leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

InteGrand · Feb 11, 2016

drsabz101 said:
How to do this question :

A) find the equation of the tangent to y= square root of x, -1. Where x=t
I differentiated this , then sub in t for x .

$\noindent Is the curve $y=\sqrt{x-1}$? And they want the equation of the tangent at $x=t$?$

leehuan · Feb 11, 2016

$\\ y=\sqrt { x } -1\\ \frac { dy }{ dx } =\frac { 1 }{ 2\sqrt { x } } \\ x=t\Rightarrow y=\sqrt { t } -1,\frac { dy }{ dx } =\frac { 1 }{ 2\sqrt { t } } \\ \therefore Tangent\quad is\\ y-\left( \sqrt { t } -1 \right) =\frac { 1 }{ 2\sqrt { t } } \left( x-t \right) \\ 2\sqrt { t } y-2t+2\sqrt { t } =x-t\\ x-2\sqrt { t } y+t=0$

But yeah, also InteGrand's question for verification please

drsabz101 · Feb 11, 2016

The correct question was leehuans interpretation

InteGrand · Feb 11, 2016

drsabz101 said:
The correct question was leehuans interpretation

To write square roots like that without LaTeX, it is probably good to write it like this: y = sqrt(x) - 1.

drsabz101 · Feb 17, 2016

Can somebody show how to solve the inequality g(x)>0, where g(x)= 1/x+3 - 2

InteGrand · Feb 17, 2016

drsabz101 said:
Can somebody show how to solve the inequality g(x)>0, where g(x)= 1/x+3 - 2

$\noindent We have $\frac{1}{x+3}-2 >0$ (by the way, if writing this without \LaTeX, it would be less ambiguous if we put brackets around the denominator, i.e. write it as $1/(x+3)$). Note that $x\neq -3$.$

$\noindent Rearranging, $\frac{1}{x+3}>2$. So $0<x+3 < \frac{1}{2}$ $\Big{(}$to see this, check where on the graph of $y=\frac{1}{t}$ we have $y>2$$\Big{)}$. So subtracting 3 from everything yields $\boxed{-3<x<-\frac{5}{2}}$.$

drsabz101 · Feb 17, 2016

how did you get : 0< x+3< 1/2

InteGrand · Feb 17, 2016

drsabz101 said:
how did you get : 0< x+3< 1/2

$\noindent If $x+3$ is \textit{not} greater than 0 (i.e. if it is less than 0, since it can't actually equal 0, as it is a denominator), there's no way $\frac{1}{x+3}$ can be greater than 2 (as it would be negative).$

drsabz101 · Feb 17, 2016

Also how would you determine the range and domain for this? What steps are taken

InteGrand · Feb 17, 2016

drsabz101 said:
Also how would you determine the range and domain for this? What steps are taken

$\noindent That's not really necessary for solving the inequation. Are you just asking out of interest? The function is $g(x):= \frac{1}{x+3}-2$. Its domain is $\mathbb{R}\backslash \{-3\}$ (i.e. the set of all real numbers excluding $-3$; for HSC purposes, it is sufficient to write this as ``$x\neq -3$''). This is because the function is clearly well-defined at any real number except $-3$, where it has a singularity. For the range, it is $\mathbb{R}\backslash \{-2\}$ (for HSC purposes, it is sufficient to write this as ``$y\neq -2$''). This is because $\frac{1}{x+3}$ can attain any real number except 0, so the function $g$ can attain any real value except $-2$.$

leehuan · Feb 17, 2016

Euclidean Geometry and locus never made any sense to me. -_-

Part (i) and (ii) are easy. How do I go about (iii) and (iv)?

drsabz101 · Feb 17, 2016

Q. A function is defined by f(x)=ax^2+bx+c. It is known f(x) is odd, and f(5)=25. Fin a,b and c

I got b = -5 , but I don't know how to find a and c ?

InteGrand · Feb 17, 2016

drsabz101 said:
Q. A function is defined by f(x)=ax^2+bx+c. It is known f(x) is odd, and f(5)=25. Fin a,b and c

I got b = -5 , but I don't know how to find a and c ?

$\noindent If $f$ is odd, then $a$ must be 0 and also, $c$ needs to be 0. So $b=5$, as $f(5)=25$.$

drsabz101 · Feb 17, 2016

How would you apply the formula : f(-x)=-f(x) to get a and c

InteGrand · Feb 17, 2016

drsabz101 said:
How would you apply the formula : f(-x)=-f(x) to get a and c

$\noindent A polynomial that is odd contains only odd-power terms. So $a=0=c$.$

leehuan · Feb 17, 2016

drsabz101 said:
How would you apply the formula : f(-x)=-f(x) to get a and c

What he's saying is that if you think about it, if you have a quadratic x^2 lying around, it can't be an odd function; has to be even.

leehuan said:
Euclidean Geometry and locus never made any sense to me. -_-

Part (i) and (ii) are easy. How do I go about (iii) and (iv)?

Drsoccerball · Feb 17, 2016

The taylor series can be derived using the mean value theorem.

$\frac{f(a)-f(b)}{(a-b)} \approx f'(a)$

$f(x) \approx f(b) +f'(x)(x-b)$

$How do we prove the general taylor series using only the mean theorem?$

InteGrand · Feb 17, 2016

leehuan said:
Euclidean Geometry and locus never made any sense to me. -_-

Part (i) and (ii) are easy. How do I go about (iii) and (iv)?

