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my solutions, all questions (2 Viewers)

failingTheHsc

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here ya go guys:

Question 1
(a) 0.0302
(b) 3+sec^2 x
(c) 95 degrees
(d) $280
(e) x^3-8x+5
(f) x=10, -4

__________________

Question 2
(a) ex + 2y - 4 + e^2 = 0
(b) i) root(2)
ii) -1
iii) 135
iv) y = -x + 10, C(10, 0)
v) is a proof
vi) 35

__________________

Question 3
(a) i) 18e^x(2e^x-4)^8
ii) x (2 sin x + x cos x)
(b) 89
(c) a graph
(d) i) ln (x^2 + 5) + C
ii) root(3) - 1

_________________

Question 4
(a) i) 55
ii) 64 degreees 28 minutes

(b) i) 1/6
ii) 1/2

c) i) A(1, -3)
ii) 4.5

________________

Question 5
(a) i) proof
ii) (0,0) inflexion and (3, -27) min t.p.
iii) graph
iv) 0 < x < 2

b) i) 63 blocks
ii) 73 rows
iii) 3738 blocks

___________________

Question 6
(a) x = 12
(b) i) alt angles, parallel lines
ii) proof, using supplementary angle and angle sum of triangle
iii) use parallel lines from a single point in ratio

(c) i) C<sub>0</sub> = 5, k = 0.5798
ii) 5.6 years

___________________

Question 7
(a) i) 2 + root(2)
ii) r = 1/(root(2) - 1), |r| < 1. Therefor no limiting sum

b) i) pi/3, 5pi/3
ii) 6ms^-1
iii) is a graph
iv) 9m

___________________

Question 8
(a) y = 2
(b) v = pi {5, 1 (ln y)^2 dy
(c) 11.9
(d) i) proof. do normal thing to see where the lines would intersect and show the discriminant is positive for all values m, so will always have a solution.
ii) m=1 or m= 2/3
iii) y = x - 3 and y = 2/3 x - 4/3

__________________

Question 9
(a) x = 7pi/6, 11pi/6
(b) i) pi/6
ii) pi r^2 / 2
iii) pi r^2 / 3
iv) (pi r^2 - 3 root(3) r^2 ) / 6

(c)I) dL/dt = v - u, v and u are constants, L = (v-u)t. Substitue with the E equation given
ii) v = 3u/2

___________________

Question 10
(a)i) 120000a^2 - Fr - F, r = 1 + 0.06/k
ii) proof.
iii) $2324.47
iv) $498

(b)i) proof. note that XA = 1 - x, PM = 2 - PA, l = XP + PM
ii) just differentiate i) (too much algebra 4 me, i skipped it)
iii) M is closest when l is largest. solve dl/dx = 0, and using the factorisation
x = 1 and some other number involving r's. so i just used 1 ;o)

edit: changed a few answers and got rid of those <sup> things

there ya go. thats wat i did. let me know any agreements disagreements.
 
Last edited:

pri

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fuk u....u seem to have killed it!!!!!!!!!!!!!!!!!
 

Beaky

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Originally posted by sepul
heh yeah, you sure the probability anwers are correct tho' ?
Yeah Im going with your first answer 1/6...

Second one 1/2 not too sure bout
 

failingTheHsc

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Originally posted by Beaky
Yeah Im going with your first answer 1/6...

Second one 1/2 not too sure bout

4 b) ii)
for the probability i did a table of possible outcomes
1 1
1 3 *
1 5
1 7
2 1
2 3 *
2 5
2 7
3 1 *
3 3 *
3 5 *
3 7 *

6 * 1/3 * 1/ 4 = 1/2

wat did u get?
 

SmokedSalmon

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The 1/2 is correct.

1 - (none stop on 3)
1 - (2/3 x 3/4)
1- 1/2
= 1/2

therefore 'failingthehsc' is correct :)

Comparing my answers to yours (of what I can remember) I seem to have everything similar or the same. A few are different but couldn't be bothered to post up.
 

Beaky

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bahhh your right :D

I can see

wat i did was P(3,x) P(x,3)
which is like 1/3 x 1/4 + 1/3 x 1/4

WTF was i thinking!!!

lol your gonna cane it both of u guys... prob 110+ good job... :D
 

PoLaRbEaR

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Question 4
(a) i) 55
ii) 64 degreees 28 minutes
i) It says the bearing is 325 degrees...so angle PQR is:
360-325 = 35 degrees
ii)You should get at most 2 out of 3 marks because of your first one being wrong
 

Winston

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Originally posted by PoLaRbEaR
i) It says the bearing is 325 degrees...so angle PQR is:
360-325 = 35 degrees
ii)You should get at most 2 out of 3 marks because of your first one being wrong
No it's not 360 dude

the whole thing is 325

so you minus the 3 quadrants


325-270 = 55
 

iambored

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Originally posted by failingTheHsc
question 8
(b) v = pi {<sup>5</sup><sub>1</sub> (ln y)<sup>2</sup> dy
( [/B]
is that wrong? because u can't integrate a log

but come to think of it i found the area!! argh!
 

Winston

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Originally posted by iambored
is that wrong? because u can't integrate a log

but come to think of it i found the area!! argh!
take a look at my solution on the the solutions thread



meh don't worry about the logic of not being able to integrate a log, in theory it can be done, just not at our level of maths, but they didn't ask us for it anyways. They just wanted the equation.
 

SoCal

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I must have made at least 15 marks worth of stupid mistakes:mad: and don't you just hate it when you cross out the right answer and put a different answer in:mad:.

How are you supposed to answer Question 9 (a):confused:?
 

spark

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I just marked mine against this and I probably got 65% more or less. How fucking fantastic!
 

Sah

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integration of ln x is
xlnx - x :) it works but we didnt have to do it and i dont know integration of (ln x) square

umm for question 10b last one i got 2 answers x=1 and x=r and i thought x=r was the correct one when i subbed it, maybe i chose wrong one cause i was presed for time.
 

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