Hey Jake,
I'm struggling in binomial theorem. I can kinda half do questions but find it really difficult to finish them off. We've been doing it for quite a while now in class and I can't seem to get the hang of it. I was wondering if you had any specific tips to binomial theorem, i'm just finding it really overwhelming to try and wrap my head around it and yeah.
I'm struggling in binomial theorem. I can kinda half do questions but find it really difficult to finish them off. We've been doing it for quite a while now in class and I can't seem to get the hang of it. I was wondering if you had any specific tips to binomial theorem, i'm just finding it really overwhelming to try and wrap my head around it and yeah.
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.
Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.
These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!
Best Regards
Happy Physics Land
Happy Physics Land, this is a behemoth of helpful tips and incredibly succinct ways to solve often difficult questions. I just wanted to personally thank you for your contribution, and encourage anyone with knowledge to always step in and lend others a hand. By teaching others, you are just proving to yourself that you understand the content inside and out.
I'd love to incorporate your tips into a future document of tips for Binomial theorum, including diagrams and practice questions. For now, though, be proud that you will be helping the literally thousands of students who will check this forum daily.
For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Hello Jake,
Thank you so much Jake, I am doing these because when I attended you lectures you very generously gave out very helpful tips for the thousands of students that have attended your lecture and Im just aspired to become a person like you! I owe you a big thanks and I can only repay this gratitude in the form of helping more HSC students around the community.
Personally however, I am a little bit stuck on a proof if you are available to help me jake. It is to use mathematical induction to prove nCr = n!/r!(n-r)!
Thank you very much Jake
Best Regards
Happy Physics Land
Hey Jake
That question wasnt exactly a past paper question but l was just curious about how to use mathematical induction to prove that nCr = n!/r!(n-r)! This way of doing it through mathematical induction is the more tedious way but l honestly had no idea where to start with this proof except for substituting r = 1. If possible would you like to kindly send me a photo of your solution? Thank you so much Jake and sorry for taking up so much of your time because of my curiosity!
Regards
Happy Physics Land
Hey Jake,
A little bit stuck on an inverse function question, may I please get a hand with part C of question 9?
(http://i.imgur.com/bZTPVvr.jpg)
Hey Happy Physics Land!
I love this question, because it looks really difficult, and then when I explain it to you I promise you'll kick yourself. I seriously think questions like this, which end up having intuitive, as opposed to learnt, answers, are the hardest to get right in an exam situation. Never the less, here is my proof. There are heaps of ways of proving this but I like this one because it's simple, and to be honest, obvious once you've thought about it enough. I don't mean to suggest that this is an easy question: it really isn't. But the proof IS easy, once you know it :)
(http://i.imgur.com/BnJvqeO.png?1)
Thanks for your questions! I really encourage everyone to have a go, post something or answer a question, because having a community like this behind you will seriously add to your learning this year.
For anyone else that wants to help contribute to this forum, both by asking questions and by answering them, you'll have to make an ATAR Notes account here. Once you've done that, a little 'reply' button will come up when you're viewing threads, and you'll be able to post whatever you want! :)
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.
Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.
These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!
Best Regards
Happy Physics Land
Thankyou so much both Jake and Happy Physics Land! Your responses have given me a new confidence that binomial isn't as intimidating as I think it is :)
Jake, if you don't mind, could you please show me the solution to question 5b from the 2001 3u paper?
Hi, I'm a bit confused with this induction question. I was wondering if you could explain to me the key steps. Thank you!
The sum of consecutive odd positive integers is divisible by 4.
Hey Phillorsm! Awesome question, and glad to hear that binomial is growing you! I'll be honest and say it is still is one my least favourite parts of teaching the course, though . That being said, I mustered the courage to tag in for Jake and write a quick solution for you! ;) this question works best by writing out a whole bunch of terms, so try doing that and following on with this quick solution. If you need a bit more detail, please let me know! :)
(http://i.imgur.com/Suhftxk.png)
Hi there IkeaandOfficeworks! I bought a wardrobe from you a few weeks back ;)
So this is an awesome question, thanks so much for posting it.
However, I'm interested, because I don't think you can prove this with induction, because it isn't necessarily true!
The sum of consecutive odd positive integers is divisible by 4. So that means 1+3 should be divisible by 4. And it is. So should 1+3+5? And already we hit an exception the rule.
I am thinking that perhaps there is something missing from your question, some other condition. Or equally likely, I'm missing something because it is 11:30 ;)
So, please get back to us! However, in the mean time, I'll leave you with a general method for induction questions.
1 - Express the question algebraically. Put it into a formula you can manipulate and change to suit your needs, usually in terms of n (EG - Divisibility induction questions like this would have some expression equal to 4M, where M is some integer, some involve inequalities, etc). This is actually sometimes the hardest part!
2- Prove the result for the lowest value you need to (usually 1, sometimes higher)
3 - Now, assume the result is true for n=k. Then, use this result to prove for n=k+1. Almost always, this involves rearranging to allow a substitution. The algebra is normally fairly easy, but it is hard to know where to go. The solution? Practice makes perfect!
4 - Conclude that since the result is true for n=1, and n=k+1, it is true for n=k+1=2, n=2+1=3, etc etc! This ending is really important, lots of people forget it, and its worth a mark by itself!
I hope this helps for your induction questions, feel free to keep throwing them at us! Sorry I couldn't be of more help with this one :)
Hey Jamon:
Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?
Best Regards
Happy Physics Land
Hey Jamon:
Jamon you pretty much covered what I did but can you check whether we should prove for n=k+2 or n=k+1? Because in this case we are dealing with CONSECUTIVE ODD NUMBERS that differ by 2, do you reckon we should prove for n=k+2?
Best Regards
Happy Physics Land
Oh, and I almost forgot, don't forget to conclude your induction proof! You've proved for n=k+2, you need to extrapolate this for further numbers. Just a simple, it is true for n=1, and n=k+2, so it is true for n=1+2=3, n=3+2=5, etc, for any sum of two consecutive odd positive integers, will suffice.
Hello Jamon:
Thank your for your recommendations and also for giving approval to my induction proof :D ! I think that the use of N notation to represent all the positive integers is quite a clever idea and I really liked the format of the conclusion, very succinct but effective as a concluding statement! But overall thank you for checking my solution!
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
I think there is a typo; 40000 and 50000 should be what is written in the question (given the answers provided), or else every 5-digit arrangement would be larger than both 4000 and 5000.
Assuming that 40000 and 50000 are respectively meant:
For part b), I find it easier to look at which arrangements are not counted, because there are fewer cases to consider. In this case, only those arrangements starting with a 3 will be not counted. So, let's assume the starting digit of the arrangement is 3, so we have 3 _ _ _ _.
We then have to place 4, 4, 5 and 6 into the remaining four slots. This can be done in 4!/2! = 12 ways, by a similar argument to what you did in part a). All other arrangements work, so the total number of successful arrangements is 60 - 12 = 48.
For part c), we can again use the same 'counting the complement' technique. This time, we exclude those arrangements starting with 5 or 6. By the same working out as in part b), we can deduce that if the starting digit is 5, there are 12 arrangements (as you then have 5 _ _ _ _ with 3, 4, 4, 6 to place), and there are also 12 arrangements if the starting digit is 6 (6 _ _ _ _ , with 3, 4, 4, 5 to place). Hence, the total number of arrangements less than 50000 is 60 - 12 -12 = 36.
It is also possible to count the arrangements that are included for part c), but you do have to be a bit careful: if the starting digit is 3, then there are 12 successful arrangements. However, note that if the starting digit is 4, then you have 4 _ _ _ _ with 3, 4, 5, 6 to place. Because the remaining digits are all distinct, there are 4! ways to place them, and hence there are 24 successful arrangements. Therefore, the number of successful arrangements is 12 + 24 = 36, which matches the result we obtained earlier.
Hi ive got a question on permutations. This is probably a simple question but my mind is twisted right now so I can't seem to get b) and c) right. Not sure either if 4000 and 5000 is a typo for 40000 and 50000 respectively as the answers are small numbers??
a) How many different arrangements can be made from the numbers 3,4,4,5 and 6? ( which i figured was 5!/2! =60)
b) how many arrangements form numbers greater than 4000? (ans: 48)
c) how many form numbers less than 5000? (ans: 36)
Hmm ok this is quite interesting. We have actually defined the question differently, you saw it as all the numbers that can be made and are greater than 4000 or less than 5000 and I saw the question as all the four-digit numbers than are great than 4000 or less than 5000. Im not too sure about the wording of this question or maybe is it just a typo of 40000 and 50000.
Hey I have a test coming up and I am having trouble with ext locus and parabola. The question is
Find the equation of the chord of contact AB of tangents drawn from an external point (x_{1}, y_{1}) to the parabola x^{2} = 12y
Thankyou :)
Hey! I was just practicing some curve sketching and I was wondering for curves that look like these:
y= 3x/(x^2+9)
How would you know the shape it forms? Cause really all I knew was that it intercepts at the origin, is odd and has a horizontal asymptote at 0.. Thanks, sorry if this sounds dumb it's just that it seems a bit dodgy how I'm assuming it looks like a sorta sideways s (sorry I can't really describe and I don't know how to draw/paste it in here).. Thank you!
Neutron
Thank you so much HPL! Yeah yeah it makes sense now.. One more thing though, how do you know whether a point is a point of discontinuity (open circle) or an asymptote?
