ATAR Notes: Forum

HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Advanced => Topic started by: RuiAce on February 16, 2016, 09:09:21 am

Title: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 09:09:21 am: Here, I will occasionally post challenge questions for the daring 2U (and above) student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their discretion.

$\text{Let } p \text{ and } q \text{ be any two real numbers.} \\ \text{Show that if } \left| p \right| +\left| q \right| =\left| p+q \right| \text{ then }pq\ge 0$

$\text{Hence or otherwise, find the set of values for } x \text{ that satisfy } \\ \left| x-a \right| +\left| x-b \right| =\left| a-b \right| \\ \text{where }a \text{ and } b \text{ are real }$

Spoiler
Required knowledge: Preliminary Basic Arithmetic & Algebra, Preliminary The Quadratic Function
Title: Re: Mathematics Challenge Marathon
Post by: brenden on February 16, 2016, 09:15:54 am: Quote from: RuiAce on February 16, 2016, 09:09:21 am
Here, I will occasionally post challenge questions for the daring 2U (and above) student to attempt.

Questions are not intended to reflect the scope of the difficulty in actual HSC questions, however may be completed using only knowledge taught in the course. Spoilers are intended to reveal what topics to draw knowledge from when a genuine, unaided attempt has been unsuccessful.

I invite everyone to also post their own questions at their discretion.

$\text{Let } p \text{ and } q \text{ be any two real numbers.} \\ \text{Show that if } \left| p \right| +\left| q \right| =\left| p+q \right| \text{ then }pq\ge 0$

$\text{Hence or otherwise, find the set of values for } x \text{ that satisfy } \\ \left| x-a \right| +\left| x-b \right| =\left| a-b \right| \\ \text{where }a \text{ and } b \text{ are real }$

Spoiler
Required knowledge: Preliminary Basic Arithmetic & Algebra, Preliminary The Quadratic Function
Awesome!!
Title: Re: Mathematics Challenge Marathon
Post by: Happy Physics Land on February 16, 2016, 10:09:46 am: Good stuff Rui, you can finally share your wisdom with this awesome community!
Title: Re: Mathematics Challenge Marathon
Post by: KoA on February 16, 2016, 05:09:45 pm: Solution
(http://i.imgur.com/NpJxu5C.jpg)
Title: Re: Mathematics Challenge Marathon
Post by: keltingmeith on February 16, 2016, 05:48:57 pm: Quote from: KoA on February 16, 2016, 05:09:45 pm
Solution
(http://i.imgur.com/NpJxu5C.jpg)

Your solution for the first isn't too bad! But, you've only proven that:

$pq\geq 0 \implies \mid p+q\mid = \mid p\mid + \mid q\mid$

You need to have another shot at proving the other way round, methinks. ;) (albeit, the other way round is a LOT more difficult, and requires more logical reasoning than straight up mathematics)

I love just how smoothly you put that into the second half of the question, though! Absolutely smashed it.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 06:19:37 pm: Quote from: KoA on February 16, 2016, 05:09:45 pm
Solution
(http://i.imgur.com/NpJxu5C.jpg)
The second part was done well.

However, whilst the first part was done correctly, it was not done in the most elegant manner:

$\left| p+q \right| =\left| p \right| +\left| q \right| \\ {\left(p+q\right)}^{2}={\left(\left|p\right|+\left|q\right|\right)}^{2} \\ {p}^{2}+2pq+{q}^{2}={p}^{2}+2\left| pq \right| +{q}^{2} \\ pq=\left| pq \right| \\ \text{As }\left| pq \right| \ge 0 \text{ by the definition of absolute value },pq\ge 0$

NEXT QUESTION
$\text{Solve } \sin ^{ 3 }{ x } -\cos ^{ 3 }{ x } =\sin { x } -\cos { x } \text{ for all } 0\le x \le 2\pi$

You may answer for 0°≤x≤360° if you have not been exposed to the radian measure of an angle.
Spoiler
Required knowledge: Preliminary Trigonometric Ratios
Title: Re: Mathematics Challenge Marathon
Post by: Happy Physics Land on February 16, 2016, 07:08:10 pm: memes

(http://i.imgur.com/UbGV0sQ.jpg)
Title: Re: Mathematics Challenge Marathon
Post by: jakesilove on February 16, 2016, 07:10:26 pm: Quote from: Happy Physics Land on February 16, 2016, 07:08:10 pm
memes

(http://i.imgur.com/UbGV0sQ.jpg)

Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 07:12:14 pm: Quote from: jakesilove on February 16, 2016, 07:10:26 pm
Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

Yeah. The case that (sin(x)-cos(x))=0 has been disregarded in this case, and thus tan(x)=1 was ignored.
Title: Re: Mathematics Challenge Marathon
Post by: Happy Physics Land on February 16, 2016, 07:18:58 pm: Quote from: jakesilove on February 16, 2016, 07:10:26 pm
Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

Ahhhhhh ok thank you so much Jake for pointing that out!!! I'm actually quite unaware of this detail and I guess its gonna be fatal in an exam (better jot this down).

ok so if l consider sinx - cosx cannot equal 0
then sinx cannot equal cos x
then tan x cannot equal 1
then x cannot equal pi/4 and that shouldnt really matter
but since we are dividing by cos x on both sides
cos x cannot equal 0
hence x cannot equal to pi/2 or 3pi/2

So the final answer for x would be x = 0, pi and 2pi?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 07:22:19 pm: Quote from: Happy Physics Land on February 16, 2016, 07:18:58 pm
Ahhhhhh ok thank you so much Jake for pointing that out!!! I'm actually quite unaware of this detail and I guess its gonna be fatal in an exam (better jot this down).

ok so if l consider sinx - cosx cannot equal 0
then sinx cannot equal cos x
then tan x cannot equal 1
then x cannot equal pi/4 and that shouldnt really matter
but since we are dividing by cos x on both sides
cos x cannot equal 0
hence x cannot equal to pi/2 or 3pi/2

So the final answer for x would be x = 0, pi and 2pi?

