Post by: RuiAce on February 28, 2016, 07:30:50 pm
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 own discretion, and for anyone who has completed 4U maths or equivalent to also answer. I invite collaboration, as from time to time, some questions may be, one will argue, ridiculous.
Spoiler
Required knowledge:
a) HSC 4U Complex Numbers
b) Preliminary 2U Tangent to a Curve and Derivative of a Function, HSC 4U Complex Numbers
c) Preliminary 2U Basic Arithmetic and Algebra, HSC 4U Complex Numbers, HSC 2U Trigonometric Functions
a) HSC 4U Complex Numbers
b) Preliminary 2U Tangent to a Curve and Derivative of a Function, HSC 4U Complex Numbers
c) Preliminary 2U Basic Arithmetic and Algebra, HSC 4U Complex Numbers, HSC 2U Trigonometric Functions
Edit: 3 years later I found a typo with part a). It has been fixed.
Post by: RuiAce on March 07, 2016, 10:30:51 am
Post by: Ali_Abbas on August 20, 2016, 10:17:26 pm
1/2*sin^-1[tan(x/2)] + C, C constant.
Is that correct ?
Post by: RuiAce on August 20, 2016, 10:25:45 pm
For the integral of the square-root of tanx I got:Looks off.
1/2*sin^-1[tan(x/2)] + C, C constant.
Is that correct ?
What was your method?
Post by: birdwing341 on August 21, 2016, 11:20:21 am
Sorry for unclear image
Post by: RuiAce on August 21, 2016, 11:31:08 am
I got something stupefyingly difficult. This method does yield an answer, but I'm not sure if it is the most effective.Method's right. Don't remember the final answer off the top of my head but it takes a similar form.
Sorry for unclear image
The integral is famous. There's plenty of ways to do it and can be found everywhere on the Internet
Post by: jamonwindeyer on August 21, 2016, 02:35:18 pm
For the integral of the square-root of tanx I got:
1/2*sin^-1[tan(x/2)] + C, C constant.
Is that correct ?
Welcome to the forums Ali! Happy to have you around ;D let me know if you need help finding anything ;D
Post by: Ali_Abbas on August 21, 2016, 03:29:35 pm
Welcome to the forums Ali! Happy to have you around ;D let me know if you need help finding anything ;D
Thanks for the welcome message Jamon :) If I require any assistance navigating my way around here I'll let you know.
Cheers
Post by: Ali_Abbas on August 22, 2016, 11:55:14 am
Suppose we have the hyperbola y = 1/x defined over the positive real numbers (first quadrant). Let P(p, 1/p) and Q(q, 1/q) be two arbitrarily fixed points along the curve, with p < q. Define M as the midpoint of the chord PQ. The line segment OM intersects the hyperbola at R(r, 1/r), where O is the origin (0,0).
Without expressing the coordinates of R in terms of p and q, i.e. without deriving the equation of the line OM, prove that the tangent to the hyperbola at R is parallel to the chord PQ.
Mod edit: Altered the language to make it a bit "easier" to comprehend with respect to the HSC 4U course
Post by: RuiAce on August 23, 2016, 08:52:47 am
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-08-23%20at%208.48.43%20AM_zpsbz1074hy.png)
Are you looking for a fully geometric proof? Because when I first glance at it, trying to prove pq=r2 is the easiest way to go about it but it feels pointless if the equation of OM is denied.
Post by: Ali_Abbas on August 23, 2016, 09:41:27 am
Gonna leave a diagram for anyone who attempts it.
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-08-23%20at%208.48.43%20AM_zpsbz1074hy.png)
Are you looking for a fully geometric proof? Because when I first glance at it, trying to prove pq=r2 is the easiest way to go about it but it feels pointless if the equation of OM is denied.
Indeed, proving r^2 = pq (of which there are four ways to do so) is the most immediate way to achieve the desired result. However, as you have alluded to, I am in fact looking for a full geometric proof (not strictly Euclidean) just as a way to make it slightly more challenging.
Post by: Paradoxica on October 05, 2016, 06:32:06 pm
Let P(k) be the number of ways you can use exactly k different letters in an n letter word.
i) Explain why
ii) Show that
Let the Score of a word, X, be defined as 1/(1+ρ(X)), where ρ(X) is the number of letters that were not used by the word X.
iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:
iv) Hence, evaluate S in closed form.
