my solutions, all questions (2 Viewers)

[shazabdazla]

i got pretty much the same as failing the hsc...except:

for question 1:

part (c): 95 degrees (nearest degree)

part (e) y = x (cubed) - 8x + c

For question 3:

part (a) (ii): y' = x (x cos x + 2 sin x)

For question 4:

part (a) (ii): i put 64 degrees (n. degree), and not in bearing format, just as a degree, dyou think ill get any marks for that n how much

For question 5:

part (a) (iv): concave down for 0 < x < 1 (there is another inflexion at (1, -3))

feckit ill check the rest later...you seem to have done really well!!!!

(i left out about 20 marks!!)

 

[eeyore]

did you go through the paper and do it all again?! all i have to say is... YOU'RE CRAZY!!

 

[True_Monolith07]

he appears to have the same answers as me most of the time....(theres 2 different which i know i got wrong)

 

[Beaky]

Originally posted by shazabdazla

For question 5:
part (a) (iv): concave down for 0 < x < 1 (there is another inflexion at (1, -3))

Yer I got this... I cant see the point of inflexion in theory
12x^2-12x = 0 seems to me like (1,-3)

But when I did it on the computer graph thingo... it reckons point of inflexion is (2,y) can someone show this to me??

 

[SoCal]

Originally posted by SmokedSalmon
Make sinx = X

therefore 2X - 3X - 2 = 0
(2X + 1) (X - 2)
X = 2 or -1/2

Sinx = 2 or -1/2
However sinx= 2 cannot work x must be between -1 < x< 1
therefore
Sinx = -1/2
x = 7pi/6, 11pi/6

I can see that now, thanks. I tried to do it having:

(sinx - 2) (2sinx + 1)

so sinx - 2 = 0 or 2sinx + 1 =0
sinx = 2 or sinx= -1/2

Then I just left it like that because I was hurrying to finish.

 

[Styled n Slick]

Youre all crazy, just get ur marks. We fukt up (well some of us), stop braggin about what u got right and wrong and just remember 2 burn your books, simple! all i can say is i put up with a load of shit in that test and i know that my mark is in the 50%, so if u got 90+, stop ur whinging and think of us!

 

[SoCal]

Originally posted by failingTheHsc
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

Also, where is pi coming from in your solutions failingTheHsc?

 

[failingTheHsc]

wat do u mean?

i used radians in calculating the angles. but technically you should get pi when using degrees aswell???

 

[Almost There]

styled and slick-yep i agree. ppl the exam is over...u never have to do taht paper again...so let it go...who gives a fuck!! i have failed so bring it on!!

 

[SoCal]

Originally posted by failingTheHsc
wat do u mean?

i used radians in calculating the angles. but technically you should get pi when using degrees aswell???

Well, obviously I have no idea what this question was going on about so don't worry.

 

[Russian]

did anyone know what aCartesian Plane was, i overlooked that question but i dont think i could have done it anyway

i think i got 65% or somthing areound that whaich is pretty good for me since im hovering around the middle of 2u rankings of maths in my year, tho ppl in our year r freaks, i hope i get a band 5

 

[failingTheHsc]

cartesian plane is just a graph

 

[Ragerunner]

for the stationary point question, you sure it was a inflexion?

I tested below 0 and above zero and i think i got a maximum turning point -_-

 

[shazabdazla]

hey beaky...i think i worked out where i went wrong...for q.5.a.iv

ok

y= x (to 4th) - 4 x (cubed)

y' = 4 x (cubed) - 12 x (square)

y'' = 12 x (square) - 36 x

= 12 x (x - 2)

so x = 0, 2....

DAMITT!!!!!!!!!!!!!!!!!!!!!!!!!!!

 

[veg]

nah, it's an inflexion point. when you test with the derivative, you get negative answers for both sides... and when you find the second derivative and make that equal zero, you get 0 as apoint.. so it's a stationary point of inflexion.

get what i mean?

 

[t-i-m-m-y]

btw that one about the volume (definite integral) without evaulating it...
i wrote the lny one.. but seeing as 2unit does not cover that

we can alternatively evaluate the area under the curve dx ways... so u do (5-whatever the curve was)^2 and times by pi to get the volume.. if u get what i mea

thats what i wrote-- well actually i wrote both

btw it was worth three marks.. so yeah

 

[Left-ism]

7 b) iv)
how did u find the distance..integrate the v? get displacement and then distance between two?...cos i dont think thats right...because the curve changes direction...not sure what i put, you maybe right. jus wanna clarify
thanks

 

[True_Monolith07]

yep.... i did 7 b) iv) and i got 2m

 

[failingTheHsc]

Originally posted by Left-ism
7 b) iv)
how did u find the distance..integrate the v? get displacement and then distance between two?...cos i dont think thats right...because the curve changes direction...not sure what i put, you maybe right. jus wanna clarify
thanks

the graph is of velocity so the distance is the area under the graph.

when you draw the graph in iii) a bit of the graph is under the x axis so i integrated in 2 parts and added them together.

 

[Left-ism]

ok good..thx people