Post by: dcc on April 11, 2009, 11:50:19 am
1.) Find
Solution 1 - Over9000
Solution 2 - dcc
2.) Show that
Solution 1 - TrueTears
Solution 2 - Neobeo
Solution 3 - dcc
3.) Show
4.) Show that
Solution 1 - /0
5.) For a real number
(Source: 1995 Hosei University entrance exam/Business administration)
Solution 1 - coblin
6.) Find the minimum area of the part bounded by the parabola
(Source: 1963 Tokyo Metropolitan University entrance exam)
Solution 1 - Neobeo
Solution 2 - TrueTears
7.) Evaluate
(Source: 2008 Miyazaki University entrance exam/Agriculture)
Solution 1 - kamil9876
8.) Evaluate
(Source: Wikipedia)
Solution 1 - TrueTears
Solution 2 - dcc
9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?
(Source: Neobeo)
Solution 1 - /0
Solution 2 - kamil9876
10.) Show (without calculus) that the minimum value of
Solution 1 - Damo17
11.) Find the maximum value of
Solution 1 - humph
12.) Prove that
(Source: TrueTears)
Solution 1 - Over9000
13.) Show that
(Source: Damo17)
Solution 1 - dcc
Solution 2 - Over9000
Solution 3 - golden
14.) Let N be the positive integer with 2008 decimal digits, all of them 1. That is,
(Source: /0 - Melbourne University/BHP Billiton Maths Competition 2008)
Solution 1 - Over9000
15.) Neobeo is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'. Neobeo notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream. Upon walking outside any particular shop, Neobeo feels a huge compulsion to purchase an ice-cream. For every shop that Neobeo visits, he is
(Source: IRC & the recesses of my brain)
Solution 1 - golden
16.) Find
Solution 1 - TrueTears
Solution 2 - kamil9876
Solution 3 - dcc
17.) Find
18.) Find
Do not use L'Hospital's rule to evaluate this, because that is boring. Try and use limit laws and properties, rather then mindless derivatives.
MORE TO COME, I WILL EDIT THIS POST.
Post by: enwiabe on April 11, 2009, 12:10:07 pm
Post by: TrueTears on April 11, 2009, 01:57:22 pm
Let
Post by: Over9000 on April 11, 2009, 03:12:49 pm
lets focus on
so
so
that means
however
so
now we add the \cos4x we got all the way back at the top and we get
so
Therefore to sum up
Post by: dcc on April 11, 2009, 04:18:41 pm
Let
Note:
Post by: dcc on April 11, 2009, 04:28:33 pm
Post by: kamil9876 on April 11, 2009, 06:14:06 pm
Post by: dcc on April 11, 2009, 08:14:59 pm
I've attatched the trick to number7, rest is trivial.
How hard would it be to just go the extra yard and finish it off? :P
Post by: /0 on April 11, 2009, 08:32:44 pm
The rest follows from the identity
All steps are reversible if we assume
For the more general case, the cheap way to solve is to simply say:
Post by: kamil9876 on April 11, 2009, 09:16:20 pm
I've attatched the trick to number7, rest is trivial.
How hard would it be to just go the extra yard and finish it off? :P
It was all in the spirit of mathematics:
"A physicist and engineer and a mathematician were sleeping in a hotel room when a fire broke out in one corner of the room. Only the engineer woke up he saw the fire, grabbed a bucket of water and threw it on the fire and the fire went out, then he filled up the bucket again and threw that bucketfull on the ashes as a safety factor, and he went back to sleep. A little later, another fire broke out in a different corner of the room and only the physicist woke up. He went over measured the intensity of the fire, saw what material was burning and went over and carefully measured out exactly 2/3 of a bucket of water and poured it on, putting out the fire perfectly; the physicist went back to sleep. A little later another fire broke out in a different corner of the room. Only the mathematician woke up. He went over looked at the fire, he saw that there was a bucket and he noticed that it had no holes in it; he turned on the faucet and saw that there was water available. He, thus, concluded that there was a solution to the fire problem and he went back to sleep."
