Author Topic: 4U Maths Question Thread (Read 665319 times) Tweet Share

RuiAce · « **Reply #1680 on:** January 27, 2018, 10:01:41 pm »

Quote from: Dragomistress on January 27, 2018, 09:46:36 pm

How should I do this?

Sketch the locus given and then deduce the max and min values of $ \arg(z) $. Recall that $ \arg(z) $ measures the angle made at 0+0i with respect to the positive real axis.

Note that the argument is optimised (i.e. maximised and minimised) whenever the ray drawn is tangential to the circle you sketched earlier.

clovvy · « **Reply #1681 on:** January 27, 2018, 10:04:38 pm »

This is a locus question, the first thing you want to do is sketch the locus.... Iz-2iI=1... in this case let z=x+iy and then square root both sides than you will notice that it is a circle with radius 1 at the centre (0,2)...... and then to find arg(z)... Rui I can't remember exactly how to do that... -_-

RuiAce · « **Reply #1682 on:** January 28, 2018, 07:24:31 pm »

Deleted a previous reply - Now that I have more time I've decided to just do this one. Having said that though, still use the 4U trial survival handout as a reference point.
$|z-2i|=1\text{ is as stated,}\\ \text{a circle centred at }(0,2)\text{ with radius 1.}$
It may be worth noting that if we had $ |z-2i|\le 1 $, for this type of question we would ignore the disc's interior. This is because the minimum and maximum of both $ |z| $ and $\arg z$ can only be attained on the disc's boundary, i.e. the actual circle. It just so happens that for the argument, we only consider when it's tangential to the circle.

$\text{Sketch tangents from }(0,0)\text{ to the circle.}\\ \text{Here, I will demonstrate how to find the min.}\\ \text{The max is left as your exercise.}$
Note, however, that the max can also be found from symmetry, once you know the min,
$\text{Let }C\text{ be the centre of the circle and}\\ P\text{ be the complex number }z\text{ at the point of intersection}\\ \text{between the circle and its right-hand tangent.}\\ \text{Put }Q\text{ as some random point on the positive }x\text{-axis just for reference.}$
$\text{Note, then, that }\arg z = \angle POQ.\\ \text{So this is the angle we must find.}$
_________________________________________________
$CP = 1\text{ as it is the length of the radius.}\\ CO = 2\text{ just because }C\text{ is two units vertically above the origin.}\\ \text{But also, }\angle CPO = \frac\pi2\text{ as the radius is perpendicular}\\ \text{to the tangent drawn to point of contact.}$
$\text{Hence, using right-angled trigonometry,}\\ \sin \angle COP = \frac{1}{2}\implies \angle COP = \frac\pi6.$
$\text{But obviously }\angle COQ = \frac\pi2.\\ \text{Hence, }\angle POQ =\boxed{\arg z = \frac\pi 3}$

RustyWasTaken · « **Reply #1683 on:** January 29, 2018, 01:32:28 am »

Cheers

RuiAce · « **Reply #1684 on:** January 29, 2018, 09:22:04 am »

Quote from: RustyWasTaken on January 29, 2018, 01:32:28 am

Cheers

$\text{This is just replacing }x\text{ with }x+k\\ \text{which gives }a(x+k)^3+b(x+k)^2+c(x+k)+d.\\ \text{Expand and simplify.}$

Jeeffffffffffffff · « **Reply #1685 on:** January 31, 2018, 01:34:45 pm »

Show that if x>0, y>0 then
1/x²+1/y²>=8/(x+y)²

Please help, fully worked solution would be appreciated.

maria.micale · « **Reply #1686 on:** January 31, 2018, 01:36:33 pm »

Using expansion (x+ y + z)³, show that (x + y + z)³ >= 27xyz

I've written out expansion but cannot solve question. Any help appreciated.

Opengangs · « **Reply #1687 on:** January 31, 2018, 01:48:43 pm »

Quote from: Jeeffffffffffffff on January 31, 2018, 01:34:45 pm

Show that if x>0, y>0 then
1/x²+1/y²>=8/(x+y)²

Please help, fully worked solution would be appreciated.

$(x - y)^2 \geq 0 \\ x^2 + y^2 \geq 2xy \text{ (*)}$
$\frac{1}{x^2} + \frac{1}{y^2} = \frac{x^2 + y^2}{(xy)^2} \\ \text{Since } x^2 + y^2 \geq 2xy \text{ then: } \\ \frac{1}{x^2} + \frac{1}{y^2} \geq \frac{2xy}{(xy)^2} \\ \frac{1}{x^2} + \frac{1}{y^2} \geq \frac{2}{xy} \\ \frac{1}{x^2} + \frac{1}{y^2} \geq \frac{2(x + y)^2}{xy(x + y)^2} \\ \frac{1}{x^2} + \frac{1}{y^2}<br />\geq \frac{2(x^2 + y^2 + 2xy)}{xy(x + y)^2}$
$\text{By (*): } \frac{2(x^2 + y^2 + 2xy)}{xy(x + y)^2} \geq \frac{2(2xy + 2xy)}{xy(x + y)^2} \\ \therefore \frac{1}{x^2} + \frac{1}{y^2} \geq \frac{2(2xy + 2xy)}{xy(x + y)^2} \\ \text{You can complete the proof from there.}$

RuiAce · « **Reply #1688 on:** January 31, 2018, 01:49:06 pm »

Quote from: Jeeffffffffffffff on January 31, 2018, 01:34:45 pm

Show that if x>0, y>0 then
1/x²+1/y²>=8/(x+y)²

Please help, fully worked solution would be appreciated.

$\text{The proof that }\boxed{\frac{a+b}{2}\ge \sqrt{ab}}\text{ for }a > 0, b > 0\\ \text{ is left as your exercise.}$
$\text{Let }a = \frac{1}{x^2}\text{ and }b = \frac{1}{y^2}.\\ \text{We first have }\boxed{\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{2}{xy}}$
$\text{Now, from }(x-y)^2\ge 0\text{ we also have}\\ \begin{align*}x^2-2xy+y^2&\ge 0\\ x^2+2xy+y^2&\ge 4xy\\ (x+y)^2&\ge 4xy\\ \frac{1}{xy}&\ge 4(x+y)^2\end{align*}\\ \text{noting that }x, y > 0.$
$\text{And hence, }\boxed{\frac{2}{xy} \ge \frac{8}{(x+y)^2}}\\ \text{so combining with the inequality from earlier, }\boxed{\frac{1}{x^2}+\frac{1}{y^2}\ge \frac{8}{(x+y)^2}}$

To provide some intuition:

Working backwards did not help at all here. As a more advanced technique, you should have proven $ a+b \ge 2\sqrt{ab} $ several times if you're up to questions of this calibre, so just assume that it's true and work with it. This was beneficial in collapsing the $x^{-2}+y^{-2} $ into $ 2(xy)^{-1} $.

Now that we have something more manageable, we can work backwards again. It turns out that THIS time, working backwards actually got something relatively nice. So we use it.

RuiAce · « **Reply #1689 on:** January 31, 2018, 01:54:16 pm »

Quote from: maria.micale on January 31, 2018, 01:36:33 pm

Using expansion (x+ y + z)³, show that (x + y + z)³ >= 27xyz

I've written out expansion but cannot solve question. Any help appreciated.

$\text{As above, we will use }\boxed{\frac{a+b}{2}\ge \sqrt{ab}}$
\begin{align*}&\quad (x+y+z)^3\\&= x^3+y^3+z^3 + 3(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y) + 6xyz\\ &= (x^3+y^3+z^3) + 3z(x^2+y^2) + 3y(x^2+z^2) + 3x(y^2+z^2) + 6xyz\\ &\ge (x^3+y^3+z^3) + 6zxy + 6yxz + 6xyz + 6xyz\\ &= x^3+y^3+z^3 + 24xyz \end{align*}
$\text{It remains now to show that }x^3+y^3+z^3 \ge 3xyz\\ \text{after which, we will be done.}$
This is really just the AM-GM inequality with three variables. The AM-GM inequality with four variables is easy to prove, and you can use that one to prove the version with three variables.

To provide some intuition: Since we have to expand a cube involving three terms to prove something with $xyz$ in the RHS, it was already extremely likely that $ a+b+c\ge 3\sqrt[3]{abc} $ would be useful (by simply letting $a = x^3$ and etc.). But this is quite tedious to prove, and there's still no guarantee that it'd do all the magic. So I started by considering every other term.

The terms $x^2y + xy^2 + x^2z+xz^2+y^2z+yz^2 $ all took the form something-squared plus something. Just like in the above case, $a+b \ge 2\sqrt{ab} $ could have been really useful. But I didn't want renegade square roots appearing because there's no square root at all in what I'm trying to prove.
Since a square and square root cancel each other out when dealing with positive numbers, it made sense to apply that on the squares instead. Hence, I used $x^2+y^2\ge 2xy $ instead of $ x+y \ge 2\sqrt{xy} $.

But of course, I had to rearrange the terms and do some factorising before I could even think of using that formula.

It was from there, I saw that I suddenly had almost all of the $xyz$ in place (I already had 24 of them, and only needed 3 more). Well, the last 3 takes effort to get at, but at least I know I could get there

Lefkiiii6 · « **Reply #1690 on:** January 31, 2018, 02:18:49 pm »

Prove that, if a>0, b>0, a⁴ + b⁴ > a³b + ab₃ by writing expansion (a-b)₄

Using x³+y³>=(x/z+y/z)xyz and similar expressions for y³+z³ and z³+x³ to deduce that x³+y³+z³>=3xyz.

These are two questions I am having trouble with. Can anyone help?

RuiAce · « **Reply #1691 on:** January 31, 2018, 02:24:20 pm »

Quote from: Lefkiiii6 on January 31, 2018, 02:18:49 pm

Prove that, if a>0, b>0, a⁴ + b⁴ > a³b + ab₃ by writing expansion (a-b)₄

Using x³+y³>=(x/z+y/z)xyz and similar expressions for y³+z³ and z³+x³ to deduce that x³+y³+z³>=3xyz.

These are two questions I am having trouble with. Can anyone help?

That second one is another famous way of proving the three-variable AM-GM inequality.
$\text{We add the following three:}\\ \begin{align*}x^3+y^3 &\ge \left(\frac{x}{z} + \frac{y}{z} \right) xyz\\ x^3+z^3 &\ge \left( \frac{x}{y} + \frac{z}{y}\right) xyz\\ y^3+z^3 &\ge \left(\frac{y}{x}+\frac{z}{x} \right)xyz\end{align*}$
$\text{and after doing some regrouping, obtain:}\\ \boxed{2(x^3+y^3+z^3) \ge xyz\left(\frac{x}{y}+\frac{y}{x}\right) + xyz\left( \frac{x}{z}+\frac{z}{x}\right) + \left( \frac{y}{z} + \frac{z}{y}\right) }$
$\text{However, observe that:}\\ \begin{align*}(x-y)^2&\ge 0\\ x^2-2xy+y^2&\ge 0\\ x^2+y^2&\ge 2xy\\ \frac{x^2+y^2}{xy}&\ge \frac{2xy}{xy}\\ \frac{x}{y}+\frac{y}{x}&\ge 2\end{align*}$
$\text{So subbing this (and similar expressions) in:}\\ 2(x^3+y^3+z^3)\ge 2xyz + 2xyz + 2xyz\\ \text{and thus }\boxed{x^3+y^3+z^3\ge 3xyz}$
_______________________________________________________________
$\text{From the binomial theorem,}\\ (a-b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4.\\ \text{And, in a similar way to how }(a-b)^2 \ge 0\\ \text{we have }(a-b)^4 \ge 0.$
$\text{Combining these two and rearranging gives}\\ \boxed{a^4+b^4 \ge 4(a^3b+ab^3) -6a^2b^2}$
$\text{Now, recalling that }x+y\ge 2\sqrt{xy},\\ \text{putting }x=a^2\text{ and }y=b^2\text{ gives}\\ a^2+b^2\ge 2ab\\ \text{and consequently, }\boxed{-6ab \ge -3(a^2+b^2)}$
$\text{So subbing that into the above:}\\ \begin{align*}a^4+b^4 \ge 4(a^3b+ab^3) - 6a^2b^2 &= 4(a^3b+ab^3) - (6ab)ab\\ &\ge 4(a^3b+ab^3) -3(a^2+b^2)ab\\ &= 4(a^3b+ab^3)-3(a^3b+ab^3)\\ &= a^3b+ab^3\end{align*}\\ \text{as required.}$

Lefkiiii6 · « **Reply #1692 on:** January 31, 2018, 09:14:40 pm »

Ok thanks heaps for all that Rui and all your help in the past! I have another question. We didn't really cover cube roots of unity or roots of unity in general so I'm unsure about this stuff.