The only thing that Jake justified was that this is true if and only if sin(x)-cos(x)≠0.

In this case, there is no exception to this.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 10:03:35 pm: NEXT QUESTION

$\text{Suppose that }S=\sum_{k=0}^{\infty}{\sin^{2k}{\theta}}\text{ and }C=\sum_{k=0}^{\infty}{\cos^{2k}{\theta}}\\ \text{If }0<\theta<\frac{\pi}{2}\text{, show that }\frac{1}{S}+\frac{1}{C}=1$

Spoiler
Required knowledge: Preliminary Trigonometric Ratios, HSC Series
Title: Re: Mathematics Challenge Marathon
Post by: nerdgasm on February 16, 2016, 10:45:56 pm: Here's my attempt at a solution (rather handwavy, but I'm lazy):

I'll first assume that we're summing over k (seeing as I can't see an 'n' in either summation expression)
Note that the restriction on the value of theta implies that both sin(theta) and cos(theta) are positive. From there, it is easy to see that their powers must be as well; therefore both S and C are non-zero.

We may recognise S and C as infinite geometric series with ratio term sin^2(theta) and cos^2(theta) respectively. Note also that the value of sin(theta) and cos(theta) is between 0 and 1, and hence the value of sin^2(theta) and cos^2(theta) is also between 0 and 1. Therefore, as our infinite geometric series has ratio term between 0 and 1, we may use the formula for sum of an infinite geometric series.

If we do this, we should end up with S = 1/(1-sin^2(theta)) and C = 1/(1-cos^2(theta)). The denominators are non-zero again by virtue of sin^2(theta) and cos^2(theta) lying strictly between 0 and 1.

Therefore, 1/S + 1/C = 2 - sin^2(theta) - cos^2(theta) = 1, as required.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 16, 2016, 10:55:48 pm: Quote from: nerdgasm on February 16, 2016, 10:45:56 pm
Here's my attempt at a solution (rather handwavy, but I'm lazy):

I'll first assume that we're summing over k (seeing as I can't see an 'n' in either summation expression)
Note that the restriction on the value of theta implies that both sin(theta) and cos(theta) are positive. From there, it is easy to see that their powers must be as well; therefore both S and C are non-zero.

We may recognise S and C as infinite geometric series with ratio term sin^2(theta) and cos^2(theta) respectively. Note also that the value of sin(theta) and cos(theta) is between 0 and 1, and hence the value of sin^2(theta) and cos^2(theta) is also between 0 and 1. Therefore, as our infinite geometric series has ratio term between 0 and 1, we may use the formula for sum of an infinite geometric series.

If we do this, we should end up with S = 1/(1-sin^2(theta)) and C = 1/(1-cos^2(theta)). The denominators are non-zero again by virtue of sin^2(theta) and cos^2(theta) lying strictly between 0 and 1.

Therefore, 1/S + 1/C = 2 - sin^2(theta) - cos^2(theta) = 1, as required.

My apologies for the misleading 'n' term. But yes, correct!

It may have however, been neater to convert 1-sin²θ to cos²θ before applying the reciprocal. However this doesn't damage the elegance of the process.
____________________________________

NEXT QUESTION:

I didn't like my original question so I have replaced it:

$\text{Show that if }x,\,y\text{ and }z\text{ form a geometric progression and}\\{x}^{\frac{1}{a}}={y}^{\frac{1}{b}}={z}^{\frac{1}{c}}\\ \text{then }a,\,b\text{ and }c\text{ are in arithmetic progression.}$

Spoiler
Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series
Title: Re: Mathematics Challenge Marathon
Post by: juliakate_99 on February 19, 2016, 06:47:19 pm: Quote from: RuiAce on February 16, 2016, 06:19:37 pm

NEXT QUESTION
$\text{Solve } \sin ^{ 3 }{ x } -\cos ^{ 3 }{ x } =\sin { x } -\cos { x } \text{ for all } 0\le x \le 2\pi$

You may answer for 0°≤x≤360° if you have not been exposed to the radian measure of an angle.
Spoiler
Required knowledge: Preliminary Trigonometric Ratios

Would the answer be 2pi/3 and 4pi/3?
( I couldn't upload a photo of my working, but basically divided both sides of the equation by sinx -cosx giving me sin^2x-cos^2x=1. Rearranged to get cos^2x-sin^2x=-1, therefore cos2x=-1, cos x=-1/2.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 19, 2016, 07:10:56 pm: Quote from: juliakate_99 on February 19, 2016, 06:47:19 pm
Would the answer be 2pi/3 and 4pi/3?
( I couldn't upload a photo of my working, but basically divided both sides of the equation by sinx -cosx giving me sin^2x-cos^2x=1. Rearranged to get cos^2x-sin^2x=-1, therefore cos2x=-1, cos x=-1/2.