Post by: Collin Li on April 11, 2009, 09:35:30 pm
When
When
When
Post by: Neobeo on April 12, 2009, 12:34:16 pm
6.) Find the minimum area of the part bounded by the parabolaand the line
.
(Source: 1963 Tokyo Metropolitan University entrance exam)
It is clear that
Also define
Minimising area, we get
Backsolving,
Then the integral for the area under the curve is simply:
Post by: Neobeo on April 12, 2009, 01:03:58 pm
2.) Show thatgiven
.
(http://img21.imageshack.us/img21/4342/superfun.png)
Post by: TrueTears on April 12, 2009, 01:33:55 pm
and
(Say
after sketching the graphs of
The intersection points in terms of a are
These are the integration limits so:
To find the minimum area we require
solving for a yields
Therefore the intersection points become
Post by: dcc on April 12, 2009, 01:51:50 pm
2.) Show that
Note that
Therefore
Post by: Flaming_Arrow on April 13, 2009, 01:23:14 am
Post by: /0 on April 13, 2009, 02:32:31 am
9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?
(Source: Neobeo)
Let the stick of length L be broken into fragments of length x, y and z.
Since the problem is continuous probability, it's best to use a graph or something like that.
Hence, all possible breaks lie on the plane
When the above graph is drawn, the area is represented by an equilateral triangle, and it is
_______________________________________
Now adding the restrictions
Since
So the allowable area is
So the triangle is equilateral, and
_______________________________________
So the probability is
fun problem
Post by: hard on April 13, 2009, 02:45:58 pm
Post by: TrueTears on April 13, 2009, 04:00:10 pm
First just some derivations:
Thus
Back to the question:
Treating the x in the question as a constant
For the definite integral with lower limit q and upper limit p:
To get the definite integral for this question let p = 1 and q = 0 , this yields:
BUT
subbing this in leads to a double integral namely:
after changing the order of integration we have:
So now we are integrating with respect to x first thus we treat t as a constant
Good learning experience for me this question.
Post by: Damo17 on April 13, 2009, 04:37:28 pm
10.) Show (without calculus) that the minimum value ofis
.
Using the theorem:
If n positive functions have a fixed product, there sum is a minimum if it can be arranged that the functions are equal.
let
Although the three terms of the sum have a constant product, we run into trouble if we apply the aforementioned theorem as there is no value of x that satisfies these equations.
But if we set aside the constant,
The minimum comes by equating:
We take the positive solution as per the restriction of the question.
(using the fact that if
Post by: dcc on April 13, 2009, 04:48:17 pm
8.) Evaluate
(Source: Wikipedia)
Post by: Over9000 on April 13, 2009, 05:01:36 pm
Prove
let x be a rational number
let
let
A rational result for this is impossible, therefore
Post by: dcc on April 13, 2009, 06:10:33 pm
13.) Show that.
(Source: Damo17)
Post by: Over9000 on April 13, 2009, 06:36:54 pm
Divide both sides by 9
Post by: Over9000 on April 13, 2009, 09:02:14 pm
We must find the 1005th digit after the decimal point in the expansion
so we begin to see a trend where an even amount of digits of 1 are of the form 3.31..... and each time two more digits of 1 are added there is a 3 added to left side of decimal point and a 3 added to right side of decimal point.
so for example, a term with 8 digits of 1 square rooted i.e
so for an integer with 2008 digits of 1, there are 1004 3's before the decimal point and 1004 3's after it. We are looking for the 1005th term and as we can see in the trend, the last 3 is always followed up by a 1, therefore the 1005th term after the decimal point in an integer with 2008 digits of 1 using expansion
This isnt a very alegbraic way of doing this question but I just used patterns
Post by: kamil9876 on April 13, 2009, 09:50:40 pm
9.) If you break a stick into 3 pieces what is the probability that the 3 pieces can form a triangle?