$\noindent For part (iii), we should do it in three parts, based on where $P$ is. Let $y$ be the angle $AXB$, where $X$ is a point in the major arc $AB$ in the diagram (note $y$ is a constant, due to ``angles in the same segment'' and as $AB$ is fixed). The geometry is slightly different based on whether $P$ is on the major arc in the diagram, $P$ is on either $A$ or $B$, or $P$ is in the minor arc.$

$\noindent Note that the proper way to write the triangle similarity from (ii) is $\triangle PCD$ is similar to $\triangle PAB$ (this way, the vertices align properly with the angles). So $\frac{CD}{PC}=\frac{AB}{PA}\Rightarrow CD = \frac{PC}{PA}\cdot AB$ (ratios of corresponding sides of similar triangles). Since $AB$ is constant, it suffices to show that $\frac{PC}{PA}$ is constant. In all three cases of where $P$ is, we indeed have that $\frac{PC}{PA}$ is (the same) constant, equal to the $\cot y$. We show this below.$

$\noindent First case: $P$ is on the major arc in the diagram. In this case, $\frac{PC}{PA}=\cot \angle APC = \cot y$, straight from right-angled triangle $PCA$.$

$\noindent Second case: $P$ is on the minor arc in the diagram. Note that in this case, triangle $APB$ is obtuse at $P$, and so the altitudes in question actually both go outside triangle $APB$ (so points $C$ and $D$ lie outside triangle $APB$; in fact, they are outside the circumcircle too). In this case, $\angle APB = 180^\circ - y$ (opposite angles of cyclic quadrilateral $AXBP$, where $X$ is a point in the major arc in the diagram), So $\angle APC = y$ (angles on straight line $CPB$). But again, $\frac{PC}{PA}= \cot \angle APC$ in right-angled triangle $APC$. So $\frac{PC}{PA}=\cot y$ again, as desired.$

$\noindent Third case, $P$ coincides with $A$ or $B$ (assume $B$ w.l.o.g.). This is basically the limit of the phenomenon where $P$ approaches $B$. In this limit, $PB$ becomes understood as the tangent at $B$, as the line segment $PB$ becomes the tangent at $B$ as $P$ approaches $B$. So $C$ is the point on this tangent such that $AC$ is perpendicular to the tangent (i.e. $\angle ACB = 90 ^\circ $). Also, the point $D$ is just the point $B = P$, since $D$ is the foot of the perpendicular from $B$ to $AP$, but $B$ lies on $AP$, so $D$ is just $B$. Now, $\angle APC = y$ again (this time by the alternate segment theorem). So again $\frac{PC}{PA}= \cot y$. This completes our proof.$

$\noindent (iv) Let $N$ be the midpoint of $CD$. Let $M$ be the midpoint of $AB$. The locus of $N$ is a circle centred at $M$, with radius $MN$, where $MN$ is a constant value (we probably don't need to worry about what it actually is since we aren't given any measurements in the diagram).$

$\noindent To see this, let $AM = s = BM$, where $s$ is just half the length of $AB$, and hence constant. Now, $DM = s$, since $M$ is the midpoint of right-angled triangle $ABD$, and similarly $CM = s$. (It is easy to verify that these lengths are still the same in the degenerate case where $P=B$, since then, the argument for $CM$ is the same, and then, $DM=s$ because $D=B$ in this case, and $BM = s$. So triangle $CDM$ has constant side lengths (recall $CD$ is constant from before). By the SSS rule of congruency for triangles, $\triangle CDM$ stays the same shape and size as $P$ varies. Hence $MN$ (a median of this triangle), is also a constant length. Hence $N$ indeed lies on a circle centred at $M$ as $P$ varies. I'm pretty sure we indeed get the full circle, but not sure how to rigorously show this right now, maybe someone else can show it.$

$\noindent Edit: oh yeah, we can probably show this easily using bashy methods. E.g. a HSC-level method to bash this out would be to let the circle be $x^2 + y^2 =1$, parameterise the point $P$ as $\left(\cos t,\sin t\right),$ $t\in \left[0,2\pi \right)$, and let $AB$ be the line $y=-k$, where $0\leq k \leq 1$ (or use strict inequalities if you want). Then we can calculate the coordinates of $N$ in terms of $t$ and use this to show that we get the full circle (if we do). I'm too lazy to try this method, so I'll leave it as an exercise for anyone curious. There should be some nice non-bash HSC way of showing it though, since I think this is an old HSC 4U Q.$

$\noindent If so, which year's HSC 4U paper is this from?$

Drsoccerball · Feb 17, 2016

ahahaha that question was the only question(That circle question) that fazed my maths teacher throughout the year. He labelled it as "The Cursed Question."

EDIT: Also that question was 2 marks the working out shouldn't of been that long...

Paradoxica · Feb 17, 2016

InteGrand said:
.

The locus can only be a full circle if the altitudes are allowed to project from the triangle extended outside the circle when the triangle is obtuse. If this is not the case, then you only get a circular arc.

leehuan's All-Levels-Of-Maths SOS thread (3 Viewers)

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