Solutions (2 Viewers)

[Ghost1788]

Ok so for all those people who did get the answers to the questions in Question 8, Question 9 and Question 10 ...post your solution for all to see..i'm particularly curious about how some people managed to get answer for Question 10 part b...

below is my solution for Question 10 part a i and ii i didnt get the rest...

Explain why α + β = m and αβ=-b

y=x²
y=mx + b
A(α,α²)
B(β,β²)

x²= mx + b
x² - mx = b

Sub x = α

α² - αm = b

Similarly

β² - βm = b

α² - αm = β² - βm
α² - β² = αm - βm
(α - β)(α + β) = m(α - β)

m = (α + β) # ------*1*

b = α² - αm

Sub in *1*

b = α² - α(α + β)
= α² - α² - αβ
b = -αβ
αβ = -b

Given "(α - β)² + (α² - β²)² = (α - β)²[1 + (α² + β²)]"

Show AB = sqrt[(m² + 4b)(1 + m²)]

AB² = (α - β)²[1 + (α + β)²]
=(α² - 2αβ + β²)[1+ α² + 2αβ + β²]
=(m² - 2αβ - 2αβ)[1 + m² - 2αβ + 2αβ]
=(m² - 4αβ)[1 + m²]
=(m² - 4(-b))[1 + m²]

AB² = (m² - 4(-b))(1 + m²)
therefore
AB = sqrt[(m² - 4(-b))(1 + m²)] # as required

-Ghost

 

[rama_v]

For 10 a i
x^2 - mx - b = 0
the solutions to this equation are x = alpha and x = beta
alpha + beta = -b/1 = (m/1) = m (sum of roots)
therefore:
alpha + beta = m
SImilarly alpha beta = c/a = -b

 

[Trippendicular]

What did people who thought they did 10b) correctly get? I got 7/16 for the probability they meet and t=60/root 2 for 50% chance

 

[Jago]

i got 7/16 as well. Shame i ran out of time for the last bit, i actually had some idea of how to do it (maybe)

 

[LostAuzzie]

What did people who thought they did 10b) correctly get? I got 7/16 for the probability they meet and t=60/root 2 for 50% chance

I too got 7/16 for the probability, didnt get to do the 50% chance, ran out of time

 

[Homercles]

This is the first time I've ever posted here, but I think people may be trying too hard for 10ai.

y = mx + b

The gradient of this line is m. The points (a,a(squared)) and (B,B(squared)) lie on the line, so:

m = {B(squared) - a (squared)} / {B - a}

= {(B-a)(B+a)/(B-a)

= B + a

Substitute this value for m and either of the points for x and y into the equation and the second part (b= -aB?) should work out fairly easily.

 

[littlebinzy]

How did you get to 7/16?

I screwed up the first part of question ten. I just said they were roots and there b/a, -c/a... at the time I thought it was wrong, and told myself it was wrong and that roots go -b/a and c/a but pfft I didn't have any other ideas...

 

[nono]

I have a few questions I need answered...

Q6) (b) ii
The questions says "after 20 minutes"...I found first derivative and sub in 20- that gives the rate AT 20 minutes...did we have to say "therefore, after 20 mins it would drain at a rate of less than 80L/minute"??

Q6 (b) iii
What the hell was this?? I got t=60 which is an endpoint and therefore couldn't prove it was a max. I then tried second derivative which was a constant which led to believe that there was no fastest rate?? Hmmm..

Q8 (c) iii
Was this 13 years?? I got 12.3something and I'm not sure if I should have rounded up for down...

Q10 (a) iv
Did we have to prove that it was a max?? And did anyone get P(2/m, 4/m^2)?

Q10 (b) all
Can someone please explain to me what this was on about???

Thanks.

 

[LostAuzzie]

This is the first time I've ever posted here, but I think people may be trying too hard for 10ai.

y = mx + b

The gradient of this line is m. The points (a,a(squared)) and (B,B(squared)) lie on the line, so:

m = {B(squared) - a (squared)} / B - a

= {(B-a)(B+a)/(B=a)

= B + a

Substitute this value for m and either of the points for x and y into the equation and the second part (b= -aB?) should work out fairly easily.

That is what I did for that part too

 

[dinipoo]

This is the first time I've ever posted here, but I think people may be trying too hard for 10ai.

y = mx + b

The gradient of this line is m. The points (a,a(squared)) and (B,B(squared)) lie on the line, so:

m = {B(squared) - a (squared)} / B - a

= {(B-a)(B+a)/(B=a)

= B + a

Substitute this value for m and either of the points for x and y into the equation and the second part (b= -aB?) should work out fairly easily.

damn. the paper is set for ppl like u. def the way to do it for a one mark q. good on ya, wish i thought like that.

 

[LostAuzzie]

Q10 (a) iv
Did we have to prove that it was a max?? And did anyone get P(2/m, 4/m^2)?

Dont think we had to prove it was a max, and from memory yeh thats what I got for that question

 

[-cathie-]

Loan Repayments

Does anybody remember what year the loan would be repayed?!

 

[Jago]

How did you get to 7/16?

I screwed up the first part of question ten. I just said they were roots and there b/a, -c/a... at the time I thought it was wrong, and told myself it was wrong and that roots go -b/a and c/a but pfft I didn't have any other ideas...

x - y >= 1/4, y - x =< 1/4

graph the lines, find the 2 triangles. 1 - that = 7/16

 

[c0jo]

i think 2017 ?

 

[Srixon]

hey did anyone get t = 30 for 10 b) iii i thought that it may be this as there are 60 minutes in an hour and if u stayed there for 30 minutes u would have a 50% chance of meeting them

 

[snickerdoodle]

Does anybody remember what year the loan would be repayed?!

I think it was 2018.

Edit: Damn, it was 2017! Stupid calculator.

 

[-cathie-]

I think it was 2017.... (around thatz)

Geez... should have read the question... I just said in the 12th year!

 

[rachey_poo]

i got all of q10a out....i found that using x^2-mx-b=0 was the best way to do (i), and then using alpha and beta as roots of the equation and using alpha + beta = -b/a and alphabeta=c/a
in part (iv) though....i have x=m/2 y=m^2/4 rather than the other way round as ive seen people post so far....
A = 1/2(mx'sqrt'm^2+4b - x^2'sqrt'm^2+4b + b'sqrt'm^2+4b)
dA = 1/2(m'sqrt'm^2+4b -2x'sqrt'm^2+4b)
dx
0= 1/2(m'sqrt'm^2+4b -2x'sqrt'm^2+4b)
0=m-2x
2x=m
x=m/2
therefore y=m^2/4

anyone????

 

[Trippendicular]

This is my answer to 10 b ii)

 

[rama_v]

yep i got m/2 and m^2/4 for that question