If the stick has length L. And the three pieces have length a,b and c and c is the longest one then:
Say L=1 unit for simplicity.
So basically we have the segment [0,1] and we must choose 2 points such that no segment has a length greater than 0.5. Say the first point we choose is in the first half. That means that the next point must be in the second half because if they would be both in the same half then we would get a piece greater than or equal to 0.5.
Example: say the first point is at point 0.2. that means the next point must be in (0.5,0.7).
Generally: if the first point is at x (and in the first half) then the next point must be in (0.5,x+0.5)
Let's turn this into a discrete probability problem
We will do this by splitting the stick up into n segments. In other words say the only possible points we can pick are 1/n, 2/n, 3/n.... n/n=1
Say we pick the kth point, in other words, x=k/n. using an analog of the general principle shown earlier, we can show that the next point must be in between the (n/2 +1)th and (n/2 + k)th point (inclusive). (assume k is in the first half)
So that means that if we the kth point, the probability that we pick a valid point is k/n. The probability that we pick the kth point is 1/n. That means that the probability that we pick the kth point followed by a valid point is
That means the probabilty of picking a point in the first half, followed by a valid point is:
Now we want a continous situation so we let n approach infinity.
This gives a probability of 1/8. However now we have to double that to take the second half into account so it is 1/4.
And well done \0 on ur shorter solution :)
Post by: /0 on April 14, 2009, 01:27:32 am
so N=1111.......1111 and so on with 2008 recurrences of the digit 1
We must find the 1005th digit after the decimal point in the expansion
so we begin to see a trend where an even amount of digits of 1 are of the form 3.31..... and each time two more digits of 1 are added there is a 3 added to left side of decimal point and a 3 added to right side of decimal point.
so for example, a term with 8 digits of 1 square rooted i.ehas 4 3's before the decimal place and 4 3's after it followed by a 1 then 6 and so on.
so for an integer with 2008 digits of 1, there are 1004 3's before the decimal point and 1004 3's after it. We are looking for the 1005th term and as we can see in the trend, the last 3 is always followed up by a 1, therefore the 1005th term after the decimal point in an integer with 2008 digits of 1 using expansion, is 1
This isnt a very alegbraic way of doing this question but I just used patterns
correct, nice observation
Post by: TonyHem on April 14, 2009, 02:10:57 am
i have no idea how to any of these questions lol
pretty much the same here
Post by: humph on April 14, 2009, 02:49:37 pm
11.) Find the maximum value ofWe have thatgiven that
.
By the Cauchy-Schwarz inequality,
So
In fact, equality holds in the Cauchy-Schwarz inequality when one vector is a multiple of the other; that is, if
As
and hence
we clearly have that equality holds when
and so
Post by: dcc on April 15, 2009, 10:20:10 pm
Post by: TrueTears on April 15, 2009, 10:57:34 pm
Let
Post by: kamil9876 on April 15, 2009, 11:02:32 pm
OR:
However:
equating this with (1) gives QED
(btw, my
Post by: kamil9876 on April 15, 2009, 11:31:03 pm
Noticing the fundamental limit:
Letyields:
lol damn that was trivial. there are many definitions of e. This limit would be less trivial if we began with a different definition of e, namely:
or that series expansion.
btw: dcc what was ur 'expected solution' since this is aimed at spec students so i tried to limit(sorry for pun, only found it when proofreading) myself to spec knowledge however by doing so we made the problem more trivial, which is not a general trend of these problems.
Post by: dcc on April 15, 2009, 11:36:12 pm
Post by: dcc on April 15, 2009, 11:41:02 pm
btw: dcc what was ur 'expected solution' since this is aimed at spec students so i tried to limit(sorry for pun, only found it when proofreading) myself to spec knowledge however by doing so we made the problem more trivial, which is not a general trend of these problems.
