HSC 2012 Maths Marathon (archive) (3 Viewers)

[LoveHateSchool]

I apologize if there is one-I couldn't see it!

To get the ball rolling...

A ball is dropped from a height of 2.4m and bounces back up to a height of 1.8m. It then falls and bounces back up to a height of 1.35m. Everytime the ball bounces back to a height three quarters of previous time. How far will the ball travel before rest?

 

[LoveHateSchool]

Re: 2012 HSC Mathematics Marathon!

^Haha troll, angry because I disagreed with your troll post?

I wish you luck in English, you insult is pretty fractured imho And don't worry, even though I live more than 20 km from UNSW, I'll still do better than you

 

[bleakarcher]

Re: 2012 HSC Mathematics Marathon!

I apologize if there is one-I couldn't see it!

To get the ball rolling...

A ball is dropped from a height of 2.4m and bounces back up to a height of 1.8m. It then falls and bounces back up to a height of 1.35m. Everytime the ball bounces back to a height three quarters of previous time. How far will the ball travel before rest?

Let the total distance covered by the ball till rest be d.
d=2(2.4+0.75*2.4+0.75*2.4*0.75+...)=2[2.4/(1-0.75)]=4.8/0.25=4.8*4=19.2m
Hence, d=19.2m

 

[LoveHateSchool]

Re: 2012 HSC Mathematics Marathon!

^Yay good job, and when you answer a Q, you get to post the next one for people to have a go at

 

[nightweaver066]

Re: 2012 HSC Mathematics Marathon!

 

[bleakarcher]

Re: 2012 HSC Mathematics Marathon!

This thread just decided to die.

i) angle QAC=angle BCA (alternate angles, AQ//CB)
angle BAC=angle BCA (base angles of isos. triangle ABC)
:. angle QAC=angle BAC
Hence, AC bisects angle QAB.
ii) Let angle BAD=x
angle BAC=90-x (adj. comp. angles)
angle BCA=angle BAC=90-x (base angles of isos. triangle ABC)
angle QAC=angle BCA=90-x (alternate angles, QA//CB)
:.angle DAP=180-90-(90-x)=180-90-90+x=x
Hence, AD bisects angle PAB

Sorry for bad format.

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

Just need to verify something, for a question like this:

5x/2x-1>3
5x(2x-1)>3(2x-1)^2
10x^2-5x>12x^2-12x+3
2x^2-7x+3<0
(2x-1)(x-3)<0

Do you use the original inequation to solve the answer, making it end as x<1/2, x>3

OR

Do you use the last step to determine the answer, making it 1/2<x<3?

Having a mind blank and really want to find out.

 

[SpiralFlex]

Re: 2012 HSC Mathematics Marathon!

Just need to verify something, for a question like this:

5x/2x-1>3
5x(2x-1)>3(2x-1)^2
10x^2-5x>12x^2-12x+3
2x^2-7x+3<0
(2x-1)(x-3)<0

Do you use the original inequation to solve the answer, making it end as x<1/2, x>3

OR

Do you use the last step to determine the answer, making it 1/2<x<3?

Having a mind blank and really want to find out.

I don't think you understand what you are actually doing when you solve the inequality. Also, I don't understand what you are asking.

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

I don't think you understand what you are actually doing when you solve the inequality. Also, I don't understand what you are asking.

If your talking about the step skips I skipped steps because I couldn't be bothered to type the whole thing, if it's something else I'm doing terribly wrong please tell me

At school, we were taught that if the equation was lower than the other side, it would be -a<x<a
Though if the LHS is bigger than the RHS then it is x>a or x<-a

To use what I got taught, do you use the last step (2x-1)(x-3)<0 which is smaller than 0 meaning it would be 1/2<x<3

OR

Do we use the original equation, 5x/2x-1>3, which is bigger than 3, to find x which means the answer would be x<1/2, x>3

 

[nightweaver066]

Re: 2012 HSC Mathematics Marathon!

Do we use the original equation, 5x/2x-1>3, which is bigger than 3, to find x which means the answer would be x<1/2, x>3

lol what..

This parts incorrect. If you sub in x = 0, doesn't LHS become 0 which is < 3?

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

If that doesn't make sense is the answer 1/2 < x< 3 or x < 1/2 ,x > 3?

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

lol what..

This parts incorrect. If you sub in x = 0, doesn't LHS become 0 which is < 3?

Ok than I guess its the other option.

 

[nightweaver066]

Re: 2012 HSC Mathematics Marathon!

If that doesn't make sense is the answer 1/2 < x< 3 or x < 1/2 ,x > 3?

x = 0 lies in x < 1/2, x > 3 and it doesn't hold for the original inequation.

If you test x = 2, you'll end up with 10/3 > 3 which is true so the answer is that domain, i.e. 1/2 < x < 3.

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

Probably explained it so retardedly no one actually understood what I was asking lol. Anyway, thanks night.

 

[nightweaver066]

Re: 2012 HSC Mathematics Marathon!

 

[Timske]

Re: 2012 HSC Mathematics Marathon!

i) 2a^2 - 7a + 3 = (2a - 1)(a - 3)

ii) 2(log2(x))^2 - 7(log2(x)) + 3 = 0

Let log2(x) = a
i.e 2a^2 - 7a + 3 = 0
(2a - 1)(a - 3) = 0
a = 1/2: a^2 = 1/4 and a = 3
(log2(x))^2 = 1/4 and log2(x) = 3

:. x = root2 and x = 8

 

[Timske]

Re: 2012 HSC Mathematics Marathon!

Consider the equation x^2 + (k - 4)x + 9 = 0

For what values of k does the equation have:
i) equal roots;
ii) distinct real roots?

 

[carpe_diem]

Re: 2012 HSC Mathematics Marathon!

Consider the equation x^2 + (k - 4)x + 9 = 0

For what values of k does the equation have:
i) equal roots;
ii) distinct real roots?

Not sure for ii) BUT for i)

equal roots when discriminant = 0
b^2-4ac = 0
(k-4)^2 - 4x1x9 = 0
k^2 - 8k + 16 - 36 = 0
k^2 - 8k - 20 = 0
(k-10)(k+2) = 0
Therefore k=10 or k=-2

If someone can do part ii) that would be appreciated!

 

[Timske]

Re: 2012 HSC Mathematics Marathon!

If someone can do part ii) that would be appreciated!

Hint: distinct real roots; discriminant > 0

 

[Examine]

Re: 2012 HSC Mathematics Marathon!

Factor as fully as possible
a^3+b^3+a+b