Firstly, you should've had sin²(x)+sin(x)cos(x)+cos²(x)-1 like HPL correctly achieved. This occurs due to the factorisation a³-b³=(a-b)(a²+ab+b²) This simplifies down to sin(x)cos(x), and he decided to take it one step further and rewrite it as sin(2x) using MX1 techniques.

Secondly, this was mentioned above. Who said it was permissible to cancel out (sin(x)-cos(x))?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 19, 2016, 07:16:02 pm: I suppose as the question was never completed correctly (despite attempted), I will post up the solution.

Quote from: RuiAce on February 16, 2016, 10:55:48 pm
NEXT QUESTION
$\text{Show that if }x,\,y\text{ and }z\text{ form a geometric progression and}\\{x}^{\frac{1}{a}}={y}^{\frac{1}{b}}={z}^{\frac{1}{c}}\\ \text{then }a,\,b\text{ and }c\text{ are in arithmetic progression.}$

Spoiler
Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series
Title: Re: Mathematics Challenge Marathon
Post by: Durnity on February 24, 2016, 09:36:54 pm: D: Oh damn these are tough. After 3 attempts I finally got the 1st one lol...
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 28, 2016, 07:21:33 pm: I don't usually post questions when there's a previous unanswered but I suppose I will on this forum.

$\text{Simplify:} \\ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}}$

Spoiler
Required knowledge: Preliminary Basic Arithmetic and Algebra, Preliminary The Quadratic Polynomial and the Parabola
Title: Re: Mathematics Challenge Marathon
Post by: Happy Physics Land on March 01, 2016, 05:20:25 pm: Quote from: RuiAce on February 28, 2016, 07:21:33 pm
I don't usually post questions when there's a previous unanswered but I suppose I will on this forum.

$\text{Simplify:} \\ \sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x+\dots}}}}}$

Spoiler
Required knowledge: Preliminary Basic Arithmetic and Algebra, Preliminary The Quadratic Polynomial and the Parabola

Im not gonna upload a picture because laptop is rigged
Let y = root x + root x + root x ...
square both sides
y^2 = x + (root x + root x + root x ....)
but y = root x + root x + root x ....
hence y^2 = x + y
So the whole thing becomes y^2 - y - x = 0
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on March 01, 2016, 06:17:55 pm: Quote from: Happy Physics Land on March 01, 2016, 05:20:25 pm
Im not gonna upload a picture because laptop is rigged
Let y = root x + root x + root x ...
square both sides
y^2 = x + (root x + root x + root x ....)
but y = root x + root x + root x ....
hence y^2 = x + y
So the whole thing becomes y^2 - y - x = 0
Basically the idea.
Title: Re: Mathematics Challenge Marathon
Post by: lazydreamer on March 05, 2016, 02:39:52 pm: they have something similar to this at BoS :D

anyway, is Happy Physics Land just missing y>= 0 ?

Quote from: RuiAce on February 16, 2016, 07:22:19 pm
Incorrect. The only thing that Jake justified was that this is true IF AND ONLY IF sin(x)-cos(x)≠0.

In this case, there is no exception to this.

so:
(sinx-cosx)(sin^2x+sinxcosx+cos^2x) = sinx - cosx
(sin^2x + sinxcosx + cos^2x) = 1 [where sinx-cosx does not = 0] <--so i would have to write this in exams yea?
because sin^2x + cos^2x = 1, therefore sinxcosx = 0
sinxsin(90-x) = 0, therefore:
sinx = 0, sin(90-x) = 0
x= 0, 180, 360
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on March 06, 2016, 10:58:17 am: Quote from: lazydreamer on March 05, 2016, 02:39:52 pm
they have something similar to this at BoS :D

anyway, is Happy Physics Land just missing y>= 0 ?

so:
(sinx-cosx)(sin^2x+sinxcosx+cos^2x) = sinx - cosx
(sin^2x + sinxcosx + cos^2x) = 1 [where sinx-cosx does not = 0] <--so i would have to write this in exams yea?
because sin^2x + cos^2x = 1, therefore sinxcosx = 0
sinxsin(90-x) = 0, therefore:
sinx = 0, sin(90-x) = 0
x= 0, 180, 360

Yes, but of course, you CAN'T write sinx-cosx does not = 0 because you were never told that!
Title: Re: Mathematics Challenge Marathon
Post by: lazydreamer on March 06, 2016, 12:28:04 pm: Quote from: RuiAce on March 06, 2016, 10:58:17 am
Yes, but of course, you CAN'T write sinx-cosx does not = 0 because you were never told that!

ooh no wonder, i thought both threads were just a coincidence ;D could you be the one who also started the BoS one?

aaah i'm sorry, idk what i'm doing :'( sooooo then i would have to take everything to the left and then factor out sinx-cosx?

and btw did you ever post up the soln. for the 'prove a,b,c are arith. progression' question? so so hard i can't even lol
Title: Re: Mathematics Challenge Marathon
Post by: pi on March 06, 2016, 04:12:34 pm: You guys might enjoy some of the threads we have in the VCE board Challenging Question Threads

Feel free to nick questions from there or bump up those threads, many have been solved but perhaps you guys can find some more elegant proofs :)
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on March 07, 2016, 07:05:42 am: Quote from: lazydreamer on March 06, 2016, 12:28:04 pm
ooh no wonder, i thought both threads were just a coincidence ;D could you be the one who also started the BoS one?

aaah i'm sorry, idk what i'm doing :'( sooooo then i would have to take everything to the left and then factor out sinx-cosx?

and btw did you ever post up the soln. for the 'prove a,b,c are arith. progression' question? so so hard i can't even lol

No, I didn't 'start' any.

