2004 CSSA solutions (1 Viewer)

[pull up]

hay people, does anyone happen to have the 2004 Catholic Trial solutions? if u do, please provide link, or email them to mo3_da_ho3@hotmail.com
thanx.

 

[ifsonotso_100]

bump this, solutions please.

 

[acmilan]

Id be willing to write up solutions if someone uploads the actual exam

 

[goan_crazy]

I have both but they arent scanned yet
Maths isn't my priority at the moment
i have 5 exams before maths this week
maths is the one after
I'll upload it maybe after friday if i get a chance

 

[FinalFantasy]

or you can just post the questions which u can't do and im sure someone would help

 

[ifsonotso_100]

well the cssa 04 2 maths is on the mirror.

 

[FinalFantasy]

Question 1:
A) x^5=5000
.: x=(5000)^(1\5) , u put it in ur calculator for the 3 sig figs lol
B)0.3+0.3'=0.3+1\3=19\30
C)tan@=3
@=72 degrees, @=252 degrees(180+72)
D)1-(a-b)\(a+b)=[(a+b)-(a-b)]\(a+b)=2b\(a+b)
E)8^x=32
ln 32=xln8
x=(ln 32)\(ln 8)=5ln2 \ 3ln2=5\3 <---Edit here
F)1\(2-sqrt 3)=a+bsqrt (3)
1\(2-sqrt 3) * (2+sqrt 3)\(2+sqrt 3)=(2+sqrt 3)\(4-3)=2+sqrt3
.: a=2 and b=1

Question 2:
A)
i) d\dx (3x+4)^7=7(3x+4)^6 * 3=21(3x+4)^6
ii)d\dx (x³e^x)=3x²e^x+e^x(x³)=x²e^x(3+x)
iii)d\dx (tan 5x)\5x=(5sec² (5x)*(5x)-5tan(5x) )\(5x)²
=5(5xsec²(5x)-tan(5x) )\25x²=(5sec²5x-tan 5x)\5x²
B)
i)int. (e^(3x) +x^(1\2))dx=(1\3)e^(3x)+(2\3)x^(3\2)+C
ii)int. (x^4+1)\x dx [from x=1 to x=2]
=int. (x³+1\x) dx [x=1 to x=2]
=[(1\4)x^4+ln |x| ] from x=1 to x=2
=(1\4)2^4+ln (2) - [ (1\4)+ln (1) ]
=15\4+ln (2)
iii)dy\dx=2x-sin x
y=int. (2x-sinx ) dx=x²+cosx+C
"y=2 when x=0"
so 2=0²+cos 0+C
.: C=1
.: y=x²+cosx+1

ok time to sleep
sum 1 else can do da rest
btw I haven't checked these so use at ur own risk lol

 

[acmilan]

I trust FF got q1 and 2 right so ill do some. Ill start from the end as thats the ones most people want.

10(a) (i) dv/dt = k so dv = k dt for some constant k. Integrating both sides gives v = kt + c₁

(ii) Integrate v = kt + c₁ to get x. x = 1/2(kt²) + c₁t + c₂

When t = 0, x = 1 so 1 = 0 + 0 + c₂ and hence c₂ = 1
When t = 1, x = 2 so 2 = 1/2k + c₁ + 1 which gives k + 2c₁ = 2
When t = 2, x = 9 so 9 = 2k + 2c₁ + 1 which gives 2k + 2c₁ = 8

k + 2c₁ = 2
2k + 2c₁ = 8

Subtract one from the other gives k = 6 and sub back in gives c₁ = -2

Hence x = 3t² - 2t + 1

(iii) v = kt + c₁ but we know k and c₁ which gives v = 6t - 2

Particle comes to rest when v = 0, ie. when 6t - 2 = 0 and t = 3 seconds.