I don't really have any 'expected solutions' for any of these problems, I just figured this thread would get more coverage in the Specialist Maths forums then in the General Mathematics forum.
Post by: kamil9876 on April 15, 2009, 11:59:07 pm
Post by: TrueTears on April 16, 2009, 12:09:11 am
(http://upload.wikimedia.org/math/7/0/6/7060db67ab58934d6c044c97ae0096aa.png)
Let
Differentiating with respect to
therefore equation becomes
limit
Post by: TrueTears on April 16, 2009, 12:23:03 am
17) Using l'hopital's theorem again
let
so
so
limit
Post by: kamil9876 on April 16, 2009, 02:03:15 pm
Term inside the outermost brackets is between -1 and 1 when x is beyond a certain value, provided that a is not between -1 and 1. Therefore that term raised to some number is also between -1 and 1.
Now take limit as x approaches infinity of all sides (aka sandwhich theorem or squeeze theorem)
Hence the thing equals 0 for all values
Btw: the above is for a>1. For a<-1 we need to reverse the inequality, which still gives the same answer.
For value a=0. The limit is obviously 0.
For values in (0,1):
Looking at the original expression, it is obvious that the
Hence we know that for all values of x beyond some number:
Now take the limit as x approaches infinity of these terms and u find that the limit of the lower bound is zero since the
the term
can be made as small as we like. Hence when x is beyond some value the term
Post by: TrueTears on April 16, 2009, 02:05:10 pm
Post by: Ahmad on June 26, 2009, 10:55:53 pm
(http://i202.photobucket.com/albums/aa132/ahmad0/geotr.gif)
These lines intersect each other to form a triangle, which is the triangle defined by the 3 red dots shown. What is the area of this triangle?
(Bonus: what if instead of dividing the opposite side into 3, you divide it into n?)
Post by: toomoo on June 29, 2009, 07:09:41 pm
Cheers :)
Post by: kamil9876 on June 29, 2009, 07:16:28 pm
Post by: NE2000 on June 29, 2009, 07:18:05 pm
Can someone explain the disadvantages of doing all three maths in there vce?
Cheers :)
Only two may count to your Primary 4. So if you are really a maths person and you 50 all your maths, then one of them will still be relegated to 10%. Other than that, the only other disadvantage is that you might get bored doing further at the same time as spesh. Ideally I would say the best way to do this would be further yr. 10, methods yr. 11, spesh yr. 12 so you avoid that potential pitfall
Post by: NE2000 on June 29, 2009, 07:18:40 pm
VCE maths is dull. Kills the creativity and appreciation of mathematical rigour that some of the questions in here and other recreational problem threads require.
Although some of the spesh integration stuff that requires thinking outside the box a bit is always good to do and gives a good sense of satisfaction at the end
Post by: dcc on June 29, 2009, 09:20:27 pm
Post by: NE2000 on June 29, 2009, 09:22:57 pm
Please try to keep on-topic.
yep was just responding to toomoo's off-topic post...... :-\
Post by: TrueTears on June 30, 2009, 02:28:44 am
lol
Post by: /0 on June 30, 2009, 06:03:33 am
Istranscendental, for algebraic a ≠ 0,1 and irrational algebraic b ?
lol
yes
Post by: Ahmad on June 30, 2009, 09:07:39 am
Post by: evaporade on June 30, 2009, 11:02:09 am
I have a triangle, and I connect each vertex of the triangle to a point on the opposite side which divides the side into 3. Like this:(http://i202.photobucket.com/albums/aa132/ahmad0/geotr.gif)
These lines intersect each other to form a triangle, which is the triangle defined by the 3 red dots shown. What is the area of this triangle?
(Bonus: what if instead of dividing the opposite side into 3, you divide it into n?)
[(n-2)^2 (n^2-n+1)] / [n^2 (n-1)^2] of the area of the given triangle.
