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

asianese · Apr 5, 2013

Re: HSC 2013 3U Marathon Thread

$$a) A circle has the equation $ x^2+y^2=1. $ A tangent is drawn from the point $P(\cos\theta,\sin\theta) $ where $\theta $ is the angle between the positive $x$-axis and the radius that meets the circle at $P.\\\\ $i) By considering $ y $ as a positive function of $x$, find the equation of the tangent at $P. \hfill{\fbox{3}}\\\\ $ii) Suppose another tangent is drawn from $ Q(\cos\theta,-\sin\theta). $ Write down the equation of this tangent and hence find the point of intersection, $T$, of the two tangents. $ \hfill{\fbox{2}}\\\\ $iii) Find the angle $ \theta $ such that $ T $ has co-ordinates $(2,0). \hfill{\fbox{1}}\\\\$

$$b) Now suppose that $ T $ moves in simple harmonic motion along the $x$-axis with its acceleration given by $ \ddot{x}=4x.\\\\ $i) Find its velocity in terms of its distance from the origin.$ \hfill{\fbox{3}}\\\\ $ii) By considering the velocity equation and the construction of $T$, find the restriction of the position of $T. \hfill{\fbox{2}} \\\\ $iii) By considering the graph of $f(\alpha)=\frac{1}{\cos\alpha}$ or otherwise, deduce the possible values of $\theta.\hfill{\fbox{2}}\\\\$

$$c) Suppose now that $T$ is free to move along $x\geq1. $ What happens to the angle $ \theta $ as $T$ approaches infinity? Comment on the physical restriction of $\theta$. \hfill{\fbox{2}}$

Sy123 · Apr 14, 2013

Re: HSC 2013 3U Marathon Thread

$$A triangle has side lengths$ \ \ a, b, c$

$By considering angles of the triangle or otherwise, prove that, the area of the triangle$ \ \ A \ \ $is given by$

$A = \sqrt{s(s-a)(s-b)(s-c)} \ \ \ $where$ \ \ s =\frac{1}{2}(a+b+c)$

Sy123 · Apr 21, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$A triangle has side lengths$ \ \ a, b, c$

$By considering angles of the triangle or otherwise, prove that, the area of the triangle$ \ \ A \ \ $is given by$

$A = \sqrt{s(s-a)(s-b)(s-c)} \ \ \ $where$ \ \ s =\frac{1}{2}(a+b+c)$

Let theta be an angle in the triangle.

$c^2=a^2+b^2-2ab\cos \theta$

$\cos theta = \frac{a^2+b^2-c^2}{2ab}$

$A=\frac{1}{2}ab \sin \theta$

$1-\cos^2 \theta = 1-\left(\frac{a^2+b^2-c^2}{2ab} \right)^2$

$\therefore \ \sin \theta = \sqrt{1-\left(\frac{a^2+b^2-c^2}{2ab} \right )^2}$

$\therefore \ A=\frac{1}{2} ab \sqrt{\frac{4a^2b^2-(a^2+b^2-c^2)^2}{4a^2b^2}}$

$A = \frac{1}{4}\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}$

$A=\frac{1}{4} \sqrt{(2ab-(a^2+b^2-c^2))(2ab+(a^2+b^2-c^2))}$

$A=\frac{1}{4} \sqrt{(c^2-(a-b)^2)((a+b)^2-c^2}$

$A=\frac{1}{4}\sqrt{(c+b-a))(c+a-b))(a+b-c)(a+b+c)}$

$A=\sqrt{0.5(c+b-a) \cdot 0.5(c+a-b) \cdot 0.5(a+b-c) \cdot 0.5(a+b+c)}$

$A=\sqrt{s(s-a)(s-b)(s-c)}$

=========================================

$$Solve for$ \ \ x, y, z$

$a^2x+ay+z=a^2$

$ax+y+bz=1$

$a^2bx+y+bz = b$

Sy123 · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

$By utilising an appropriate substitution, Find$

$\int \sqrt{x -2x\sin^2x + x\sin^4 x} \ dx$

HeroicPandas · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$By utilising an appropriate substitution, Find$

$\int \sqrt{x -2x\sin^2x + x\sin^4 x} \ dx$

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?

