HSC 2015 MX2 Marathon (archive) (2 Viewers)

ymcaec · Dec 11, 2014

Re: HSC 2015 4U Marathon

Sy123 said:
$\\ $Describe the locus of$ \ z \ $in the Argand plane if$ \ \frac{1}{z} + \frac{1}{\bar{z}} = 1$

$$Let $z = x + iy,\,\,z \ne 0.\\\\\,\,\frac{1}{{x + iy}} + \frac{1}{{x - iy}} = 1\\\frac{{x - iy + x + iy}}{{\left( {x + iy} \right)\left( {x - iy} \right)}} = 1\\\\2x = {x^2} + {y^2}\\{\left( {x - 1} \right)^2} + {y^2} = 1$

Circle with centre (1,0), radius 1, excluding (0,0).

Axio · Dec 15, 2014

Re: HSC 2015 4U Marathon

$The equation$ \ x^3+3x+2 \ $has roots \alpha , \beta,\gamma .$

$$Find the equation with roots \alpha +\frac{1}{\alpha }, \beta +\frac{1}{\beta }, \gamma +\frac{1}{\gamma }.$

porcupinetree · Dec 19, 2014

Re: HSC 2015 4U Marathon

Axio said:
$The equation$ \ x^3+3x+2 \ $has roots \alpha , \beta,\gamma .$

$$Find the equation with roots \alpha +\frac{1}{\alpha }, \beta +\frac{1}{\beta }, \gamma +\frac{1}{\gamma }.$

http://imgur.com/w1DCp1x

I can't use the program (forget what it's called) that everyone uses for writing/answering questions, so:

If P(x) = x^4 - 8x^3 + 30x^2 - 56x + 49 has a non-real double zero, solve the equation P(x) = 0 over C and factorise P(x) fully over R

Axio · Dec 20, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
http://imgur.com/w1DCp1x

I can't use the program (forget what it's called) that everyone uses for writing/answering questions, so:

If P(x) = x^4 - 8x^3 + 30x^2 - 56x + 49 has a non-real double zero, solve the equation P(x) = 0 over C and factorise P(x) fully over R

$$Let$ \ P(x)=(x-z)^2(x-\bar{z})^2$
$=(x^2-2Re(z)+|z|^2)^2$
$$By inspection$, P(x)=(x^2-4x+7)^2 \ $over$ \ \mathbb{R}$
$\therefore \ $By evaluating the real part and modulus of z,$ \ Re(z)=2, \ Im(z)=\pm \sqrt{3}$
$\therefore \ $Roots are:$ \ x=2\pm \sqrt{3}i$

--

$$Describe the locus of the equation:$ \ z=i\bar{z}$

InteGrand · Dec 20, 2014

Re: HSC 2015 4U Marathon

Let z = x + iy, where x, y ∈ ℝ.

Then $z = i\bar{z}\Rightarrow x+iy = i(x-iy)$

$\Rightarrow x+iy = ix + y$
$\Rightarrow (x-y) +i(y-x) = 0$

Equating real and imaginary parts, x - y = 0 and y - x = 0, so x = y, i.e. the locus is the straight line y = x, which is a straight line passing through the origin with slope 1.

Axio · Dec 21, 2014

Re: HSC 2015 4U Marathon

$$Suppose$ \ 0\leq x\leq \frac{\sqrt{2}}{2}$

$$i) Show that$ \ 0\leq \frac{2x^2}{1-x^2}\leq 4x^2$
$$ii) Hence show that$ \ 0\leq \frac{1}{1-x}+\frac{1}{1+x}-2\leq 4x^2$

porcupinetree · Dec 21, 2014

Re: HSC 2015 4U Marathon

Axio said:
$$Suppose$ \ 0\leq x\leq \frac{\sqrt{2}}{2}$

$$i) Show that$ \ 0\leq \frac{2x^2}{1-x^2}\leq 4x^2$
$$ii) Hence show that$ \ 0\leq \frac{1}{1-x}+\frac{1}{1+x}-2\leq 4x^2$

Here's my working, I feel as though there's a much easier solution to part i that's right under my nose, however this is the way that I did it.

https://imgur.com/UoF6Kk9

Axio, may I ask a question? In your answer to my previous question, one of your lines of working was:

Axio said:
$$By inspection$, P(x)=(x^2-4x+7)^2 \ $over$ \ \mathbb{R}$

Can I ask how you concluded that P(x) was that after just looking at it? I can understand that the constant of P(x) must be the square of the constant inside the brackets, but did you use any particular method for concluding that the coefficient of x (in the brackets) was -4?