Post by: toomoo on June 30, 2009, 07:04:28 pm
Although some of the spesh integration stuff that requires thinking outside the box a bit is always good to do and gives a good sense of satisfaction at the end
[/quote]
Cheers brother
Post by: Ahmad on July 01, 2009, 11:20:46 am
Post by: evaporade on July 01, 2009, 01:12:40 pm
1/7
(n-2)^2 /(n^2-n+1)
Post by: Ahmad on July 01, 2009, 02:18:29 pm
It would be nice (and possibly helpful to others) if you could briefly outline how you did it
Post by: evaporade on July 01, 2009, 05:52:41 pm
Post by: /0 on July 01, 2009, 08:59:43 pm
Post by: evaporade on July 03, 2009, 09:04:08 am
(http://img13.imageshack.us/img13/4088/vcenotesforumgeometry.gif)
You can extend to a different polygon.
Post by: Cthulhu on July 05, 2009, 07:04:00 pm
Consider the function of complex variable s
Show that all non-trivial zeroes have real part 1/2
Post by: evaporade on July 05, 2009, 07:48:30 pm
Post by: zzdfa on July 05, 2009, 08:34:05 pm
the other threadz are dead so ill post here:
(http://img21.imageshack.us/img21/5870/analysis.jpg)
b) and c) are quite simple, but my answer for a) is looooooooong . im wondering if i missed something simple.
any ideas?
Post by: kamil9876 on July 05, 2009, 08:35:54 pm
BONUS QUESTION FOR 1e1000000 points.
Consider the function of complex variable s
Show that all non-trivial zeroes have real part 1/2
Haha some people think they can:
http://www.google.com.au/search?hl=en&q=proof+of+riemann+hypothesis
Fermat's last theorem is funny too:
http://www.fermatproof.com/
Post by: Cthulhu on July 05, 2009, 08:53:46 pm
I'm glad someone caught on. ;)BONUS QUESTION FOR 1e1000000 points.
Consider the function of complex variable s
Show that all non-trivial zeroes have real part 1/2
Haha some people think they can:
http://www.google.com.au/search?hl=en&q=proof+of+riemann+hypothesis
Fermat's last theorem is funny too:
http://www.fermatproof.com/
It's always funny when people make websites with "proofs" of things like the Riemann Hypothesis or when they have a Theory of Everything.
If you're interested here is a documentary about the proof of Fermat's Last Theorem. IIRC the final proof was over 100 pages long.
Interesting fact: Andrew Wile's was knighted for the proof.
Edit: Here: Have the article as well.
Post by: /0 on July 05, 2009, 10:32:47 pm
Post by: zzdfa on July 05, 2009, 10:42:25 pm
(http://www.hoboes.com/library/graphics/NetLife/POV/Die10/lens1.png)
i think you passed over the fact that r is fixed :)
Post by: d0minicz on July 05, 2009, 10:58:49 pm
the guy in the vid was fcking crazyI'm glad someone caught on. ;)BONUS QUESTION FOR 1e1000000 points.
Consider the function of complex variable s
Show that all non-trivial zeroes have real part 1/2
Haha some people think they can:
http://www.google.com.au/search?hl=en&q=proof+of+riemann+hypothesis
Fermat's last theorem is funny too:
http://www.fermatproof.com/
It's always funny when people make websites with "proofs" of things like the Riemann Hypothesis or when they have a Theory of Everything.
If you're interested here is a documentary about the proof of Fermat's Last Theorem. IIRC the final proof was over 100 pages long.
Interesting fact: Andrew Wile's was knighted for the proof.
Edit: Here: Have the article as well.
Post by: ryley on July 05, 2009, 11:34:44 pm
Post by: evaporade on July 06, 2009, 12:21:51 am
^lol
the other threadz are dead so ill post here:
(http://img21.imageshack.us/img21/5870/analysis.jpg)
b) and c) are quite simple, but my answer for a) is looooooooong . im wondering if i missed something simple.
any ideas?