Post by: Paradoxica on October 05, 2016, 06:44:33 pm
Hint: It's a logarithm
;D
Post by: RuiAce on October 05, 2016, 07:15:14 pm
'
Hint: It's a logarithm
;D
Spoiler
Moderator Edit: Added spoiler to solution
Post by: jamonwindeyer on October 05, 2016, 07:17:39 pm
Post by: RuiAce on October 05, 2016, 07:27:38 pm
Jesus Rui that was quick, save some for the students ;)So not touching perms and combs lel
Eh, I've seen integrals way worse than that, and I couldn't get them out. This one was only barely in my doable margin because of the hint. So I was just like 'why not'.
Post by: Paradoxica on October 05, 2016, 07:28:48 pm
Hint: Divide.
Post by: RuiAce on October 06, 2016, 11:34:54 pm
I considered the hint, but it took me way too long to figure out what happens after the hint is applied, hence all of this.
I won't put up the solution to the 3x quicker method just yet.
Unnecessarily over complicated non-HSC solution
(http://uploads.tapatalk-cdn.com/20161006/1273c354f7b7aa54fa08ea9caf390900.jpg)
Post by: Paradoxica on October 06, 2016, 11:43:30 pm
Post by: Mahan on November 08, 2016, 10:49:28 pm
By dividing throughout by, or otherwise, evaluate:
Since, this is a relatively old post, I thought it would be useful to give a solution for it.
This method doesn't use the dividing trick:
before I start the proof it is useful to prove:
let
by back substitution we can write it in terms of x.
the answer is
Post by: RuiAce on November 08, 2016, 10:56:34 pm
Post by: Mahan on November 08, 2016, 11:03:46 pm
Yes, using the hint makes the question pretty simple.Just for the fun of it, I just presented a different proof. :)
Post by: RuiAce on November 08, 2016, 11:16:36 pm
Hint: Divide.
Post by: RuiAce on November 21, 2016, 11:24:12 am
Consider an alphabet with n different available letters.To help explain part (i) (because I took a while understanding what was going on)
Let P(k) be the number of ways you can use exactly k different letters in an n letter word.
i) Explain why
ii) Show that
Let the Score of a word, X, be defined as 1/(1+ρ(X)), where ρ(X) is the number of letters that were not used by the word X.
iii) Show that the sum of all the Scores, S, over all possible n letter words, is given by:
iv) Hence, evaluate S in closed form.
Explanation
Consider a three letter alphabet: {a, b, c}
Then, P(1) is the number of ways we can make three letter words, out of just one letter of the alphabet. If that letter was a, then aaa is the only word.
P(2) is the number of ways we can make three letter words, out of any two letters of the alphabet. If the letters are a and b, then the words are:
aab, aba, baa, bba, bab, abb
P(3) is the number of ways we can make three letter words using all the letters. So that's the easy 3!
The point of introducing the nCk is to quantify the fact we could've chosen any k of the 3 letters. In P(1), we could've chosen a, b or c to be our letter. (And indeed, 3Ck=3 possible letters.)
And now for the question
Then, P(1) is the number of ways we can make three letter words, out of just one letter of the alphabet. If that letter was a, then aaa is the only word.
P(2) is the number of ways we can make three letter words, out of any two letters of the alphabet. If the letters are a and b, then the words are:
aab, aba, baa, bba, bab, abb
P(3) is the number of ways we can make three letter words using all the letters. So that's the easy 3!
The point of introducing the nCk is to quantify the fact we could've chosen any k of the 3 letters. In P(1), we could've chosen a, b or c to be our letter. (And indeed, 3Ck=3 possible letters.)
And now for the question
_________________________________
_________________________________
If a 4U student has not seen that Greek letter before, that is rho.
Explanation
Going back to the case n=3
If the words satisfied k=1 (e.g. aaa), then ρ(X) = 2 (Note: 1+2=3)
If the words satisfied k=2 (e.g. aab), then ρ(X) = 1 (Note: 2+1=3)
If the words satisfied k=3 (e.g. abc), then ρ(X) = 0 (Note: 3+0=3)
This can be seen by just realising how man letters we did not use, in each case.
The important thing to realise is that for each value of k, ρ(X) differs. As a matter of fact, ρ(X) always goes down by 1.
In fact, ρ(X) = 3 - k
If the words satisfied k=1 (e.g. aaa), then ρ(X) = 2 (Note: 1+2=3)
If the words satisfied k=2 (e.g. aab), then ρ(X) = 1 (Note: 2+1=3)
If the words satisfied k=3 (e.g. abc), then ρ(X) = 0 (Note: 3+0=3)
This can be seen by just realising how man letters we did not use, in each case.
