Some solutions? (1 Viewer)

[zahmad]

Could someone help me out with these questions?
I can't figure them out! Thanks for all your help:

Q5C
Q5D(i)
Q7A(ii)
Q7C(i)(ii)
Q8B
Q8C
Q9A
Q10B

Word file attached

 

[Epistemophobia]

dam.

 

[zahmad]

I've done them all, but need to check answers!

 

[lyounamu]

Could someone help me out with these questions?
I can't figure them out! Thanks for all your help:

Q5C
Q5D(i)
Q7A(ii)
Q7C(i)(ii)
Q8B
Q8C
Q9A
Q10B

Word file attached

5c. y = x . ln(1-3x)
y' = x . -3/(1-3x) + ln(1-3x)
= -3x/(1-3x) + ln(1-3x)

5di
when x = 1, y = e^-2
when x = 0, y = 1
when x = -1, y = e^2

A = (approximately) 2/3(e^-2 + 4 + e^2) = 2/3e^2 + 2e^2/3 + 4

7aii

dT/dx = (1/2 . (x^2+4)^-1/2 . 2x)/5 - 1/13
= 2x/10(x^2+4)^1/2 - 1/13
Minimum point is when dT/dx = 0

i.e. 2x/10(x^2+4)^1/2 - 1/13 = 0
2x/10(x^2+4)^1/2 = 1/13
2x = (10(x^2+4)^1/2)/13
26x = 10(x^2+4)^1/2
(26x)^2 = 100(x^2+4)
676x^2 = 100x^2 + 400
576x^2 = 400
x^2 = 25/36
x = 5/6 #

Rest should be easy - where you prov that x=5/6 is indeed a minimum point.

7ci
1st trough:

2.5 = 2.5 + 4(1-1)

2nd trough:

2.5 + 4 = 2.5 + 4(2-1)

nth trough:

2.5 + 4(n-1)

7cii

Sum of distance travelled = 1314 = 2Sn because you need to keep coming back to tap. So Sn = 657

Sn = n/2(2a+(n-1)d) where a = 2.5 and d=4
657 = n/2(5+(n-1)4)
657 = n/2(1+4n)
657 = n/2 + 2n^2
1314 = n + 4n^2
4n^2 + n -1314 = 0

Using quadratic formula:
n = 18

 

[lyounamu]

Could someone help me out with these questions?
I can't figure them out! Thanks for all your help:

Q5C
Q5D(i)
Q7A(ii)
Q7C(i)(ii)
Q8B
Q8C
Q9A
Q10B

Word file attached

8b

I cannot draw here and I don't have a scanner to show you my drawing but I will describe the graph here. The graph will be DECEASING but at decreasing rate.

8ci and ii

Let the terms be a (the first term), b (a+d) and c (a+2d) AND d is the difference between them

d = b-a = c - b
so b-a = c-b
a = 2b - c


a+b+c = 24

2b - c + b +c = 24

3b = 24

b = 8

Now

a^2 + b^2 + c^2 = 200

(2b-c)^2 + 8^2 + c^2 = 200
(16-c)^2 + 64 + c^2 = 200
256 - 32c + c^2 + 64 + c^2 = 200
2c^2 - 32c +120 = 0

c=10 so a = 6

Three terms are 6, 8 and 10

It can also be the other way around. 10, 8 and 6.

 

[zahmad]

Thanks, what does it mean by use one application in Q5d(i)?

 

[Aerath]

Hey, sorry they're so messy:

 

[Aerath]

Hmmmph, Lyounamu beat me to it (again).

 

[lyounamu]

For 9 & 10 questions, I will just post up answers. Doing them and showing them here will just be waste of time as you only want the answers...

9ai1
18000 - M

2

18000 x 1.0055^3 - M(1+1.0055+1.0055^2+1.0055^3)

ii

M = $425.36...

iii

TC = 48 . M = $20417.29693...

EDIT: I am not doing 10, as Aerath already kindly did it for us.

 

[lyounamu]

Thanks, what does it mean by use one application in Q5d(i)?

You use it once.

You only have 3 function values.