10 (b) (i) (Note: alpha = A and beta = B)
In triangle ACD, cosB = CD/AC = h/b so h = bcosB
In triangle BCD, cosA = CD/BC = h/a so h = acosA
Hence h = bcosB = acosA

(ii) General area of a triangle formula: A = 1/2absinC

In triangle ACD, Area = 1/2*AC*CD*sinB = 1/2*b*h*sinB = 1/2absinBcosA (since h = acosA from (i))

(iii) Area = 1/2*CD*BC*sinA = 1/2ahsinA = 1/2absinAcosB (since h = bcosB)

(iv) Area = 1/2*AC*BC*sin@ (where @ = angle ABC) = 1/2absin(A + B)

(v) Area (ABC) = Area (ACD) + Area (BCD)
1/2absin(A + B) = 1/2absinBcosA + 1/2absinAcosB

You can divide by 1/2ab
so sin(A + B) = sinBcosA + sinAcosB


I just went through this quickly, please tell me if I made any errors. Ill try do some of the others when I get back. If anyone else wants to try some the link to the paper is here: http://www.boredofstudies.org/mirror/2004/2004_Maths2u_T_CSSA_q.pdf

 

[acmilan]

Questions 3-6

Like before, possibility of errors. Let me know if you find any and ill fix it

edit: fixed up q3(a)

 

[sunjet]

acmilan you uai machine =)

 

[reemz]

any1 got 2005 papers answeres?

 

[ifsonotso_100]

can you post the paper, and someone will give answers.

 

[FinalFantasy]

yes, post the paper, the 3unit paper as well plz

 

[reemz]

i wish i had the paper....... has any1 got it yet? cum on pp help each other out ....

 

[FinalFantasy]

Question 7:
A)
i)
Area=int. x^-2 dx from x=1 to x=3
=[-1\x] from 1 to 3
=-1\3-(-1)=2\3 units²
ii) V=pi int. y² dx=pi int. x^-4 dx from x=1 to x=3
=pi [-1\3x³] from 1 to 3
=pi (-1\81-(-1\3) )=26pi\81 units ³
B)
i) *sketch*
range of f(x)=e^x is y>0
ii)
V=pi int. x² dy from y=3 to y=5

y=e^x
.: ln (y)=x
x²=(ln y)²
.: V=pi int. (ln y)² dy from y=3 to y=5

iii) y: 3, 3.5, 4, 4.5, 5
[ln y]: ln (3), ln (3.5), ln (4), ln (4.5), ln (5)
[ln y]² (ln 3)², (ln 3.5)², (ln 4)², (ln 4.5)², (ln 5)²
area=pi*1\6[(ln 3)²+4((ln 3.5)²+(ln 4.5)²)+2(ln4)²+(ln5)²]
=calculator

Just a quick question before I leave again
Now only question's 8 and 9 left...
Might have mistakes, so tell me if you find any

 

[acmilan]

Q8-9

Im a bit iffy with 9(b). I remember interpreting it wrong when i actually did the exam and arguing with my teachers about the wording (they actually wrote this exam). But if i remember right, thats the way they wanted to do it.

 

[Alexluby]

acmillian.... just wonderin.. for question 3 c) v) .... is ur solution wrong?? i think it's 16 isn't it... if not can u explain ur answer?

thx

 

[acmilan]

I dont know. Not drawing it makes it hard to imagine. I took PC as the base of the triangle, and since P and C have both the same y-coordinates, they lie on a horizontal line of length 5 - (-1) = 6 units. So you need the height, which is the perpendicular distance from A to the line PC. But since PC is a horizontal line, the perpendicular distance would just be the distance between the y-coordinates of A and points on the line PC, ie. 7 - 5 = 2 units. It could be wrong though. How did you do it?

edit: to get area you multiply the above two numbers and half it, ie. A = 6 x 2 x 0.5 = 6 units²

 

[nit]

3c(v) is 6.

 

[Alexluby]

Acmillan, not to insult u or anything, but i think maybe coz you were doing ans for Qs 3-6, there's quite some careless mistakes in there...