The important thing to realise is that for each value of k, ρ(X) differs. As a matter of fact, ρ(X) always goes down by 1.
In fact, ρ(X) = 3 - k
_________________________________
I don't fully trust what I say from here due to how P(k) has been defined.
Post by: Paradoxica on March 16, 2017, 10:23:44 pm
Here's a question for people to try:
Suppose we have the hyperbola y = 1/x defined over the positive real numbers (first quadrant). Let P(p, 1/p) and Q(q, 1/q) be two arbitrarily fixed points along the curve, with p < q. Define M as the midpoint of the chord PQ. The line segment OM intersects the hyperbola at R(r, 1/r), where O is the origin (0,0).
Without expressing the coordinates of R in terms of p and q, i.e. without deriving the equation of the line OM, prove that the tangent to the hyperbola at R is parallel to the chord PQ.
Mod edit: Altered the language to make it a bit "easier" to comprehend with respect to the HSC 4U course
Consider an arbitrary chord AB parallel to PQ using the following parameters: A(p/r,r/p), B(qr,1/qr) where r is a positive scaling factor. Construct tangents at A and B and denote the point of intersection I.
With some algebra, it is verified that OI and OM share the same gradient. Thus, I always lies on OM for any and every positive value of r.
Taking the limit as
The tangent at point P is unique to P, so conversely, the point I degenerates and becomes P.
Post by: wu345 on April 06, 2017, 07:33:58 pm
The root locus of a complex quadratic is the set of all points on the Argand Diagram that could be a root of the quadratic.
Sketch the root locus of
Post by: wu345 on April 06, 2017, 07:54:23 pm
i) Show that
ii) Show that
iii) Hence show that
iv) Hence show that
Post by: RuiAce on April 07, 2017, 04:49:14 pm
Letbe a root of the equation
, where
The root locus of a complex quadratic is the set of all points on the Argand Diagram that could be a root of the quadratic.
Sketch the root locus of
________________________________________
Note that through these computations, the points (1,0) and (-1,0) are technically excluded. Coincidentally, the other case(s) brings it back.
________________________________________
________________________________________
Note that the above line was an abuse of notation. Infinity is not a number.
Proof of the limit used on g(p)
(Brief) Explanation as to why the functions decrease/increase from -1 respectively
In a similar way, the other case of \( p\le -2\) shows that \( y=0\) is a part of the locus for all \( x > 0\).
________________________________________
Post by: wu345 on April 07, 2017, 07:22:23 pm
Post by: RuiAce on June 28, 2017, 05:13:03 pm
Hint
You only have to consider when \(|\alpha|=1\). After you prove it holds for \( |\alpha|=1\), the case \( |\alpha| < 1 \) falls out pretty easily from it
Spoiler
You need to know circle geometry as well as have a substantial understanding of complex number tricks.
Post by: Ali_Abbas on July 02, 2017, 03:10:41 pm
A classic complex numbers question I just did a few minutes ago.
I believe I have a solution that differs from that of Rui's and is as follows.
Post by: RuiAce on July 02, 2017, 03:35:52 pm
(http://i.imgur.com/raQnW5n.png)
Post by: RuiAce on July 02, 2017, 03:44:07 pm
Hint
Can be made much easier using one of the tricks used in the previous problem
Spoiler
Find a contradiction!
Post by: Ali_Abbas on July 02, 2017, 08:23:21 pm
Tbh all I did was this.
(http://i.imgur.com/raQnW5n.png)
You need to be careful when you apply the triangle inequality on a difference of terms as opposed to a sum of terms (occurring within the modulus signs). Although it will still give you a correct upper bound, it doesn't always give the true maximum of the expression. To see this, we can apply the triangle inequality on |alpha-1| + |alpha+1| giving:
But clearly 4 is not the true maximum so I'm not sure if your application of the triangle inequality qualifies as a concrete proof or if it merely gave the true maximum of the sum by shear coincidence.
Post by: RuiAce on July 02, 2017, 08:24:35 pm
You need to be careful when you apply the triangle inequality on a difference of terms as opposed to a sum of terms (occurring within the modulus signs). Although it will still give you a correct upper bound, it doesn't always give the true maximum of the expression. To see this, we can apply the triangle inequality on |alpha-1| + |alpha+1| giving:The question did not require the least upper bound. I had explicitly stated that you need only prove this for 2sqrt(2).
But clearly 4 is not the true maximum so I'm not sure if your application of the triangle inequality qualifies as a concrete proof or if it merely gave the true maximum of the sum by shear coincidence.