Hey foodmood16,
I did get an answer to your question, it was simply a matter of substituting into the equation of chord of contact. Just double check with your answer just in case I got it wrong (when questions are too easily solved I feel a bit intimidated). But yes great question doe because just about everyone tends to forget about this chord of contact equation.
(http://i.imgur.com/36BiedQ.jpg)
(sorry for the dodgy diagram l drew)
Best Regards
Happy Physics Land
Thankyou Happy Physics Land! That helps a heap as I didn't quite understand it when it was explained in class :)
For those who didn't know: HPL is a legend and should be revered.
Thanks man!
Jake
in what kind of sequence will you use to sketch y=4sin[3(x-(/6))] 0<=x<=2
in what kind of sequence will you use to sketch y=4sin[3(x-(/6))] 0<=x<=2
Is the question y = 4sin[3(x-1/6)] or something else??
sorry the pie symbol didnt appear
y=4sin[3(x-pie/6)]-2
sorry the pie symbol didnt appear
y=4sin[3(x-pie/6)]-2
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D
Solve the following equation:
2cos2ϴ=1-3sin2ϴ 0≤ϴ≤360
Thank you!
Neutron
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :DThe presence of the 2cos(2θ) is not able to eliminate the 1 as no matter what expansion of the double angle formula is used, a 2 will pop out instead of a 1. This makes expanding all double angles, unfortunately, a folly.
Solve the following equation:
2cos2ϴ=1-3sin2ϴ 0≤ϴ≤360
Thank you!
Neutron
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)
P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)
P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.
Hey Jake, or whoever answers this...
So In your lectures you mentioned that with Induction you are allowed to skip steps in the proof to save time. I was just wanting to check with you if this is what you meant by that, and if it would get all the marks :)
P.S. Sorry if the image is a bit small, it wouldnt let me upload anything over 512kB.
how to sketch y=inverse cos(x^2)
how to solve lim tan3x/tan2y
x->0
Hello Im also a year 12 student doing my maths extension I, and I have encountered similar issues beforehand as you, because binomial is very tedious and the sigma notations always scare me because I am unfamiliar with it. So I began doing questions first from maths in focus which provides easy questions on binomial theorem and then moved on to do more exercises in the cambridge book. It is very challenging however beneficial to do those development and extension questions as well because it is likely that your teacher will confront you with a similar style question.
Here are several tips that I think has helped me a lot with binomial:
1. It is very important to remember that the binomial co-efficients always starts with nC0, not nC1
2. When proving binomial identities, a lot of those can be related back to pascal's triangle. So if you are confused about whats the significance of the proof or what you are trying to prove, write down the first few rows of pascal's triangle to give you a better observation of what exactly you are trying to prove
3. DO NOT BE SCARE OF SIGMA NOTATIONS! A lot of my friends instantly give up as soon as they seen sigma notation because its such a weird representation of a series of numbers. Sometimes we dont think of sigma notation as normal maths, but rather, some "alien language". But its VERY CRUCIAL TO REMEMBER that SIGMA NOTATIONS ARE OUR FRIENDS. It is just A SERIES, nothing more, just A SERIES OF NUMBERS. It is helpful for us because instead of having to tediously look a long, boring chain of numbers, a simple sigma notation essentially summarise it for us in simple expressions. On the bottom of sigma notation there is r= some number or k = some number, this just means that for the expression next to the sigma sign, the initial variable is what r or k represents. E.g. for 3^r, r = 0, that means we start with 3^0. On the top of the sigma sign there is usually "n", which is indicative that the series terminates at r = n, whatever that n value maybe. E.g. for 3^r, we terminate at 3^n.
4. It is beneficial sometimes when solving binomial questions to expand the binomial out. If it is too long an expansion, just write out the first 3-4 terms and the last 3 terms. This helps us to find patterns that can help us to solve the question.
5. In Binomial questions associated with integration or differentiation, we almost always find a value for x (i.e. let x = something) to make our solution look more similar to what the question requires for us to prove/find. A sneaky tip is that HSC examiners would usually write the question in a way that students will let x = 0 or 1.
6. When we are proving an identity in binomial theorem, its not always compulsory to start with the side thats more complicated. This is counter-intuitive to what we have always been learning because we are always used to solving something thats looks more intimidating because there is a higher chance that we can somehow manipulate it to make it look more neat/tidy, and resemble the other side of the equation. In binomial theorem, this is not always the case. For example, consider the proof for "Sum of nCr from r=0 to r=n) = 2^n". Logically, we would begin with the left hand side because it is more complicated and we would hope for a neat result to come out in the end. However, if we begin with the right hand side it will be much easier because RHS = 2^n = (1+1)^n = sum of (nCr x 1^r) from r=0 to r=n. Since 1^r is always 1, we can effectively prove that 2^n = sum of nCr from r=0 to r=n.
7. It almost always helpful that when you are stuck on a binomial proof question to go back to the basics of expanding (1+x)^n, or remembering that (1+x)^n = the sum of (nCr x x^r ) from r= 0 to r=n.
8. When finding the constant term that involves expanding two binomials, expand both and select one term from each binomial expansion that will cancel each other's variable out when multiplied together, leaving us with just a number.
9. Transformations of (1+x)^n will always change the position of the greatest co-efficient in the expansion. (1+x)^n will have its greatest co-efficient at the centre, (1+3x)^n will have its greatest co-efficient shifted to the right and (1+5x)^n will have its greatest co-efficient shifted even further to the right. Adversely, (3+x)^n will have its greatest co-efficient shifted to the left and (5+x)^n will have its greatest co-efficient shifted even more to the left and so on.
These are all just some of my tips that l found very helpful to know. Im not sure how much this will help you but yeah good luck in everything this year!
Best Regards
Happy Physics Land
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?
Hey atarz!
If possible can you please supply me with the actual question? If the question only asks for the integral of sin^{-1}(x), then I have provided a solution below. It is quite infrequent for school exams to ask you such questions, especially when 4 unit candidates can solve it relatively easily using integration by parts. I cant see easier ways of doing this except for using Integration by parts which definitely isnt a 3 unit concept. You can try doing this question also with areas (i.e. the area under sin^{-1}x). I'm suspecting that there may be a part beforehand in the question that may help with integrating inverse of sine using 3u methods?
Anyways, I have posted my solution below through integration by parts:
(http://i.imgur.com/wOiKW9q.jpg)
Best Regards
Happy Physics Land
Hi, just wondering,
we have only just done inverse trigonometry but there was a question in our test that we had not learnt how to attempt. We had to find the integral of the inverse of sin. How would you attempt this question?
As this is a 3U forum, if you were asked such a question in the 3U exam there is a probability of 100% that the integral came with BOUNDARIES and was a DEFINITE integral
You need to draw out the graph of the inverse sine function, and then find
A = area of rectangle - integral of sin(y)dy for certain boundaries.
Integration by parts is NOT in the 3U course, as already mentioned.
Hi,
I'm having trouble with an Integration By Substitution question:
∫(x-2)/(√x+2) dx
using u^2 =x+2
Can you help me out? Thanks very much :)
OMG yes.... after all inverse sine x comes from the original curve sine y..... of course... this way the question is actually a lot easier. We are essentially just finding the area of sin y within the domain 0 <= x <= pi/2 and then using a rectangle to minus that area. And then because arcsin x is defined from -pi/2 to pi/2, just times the resultant area by 2. Am I right in saying this rui?
Hi,
I'm having trouble with an Integration By Substitution question:
∫(x-2)/(√x+2) dx
using u^2 =x+2
Can you help me out? Thanks very much :)
Not too sure why HPL used that substitution when you were given that specific one.
Wait what I thought I did use that substitution? Its just with the numerator I subtracted 4 from u to make it the same as the numerator.
You used:
RuiAce used:
Both work quite well actually!
Yeah I agree Jamon, I realised that only shortly after I posted my comment. I guess at the end of the day it doesnt matter what you substitute it with. After all, substitution is only a technique, and if that technique helps you to solve the question then I guess regardless which mechanism you adopt you would get the same outcome.
Hey guys!
I was wondering whether you could show me how to sketch the function y=tan^{-1}(tanx) over the domain -π≤x≤π <3 thank youu
Neutron
hey, i've been stuck on this question. i keep getting the answer 30.
In a tennis club, there are five married couples available to play a "mixed doubles" match, that is, a match in which a combination of one man and one woman play against a combination of another man and another woman. In haw many ways can a group of four persons be chosen for this match if:
(b) a man and his wife may not play in the match either as partners or as opponents.
Answer is 60
Hey Neutron!
Tan^{-1}(tanx) will cancel each other out, leaving us simply with y = x. So all you need to do is plot the points (pi, pi) and
(-pi, -pi), and then connect these two points (essentially just the linear graph y=x).
If you really need me to show you the sketch just tell me, but Im sure you are capable of doing it! :)
Best Regards
Happy Physics Land
Yeah I know that but the answer gave me something very trippy :'( I'll see if I can upload a pic later D:
Your hint is to think about both the domains AND ranges of both the tangent function, and it's inverse.
Yeah I know that but I was assuming a -pi <= x <= pi
Tangent discontinues at pi/2. Not pi.
How do you do the last part for the first question ?
please help thanks
How do you do the last part for the first question ?
please help thanks
How do you integrate cosec3xcot3x?
Mind posting the solutions to the previous parts? They seem to be useful for the answer to d)
Not sure if the perms/combs question was intentionally placed there but that one isn't too bad.
How do you integrate cosec3xcot3x?
Hey Xus!
Interesting question! Ok to do this question, we would use 3 unit integration by substitution, I have posted my solution below, if you have any further questions please dont hesitate to ask! :) :)
(Sorry if the image is a bit blurry, still using brick phones hehe :D )
(http://i.imgur.com/2qcrkrq.jpg)
Best Regards
Happy Physics Land
these are the solutions for previous parts
Hi all,
Really struggling with part C of this 3D trig question, I just can't seem to get it out...
Help would be very much appreciated!
Thankyou!! :)
Hi all,
Really struggling with part C of this 3D trig question, I just can't seem to get it out...
Help would be very much appreciated!
Thankyou!! :)
Not sure how to get C
why is it 10C7(1/4)^7(3/4)3 not just (1/4)^7(3/4)3
C=combination
Ive attached the question and the answer to part ii
Im not sure why rx=t
part iii) please thanks
Hi could i please have help with this question:
DIfferentiate:
Thanks a lot in advance.
Hey:)
Just wondering if someone could give me a hand with the attached question?
Thank you so much :D
Hey,
Could I please have help on part iii of this question? It's really confusing me. Thanks in advance.
Thanks Jake! Unfortunately I've also got another question that I'm finding difficult and I'd be grateful for help... Hopefully this one is easier to solve. Could I have help just with part i? Part ii I'll attempt later.
Hi there, I'm having trouble with these two questions. Any help would be appreciated. Thanks
1. Show that f(x)= x^3+2x-4 has only 1 root
2. Given log(xy^3)=m and log(x^3y^2)=p, find log(sqrt(xy)) in terms of m and p (all logs are base 'b')
Should i be taken the positive velocity since v>0 when t=0
ans for part ii)1-e^-t
Hey! My proof is below :) this question is mainly about applying the relevant velocity/displacement formulas, and then ensuring you integrate correctly. I hope the solution helps!
(http://i.imgur.com/brhWLpH.png?1)
Jake
Actually I had almost the same question in my 3u exam, the hard part in this question for a lot of my mates was actually to realise the fact that you have to implement natural log in the integral
a particle moves in SHM of period 8 hours and amplitude 6 metres. When t=3 hours, x=4m by using x-b=asin(wt) find centre of motion
Hi, need some help with this differentiating trig func. question :)
wasn't sure which thread this question should go but...here~
thanks in advance!
Hey!
This was quite a tough one: I did it wrong twice, working through methods that ended up being way easier, but gave you a different result that became even more difficult to prove equaled the RHS. In the end, brute forcing the entire thing was the easiest way to go! I hope my answer below is helpful: If you're ever unsure, just attack the question using the rules you know!
(http://i.imgur.com/eMi5Uy7.png?1)
Jake
Hey!oh i didn't try to approach it from the RHS...my bad haha
This was quite a tough one: I did it wrong twice, working through methods that ended up being way easier, but gave you a different result that became even more difficult to prove equaled the RHS. In the end, brute forcing the entire thing was the easiest way to go! I hope my answer below is helpful: If you're ever unsure, just attack the question using the rules you know!
(http://i.imgur.com/eMi5Uy7.png?1)
Jake
Mind=blown xD
Hi!, how do you find the inverse of y=x^{3}+3x. Thank you! :D
Hey! This would require factorisation of cubic equations, which is immensely complicated and way beyond any curriculum dotpoint (you can check it out here http://www.math.vanderbilt.edu/~schectex/courses/cubic/). Is it possible that you have recorded the question wrong? Perhaps I have missed something, but I can't think of a way to solve this.
Sorry!
Jake
not sure how to do this
for eg. (1+x)^3(1+x)^5
why is the coefficient
(3C3)(5C0) + (3C2)(5C1) + (3C1)(5C2) + (3C0)(5C3)
not (3C0)(5C0) + (3C1(5C1).....
how do i find the coefficient of x^(-8)
in the expansion of (x-1/x)^6 times (x+1/x)^8
when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1
since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80
how do i find the coefficient of x^(-8)
in the expansion of (x-1/x)^6 times (x+1/x)^8
when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1
since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80
is there a faster way to do this instead of sin both sides , let α=cos^-1(4/5) and β=cos^-1(3/5) and prove sin(β+α)=1
inverse cos^-1 (4/5) + inverse cos^-1(3/5) = pie/2
Hey hey!Food for thought
I've attached my answer below: For questions like this, you generally want to set up some right angled triangles and figure it out from there. It usually isn't this conveniently set up. Generally, you'll create the triangles, find sine of alpha and sine of beta, cosine of alpha and cosine of beta, expand whatever you're trying to solve (using double angle formulas) and just plug in the values you've found.
(http://i.imgur.com/xalJTa0.jpg)
Jake
how do i find the coefficient of x^(-8)
in the expansion of (x-1/x)^6 times (x+1/x)^8
when im asked to find the greatest coefficient eg. (1-2x)^9 am i supposed to exclude the negative when im trying to find k from T(k+1)/T(k) >1
since the greatest coefficient is the number largest in magnitude how do i know if there r no 2 greatest coefficient with different sign eg. -80 and 80
Note: Greatest coefficient has not been examined in the HSC since the 1980s or something. You must still study it, but it can be the least of your worries.
help with this ques thanks
let n be a positive even integer
expand and simplify
(a+b)^n + (a-b)^n
what i got was
2(nC0)a^n + 2(nC2)(a^(n-2))b^2 + 2(nC2)(a^(n-4))(b^4) ....+ 2(nCn)(b^n)
Hi guys! I need help with this question:
tan^-1 (3/4) = 2tan^-1 (1/3)
Thank you!
how to find coefficient of x^8 in the expansion of (x^2-2x+3)^5
(http://uploads.tapatalk-cdn.com/20160517/592faed7af0f73f148da4f3132e865a6.jpg)
ques: 7 letter words are formed using the letters of UNUSUAL. One of the words is selected at random. find probability that the word had none of the U together
im not sure what i did wrong but ans is 2/7
Hey amandali! Your approach was definitely on the right track, you are very close!! The issue is that I think there are subtle things at play when considering 2U's together, I want to try considering it a little bit differently:
Focusing on two U's together. Let us first consider the possibility that the two u's must be at either end. This can occur two ways. Then, we need another letter next to them (to block the 3rd U). We can choose from 4 remaining non-U's, so we then multiply by 4. Then, we arrange the remaining 4 letters, 4! Now, we actually haven't done anything that would require dividing by 3! here (we've not changed the order of the u's with respect to each other), so the answer is simply:
Now consider the two U's somewhere in the middle, as you have done. Stick the u's in the middle of the packet, that's where they must be, no probability involved yet. Now, we can pick two letters from four remaining options to stick either side. That is a combination. Then, we can swap their place, multiply by two (you could also just consider a permutation in the previous step). Then, we can order the word in 4! ways. That gives:
Add these, plus the 120 you got in the last part of your solution, and we end up with 600 total ways the letters can be arranged so that U's are together. Therefore:
I hope this helps! That was a tricky one, stumped me for a bit there! ;D
This is the first time I've had a problem with factorials being donated as '!'. Reading through your (well written) answer, I kept seeing a super enthusiastic Jamon explaining how probability works. 'Now, we just sub this in! Then, divide by this! Isn't maths just so very fun!'.
Confession: I purposely didn't answer that question because I don't want to associate with perms and combs again until next semester when I'm doing it in discrete maths.
Ahaha! Well if you handle the probability questions next semester I'll handle the integrals ;)
(http://uploads.tapatalk-cdn.com/20160520/80af3be4933c70eec30dd468e0d2ddde.jpg)
need help with this ques thanks
Worthwhile mention: The harder binomial coefficient proofs always include at least one of the following:
Symmetry property of binomial coefficient: ^{n}C_{k}=^{n}C_{n-k}
Pascal's identity:^{n+1}C_{k+1}=^{n}C_{k}+^{n}C_{k+1}
I remember the first time I tried this proof in my HSC I tried to do it with the full factorial expansions of each term, ever try it that way? I mean, not that you ever would, my teacher quickly steered me right ;DWell...
Well...
You'd end up having to prove THAT ugly thing somehow, oh dear
Not too sure if symmetry in the sum is required to get there without reverting steps and reintroducing the binomial coefficient but LOL not attempting that.
Ahaha I think I got roughly to your third line before getting help back in 2014, definitely looks as disgusting as I remember it. If you are at UNSW and do Math 1131/41 next semester, you'll do stuff on series (mostly convergence tests though), so who knows, it might actually be doable :P*51, taking maths with actuarial :P
*51, taking maths with actuarial :P
Too bad it's not an infinite sum or it would've been tempting to play around with a Taylor approximation. Finite sums are so different to limiting sums.
*51, taking maths with actuarial :P
Too bad it's not an infinite sum or it would've been tempting to play around with a Taylor approximation. Finite sums are so different to limiting sums.
I didn't realise you were at UNSW!
Ah that's right I forgot they separated you guys
Aha check signature :P
I miss the days when maths used to be my easiest subject.