Yep. (sinx-cosx) is factored.

Hint for that question: There's a bit of simultaneous equations going on
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on March 07, 2016, 07:11:34 am: Quote from: pi on March 06, 2016, 04:12:34 pm
You guys might enjoy some of the threads we have in the VCE board Challenging Question Threads

Feel free to nick questions from there or bump up those threads, many have been solved but perhaps you guys can find some more elegant proofs :)
Coolio. Will check when I run out of questions
Title: Re: Mathematics Challenge Marathon
Post by: lazydreamer on March 07, 2016, 06:39:44 pm: Quote from: RuiAce on March 07, 2016, 07:05:42 am
No, I didn't 'start' any.

Yep. (sinx-cosx) is factored.

Hint for that question: There's a bit of simultaneous equations going on

ooh are you lee? haha

(sinx-cosx)(sinxcosx+1) - (sinx-cosx) = 0
(sinx-cosx)(sinxcosx) = 0

sinx - cosx = 0, sinxcosx = 0

hence x = pi/4, 180+pi/4, or x = 0, pi/2, pi, 3pi/2, 2pi

?? hope that's it...
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on March 17, 2016, 09:53:56 am: $\text{Evaluate}\\ \sum_{n=0}^{80}{\frac{1}{\sqrt{k+1}+\sqrt{k}}}$

Spoiler
Required knowledge: Preliminary Basic Arithmetic and Algebra, HSC Series
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on January 11, 2017, 10:48:06 pm: I feel as though my tone was too toxic when I first signed up for these forums.

Anyway, here's another question. I originally wrote it for a treat but it was considered 'too hard'. So I decided to put it here. Warning: It is long.
(http://i.imgur.com/wgXeDo1.png)
Required knowledge
Preliminary Basic Arithmetic and Algebra, HSC Exponential and Logarithmic Functions, HSC Series, HSC Geometrical Applications of Differentiation

Hint for d)
It says to use part b). But avoiding using the sum of a G.P. MAY prove useful.
Title: Re: Mathematics Challenge Marathon
Post by: jamonwindeyer on January 11, 2017, 10:49:15 pm: Quote from: RuiAce on January 11, 2017, 10:48:06 pm
I feel as though my tone was too toxic when I first signed up for these forums.

Anyway, here's another question. I originally wrote it for a treat but it was considered 'too hard'. So I decided to put it here. Warning: It is long.
(http://i.imgur.com/8fKDf2R.png)
Required knowledge
Preliminary Basic Arithmetic and Algebra, HSC Exponential and Logarithmic Functions, HSC Series, HSC Geometrical Applications of Differentiation

Hint for d)
It says to use part b). But avoiding using the sum of a G.P. MAY prove useful.

OMG IT HAS SURFACED!!!
Title: Re: Mathematics Challenge Marathon
Post by: Rathin on January 11, 2017, 11:13:43 pm: Whenever I see a finincial maths series question I get a little depressed lol..so glad I'm not sitting the 2u hsc
Title: Re: Mathematics Challenge Marathon
Post by: wyzard on January 11, 2017, 11:43:18 pm: Quote from: RuiAce on January 11, 2017, 10:48:06 pm
I feel as though my tone was too toxic when I first signed up for these forums.

Anyway, here's another question. I originally wrote it for a treat but it was considered 'too hard'. So I decided to put it here. Warning: It is long.
(http://i.imgur.com/wgXeDo1.png)
Required knowledge
Preliminary Basic Arithmetic and Algebra, HSC Exponential and Logarithmic Functions, HSC Series, HSC Geometrical Applications of Differentiation

Hint for d)
It says to use part b). But avoiding using the sum of a G.P. MAY prove useful.

Here's my train of thoughts.

"Yeap first question showing some stuff in calculus... pretty straight forward... Next question financial math...yeah nah :'("
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 16, 2017, 01:44:16 pm: $\text{It's actually a no-brainer (beyond trivial) 4U question but I decided to throw it here for the lols.}\\ \int_{-100}^{-10} \frac{30}{x}dx$
Remember, you can't log a negative number (until you do complex analysis).
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 16, 2017, 03:43:58 pm: A nice work-around given the even power which I didn't anticipate. It wouldn't be "too" valid if the power ended up being odd.

Also, I very much prefer an exact form; not a calculator value
Title: Re: Mathematics Challenge Marathon
Post by: pikachu975 on July 16, 2017, 04:01:07 pm: Quote from: RuiAce on July 16, 2017, 01:44:16 pm
$\text{It's actually a no-brainer (beyond trivial) 4U question but I decided to throw it here for the lols.}\\ \int_{-100}^{-10} \frac{30}{x}dx$
Remember, you can't log a negative number (until you do complex analysis).