Post by: Ali_Abbas on July 02, 2017, 08:30:31 pm
The question did not require the least upper bound. I had explicitly stated that you need only prove this for 2sqrt(2).
But I thought 2sqrt(2) is the least upper bound...isn't it?
Post by: RuiAce on July 02, 2017, 08:32:05 pm
But I thought 2sqrt(2) is the least upper bound...isn't it?Yes, but the identification of this was not necessary for this question. Therefore my proof still holds validity.
I did not see the point in forcing the requirement of the least upper bound when giving this question to Ext 2 students, nor did the original question specify this either.
Post by: Ali_Abbas on July 02, 2017, 08:40:37 pm
Yes, but the identification of this was not necessary for this question. Therefore my proof still holds validity.
I did not see the point in forcing the requirement of the least upper bound when giving this question to Ext 2 students, nor did the original question specify this either.
Alright I see what you're saying but it is my opinion that had the question asked for some other upper bound (e.g. 4), then although one can easily show it, it would be a silly question and a bit misleading to overstate the range of values the sum can take. It seems common sense to assume it to be the least upper bound so that we can be certain that the equation |alpha-1| + |alpha+1| = k will always have a solution when |alpha| <= 1 and k <= 2sqrt(2).
Post by: RuiAce on July 02, 2017, 08:44:36 pm
Post by: RuiAce on July 21, 2017, 12:17:55 am
Post by: RuiAce on July 21, 2017, 07:27:14 pm
Post by: RuiAce on July 21, 2017, 07:37:34 pm
Post by: Ali_Abbas on July 22, 2017, 12:32:55 am
Post by: ellipse on July 22, 2017, 12:58:17 pm
Post by: RuiAce on July 22, 2017, 01:54:18 pm
Post by: RuiAce on July 29, 2017, 05:08:27 pm
Post by: pikachu975 on August 01, 2017, 11:47:44 pm
It's too hard for me to type out since I can't use latex, but would it be right to expand the brackets using binomial expansion then integrate and simplify?
Post by: RuiAce on August 02, 2017, 12:02:08 am
It's too hard for me to type out since I can't use latex, but would it be right to expand the brackets using binomial expansion then integrate and simplify?I just tried doing that. Pretty much gave up after line 3. (Which means it may actually work, but it's going to be really unpleasant.)
Think about what you normally do when you see random n's appearing in integrals.
Post by: pikachu975 on August 02, 2017, 01:29:38 am
I just tried doing that. Pretty much gave up after line 3. (Which means it may actually work, but it's going to be really unpleasant.)
Think about what you normally do when you see random n's appearing in integrals.
Integration by parts then? I'll try it tomorrow I've got the CSSA 4u on thursday so this will be a cool question.
Post by: RuiAce on September 08, 2017, 10:53:36 pm
I had a go, and I promise it's doable within MX2 boundaries.
Post by: ellipse on September 10, 2017, 05:01:24 pm
This question first appeared as a meme, ironically enough. At first glance, it looks ridiculous. But it's all about finding the right starting point...
I had a go, and I promise it's doable within MX2 boundaries.
its a telescoping series,
hence you'll end up with the first term (J_0/I_0) + the last term (J_n)/(I_n)
as n goes to infinity, the last term disappears (ill try to prove this later) so evaluating J_0 and I_0, we get pi^2/6
its interestingly the basel problem, and also appeared as the last q in 2010 mx2
Post by: RuiAce on September 10, 2017, 05:15:02 pm
its a telescoping series,Of course, the right idea.
hence you'll end up with the first term (J_0/I_0) + the last term (J_n)/(I_n)
as n goes to infinity, the last term disappears (ill try to prove this later) so evaluating J_0 and I_0, we get pi^2/6
its interestingly the basel problem, and also appeared as the last q in 2010 mx2
Care to show some working out now? You wouldn't be able to say in an exam, by inspection, the last term vanishes.
Edit: Had a look, because I definitely remembered the π^2/6 in the 2010 exam. Didn't realise the final result was the same.
But pretty much as you alluded to there's a faster way than the 2010 exam method.
Post by: Paradoxica on October 24, 2017, 11:30:18 pm
*briefly tries bashing*
*realises polar co-ordinates is faster*
*facepalms*
Post by: Shlomo314 on October 30, 2017, 02:59:29 pm
Post by: RuiAce on October 30, 2017, 04:59:18 pm
i guessing this is wrong because pythag can only be used on the real plane and the imaginary doesn't exist on the im plane?Lol I saw this as a meme.