(http://img199.imageshack.us/img199/7790/vcenotesforum4.gif)
Post by: zzdfa on July 06, 2009, 12:37:56 am
but yeah, pretty much my proof was:
(1) that if k>=3 then there are infinite number of unit vectors perpendicular to x-y
(2) that for every unit vector perpendicular to x-y there is a z that satisfies |z-x|=|z-y|=r
looking back over my proof i realize most of it was showing (1). probably shouldve just prepended 'clearly' to the statement and be done with it.
surprisingly I don't think i used the triangle inequality for a) b) or c)
Post by: evaporade on July 06, 2009, 12:42:02 am
Post by: zzdfa on July 06, 2009, 12:57:22 am
for any given r>d/2,
there exists infinitely many z such that |z-x|=|z-y|=r
This is what I think you think the question asks:
there exists infinitely many z such that |z-x|=|z-y|>d/2
they are different.
(http://img22.imageshack.us/img22/3995/rarx.jpg)
look at the red lines. their lengths are greater than r. so they don't satisfy the condition that |z-x|=|z-y|=r.
edited to make the point i was trying to make clearer
Post by: evaporade on July 06, 2009, 09:02:04 am
Don't know what you meant by 'there exists infinitely many z such that |z-x|=|z-y|>2d'
The diagram clearly shows 2r > d
Post by: evaporade on July 06, 2009, 09:06:34 am
Post by: zzdfa on July 06, 2009, 11:18:47 am
Post by: evaporade on July 06, 2009, 01:15:41 pm
r is the 'distance' from any z on the plane to x or y.
r is not a constant as you tended to suggest in quote "look at the red lines. their lengths are greater than r. so they don't satisfy the condition that |z-x|=|z-y|=r".
Post by: kamil9876 on July 06, 2009, 04:31:56 pm
Post by: evaporade on July 06, 2009, 05:05:10 pm
Post by: kamil9876 on July 06, 2009, 05:23:44 pm
It's possible, generally speaking, for a set to have finite members, but a union of infinite such sets to have infinite members, hence it's important for this problem to show that each subset(each selected r value) has infinite members. But I think your diagram shows that (the disk is a set of circles(subsets) and each circle contains infinite points) so all good :)
Post by: evaporade on July 06, 2009, 06:38:24 pm
Post by: evaporade on July 06, 2009, 07:14:25 pm
Post by: zzdfa on July 06, 2009, 07:40:12 pm
r > 0 means any r > 0, not a constant r > 0. I used a dotted circle (look at the picture) to mean the plane is infinite, not because I was thinking about a constant r.
So for a question such as:
the answer would be infinity?
By the way could you please name the set with finite number of members, which you referred to?
i think he was just using it as an example,
that it is possible to form a set with an infinite number of elements, by taking the union of an infinite number of finite sets, thus if one wanted to prove that each individual set was infinite, it does not suffice to show that the union of them is infinite.
e.g.
The natural numbers = {1} U {2,3,4.....}
the union is infinite but one of the sets is finite.
Post by: evaporade on July 06, 2009, 07:52:56 pm
Post by: kamil9876 on July 06, 2009, 07:54:16 pm
evaporade: I never said "finite sets" in this particular problem, but in general terms(even said "generally speaking"). However a proof of such a statement would have to rule out the possibility of an infinite set of finite set. I was going to ask you what is your answer to "how would this statement(a) need to be modified for k=2?" because with your logic the answer would be "this statement is also true for k=2" but according to mine and zzdfa's understanding it is not as it a case of an infinite set of finite sets. (a line being a set of points).
Post by: kamil9876 on July 06, 2009, 07:57:34 pm
So what is the point in considering the union of sets when you can simply name the infinite set?
thus if one wanted to prove that each individual set was infinite, it does not suffice to show that the union of them is infinite.
e.g.