Next question:

If P(x) = x^6 + x^4 + x^2 + 1, show that the solutions of the equation P(x) = 0 are among the solutions of the equation x^8 - 1 = 0. Hence factorise P(x) fully over R.

Axio · Dec 21, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
Can I ask how you concluded that P(x) was that after just looking at it? I can understand that the constant of P(x) must be the square of the constant inside the brackets, but did you use any particular method for concluding that the coefficient of x (in the brackets) was -4?

I did it by equating the coefficients of the x terms. So if you just let -2Re(z)=n, then by looking at the brackets: (x^2 +nx+7)(x^2+nx+7) you know that the sum of the x terms will be 7nx+7nx which =-56x. Then 14n=-56, n=-4.

porcupinetree said:
Here's my working, I feel as though there's a much easier solution to part i that's right under my nose, however this is the way that I did it.

Your method is good

. An alternative method would be:

$x\leq \frac{\sqrt{2}}{2}$
$x^2\leq \frac{1}{2}$
$1-x^2\geq \frac{1}{2}$
$\frac{1}{1-x^2}\leq 2$
$\frac{2x^2}{1-x^2}\leq 4x^2...$

Axio · Dec 22, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
Next question:

If P(x) = x^6 + x^4 + x^2 + 1, show that the solutions of the equation P(x) = 0 are among the solutions of the equation x^8 - 1 = 0. Hence factorise P(x) fully over R.

$0=x^8-1$
$0=(x^4-1)(x^4+1)$
$0=(x^2-1)(x^2+1)(x^4+1)$
$0=(x-1)(x+1)(x^6+x^4+x^2+1)$
$\therefore \ $Solutions of P(x)=0 are the solutions of$ \ x^8=1 \ $where$ \ x\neq \pm 1$
$\therefore \ $Solutions are:$ \ x=cis\pm \frac{\pi }{4}, cis\pm \frac{\pi }{2}, cis\pm \frac{3\pi }{4}$
$\therefore P(x)=(x^2+1)(x^2-\sqrt{2}x+1)(x^2+\sqrt{2}x+1)$

---

$$Solve$ \ x^2-4x+1-4i=0$

porcupinetree · Dec 22, 2014

Re: HSC 2015 4U Marathon

Axio said:
$$Solve$ \ x^2-4x+1-4i=0$

https://imgur.com/AbcGhk4

---

z satisfies the equation |z - 2 - 2i| = √2.
a) Sketch the locus of the point P representing z on an Argand diagram.
b) Find the maximum and minimum values of |z| and the values of z for which these extremes are attained.
c) Find the range of possible values of arg z.

InteGrand · Dec 23, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
https://imgur.com/AbcGhk4

---

z satisfies the equation |z - 2 - 2i| = √2.
a) Sketch the locus of the point P representing z on an Argand diagram.
b) Find the maximum and minimum values of |z| and the values of z for which these extremes are attained.
c) Find the range of possible values of arg z.

a) The locus is |z - (2+2i)| = √2. This is a circle with centre (2,2), radius √2, sketched here:

http://graphsketch.com/parametric?m..._lines=1&line_width=4&image_w=850&image_h=850

b) Closest and furthest points on the locus to and from the origin lie on the line through the origin and the circle's centre, which is the line y = x. The circle's equation is (x - 2)² + (y - 2)² = 2. Sub. y = x and solve for x:

(x - 2)² + (x - 2)² = 2
⇒ (x - 2)² = 1
⇒ (x - 2) = -1, +1
⇒ x = 1, 3.

So as y = x, the closest point on the locus to the origin is (1,1), and the furthest is (3,3).
Hence the minimum value of |z| is $\sqrt{1^2 + 1^2} = \sqrt{2}$ , occurring when z = 1+i, and the maximum value of |z| is $\sqrt{3^2 + 3^2} = 3\sqrt{2}$ , occurring when z = 3+3i.

c) Clearly, arg z is acute. Its extreme values occur when OP is tangent to the circle. Let the equations of the tangents be y = mx (m, the slope, will have two values, as there are two tangents).

When the tangent touches the circle, substituting y = mx into (x - 2)² + (y - 2)² = 2, we have

(x - 2)² + (mx - 2)² = 2
⇒ x² - 4x + 4 + m²x² - 4mx + 4 = 2
⇔ (m²+1)x² - 4(m+1)x + 6 = 0.

For tangents, set discriminant to 0:

16(m+1)² - 4(m²+1)(6) = 0

Solving this quadratic equation for m gives $m = 2\pm\sqrt{3}$

Since m = tan(arg z), arg z = tan^-1 m

⇒ the range of possible values for arg z is that $\text{arg }z \in\left [ \tan ^{-1} (2-\sqrt{3}), \tan^ {-1} (2+\sqrt{3}) \right ]$

One can further show using compound angle formulae that $\tan 15 ^\circ = 2-\sqrt{3} \Rightarrow \tan ^{-1} (2-\sqrt{3}) = 15 ^\circ = \frac{\pi}{12}$ , and by symmetry of the situation, $\tan^ {-1} (2+\sqrt{3}) = 90 ^\circ - 15 ^\circ = \frac{5\pi}{12}$ , so $\text{arg }z \in\left [ \frac{\pi}{12}, \frac{5\pi}{12} \right ]$

FrankXie · Dec 23, 2014

Re: HSC 2015 4U Marathon

Alternatively, denoting the centre of the circle be C, point of contact be P, then

$OP=2\sqrt2, CP=\sqrt2, \sin\angle{COP}=\frac{CP}{OP}=\frac12, \angle{COP}=\frac{\pi}{6}, $ therefore, arg$z=\frac{\pi}{4}-\frac{\pi}{6}=\frac{\pi}{12}$

FrankXie · Dec 23, 2014

Re: HSC 2015 4U Marathon

NEXT QUESTION

Prove that all the roots of the equation

$z^n\cos n\alpha+z^{n-1}\cos (n-1)\alpha+\cdots+z\cos\alpha=1$

$$ where $ \alpha $ is real, lie outside the circle $ |z|=\frac12.$

InteGrand · Dec 26, 2014

Re: HSC 2015 4U Marathon

FrankXie said:
NEXT QUESTION

Prove that all the roots of the equation

$z^n\cos n\alpha+z^{n-1}\cos (n-1)\alpha+\cdots+z\cos\alpha=1$

$$ where $ \alpha $ is real, lie outside the circle $ |z|=\frac12.$

Denote the equation $\sum_{k=1}^{n}z^k \cos k\alpha = z^n\cos n\alpha + z^{n-1}\cos (n-1)\alpha+\cdots z\cos \alpha =1$ by $(\ast)$ . Let $\omega =r.\text{c}i\text{s}(\theta)$ be a root of $(\ast)$ , where $r,\theta \in \mathbb{R}$ and $r>0$ (so $r = |\omega |$ ). We show that r must be greater than ½.

As $\omega$ is a root, we have $\sum_{k=1}^{n}\omega ^k \cos k\alpha = 1$ .