The fallacy is in that Pythagoras theorem uses lengths, so you need to take absolute values (moduli). It's a good joke though
Post by: RuiAce on November 11, 2017, 03:45:43 pm
Extra credit: Rinse and repeat using \( \ge \) instead.
Post by: RuiAce on November 13, 2017, 10:08:40 pm
Post by: RuiAce on November 21, 2017, 03:23:26 pm
Post by: ellipse on December 03, 2017, 08:33:03 pm
Post by: RuiAce on December 04, 2017, 05:41:30 pm
Post by: RuiAce on December 10, 2017, 03:12:12 pm
We prove this, by demonstrating that in every set of \(n\) objects, all \(n\) of them have the same colour. Note that this is an equivalent statement, and thus is also valid to prove.
When \(n=1\) this is trivially true, because any set that has only has one object will only have one colour, so all objects must have the same colour.
Assume that every set with \(k\) objects have the same colour. Consider what happens when \(n=k+1\).
For any arbitrary set with \(k+1\) elements, we may label the elements in it, so that the set takes the form \( \{1,2,3,\dots,k,k+1\} \). A subset of this set would just be the first \(k\) elements, i.e. the set \( \{1,2,3,\dots,k\} \). Since this is one of the many sets that have \(k\) elements, from our inductive assumption all elements in this set must have the same colour.
Now, another subset of this set would be the last \(k\) elements, i.e. the set \( \{ 2,3,\dots,k,k+1\} \). Again, this is one of many sets that have \(k\) elements, so all the elements in this set must also have the same colour.
But these two sets have overlaps, so by consequence all elements in \( \{1,2,3,\dots,k,k+1\} \) must have the same colour. Hence, all elements in an arbitrary set with \(k+1\) elements all have the same colour.
So it follows by induction that in every set of \(n\) objects, all colours of the objects in said sets are the same, and hence every colour must be the same.
This is obviously wrong. But why?
Post by: RuiAce on January 21, 2018, 06:05:15 pm
(Won't provide hints unless someone actually attempts it.)
Post by: Opengangs on April 24, 2018, 09:40:03 pm
About the Putnam Competition
The William Lowell Putnam Mathematical Competition (often just abbreviated to Putnam) is an annual Mathematics competition geared towards undergraduates, who all compete for prize money and scholarships to some of the most prestigious and highly regarded universities in the world. Examples include: Harvard University, MIT, Princeton, Carnegie Mellon, etc.
The test is divided up into halves (A and B). It's a typical 6 hour long paper (3 each), with only 12 questions. The maximum amount of marks awarded for each question is 10 marks. Even though there are 120 points to be awarded, the median is only around 1-5 points. The top ever score was 63/120 by students at Harvard University.
The test is divided up into halves (A and B). It's a typical 6 hour long paper (3 each), with only 12 questions. The maximum amount of marks awarded for each question is 10 marks. Even though there are 120 points to be awarded, the median is only around 1-5 points. The top ever score was 63/120 by students at Harvard University.
The very last question was a Question 6 from a previous Putnam paper, and the goal here is to walk you through the process by first depicting an easier way in approaching mathematical thinking.
Post by: TheAspiringDoc on May 16, 2018, 06:22:26 pm
This may be hard to believe, but I promise this question is doable via only 4U methods.I’m gonna guess the integrand is an odd function and therefore the definite integral is 0?
(Won't provide hints unless someone actually attempts it.)
Post by: RuiAce on May 16, 2018, 06:24:09 pm
I’m gonna guess the integrated is an odd function and therefore the definite integral is 0?You'll find that \(f(-x) \neq -f(x)\).
Edit: I had a go at this question again. I will post the answer, but not any solution yet.
\[ \frac\pi{8072} \]
Seeing as though this is being used as a challenge, I'll just explain a bit about what's going on with the notation.
Not sure if this is a challenge question or not (depends on whether there's a simpler solution than mine which there probably is), but I'll post it here anyways.
I'm only like 85-90% sure my proof of part (i) is correct as it's quite lengthy and slightly complicated but I did put a lot of thought into it so hopefully it's fine.
\( \mathbb{N} \backslash \{ 2\} \) is just asking for all natural numbers, i.e. 0, 1, 2, 3, 4, 5, 6, ..., but except 2. So if you read on, we're really just assuming \(n\) is a prime number not equal to 2.
Part i is easy enough to visualise. But be careful how you argue it........that might be harder than actually doing parts ii and iii. In part ii, basically you know that \( \sum \) means to add all the terms. \( \prod\) means to multiply them instead.