The natural numbers = {1} U {2,3,4.....}
the union is infinite but one of the sets is finite.
Post by: zzdfa on July 06, 2009, 07:59:00 pm
The question asks that,
given any r, is there an infinite number of points such that |z-x|=|z-y|=r
there's a difference
Post by: evaporade on July 06, 2009, 08:10:34 pm
Post by: evaporade on July 06, 2009, 08:17:38 pm
No modification required for (a), (b) and (c). The infinite set of z in this case is the perpendicular bisector of xy.
Post by: kamil9876 on July 06, 2009, 08:32:55 pm
"if k=2 then there are only 2 such z's for any r".
Maybe my english is bad, but I think the question does mean 'for any given r' since it is introduced in the opening sentence just like x,y and d are and we do consider x,y and d as constant hence same for r. This is just a matter of understanding the english.
But as to whether there is a difference between the two interpretations, there certainly is for reasons reiterated many times above.
In fact you could just as easily say that x and y are not constant(introduced in the exact same sentence as r) and then the answer would be so trivial.
Btw: Interpreting it as "for any given r" is a much more fun problem(hence the thread name) and the statements then do indeed require modifications for k=2.
Post by: evaporade on July 06, 2009, 10:02:59 pm
(http://img21.imageshack.us/img21/5870/analysis.jpg)
Let us make things simple, consider the following:
Suppose x,y E R, |x - y| = d > 0. Are you suggesting that the value of x and the value of y are given constants?
Post by: kamil9876 on July 06, 2009, 10:19:48 pm
Post by: evaporade on July 06, 2009, 10:21:43 pm
Suppose x,y E R, y = 2x + d. Do you interpret the value of x and the value of y are given constants?
Post by: evaporade on July 06, 2009, 10:24:11 pm
Forget the word 'constant' then. All I mean is that the question can be stated as "prove that for every given x, y and r (such that 2r>|x-y|) there are infinitely many z such that...". This is what I meant by constant, just like in your diagram x and y are constant and fixed points because you are proving that 'for every given x and y...'
But the question was not stated like that. You chose to interpret it that way.
Post by: zzdfa on July 06, 2009, 10:25:21 pm
Post by: evaporade on July 06, 2009, 10:27:31 pm
Post by: zzdfa on July 06, 2009, 10:32:46 pm
a better one would be the one i posted earlier.
sure you could say there are an infinite number of cases where r=m
but almost always it means 'for any particular r, how many m where m=r'
and in response to your example, i would say, there is exactly one d such that y=2x+d.
Post by: evaporade on July 06, 2009, 10:34:54 pm
Post by: evaporade on July 06, 2009, 10:35:47 pm
Post by: humph on July 08, 2009, 02:54:35 pm
Post by: brightsky on January 02, 2010, 06:15:48 pm
13. Prove that
Let
Then
Hence,
From the evaluation above:
and
So,
Post by: brightsky on January 02, 2010, 09:37:40 pm
17.
This is an indeterminate form 0/0, as if you x = 0 is undefined when you sub it in. So using L'Hospital's rule, where
Write
Sub x = 0 in:
Write
Write
Sub x = 0 in:
The limit of
Sub x = 0 in:
Post by: TrueTears on January 02, 2010, 09:40:03 pm
Maybe try sandwich it, play around.
Post by: brightsky on January 02, 2010, 09:48:10 pm
Post by: TrueTears on January 02, 2010, 09:50:49 pm
Find
17) Using l'hopital's theorem again
let
so
so
limityields
Post by: brightsky on January 02, 2010, 09:52:46 pm
Post by: TrueTears on January 02, 2010, 09:55:26 pm
Post by: zzdfa on January 02, 2010, 10:12:47 pm
=
=
remember doing derivatives by first principles? let
but we also know that
Post by: taiga on March 29, 2010, 09:42:17 pm
I would have just said
1/3 + 1/3 + 1/3 = 1
therefore
0.3333 + 0.3333 + 0.3333 = 1
Post by: Martoman on May 03, 2010, 04:50:52 pm
I know its been answered before but not my way.
