HSC 2012 MX2 Marathon (archive) (2 Viewers)

Aesytic · May 5, 2012

Re: 2012 HSC MX2 Marathon

1.since the equation is a cubic, there must be at least 1 real root, so therefore the last root must be real since alpha and beta are complex roots
also, since the coefficients of the polynomial are all real, the 2 complex roots alpha and beta must be conjugates of one another because the other root is real, and the conjugate of a real number is the same, leaving alpha and beta to form a conjugate pair
using the product of roots, and letting the last root be gamma, alpha*beta*gamma = -1
gamma = -1/alpha*beta
when a complex number and its conjugate is multiplied together, the result is the modulus squared
.'. gamma = -1/|beta|^2

2. subbing gamma into the equation x^3 + 5x + 1,
-1/(|beta|^2)^3 - 5/|beta|^2 + 1 = 0 since -1/|beta|^2 is a root of this equation
mutiplying everything by |beta|^6,
-1 - 5|beta|^4 + |beta|^6 = 0
(|beta|^2)^3 - 5(|beta|^2)^2 - 1 = 0
looking at this equation, we can therefore see that |beta|^2 is a root of the equation x^3 - 5x^2 - 1 = 0, and from before it was established that alpha*beta = |beta|^2
.'. alpha*beta is a root of the equation x^3 -5x^2 - 1 =0

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

good stuff, someone revised this thread

AAEldar · May 5, 2012

Re: 2012 HSC MX2 Marathon

I'll add a question into the loop...

$Find without using a trigonometric substitution:$

$\int \frac{1}{\sqrt{1+x^2}}\;dx$

Want to see the working and I don't want to see any hyperbolic functions being thrown in there!

Also I doubt you'd ever get something like that, you'd probably be given the first step or a substitution but it's still fun!

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

AAEldar said:
I'll add a question into the loop...

$Find without using a trigonometric substitution:$

$\int \frac{1}{\sqrt{1+x^2}}\;dx$

Want to see the working and I don't want to see any hyperbolic functions being thrown in there!

Also I doubt you'd ever get something like that, you'd probably be given the first step or a substitution but it's still fun!

this is on the standard integrals sheet lol

AAEldar · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
this is on the standard integrals sheet lol

Yea but a good exercise to show it

lolcakes52 · May 5, 2012

Re: 2012 HSC MX2 Marathon

Okay.
Question 1: The ellipse (x^2)/2 +y^2 = 1 is intercepted at two distinct points P(x1,y1) and Q(x2,y2) by the line y=mx+3. Show that points x1 and x2 are the solutions to the equation (1+2m^2)x^2 +12mx +16 =0
Question 2: Find the values of m for which x1 and x2 are real distinct points. (sorry for the poor phrasing)
Question 3: Find the equations of the tangents to the ellipse from the point (0,3).

Aesytic · May 5, 2012

Re: 2012 HSC MX2 Marathon

1. solving simultaneously with the ellipse and the line,
(x^2)/2 + (mx+3)^2 = 1
x^2 + 2(mx+3)^2 = 2
x^2 + 2m^2x^2 + 12mx + 18 = 2
(1+2m^2)x^2 + 12mx + 16 = 2
since the line and the ellipse intersect when x = x1 and when x=x2, then x1 and x2 are solutions to this equation
2. in order for x1 and x2 to be real distinct points, the discriminant of the equation must be >0
.'. 144m^2 - 4*16(1+2m^2) > 0
144m^2 - 64 - 128m^2 > 0
16m^2 - 64 > 0
m^2 - 4 > 0
(m-2)(m+2)>0
.'. m>2 , m<-2
3. the line y=mx + 3 is a tangent to the ellipse that passes through (0,3) when the points x1 and x2 are the same
x1 and x2 are the same when the discriminant equals 0
replacing the inequality in question 2 with an equality sign,
m=2 or -2
.'. y = 2x + 3 or y = -2x + 3

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

karnbmx · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

basically, you find the equation of the tangent at P using parametric coordinates (root(32)t,root(32)/t), then using X and Y intercepts, calculate the area (the parameters should cancel out and give a constant value).

I'll type up the full solution soon.

Just out of interest Barbernator, where did you get this question from?

EDIT:

here you go:

x = 4root(2)t y = 4root(2)/t, where t is a parameter

m = -1/t^2

therefore, y -4root(2)/t = -1(x-4root(2)/t)/t^2

Therefore x int: (8root(2)t, 0) y-int: (0,8root(2)/t)

Hence Area of Triangle = 8root(2)t * 8root(2)/t * 1/2 = 32root2 units squared

As there are no parameters present, the area must be a constant, with a value of 32root(2).

