Extension One Revising Game (2 Viewers)

[lolokay]

K was found by finding a value such that 27K³/9K = 4/3, so we could turn it into a multiple of the cos3A expansion. This gives K = 2/3.
When we substitute x=2/3cosA in we get 2(cos3A) = 1, so cos3A = 1/2 = cos60, cos300, cos420 etc. (you don't need any more values, since they give the same as those 3 for their respective cosA values).
So A = 20, 100, 140. The roots are 2/3cos20, 2/3cos100, 2/3cos140
From polynomials, you know that the product of the roots = 1/27 (by subtracting 1 from each side, and you get abc = -(-1) where a, b and c are the roots)
so 2/3cos20.2/3cos100.2/3cos140 = 1/27
cos20.cos100.cos140 = 1/8
cos20.cos60.cos100.cos140 = 1/16

 

[bored of sc]

4(x+6) > (x+6)²
0 > (x+6)²-4(x+6)
0 > (x+6)[x+6-4]
0 > (x+6)(x+2)

-6 < x < -2

Correct.

cos75 = cos(45+30) = cos45.cos30 - sin45.sin30 = sqrt3/2sqrt2 - 1/2sqrt2 = sqrt3-1/2sqrt2

Correct. You can rationalise the denominator.

m1 = 3, m2 = 3/2
tan@ = |(3-3/2)/1+9/2| = 3/11
@ = 15*15'18.43'' = 15*15' degrees.

Correct.

let t = tanx for simplicity.

LHS
= 1/1+t + 1/1-t
= 1-t + 1+t/(1-t²)
= 2/(1-t²)
= 2/(1-t²)*t/t
=2t/(1-t²)*1/t
= tan2x/tanx

Wow. Never thought of it that way.

i) -1/2
ii) -1/4
iii) 3/4

Correct.

LHS -
= sinx.cosy + siny.cosx/cosx.cosy+sinx.siny -> divide through by cosx.cosy
= tanx+tany/1+tanx.tany

Correct.

2*4! = 48

2! of Zac and joanne.
7 groups.
2!*6! = 1440

Incorrect. It's 2!7!. There are 8 groups as there are 9 (Sophia and 8 guests) people with one group of 2.

 

[bored of sc]

2*4! = 48

As already been said it's 120.

Total = 6!/(2!2!) = 180

P's together = 5!/2! = 60

P's separated = total - together = 180-60 = 120.

But you're one smart cookie.

 

[midifile]

Lets get this going again

Qu:
The region enclosed between y=sinx and y=coxs and the vertical lines x=0 and x=pi/2 is revolved around the x axis. Find the volume of the solid generated.

 

[danz90]

Lets get this going again

Qu:
The region enclosed between y=sinx and y=coxs and the vertical lines x=0 and x=pi/2 is revolved around the x axis. Find the volume of the solid generated.

lol cbf to put up working

but i got V = (2pi/3) units cubed

most probably wrong.. I SUCK


QUESTION ( an easier one):

 

[midifile]

lol cbf to put up working

but i got V = (2pi/3) units cubed

most probably wrong.. I SUCK


QUESTION ( an easier one):

The answer is actually pi u³ = [
The trick is that you actually use the limits pi/4 and 0 coz if you draw a diagram, you can see that the area enclosed by the graphs and the y axis ends at x=pi/4

Anyway for your question

T_k+1 = ¹²C_k(2x)^12-k(1/x²)^k
independent of x when 12-k = 2k
3k = 12
so k = 4

Therefore T₅ = ¹²C₄(2x)⁸(1/x²)⁴
= 495 x 256 = 126720

Qu: Given that a root of the equation e^x -x -2 = 0 is close to x = 1.2, use one application of Newtons Method to find a second approximatino correct to 2dp

 

[danz90]

The answer is actually pi u³ = [
The trick is that you actually use the limits pi/4 and 0 coz if you draw a diagram, you can see that the area enclosed by the graphs and the y axis ends at x=pi/4

Anyway for your question

T_k+1 = ¹²C_k(2x)^12-k(1/x²)^k
independent of x when 12-k = 2k
3k = 12
so k = 4

Therefore T₅ = ¹²C₄(2x)⁸(1/x²)⁴
= 495 x 256 = 126720

Qu: Given that a root of the equation e^x -x -2 = 0 is close to x = 1.2, use one application of Newtons Method to find a second approximatino correct to 2dp

CORRECT

If that question you gave me with the volume was 3 marks.. if I used the incorrect limits.. but integrated correctly with the wrong limits.. how many marks would I get?