(http://uploads.tapatalk-cdn.com/20160521/08452256328b16bf7ee1ee08f11e3ab0.jpg)
i cant do part a) and part b (iii)
for part b (i) i got [ (1+x)^(n+1)-1 ] / n
(http://uploads.tapatalk-cdn.com/20160521/08452256328b16bf7ee1ee08f11e3ab0.jpg)
i cant do part a) and part b (iii)
for part b (i) i got [ (1+x)^(n+1)-1 ] / n
for part a) i got prob(win1st or 2nd turn)= p+rq and im not sure how to progress from there
for part a) i got prob(win1st or 2nd turn)= p+rq and im not sure how to progress from there
Simultaneous solutions it seems ;D Virtually identical for you Katherine so either or ;D
You know I sat your MX1 paper right :P
(http://uploads.tapatalk-cdn.com/20160527/dcbb8ed6d88c7b128d76746d4e8c5a9f.jpg)h
how to do ques 2 thanks
question: a die is rolled 100 times. What is the most likely number of sixes that will be thrown?
Ans: 16
am i supposed to use the greatest coefficient for these kind of ques cuz the answer i got was wrong using this method
nCr*(1/6)^r*(5/6)^(n-r) divided by nC(r-1)*(1/6)^(r-1)*(5/6)^(n-r+1)
or am i supposed to do (1/6)*100
Hii!
I'm having trouble with the following question, any help would be much appreciated :)
Thanks!
1. particle has a=2-2x, initially v=4m/s , x=0
a)find v in terms of x
what i got was v=(16+4x-2x^2)^1/2 how do i know whether v is positive or negative
b) find the greatest v
do i normally just let a=0 for min/max velocity
what do i have to do if the ques asks for max/min acceleration
particle starts from the origin and has
v=cos^2(x)
a= -2sin(x)*cos^3(x)
x= tan^-1(t) inverse
where 0<=x<π
Describe the motion of particle from its initial position to its limiting position (2 marks)
the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2
Im not sure what's required for questions that ask me to describe the motion
the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate
Hey!
I've attached my solution before. This is just a straightforward application of the derivative formula for inverse trigonometric functions: you need to be extremely comfortable with these kinds of questions!
Hope this helps :)
(http://i.imgur.com/bJCnpGQ.png?1)
Jake
ok, thanks. I understand this much, but i'm having trouble further simplifying it. The final answer provided in the textbook is -1/√ (9-x^{2}). How did the 3 in the denominator cancel out?
By simply expanding it in.
1. particle has a=2-2x, initially v=4m/s , x=0
a)find v in terms of x
what i got was v=(16+4x-2x^2)^1/2 how do i know whether v is positive or negative
b) find the greatest v
do i normally just let a=0 for min/max velocity
what do i have to do if the ques asks for max/min acceleration
particle starts from the origin and has
v=cos^2(x)
a= -2sin(x)*cos^3(x)
x= tan^-1(t) inverse
where 0<=x<π
Describe the motion of particle from its initial position to its limiting position (2 marks)
the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2
Im not sure what's required for questions that ask me to describe the motion
the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate
why is the answer 2/5
Hey:)
I'm struggling a bit with the attached question from the Binomial Theorem topic, wondering if I could have a hand?
Thanks so much 8)
i dont get part (iii) for question 1
ques2 : the answer is 35/72
i dont get part (iii) for question 1
ques2 : the answer is 35/72
I distinctly remember doing this question in the lead up to my Extension 1 exam, it was nasty, very grateful to have not seen this question under pressure :o
how to do this question ?
the region in the first quadrant bounded by x=y-y^3, x=1, y=1 is rotated about y axis. Find the volume of solid generated
(http://uploads.tapatalk-cdn.com/20160610/67eee88c7355af3076a061cbd31a15d9.jpg)
need help with part c)
(http://uploads.tapatalk-cdn.com/20160610/fb24f3c84c059b3547637cbc2e071e50.jpg)
need help with this ques thanks
how to do part b) and c)
ans for b = 101/144 and ans for c =0.113)
for part c)
what i did was 20C12*(5/8)^12*(3/8)^8 + 20C12*(7/9)^12*(2/9)^8
(http://uploads.tapatalk-cdn.com/20160611/4ad4e133b5469c87d70cb6793c70502b.jpg)(http://uploads.tapatalk-cdn.com/20160611/1c1b825b3e9315d84c60ed406d4ba3d5.jpg)
for part c ii) i dont get how the last constant becomes (pCp)(qCp)
for part b) what is the reason that the velocity is negative
i dont understand part c)iii) ive attached the answer it below
for part b) what is the reason that the velocity is negative
i dont understand part c)iii) ive attached the answer it below
Hey guys! can you help me with these motion questions? Thank you!
Hey guys! can you help me with these motion questions? Thank you!
i did integration and got stucked
i did integration and got stucked
Sorry that it's hand written; Rui and Jamon still haven't taught me LaTeX.
Oi. I read this :P
That was the idea ;)
Hey Amandali!
I've attached the solution before. Using a method like this is pretty standard, however usually in the 3U course they lead your through it a bit more. Still, if you ever get a question like this, just think about what you are TRYING to prove and basically brute-force your way there!
Sorry that it's hand written; Rui and Jamon still haven't taught me LaTeX.
(http://i.imgur.com/DLB1Nq2.jpg?1)
Jake
Oh wait, slight correction after looking at this again
You forgot the (annoying) +C when integrating
(The final answer is unaffected though because when you subtract the two resultant equations, the C cancels out)
Yeah I thought about including that, but thought it trivial (if one integral equals another, obviously any constant is going to cancel, so generally I just exclude it from the start).Well in the second line it's kinda important because otherwise, that statement at that step is false
Well in the second line it's kinda important because otherwise, that statement at that step is false
Well in the second line it's kinda important because otherwise, that statement at that step is false
how to do last part thanks
It's funny, at HSC level they stress this sort of thing, my Complex Analysis lecturer is happy for us to omit it in very particular circumstances where it is apparent that the constant is irrelevant. Interesting how the frame of thought changes (though I doubt a marker would take away anything in a HSC exam) ::)HSC students only know so much so what can you emphasise :P
ques: How many different arrangements of the word MAMMOTH can be made if only fiveI'm pretty sure I explained this question recently in the 4U question thread
letters are used?
i dont get the answer
is there an alternative way of doing this
ques: How many different arrangements of the word MAMMOTH can be made if only five
letters are used?
i dont get the answer
is there an alternative way of doing this
Disclaimer: I thoroughly dislike combinatorics. My explanations for them are usually suboptimal to those for other topics. If the explanation is insufficient, please allow someone else to supplement my answer.
My explanations for combinatorics will improve after I do discrete maths, which may be a while.
I wasn't able to justify the method you used either.
Can you help me with this motion question? Thanks guys! :D
A ball is projected so that its horizontal range is 45 metres and it passes through a point 22 1/2 metres horizontally from and 11 1/4 metres vertically above the point of projection. Find the angle of projection and the speed of projection.
Can i please have help with this question? :(
Hey there! You sure can, the first bit is to handle the integrand:Hey! Thanks for the quick reply, but I still don't quite get it. I understand YOUR working method, but the answers say it's pi( 3pi + 8 ) units3 :(? [or the answers could be completely wrong that works too]
Now we complete the integration. We'll use our double angle result to handle the cos squared, which is something you may have seen before, let me know if not!
Let me know if anything here doesn't quite make sense!! ;D
Hey! Thanks for the quick reply, but I still don't quite get it. I understand YOUR working method, but the answers say it's pi( 3pi + 8 ) units3 :(? [or the answers could be completely wrong that works too]
Is this for prelim or HSC?
Having a bit of a struggle over here...
Grain is ejected from a chute at the rate of 0.1 metres cubed a minute and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone i s increasing at the instant 3 minutes after the opening of the chute. Answer is 0.0732.
Any help would be appreciated :)
Having a bit of a struggle over here...
Grain is ejected from a chute at the rate of 0.1 metres cubed a minute and is forming a heap on a flat horizontal floor. The heap is in the form of a circular cone of semi-vertical angle 45 degrees. Find the rate, in metres per minute, at which the height of the cone i s increasing at the instant 3 minutes after the opening of the chute. Answer is 0.0732.
Any help would be appreciated :)
Edit: Jake just beat me to a solution, but I finished it, so I'll post it anyway. Maybe it will be equally useful, we used the same method ;)It was about time he got a question :P
It was about time he got a question :P
Yeah was pretty happy with myself, sorry matey
Yeah was pretty happy with myself, sorry matey
Many thanks, it seems like a classic case of calculator mistake for me, but that's alright :).
Glad someone could get a self-confidence boost!!
The race continues ;) or we just need to actually make a system like we always say we are going to aha
Calculator mistakes are the worst, especially for a question like this where you've done a solid piece of Mathematics to get to the answer. Rui and Jamon are just too good at the typing-up-maths-on-the-forums-thing, it's difficult to compete.
But I mean if I see a question unanswered for 3 hours and you guys are MIA... :P
But I mean if I see a question unanswered for 3 hours and you guys are MIA... :P
Actually I would've written this one out just cause of the triangle haha
If f(x) = 2 - (x)1/2, x ≥ 0 and g(x) = (x-2)2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)]
Moderator Action: Added LaTex formatting for easier reading.
Find all values for x for which f[g(x)] = x = g[f(x)].
If f(x) = 2 - (x)^1/2, x ≥ 0 and g(x) = (x-2)^2, for all x. Find all values for x for which f[g(x)] = x = g[f(x)]
the (1/2) and 2 are both powers I forgot to put them in sorry
That's what I got but the questions asks for values (plural) which made me unsure
I understand how you got to where you are, I do not however know how to express the answer that is what is troubling me.