Post by: stolenclay on May 31, 2018, 02:55:46 am
This may be hard to believe, but I promise this question is doable via only 4U methods.
(Won't provide hints unless someone actually attempts it.)
It's been a while since I did maths on AN, but I spent too long on this to not cash in ::) Hopefully there aren't any mistakes!
Let \(k = 1009\). Then the integral becomes
\[
\int_{-\pi/(4k)}^{\pi/(4k)} \frac{1}{(2k)^{2kx}+1} \frac{\cos^{2k}(2kx)}{\sin^{2k}(2kx) + \cos^{2k}(2kx)} \, dx.
\]
Make the change of variables \(x = r/(2k)\). Then \(dx/dr = 1/(2k)\) and the integral becomes
\[
\int_{-\pi/2}^{\pi/2} \frac{1}{(2k)^{r}+1} \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r} \frac{1}{2k} \, dr.
\]
Define the functions \(f\colon \mathbb{R} \to \mathbb{R}\) and \(g\colon \mathbb{R} \to \mathbb{R}\) by
\[
f(r) = \frac{1}{(2k)^r + 1}
\quad \text{and} \quad
g(r) = \frac{\cos^{2k}r}{\sin^{2k}r + \cos^{2k}r}
\quad \forall r \in \mathbb{R}.
\]
Then the integral we wish to compute is
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx.
\]
Hint 1 (for my particular solution)
Try to use \(f(x) g(x) = [1 - f(x)] [1 - g(x)] + f(x) + g(x) - 1\).
Hint 2 (for my particular solution)
As is usual with these types of questions, we wish to make use of identities/symmetries satisfied by \(f\) and \(g\). For this particular problem, we will use the result that for every \(x \in \mathbb{R}\) we have
 &= 1 - f(-x), \\<br />g(x) &= g(-x), &\quad& \text{and} \\<br />g(x) &= 1 - g(x + \pi/2).<br />\end{alignat*}<br />)
I leave it to the reader to verify that these hold. It is perhaps a bit arbitrary to begin with these, and in fact these are not identities that come out of thin air; rather they are motivated by the observation in Hint 1.
I leave it to the reader to verify that these hold. It is perhaps a bit arbitrary to begin with these, and in fact these are not identities that come out of thin air; rather they are motivated by the observation in Hint 1.
Solution
We integrate the equation in Hint 1 in \(x\) over the interval \([-\pi/2, \pi/2]\) to obtain
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \underbrace{\int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx}_{I_1} + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Let \(I_1\) be the integral as indicated above. We use a change of variables \(x = -r\). Then \(dx/dr = -1\) and \(I_1\) becomes
\[
I_1
= \int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx
= \int_{\pi/2}^{-\pi/2} \underbrace{[1 - f(-r)]}_{{} = f(r)} [1 - \underbrace{g(-r)}_{{} = g(r)}] (-1) \, dr
= \int_{-\pi/2}^{\pi/2} f(r) [1 - g(r)] \, dr.
\]
Note that we have used some of the identities in Hint 2. We now have
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \int_{-\pi/2}^{\pi/2} f(x) [1 - g(x)] \, dx + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Adding \(\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx\) to both sides, we obtain
\[
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= 2\int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi
= 2\int_{-\pi/2}^{\pi/2} \underbrace{\bigg[f(x) - \frac{1}{2}\bigg]}_{\text{odd in \(x\)}} \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx
= \int_{-\pi/2}^{\pi/2} g(x) \, dx.
\]
Note that the fact that \(f(x) - 1/2\) is odd in \(x\) follows from an identity in Hint 2. We now wish to compute \(\int_{-\pi/2}^{\pi/2} g(x) \, dx\). By evenness in \(x\) of \(g(x)\), this is the same as \(2\int_{0}^{\pi/2} g(x) \, dx\), which is also the same as \(2\int_{-\pi/2}^{0} g(x) \, dx\). Since \(g(x) = 1 - g(x + \pi/2)\) for all real \(x\), making the substitution \(x = r - \pi/2\), we have
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= 2\int_{-\pi/2}^{0} g(x) \, dx
= 2\int_{-\pi/2}^{0} [1 - g(x + \pi/2)] \, dx
= 2\int_{0}^{\pi/2} [1 - g(r)] \, dr.
\]
We now write
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= \frac{1}{2} \Bigg[\int_{-\pi/2}^{\pi/2} g(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx\Bigg]
= \frac{1}{2} \Bigg[2\int_{0}^{\pi/2} g(x) \, dx + 2\int_{0}^{\pi/2} [1 - g(x)] \, dx\Bigg]
= \frac{\pi}{2}
\quad \implies \quad
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{2}.
