Author Topic: 4U Maths Question Thread (Read 665491 times) Tweet Share

RuiAce · « **Reply #270 on:** June 23, 2016, 01:23:54 pm »

Quote from: katherine123 on June 23, 2016, 11:41:13 am

how do i do part iii)

i got
Part i) h=[(root3)*a]/2
part ii) ZY=xa/b

$\textbf{1998 HSC 4U Additional}$
$\text{Regard the SIDE-view of the sandstone.}$

$\text{Regard the right-angled triangular portion of the trapezium. By a similar triangle analysis:}\\ \begin{align*}\frac{h-\frac{\sqrt{3}}{{2}}a}{a-\frac{\sqrt{3}}{2}a}&=\frac{x}{b}\\ \iff h-\frac{\sqrt{3}}{{2}}a&=\frac{x}{b}\left(\frac{(2-\sqrt{3})a}{2}\right)\\ \iff h&=\frac{a}{2}\left(\sqrt{3}+(2-\sqrt{3})\frac{x}{b}\right)\end{align*}\\ Q.E.D.$

RuiAce · « **Reply #271 on:** June 23, 2016, 01:49:22 pm »

Quote from: jamonwindeyer on June 23, 2016, 10:35:51 am

I'll let Rui correct me if this isn't quite 4 Unit friendly, but this is actually very simple if we know the formula for a sphere in three dimensions (x,y,z):

$x^2+y^2+z^2=R^2$

At the cross section, the height above the xy plane (the z-value) is h, so we can substitute:

$x^2+y^2+h^2=R^2 \quad \therefore x^2+y^2=R^2-h^2$

What we notice here is that this is actually the formula for a circle. We've just found the formula for the outer circle of the cross section, with that term on the right being the square of the radius like normal. So:

$A=\text{Area of Bigger Circle} - \text{ Area of Smaller Circle} \\ = \pi(R^2-h^2)-\pi r^2 \\ = \pi(R^2-h^2-r^2)$

Let me know if this makes sense! And Rui might come along and give a better method, I'm not sure if 3-dimensional surface equations are what would be the normal approach

Haha don't even need a z-axis
$\textbf{2001 HSC Mathematics Ext 2}$
$\text{Analyse the following right-angled triangle}$

$\text{Keep in mind that }R\text{ is the radius of the actual sphere.}\\ \text{On the contrary, }a\text{ is the length involved in the area.}$
$\text{Note that }a>r\\ \text{The area of the annulus cross-section is as Jamon pointed:}\\ \begin{align*}A&=\pi a^2 - \pi r^2\\ &= \pi (R^2-h^2) - \pi r^2 \tag{Pythagoras' Theorem}\\ &= \pi (R^2-h^2-r^2)\end{align*}$

jamonwindeyer · « **Reply #272 on:** June 23, 2016, 02:14:51 pm »

Quote from: RuiAce on June 23, 2016, 01:49:22 pm

Haha don't even need a z-axis

Cheers Rui, that's much easier, too much 3-dimensional calculus has spoiled me

jakesilove · « **Reply #273 on:** June 23, 2016, 03:45:57 pm »

Quote from: jamonwindeyer on June 23, 2016, 02:14:51 pm

Cheers Rui, that's much easier, too much 3-dimensional calculus has spoiled me

Yeah, for better or worse you don't do any 3-dimensional work in Extension 2 (probably better, the course is hard enough)

amandali · « **Reply #274 on:** June 24, 2016, 12:32:20 pm »

does this require comparison of areas

RuiAce · « **Reply #275 on:** June 24, 2016, 01:25:19 pm »

Quote from: amandali on June 24, 2016, 12:32:20 pm

(Image removed from quote.)

does this require comparison of areas

Can't see what areas to compare. I would've just attacked it with an analysis of monotonic behaviour.