 

[zahmad]

Um, two different answers for Q5d, whose right?
Could you please kindly show working for Q9....bit unsure, got it wrong

 

[Timothy.Siu]

Q5C
they want u to find the derivative, so its just y^'= (-3x²)/(1-3x) + ln(1-3x)
Q5D(i)
y=e]sup]-2x[/sup] area of that from -1 to 1 is approx. = 1/3(e²+4+e^-2
Q7A(ii)
u know what T is, so you just differentiate it
T^'= x/5[SQRT(x²+4)] - x
cudn't do it, my differentiation must be wrong

Q7C(i)(ii)

Q8B
it just means the gradients negative less than 0 but concave up greater than zero
Q8C
a+a+n+a+2n = 24 3a+3n=24 a+n=8
a^2+a^2+2an+n^2+a^2+4an+4n^2=200
3a^2+6an+2n^2=200 a=8-n from above
3(64-16n+n^2)+6(8n-n^2)+2n^2=200
192-48n+3n^2+48n-6n^2+2n^2=200
192-n^2=200 n^2=-8 LOL i did something wrong

Q9A
let R= 1+0.066/12
first repayment - 18000-$m
2. 2nd = (18000-$m)R-m = 18000R-MR-M
3rd = 18000R^2-MR^2-MR-M
4th = 18000R^3-MR^3-MR^2-MR-M

ii)nth repayment = 18000R^(n-1) - [M(R^n-1)/r-1]
theres 48 repayments in total
18000R^(n-1) - [M(R^n-1)/r-1]=0 and then u solve for M i got m=425.36$
iii) multiplay M by 48 = 20417.30$$$

Q10B

 

[lyounamu]

Um, two different answers for Q5d, whose right?
Could you please kindly show working for Q9....bit unsure, got it wrong

Follow Aerath's that's better method.

My earlier post on the function value edited.

 

[Timothy.Siu]

Follow Aerath's that's better method.

My earlier post on the function value edited.

lol aren't both the methods the same
lol and i got a different answer to u for 9 :S dont like these consumer arithmetic

 

[Mothers]

Namu, are you doing 4U next year?

 

[lyounamu]

lol aren't both the methods the same
lol and i got a different answer to u for 9 :S dont like these consumer arithmetic

His one is better because he has a diagram on it - jks.

Nah, I actually didn't do one value.

 

[lyounamu]

Um, two different answers for Q5d, whose right?
Could you please kindly show working for Q9....bit unsure, got it wrong

How do you know that you got it wrong? Have you got solution? If that's so please check if I am right or not. If I am right, I will post mine up because there isn't much point in doing so because mine is similar to timothy's.

 

[Timothy.Siu]

Namu, are you doing 4U next year?

i think so although i do know someone that did acc. mx1 last yr and did well and didn't do 4unit maths this year

 

[lyounamu]

Q5C
they want u to find the derivative, so its just y^'= (-3x²)/(1-3x) + ln(1-3x)
Q5D(i)
y=e]sup]-2x[/sup] area of that from -1 to 1 is approx. = 1/3(e²+4+e^-2
Q7A(ii)
u know what T is, so you just differentiate it
T^'= x/5[SQRT(x²+4)] - x
cudn't do it, my differentiation must be wrong

Q7C(i)(ii)

Q8B
it just means the gradients negative less than 0 but concave up greater than zero
Q8C
a+a+n+a+2n = 24 3a+3n=24 a+n=8
a^2+a^2+2an+n^2+a^2+4an+4n^2=200
3a^2+6an+2n^2=200 a=8-n from above
3(64-16n+n^2)+6(8n-n^2)+2n^2=200
192-48n+3n^2+48n-6n^2+2n^2=200
192-n^2=200 n^2=-8 LOL i did something wrong

Q9A
let R= 1+0.066/12
first repayment - 18000-$m
2. 2nd = (18000-$m)R-m = 18000R-MR-M
3rd = 18000R^2-MR^2-MR-M
4th = 18000R^3-MR^3-MR^2-MR-M

ii)nth repayment = 18000R^(n-1) - [M(R^n-1)/r-1]
theres 48 repayments in total
18000R^(n-1) - [M(R^n-1)/r-1]=0 and then u solve for M i got m=425.36$
iii) multiplay M by 48 = 20417.30$$$

Q10B

Highlighted part is wrong. It should just be 48 because you pay 48 times.

 

[Timothy.Siu]

Highlighted part is wrong. It should just be 48 because you pay 48 times.

lol nah thats rⁿ -1 not r^n-1 :uhhuh: