Author Topic: 4U Maths Question Thread (Read 665221 times) Tweet Share

kaustubh.patel · « **Reply #1650 on:** January 17, 2018, 07:26:05 pm »

Hey Jake and all 4U lovers. Could I please get some help with some 4U graphing.
I'm quite confused on approaching the absolute value one (| |y|-|x| |=1) and I'd prefer to learn how to do the second one (y=x^3/(x^2-9)) without using calculus.

RuiAce · « **Reply #1651 on:** January 17, 2018, 07:36:13 pm »

Quote from: kaustubh.patel on January 17, 2018, 07:26:05 pm

Hey Jake and all 4U lovers. Could I please get some help with some 4U graphing.
I'm quite confused on approaching the absolute value one (| |y|-|x| |=1) and I'd prefer to learn how to do the second one (y=x^3/(x^2-9)) without using calculus.

$\text{By inspection we have an }x\text{-intercept at the origin}\\ \text{and vertical asymptotes at }x=\pm 3.$
$\text{We have}\\ \begin{align*}\frac{x^3}{x^2-9}&=\frac{x^3-9x}{x^2-9}+\frac{9x}{x^2-9}\\ &= x + \frac{9x}{x^2-9}\end{align*}$
$\text{As }x\to \infty, \frac{9x}{x^2-9}\to 0.\\ \text{Hence there will be an oblique asymptote in }y=x.$
This should be enough information to sketch the curve.
_______________________________________________
$\text{Alternatively, if one wishes to stick exclusively to 4U techniques,}\\ \text{it is easy to sketch }y=x^2-9.\\ \text{From there, use the transformation to sketch }y=\frac{1}{x^2-9}$
Recall key aspects of the graph of $y = \frac{1}{f(x)} $. $x$-intercepts and vertical asymptotes interchange, the nature of stationary points are inverted and etc.
$\text{It is also easy to sketch }y=x^3.\\ \text{From there, you would use multiplication of ordinates to sketch }y=\frac{x^3}{x^2-9}$
Of course, multiplication of ordinates will not guarantee a fully accurate graph. But it is the best compromise in only 4U methods.

RuiAce · « **Reply #1652 on:** January 17, 2018, 07:52:30 pm »

$\text{Recall the definition of the absolute value.}\\ \text{We will be using this throughout the entire question.}$

Definition

$|a| = \begin{cases}a&\text{if }a\ge 0\\ -a &\text{if }a < 0\end{cases}$

$\text{One may explicitly check that the }x\text{ and }y\text{ intercepts}\\ \text{will be at the points }(0,\pm1)\text{ and }(\pm1,0)$
$\text{Consider a quadrant-by-quadrant analysis.}$
$\text{In the first quadrant, }x > 0\text{ and }y > 0\\ \text{Hence}\\ \begin{align*}\big| |y|-|x|\big|&=1\\ |y-x|&=1\\ y-x&=\pm 1\\ y&=x\pm 1\end{align*}$
So in the first quadrant, you will sketch the pair of parallel lines $ y = x+1$ and $y = x-1$.
$\text{In the second quadrant, }x<0\text{ and }y>0\\ \text{Hence}\\ \begin{align*}\big| |y|-|x| \big|&= 1\\ |y+x|&=1\\ y+x&=\pm 1\\ y&=-x\pm 1\end{align*}$
So in the second quadrant, you will sketch the pair of parallel lines $y=-x-1$ and $y=-x+1$
$\text{In the third quadrant, }x<0\text{ and }y<0\\ \text{Hence}\\ \begin{align*}\big| |y|-|x|\big| &= 1\\ |-y+x|&=1\\ -y+x&=\pm 1\\ y&=x\mp 1\end{align*}$
So in the third quadrant, sketch the same pair of parallel lines as in the first quadrant. i.e. $y=x+1$ and $y=x-1$.
$\text{A similar analysis shows that the fourth quadrant case}\\ \text{also coincides with the second quadrant case.}$
$\text{After sketching everything, you should end up with an X-shape graph.}$

kaustubh.patel · « **Reply #1653 on:** January 19, 2018, 08:33:36 pm »

Wow Rui your breakdown of the absolute value one and treating them for 4 separate quadrants was really helpful, I actually didn't use this method but I realise now how working backwards from the definition can make it much easier. The other question I prefer your method of finding the reciprocal of the parabola and multiplying y=x^3 thats much easier to visualise and do on the spot. Thank you heaps this will be pretty darn useful so much appreciated.

philgee · « **Reply #1654 on:** January 20, 2018, 06:00:31 pm »

Hey Rui! Not sure how to prove that Re(alpha)=1.

RuiAce · « **Reply #1655 on:** January 20, 2018, 06:03:41 pm »

Quote from: philgee on January 20, 2018, 06:00:31 pm

Hey Jake! Not sure how to prove that Re(alpha)=1.

$\text{Assume that one complex root is }\alpha\\ \text{and one rational root is }\beta.$
$\text{From the given information, another root must be }-\beta.\\ \text{Also, }\overline{\alpha}\text{ is the last root as the coefficients}\\ \text{of }P(x)\text{ are all real.}$
$\text{So by the sum of roots:}\\ \begin{align*}\beta + (-\beta) + \alpha + \overline{\alpha}&=\frac{32}{16}\\ 2 \text{Re}(\alpha)&=2\\ \text{Re}(\alpha)&=1\end{align*}$
Note: $ z + \overline{z} = 2 \text{Re}(z) $ is a known result from complex numbers. It can easily be proven by letting $ z = x+iy$.

philgee · « **Reply #1656 on:** January 20, 2018, 06:27:27 pm »

Thanks Rui! Your explanation was clear and concise!

philgee · « **Reply #1657 on:** January 20, 2018, 08:27:00 pm »

Hi rui!
I couldnt find the roots for this question

RuiAce · « **Reply #1658 on:** January 20, 2018, 09:56:50 pm »

Quote from: philgee on January 20, 2018, 08:27:00 pm

Hi rui!
I couldnt find the roots for this question

deleted materials

I got onto this later than when I first saw it because I was in the middle of an online meeting - soz

I'm still working on it, and having done some backtracking I realised that $k=8$ and the roots are $\pm \frac12, 1\pm \frac{i}2$. But all the ways that actually got me to the answer are pretty disastrous and not elegant, so I don't wanna post them just yet. Before I give up and just post something inelegant, can I get the source of the question?

I ended up finding a solution that's still not elegant, but at least it's less handwavy. I don't believe this was the intended approach but it will suffice.
$\text{Backtrack a bit.}\\ \text{When we did the product of roots, we found out that}\\ \boxed{|\alpha|^2\beta^2 = \frac5{16}}\implies |\alpha|^2 = \frac{5}{16\beta^2}$
$\text{Now, if we consider the sum of two roots at a time,}\\ \begin{align*}\alpha\overline{\alpha}+\alpha\beta-\alpha\beta+\overline{\alpha}\beta-\overline{\alpha}\beta-\beta^2&=1\\ |\alpha|^2-\beta^2&=1\end{align*}$
$\text{Of course, this is now just a system of simultaneous equations.}\\ \text{Subbing the earlier one into the new equation,}\\ \begin{align*}\frac{5}{16\beta^2}-\beta^2&=1\\ 16\beta^4+16\beta^2-5&=0\\ (4\beta^2-1)(4\beta^2+5)&=0\\ \beta^2&=\frac14, -\frac54\end{align*}$
$\text{But since }\beta\text{ is rational, it must obviously be real.}\\ \text{Hence }\beta^2=\frac14\text{ and we may without loss}\\ \text{of generality take }\beta = \frac12.$
$\text{Let }\alpha = 1+iy\text{ for some real }y.\\ \text{Subbing back into anything above gives}\\ |\alpha|^2 = \frac{1}{16} \implies 1+y^2 = \frac{1}{16}.\\ \text{This eventually gives us that }\alpha = 1+\frac{i}{2}\text{ or }\alpha = 1-\frac{i}{2}.$
From here, it's easy to finish the question off.

This particular approach is nice in that we don't have to find $k$. But I'm somewhat unsure on if we're not meant to find $k$ or if we actually were.

Jeeffffffffffffff · « **Reply #1659 on:** January 24, 2018, 01:42:37 pm »

Use the fact that a²+b²>=2ab to prove that:
i) (ab+xy)(ax+by)>=4abxy
ii) ax+by<=1 if a²+b²=1 and x²+y²=1

Just starting out with this stuff still and I think our teacher is just as confused as us, any help would be appreciated, thanks

RuiAce · « **Reply #1660 on:** January 24, 2018, 02:03:51 pm »

Quote from: Jeeffffffffffffff on January 24, 2018, 01:42:37 pm

Use the fact that a²+b²>=2ab to prove that:
i) (ab+xy)(ax+by)>=4abxy
ii) ax+by<=1 if a²+b²=1 and x²+y²=1

Just starting out with this stuff still and I think our teacher is just as confused as us, any help would be appreciated, thanks

$\text{Expand and regroup:}$
\begin{align*}(ab+xy)(ax+by)&=a^2bx+bxy^2+ab^2y+ayx^2\\ &= (a^2bx+bxy^2) + (ab^2y+ayx^2)\\ &= bx(a^2+y^2)+ay(b^2+x^2)\\ &\ge bx(2ay) + ay(2bx)\tag{using the given inequality}\\ &= 4abxy\end{align*}
Intuition behind this method - To use the identity, somehow I had to force out a bunch of squares. But it wasn't obvious what I could do until I had expanded it out in the first place.
__________________________________________

Note that I was only able to generate this proof after working backwards on scrap paper.
$\text{The given inequality tells us that also, }a^2+x^2\ge 2ax.$
$\text{From here, fudge in useful terms as follows.}\\ \begin{align*}-a^2-x^2&\le -2ax\\ 1-a^2-x^2+a^2x^2&\le 1-2ax+a^2x^2\\ (1-a^2)(1-x^2)&\le (1-ax)^2\end{align*}$
$\therefore b^2y^2\le (1-ax)^2\\ \text{Taking positive square roots on both sides,}\\ by\le 1-ax \implies \boxed{ax+by\le 1}$

Dragomistress · « **Reply #1661 on:** January 24, 2018, 04:34:41 pm »

May someone explain what has happened?

RuiAce · « **Reply #1662 on:** January 24, 2018, 04:37:00 pm »

Quote from: Dragomistress on January 24, 2018, 04:34:41 pm

May someone explain what has happened?

They just used the compound angle identities.
$\text{What they really mean is }\cos [(k+1)\theta]\\ \text{so just consider the compound angle expansion}\\ \cos[(k+1)\theta]= \cos(k\theta+\theta) = \cos k\theta\cos\theta-\sin k\theta\sin\theta$

maria.micale · « **Reply #1663 on:** January 24, 2018, 06:10:54 pm »

I've got this question from Fitzpatrick that I can't solve.

Given that x² + y² + z² >= xy + yz + zx

Show that:

1) a²b² + b²c² + c²a² >= abc(a +b +c)
2) (x + y + z)² >= 3(xy + yz + zx)

Thanks in advanced!!!

RuiAce · « **Reply #1664 on:** January 24, 2018, 06:16:00 pm »

Quote from: maria.micale on January 24, 2018, 06:10:54 pm

I've got this question from Fitzpatrick that I can't solve.

Given that x² + y² + z² >= xy + yz + zx

Show that:

1) a²b² + b²c² + c²a² >= abc(a +b +c)
2) (x + y + z)² >= 3(xy + yz + zx)

Thanks in advanced!!!

$\text{Let }x=ab, y=bc, z=ca.\\ \begin{align*}x^2+y^2=z^2&\ge xy+xz+yz\\ a^2b^2+b^2c^2+c^2a^2&\ge ab^2c+a^2bc+abc^2\\ &=abc(a+b+c)\end{align*}$
___________________________________
$\text{Recalling that }(x+y+z)^2 = x^2+y^2+z^2+2(xy+xz+yz)$
(this is actually a formula you used several times in the polynomials topic: \( \alpha^2+\beta^2+\gamma^2 = (\alpha+\beta+\gamma)^2-2(\alpha\beta+\alpha\gamma+\beta\gamma) )
$\text{subbing the given inequality in immediately gives}\\ (x+y+z)^2 = (xy+xz+yz)+2(xy+xz+yz)$

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Author Topic: 4U Maths Question Thread (Read 665221 times) Tweet Share

kaustubh.patel

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