Solutions (2 Viewers)

[frootloop]

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????

Yup got that! YAY.

 

[LostAuzzie]

I put 2017, wasnt sure whether to round up or not

 

[Phantasmagoria]

It's a bit hard to explain what I did for Question 10b ...

10b ii) I got 7/16 too! Basically, you draw the two inequalities and have to find the area in between. Sub 1/4 into the inequalities, to find that they each hit x=1 and y=1 at 3/4. Then find the area of the two triangles (each is 3/4 x 3/4 x 1/2), and subtract them from the area of the square. i.e. 1 - (3/4)^2 = 7/16

iii) It's basically the same thing, only you have t/60 rather than 15/60. Sub t/60 into the inequalities to find that they hit x=1 and y=1 at 1- (t/60). Which means the two triangles each have the area of 1/2[1-(t/60)]^2. Therefore you have the equation: [1 - (t/60)]^2 = 1/2. Expand, factorise, and voila! I got 17 minutes 34 seconds, by the way.

I hope my explanation actually makes sense ...

I got 2018 for 8c)iii. Since you have to make the payment at the END of each year, you would only have paid 12 payments by the END of 2017. Therefore that tiny little bit would have to be paid in 2018. (Gah, am getting confused.)

 

[-cathie-]

9a

Answers...? without the graph of course....

 

[Trippendicular]

Your 10 b iii) sounds right, I think I missed taking the t/60 value away from one to find the sides of the triangles and therefore i got the opposite result to you, 42 mins and 26 seconds

 

[Trippendicular]

Answers...? without the graph of course....

ii) t=pi

iii) D=2 pi

from memory

 

[rachey_poo]

damn....i would have got 2018, if i'd read the question right....still, i did put in the 13th year after the loan began, so that would give me 2018....spose it depends how many marks they allocate for the actual year....
any one get t=4 for q7bii) ????

 

[sikeveo]

I got 2018 for 8c)iii. Since you have to make the payment at the END of each year, you would only have paid 12 payments by the END of 2017. Therefore that tiny little bit would have to be paid in 2018. (Gah, am getting confused.

It was 12.23~ years, thus it was 2017. You rounded up a year too much. The first payment was at the end of 2005.

 

[-cathie-]

I got 2018 for 8c)iii. Since you have to make the payment at the END of each year, you would only have paid 12 payments by the END of 2017. Therefore that tiny little bit would have to be paid in 2018. (Gah, am getting confused.

It was 12.23~ years, thus it was 2017. You rounded up a year too much. The first payment was at the end of 2005.

WHAT! It is 2017 or 2018?!

 

[frootloop]

damn....i would have got 2018, if i'd read the question right....still, i did put in the 13th year after the loan began, so that would give me 2018....spose it depends how many marks they allocate for the actual year....
any one get t=4 for q7bii) ????

I got t=4.

Excellent.

 

[Zali]

I got 2018 for 8c)iii. Since you have to make the payment at the END of each year, you would only have paid 12 payments by the END of 2017. Therefore that tiny little bit would have to be paid in 2018. (Gah, am getting confused.

It was 12.23~ years, thus it was 2017. You rounded up a year too much. The first payment was at the end of 2005.

same except like cathie i didn't read the question and just put in the 13th year even tho i have 'read the question' written in white out across my calculator

 

[Jago]

yep at t=4 the area above the t axis = the area below.

 

[rama_v]

I put down 2017 because 12.25 was 12 years and a bit
I think the year is still 2017 even though its the 2018th year, liek for example this is the 2006th year even though we say its 2005. I think a similar question was in last years paper and u had to similalrly round down.

 

[rachey_poo]

yay for t=4!

 

[rachey_poo]

ok...new question....can someone explain how they worked q6c? i did think i worked it out...but the more i look at it, the more i think im wrong....
if you use their hint of splitting it into two regions...the top one would be (correct me if im wrong..please!):
y=12-2x^2-x^2
y=12-3x^2
3x^2=12-y
x^2=(12-y)/3
and using V= pi times the integral of (12-y)/3 dy

???

 

[rachey_poo]

lol mathewm same question!!

 

[rama_v]

You had to split it up by rotating the y = x^2 around the y axis from 0 to 4 or whatever and then teh other curve around teh y axis from 4 to 12.

 

[-cathie-]

You had to split it up by rotating the y = x^2 around the y axis from 0 to 4 or whatever and then teh other curve around teh y axis from 4 to 12.

Anybody remember the final answer?!

 

[botho]

10) A) I) the easy way

m = Y2 – Y1
X2 – X1

= β2 –α2
β – α

= (β + α)(β – α)
β – α

m = (β + α)


Y –Y1 = m(x – x1)
Y - α2 = (β + α)(x – α)
Y = βx + α x - α β - α2 + α2
= mx - α β

Y = mx + b
-αβ = -b

HOPE THIS IS RIGHT

 

[redw0lf]

re : 8 b) iii)

At the end of 2005, ie. A1, the first payment is made.
It then follows, that at A12, the time will be the the end of 2017. sub in n = 12. There is still $104,024 left owing.
Therefore the final payment will be made in 2018