= [30ln|x|] from -100 to -10
= 30(ln10-ln100)
= 30ln(1/10)

Is this valid considering you put absolute values when integrating to log?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 16, 2017, 04:10:47 pm: Quote from: pikachu975 on July 16, 2017, 04:01:07 pm
= [30ln|x|] from -100 to -10
= 30(ln10-ln100)
= 30ln(1/10)

Is this valid considering you put absolute values when integrating to log?
Pretty much yeah.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 21, 2017, 07:16:37 pm: $\text{Sketch }y=x|x|$
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 01:44:05 pm: Quote from: RuiAce on July 21, 2017, 07:16:37 pm
$\text{Sketch }y=x|x|$

I understand this graphically, but I'm afraid I don't get the maths behind it :/
(https://uploads.tapatalk-cdn.com/20170722/8e2e89a4a94e9139958000892f8d1445.jpg)
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 01:49:07 pm: $\text{For the maths behind it, use the definition of the absolute value}$
$|x|=\begin{cases}x&\text{ if }x\ge 0\\ -x&\text{ if }x < 0\end{cases}$
$\text{Hence}\\ x|x| = \begin{cases}x^2 &\text{ if }x\ge 0\\ -x^2 &\text{ if }x < 0\end{cases}$
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 01:57:41 pm: Quote from: RuiAce on July 22, 2017, 01:49:07 pm
$\text{For the maths behind it, use the definition of the absolute value}$
$|x|=\begin{cases}x&\text{ if }x\ge 0\\ -x&\text{ if }x < 0\end{cases}$
$\text{Hence}\\ x|x| = \begin{Cases}x^2 &\text{ if }x\ge 0\\ -x^2 &\text{ if }x < 0$

Yeah I understand that, what I don't get is why it's a cubic. I understand visually, but I can't prove what makes it not be a positive parabola above a negative one.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 01:58:56 pm: Quote from: georgiia on July 22, 2017, 01:57:41 pm
Yeah I understand that, what I don't get is why it's a cubic. I understand visually, but I can't prove what makes it not be a positive parabola above a negative one.
It actually ain't a cubic at all ;)

Being two separate halves of a parabola might make it appear as though it were a cubic, but it most certainly is not a cubic.
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 02:00:18 pm: Quote from: RuiAce on July 22, 2017, 01:58:56 pm
It actually ain't a cubic at all ;)

Being two separate halves of a parabola might make it appear as though it were a cubic, but it most certainly is not a cubic.

I meant it looks like one ahahaha!
How does this work? do you not post the next question until the previous one is completed?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:02:59 pm: Also, why it's half a 'positive' parabola and half a 'negative' parabola is really just a consequence out of the algebra. Basically you ended up with a piece-wise defined function
Quote from: georgiia on July 22, 2017, 02:00:18 pm
I meant it looks like one ahahaha!
How does this work? do you not post the next question until the previous one is completed?
Used to be like that, but now I don't care too much and I'm happy for questions to be thrown in whenever someone feels like it.
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 02:07:23 pm: For Q.a did you mean

a^x ln(x) ?

Maybe I'm just doing something wrong?
(https://uploads.tapatalk-cdn.com/20170722/7d96aa61770e1cc07551d42ec761e447.jpg)
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:09:09 pm: Quote from: georgiia on July 22, 2017, 02:07:23 pm
For Q.a did you mean

a^x ln(x) ?

Maybe I'm just doing something wrong?
(https://uploads.tapatalk-cdn.com/20170722/7d96aa61770e1cc07551d42ec761e447.jpg)
$\text{Not sure what happened at the start}\\ \frac{d}{dx} a^x = \frac{d}{dx} a^{e \ln x}\\ \text{Remember that }\ln x\text{ is in the power}$
Hint: Whilst it's doable that way, that isn't the best way to apply the identity given
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 02:19:25 pm: Quote from: jamonwindeyer on January 11, 2017, 10:49:15 pm
OMG IT HAS SURFACED!!!
(https://uploads.tapatalk-cdn.com/20170722/7f9466656b3214deb9fb0d11a39eb9d9.jpg)

Sorry I'm very messy :-\
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:23:22 pm: $\text{1. Fill in the blanks:}\\ \text{A real number }x\text{ is said to be rational if}\\ \text{we may write }x = \frac{p}q,\text{ where }p\text{ and }q\text{ are }\_\_\_\_\_\_\_\text{, and }q \_ \_ \_\_$
$\text{2. In this question, we will prove that }\log_2 3\text{ is irrational.}\\ \text{Suppose, under the similar conditions above, we have }\log_2 3 = \frac{p}{q}$
$\text{a) Prove that }2^p=3^q\\ \text{b) Hence explain why }\log_23\text{ is irrational}$
Spoiler
Find a contradiction!
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:28:19 pm: $\text{Find }\textbf{ALL}\text{ values of }\theta\text{ such that the limiting sum}\\ 1+\tan \theta + \tan^2\theta + \tan^3\theta + \dots \\ \text{exists.}$
Hint
You may like to use a graph.
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 22, 2017, 02:32:25 pm: Quote from: georgiia on July 22, 2017, 02:19:25 pm
(https://uploads.tapatalk-cdn.com/20170722/7f9466656b3214deb9fb0d11a39eb9d9.jpg)

Sorry I'm very messy :-\
What have I done wrong?(https://uploads.tapatalk-cdn.com/20170722/ce83ef7e2f2f2e39a8c0393daf8d6e62.jpg)
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:33:57 pm: Lol, sorry, that was actually my fault. I'll fix that up now.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 22, 2017, 02:38:27 pm: $\text{1. Find the maximum value of }f(x) = \frac{1}{1+x^2}$
$\text{2. Does }f(x)\text{ have a minimum value?}\\ \textit{Carefully}\text{ explain your answer}$
Title: Re: Mathematics Challenge Marathon
Post by: georgiia on July 24, 2017, 08:27:55 pm: Quote from: RuiAce on July 22, 2017, 02:38:27 pm
$\text{1. Find the maximum value of }f(x) = \frac{1}{1+x^2}$
$\text{2. Does }f(x)\text{ have a minimum value?}\\ \textit{Carefully}\text{ explain your answer}$