Hey was wondering if someone could please help me out with this strange binomial question for the 2009 HSC.Pulling strings here having me do a whole page...!
I even had a look at the answers and had absolutely no idea how to do it :P
I know its long...sorry and thanks in advance :)
Even when asking for all values, one is still acceptable!! Let me take another look though...
Oh, I did forget to take into account the fact that the square root of the first line can be positive or negative!! This would create a second option:
So actually, taking into account that first option, any value of x greater than or equal to zero will work!! Wow, what a difference a little sign error can make (I can never go more than a few weeks without making one of these) ;) sorry about that! Does this make sense? :)
I understand how you got to where you are, I do not however know how to express the answer that is what is troubling me.
Sure! So having reached that point, you would likely write something like this:I disagree
Does that help? ;D
If dy/dx = 5x and dx/dt = -2
Find: dy/dt and d^2y/dt^2
To find dy/dt, we can use the relationship that we learn in 3U:Lol that's just the chain rule.
Lol that's just the chain rule.
Also I'm not too sure if it works like that for the second derivative... https://en.wikipedia.org/wiki/Chain_rule#Higher_derivatives
Which makes me also question the validity of this problem at the Extension 1 level.
To be honest, I couldn't remember whether they call it the Chain rule in High School or they just told you it was a thing that was true. I also assumed they couldn't do the method you wrote out above, so assumed mine had to be true. I was probably/definitely wrong with that second half, sorry!Haha they explicitly teach the chain, product and quotient rules in 2U for a reason.
Prove: cos^-1(-x) = pi - cos^-1(x)
Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right.Sure thing. (Also, lol I made a mistake you don't really need compound angles; you just need ASTC for this particular one which eases out the proof)
Would just like to check
Compound angles please(it is withing inverse functions), I have proved it but not sure if it is right.
Would just like to check
If it's within inverse functions, I think the easiest (and neatest) way to prove this relationship is to create a functionYeah I agree that calculus is the faster method here. He just wanted a compound angles proof.
Differentiate the thing, using the formulas you've learnt in the inverse function section of the course. You'll see it is equal to zero, suggesting that the gradient of this function is zero (first derivative = gradient). If the gradient is zero, then the original function can only be a horizontal line (ie. for any value of x, there is only one value for y). From there, proving it equals Pi (as the question asks) is simple: sub in any value for x. Whether you put x=1, x=3.14 or x=1.6893*10^8, you'll get Pi out as the y value. As the function has a gradient of 0, this solution holds true for any value of x, therefore the original relationship must be true!
Jake
Pulling strings here having me do a whole page...!
___________________________
_____________________________
Thanks so much for the previous answerSome clarity needed.
Thanks for the help, this should be my last one in a while:
an inverted cone of radius 20cm and depth is filled with water. The water flows out through the apex of the cone at a constant rate of 10pi cm^3s^-1. Find the rate at which the water is falling when the depth of the water is 20cm.
Just like my other questions i have an answer, but it is a take home task and i would just like to check my answer.
Because the vessel is losing water that means the rate of change is negative right?
Thanks heaps man, Sorry for all the dumb questions (in another setting where it is not an assessment I would not be second guessing myself)
Just quickly how was the most effective way you found studying for ext 1, just do past papers or was there something else.
first ques: not sure how to do last part ie finding tAs always, keep in mind that questions before the year 2001 do not accurately reflect the nature of the current HSC.
second ques: not sure how to do last part
first ques: not sure how to do last part ie finding t
second ques: not sure how to do last part
A friend of mine needed help with this question:
1. a. Consider the function f(x)=e^x/(1+e^x). Find f'(x) and deduce that f(x) is increasing for all x
b. State the range of f(x)
c. Find the inverse function f^-1(x)
d. Draw y-f(x) and y=f^-1(x) on the same diagram
Hey!
My solution is below
(http://i.imgur.com/bax8bCK.png?1)
(http://i.imgur.com/srOuRoe.png?1)
Jake
A friend of mine needed help with this question:
1. a. Consider the function f(x)=e^x/(1+e^x). Find f'(x) and deduce that f(x) is increasing for all x
b. State the range of f(x)
c. Find the inverse function f^-1(x)
d. Draw y-f(x) and y=f^-1(x) on the same diagram
Hi Guys! Im having problems with this question:
The perimeter of a circular sector is 20 cm. The radius is increasing at the rate of 5cm/s . At what rate is the angle of the sector changing when the radius length is 10 cm? Also at what rate is the area changing when the radius length is 10 cm.
Thanks guys!
Just in case anyone's interested feel free to try out this question
1. a. SKetch f(x)=(e^x)-4, showing clearly the points of intersection with the axes and the equations of any asymptotes
b. On the same diagram, sketch the graph of the inverse function f^-1(x), showing clearly any important features
c. Explain why the x-coordinate of any points of the intersection y=f(x) and y=f^-1(x) satisfies e^x - x - 4=0
d. Show the equation e^x - x - 4 =0 between x=0 and x=2 and use the method of 'halving by intervals' to find this root correct to the nearest whole number
Also anyone here wishing to try a few challenge circle geometry questions feel free to
If you're looking to post challenge questions, just for your own and our edification, you can find the challenge question thread here. If you actually need help with these questions, because you're not sure how to do them, then this is the place to post them; but no point posting them here just because they are 'hard'.
Happy to write up a solution, if you want one :)
Jake
Oh whoops, my bad ahaha but what if they're not "hard" in that sense but you still want to test someone?Please reserve this thread for SOS pleas, not for tests.
Oh whoops, my bad ahaha but what if they're not "hard" in that sense but you still want to test someone?
Please reserve this thread for SOS pleas, not for tests.
Not sure how obvious it actually was but to me it was very clear that you just took the circle geometry questions straight out of maths in focus.
What is SOS? SO for tests I put it in the challenge threadSave our souls
Yes it was from MIF but I had to crop the image
Save our souls
In other words a plea for help.
Hi! This is a very generic question compared to the others, but I'm doing Prelims next term, and i was wondering for all exams (HSC Trials,Prelims) involving 3D trigonometry, are diagrams always given?? I have done a lot of practice questions involving them but there are certain ones without diagrams and require you to draw them, and i really struggle at drawing the correct diagram whilst being crammed with tons of info on bearings, numbers and letters.Generally, yes.
Hi! This is a very generic question compared to the others, but I'm doing Prelims next term, and i was wondering for all exams (HSC Trials,Prelims) involving 3D trigonometry, are diagrams always given?? I have done a lot of practice questions involving them but there are certain ones without diagrams and require you to draw them, and i really struggle at drawing the correct diagram whilst being crammed with tons of info on bearings, numbers and letters.
for part iv) is it okay to say:Not written with as much sophistication as my picky self would prefer e.g. <AQT = <AQC = <CRA however enough has been covered to say it is most certainly correct.
since <TAQ=<QCB (from part iii)
therefore converse of equal chord RQ subtend equal angle
therefore CAQR is a cyclic quad
given <AQT=90
therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR)
therefore AR is perpendicular to CB
for part iv) is it okay to say:
since <TAQ=<QCB (from part iii)
therefore converse of equal chord RQ subtend equal angle
therefore CAQR is a cyclic quad
given <AQT=90
therefore <AQT= <CRA=90 (equal chord subtend equal angle in cyclic quad CAQR)
therefore AR is perpendicular to CB
Hi Jake (and whoever else reads this),
Just to get straight to the point,
I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012)
(Image attached, because I don't know how to paste it into this textbox LMAO much technology yay)
And alsoooo,
Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS
Hi Jake (and whoever else reads this),Firstly, exactly what Jake said.
Just to get straight to the point,
I don't quite understand the answer to this binomial question (Taken from CSSA Trial 2012)
(Image attached, because I don't know how to paste it into this textbox LMAO much technology yay)
And alsoooo,
Isn't it so much easier to just belt out factorial notation, and slay LHS=RHS
Hi,
Could someone please attempt this question and tell me which answer they get? I got:
I was just wondering whether or not I got hopelessly lost somewhere... It's a super hard question. :-\
Thanks in advance!
P.S. Sorry if the way I wrote my solution is confusing. Not that good with the LaTex stuff.
Hey guysJamon gave some tips which will be on his lecture slides when they're released.
Just curious but how would you tackle 3D triginometry?
Hey guys
Just curious but how would you tackle 3D triginometry?
(http://uploads.tapatalk-cdn.com/20160715/93dc3fb158b7ae7f1d4dca6284a5925c.jpg)Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC.
how to do part iv )
Do you have a solution to this question, or the source of it? I requested a solution of this question and it used concepts WELL beyond that of the HSC.
Totally agree: I can prove that it equals e, but only using university level maths. I'm sure you can prove the relationship somehow using iii) but honestly it doesn't fall out as easily as I would have expected.The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < n^{k}.to prove the < 3 component.
The proof I got given demonstrated how by the monotone sequence theorem the binomial expansion as a sum is greater than 2 upon inspection. It also assumes n!/(n-k)! < n^{k}.to prove the < 3 component.
Actually, I think that that's fine. Once you get to the Taylor serious of e, you can easily calculate that the number will be between 2 and 3 by summing up some terms. Maybe that's the best way?Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right?
Hey Zoe,
First of all, your description of a potential method of answering this question is absolutely gold. Keep at it, love your work.
