HSC Mathematics Marathon (1 Viewer)

ohexploitable · Mar 14, 2011

rawrence said:
^ cool story

$\\\\a)\text{ The polynomial } P(x) = ax^{n+1} + bx^n + 1\text{ is divisible by }(x-1)^2\\ \text{Show that }a = n \text{ and } b = -(1+n) \\\\b)\text{ Show that } P(x) = \frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1\text{ has no double roots}$

$\\P(1)=a+b+1=0\\a+b=-1\\\\P'(1)=a(n+1)+bn=0\\n(a+b)+a=0\\\\a=n\\\\a=-1-b\\-1-b=n\\b=-(1+n)$

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$\\\text{Suppose a double root existed, the root would be the solution to the following:}\\\\P(x)=P'(x)=0\\\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1=\frac{x^3}{3!} + \frac{x^2}{2!} + x + 1\\\frac{x^4}{4!}=0\\x=0\\\\\text{However, }P(0)\neq0,\text{ this is a contradiction}\\\therefore \text{no double root exists}.$

rawrence · Mar 14, 2011

ohexploitable said:
$\\P(1)=a+b+1=0\\a+b=-1\\\\P'(1)=a(n+1)+bn=0\\n(a+b)+a=0\\\\a=n\\\\a=-1-b\\-1-b=n\\b=-(1+n)$

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$\\\text{Suppose a double root existed, the root would be the solution to the following:}\\\\P(x)=P'(x)=0\\\frac{x^4}{4!} + \frac{x^3}{3!} + \frac{x^2}{2!} + x + 1=\frac{x^3}{3!} + \frac{x^2}{2!} + x + 1\\\frac{x^4}{4!}=0\\x=0\\\\\text{However, }P(0)\neq0,\text{ this is a contradiction}\\\therefore \text{no double root exists}.$

Very good

Drongoski · Mar 14, 2011

hscishard said:
Differentiate x+y+z = 0

wrt ?

hscishard · Mar 14, 2011

what's wrt?

hscishard · Mar 14, 2011

and yea, that wasnt really a question lol

rawrence · Mar 14, 2011

hscishard said:
what's wrt?

With Respect To

Omnipotence · Mar 15, 2011

rawrence said:
With Respect To

x

rawrence · Mar 15, 2011

$\text{The quadratic equation }x^2 - x + K = 0\text{ where K is a real number }\\\text{ has two distinct positive real roots, } \alpha \beta \\\\a)\text{ Show that }0<K<\frac{1}{4}\\\\b)\text{ Show that }\alpha^2 + \beta^2 = 1-2K\text{ and deduce that }\alpha^2 + \beta^2 > \frac{1}{2}\\\\c)\text{ Show that }\frac{1}{\alpha^2} + \frac{1}{\beta^2} > 8$

lalalala

muzeikchun852 · Mar 18, 2011

rawrence said:
$\text{The quadratic equation }x^2 - x + K = 0\text{ where K is a real number }\\\text{ has two distinct positive real roots, } \alpha \beta \\\\a)\text{ Show that }0<K<\frac{1}{4}\\\\b)\text{ Show that }\alpha^2 + \beta^2 = 1-2K\text{ and deduce that }\alpha^2 + \beta^2 > \frac{1}{2}\\\\c)\text{ Show that }\frac{1}{\alpha^2} + \frac{1}{\beta^2} > 8$

lalalala

$\text{is this even 4 unit? anyway.}\\\\\text{1)} x^2-x+K=0 \\\\b^2-4ac>0 \\\\ 1-4K>0 \\\\ K<\frac{1}{4}\\\\\text{2)}\alpha +\beta =1 \\\\\alpha \beta =K \\\\\alpha^2 + \beta^2\\\\=(\alpha+\beta )^2-2\alpha \beta \\\\=1-2K\\\\>1-2\cdot \frac{1}{4}\\\\>\frac{1}{2}\\\\\text{3)}\frac{1}{\alpha^2} + \frac{1}{\beta^2}\\\\=\frac{\alpha ^2+\beta ^2}{(\alpha \beta )^2}\\\\=\frac{1+2K}{K^2}\\\\>\frac{1-2\cdot \frac{1}{4}}{\frac{1}{16}}\\\\>\frac{1}{2}\cdot 16\\\\>8$

muzeikchun852 · Mar 18, 2011

rawrence · Mar 18, 2011

muzeikchun852 said:
$\text{my turn: easy one.} P (a\: cos\beta ,b\: sin\beta ) \text{lies on the ellipse}\: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.\\\\\text{The normal at P cuts the x-axis at X and the y-axis at Y.}\\\\\text{Show that}\frac{PX}{PY}=\frac{b^2}{a^2}$

