Extension One Revising Game (1 Viewer)

[midifile]

ive posted up my solution though i must say that maths isnt going to be important to most people until monday 12-ish, then we'll see people flooding these forum pages.

my question, rather easy.

Show that (x-1)(x-2) is a factor of

P(x)=x^n(2^m-1)+x^m(1-2^n)+(2^n-2^m)

P(x)=xⁿ(2^m-1)+x^m(1-2ⁿ)+(2ⁿ-2^m)

P(1) = 2^m-1+1-2ⁿ+2ⁿ-2^m = 0
therefore 1 is a root, therefore (x-1) is a factor

P(2) = 2ⁿ(2^m-1)+2^m(1-2ⁿ)+(2ⁿ-2^m)
= 2ⁿ2^m-2ⁿ+2^m-2^m2ⁿ+2ⁿ-2^m
= 0
therefore 2 is a root therefore (x-2) is a factor

Therfore (x-1)(x-2) is a factor of p(x)

 

[midifile]

My question (or rather questions):

There is a row of five coloured lights each with its own switch. When switched on, each light is equally likely to show green, red or amber.

a. If all the lights are switched on how many different colour patterns are possible?

b. What is the probability that:
i) The first three lights from the left are green, but not the fourth?
ii) The first three lights from the left are teh same colour, but not the fourth?
ii) Exactly three of the lights are green?

c. If all the lights are switched on five times, find, as a decimal correct to three places, the probability that exactly three lights will be the same colour on two or three occasions.

 

[duy.le]

my solution:

a) 243
b)i)2/81
ii)2/27
iii)40/243
c)0.189

there might be a mistake im not sure, please confirm (ive seen this before so i have a slight advantage, hehe)

my question:

i) prove that : sqrt[ 1-x / 1+x ] = 1-x / sqrt (1-x^2)

ii) hence or otherwise evaluate

/(1/2)
|
| sqrt[ 1-x / 1+x ] dx
|
0/

(integrate from 0->1/2)

 

[midifile]

my solution:

a) 243
b)i)2/81
ii)2/27
iii)40/243
c)0.189

there might be a mistake im not sure, please confirm (ive seen this before so i have a slight advantage, hehe)

They are all right except the last one =]

c. P(3 green) = 40/243 (part b i)
therefore P (3 of the same colour) = 120/243 = 40/81
P (all 3 not the same) = 123/243 = 41/81
Therefore P(2 or 3 times 3 are the same) = ⁵C₂(41/81)³(40/81)² + ⁵C₃(41/81)²(40/81)³
=0.625

my question:

i) prove that : sqrt[ 1-x / 1+x ] = 1-x / sqrt (1-x^2)

ii) hence or otherwise evaluate

/(1/2)
|
| sqrt[ 1-x / 1+x ] dx
|
0/

(integrate from 0->1/2)

i) sqrt[1-x/1+x] = sqrt[1-x/1+x] x sqrt[1-x/1-x]
= 1-x/sqrt[(1-x)(1+x)
= 1-x/sqrt(1-x²)

ii) Int (0-->1/2) sqrt[ 1-x / 1+x ] dx = Int (0-->1/2) 1-x/sqrt(1-x²) dx
= Int (0-->1/2) 1/sqrt(1-x²) dx - Int (0-->1/2) x/sqrt(1-x²) dx
= [sin^-1x](1/2 --> 0) - Int (0-->1/2) x/sqrt(1-x²) dx

For the integral let u = 1-x²
du = -2x
u₁ = 3/4
u₂ = 1

Therefore integral = pi/6 - 0 + 1/2Int (1-->3/4) u^-1/2du
= pi/6 + 1/2[2u^1/2](1-->3/4)
= pi/6 + 1 - rt3/2

 

[midifile]

New qu:

When (1 + ax)⁵ + (1 + bx)⁵ is expanded in ascending powers of x, the expansion begins: 2 + 30x + 220x²...
i) show that a + b = 6 and a² + b² = 22
ii) deduce the value of ab
iii) find the coefficient of x³

 

[bored of sc]

New qu:

When (1 + ax)⁵ + (1 + bx)⁵ is expanded in ascending powers of x, the expansion begins: 2 + 30x + 220x²...
i) show that a + b = 6 and a² + b² = 22
ii) deduce the value of ab
iii) find the coefficient of x³

(1+ax)⁵ + (1+bx)⁵ = 1+5ax+10a²x²+10a³x³+5a⁴x⁴+a⁵x⁵ + 1+5bx+10b²x²+10b³x³+5b⁴x⁴+b⁵x⁵ = 2+30x+220x²...

Equating coefficients:
i) 5a+5b = 30
a+b = 6
10a²+10b² = 220
a²+b² = 22

ii) (a+b)²-2ab = 22
36-2ab = 22
2ab = 14
ab = 7

iii) a+b = 6 (1)
ab = 7 (2)
therefore a²-6a+7 = 0 (subbing b = 7/a into (1))
a = 3+rt2 (quadratic formula)
sub into (1)
b = 3-rt2 when a = 3+rt2 and 3+rt2 when a = 3-rt2
thus a and b are conjugate surds (just a bit of maths jargon)
10a³+10b³ = coefficient of x³
sub in a and b
10(3+rt2)³+10(3-rt2)³
= 10(27+27rt2+18+2rt2)+10(27-27rt2+18-2rt2)
= 2*10*(27+18)
= 900

If I'm not right I'm gunna cry.