Your method, and the method in the answers, is actually the same. If you do what you were planning on doing, that's great, but you would have to equate coefficients as there isn't another way to compare the RHS to the LHS and prove that they are equal. So, you would equate coefficients, and get exactly the same answer as they did. They're answer probably isn't comprehensive enough anyway; if your method gets you the right relationship, you're absolutely doing the right thing.
If you wanted a more comprehensive proof, though, let us know, however as far as I can tell your "do shit get marks" methodology is spot on.
Jake
Well I mean, in 3U we can't assume the Taylor series either. But I mean, we can't assume that the sequence will be monotone without proof right?
Whilst we're here I want to learn them already. Slight motivation to want to go back to uni to learn them next sem.
So would this cut it?
I'm with Jake; it's fine.
So would this cut it?
Can someone here please help me with this
cos^-1[cos1050].
This is what I did:
cos1050=cos330
cos^-1[cos-30] (answer can't be -30 because it is not in the restriction)
Thanks
Can someone here please help me with this
cos^-1[cos1050].
This is what I did:
cos1050=cos330
cos^-1[cos-30] (answer can't be -30 because it is not in the restriction)
Thanks
Thanks Jake and Rui
How do I do this: if a=tan^-1 (-1/3) find cosa and sina
The reason why I am confused is because you can't have negative dimensions
Does this mean I'll need o draw it on a coordinate geometry graph?
Quick Question on Parametrics:
Find the equation of the tangent to the parabola x=4t, y=2t^2 at the point where t=3.
Thank you :)
dx/dt= 4, dy/dt=4t
dy/dx= dy/dt * dx/dt = 4t * 1/4 = t --> t=3, dy/dx=3
dx/dt= 4, dy/dt=4t
dy/dx= dy/dt * dx/dt = 4t * 1/4 = t --> t=3, dy/dx=3
Thanks High Tide! That's the other method I alluded to at the end of my response; clearly quicker if you are more comfortable with this methodology.Could've just asked. It does have a name you know :P
Could someone here please help me with this question:
I've seen the working out for this question but I'm struggling to understand the theory behind it. It would be lovely if someone could solve it for me and explain to me the theory behind it (because that's what I'm struggling with)
Greatly appreciated
Thanks ;D
Um, did you take the restriction into account because that's what confused me the most about this question?Mathematicians call these types of proofs "Proofs without loss of generality"
(http://uploads.tapatalk-cdn.com/20160718/570133fec414d0cb8f9b4840f48dd1b9.jpg)
how to do ques 4a)
ans is 5:28pm to 9:12 pm
(http://uploads.tapatalk-cdn.com/20160718/570133fec414d0cb8f9b4840f48dd1b9.jpg)
how to do ques 4a)
ans is 5:28pm to 9:12 pm
So, the overall equation isI thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet
From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way!
Jake
Don't forget arccos just means cos^{-1}
I thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet
Your working is fine. And we seriously need a system...He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air
He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air
(http://uploads.tapatalk-cdn.com/20160718/65e4b1f299235eb64ebf2dfe16ecff82.jpg)
i dont get ques 4ii)
the ans is written below
(http://uploads.tapatalk-cdn.com/20160715/93dc3fb158b7ae7f1d4dca6284a5925c.jpg)
how to do part iv )
Can you please tell us where you got this question from?
i got this from 2015 independent trial exam paper
Hey! So we need to start out by recognising the general form of an equation like this.
Now, we spend some time filling in all the blanks. Firstly, we know that A is the amplitude, which will just be half the distance from high tide to low tide. So,
Now, we know that period is calculated by
so
Taking period in hours (and therefore time in hours). We solve this, and plug in what we have
Now, we can see that the center of motion is halfway between high and low tide. This is the value for c, so we can plug that in easily
To find b, we have to sub stuff in. It's up to you how you do this part; I think it's easiest to let t=0 at low tide (8.2m) and solve from there. You just need to remember that, at high tide, t doesn't equal 7:20pm, it will equal 6 hours of 15 minutes.
So, the overall equation is
From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way!
Jake
I think it is the 13.3 bit that was incorrect, the actual cosine function itself is fine! Where did the 13.3 come from for you? :DOhh right wrong equilibrium
how do we know that it's a sin or cos function
i assumed it starts at low tide so i put my function as x=-acos(nt+alpha)+13.3 but it didnt work out
Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question?
The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E,
(b) the letters I and P are next to one another,
(c) there are three letters between N and T,
(d) there are at least three letters between N and T.
how to do ii)I don't like this question
What if you did the triangle method for this question?Oh, the ratio definition proof is the triangle proof. I just gave it a different name there.
-tan(alpha)=tan(beta)
-Alpha=beta
Is this possible since they both represent angles?
Bit confused here. What do you mean by that expression on the LHS?
Or was the square root not meant to be there
Notes:
Lol fair enough.
But that solution lacks rigour in my eyes. Without proper justification that both alpha and beta lie between -pi/2 and pi/2, cancelling out the tan's is not properly justified.
(Or it could be I'm too used to uni maths and everything has to be presented with 100% precision)
I'd be leaning towards saying that the solution would probably be passable at this level of study, but I could be mistaken. The train of thought is solid enough ;D that being said, I recommend getting in the habit of writing all restrictions in the HSC, even those which are seemingly inconsequential or even obvious, to make the marker happy and to make sure your reasoning is clear, catch tricks, etc etc ;D you'd almost definitely get away with it, but better safe than sorry :DIt's as you said in the lecture
It's as you said in the lecture
I'm the type of person who will get you out for anything in maths :P
Hey everyone,
I found this hectic induction question but I couldn't find the solutions, if anyone can do it please reply!!!
Also the image was too big to attach so I just put it onto a document...
Hey guys this is my first time on here, I'm struggling with probability right now (mainly permutations and combinations). Could someone please help me with this question?Hi Tim,
The letters of the word PRINTER are arranged in a row. Find the probability that: (a) the word starts with the letter E,
(b) the letters I and P are next to one another,
(c) there are three letters between N and T,
(d) there are at least three letters between N and T.
Hi Tim,
I think the answer are 1/7 , 2/7,1/7, and 2/7.
Please correct me if i am wrong thanks!
how to find domain of y= inverse sin(1-x^2)
ThanksWhat do you mean?
Do I chain rule everything with a coeffiient in front of the inverse trig?
I tend to get confused with coefficients and differentating inverse trig
Similar to how I think 2tantheta=tan2theta
What do you mean?
I used the chain rule because of the stupid square root inside the inverse sine, not the coefficient
So if you had y=2 squareroot of (sin^-1x) would you still have to use the chain rule?
Could they ask us to sketch a graph of that function^In Extension 2, of course
And how would you even sketch that graph??
For this question (that I have attached below) why do we take 3 outside of the integral rather than -3 (look at 2nd step)
The reason they take 3 instead of -3 is because integration to inverse cos requires a negative coefficient.
They could just have easily taken out -3, and integrated to sin instead of cos. Hope that makes sense!
Jake
For clarity conic, both answers are correct, just depends what you want to do with it ;D
Thanks Jamon
Anyways I am confused with the following inverse trig integration questions
For the first Integral I attached below, why do we remove 1/2 out to the front?
Ignore the second because I believe that this is a similar principle to the first
For the third why do we use polynomial long division?
Also could someone please help me with the second attached file and explain the theory behind this because I'm not too sure what the underlying theory behind this is (I mean I've seen the working to it but I don't understand what's going on)
Thanks
Thanks JamonWait woah woah woah woah woah.
Anyways I am confused with the following inverse trig integration questions
For the first Integral I attached below, why do we remove 1/2 out to the front?
Ignore the second because I believe that this is a similar principle to the first
For the third why do we use polynomial long division?
Also could someone please help me with the second attached file and explain the theory behind this because I'm not too sure what the underlying theory behind this is (I mean I've seen the working to it but I don't understand what's going on)
Thanks
Wait woah woah woah woah woah.
That Q1 is not a 3U question.
Too true, fell into my pattern there aha woops ;) you'll need partial fractions conic!! That's a 4U concept ;D
This was the working to it (attached below)Explain how your minus became a plus
Explain how your minus became a plus
Hey guys, i just have a quick question; many application to the physical world questions have a question where they basically say "describe the movement of the particle". I was just wondering how you interpret the displacement, velocity and acceleration to describe its motion? Like if one is positive and another is negative, or if all are negative or all are positive, and combinations like that. Also what is a limiting positionYou need to understand what motion actually is in the context of our course.
Thanks heaps!!
Hey guys, how do you do part ii and iii.(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_zpsizhyzxyf.jpg)(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_0001_zpsdosyzk5b.jpg)
(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_zpsizhyzxyf.jpg)(http://i1164.photobucket.com/albums/q566/Rui_Tong/CCI01082016_0001_zpsdosyzk5b.jpg)
Am I wrong or are the answers wrong?
You are correct, but the answer is correct too!
Using a=1/3 in your formula, the answer comes out! That's where the 1/3 out the front comes from ;D
Hey thanks so much for the reply. I was just wondering, is the time of flight 20 seconds for both particles. because the answers say that for particle A its 20s but for particle B its 10s. You think the answers are wrong?It would be 10s for particle B simply because it started ten seconds later.
I am still confused ???Don't see your point.
I am still confused ???