(Note that this proof is a bit informal. You may have to verify it's suitability with your teacher)
$\text{First part: Let }f(x)=\arctan{x}-x+\frac{x^3}{2}\\ \implies f^\prime (x) = \frac{1}{1+x^2}-1+\frac{3x^2}{2}> 0 \text{ for all }x>0\\ \text{Hence }f(x)\text{ is monotonic increasing}$
$f(0)=\arctan{0}-0+\frac{0^3}{2}=0\\ \text{Hence, excluding the point }x=0, f(x)\text{ increases from 0}\\ \therefore f(x)>0 \, \forall \, x>0$
$\text{That is, when }x>0, \arctan{x}-x+\frac{x^3}{2}>0 \iff x+\frac{x^3}{2} < \arctan{x}$

The other side does not look immediately obvious to me, so I will have a closer review of it later
_______________________________________
Edit: Suggestion
$\text{When }g(x)=\arctan{x}-x+\frac{x^3}{3}-\frac{x^5}{5}\text{ it can be shown that }\\ g^\prime (x) = \frac{-(5x^6+2x^4)}{3x^2+3}\\ \text{If you think about it, it can be shown that this is always less than 0.}$

Pro tip (aka. Giveaway)

We're full of even powers here...
$x^{2n}\ge 0 \text{ for ALL }x$

$\text{Thus, consider a monotonic DECREASING function here.}$

katherine123 · « **Reply #276 on:** June 28, 2016, 01:32:30 am »

questions for harder ex 1
1. For any real x,y,z,u prove that (x^2+y^2)(z^2+u^2)>and equal to (xz+yu)^2
2.for any real numbers x,y,z,a,b,c, prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>and equal to (ax+by+cz)^2

jamonwindeyer · « **Reply #277 on:** June 28, 2016, 09:18:58 am »

Quote from: katherine123 on June 28, 2016, 01:32:30 am

questions for harder ex 1
1. For any real x,y,z,u prove that (x^2+y^2)(z^2+u^2)>and equal to (xz+yu)^2
2.for any real numbers x,y,z,a,b,c, prove that (x^2+y^2+z^2)(a^2+b^2+c^2)>and equal to (ax+by+cz)^2

Hey Katherine! The first one is fairly okay once you see it:

$(x^2+y^2)(z^2+u^2) \ge (xz+yu)^2 \\\implies x^2z^2+x^2u^2+y^2z^2+y^2 u^2 \ge x^2z^2+y^2u^2+2uxyz \\\implies x^2u^2+y^2z^2 \ge 2uxyz$

This is a result from Extension 2 I believe, but you can move the RHS over to the left and obtain this using by factorisation if you like:

$\implies (xu-yz)^2\ge0$

Either way, the result is proven

Your second question is very very similar to your first! Expand both sides, cancel like terms, then move everything over to one side and factorise:

$(x^2+y^2+z^2)(a^2+b^2+c^2)\ge(ax+by+cz)^2 \\\implies x^2a^2+x^2b^2+x^2c^2+y^2a^2+y^2b^2+y^2c^2+z^2a^2+z^2b^2+z^2c^2 \ge a^2x^2+2abxy+2acxz+b^2y^2+2bcyz+c^2z^2 \\\implies x^2b^2+x^2c^2+a^2y^2+c^2y^2+a^2z^2+b^2z^2-2abxy-2acxz-2bcyz \ge 0 \\\implies (xb-ay)^2+(xc-az)^2+(cy-bz)^2 \ge 0$

Which is definitely true

that last factorisation might be a little tricky to see, let me know if this doesn't make sense

RuiAce · « **Reply #278 on:** June 28, 2016, 09:43:13 am »

Here is one massive take out from Jamon's answer:

Where possible, you should feel safe to work BACKWARDS. Sometimes trying to work forwards just doesn't produce an obvious result, and you must start from somewhere else.

@Jamon however, because we're working backwards we don't want to use \implies, because that's for working forwards! We can always use \iff wherever appropriate, however preferably use \impliedby (left sided implication)

jamonwindeyer · « **Reply #279 on:** June 28, 2016, 10:17:26 am »

Quote from: RuiAce on June 28, 2016, 09:43:13 am

Here is one massive take out from Jamon's answer:

Where possible, you should feel safe to work BACKWARDS. Sometimes trying to work forwards just doesn't produce an obvious result, and you must start from somewhere else.