 

[midifile]

thats right =]

 

[bored of sc]

Next question:

Trigonometric Functions (my favourite):

(a) Prove (1+cosx+cos2x)/(sin2x+sinx) = cotx

(b) Prove sqrt[(1-cosA)/(1+cosA)] = tan(A/2) and hence show that tan(pi/8) = sqrt(2)-1

 

[midifile]

Next question:

Trigonometric Functions (my favourite):

(a) Prove (1+cosx+cos2x)/(sin2x+sinx) = cotx

(b) Prove sqrt[(1-cosA)/(1+cosA)] = tan(A/2) and hence show that tan(pi/8) = sqrt(2)-1

a. (1+cosx+cos2x)/(sin2x+sinx) = (1 + cosx + 2cos²x - 1)/(2sinxcosx + sinx)
= [cosx(2cosx + 1)]/[sinx(2cosx + 1)]
= cosx/sinx
= cotx

b. Let t=tan(A/2)
sqrt[(1-(1-t²/1+t²))/(1+(1-t²/1+t²))]
= sqrt[(1+t²-1+t²)/(1+t²+1-t²)
= sqrt(2t²/2)
= t
= tan(A/2)

tan(pi/8) = sqrt[(1-cos(pi/4))/(1+cos(pi/4))]
= sqrt[(1-(1/sqrt2)/(1+(1/sqrt2))]
= sqrt[(sqrt2-1)/(sqrt2+1)] x sqrt[(sqrt2-1)/(sqrt2-1)]
= (sqrt2 -1)/sqrt(2-1)
= sqrt2 -1

 

[Azreil]

a)LHS = (1+cosx+cos2x)/(sin2x+sinx)
= (1+cosx+2(cosx)^2-1)/(2sinxcosx+sinx)
= (cosx+2(cosx)^2)/(2sinxcosx+sinx)
= (cosx[1+2cosx])/(sinx[2cosx+1)
= cosx/sinx
= cotx
= RHS.
b) LHS = sqrt[(1-cosA)/(1+cosA)]
= sqrt(1-cosA) / sqrt(1+cosA)
= sqrt[(1-cosA)/2] / sqrt[(1+cosA)/2]
= sin(A/2) / cos (A/2)
= tan(A/2)
= RHS.

Question:
Simplify n*(n-1)C(1) + n*(n-1)C(2) +...+ n*(n-1)C(n-2)
Hence find the smallest positive integer n such that
n*(n-1)C(1) + n*(n-1)C(2) +...+ n*(n-1)C(n-2) > 20 000.

 

[gurmies]

lyonamu, nobody has posted the answer to your question so I gave it a go:

cos (x + nPI) = (-1)^n . cos x

For n = 1

LHS = cos (x + 180)
= -cos x
RHS = -1(cos x)
= LHS

Therefore the formula is true for n = 1.

Assume the formula is true for n = k

cos (x + kPI) = (-1)^k . cos x

For n = k + 1

cos (x + (k+1)PI) = cos (x + kPI + PI)
= -cos (x + kPI)
= -((-1)^k . cos x)
= (-1)^(k+1) . cos x

Which is of the same form as for n = k

Therefore if the formula is true for n = k, it is true for n = k+1.

BUT the formula is true for n = 1
" " " " " " n = 2
and so on...

Therefore the formula is true for all positive integral n.

Find the number of arrangements of the letters in the word PENCILS if "e" preceeds "i".

 

[alez]

(1^2+1)1!=2 1(1+1)!=2
true for n=1
assume true for n=k
2.1!+5.2!+........+(k^2+1)k!=k(k+1)!
let n=k+1
2.1!+5.2!+......+(k^2+1)k!+((k+1)^2+1)(k+1)!=(k+1)(k+2)!
k(k+1)!+((k+1)^2+1)(k+1)!=(k+1)![k+(k+1)^2+1]
=(k+1)![k^2+3k+2]
=(k+1)!(k+2)(k+1)
=(k+1)(k+2)!
therefore, true for n=k+1
etc etc

 

[Trebla]

Find the angle between the lines y = 7x + 1 and x + 7y + 1 = 0

 

[Zeber]

90* by observation.

 

[minijumbuk]

1=5
2=10
3=30
4=120
5=?

 

[Zeber]

600

 

[midifile]

1=5
2=10
3=30
4=120
5=?

600?

EDIT: Zeber beat me =[

 

[bored of sc]

1=5
2=10
3=30
4=120
5=?

5 = 1 since 1 = 5

 

[Zeber]

hahahaha, oh wow - nice.

 

[midifile]

lol nice qu

okay new qu:

A term in the expansion of (2 + mx)¹² is 24057x⁵. Find the value of m