Completed Solutions (1 Viewer)

[tim_dawborn]

Total SOLUTIONS FOR 2005 HSC PAPER

this is my solutions - possibly some mistakes - the corrections are in trhe latter end of the other solutions thread

plz post some feedback and comments on your thorts and results of the paper

 

[EvilDude]

3aii - it is between 0.7 and 0.75 therefore to 1dp is just 0.7
dii - you want the smallest l for x is still real. That is b^2 - 4ac > 0
so b^2 > 4ac
but we want smallest b^2, occurs when b^2 = 4ac
l^2 = 4(12)
l = root(48) = 4 root(3)
7bii
I did this graphically.
Drew a general cubic (with A>0) and then you can see that if the second stationary point is above the y axis, then the cubic will only have one real root.

From here it's some working based on proving that the stationary point (x = root(3)/3) lies above y axis for A<3root3/2

Thanks for solutions, helped me find a few of my stupid mistakes, so now no high expectations.

 

[HSCExamMarker]

My answers are all same except for the question 7(a)
coz i could not do it..

omg althought i am first in 4unit and got like 92in trials...suddenly my mind was blocked at that question...I feel so bad that i missed easy stuff..

otherwise, most of them are same and the one with x^2-lx+12=0
u simply differentiate to find a minimum point then sub in
although i made STUPID arithmetic mistake in this

I am not sure about that Q7(b) too..coz i could not figure out..
I really liked to hav hard but straightfoward questions..oh well..not happened.

Any find difficulty with 7(a) ? or it's only me -_- lol

 

[HSCExamMarker]

By the way for Q7a. ii) i just wrote down
dr/da = da/dt * dr/da
and therefore dr/da = 0.02 * dr/da
but not caculated coz i could not do (i)
would i get ANY mark for writing this down??

 

[EvilDude]

By the way for Q7a. ii) i just wrote down
dr/da = da/dt * dr/da
and therefore dr/da = 0.02 * dr/da
but not caculated coz i could not do (i)
would i get ANY mark for writing this down??

I think you'd need more work for a mark, as it's only worth 2 and one mark might be for answer while the other for the differentiation etc.

 

[justchillin]

yes u will...

 

[tim_dawborn]

yea i agree - prob need some more for a mark

 

[haboozin]

Q7b ii

0 < A(1/3^(3/2)) - A/3^(1/2) + 1
A(1/3^(1/2) - 1/3^(3/2) < 1
A((3 - 1)/3^3/2) < 1
A(2/3^(3/2)) < 1
A < 3^(3/2)/2
A < 3.3^(1/2) /2


theres actually 2 ways to do this to get u the answer!!
either minimum value is bigger than or max is less than
(i did the max is less than version in the above working)

wow if onyl that question was worth more marks...

why couldnt anyone picture this?
if the minimum is above the root then there will only be one real root..

 

[datacore]

for the volumes question i think i had something like this


...

= pi/2 (pi/8 - 1/4) <-- which is the correct answer
but then i decided to simplify it

= pi^2 / 18 - pi/8

so do you think i would lose a mark

 

[EvilDude]

I think I did the same, just worked in a different direction. I used a substitution in A to show it had to be minimum value instead of proving it like that though.

 

[gemblack88]

By the way for Q7a. ii) i just wrote down
dr/da = da/dt * dr/da
and therefore dr/da = 0.02 * dr/da
but not caculated coz i could not do (i)
would i get ANY mark for writing this down??

yeh thats exactly what i did for that question. i reckon there's a chance we could get a mark for that...

 

[tez0r]

hmmm..but don't they take all of your working into consideration?

 

[mnanda88]

hey it says ur file is like zippd

ahh i want da answers ... i culdnt get in2 ur file phoenix88 ...
yay no more maths ... awww i failed

 

[oneeye]

Q7b ii

0 < A(1/3^(3/2)) - A/3^(1/2) + 1
A(1/3^(1/2) - 1/3^(3/2) < 1
A((3 - 1)/3^3/2) < 1
A(2/3^(3/2)) < 1
A < 3^(3/2)/2
A < 3.3^(1/2) /2


theres actually 2 ways to do this to get u the answer!!
either minimum value is bigger than or max is less than
(i did the max is less than version in the above working)

wow if onyl that question was worth more marks...

why couldnt anyone picture this?
if the minimum is above the root then there will only be one real root..

What about
0 < A(-(1/3^(3/2))) - A/-(3^(1/2)) + 1
Giving
A >- 3.3^(1/2) /2
and when A=0 there are on zeros as f(x) =1.
Therefore one zero only when 0< absolute A <3.3^(1/2) /2.

Also, shouldn't the question have said one REAL zero?

 

[Slidey]

Observe that a cubic has only one root when the local minimum occurs above y=0 (that is both stat points occur above y=0), or when the local maximum occurs below y=0.

f(sqrt3/3)= Asqrt3/9 – Asqrt3/3 + 1 = 1 - 2Asqrt3/9
f(-sqrt3/3)= 1 + 2Asqrt3/9

f(x) is cubic and leading coefficient is positive, thus f(-sqrt3/3) is the max and f(sqrt3/3) is the min.

Now from the observation above:
1 + 2Asqrt3/9 < 0 or
1 - 2Asqrt3/9 > 0
That is:
A < -3sqrt3/2 or
A < 3sqrt3/2
Since A>0 (given), it must be 0 < A < 3sqrt3/2. #

 

[haboozin]

Also, shouldn't the question have said one REAL zero?

no because the 3unit syllabus only deals with real numbers.
So to them complex numbers are non existant and they wouldnt confuse students because of that.

if this was a 4unit exam then it would defenatly say REAL.

 

[haboozin]

also for Q6 ii - it should be tan ( 180 - @) not 360 - @... it gives the right answer but the way gone about it is wrong

 

[tim_dawborn]

also for Q6 ii - it should be tan ( 180 - @) not 360 - @... it gives the right answer but the way gone about it is wrong

u sure ?? if it was tan(180 - @) that would imply that the particle was moving upwards and to the left; whereas teh particle is moving downwards and to the right

 

[haboozin]

u sure ?? if it was tan(180 - @) that would imply that the particle was moving upwards and to the left; whereas teh particle is moving downwards and to the right

but does direction matter?
gradient is done by the angle the line makes with the horisontal...

can either be obtuse or accute... in this case its obtuse.

so the angle is 180 - @ which give u the obtuse angle the line makes to the horisontal.

 

[justchillin]

You just make it accute with the verticle velocity negative...