HSC 2013 MX2 Marathon (archive) (16 Viewers)

seanieg89 · Oct 18, 2013

Re: HSC 2013 4U Marathon

Harder 3U (Combinatorics):

In the board game Risk, if there are at least three attacking units and at least two defending units, then the attacker rolls three dice and the defender rolls two dice. The top two dice of the attacker are paired up with the two dice of the defender, and ties are awarded to the defender.

Eg if the attacker rolls 3,5,4 and the defender rolls 4,4 then each side loses one unit (as the attackers 5 beats the defenders 4 and the attackers 4 loses to the defenders other 4.)

Find the probability of:

i) Attacker loses two units.

ii) Both sides lose one unit.

iii) Defender loses two units.

seanieg89 · Oct 18, 2013

Re: HSC 2013 4U Marathon

Volumes/Integration:

$Consider the surface $S$ above the $xy$-plane with height given by $h(x,y)=e^{-(x^2+y^2)}$.\\ \\ i) Find the volume $V_1$ of the region bounded by the $xy$-plane, the surface $S$, and the infinitely long rectangular prism $\max\{|x|,|y|\}=R$ for some fixed $R>0$ by taking suitable slices.\\ \\ ii) Find the volume $V_2$ of the region bounded by the $xy$-plane, the surface $S$, and the infinitely long cylinder $x^2+y^2=R^2$ by cylindrical shells.\\ \\iii) Hence compute $\lim_{R\rightarrow\infty}\int_{-R}^R e^{-t^2}\, dt.$$

spyris · Oct 18, 2013

Re: HSC 2013 4U Marathon

Hi, I have a problem with a question, its question 1996 3a the volume, here's the link:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/96MAT4U.PDF

I have no idea why the outer radius of the annulus is 4-x rather than 4+x

alexandred · Oct 18, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Volumes/Integration:

$Consider the surface $S$ above the $xy$-plane with height given by $h(x,y)=e^{-(x^2+y^2)}$.\\ \\ i) Find the volume $V_1$ of the region bounded by the $xy$-plane, the surface $S$, and the infinitely long rectangular prism $\max\{|x|,|y|\}=R$ for some fixed $R>0$ by taking suitable slices.\\ \\ ii) Find the volume $V_2$ of the region bounded by the $xy$-plane, the surface $S$, and the infinitely long cylinder $x^2+y^2=R^2$ by cylindrical shells.\\ \\iii) Hence compute $\lim_{R\rightarrow\infty}\int_{-R}^R e^{-t^2}\, dt.$$

(i) Slice is perpendicular to xy plane and x-axis, gives integral of e^-(x^2+y^2) from -R to R wrt y; integrating a second time wrt to x over same interval gives a total volume of (integral of e^-(t^2)dt from -R to R)^2 since we can separate x and y and change dummy variable
(ii) Volume is integral from 0 to R of 2pi*t*e^-(t^2)dt which gives pi(1-e^-(R^2))
(iii) as r --> infinity, v1 --> v2 and v2 --> pi so the limiting value of the integral is sqrt(pi)?
Sorry for no LaTex, on iPad.

RealiseNothing · Oct 18, 2013

Re: HSC 2013 4U Marathon

spyris said:
Hi, I have a problem with a question, its question 1996 3a the volume, here's the link:

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/96MAT4U.PDF

I have no idea why the outer radius of the annulus is 4-x rather than 4+x

Because $x<0$ so the negatives will need to cancel out.

Sy123 · Oct 18, 2013

Re: HSC 2013 4U Marathon

$$Let$ \ \ \lfloor z \rfloor \ \ $be the greatest integer less than or equal to$ \ z$

$$That is$ \ \ \lfloor \pi \rfloor = 3$

$Prove that$

$\int_1^a \lfloor x \rfloor f'(x) \ dx = \lfloor a \rfloor f(a) - (f(1) + f(2) + \dots + f(\lfloor a \rfloor) )$

alexandred · Oct 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$$Let$ \ \ \lfloor z \rfloor \ \ $be the greatest integer less than or equal to$ \ z$

$$That is$ \ \ \lfloor \pi \rfloor = 3$

$Prove that$

$\int_1^a \lfloor x \rfloor f'(x) \ dx = \lfloor a \rfloor f(a) - (f(1) + f(2) + \dots + f(\lfloor a \rfloor) )$