HeroicPandas · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

asianese said:
$$b) Now suppose that $ T $ moves in simple harmonic motion along the $x$-axis with its acceleration given by $ \ddot{x}=4x.$

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

Sy123 · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
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.

HeroicPandas · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

Definately a 3U question:

$$The polynomial P(x) = x^5 - 5cx + 1, $ $ where c is a real number. \\ \\ $$(a) By considering the turning points, prove that if c $< 0, $ P(x) has just one real root, which is negative. \\ \\ $$(b) Prove that P(x) has three distinct real roots if and only if c $> \left ( \frac{1}{4} \right )^\frac{4}{5}.$

asianese · Apr 27, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
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 · Apr 28, 2013

Re: HSC 2013 3U Marathon Thread

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

HeroicPandas · Apr 28, 2013

Re: HSC 2013 3U Marathon Thread

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

shongaponga · Apr 28, 2013

Re: HSC 2013 3U Marathon Thread

$Island 1 and Island 2 are positioned as such that there perpendicular distances from a straight shore line are 1km and 2km respectively, and along the shore line these two islands are 6km apart. A pier is to be built on the shore of the lake and a straight road from each island to the pier is to be constructed. Where should the pier be positioned to minimize the sum of the distances from each island to the pier?$

study1234 · Apr 28, 2013

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 · May 2, 2013

Re: HSC 2013 3U Marathon Thread

$Find without integrating$

$\int_b^a \sqrt{a^2-x^2} \ dx$

$a > b > 0$

omgiloverice · May 8, 2013

Re: HSC 2013 3U Marathon Thread

shongaponga · May 9, 2013

Re: HSC 2013 3U Marathon Thread

omgiloverice said:

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.

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!

Makematics · May 10, 2013

Re: HSC 2013 3U Marathon Thread

HeroicPandas said:
Definately a 3U question:

$$The polynomial P(x) = x^5 - 5cx + 1, $ $ where c is a real number. \\ \\ $$(a) By considering the turning points, prove that if c $< 0, $ P(x) has just one real root, which is negative. \\ \\ $$(b) Prove that P(x) has three distinct real roots if and only if c $> \left ( \frac{1}{4} \right )^\frac{4}{5}.$

(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 · May 10, 2013

Re: HSC 2013 3U Marathon Thread

Makematics said:
(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

Sy123 · May 10, 2013

Re: HSC 2013 3U Marathon Thread

$$Define the function$ \ \ I$

$y = I(x)$

$I \ \ $has the property that any real value input will output the integer closest to that value, in the case that it has decimal part$ \ \ .5 \ \ $it will take the number below it$$

$$i) Find$ \ \ I(\pi) \ \ $and$ \ \ I(3.5)$

$ii) Prove$

$\int_0^n I(x) \ dx = \frac{n^2}{2}$

$For some integer n$

RealiseNothing · May 10, 2013

Re: HSC 2013 3U Marathon Thread

Sy123 said:
$$Define the function$ \ \ I$

$y = I(x)$

$I \ \ $has the property that any real value input will output the integer closest to that value, in the case that it has decimal part$ \ \ .5 \ \ $it will take the number below it$$

$$i) Find$ \ \ I(\pi) \ \ $and$ \ \ I(3.5)$

$ii) Prove$

$\int_0^n I(x) \ dx = \frac{n^2}{2}$

$For some integer n$

$I(\pi)=3$

$I(3.5)=3$

Split the integral into:

$\int_{0}^{0.5} I(x)~dx + \int_{0.5}^{1.5} I(x)~dx + \int_{1.5}^{2.5} I(x)~dx + ... + \int_{n-1.5}^{n-0.5} I(x)~dx + \int_{n-0.5}^{n}$

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

Hence we arrive at:

$\int_{0}^{0.5}0~dx + \int_{0.5}^{1.5}1~dx + ... + \int_{n-0.5}^{n} n~dx$

Which upon evaluation becomes:

$0+1+2+3+...+(n-1)+\frac{n}{2}$

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

$\frac{n(n-1)}{2} + \frac{n}{2}$

$\frac{n^2 - n + n}{2}$

$\frac{n^2}{2}$

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

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