For your last Q, i got x2 = 1.15 (2dp)

Next Q:

 

[tommykins]

1/3[sin^3 x]pi/4 -> 0 = 1/3[1/sqrt2]^3 = 1/6sqrt2.

 

[Kujah]

let u = sin x
du = cos dx

When x=pi/4, u= 1/root 2
When x=0, u= 0

Therefore,

I= Integral of u^2 du [with limits from above]

= [u^3/3]
= [ (1/root2)^3- 0 ]
= root 2/12

Mr and Mrs Roberts and their four children go to the theatre. They are randomlyal
located six adjacent seats in a single row.

What is the probability that the four children are allocated seats next toeachother?

 

[danz90]

tommykins and Kujah; both correct.

i suck at 3u maths

 

[tommykins]

let u = sin x
du = cos dx

When x=pi/4, u= 1/root 2
When x=0, u= 0

Therefore,

I= Integral of u^2 du [with limits from above]

= [u^3/3]
= [ (1/root2)^3- 0 ]
= root 2/12

Mr and Mrs Roberts and their four children go to the theatre. They are randomlyal
located six adjacent seats in a single row.

What is the probability that the four children are allocated seats next toeachother?

4!3!/6! = whatever

 

[hopethisworks]

which bear is best?

 

[midifile]

CORRECT

If that question you gave me with the volume was 3 marks.. if I used the incorrect limits.. but integrated correctly with the wrong limits.. how many marks would I get?

I think you'd get 2

 

[vds700]

Question: Determine the coefficient of x^5 in the expansion of

(1 - 3x + 2x^3)(1 - 2x)^6

 

[midifile]

Question: Determine the coefficient of x^5 in the expansion of

(1 - 3x + 2x^3)(1 - 2x)^6

Coefficient of x⁵ = ⁶C₅(-2)⁵ -3 x ⁶C₄(-2)⁴ + 2 ⁶C₂(-2)²
= -192 -720 + 120
= -792

Qu. Bob tosses 3 coins 10 times in a row. Calculate the probability of obtaining 2 heads and a tail at least two times.

 

[vds700]

Coefficient of x⁵ = ⁶C₅(-2)⁵ -3 x ⁶C₄(-2)⁴ + 2 ⁶C₂(-2)²
= -192 -720 + 120
= -792

Qu. Bob tosses 3 coins 10 times in a row. Calculate the probability of obtaining 2 heads and a tail at least two times.

correct

P(2H and a tail ) = 3/8 = p (assuming any order)
, q = 5/8

P = P(X=2) + P(X=3) + ... + P(X=10) where X is the no. of times he gets 2 heads and a tail

=1 - [P(X=0) + P(X=1)]
=1-[(5/8)^10 + (10C1)(3/8)(5/8)^9]
=1-(0.0091 + 0.055)
=0.94

New q
Use mathematical induction to prove that, for all positive integers, n

SIGMA(r=1 -> n) r^2/[(2r-1)(2r+1)] = n(n+1)/2(2n+1)

 

[azureus88]

for n=1, LHS=1/3 RHS=1/3 so true for n=1

assume true for n=k

for n=k+1,
LHS = [k(k+1)]/[2(2k+1)] + [(k+1)^2/(2k+1)(2k+3)]
= [k(k+1)(2k+3)+2(k+1)^2]/[2(2k+1)(2k+3)]
= [(k+1)(k+2)(2k+1)]/[2(2k+1)(2k+3)]
= [(k+1)(k+2)]/[2(2k+3)] as required

true for n=1, n=k and n=k+1, so true for all n>1

Next Q, prove n^2 - 11n + 30 >or equal to 0 for n>or equal to 1

 

[u-borat]

err that question's wrong man; sub in n=5.5 and you get a negative answer....

 

[vds700]

err that question's wrong man; sub in n=5.5 and you get a negative answer....

Holds for integers only. Sorry, i didnt copy the original question exactly coz i assumed people would know its integers. I fixed it up

 

[u-borat]

na man, my queries with azureus' new question, yours is perfectly fine.