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HSC 2013-14 MX1 Marathon (archive) (1 Viewer)

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Sy123

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Sy123

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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

Let us factorise the x out and notice a perfect square
= ∫√x √sin^4 x - 2sin^2 x + 1 .dx
= ∫√x √(sin^2 x -1)^2
= ∫√x √(-cos^2 x)^2
= ∫√x |cos^2 x|.dx

There are no limits/borders, hence we ignore the absolute sign

∫√x cos^2 x.dx

Is this possible?
 
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Sy123

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Re: HSC 2013 3U Marathon Thread

Let us factorise the x out and notice a perfect square
= ∫√x √sin^4 x - 2sin^2 x + 1 .dx
= ∫√x √(sin^2 x -1)^2
= ∫√x √(-cos^2 x)^2
= ∫√x |cos^2 x|.dx

There are no limits/borders, hence we ignore the absolute sign

∫√x cos^2 x.dx

Is this possible?
Lol don't worry about that question, I stuffed it up.
 
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Re: HSC 2013 3U Marathon Thread

In a simple harmonic motion(SHM), acceleration is defined as x(dot dot) = - n^2 x

Ur equation has no negative sign, hence its not a SHM
Oh shet. I fixed that up afterward but ceebs to edit since no one did it. thanks for the note.
 

omgiloverice

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Re: HSC 2013 3U Marathon Thread



I'll do part b later when I have time to teach myself SHM
 

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HeroicPandas

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Re: HSC 2013 3U Marathon Thread

Change cotθ into cosθ/sinθ, it'll make everything neater! ^^
 

study1234

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Re: HSC 2013 3U Marathon Thread

ABCD is a cyclic quadilateral in which the opposite sides and DC are equal. Prove that the diagonals AC and BD are equal.
 

Sy123

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shongaponga

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Re: HSC 2013 3U Marathon Thread

Well done, was quiet a straight forward question i must admit and the only reason i posted it was because i was hoping someone would post a non-calculus solution. It can be done very elegantly as follows:

Note that the shortest distance between 2 points is a straight line.

Reflecting Island 1 as shown yields 2 triangles that are similar.

diagram.png

Therefore, using properties of similar triangles:

2/(6-x) = 1/x => 2x = 6-x => x = 2. And this must be the smallest value for x due to the above note. No calculus required!
 

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Makematics

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Re: HSC 2013 3U Marathon Thread

Definately a 3U question:


(a) P'(x)=5x^4 -5c
To find any stationary points, solve P'(x)=0
5x^4 -5c=0
x^4 =c
x = ±∜c
But since c<0, ∜c is undefined and hence there are no stationary points.
P(0)=0-0+1
= 1
Also since c<0, we notice that P'(x)>0 for all real x, and hence P(x) is monotonic increasing.
By sketching a possible graph of P(x), we can clearly see that P(x) will have only one root, and that this root will be negative.

(b) The abscissae of the two stationary points are ∜c and -∜c.
P(∜c)=(∜c)^5 -5c(∜c) + 1
=c^(5/4) -5c^(5/4) + 1
= 1 - 4c^(5/4)

P(-∜c)=(-∜c)^5 -5c(-∜c) + 1
=-c^(5/4) +5c^(5/4) + 1
= 1 + 4c^(5/4)

By considering a graph of P(x), we can tell that there will be 3 distinct real roots if the product of the y coordinates is negative. i.e. one turning point is above the x-axis, one turning point is below the x-axis.

∴P(x) has three distinct real roots iff P(∜c).P(-∜c) < 0
P(∜c).P(-∜c)= [1 - 4c^(5/4)].[1 + 4c^(5/4)]
= 1 - 16c^(5/2)
= 1 - [4c^(5/4)]^2
< 0
1 - [4c^(5/4)]^2 < 0
[4c^(5/4)]^2 > 1
4c^(5/4) > 1
c^(5/4) > 1/4
c^5 > (1/4) ^4
c > (1/4) ^(4/5)
∴ P(x) has three distinct roots if and only if c > (1/4) ^(4/5).

fuuuu someone teach me how to use latex...
 

HeroicPandas

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Re: HSC 2013 3U Marathon Thread

(a) P'(x)=5x^4 -5c
To find any stationary points, solve P'(x)=0
5x^4 -5c=0
x^4 =c
x = ±∜c
But since c<0, ∜c is undefined and hence there are no stationary points.
P(0)=0-0+1
= 1
Also since c<0, we notice that P'(x)>0 for all real x, and hence P(x) is monotonic increasing.
By sketching a possible graph of P(x), we can clearly see that P(x) will have only one root, and that this root will be negative.

(b) The abscissae of the two stationary points are ∜c and -∜c.
P(∜c)=(∜c)^5 -5c(∜c) + 1
=c^(5/4) -5c^(5/4) + 1
= 1 - 4c^(5/4)

P(-∜c)=(-∜c)^5 -5c(-∜c) + 1
=-c^(5/4) +5c^(5/4) + 1
= 1 + 4c^(5/4)

By considering a graph of P(x), we can tell that there will be 3 distinct real roots if the product of the y coordinates is negative. i.e. one turning point is above the x-axis, one turning point is below the x-axis.

∴P(x) has three distinct real roots iff P(∜c).P(-∜c) < 0
P(∜c).P(-∜c)= [1 - 4c^(5/4)].[1 + 4c^(5/4)]
= 1 - 16c^(5/2)
= 1 - [4c^(5/4)]^2
< 0
1 - [4c^(5/4)]^2 < 0
[4c^(5/4)]^2 > 1
4c^(5/4) > 1
c^(5/4) > 1/4
c^5 > (1/4) ^4
c > (1/4) ^(4/5)
∴ P(x) has three distinct roots if and only if c > (1/4) ^(4/5).

fuuuu someone teach me how to use latex...
Very nice! Excellently done, same method as mine

 

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Sy123

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RealiseNothing

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Re: HSC 2013 3U Marathon Thread





Split the integral into:



Now in each of these intervals, is 0, 1, 2, ..., n respectively for each term. I.e. the first term is integral , the second integral up to the last integral where

Hence we arrive at:



Which upon evaluation becomes:



Note the last term is divided by 2 as the interval we are integrating over is halved. Summing this gives:







#
 
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