Split into separate integrals over each interval from n to n+1, last is from [a] (where [x] is floor fn) to a, all the [n]s can be treated as constants over those intervals so so we have (sum from 1 to ([a]-1) of (n(f(n+1)-f(n))) + [a](f(a)-f([a]-1)) which telescopes to give the result. Again sorry for poor format.

hit patel · Oct 18, 2013

Re: HSC 2013 4U Marathon

Hope the 2014 one comes soon for both 3U and 4U

Sy123 · Oct 18, 2013

Re: HSC 2013 4U Marathon

Why don't we just add 2014 posts on here

I think its better to have a megathread rather than individual threads for each year

Can someone post a question

hit patel · Oct 18, 2013

Re: HSC 2013 4U Marathon

WEll stuff u learnt I havent learnt a lot of it... I am a newbie

. Will get crushed on this thread.

ForOneEon · Oct 18, 2013

Re: HSC 2013 4U Marathon

'Kay, here's a volumes question!

A glass vase is made by rotating the region between the curves y = x^0.5, (0 < x < 9)and y = 0.75 (x-1)^0.5, (1 < x < 9) (They're actually equal to 0-9 and 1-9 respectively)

a) Find the area by the Washer method, hence find the volume.

b) Rotate the graphs so that the x and y values are interchanged. Find the volume by cylindrical shells.

ForOneEon · Oct 18, 2013

Re: HSC 2013 4U Marathon

God, I need to learn how to use Latex. And find a good graphing tool.

hit patel · Oct 18, 2013

Re: HSC 2013 4U Marathon

But from the stuff I know here:

a) If s is a non real 7th root of 1 of smallest positive argument, show s= cis (2pi/7) 1

b) Show that 1+s + s^2 + s^3 + s^4 + s^5 + s^6=0 1

c) Find value of s + 1/s , hence find exact value of cos(pi/7) x cos (2pi/7) x cos(3pi/7) 3

d) Known that z1= s+ s^2 + s^4 and z2= s^3+s^5+s^6 are roots of a quadratic equation. Find the equation 3

e) State Exact Value of sin 2pi/7 + sin 3pi/7 - sin pi/7 4

Mark: /12

Yep thats it sorry for the no latex question. Marks are awarded By Sy123/ HeroicPanda. Depending on Who is online and who answers it.

Sy123 · Oct 18, 2013

Re: HSC 2013 4U Marathon

$Let$ \ \ f(x) \ \ $be a differentiable function that satisfies$

$f(x+y) = f(x) f(y) \ \ $for all$ \ \ x,y \in \mathbb{R}$

$f'(0) = 1$

$$Find$ \ \ f(x)$

alexandred · Oct 18, 2013

Re: HSC 2013 4U Marathon

Sy123 said:
$Let$ \ \ f(x) \ \ $be a differentiable function that satisfies$

$f(x+y) = f(x) f(y) \ \ $for all$ \ \ x,y \in \mathbb{R}$

$f'(0) = 1$

$$Find$ \ \ f(x)$

Differentiating by first principles, use l'hopitals rule and f'(0)=1, integrate to get e^x

seanieg89 · Oct 18, 2013

Re: HSC 2013 4U Marathon

alexandred said:
(i) Slice is perpendicular to xy plane and x-axis, gives integral of e^-(x^2+y^2) from -R to R wrt y; integrating a second time wrt to x over same interval gives a total volume of (integral of e^-(t^2)dt from -R to R)^2 since we can separate x and y and change dummy variable
(ii) Volume is integral from 0 to R of 2pi*t*e^-(t^2)dt which gives pi(1-e^-(R^2))
(iii) as r --> infinity, v1 --> v2 and v2 --> pi so the limiting value of the integral is sqrt(pi)?
Sorry for no LaTex, on iPad.

Your answer to iii) is correct, but I was looking for some justification as to why "v1->v2".

alexandred · Oct 18, 2013

Re: HSC 2013 4U Marathon

seanieg89 said:
Your answer to iii) is correct, but I was looking for some justification as to why "v1->v2".