\]
Dividing both sides by \(4k\), we obtain
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{8k}.
\]
Recalling that \(k = 1009\), we have
\[
\int_{-\pi/4036}^{\pi/4036} \frac{1}{2018^{2018x}+1} \frac{\cos^{2018}(2018x)}{\sin^{2018}(2018x) + \cos^{2018}(2018x)} \, dx
= \frac{\pi}{8072}.
\]
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \underbrace{\int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx}_{I_1} + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Let \(I_1\) be the integral as indicated above. We use a change of variables \(x = -r\). Then \(dx/dr = -1\) and \(I_1\) becomes
\[
I_1
= \int_{-\pi/2}^{\pi/2} [1 - f(x)] [1 - g(x)] \, dx
= \int_{\pi/2}^{-\pi/2} \underbrace{[1 - f(-r)]}_{{} = f(r)} [1 - \underbrace{g(-r)}_{{} = g(r)}] (-1) \, dr
= \int_{-\pi/2}^{\pi/2} f(r) [1 - g(r)] \, dr.
\]
Note that we have used some of the identities in Hint 2. We now have
\[
\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \int_{-\pi/2}^{\pi/2} f(x) [1 - g(x)] \, dx + \int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi.
\]
Adding \(\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx\) to both sides, we obtain
\[
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= 2\int_{-\pi/2}^{\pi/2} f(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx - \pi
= 2\int_{-\pi/2}^{\pi/2} \underbrace{\bigg[f(x) - \frac{1}{2}\bigg]}_{\text{odd in \(x\)}} \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx
= \int_{-\pi/2}^{\pi/2} g(x) \, dx.
\]
Note that the fact that \(f(x) - 1/2\) is odd in \(x\) follows from an identity in Hint 2. We now wish to compute \(\int_{-\pi/2}^{\pi/2} g(x) \, dx\). By evenness in \(x\) of \(g(x)\), this is the same as \(2\int_{0}^{\pi/2} g(x) \, dx\), which is also the same as \(2\int_{-\pi/2}^{0} g(x) \, dx\). Since \(g(x) = 1 - g(x + \pi/2)\) for all real \(x\), making the substitution \(x = r - \pi/2\), we have
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= 2\int_{-\pi/2}^{0} g(x) \, dx
= 2\int_{-\pi/2}^{0} [1 - g(x + \pi/2)] \, dx
= 2\int_{0}^{\pi/2} [1 - g(r)] \, dr.
\]
We now write
\[
\int_{-\pi/2}^{\pi/2} g(x) \, dx
= \frac{1}{2} \Bigg[\int_{-\pi/2}^{\pi/2} g(x) \, dx + \int_{-\pi/2}^{\pi/2} g(x) \, dx\Bigg]
= \frac{1}{2} \Bigg[2\int_{0}^{\pi/2} g(x) \, dx + 2\int_{0}^{\pi/2} [1 - g(x)] \, dx\Bigg]
= \frac{\pi}{2}
\quad \implies \quad
2\int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{2}.
\]
Dividing both sides by \(4k\), we obtain
\[
\frac{1}{2k} \int_{-\pi/2}^{\pi/2} f(x) g(x) \, dx
= \frac{\pi}{8k}.
\]
Recalling that \(k = 1009\), we have
\[
\int_{-\pi/4036}^{\pi/4036} \frac{1}{2018^{2018x}+1} \frac{\cos^{2018}(2018x)}{\sin^{2018}(2018x) + \cos^{2018}(2018x)} \, dx
= \frac{\pi}{8072}.
\]
Comments
This holds for all positive integers \(k\). Integrality of \(k\) is used when the identities regarding the function \(g\) in Hint 2 are derived.
I changed the integration variable from \(r\) to \(x\) multiple times without comment. I'm not sure if it's possible for marks to be lost for this under a HSC marking scheme, so be careful.
I changed the integration variable from \(r\) to \(x\) multiple times without comment. I'm not sure if it's possible for marks to be lost for this under a HSC marking scheme, so be careful.
Post by: Ali_Abbas on May 31, 2018, 10:03:26 pm
This may be hard to believe, but I promise this question is doable via only 4U methods.
(Won't provide hints unless someone actually attempts it.)