1. Max. value at y=1

2. There is no min. as f(x) is a bell curve (i think?)
∴ as x → +∞, y → 0
ans as x → -∞, y → 0
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 25, 2017, 08:35:52 am: Quote from: georgiia on July 24, 2017, 08:27:55 pm
1. Max. value at y=1

2. There is no min. as f(x) is a bell curve (i think?)
∴ as x → +∞, y → 0
ans as x → -∞, y → 0
Right idea with the bell curve (it isn't actually one, it just looks like it; a bell curve would be $ y=e^{-\frac12x^2}$ but it's better to make it explicit that there's an asymptote going on.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 27, 2017, 10:32:03 pm: This isn't a computational question but rather just testing your understanding of terminology. Because a lot of 2U students get confused over this.

Explain the difference between a sequence and a series.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on August 05, 2017, 06:12:22 pm: $\text{1. Prove from first principles that }f(x)=|x|\\ \text{is not differentiable at }x=0$
$\text{2. Explain why this is the case intuitively.}$
$\text{3. Find }f^\prime(x)\text{ for all }x\neq 0$
Title: Re: Mathematics Challenge Marathon
Post by: Opengangs on August 06, 2017, 12:21:26 am: 1. From first principles, we get:
lim(h -> 0) (|0 + h| + |0|)/h, which can be broken into two smaller cases: when h tends towards 0+ and 0-
From the right side (0+), we get lim(h -> 0) (h/h) = 1, since h > 0
From the left side (0-), we get lim(h -> 0) (-h/h) = -1, since h < 0

Since lim(h -> 0+) is not the same as lim(h -> 0-), at the point x = 0, it is not differentiable.

2. Explaining why it's not differentiable at the point, x = 0.
As f(x) tends towards 0 from the positive side, its differential is incrementally tending towards 1, while from the negative side (0-), it is incrementally tending towards -1. This means that a differential cannot exist at the point x = 0, as the change in x (delta x) is not defined at that particular point.

3.
$f'(x) = \lim_{h \to 0} \frac{|x + h| - |x|}{h}$
$= \lim_{h \to 0} \frac{(|x + h| - |x|)(|x + h| + |x|)}{h(|x + h| + |x|)}$
$= \lim_{h \to 0} \frac{(x + h)^{2} - x^{2}}{h(|x + h| + |x|)}$
$= \lim_{h \to 0} \frac{x^{2} + 2xh + h^{2} - x^{2}}{h(|x + h| + |x|)}$
$= \lim_{h \to 0} \frac{h(2x + h)}{h(|x + h| + |x|)}$
$= \lim_{h \to 0} \frac{2x + h}{|x + h| + |x|}$
$= \frac{2x}{2|x|}$
$= \frac{x}{|x|}$

$f'(x) = \frac{|x|}{x}$
Title: Re: Mathematics Challenge Marathon
Post by: pikachu975 on August 06, 2017, 12:23:03 am: Quote from: RuiAce on August 05, 2017, 06:12:22 pm
$\text{1. Prove from first principles that }f(x)=|x|\\ \text{is not differentiable at }x=0$
$\text{2. Explain why this is the case intuitively.}$
$\text{3. Find }f^\prime(x)\text{ for all }x\neq 0$

1) f(x+h) = |x+h|
f(x) = |x|
f'(x) = lim(h->0) [|x+h| - |x|]/h
at x = 0, f'(x) = lim(h->0) h/h and 0/0 is undefined.

2) There is no tangent at the corner part of the absolute value function hence no gradient/derivative at x = 0.

3) For x>0, f'(x) = 1, for x<0, f'(x) = -1, which can be generalised as f'(x) = x/|x|
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on August 06, 2017, 12:32:47 am: Quote from: pikachu975 on August 06, 2017, 12:23:03 am
1) f(x+h) = |x+h|
f(x) = |x|
f'(x) = lim(h->0) [|x+h| - |x|]/h
at x = 0, f'(x) = lim(h->0) h/h and 0/0 is undefined.

2) There is no tangent at the corner part of the absolute value function hence no gradient/derivative at x = 0.

3) For x>0, f'(x) = 1, for x<0, f'(x) = -1, which can be generalised as f'(x) = x/|x|
Just be careful of this. Everything was right except for this part, which Opengangs correctly answered.

If this were the case, because I'm taking a limit I'm allowed to cancel out the h's to just get 1. That would imply f'(0)=1, which is definitely not what we wanted to see.

The problem was in that when x=0, you should've left the numerator as |h|, not take off the absolute values without breaking the cases.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on August 10, 2017, 01:56:13 pm: $\text{In this question, we attempt to differentiate by a squared variable.}$
$\text{In practice, the normal distribution is defined by its mean and variance}\\ \text{but we choose the variance over the standard deviation.}$
(Note: Stddev^2 = Var)
$\text{Given a random sample from a normal distribution with mean }0\\ \text{ and (unknown) variance }\sigma^2\\ \text{we may compute the 'log-likelihood' function to be}\\ \boxed{\ell = -\frac{n}{2}\ln \left(2\pi \sigma^2\right) - \frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 }$
_______________________________
$\text{Question 1.}\\ \text{a) Find }\frac{d\ell}{d\sigma}\\ \text{b) Find the value of }\sigma\text{ such that }\ell\text{ is maximised}$
$\text{Question 2.}\\ \text{a) Now, attempt to find }\frac{d\ell}{d\sigma^2}\\ \text{You will need to treat }\sigma^2\text{ as what you're differentiating}\\ \text{with respect to}\\ \text{b) Set }\frac{d\ell}{d\sigma^2}=0\text{. Do you get the same value as in Q1?}$
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on October 11, 2017, 01:46:56 pm: Not hard; it just looks pretty
$\text{Prove that every equilateral triangle is similar to each other}$
Title: Re: Mathematics Challenge Marathon
Post by: winstondarmawan on October 11, 2017, 03:21:02 pm: Quote from: RuiAce on October 11, 2017, 01:46:56 pm
Not hard; it just looks pretty
$\text{Prove that every equilateral triangle is similar to each other}$