Don't know how much justification is required, or if this is legit, but: 0 < v1-v2 < limit as R' goes to infinity of integral of 2pi*t*e^-t^2 from R to R' = pi*e^-R^2
So as R goes to infinity v1-v2 goes to 0.

ForOneEon · Oct 18, 2013

Re: HSC 2013 4U Marathon

hit patel said:
But from the stuff I know here:

a) If s is a non real 7th root of 1 of smallest positive argument, show s= cis (2pi/7) 1

b) Show that 1+s + s^2 + s^3 + s^4 + s^5 + s^6=0 1

c) Find value of s + 1/s , hence find exact value of cos(pi/7) x cos (2pi/7) x cos(3pi/7) 3

d) Known that z1= s+ s^2 + s^4 and z2= s^3+s^5+s^6 are roots of a quadratic equation. Find the equation 3

e) State Exact Value of sin 2pi/7 + sin 3pi/7 - sin pi/7 4

Mark: /12

Yep thats it sorry for the no latex question. Marks are awarded By Sy123/ HeroicPanda. Depending on Who is online and who answers it.

Well, here goes! Still working on (e).

a) s^7 = 1
= cis (0 + 2k[pi]), k = 0,1,2,3
s = cis (2[pi]/7) [smallest argument]

b) s^7 - 1 = 0
(s - 1)(1 + s + s^2 + s^3 + s^4 + s^5 + s^6) = 0
but s =/= 1
therefore (1 + s + s^2 + s^3 + s^4 + s^5 + s^6) = 0

c) s + 1/s = cis (2[pi]/7) + cis (-2[pi]/7)
= cos (2[pi]/7) + i sin (2[pi]/7) + cos (2[pi]/7) - i sin (2[pi]/7) [cos even, sin odd]
= 2 cos (2[pi]/7)

( s + 1/s )( s^2 + 1/s^2 )( s^3 + 1/s^3 ) = 8 cos(2pi/7) x cos (4pi/7) x cos(6pi/7)
= 1
cos(2pi/7) x cos (4pi/7) x cos(6pi/7) = 1/8
= cos(2pi/7) x cos (-3pi/7) x cos(-pi/7)
= cos(2pi/7) x cos (3pi/7) x cos(pi/7) [cos even]

d) z1= s + s^2 + s^4 and z2= s^3 + s^5 + s^6

Sum of roots = z1+ z2
= s^3 + s^5 + s^6 + s + s^2 + s^4
= -1

Product of Roots = z1 x z2
= ( s + s^2 + s^4 )( s^3 + s^5 + s^6 )
= s^4 + s^6 + s^7 + s^5 + s^7 + s^8 + s^7 + s^9 + s^10
= 3 + s^-3 + s^-1 + s^-2 + s + s^2 + s^3 [ divide by s^7 ]
= 3 + ( s + 1/s ) + ( s^2 + 1/s^2 ) + ( s^3 + 1/s^3 )
= 3 + 2 {cos(2pi/7) + cos (4pi/7) + cos(6pi/7)}

but
2cos(2pi/7) + 2cos (4pi/7) + 2cos(6pi/7) + 1 = 0
therefore
cos(2pi/7) + cos (4pi/7) + cos(6pi/7) = -1/2

so
= 2
thus, equation is X^2 - X + 2 = 0

e) Still working on it.

Still can't lay stuff out properly.

gustavo28 · Oct 18, 2013

Re: HSC 2013 4U Marathon

psychotropic said:
Given four concyclic points. For each subset of three points take the incenter. Show that the four incenters from a rectangle.

Let the four points be A,B,C,D. Let the incentre of ABC be I and the incentre of DBC be J. Let the midpoint of the arc BC be M.
Sketch: 1) Prove IM=CM=BM and similarly JM=CM=BM.
2) Hence BCIJ is cyclic
3) Then fill some angles in to get the right angles

gustavo28 · Oct 18, 2013

Re: HSC 2013 4U Marathon

Prove that for all non-negative a,b,c

a(a-b)(a-c)+b(b-a)(b-c)+c(c-a)(c-b) >= 0

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