If 1, w1 and w2 are the cube roots of unity, prove that:

w1 = conjugate of w2 = conjugate of (w2)²
w1 + w2 = -1
w1w2=1

RuiAce · « **Reply #1693 on:** February 01, 2018, 01:26:32 pm »

Quote from: Lefkiiii6 on January 31, 2018, 09:14:40 pm

Ok thanks heaps for all that Rui and all your help in the past! I have another question. We didn't really cover cube roots of unity or roots of unity in general so I'm unsure about this stuff.

If 1, w1 and w2 are the cube roots of unity, prove that:

w1 = conjugate of w2 = conjugate of (w2)² - Also, this bit is a mistake. It is equal to just $w_2\,^2$; no conjugate.
w1 + w2 = -1
w1w2=1

$\text{Sticking within the complex numbers topic (and not jumping into polynomials)}\\ \text{the first one is really best done by actually solving the equation }\boxed{z^3 = 1}$
Essentially, what the "cube roots of unity" means is just the "cube roots of 1". Hence the equation above.
$\text{Solving will give us the three roots}\\ 1, \text{ cis }\frac{2\pi}{3}, \text{ cis }\frac{4\pi}{3}.\\ \text{You should be familiar with how that process works at least.}$
$\text{But then you can just recognise that}\\ \begin{align*}\text{ cis }\frac{2\pi}{3} &= -\frac{1}{2} + \frac{\sqrt3}{2}i\\ \text{ cis }\frac{4\pi}{3} &= -\frac{1}{2}-\frac{\sqrt3}{2}i\end{align*}\\ \text{which are clearly conjugates of each other.}$

Alternately, if we choose to use polynomials, and thus not explicitly evaluate w1 and w2

$\text{The equation }z^3-1=0\text{ can be factorised to give}\\ (z-1)(z^2+z+1) = 0.$
$\text{Since }1, w_1\text{ and }w_2\text{ are the roots of }z^3-1=0,\\ \text{because }1\text{ is the only root of }z-1=0,\\ w_1\text{ and }w_2\text{ must both be the roots of }z^2+z+1=0.$
$\text{But }z^2+z+1=0\text{ obviously has no real roots (can check by the determinant)}\\ \text{so by the conjugate root theorem, }w_1 = \overline{w_2}.$

Also, you can just square $w_2$ by hand, and you will see that it equals to $w_1$.
________________________________________
$\text{Now, once we know that }w_2^2 = w_1,\\ \text{we can just rewrite the roots as }1, w\text{ and }w^2$
(where $w = w_1$)
$\text{Of course, since }w\text{ is a root of }z^3-1,\\ \text{satisfying the added property that }w\neq 1,\\ \text{we can sub that in to obtain }\boxed{w^3 - 1= 0}$
$\text{Factorising this gives }\boxed{(w-1)(w^2+w+1)=0}\\ \text{Using the fact that }w\neq 1,\text{ we may then divide by }w-1\text{ to obtain}\\ \boxed{w^2+w+1 = 0}$
$\text{Moving the }-1\text{ to the other side gives }\boxed{w^2 + w = -1}\\ \text{which is actually exactly what we needed.}$

Again, alternatively if we choose to use polynomials

This is easy to prove by just using the sum of roots.

The same goes for the product of roots for part 3.

The third part can be done by observing that $w_1 w_2 = w w^2 = w^3 = 1$

Edit: Tbh some bits of that might’ve been explained poorly. The thing to realise, however, is that w2^2 = w1 underlies a lot of the things we do with the cube roots of unity. This is why we usually never use w1 and w2, but prefer w and w^2 instead.

stesoo · « **Reply #1694 on:** February 01, 2018, 09:20:53 pm »

Hi i need help with this q,
for a real number r, the polynomial 8x^3-4x^2-42x+45 is divisible by (x-r)^2. Find the value of r.

thanks

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