Isn't this just equiangular?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on October 11, 2017, 03:43:33 pm: Quote from: winstondarmawan on October 11, 2017, 03:21:02 pm
Isn't this just equiangular?
Yes
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on November 11, 2017, 03:29:54 pm: Another one that's not hard but just looks pretty.
$\text{Prove (algebraically) that provided }f(x)\text{ is well defined at }0\\ \text{if }f(x)\text{ is an odd function, then the graph }y=f(x)\\ \text{must pass through the origin.}\\ \text{Also interpret your result geometrically}$
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on December 09, 2017, 11:52:01 pm: $\text{Prove that the line through any two arbitrary points}\\ \text{is }\textbf{unique}.$
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on January 01, 2018, 12:36:22 pm: $\text{New year's special - 2U}\\ \text{The hardest milestone a man wishes to break in 2018 has a success rate of }\frac{1}{2018}.\\ \text{Prove that if he keeps attempting to break the milestone forever, he will ultimately succeed.}$
Title: Re: Mathematics Challenge Marathon
Post by: zhen on January 01, 2018, 01:15:52 pm: Quote from: RuiAce on January 01, 2018, 12:36:22 pm
$\text{New year's special - 2U}\\ \text{The hardest milestone a man wishes to break in 2018 has a success rate of }\frac{1}{2018}.\\ \text{Prove that if he keeps attempting to break the milestone forever, he will ultimately succeed.}$
$\text{Let n be the number of attempts the man takes}\\ \text{Probability he has at least one success}=1-(\frac{2017}{2018})^n\\ \text{As n approaches infinity, the probability of success approaches 1, as } (\frac{2017}{2018})^n \text{ approaches zero} \\\text{Therefore, if he keeps attempting to break the milestone forever, he will ultimately succeed}$
I’m not sure if I’ve structured this proof correctly. I’m honestly just having a bit of fun.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on February 15, 2018, 11:56:43 pm: Let $ABCD$ be a parallelogram. Prove that its diagonals bisect each other.
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on July 14, 2018, 12:28:20 am: In this question, we fill in the gaps left in high school and prove properly that $ \frac{d}{dx} e^x = e^x$ and $ \frac{d}{dx} x^r = rx^{r-1} $ for any real $r$. (It is adapted from sources easily obtainable online.)
$\text{1. Consider the limit }\lim_{u\to 0} \frac{\ln (u+1)}{u}.\\ \text{Demonstrate with the substitution }v = \frac{1}{u}\text{ and other tools}\\ \text{that this limit is equal to }\lim_{v\to \infty} \ln \left( \left(1+\frac{1}{v}\right)^v \right).$
$\text{2. By convention, we actually define }e\text{ to be the constant such that}\\ \boxed{e = \lim_{n\to \infty} \left( 1 + \frac1n \right)^n}.\\ \text{Assuming this is true, what must the limit above be equal to?}$
$\text{3. Let }g(x) = e^x.\text{ Then by first principles,}\\ g^\prime(x) = \lim_{h\to 0} \frac{e^{x+h}-e^x}{h}.\\ \text{With the aid of the substitution }u = e^h - 1,\text{ prove that }g^\prime(x) = e^x.$
$\text{4. Let }f(x) = x^r\text{, for some real number }r.\\ \text{Recall that for any }y > 0, \, e^{\ln y} = y.$
$\text{First consider the case }x>0.\\ \text{Show that under these conditions, }f^\prime(x) = rx^{r-1}.$
Remember, the only derivative you know after doing Q1-Q3 is $ \frac{d}{dx} e^x = e^x$. You cannot assume anything about the derivative of log or any other function.
$\text{5. Now consider the case }x<0.\\ \text{Note that }x = -1 \times -x\text{, so therefore }\boxed{-x > 0}.\\ \text{Further, note that }x^r = (-1)^r \times (-x)^r.$
$\text{Carefully demonstrate how }f^\prime(x) = rx^{r-1}\text{ for this case as well.}$
Hint for this part
Remember: $ (-1)^r$ is just a constant. Somehow borrow what you've used in Q4, or just regurgitate what you did in Q4 in a different manner.
$\text{6. The last case to consider is }x=0.\text{ Of course, this doesn't matter if }r < 0.\\ \text{Use the first principles definition of the derivative to show that }f^\prime(x) = r x^{r-1}\\ \text{when }x=0\text{ as well, for all }r\geq 0.$
Title: Re: Mathematics Challenge Marathon
Post by: fun_jirachi on August 29, 2018, 09:25:59 pm: Quote from: RuiAce on February 15, 2018, 11:56:43 pm
Let $ABCD$ be a parallelogram. Prove that its diagonals bisect each other.
Just gonna bump this thread (because i just found it) with the solution to this question :)
(https://i.imgur.com/vTunV6E.png)