@Jamon however, because we're working backwards we don't want to use \implies, because that's for working forwards! We can always use \iff wherever appropriate, however preferably use \impliedby (left sided implication)

Aha mate, I just needed an arrow

amandali · « **Reply #280 on:** June 28, 2016, 11:08:01 pm »

help with this ques thanks

RuiAce · « **Reply #281 on:** June 29, 2016, 09:37:12 am »

Quote from: amandali on June 28, 2016, 11:08:01 pm

help with this ques thanks

$\textbf{2000 HSC 4U Additional}$
$\text{The equation of the parabolic base is }y^2=4x\\ \text{Therefore we consider the }y\text{-coordinates when }x=h$
$y^2=4h \iff y=\pm 2\sqrt{h}$
$\text{Clearly, }y=2\sqrt{h}\text{ corresponds to the top coordinate, and }y=-2\sqrt{h}\text{ corresponds to the bottom}\\ \text{Hence the semi-major axis of the semi-elliptical cross section is }a=2\sqrt{h}$
$\text{Because the length of the semi-minor axis is half that of the semi-major axis, }b=\sqrt{h}$
$\text{Hence, the area of the SEMI-elliptical cross section is}\\ \begin{align*}A&=\frac{1}{2}\pi ab\\ &= \pi h \end{align*}$

amandali · « **Reply #282 on:** June 30, 2016, 12:22:02 am »

1)for a,b>0 show that (a+b)/2 greater than and equal to [(a^2+b^2)/2]^1/2

2) if 0<t<1, show that 1/2<1/(1+t)<1

3) 1/2(x^3+y^3)greater than equal to [(x+y)/2]^3

RuiAce · « **Reply #283 on:** June 30, 2016, 10:25:44 am »

Quote from: amandali on June 30, 2016, 12:22:02 am

1)for a,b>0 show that (a+b)/2 greater than and equal to [(a^2+b^2)/2]^1/2

2) if 0<t<1, show that 1/2<1/(1+t)<1

3) 1/2(x^3+y^3)greater than equal to [(x+y)/2]^3

$\text{Nothing at all is obvious in any of these, so attempt to work }\textbf{backwards}\text{ in the hope of finding something good.}$
$\text{And by doing so, you will find Q1 }\textbf{is wrong. The inequality should be LESS than or equal to}$
$\begin{align*}\frac{a+b}{2}&\le \sqrt{\frac{a^2+b^2}{2}}\\ \impliedby \frac{a^2+2ab+b^2}{4}&\le \frac{a^2+b^2}{2}\\ \impliedby a^2+2ab+b^2&\le 2a^2+2b^2\\ \impliedby 0 &\le a^2-2ab+b^2 \\ \iff (a-b)^2 &\ge 0 \end{align*}\\ \text{Which is an allowable first step for obvious reasons}\\ \text{(All reversal steps allowed given the positivity of }a, b$
_________________
$\text{Given }0 < t < 1, 1+t\text{ is positive}\\ \begin{align*}\frac{1}{2}&< \frac{1}{1+t} < 1 \\ \impliedby t+1 &< 2 < 2(t+1)\\ \impliedby t-1 &< 0 < 2t\end{align*}$
$\text{We arrive at a dead-end, so we attempt to split the inequality}$
$t-1<0 \iff t <1\text{ which is true by definition.}\\ 0 < 2t \iff 0 < t \text{ which is true by defintion.}\\ \text{So by reversing steps appropriately, we can achieve a desired proof.}$
One way to set this one out would be:
t < 1 therefore t-1 < 0
0 < t therefore 0 < 2t
Therefore t-1 < 0 < 2t
t+1 < 2 < 2t+2
and etc.
___________________
$\text{The formula is FALSE for when }x\text{ and }y\text{ are both negative, so for simplicity I will assume }x,y>0$
$\begin{align*}\frac{x^3+y^3}{2}&\ge \left(\frac{x+y}{2}\right)^3\\ \impliedby \frac{(x+y)(x^2-xy+y^2)}{2}&\ge (x+y)\left(\frac{x^2+2xy+y^2}{8}\right)\\ \impliedby \frac{x+y}{2}\left( (x^2-xy+y^2) - \frac{x^2+2xy+y^2}{4} \right) &\ge 0 \\ \impliedby \frac{x+y}{2}\left(\frac{3}{4}(x^2-2xy+y^2)\right)&\ge 0 \\ \impliedby \frac{3}{8}(x+y)(x-y)^2 &\ge 0 \\ \impliedby (x+y)(x-y)^2 &\ge 0 \end{align*}$
$\text{So if }x\text{ and }y\text{ are positive, obviously }x+y\text{ is positive and }(x-y)^2\text{ is ALWAYS positive or }0\\ \text{So we can just rewrite the proof in the reverse order.}$
$\textbf{Note that working backwards is strictly to formulate a forwards proof.}\\ \text{The actual answer always involves working forwards, not in reverse.}$

amandali · « **Reply #284 on:** July 03, 2016, 02:18:32 am »

need help with these inequality ques thanks

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