Alright well I have an alternative solution to that of stolenclay's and is as follows:
Post by: RuiAce on May 31, 2018, 10:18:42 pm
The intended method basically involved three substitutions, the first one being the obvious \( u = 2018x\)
Post by: Opengangs on October 23, 2018, 09:50:57 am
\[ \text{(i) Show that:} \int_0^{\frac{\pi}{2}}\ln(\sin(x))\,dx = \int_0^{\frac{\pi}{2}}\ln(\cos(x))\,dx. \]\[ \text{(ii) By considering } 2I = \int_0^{\frac{\pi}{2}} \ln(\sin(x)\cos(x))\,dx, \text{ show that: } 2I = -\frac{\pi}{2}\ln2 + I.\]\[ \text{And hence, evaluate } \int_0^{\frac{\pi}{2}}\ln(\sin(x))\,dx.\]
Post by: Opengangs on May 12, 2020, 11:36:06 am
Post by: RuiAce on May 13, 2020, 11:00:15 pm
Prove that among any ten points located on a circle with diameter 5, there exist at least two at a distance less than 2 from each other.
Spoiler
Note that the radius of the circle is \(r = \frac{5}{2}\).
Partition the circle into nine congruent sectors. Each of these sectors will be formed by an angle of \( \frac{2\pi}{9}\) at the circle's centre.
By the pigeonhole principle, at least two such points must lie on one of the nine arcs corresponding to the sectors. Consider any such arc that has two points lying on it.
Form a triangle between these two points on the arc and the centre of the circle. Note that two of the triangles' edges will be radii of the circle, and the third is the distance between the two points.
Let the angle made at the centre of the circle be \(\theta\). On one hand, it now follows that \( \theta \in \left[0, \frac{2\pi}{9}\right]\). On the other hand, from the cosine rule,
\begin{align*}
D &= \sqrt{\left( \frac52 \right)^2 + \left( \frac52\right)^2 - 2 \left( \frac52\right) \left( \frac52\right) \cos \theta}\\
&= \frac{5}{2} \sqrt{2} \sqrt{1-\cos\theta}
\end{align*}
where \(D\) is the distance between the two points.
Since \(\theta\in \left[ 0,\frac{2\pi}{9}\right] \), we can simply plug in the values \(\theta = 0\) and \(\theta = \frac{2\pi}{9}\) to find that the boundary values of \(D\) are \(0\) and \(1.7101\), rounded to 4 decimal places. Note that plugging in was all that was required, because \(f(x) = \cos x\) is monotone on the open interval \( \left( 0, \frac{2\pi}{9}\right) \). (In particular, monotonic decreasing.)
But in any case, we see that the possible values for \(D\) are all strictly less than \(2\). So indeed, these two points of interest are always at a distance of less than 2 from each other.
Partition the circle into nine congruent sectors. Each of these sectors will be formed by an angle of \( \frac{2\pi}{9}\) at the circle's centre.
By the pigeonhole principle, at least two such points must lie on one of the nine arcs corresponding to the sectors. Consider any such arc that has two points lying on it.
Form a triangle between these two points on the arc and the centre of the circle. Note that two of the triangles' edges will be radii of the circle, and the third is the distance between the two points.
Let the angle made at the centre of the circle be \(\theta\). On one hand, it now follows that \( \theta \in \left[0, \frac{2\pi}{9}\right]\). On the other hand, from the cosine rule,
\begin{align*}
D &= \sqrt{\left( \frac52 \right)^2 + \left( \frac52\right)^2 - 2 \left( \frac52\right) \left( \frac52\right) \cos \theta}\\
&= \frac{5}{2} \sqrt{2} \sqrt{1-\cos\theta}
\end{align*}
where \(D\) is the distance between the two points.
Since \(\theta\in \left[ 0,\frac{2\pi}{9}\right] \), we can simply plug in the values \(\theta = 0\) and \(\theta = \frac{2\pi}{9}\) to find that the boundary values of \(D\) are \(0\) and \(1.7101\), rounded to 4 decimal places. Note that plugging in was all that was required, because \(f(x) = \cos x\) is monotone on the open interval \( \left( 0, \frac{2\pi}{9}\right) \). (In particular, monotonic decreasing.)
But in any case, we see that the possible values for \(D\) are all strictly less than \(2\). So indeed, these two points of interest are always at a distance of less than 2 from each other.
Post by: Opengangs on May 22, 2020, 03:40:11 pm
For any positive integer \(a\) and \(b\), if \(ab + 1\) or more objects are placed into \(b\) boxes, then prove that one box will contain more than \(a\) objects.
Hint
Look for a contradiction.