$\text{Let the point of intersection of DB and AC be the point E.}$
$\text{In} \triangle\text{'s DEC and BEA} \\ \angle DEC = \angle BEA (\text{vertically opposite angles are equal}) \\ DC = BA (\text{opposite sides of a parallelogram are equal}) \\ \angle DCE = \angle BAE (\text{alternate angles, DC} \parallel \text{BA, opposite sides in a parallelogram are parallel}) \\ \text{Therefore,} \triangle \text{DEC} \equiv \triangle \text{BEA (AAS)} \\ \text{Therefore, DE = BE and CE = AE (corresponding sides in congruent triangles)} \\ \text{Therefore, the diagonals of a parallelogram bisect each other.}$
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on November 03, 2018, 11:22:00 pm: \[ \int_{-2}^3 \frac{1}{x^2}\,dx \]
Title: Re: Mathematics Challenge Marathon
Post by: fun_jirachi on November 03, 2018, 11:42:16 pm: I swear I'm going to get baited
Jumping straight in without thinking gets you this which seems a little off
$\int_{-2}^3\frac{1}{x^2}dx \\ = \left[-\frac{1}{x}\right]_{-2}^3 \\ = -\frac{1}{3}+\frac{1}{2} \\ = \frac{1}{6}$
But it seemed a bit too easy doing it like that, so I graphed the function on the cartesian plane and the area under the graph is infinite/indefinite/undefined because x is undefined as it approaches zero from both ends. so uh can someone that isnt braindead explain?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on November 03, 2018, 11:44:43 pm: Quote from: fun_jirachi on November 03, 2018, 11:42:16 pm
I swear I'm going to get baited
Jumping straight in without thinking gets you this which seems a little off
$\int_{-2}^3\frac{1}{x^2}dx \\ = \left[-\frac{1}{x}\right]_{-2}^3 \\ = -\frac{1}{3}+\frac{1}{2} \\ = \frac{1}{6}$
But it seemed a bit too easy doing it like that, so I graphed the function on the cartesian plane and the area under the graph is infinite/indefinite/undefined because x is undefined as it approaches zero from both ends. so uh can someone that isnt braindead explain?
Maybe wolframalpha can show you the way ;)
Title: Re: Mathematics Challenge Marathon
Post by: fun_jirachi on November 03, 2018, 11:49:59 pm: So do you just have two options:
- Give up and cry because the function isnt defined at x=0
- Use an approximation
Because that seems really sad
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on November 04, 2018, 12:15:58 am: Pretty much the fact that the function being undefined at 0, in fact asymptotically approaching 0 is heavily related to this problem. The integral is said to be divergent because the area under the graph is actually approaching $\infty$ as $x\to 0$, both in the positive and in the negative direction. The interval $-2\leq x \leq 3$ clearly contains the value $x=0$, so when we try to compute the area under the curve we end up crossing this border, which leads to big problems.

That's why the question was a bait. A student who only thinks in terms of antiderivatives and not in terms of what the integral actually represents would obtain the wrong answer.

However, it may be worth noting that not all functions that have asymptotes will have divergent integrals. For example, $ \int_{-2}^3 \frac{1}{x^{2/3}}\,dx $ converges (it equals to some finite real number). Wolfram won't give the exact correct answer for this since it combines complex analysis into the mixture and gives the principal value of $ (-2)^{1/3} $, which is complex, but plugging into this integral calculator will give a value that's clearly not undefined.
Title: Re: Mathematics Challenge Marathon
Post by: 3.14159265359 on November 04, 2018, 01:23:43 pm: Quote from: RuiAce on November 04, 2018, 12:15:58 am
Pretty much the fact that the function being undefined at 0, in fact asymptotically approaching 0 is heavily related to this problem. The integral is said to be divergent because the area under the graph is actually approaching $\infty$ as $x\to 0$, both in the positive and in the negative direction. The interval $-2\leq x \leq 3$ clearly contains the value $x=0$, so when we try to compute the area under the curve we end up crossing this border, which leads to big problems.

That's why the question was a bait. A student who only thinks in terms of antiderivatives and not in terms of what the integral actually represents would obtain the wrong answer.

However, it may be worth noting that not all functions that have asymptotes will have divergent integrals. For example, $ \int_{-2}^3 \frac{1}{x^{2/3}}\,dx $ converges (it equals to some finite real number). Wolfram won't give the exact correct answer for this since it combines complex analysis into the mixture and gives the principal value of $ (-2)^{1/3} $, which is complex, but plugging into this integral calculator will give a value that's clearly not undefined.

since at x=0 the graph is undefined, does that mean the area doesn't exist?
Title: Re: Mathematics Challenge Marathon
Post by: RuiAce on November 04, 2018, 06:31:34 pm: Quote from: 3.14159265359 on November 04, 2018, 01:23:43 pm
since at x=0 the graph is undefined, does that mean the area doesn't exist?
The thing is, interestingly enough the fact that $x=0$ is undefined by itself does not imply that the area does not exist. The area certainly does exist, and can be computed by manually evaluating the integral.

This is more or less because $x=0$ is a single point. If the function was undefined on a whole interval instead, then the notion of area falls apart entirely. But if the function is undefined at a single point, the area may or may not exist. This is why mathematicians study convergence - it helps us investigate under what conditions should that area exist and pose no problems.