HSC 2015 MX2 Marathon (archive) (2 Viewers)

FrankXie · Feb 5, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION

$Find any straight lines that are tangent to the curve $ y=x^4+4x^3-2x^2 $ at two distinct points.$

braintic · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$Find any straight lines that are tangent to the curve $ y=x^4+4x^3-2x^2 $ at two distinct points.$

Answer with no working, if anyone wants to see if they are correct:

y=12x-9

VBN2470 · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

$Find any straight lines that are tangent to the curve $ y=x^4+4x^3-2x^2 $ at two distinct points.$

$$ Need Frank to confirm this, but I got the equation of the tangent to be $y=12x-9$. Idea is that you need to realise that since there exists two distinct points at which the tangent touches the curve, there exists two double roots for the equation $f(x)-mx-b=0$ where $y=mx+b$ is the equation of the tangent and $f(x)$ is the given function. Then you need to use algebra (i.e. sum and product of roots) to find values for $m$ and $b$ to find the tangent line $.$

EDIT: Just realised braintic got the same answer so I am guessing it is correct. Good question though!

FrankXie · Feb 5, 2015

Re: HSC 2015 4U Marathon

well done guys. next one:

$$ Show that the modulus of each root of the equation $ z^3+z+12=0 $ lie between $ 2 $ and $ \sqrt{6}.$

VBN2470 · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
well done guys. next one:

$$ Show that the modulus of each root of the equation $ z^3+z+12=0 $ lie between $ 2 $ and $ \sqrt{6}.$

$Consider the polynomial $P(x)=x^3+x+12$, then $P'(x)=3x^2+1>0$ for all real $x$, thus the function is always increasing along the reals and so there only exists one real root to $P(x)=0$. Since the coefficients of $P(x)$ are real, then the other two roots are complex conjugates (Complex conjugate root theorem). For the real root say $\alpha$, test $P(-2)$ and $P(-\sqrt6)$ and you will see that $2<|\alpha|<\sqrt6$, since $P(-\sqrt6)<0<P(-2)$. Setting the other two roots as $\beta$ and $\bar{\beta}$, you will see that by taking product of roots and manipulating the inequalities you should find that the modulus of all roots lie in the given interval.$

Drsoccerball · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
well done guys. next one:

$$ Show that the modulus of each root of the equation $ z^3+z+12=0 $ lie between $ 2 $ and $ \sqrt{6}.$

$$Let roots$= a+ib, a-ib, c$
$\sum\alpha=0 \\ \therefore c=-2a$
$\sum\alpha\beta=1$
$\therefore 1-2ac=a^2 +b^2$
$\sum\alpha\beta\gamma =-12$
$\therefore -12=(a^2 + b^2)c$
$\alpha= \frac{-1\pm\sqrt{97}}{8}$
$c=\frac{1\pm\sqrt{97}}{4}$
$$Modulus$ =\sqrt{a^2+b^2}=\sqrt{1-2ac}$
$when subbed in the modulus is between domain$

VBN2470 · Feb 5, 2015

Re: HSC 2015 4U Marathon

NEXT QUESTION:

$$ By considering $z=\cos\theta+i\sin\theta$, show that \\ $\sin\theta-\frac{\sin3\theta}{3^2}+\frac{\sin5\theta}{3^4}-\frac{\sin7\theta}{3^6}+...=\frac{72\sin\theta}{82+18\cos2\theta}$

FrankXie · Feb 5, 2015

Re: HSC 2015 4U Marathon

VBN2470 said:
NEXT QUESTION:

$$ By considering $z=\cos\theta+i\sin\theta$, show that \\ $\sin\theta-\frac{\sin3\theta}{3^2}+\frac{\sin5\theta}{3^4}-\frac{\sin7\theta}{3^6}+...=\frac{72\sin\theta}{82+18\cos2\theta}$

limiting sum of GP, with r=-z^2/9, and take imaginary part

VBN2470 · Feb 5, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
limiting sum of GP, with r=-z^2/9, and take imaginary part

Yep, the rest falls in place.

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon

NEW QUESTION

$A, B $ and $ C $ are the points that represent the complex numbers $ z_1, z_2 $ and $ z_3 $ on the Argand diagram. Prove that if $ \frac{z_2-z_3}{z_1-z_3}=\frac{z_1-z_3}{z_1-z_2}, $ then $ \triangle{ABC} $ is equilateral. $$

Drsoccerball · Feb 6, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEW QUESTION

$A, B $ and $ C $ are the points that represent the complex numbers $ z_1, z_2 $ and $ z_3 $ on the Argand diagram. Prove that if $ \frac{z_2-z_3}{z_1-z_3}=\frac{z_1-z_3}{z_1-z_2}, $ then $ \triangle{ABC} $ is equilateral. $$

Very nice question
$(z_2 - z_3)(z_1 - z_2)= (z_1 - z_3)^2$

$$Simplify:$ z_1z_2 +z_1z_3 +z_2z_3 = z_1^2 + z_2^2 +z_3^2$

$$If ABC is an equilateral triangle then let $ W=cis\frac{\pi}{3}$

$$By diagram$ \therefore z_2 - z_1 = w(z_3 - z_1) $and$ z_3 -z_2 =w(z_1 - z_2)$

$\frac{z_2 -z_1}{z_3 -z_1} = \frac{z_3 - z_2}{z_1 -z_2}$

$By Simplifying the above :$

$z_1z_2 +z_1z_3 +z_2z_3 = z_1^2 +z_2^2 +z_3^2$

$\therefore $ABC is an equilateral when$ \frac{z_2 -z_3}{z_1 -z_3} =\frac{z_1 -z_3}{z_1 -z_2}$

braintic · Feb 6, 2015

Re: HSC 2015 4U Marathon

New Question:

Find the range of values of arg z for complex numbers z satisfying:

| z - (1+i) | = Re [ z - (-1+i) ]

(The answer involves simple values, so working is necessary)

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon

Drsoccerball said:
Very nice question
$(z_2 - z_3)(z_1 - z_2)= (z_1 - z_3)^2$

$$Simplify:$ z_1z_2 +z_1z_3 +z_2z_3 = z_1^2 + z_2^2 +z_3^2$

$$If ABC is an equilateral triangle then let $ W=cis\frac{\pi}{3}$

$$By diagram$ \therefore z_2 - z_1 = w(z_3 - z_1) $and$ z_3 -z_2 =w(z_1 - z_2)$

$\frac{z_2 -z_1}{z_3 -z_1} = \frac{z_3 - z_2}{z_1 -z_2}$

$By Simplifying the above :$

$z_1z_2 +z_1z_3 +z_2z_3 = z_1^2 +z_2^2 +z_3^2$

$\therefore $ABC is an equilateral when$ \frac{z_2 -z_3}{z_1 -z_3} =\frac{z_1 -z_3}{z_1 -z_2}$

it seems to me you proved that if ABC is equilateral then the given equation is true, but the question asks you to prove its converse: given the equation, prove ABC is equilateral.

braintic · Feb 6, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
it seems to me you proved that if ABC is equilateral then the given equation is true, but the question asks you to prove its converse: given the equation, prove ABC is equilateral.

Actually he's done neither ... he met in the middle.
But every step is reversible, so all required logic is there.

RealiseNothing · Feb 6, 2015

Re: HSC 2015 4U Marathon

FrankXie said:
NEW QUESTION

$A, B $ and $ C $ are the points that represent the complex numbers $ z_1, z_2 $ and $ z_3 $ on the Argand diagram. Prove that if $ \frac{z_2-z_3}{z_1-z_3}=\frac{z_1-z_3}{z_1-z_2}, $ then $ \triangle{ABC} $ is equilateral. $$

$\arg(\frac{z_2-z_3}{z_1-z_3}) = \arg(\frac{z_1-z_3}{z_1-z_2})$

This implies that $\angle BAC = \angle BCA$ and hence $|z_1-z_2|=|z_2-z_3|$

Re-arranging the original equation and using the above:

$|z_1-z_2|^2 = |z_1-z_3|^2$ and thus $|z_1-z_2|=|z_1-z_3|=|z_2-z_3|$

>equilateral

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon

RealiseNothing said:
$\arg(\frac{z_2-z_3}{z_1-z_3}) = \arg(\frac{z_1-z_3}{z_1-z_2})$

This implies that $\angle BAC = \angle BCA$ and hence $|z_1-z_2|=|z_2-z_3|$

Re-arranging the original equation and using the above:

$|z_1-z_2|^2 = |z_1-z_3|^2$ and thus $|z_1-z_2|=|z_1-z_3|=|z_2-z_3|$

>equilateral

This is exactly what I would do.

braintic · Feb 6, 2015

Re: HSC 2015 4U Marathon

braintic said:
New Question:

Find the range of values of arg z for complex numbers z satisfying:

| z - (1+i) | = Re [ z - (-1+i) ]

(The answer involves simple values, so working is necessary)

Now that the other question is sorted .... BUMP

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon

braintic said:
Actually he's done neither ... he met in the middle.
But every step is reversible, so all required logic is there.

are you sure? I think not all steps are reversible. if one thinks the logic is all good in that proof, can you please rewrite the proof step by step in correct order?

The key point is: from $\frac{z_2 -z_1}{z_3 -z_1} = \frac{z_3 - z_2}{z_1 -z_2}$ to $z_2 - z_1 = w(z_3 - z_1)$ and $z_3 -z_2 =w(z_1 - z_2)$ is fine, but you need to prove $w=cis\frac{\pi}{3}$ , which is the most important step in the proof you can't skip.

FrankXie · Feb 6, 2015

Re: HSC 2015 4U Marathon

braintic said:
New Question:

Find the range of values of arg z for complex numbers z satisfying:

| z - (1+i) | = Re [ z - (-1+i) ]

(The answer involves simple values, so working is necessary)

sorry have to bump again lol

Drsoccerball · Feb 7, 2015

Re: HSC 2015 4U Marathon

braintic said:
New Question:

Find the range of values of arg z for complex numbers z satisfying:

| z - (1+i) | = Re [ z - (-1+i) ]

(The answer involves simple values, so working is necessary)

Not sure if this is right but :
$1+i = \sqrt{2}cis(\frac{\pi}{4}) 1-i = \sqrt{2}cis(\frac{-\pi}{4})$

$\therefore $range of argz$ = -\frac{\pi}{4} \grq arg(z) \grq \frac{\pi}{4} ?$

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what is that?It is Cowpea

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