I sort of did this in a rush, so sorry if the answer is wrong.

nightweaver066 · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
prove that the area subtended by the tangent P to the curve xy=32, the X axis, and the Y axis is constant. Find the value of this constant

Diagram:
http://i.imgur.com/kEmRg.jpg

$h: xy = 32$

$Let~P(ct, \frac{c}{t})~be~a~point~on~h~where~c^2 = 32$

$$Differentiating both sides w.r.t. x$

$x\frac{dy}{dx} + y = 0$

$\frac{dy}{dx} = \frac{-y}{x}$

$= -\frac{1}{t^2}~at~P(ct, \frac{c}{t})$

$\therefore $ Equation of tangent at P:$$

$y - \frac{c}{t} = -\frac{1}{t^2}(x - ct)$

$$At Y, x = 0$, \therefore y - \frac{c}{t} = \frac{c}{t}$

$y = \frac{2c}{t} => Y(0, \frac{2c}{t})$

$$At X, y = 0$, \therefore \frac{c}{t} = \frac{1}{t^2}(x - ct)$

$x - ct = ct$

$x = 2ct => X(2ct, 0)$

$Area~of~\Delta XPY = \frac{1}{2} \times 2ct \times \frac{2c}{t}$

$= 2c^2$

$= 64$

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

karnbmx said:
basically, you find the equation of the tangent at P using parametric coordinates (root(32)t,root(32)/t), then using X and Y intercepts, calculate the area (the parameters should cancel out and give a constant value).

I'll type up the full solution soon.

Just out of interest Barbernator, where did you get this question from?

i made it up, but i have seen questions like this before. but there is the geometric property of the rectangular hyperbola that states that the area is constant.

Carrotsticks · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
i made it up, but i have seen questions like this before. but there is the geometric property of the rectangular hyperbola that states that the area is constant.

Could have generalised the problem by using the equation xy = k^2 instead =p

Oh and there is actually a nice and cheap way of calculating (or checking) the Implicit Derivative.

$\frac{dy}{dx} = - \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$

So for the equation xy= k^2, we have:

$\\ f(x,y) = xy - k^2 \\\\ \frac{\partial f}{\partial x} = y \\\\ \frac{\partial f}{\partial y} = x \\\\ \frac{dy}{dx} = - \frac{y}{x}$

I know it doesn't seem worth it for this particular example, but it really helps when it comes to implicitly differentiating things like:

$3x^2 + 2x^2y + 2xy^2 + y^2 + xy = 1$

juantheron · May 5, 2012

Re: 2012 HSC MX2 Marathon

$\hspace{-16}\bf{\int\frac{1}{\sqrt{1+x^2}}dx}$\\\\\\ Let $\bf{1+x^2=y^2\Leftrightarrow 2xdx=2ydy}$\\\\\\ $\bf{xdx=ydy\Leftrightarrow \frac{dx}{y}=\frac{dy}{x}=\frac{d(x+y)}{x+y}}$\\\\\\ Now Integral Convert into \\\\\\ $\bf{=\int \frac{dx}{y}=\int \frac{d(x+y)}{(x+y)}}$\\\\\\ $\bf{=\ln \mid x+y \mid+C}$\\\\\\ So $\bf{\int \frac{1}{\sqrt{1+x^2}}dx = \ln \mid x+\sqrt{1+x^2} \mid+C}$$

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

post questions once you answer guys

nightweaver066 · May 5, 2012

Re: 2012 HSC MX2 Marathon

$$P is any point on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1~,~ (a > b); $ A, A' are the ends of the major axis; AP meets the minor axis at Q. Prove that the line through Q parallel to A'P is a tangent to one of the parabolas $ ay^2 \pm 4b^2x = 0.$

karnbmx · May 5, 2012

Re: 2012 HSC MX2 Marathon

Hmmm.....

$integrate: \int \frac{1}{1 + \sqrt{x}} \, dx$

nightweaver066 · May 5, 2012

Re: 2012 HSC MX2 Marathon

karnbmx said:
Hmmm.....

Find the indefinite integral of 1/1+squareroot(x)

$\int \frac{1}{1 + \sqrt{x}}dx$

$Let~u^2 = x$

$2udu = dx$

$\therefore \int \frac{2u}{1 + u}du$

$= \int 2 - \frac{2}{1 + u}du$

$= 2u - 2ln|1 + u|$

$= 2\sqrt{x} - 2ln|1 + \sqrt{x}| + c$

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

here is a nice graphs question. Graph the function <a href="http://www.codecogs.com/eqnedit.php?latex=y^2=x^3(x-2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y^2=x^3(x-2)" title="y^2=x^3(x-2)" /></a> noting the behaviour of the curve near its intercepts. (you dont need to mark inflection points) (don't cheat and use wolfram)

also, nightweavers question takes ages fuuuuuuuuuuuu

nightweaver066 · May 5, 2012

Re: 2012 HSC MX2 Marathon

barbernator said:
here is a nice graphs question. Graph the function <a href="http://www.codecogs.com/eqnedit.php?latex=y^2=x^3(x-2)" target="_blank"><img src="http://latex.codecogs.com/gif.latex?y^2=x^3(x-2)" title="y^2=x^3(x-2)" /></a> noting the behaviour of the curve near its intercepts. (you dont need to mark inflection points) (don't cheat and use wolfram)

also, nightweavers question takes ages fuuuuuuuuuuuu

Have you finished it?

barbernator · May 5, 2012

Re: 2012 HSC MX2 Marathon

nightweaver066 said:
Have you finished it?

yes, im definitely not scripting the solution though haha. I used cartesian co-ordinates, parametric might have made it easier.

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