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Solutions (1 Viewer)

Alex222

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Oct 24, 2003
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Ok i probably shouldnt be the one to be doing this, because i'm not expecting great great great things - but here are my solutions. Pleease tell me what you guys got the same or different!

Question 1 -
a) -1, 4.5
b) x<2, x>=5
c) 1.5
d) y = 12xsquared
e) 6/25

Question 2 -
a) Graph
b) x/(1 + x squared) + tan`x
c) 1
d) 80
e)i) root2cos(x + pi/4)
ii) Graph

Question 3 -
a) 30240
b) i) because its in the form - n squared x
ii) y=4
iii) pi
c) i) explanation
ii) .01784
d) Proof

Question 4 -
a) 3003
b) Not sure what i wrote?
c) 19
d) i) because it has four verteces which are concyclic
ii) because its opposite angles are supplementary
iii) proof
iv) proof

Question 5 -
a)1/12 sin6x + .5x + c
b)i) not pass horizontal line test
ii)graph
iii) x>1 (or equal to!)
iv) ????
c)i) proof
ii) b = 480
iii) 500 degrees

Question 6 -
a) i) proof
ii) pi/10
b) ???????

Question 7 -
a) distance is 2798.96 and bearing is 208 degrees
b) proof/proof/ ???


No laughing please guys! haha
 

Constip8edSkunk

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5c)iv) solve the quadratic formed when u switch the y with the x, take the value thats smaller than 2
6 b) i got something like root2 - 1 and then 2 - root 2
7a) i got 2000 something meters and 256 degrees i think
b)iii) dun think many ppl got the whole thing out, i only got the last part not the 1st eqn... time ran out
 

nerdd

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4c i got k to be 13....
and 7a i got the distance to be 1703 m and the bearing to be like 182 degrees or something like that..
 

Constip8edSkunk

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hahaha

for q4 c), i had a*(1/a)*b = -6/2 .'. b = -2 :rolleyes: ...2 marks down the drain....:(

but yeah b=-3 --> 2*-27 + 9+3k+6=0
.'.3k = 39 --> k=13
 

iambored

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Originally posted by Alex222
Question 1 -
c) 1.5
d
sis u get that with 3/2 x/sinxcosx
then transfering it to x/sinx * 1/cosx
which then gave 3/2 *1*1?
 

iambored

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Originally posted by iambored
sis u get that with 3/2 x/sinxcosx
then transfering it to x/sinx * 1/cosx
which then gave 3/2 *1*1?
i got the same answer as you, but i am wondering if my working was right

so what i did was changed the bottom to 2sinxcosx
then i took out the 2 (and the 3 on the top)
and then i changed the x/sinxcosx to x/sinx multiplied by 1/cosx
and therefore, x/sinx = 1, and 1/cosx =1
so i was left with 2/3 as my answer

i can't remember what i got for the others without looking at the paper properly, but i know i got the same for 1b)
 
Last edited:
N

ND

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It was lim(3x/sin2x), so all you have to do is take 3/2 outside, i.e. lim((3/2(2x/sin2x)), and remember that lim(ax/sinax)=1, so lim(3/2(2x/sin2x))=3/2.

note: i wrote just 'lim' instead of 'lim x->0'.
 

smeyo

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the idea of that question is as x approaches 0, sinx approaches x therefore 3/2*2x/sin2x will give 3/2*2x/2x = 3/2*1 = 3/2
 

iambored

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Originally posted by smeyo
the idea of that question is as x approaches 0, sinx approaches x therefore 3/2*2x/sin2x will give 3/2*2x/2x = 3/2*1 = 3/2
!! and u know, i looked at that question to do it like that for so long. well mine still worked, in a way, i hope i get the marks
 

braindrainedAsh

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Yay I got the Newton's Law one right and my 4 unit friend didn't get it! I feel smart-ish!
 

deyveed

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What did you get for the Newton's method one?
I had:

f(x) = sinx - 2x/3
f(1.5) = -0.974
f'(x) = cosx - 2/3
f'(1.5) = 0.333

x_2 = 1.5 + 0.974/0.333
x_2 = 4.4

How is that close to 1.5?
 

walla

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yeah i got something really close to 1.5 for newton's
 

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