Oh man that's heavy algebra. I get up to an expression like:
$\frac{PX}{PY} = \frac{b}{a}\sqrt{\frac{\frac{b^4}{a^2sec^2\beta}+sin^2\beta}{cos^2\beta + \frac{a^4}{b^2cosec^2\beta}}}$

Is that even looking right so far?

deterministic · Mar 19, 2011

Try doing it using a diagram and similar triangles... its much easier

rawrence · Mar 19, 2011

deterministic said:
Try doing it using a diagram and similar triangles... its much easier

Could you do it and show that? : S

hscishard · Mar 19, 2011

That method sounds much better.
First you'd show that the triangles are similar. I got corresp angles and 90 = 90.
Then you find the side lengths of each triangle. It isn't hard, just use pythag. And then yea
I can't draw up a diagram unfortunately..

deterministic · Mar 19, 2011

refer to attached diagram:
AP is a line parallel to x axis
You should be able to find X from the equation of the normal.
noting that PAY and OXY are similar triangles, use the ratios of sides to get:
Noting
OX= x coordinate of X
AP= x coordinate of P

$\frac{PY-PX}{PY}=\frac{OX}{AP} \\ 1-\frac{PX}{PY}=\frac{\frac{(a^2-b^2)cos\beta}{a}}{acos\beta}\\1-\frac{PX}{PY}=\frac{(a^2-b^2)}{a^2}$

rawrence · Mar 19, 2011

deterministic said:
View attachment 22753
refer to attached diagram:
AP is a line parallel to x axis
You should be able to find X from the equation of the normal.
noting that PAY and OXY are similar triangles, use the ratios of sides to get:
Noting
OX= x coordinate of X
AP= x coordinate of P

$\frac{PY-PX}{PY}=\frac{OX}{AP} \\ 1-\frac{PX}{PY}=\frac{\frac{(a^2-b^2)cos\beta}{a}}{acos\beta}\\1-\frac{PX}{PY}=\frac{(a^2-b^2)}{a^2}$

Oh, I see thanks, haven't done conics like this before

muzeikchun852 · Mar 19, 2011

rawrence said:
Oh, I see thanks, haven't done conics like this before

$\text{or you could just find the co-ordinate of X and Y first.}\\\\\text{i just quote the normal equation}\\\\\frac{ax}{cos\theta }-\frac{by}{sin\theta}=a^2-b^2\\\\X:(\frac{(a^2-b^2)cos\theta}{a},0)\\\\Y:(0,\frac{(b^2-a^2)sin\theta}{b})\\\\PX^2=(a-\frac{a^2-b^2}{a})^2cos^2\theta+b^2sin^2\theta\\\\=\frac{b^4}{a^2}cos^2\theta+b^2sin^2\theta\\\\PY^2=a^2cos^2\theta+(b-\frac{b^2-a^2}{b})^2sin^2\theta\\\\=a^2cos^2\theta+\frac{a^4}{b^2}sin^2\theta\\\\\text{try it from here.}$

blackops23 · Mar 19, 2011

deterministic said:
Try doing it using a diagram and similar triangles... its much easier

^fuaarrkin genius, can't believe it is soo much easier with similar triangles...

blackops23 · Mar 19, 2011

jm01 said:
New question: Find the locus of z if arg ((z-1)/(z+1))= pi/4

I know how to sketch the locus. But how do you find the exact equation of the locus?

ohexploitable · Mar 19, 2011

blackops23 said:
I know how to sketch the locus. But how do you find the exact equation of the locus?

Use basic geometry. The locus is a circle, contruct a triangle joining the centre, (0,1) and (0,-1) and make a chocolate cake. The length of the base is obvious (2), the angle subtended at the centre is twice the angle at the circumference (pi/2). Now find the height of the circle to get the coordinates of the centre then find the radius. The cake is a lie.

HSC Mathematics Marathon (1 Viewer)

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