3u Mathematics Marathon V 1.1 (3 Viewers)

[pLuvia]

Spoiler

For f(x)
(3-x)/(5-x)>0
(3-x)(5-x)>0
x<3, x>5

Next question
Find the general solution of the equation cos2x=cosx

 

[zeek]

cos2x=cosx
2cos²x -1 -cosx =0
.:cosx=[1+-sqrt(1-4.2.-1)]/4
=(1+-3)/4
.:cosx=1 or -0.5
.: x =2pi/3 or 0


EDIT: WOOPS LOL

cos2x=cosx
.: 2x=2pi.n+-x
.:3x=2pi.n OR x=2pi.n
.:x=2pi.n/3 OR x=2pi.n
i hope thats rite

Next question
Prove that d(3^x)/dx = 3^xln3

 

[SoulSearcher]

That doesn't look like a general solution

 

[followme]

Next question:

Assuming the tide acts in simple harmonic motion, if the sea level was 2 metres above a dock at high tide at 7:45 a.m. and 1.2 metres below the same dock at low tide at 1:50p.m. When will the sea level be equal with the height of the dock between these times?

Spoiler

shm equation:
X=Acos(nt + a)
A= (1.2+2)/2 = 1.6
period:
13hr50min - 7hr45min = 6hr5min
6hr5min * 2=12hr10min
Period=2pi/n ie. 73/6=2pi/n n=12pi/73

X=1.6 cos (12pi/73 *t +a)
t=0 X=1.6 so a=0
X=1.6 cos (12pi/73 *t)

height of dock: 1.6-2=-0.4
-0.4 = 1.6 cos (12pi/73 *t)
cos (12pi/73 *t) =-0.25
(12pi/73) t = 1.82347....
t=3.53=3hr32min
3hr32min+7hr45min =11hr17min

ie. sea level = dock is at 11:17 am

 

[followme]

next Q:
a coin is biased so that in any one throw there is a constant probability P (where P =/= 0.5) that the coin shows heads. In 6 throws of the coin the probability of 3 heads is twice the probability of 2 heads. Find the value of P

 

[pLuvia]

Spoiler

P(H)=p, P(T)=1-p
Let x be the number of heads
P(x=3)=2P(x=2)
6C3*p³*(1-p)³=2[6C2*p²*(1-p)⁴]
6C3*p=2[6C2*(1-p)]
20p=30(1-p)
p=3/5

Next question
Find lim{x->0}[sin3x/tan(x/2)]

 

[Sober]

Next question
Find lim{x->∞}[sin3x/tan(x/2)]

Surely you mean lim{x->0} ?

 

[pLuvia]

Opps, yeh sorry fixed

 

[zeek]

for x->0:
sinx -> x
cosx -> 1
tanx -> x

.: sin3x/tan(x/2) ->3x/(x/2) -> 6

Next question:
Show that ⁴C₀⁹C₄ + ⁴C₁⁹C₃ + ⁴C₂⁹C₂ + ⁴C₃⁹C₁ + ⁴C₄⁹C₀ = ¹³C₄

 

[Riviet]

(1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿC_nxⁿ

Let x = 1

Equation 1:

2ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿC_n

Let x = -1

Equation 2:

0 = ⁿC₀ - ⁿC₁ + ⁿC₂ + ... + (-1)^{n n}C_n

Hence, moving the odd terms to the other side gives:

When n is even

ⁿC₀ + ⁿC₂ + ... + ⁿC_n = ⁿC₁ + ⁿC₃ + ... + ⁿC_n-1

When n is odd

ⁿC₀ + ⁿC₂ + ... + ⁿC_n-1 = ⁿC₁ + ⁿC₃ + ... + ⁿC_n

And since Equation 1 + Equation 2:

When n is even or odd

2ⁿ = 2(ⁿC₀ + ⁿC₂ + ... + ⁿC_n)

Divide by 2:

2^n-1 = ⁿC₀ + ⁿC₂ + ... + ⁿC_n

But ⁿC₀ + ⁿC₂ + ... + ⁿC_n

= ⁿC₁ + ⁿC₃ + ... + ⁿC_n-1 when even

and

= ⁿC₁ + ⁿC₃ + ... + ⁿC_n when odd

Therefore:

ⁿC₁ + ⁿC₃ + ... = 2^n-1

Man that took ages.

 

[zeek]

Next question:
Prove that 2.2 = 4

 

[Riviet]

I thought we were over these type of questions.

 

[SoulSearcher]

We were

 

[pLuvia]

Show that 4C09C4 + 4C19C3 + 4C29C2 + 4C39C1 + 4C49C0 = 13C4

Spoiler

Using, (x+1)⁴(x+1)⁹=(1+x)¹³
Expand it all out
(4c4*x^4+4c3*x^3+..+4c0)(9c9*x^9+9c8*x^8+..+9c0)=(13c13*x^13+13c12*x^12+..+13c0)
Equate coefficients of x^4 for both sides
4c0*9c4+4c1*9c3+4c2*9c2+4c3*9c1+4c4*9c0=13c4

Next question
Consider the graph of y=1/x, show that 1/(k+1) < int.{[k+1] to k}(dx/x) < 1/k

 

[zeek]

I thought we were over these type of questions.

I can't help it

Okay im not sure what i have to do with the y=1/x graph but i can show the other bit ...

1/(k+1) < [ln x]^k+1_k < 1/k {after integration}
.:e^1/(k+1)< 1 + 1/k < e^1/k

by drawing the graphs you see that this is true .: the intial statement is true.

 

[zeek]

hmmmm ok next question...

In a flock of 1000 chickens, the number P infected with a disease at time t years is given by P=1000/(1+ce^-1000t) where c is a constant.
i) Show that, eventually, all the chickens will be infected.
ii) Suppose that when time t=0, exactly one chicken was infected. After how many days will 500 chickens be infected.
iii)Show that dP/dt= P(1000-P).

 

[onebytwo]

hmmmm ok next question...

In a flock of 1000 chickens, the number P infected with a disease at time t years is given by P=1000/(1+ce^-1000t) where c is a constant.
i) Show that, eventually, all the chickens will be infected.
ii) Suppose that when time t=0, exactly one chicken was infected. After how many days will 500 chickens be infected.
iii)Show that dP/dt= P(1000-P).

i seem to recall ive done this question before....somewhere
i) as t -> infinity, ce^-1000t->0, then P->1000

ii) if t=0 and P=1, then c=999. put P=500, then 999e^-1000t=1,
then t=[ln(999)]/100=0.0690675...years=25.2days. so after 26 days 500 chivkens will be infected
iii) on differentiating, using the chain rule, dP/dt={1000000ce^-1000t}/{1+ce^-1000t}²
taking out the common factor, which happens to be P, we end up with, dP/dt=P(1000-P)
EDIT: missed the last part

 

[pLuvia]

I can't help it

Okay im not sure what i have to do with the y=1/x graph but i can show the other bit ...

1/(k+1) < [ln x]^k+1_k < 1/k {after integration}
.:e^1/(k+1)< 1 + 1/k < e^1/k

by drawing the graphs you see that this is true .: the intial statement is true.

Spoiler

You use the graph, and then using rectangles and the area under the curve, you prove that result

 

[onebytwo]

NEXT QUESTION:
this one had me thinking for a while......
Q - five persons play a card game in which one pair play against the other pair. the fifth person acts as referee. in how many ways can this occur?

btw - how do you make a spoiler?

 

[Elvin]

5C1? lol