That's alright! Let me do the integral for you. Do you understand the algebra I applied above? Basically when we factorise out the 9 from the square root symbol, moving it outside that symbol means we must instead write 3 (the square root of 9). Let me know if that's the issue for you and I'll do the full working. Assuming it's the integral itself:
Does that help? What specifically is troubling you?
hey guys for part b, the answers say that the max value occurs when t=1, how the heck do you figure that out??
how do you show part ii?? thnx!
how do you show part ii?? thnx!
how do you show part ii?? thnx!
Hi again! Thanks for the help last time :) , um, I'm having an issue with permutations and combinations. I don't really get the some number questions and "alternating" questions at all and was wondering if you guys could help me again.
The numbers 1,2,3,4,5,6,7,8,9, are arranged in a straight line. Find the number of possible arrangements if:
i) All the odd numbers occur next to each other.
ii) All the odd numbers occur in increasing order, reading from left to right.
10 people (5 boys and 5 girls) have a birthday party. In how many ways can they be seated if the birthday boy is to be seated between two particular girls.
Sorry for the trouble.
How do you do part iii, iv, and v ???I've noticed that a lot of your questions are projectile motion based. Is there anything about the topic that troubles you?
I've noticed that a lot of your questions are projectile motion based. Is there anything about the topic that troubles you?
Most of the questions just require knowing what means what (e.g. time of flight means y=0) and braving heavy algebra.
Hi! So i was wondering when using Newton's method of approximation, whether i need my calculator in radians or degrees such as in the question attached :)
Hi! So i was wondering when using Newton's method of approximation, whether i need my calculator in radians or degrees such as in the question attached :)
Hey! Unless you're working with angles which are specifically in degrees, you should ALWAYS be using radians.
Hey dream dog, radians!! Always radians unless you see the degree symbol (and you'll never see it when doing things with Calculus) ;DBrilliant :)
This is another attempt at Q2.
How do you do part iii, iv, and v ???
Hey how do you do part iii, that's the only part i don't get. (btw i'm sorry if im annoying you with all these projectile questions, i just have a few that I dont get :-\ )
Am I doing something wrong?From line 3 to line 4 do not take out the negative and complete the square in the form A - (Bt+C)^{2}
From line 3 to line 4 do not take out the negative and complete the square in the form A - (t+C)^{2}
Line 4 to line 5 made no sense. How did -3 and 6 become -1/2 and 1
I factorised the whole thing by 6If you factorised the whole thing by 6 that should be 18 there not 1/2.
I think I'd swing at Part (v) a little differently, I interpret the question differently. When it says precisely two distances from O, doesn't that mean it has two solutions? The wording is really weird, but I don't think it relates to the distance d given earlier in the question, nothing suggests that. So my interpretation would be that we just need to prove that, provided that the condition on U holds, then there are two values of d which satisfy the other conditions given.Yeah seems more legit that way aha I just got completely lost out of the rigour of that question in my fatigue
This would make sense, provided the target is moving slow enough, it can hit it on the way up (one distance), or on the way down (another distance). At least I think. I just don't think it relates to the distance d. What do you reckon Rui? ::)
Yeah seems more legit that way aha I just got completely lost out of the rigour of that question in my fatigue
Guys i don't get part d and e. btw i really appreciate the help ;DI have to ask this again. What is the source of these questions
Can we please refrain from too many past HSC questions before the year 2001? The difficulty/rigour of these questions is seriously getting WELL beyond that of the current HSC.
In saying that, here are solutions. THSC.
Always feel free to comment further if any part of the provided solutions don't make sense though.
Isn't it good that you're doing it because you're exposing yourself to questions harder than the HSC?Sure, if you want to aim that high.
Sure, if you want to aim that high.
I never went before 2001 and I still did reasonably well anyway. It's becoming really hard to provide concise solutions, or sometimes you just get lost lol.
Sometimes it can be nearing impossible to provide solutions so easily because it's reaching a point where the lateral thinking required is not that prevalent anymore, and demanded by the students today.
Hi, I'm having trouble finding the asymptotes, turning points, inflection points etc to draw these graphs
Yeah they're pretty hard, but they're pretty fun at the same time! :P Anyway, how do you find the amplitude for the attached equation (its Simple Harmonic Motion)
'Just'. Toughest Amplitude question I've seenI did the exact same question around a month ago
I did the exact same question around a month ago
Plus you're hard to impress xWow ok
Thanks Rui!!! Where would I be without you.
Also for the attached equation (SHM) i have to find the speed of the particle when x=2. Do i just sub x=2 into the displacement eqn to get time and then sub that time into the velocity eqn, or is there a better way of doing it? thanks!
:O Neutron has something other than a physics question (surprise, i finished my trials :P)I'm in a lecture right now so I'll just address the binomial theorem question.
Okay, so for some reason I can't get part ii of this, I feel like I'm just tired and not processing properly, but if one of you kind souls could help that would be great!!
b) A rescue plane is travelling at an altitude of 120 metres above sea level and a constant speed of 216 km/h towards a stranded sailor. A canister containing a life raft is dropped from the plane to the sailor.
i) How long will it take for the canister to hit the water? (Take g=10ms^-2)
A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped when the horizontal distance from the plane to the sailor is D metres.
ii) What values can D take if canister lands at most 50 metres away from the sailor?
And also, one more! I attached a screenshot :D
Thank you guys!!
:O Neutron has something other than a physics question (surprise, i finished my trials :P)
Okay, so for some reason I can't get part ii of this, I feel like I'm just tired and not processing properly, but if one of you kind souls could help that would be great!!
b) A rescue plane is travelling at an altitude of 120 metres above sea level and a constant speed of 216 km/h towards a stranded sailor. A canister containing a life raft is dropped from the plane to the sailor.
i) How long will it take for the canister to hit the water? (Take g=10ms^-2)
A current is causing the sailor to drift at a speed of 3.6 km/h in the same direction as the plane is travelling. The canister is dropped when the horizontal distance from the plane to the sailor is D metres.
ii) What values can D take if canister lands at most 50 metres away from the sailor?
And also, one more! I attached a screenshot :D
Thank you guys!!
Hey guys how do you determine whether a particle will return to the origin or not??Does setting x=0 yield a solution?
Hey guys how do you determine whether a particle will return to the origin or not??
(http://uploads.tapatalk-cdn.com/20160805/4a76b9eb0f24662e0c40e86bd778796d.jpg)Yes it is.
(http://uploads.tapatalk-cdn.com/20160805/a84a1ded04b84c317970903fb0ee63ea.jpg)
I have attached my pathetic attempt as well, just for the lols. And my flipped photos are...wow...
Is t formulae question merely just subbing stuff in...?
Correct?(http://uploads.tapatalk-cdn.com/20160805/0c1433fc26ba8d8300c01321709c9987.jpg)Yes.
what about something like this?
Thanks Jamon!
Guys for the question attached, in part iii, I know that we have to differentiate the eqn of part ii, but do you differentiate with respect to theta or respect to alpha?
Hi i was wondering if i could get help with these two questions
If you factorised the whole thing by 6 that should be 18 there not 1/2.
Anyway, regardless of if you factorise the 6 out, put the negative back in there and complete the square.
Just curious but why specifically take out 1/squareroot of 18To use the standard form of the integral, the coefficient on t must be 1.
What am I supposed to do after?
I don't get the bit where you went to tan? (from how the integral was converted into tan)
What did I do wrong here?Two mistakes in the last line.
Two mistakes in the last line.
1. You dropped off the 1/16
2. The x went missing inside the inverse tangent
ThanksRequires partial fractions. Not 3U material.
How do you do this (I think there's no inverse trig integration rule for it)
Requires partial fractions. Not 3U material.
Thanks
What am I doing wrong
Thanks
How would I do this?
Factorise!
That's an inverse sine integral ;D you pull the 1 on root 3 out the front and then use your regular rule ;D
What happened to the dx?
Hey there, could some help explain to me how to graph these functions.
1. Cos (cos inverse x)
2. Cos (sin inverse x)
3.. Cos inverse (cos x)
4. Cos inverse (sinx)
5.. Tan inverse (tan x)
6. Tan (tan inverse x)
Sorry these are alot of graphs but I get really confused and I find it hard to wrap my head around these trigonometric inverse questions especially with all the restrictions. Any other general tips in dealing with these questions will also be appreciated :)
Thanks
Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one:
In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by $250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed $5 000 000 when it is finally won.
The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D:
Thank you legends!!
Hey hey! Thank you so much for the help with the other question ^^ I have returned with yet another one:
In a given lottery the probability that the jackpot prize is won is 0.013. Successive lottery draws are independent. The jackpot prize is initially 1, 000, 000 and increases by $250 000 each time the prize is not won. Find, correct to 5 decimal places, the probability that the jackpot prize will exceed $5 000 000 when it is finally won.
The answer just found the probability for it being won on the 17th draw (First draw when the prize is above 5 000 000) but don't you have to take into account the fact that the prize could be on on the 18th draw, 19th draw, 20th draw etc all the way till infinity? I don't understand how just finding the probability of it being won on the 17th draw is equal to finding the probability that the jackpot prize will exceed 5000 000 when it is finally won D:
Thank you legends!!
It's from one of my school's past papers :/
yoo guys, howdu do part ii?
Ahh thank you so so much!! One more thing from me (sorry) :o This is the question:Match it up:
The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB?
How do you know whether it's 3:10 or 10:3? :O Thank you!
Ahh thank you so so much!! One more thing from me (sorry) :o This is the question:
The point P divides the interval AB in the ratio 3:7. In what external ratio does the point A divide the interval PB?