Q7)
Let
So that
Now subbing in:
We know from u that
Meaning
Cracking out the partial fractions in terms of u tells us that in the form
Changing terminals and combining all our info:
Crunching out to
Post by: golden on December 19, 2010, 10:59:29 am
Maths:
13.) Show that
(Source: Damo17)
0.9999.... = 0.9 reoccurring (9).
0.9 reoccurring (9) = 0.9 + 0.09 + 0.009 ...
0.9 x 1/10 = 0.09
0.09 x 1/10 = 0.009 (etc.)
Therefore r = 1/10.
S = a(1-r^n)/(1-r)
As n approaches infinity, r^n approaches 0.
S = a(1)/(1-r)
S = a/(1-r)
a = 'first number', i.e. 0.9
r = 1/10
S = 0.9/(1-1/10)
S = 0.9/0.9
S = 1
As required.
Post by: golden on December 19, 2010, 11:22:13 am
Post by: onur369 on December 19, 2010, 11:26:04 am
May I ask to what equivalence (in terms of what year in VCE or perhaps beyond) are the questions?
it looks like further maths to me, number sequences
Post by: golden on December 19, 2010, 11:45:38 am
I'm not too sure about this question. I got 23.31%.
Pr(2nd) = 0.63y
Pr(added totals) = 100y/37
0.63y/(100y/37)
0.2331
Post by: golden on December 19, 2010, 11:46:13 am
Post by: onur369 on December 19, 2010, 11:47:21 am
I get the feeling that some of the other questions aren't Further Mathematics equivalent.
I did general maths(standard) in year 11 and the difficulty is the same lol.
Post by: golden on December 19, 2010, 11:49:17 am
Post by: dcc on December 19, 2010, 03:16:11 pm
15.) Neobeo is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'. Neobeo notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream. Upon walking outside any particular shop, Neobeo feels a huge compulsion to purchase an ice-cream. For every shop that Neobeo visits, he is 37\% less likely to purchase an ice-cream then at the previous shop. After purchasing an ice-cream, Neobeo leaves Luna Park. What is the probability of Neobeo purchasing an ice-cream at the second shop in 'Infinite Ice Cream'?
I'm not too sure about this question. I got 23.31%.
Pr(2nd) = 0.63y
Pr(added totals) = 100y/37
0.63y/(100y/37)
0.2331
I'm not entirely sure what you've done here, but the answer is correct.
Post by: brightsky on December 19, 2010, 05:13:56 pm
15.) Neobeo is walking around in Luna Park, and notices an alleyway called 'Infinite Ice Cream'. Neobeo notes that the 'Infinite Ice-Cream' appears to possess an infinitely large number of people selling ice-cream. Upon walking outside any particular shop, Neobeo feels a huge compulsion to purchase an ice-cream. For every shop that Neobeo visits, he is 37\% less likely to purchase an ice-cream then at the previous shop. After purchasing an ice-cream, Neobeo leaves Luna Park. What is the probability of Neobeo purchasing an ice-cream at the second shop in 'Infinite Ice Cream'?
I'm not too sure about this question. I got 23.31%.
Pr(2nd) = 0.63y
Pr(added totals) = 100y/37
0.63y/(100y/37)
0.2331
I'm not entirely sure what you've done here, but the answer is correct.
dcc, would you be able to post some new questions?
Post by: dcc on December 20, 2010, 12:46:21 pm
Post by: dcc on December 20, 2010, 12:56:11 pm
Post by: kamil9876 on December 20, 2010, 01:11:56 pm
Post by: liam_103 on May 31, 2011, 10:22:43 am
Keep up the good work!