solutions to hsc 2000 (1 Viewer)

[Jago]

i've tried searching this but my thread-searching abilities suck, so does anyone have a copy of the full solutions to that year's hsc?

 

[rama_v]

I reckon if someone posted them up then theyd be infringing on MANSW copyright (presuming they use that book). I say lets write up our own solutions (as was done for the CSSA trial)

I will start off, its good revision for me
1. (a)2.08
(b) x+7=>3
.: x =>-4 (on the number line coloured circle at -4, arrow going right)
(c) (sqrt3)/2
(d) P(blue) = 3/(2+3) = 3/5

(e) x = 2 + y
3(2+y) + 2y = 1
6 + 5y = 1
5y = -5
y = -1
x = 2 + -1 = 1
(f) x - 5 = 3 or x - 5 = -3
x = 8 or x = 2
(g) y = 2x + 3
x-int: -3/2
y-int: 3

 

[Jago]

that could be arranged, let me have a go at doing the paper, i'll post my answers and everyone here can correct me. Gimme 3 hours

 

[rama_v]

that could be arranged, let me have a go at doing the paper, i'll post my answers and everyone here can correct me. Gimme 3 hours

ok no worries

 

[Jago]

here goes:

1.
a) 2.08
b) x > = -4
c) (sqrt3) / 2
d) 3/5
e) x=1 , y=-1
f) x = 8, 2
g) graph

2.
a) proof
b) M(2,1)
c) y = 2x - 3
d) proof
e) 5 units
f) x² + (y+3)² = 25
g) proof
h) R(4, 5)

3.
a)
i) 3xe^x + 3e^x
ii) 2x cos(x²+1)
b) 5 1/4cm
c)
i) tan5x/5 + c
ii) 4ln2 + c
d) y = 2x-2

4.
a) proof
b)
i) 2.3km
ii) 32
iii) 161.6km

5.
a) x = 1.11, 4.25
b)
i) diagram
ii) 6/15
iii) 2/3
c)
i) k = (ln2) / 3
ii) 3200
iii) 92.42 insects per week

6.
a) graph
b)
i) 175
ii) 0
iii) 256
iv) sqrt 3
v) graph

7.
a) 2pi
b) 0.9244
c)
i) P(-1,-1) Q(2,2)
ii) 7.5u²

8.
a)
i) V = 2 - 3/(t+1)
ii) v = -1
iii) t = 1/2
iv) 4.32m
b) proof

9.
a)
i) graph
ii) 2 solutions
b) graph
c) (part 1-3) proof
iv) x= sqrt 40

10.
a)
i) A3 = 5000 - 3M
ii) proof
iii) A36 = (5000 - 3M)1.01³³ - M((1.01³³ - 1)/0.01)
iv) $161.34
b) no idea on the snowplough...


yep that's about it, oh and it'd be a great help if someone could sketch up the correct graphs, thanks.

 

[rama_v]

Alright guys here are the graphs, can someone please check them?

 

[rama_v]

And for 10 b:

(i) It is known that the speed, v, is inversely proportional to the depth of the snow. i.e.
v = A/h where A is constant.
this also means that dx/dt = v = A/h
Since the depth of the snow, h, increases at a constant rate, let the depth h be represented by mt, where m is constant
i.e. h = mt

Now, dx/dt = A / mt
= A/m (1/t)
since A and m are both constants, replace A/m with k, where k is another constant
so dx/dt = k/t as required

Im not sure if this is correct reasoning - can someone please check, thanks

10 b ii
Int (dx) = k Int (from T to T+2) 1/t dt = 1
therefore 1 = k log ((T+2) / T)
similarly,
Int (dx) = k Int ((T+2) to (T+5.5)) = 1
therefore 1 = k log ((T+5.5)/(T+2))

Combining these two equations:
k log ((T+5.5)/(T+2)) = k log ((T+2) / T)

(T+5.5)/(T+2) = (T+2)/T
solving for T, T = 2.66667 = 2 hours 40 minutes
6:00 am - 2 hours 40 minutes = 3:20 a.m.

Therefore it began snowing at 3:20 a.m.

 

[Jago]

i think i got those graphs, looks right to me.

will decypher 10b(ii) later.