Author Topic: 4U Maths Question Thread (Read 665259 times) Tweet Share

maria.micale · « **Reply #1620 on:** December 29, 2017, 07:18:30 am »

Quote from: RuiAce on December 28, 2017, 09:20:40 pm

Hint: The fact that $k$ is real means that you can do some rearranging, and then equate the real and imaginary parts. The real parts should give you a quadratic...

Alright thanks heaps Rui!!! I got it an everything now from doing what you said but now Im wondering how would you go about solving that polynomial I had before with complex coefficients where I just subbed in ki where there was a z? Could you tell me how you would solve it because I don't think I have ever tried one with complex coefficients that wasn't a quadratic or quartic that could be reduced to a quadratic.

RuiAce · « **Reply #1621 on:** December 29, 2017, 10:04:33 am »

Quote from: maria.micale on December 29, 2017, 07:18:30 am

Alright thanks heaps Rui!!! I got it an everything now from doing what you said but now Im wondering how would you go about solving that polynomial I had before with complex coefficients where I just subbed in ki where there was a z? Could you tell me how you would solve it because I don't think I have ever tried one with complex coefficients that wasn't a quadratic or quartic that could be reduced to a quadratic.

Oh, I wouldn't have.

There's no giveaway at all when it's just some arbitrary cubic or quartic. The HSC always gives you some kind of hint to make sure everything else falls out nicely, but in general there's no 'clean' way of solving a quartic whatsoever (try Google searching the quartic formula, it's bizarre).

stesoo · « **Reply #1622 on:** December 29, 2017, 10:49:38 am »

hi i have a question

how can I prove that (1+sintheta +icostheta)/(1+sintheta-icostheta) = sintheta + icostheta
thank you

RuiAce · « **Reply #1623 on:** December 29, 2017, 11:17:53 am »

Quote from: stesoo on December 29, 2017, 10:49:38 am

hi i have a question
how can I prove that (1+sintheta +icostheta)/(1+sintheta-icostheta) = sintheta + icostheta
thank you

\begin{align*}\frac{1+\sin\theta+i\cos\theta}{1+\sin\theta-i\cos\theta}&= \frac{1+i\text{ cis }(-\theta)}{1-i\text{ cis }\theta}\\ &= \frac{\text{ cis }\left(-\frac{\theta}{2}\right)}{\text{ cis }\frac{\theta}{2}}\times \frac{\text{ cis }\frac{\theta}{2}+i\text{ cis }\left(-\frac\theta2\right)}{\text{ cis }\left(-\frac\theta2\right)-i\text{ cis }\frac\theta2}\\ &= \text{cis }(-\theta)\times \frac{\cos \frac\theta2+i\sin \frac\theta2 + i \left(\cos \frac\theta2-i\sin \frac\theta2\right)}{\cos \frac\theta2-i\sin \frac\theta2-i\left(\cos\frac\theta2+i\sin \frac\theta2\right)}\\ &= \text{cis }(-\theta) \times \frac{(1+i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}{(1-i)\left(\cos \frac\theta2+\sin \frac\theta2\right)}\\ &= (\cos\theta - i\sin \theta) \times i\\ &= \sin \theta + i \cos \theta\end{align*}

Jeeffffffffffffff · « **Reply #1624 on:** December 29, 2017, 01:30:20 pm »

Find constants a,b,c such that the polynomial p(x) = x³-6x²+11x-13 is expressible as X³+aX+b where X=x-c. Hence show that the equation p(x)=0 has only one root.
I found a, b, and c quite easily but unsure of how to show that the equation has only one root?

RuiAce · « **Reply #1625 on:** December 29, 2017, 01:46:39 pm »

Quote from: Jeeffffffffffffff on December 29, 2017, 01:30:20 pm

Find constants a,b,c such that the polynomial p(x) = x³-6x²+11x-13 is expressible as X³+aX+b where X=x-c. Hence show that the equation p(x)=0 has only one root.
I found a, b, and c quite easily but unsure of how to show that the equation has only one root?

$\text{In general, the number of roots of }p(x) = 0\\ \text{will be the same as the number of roots of }p(x-c) = 0\\ \text{because the graph of }y=p(x-c)\text{ is just a horizontal translation of that of }y=p(x).$
You should be well aware of the fact that this is not 4U material as it requires you to pick the values for $a, b$ and $c$ yourself, without any guidance whatsoever.
$\text{If you applied Cardano's method correctly, you would have let }X = x+2\text{, i.e. taken }c=-2\\ \text{which gives you }p(X) = X^3-X-7.\\ \text{It now falls to determine the number of real roots this polynomial has.}$
________________________________________________________
$\text{Consider }f(x) = x^3-x-7 \implies f^\prime(x) = 3x^2 - 1.\\ \text{We obtain a local maximum at }\left(-\frac{1}{\sqrt{3}}, \frac{2}{3\sqrt{3}}-7\right)\\ \text{and a local minimum at }\left(\frac{1}{\sqrt3}, -\frac{2}{3\sqrt3} - 7\right)$
$\text{A quick sketch is then sufficient to show that }f(x)=0\text{ only has one root,}\\ \text{which is a direct consequence of the }y\text{-coordinate of the local maximum being negative.}$
$\text{Hence }p(X) = 0\text{ has only one solution}\\ \text{and thus from the result at the start, so does }p(x)=0$

maria.micale · « **Reply #1626 on:** December 29, 2017, 03:10:40 pm »

Hi I am not sure where to begin with these questions.

RuiAce · « **Reply #1627 on:** December 29, 2017, 03:39:59 pm »

Deleted materials

$\text{If }\alpha,\beta,\gamma,\delta\text{ are the roots to }ax^4+bx^3+cx^2+dx+e=0\\ \text{Then by the sum and product of roots we have}\\ \begin{align*}\alpha+\beta+\gamma+\delta &= -\frac{b}{a}\\ \alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta &= +\frac{c}{a}\\ \alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta &= -\frac{d}a\\ \alpha\beta\gamma\delta &= +\frac{e}{a}\end{align*}$
$\text{We now }\textbf{build}\text{ the polynomial whose roots are }\frac1\alpha, \frac1\beta, \frac1\gamma, \frac1\delta\\ \text{by }\textbf{reverse-engineering}\text{ these identities.}$
$\text{Sum of 1 root at a time:}\\ \begin{align*}\frac1\alpha+\frac1\beta+\frac1\gamma+\frac1\delta &= \frac{\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta}{\alpha\beta\gamma\delta}\\ &= \frac{-d/a}{e/a}\\ &= -\frac{d}{e}\end{align*}$
$\text{In a similar way, the sum of 2 roots at a time is }\frac{c}{e}\\ \text{The sum of 3 roots at a time is }-\frac{b}{e}\\ \text{and the product of roots is }\frac{a}{e}$
$\text{Hence the degree-4 polynomial with these roots will be}\\ ex^4+dx^3+cx^2+bx+a=0$
$\text{Dividing both sides by }x^4,\\ \text{the equation }e+\frac{d}{x}+\frac{c}{x^2}+\frac{b}{x^3}+\frac{a}{x^4}=0,\text{ which is exactly }P\left(\frac1x\right)=0,\\ \text{also has roots }\frac1{\alpha}\text{ etc. as required.}$
The rest should be doable similarly (except maybe part d) - I'm not sure if that requires a different approach but I don't have time to think about it just yet). A remark that for part b), an assumption is made that none of the roots are 0. This won't be necessary for the other parts.

$\text{Actually, scratch that.}$
$\text{For b), we can just do it like this.}\\ \text{If the roots of }ax^4+bx^3+cx^2+dx+e=0\text{ are }x=\alpha,\beta,\gamma,\delta$
$\text{Then the roots of }a\left(\frac1x\right)^4+b\left(\frac1x\right)^3+c\left(\frac1x\right)^2+d\left(\frac1x\right)+e=0\\ \text{must therefore be }\frac{1}{x}=\alpha,\beta,\gamma,\delta$
$\text{Which consequently implies }x=\frac1\alpha,\frac1\beta,\frac1\gamma,\frac1\delta\\ \text{as required.}$
This approach should certainly be adaptable for part d). It is essentially equivalent to performing a $u$-substitution, where $u = \frac1x$, and then working from there.

maria.micale · « **Reply #1628 on:** December 29, 2017, 07:51:25 pm »

Quote from: RuiAce on December 29, 2017, 01:46:39 pm

$\text{In general, the number of roots of }p(x) = 0\\ \text{will be the same as the number of roots of }p(x-c) = 0\\ \text{because the graph of }y=p(x-c)\text{ is just a horizontal translation of that of }y=p(x).$
You should be well aware of the fact that this is not 4U material as it requires you to pick the values for $a, b$ and $c$ yourself, without any guidance whatsoever.
$\text{If you applied Cardano's method correctly, you would have let }X = x+2\text{, i.e. taken }c=-2\\ \text{which gives you }p(X) = X^3-X-7.\\ \text{It now falls to determine the number of real roots this polynomial has.}$
________________________________________________________
$\text{Consider }f(x) = x^3-x-7 \implies f^\prime(x) = 3x^2 - 1.\\ \text{We obtain a local maximum at }\left(-\frac{1}{\sqrt{3}}, \frac{2}{3\sqrt{3}}-7\right)\\ \text{and a local minimum at }\left(\frac{1}{\sqrt3}, -\frac{2}{3\sqrt3} - 7\right)$
$\text{A quick sketch is then sufficient to show that }f(x)=0\text{ only has one root,}\\ \text{which is a direct consequence of the }y\text{-coordinate of the local maximum being negative.}$
$\text{Hence }p(X) = 0\text{ has only one solution}\\ \text{and thus from the result at the start, so does }p(x)=0$

How do we know that p(x-c) is simply a horizontal translation of p(x)?

RuiAce · « **Reply #1629 on:** December 29, 2017, 08:02:54 pm »

Quote from: maria.micale on December 29, 2017, 07:51:25 pm

How do we know that p(x-c) is simply a horizontal translation of p(x)?

$\text{This is just algebra.}\\ \text{Suppose we have the graph of }y=f(x).\\ \text{Any arbitrary point }x_0\text{ as an }x\text{-coordinate goes with }f(x_0)\text{ as its corresponding }y\text{-coordinate.}$
And thus, the point $( x_0, f(x_0) )$ lies on $y=f(x)$.
$\text{Now, suppose we wish to obtain the graph of }y=f(x-c).\\ \text{If we wanted to get the same }y\text{-coordinate as above, we can't just let }x=x_0\text{ anymore}\\ \text{because then }y=f(x_0 - c)$
$\text{But if we let }\boxed{x=x_0+c}\\ \textbf{which is just the point }c\textbf{ units to the right of }x_0,\\ y=f(x_0 + c - c) = f(x_0)\text{ at that point.}$
Hence, the point $ (x_0+c, f(x_0) ) $ lies on $y=f(x-c) $.
$\text{And because we can do this for }\textbf{any}\text{ arbitrary choice of real number }x_0\\ \text{it follows that }\textbf{for all}\text{ real numbers }x_0,\\ \text{the point with }y\text{-coordinate }f(x_0)\text{ ends up translated to the right by }c\text{ units.}$
$\text{So }y=f(x - c)\textbf{ must}\text{ therefore be the translation of }y=f(x)\text{ to the right by }c.$

maria.micale · « **Reply #1630 on:** January 03, 2018, 11:47:10 am »

Hi, I've done part a) and the answer is c = 1/27 a(9b -2a^2) but I really don't understand how to go about b). The answer is C - k = A/27 (9(B-m) -2A^2 ) fixed point has x-coordinate -A/3. Thanks in advanced!!!

RuiAce · « **Reply #1631 on:** January 03, 2018, 12:03:17 pm »

Quote from: maria.micale on January 03, 2018, 11:47:10 am

Hi, I've done part a) and the answer is c = 1/27 a(9b -2a^2) but I really don't understand how to go about b). The answer is C - k = A/27 (9(B-m) -2A^2 ) fixed point has x-coordinate -A/3. Thanks in advanced!!!

I'll start you off. This certainly involves using part a) somehow.
$\text{Consider the significance of the points being 'equidistant'.}\\ \text{The distance from the first to the second point, must be equal to}\\ \text{the distance from the second to the third point.}$
$\text{This means that the second point}\\ \textbf{must be the midpoint}\text{ of the first two.}\\ \text{Hence, the }x\text{ coordinates of the }\textbf{points of intersection}\\ \text{are what's in arithmetic progression.}$
$\text{If we tried to locate the points of intersection, upon equating we would arrive at}\\ x^3+Ax^3+(B-m)x+(C-k) = 0.\\ \text{Now try applying the previous part to this.}$

maria.micale · « **Reply #1632 on:** January 04, 2018, 12:44:22 pm »

Quote from: RuiAce on January 03, 2018, 12:03:17 pm

I'll start you off. This certainly involves using part a) somehow.
$\text{Consider the significance of the points being 'equidistant'.}\\ \text{The distance from the first to the second point, must be equal to}\\ \text{the distance from the second to the third point.}$
$\text{This means that the second point}\\ \textbf{must be the midpoint}\text{ of the first two.}\\ \text{Hence, the }x\text{ coordinates of the }\textbf{points of intersection}\\ \text{are what's in arithmetic progression.}$
$\text{If we tried to locate the points of intersection, upon equating we would arrive at}\\ x^3+Ax^3+(B-m)x+(C-k) = 0.\\ \text{Now try applying the previous part to this.}$

Ok thanks Rui, that makes sense, any further hints?

RuiAce · « **Reply #1633 on:** January 04, 2018, 01:05:34 pm »

Quote from: maria.micale on January 04, 2018, 12:44:22 pm

Ok thanks Rui, that makes sense, any further hints?

So that gives you the weird expression involving $C-k$ and according to your answers we can stop there. Which is reasonable, because we now have something that links $m$ and $k$ together.

The second point is actually the fixed point. If you let the first, second and third points have $x$-coordinate $\alpha-d, \alpha, \alpha+d$, you'll see that $\alpha = -\frac{A}{3} $

Jeeffffffffffffff · « **Reply #1634 on:** January 04, 2018, 05:46:20 pm »

Use De Moivre’s theorem to express cos5theta and sin5theta in powers of costheta and sintheta. Hence, express tan5theta as a rational function of t where t=tantheta and deduce that tan(Pi/5)tan(2pi/5)tan(3pi/5)tan(4pi/5)=5.

I’ve gotten up to having
Tan5theta= (t⁵-10t³+5t)/(5t⁴-10t²+1)
But I don’t really know how I’m supposed to use that to deduce the last line in the question. Any help would be appreciated, thanks.

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