
of them owns a mobile phone?

0.15 x 0.15 = 0.0225 => 2.25%

(ii) Based on a recent survey, 20% of the students who own a mobile phone
have used their mobile phone during class time. A student is selected at
random. What is the probability that the student owns a mobile phone
and has used it during class time?

0.85 x 0.2 = 0.17 => 17%





Solve tan2x = cot40

cot 40= tan50 (complementary angles)

therefore tan2x=tan(50+360n) {angles of any magnitude} for n=0,1,2,3.....

2x=50 +360n
x= (25 +180n) degress

god knows what the hell untouchable did

 

[random-1005]

tan2x=cot40=tan(5π/18)


.'. 2x=πk+(5π/18) => x=0.5(πk+(5π/18)) i.e. x=(π/36)(18k+5)

where the hell did this come from ? so this 5n/18 just magically appears.

im assuming this is general solution of tan in the second line, where did the k come from??

its n(pi)+Tan^-1 (B)

 

[ForrestGump]

LOL theres like 20 questions going hahah

 

[untouchablecuz]

where the hell did this come from ? so this 5n/18 just magically appears.

im assuming this is general solution of tan in the second line, where did the k come from??

its n(pi)+Tan^-1 (B)

cot(x)=tan(90-x)

when x=40,

cot(40)=tan(90-40)=tan(50)=tan(5π/18)

now, tan2x=cot(40)=tan(5π/18)

let tan(5π/18)=y

tan2x=y

2x=πk+tan^-1y=πk+tan^-1(tan(5π/18))=πk+5π/18

x=(π/36)(18k+5)

 

[random-1005]

cot(x)=tan(90-x)

when x=40,

cot(40)=tan(90-40)=tan(50)=tan(5π/18)

now, tan2x=cot(40)=tan(5π/18)

let tan(5π/18)=y

tan2x=y

2x=πk+tan^-1y=πk+tan^-1(tan(5π/18))=πk+5π/18

x=(π/36)(18k+5)

yes, that makes perfect sense!

i love how the people doing 4 unit always have to make something appear so much harder than it actually is.

how about the soln is x=25 +180n n=0,1,2,3.....

not your rubbish with ks and shit

 

[untouchablecuz]

yes, that makes perfect sense!

i love how the people doing 4 unit always have to make something appear so much harder than it actually is.

how about the soln is x=25 +180n n=0,1,2,3.....

not your rubbish with ks and shit

k is just a dummy variable; it can be interchanged with x, y, z, m, Φ, ω, θ

point being, k (or n) is simply representative integer values

also, your solution is wrong

the solution is x=(π/36)(18k+5)=90k+25=90n+25=90ω+25=90θ+25 where k or n or ω or θ is integral

 

[aero135]

Hey guys, anyone got any question 10 type questions. Ive done the WHOLE success one boook. Any one got any other good ones?

 

[star of the sea]

Hey guys, anyone got any question 10 type questions. Ive done the WHOLE success one boook. Any one got any other good ones?

given that d/dx(b^x)=b^x.loge.b

evaluate: integrate pi^x dx where x=pi and x=0

 

[hiddimp]

given that d/dx(b^x)=b^x.loge.b

evaluate: integrate pi^x dx where x=pi and x=0

im not sure, let me noe if im wrong:

let u = pi^x
du/dx = pi^x.loge(pi)
du/loge(pi) = pi^x.dx
if x= pi, u = pi^pi
if x=0, u = 1

Integrating pi^x where u = pi^pi and u = 1

= 1/loge(pi) from u = pi^pi to 1
= 1/loge(pi). (pi^pi -1)