Taking the modulus of both sides,

$|RHS|=|LHS|$

$\Rightarrow 1 = \left | \sum_{k=1}^{n}\omega ^k \cos k\alpha \right |$

$\leq \sum_{k=1}^{n}|\omega ^k \cos k\alpha|$ (by the extended triangle inequality)

$= \sum_{k=1}^{n}|\omega| ^k .|\cos k\alpha|$

$\leq\sum_{k=1}^{n}|\omega| ^k$ , equality iff $|\cos k\alpha|=1 \text{ }\forall k\in \left \{ 1,2,3,\cdots,n \right \}$ (since $|\cos k\alpha \leq 1|$ )

$=\sum_{k=1}^{n}r^k .$

So we have $\sum_{k=1}^{n}r^k\geq 1$ , equality iff $|\cos k\alpha|=1 \text{ }\forall k\in \left \{ 1,2,3,\cdots,n \right \}$ .

Using the geometric series sum formula on the LHS, and noting that the LHS is strictly less than the infinite sum of powers of r (since r > 0), this inequality yields

$\frac{r}{1-r}>\frac{r(1-r^n)}{1-r}\geq 1$ . (The left-most expression here is the value of the aforementioned infinite sum)

i.e. $\frac{r}{1-r}>1\Leftrightarrow \frac{1}{-1+\frac{1}{r}}>1$ .

Now, we can see that if $r\leq\frac{1}{2}$ , the above inequality does not hold, as when $r = \frac{1}{2}$ , the LHS is equal to 1 (not greater than it), and the LHS is monotonic increasing for r > 0. Thus we cannot have $r\leq\frac{1}{2}$ , i.e. r = |ω| must be greater than ½, as required. ■

InteGrand · Dec 26, 2014

Re: HSC 2015 4U Marathon

$\textsc{Next Question}$

If the expression $\frac{\sin\frac{x}{2}+\cos \frac{x}{2} +i\tan x}{1+2i\sin \frac{x}{2}}$ is real, find the set of all possible values for $x$ .

porcupinetree · Dec 29, 2014

Re: HSC 2015 4U Marathon

InteGrand said:
$\textsc{Next Question}$

If the expression $\frac{\sin\frac{x}{2}+\cos \frac{x}{2} +i\tan x}{1+2i\sin \frac{x}{2}}$ is real, find the set of all possible values for $x$ .

I have got as far as saying that tanx - sinx + cosx = 1, but don't really know where to go from there

Sy123 · Dec 29, 2014

Re: HSC 2015 4U Marathon

porcupinetree said:
I have got as far as saying that tanx - sinx + cosx = 1, but don't really know where to go from there

$\\ \tan x - \sin x + \cos x = 1 \\ \\ \sin x \left(\frac{1}{\cos x} - 1 \right) = 1 - \cos x \\ \\ \frac{\sin x}{\cos x} (1 - \cos x) = 1 - \cos x \\ \\ \therefore \ $Solutions$ \ \cos x = 1 , \tan x = 1$

$\\ \therefore \ x = 2k\pi \ $and$ \ x = k\pi + \frac{\pi}{4} \ $for$ \ k \ $integer$$

Ton5698 · Dec 29, 2014

Re: HSC 2015 4U Marathon

just wondering but what method did you use to input mathematics online like that Sy123?

Crisium · Dec 29, 2014

Re: HSC 2015 4U Marathon

Ton5698 said:
just wondering but what method did you use to input mathematics online like that Sy123?

LaTeX Bruh

Sy123 · Dec 29, 2014

Re: HSC 2015 4U Marathon

Ton5698 said:
just wondering but what method did you use to input mathematics online like that Sy123?

This forum has built-in LaTeX generator

LaTeX is a code for doing math writing, on this forum if you put your LaTeX code in between [.tex] and [./tex] tags (NOT including the dots)

You can look at this to get familiar with/copy paste the LaTeX code: http://www.codecogs.com/latex/eqneditor.php

HSC 2015 MX2 Marathon (archive) (2 Viewers)

Member

=o

not actually a porcupine

=o

Well-Known Member

=o

not actually a porcupine

=o

=o

not actually a porcupine

Well-Known Member

Active Member

Active Member

Well-Known Member

Well-Known Member

not actually a porcupine

This too shall pass

New Member

Pew Pew

This too shall pass

Users Who Are Viewing This Thread (Users: 0, Guests: 2)