How do you know whether it's 3:10 or 10:3? :O Thank you!
Match it up:
A---P-------B
3 : 7
|-----10------|
So A is closer to P than it is to B. So since P comes first:
So 3:10
But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question:
Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective?
Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said:
P(exactly one computer defective)
=P( 1 defective from A, 0 from B) + P(0 from A, 1 from B)
3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095
Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ )
But when you look at questions where you divide externally, normally you go from B to A and then back to P? Ahh another thing (sorry i keep doing more questions and running into more problems) With this question:You do? I always just matched up the ordering.
Warehouse A has 100 computers and the probability that one of these computers is defective is 0.02. Warehouse B has 100 computers, two of which are defective. Joe buys three computers from Warehouse A and three computers from Warehouse B. What is the probability that exactly one of the computers he has bought is defective?
Why aren't you allowed to just use binomial probability for both warehouse A and B? The answer used binomial probability for Warehouse A and normal probability for Warehouse B? This is what the solution said:
P(exactly one computer defective)
=P( 1 defective from A, 0 from B) + P(0 from A, 1 from B)
3C1 (0.02) ^1 (0.98)^2 x 98/100 x 97/99 x 96/98 + 3C0 (0.98)^3 x 2/11 x 98/99 x 3= 0.1095
Why do you have to times by 3 for the second part of the addition ( for the P(0 from A, 1 from B) ) sorry, I think my brain's fried, but tysm Rui!! (Sorry for having to type it up, my school website's down again :/ )
Did this in my 3U lecture ;)I thought the question looked familiar..
Find domain and range of the following functions and sketch their graphs:
1. y=-3sin^-1 x
For domain I got -1≤x≤1
For range I was not too sure???
Find domain and range of the following functions and sketch their graphs:
2. y=cos^-1 2x
For domain, I got -1/2≤x≤1/2
Range I'm not too sure. Is it 0≤cos^-1 2x≤π?
Find domain and range of the following functions and sketch their graphs:
3. y=3tan^-1 (x/2)
Domain: all x
Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong)
Point of inflection: let x/2=0
x=0
Sub in LHS point (and I'm not sure of what I need to do)
Find domain and range of the following functions and sketch their graphs:
1. y=-3sin^-1 x
For domain I got -1≤x≤1
For range I was not too sure???
2. y=cos^-1 2x
For domain, I got -1/2≤x≤1/2
Range I'm not too sure. Is it 0≤cos^-1 2x≤π?
3. y=3tan^-1 (x/2)
Domain: all x
Range: -Ω/2≤3tan^-1 (π/2) ≤pi/2 (I think I got this wrong)
Point of inflection: let x/2=0
x=0
Sub in LHS point (and I'm not sure of what I need to do)
Could someone here please help me
Thanks
1. y=-3sin^-1 x
For domain I got -1≤x≤1
For range I was not too sure???
hey, how do you do this guys?(http://i1164.photobucket.com/albums/q566/Rui_Tong/Screen%20Shot%202016-08-08%20at%2011.57.17%20AM_zps1rqgkwn9.png)
How would you do thisTheorems to use:
for i and ii would you have to use your circle geometry theorems
_____________
hey guys howdu do part c, ive been trying for ages idk why i cant get it >:(
Also how do you do this?? (part ii)This is what I think:
This is what I think:
You can either include the digit in the group, or you cannot.
There are 9 digits. You can say yes to that digit, or no to that digit. (And note there can't be repetition.)
So that's a total of 2^{9} outcomes.
(But, if you have to have at LEAST one digit, you can't have no no no no no no no no no)
So in total 2^{9} - 1
______________
If you were to use part (i), the trick is that you want to find 9C1 + 9C2 + 9C3 + ... + 9C1
I.e. 2^{9} MINUS 9C0
hey guys how do you do part b?
Also, in general, how do you prove that a coefficient(s) is/are the greatest in an expansion. do you just have to proves the the term in front divided by that term have a value of less than one, thus it is the largest?
guys for this i got 124, but the answer is 9420??!?!
guys for this question how do you do it? i just cant figure out the last lines.
guys,how do you do part ii??Yes. (Technically equating coefficients)
wait i think i got it; do you just expand and equate constants?
I have a feeling you may need the symmetry property ^{n}C_{k} = ^{n}C_{n-k} though.
Whats this :O. Can you teach me what it is? (please)
Guys for part ii, i can see that u have to integrate, but why are there the negative and positive signs. If u let x= -1 you'll get that but opposite to what you have to prove :S
Note that I use a definite integral to save me from the need of finding a constant of integration, and then just subbing in a new value for x
but how do you get the alternating positive and negative signs? for e.g. when u integrate, the second term is x^2, how can that be negative :SWhat do you mean?
What do you mean?
By putting -1 on the bottom you're doing F(0) MINUS F(-1), where F(-1) is the antiderivative of that expansion
guys how do you do part ii?
I don't know how to do this either :/
Thanks Jake!
Guys I just have one more binomial question (hopefully)
I don't know how to do both parts
Thanks!
Thanks Jake!From memory this was in the 2004 HSC. Like previously said, please consult the answers before you post these up.
Guys I just have one more binomial question (hopefully)
I don't know how to do both parts
Thanks!
I'm not going to be able to have a crack at this one tonight, but by the looks of it you need to expand the binomial, set it equal to it's un-expanded form, then differentiate it. You'll either have to differentiate again, or just set x=1, I'm not entirely sure by the looks of it. Give it a go, show us your working out, and we can have another go in to morning!First part's differentiating. Second part is integrating though
I'm not going to be able to have a crack at this one tonight, but by the looks of it you need to expand the binomial, set it equal to it's un-expanded form, then differentiate it. You'll either have to differentiate again, or just set x=1, I'm not entirely sure by the looks of it. Give it a go, show us your working out, and we can have another go in to morning!Yeah it's correct.
Hi!
Could someone please explain how to work out this multiple choice question? Thanks! (Answer: C)
This question looks simple but its pretty deadly, no one in my class got it, nor my teacher; help please!I only know how to do this one cause I've seen it before,
Hey! Could I please have an explanation for solution for (a)(ii). I'm a bit confused as to why k=-6
guys what is this question even?
The range for this is -3pi≤y≤0
The range for this is just the normal 0≤y≤pi
The range for this is -3pi/2<y<3pi/2
im not really sure how you get point of inflexion (sorry) but im pretty sure its still at the origin for this.
could you please explain how/why they are the range(s)?
Could I please have some help with this mathematical induction Q
Guys, for part i- how do i show that a limiting sum exists (i got s=cos^2x)
For part ii- how do you do it?
Thanks!
Guys i just don't get the last part, where is says "find the equations of the tangent lines to the curve..."
Guys again for this I don't get the last part where it says to find the rate at which theta is changing. What i did was find x when theta=pi/4 and sub it into the eqn of part ii and i got 0.2 but the answer is 11 degrees/sec :SThe answer wants it in degrees/sec. Just convert 0.2 rads into degrees by typing in 0.2 * 180/π
Hi again,
I didn't quite understand part ii) of this question... how did the answer get ln? I thought when integrating (x^2+3)^-0.5 you could use the reverse chain rule so t = (x^2+3)^0.5/(0.5 x 2x). Could someone please explain? Thank you!! :)
Hey!
The thing to note, when doing the reverse chain rule, is that you cannot do it when, inside the brackets, there is a non-linear function. So, you can integrate
But you cannot integrate
using the reverse chain rule. You have to use alternate means, such as the table of standard integrals (as is the case in this question). So, knowing that you can't actually integrate using the reverse chain rule in this case, does the answer described make sense? Like you understand their application of the table of standard integrals?
Great question though!
Jake
Ohhhh, yes that's why they used ln!! I didn't know to do this as they don't provide us with the standard table of integrals anymore (since this was from a 2014 paper, before the formula sheet got introduced). But how would you do it without using the table of integrals?
Ohhhh, yes that's why they used ln!! I didn't know to do this as they don't provide us with the standard table of integrals anymore (since this was from a 2014 paper, before the formula sheet got introduced). But how would you do it without using the table of integrals?
guys how do you do part ii?
how do you do this q guys??
Hi again,See post #429 here. I have already done this question.
Another question I couldn't quite grasp - part iv). Could someone please explain? Thank you!!
guys how do you part ii :S?
guys how do you part ii :S?
Consider the girth first, it is the shortest distance around the parcel. And then L is the longest side. Based on the dimensions, this gives us two sets of possible equations. The first, if x is the longest side, gives us this:
Or, if y is the longest:
woah i completely missed the km/HOUR, Thanks Rui and Jake!!
Just another question, I can't get part iii, i keep getting 24 but the answer is 37 :/
How would you do this:At this point, you need to read the textbook and make sure you understand what inverse sine is.
1. Let f(x)=sin^-1 x
a. State the domain and rage of f(x)
I got "all x" for this and for the range I got -pi/2 =<y=<pi/2
Just came across a geometry question and was wondering if, just as with similar triangles their sides are in corresponding ratios, can this be applied to a parallelogram?Similar polygons with number of sides greater than 3 are not in the course. Hence, no.
Just curious. Cheers.
When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1)From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5.
Always been confused with these types of questions. Cheers.
When a polynomial Px is divided by (x-2) and (x+1) the respective remainders are 5 and 8. What is the remainder when px is divided by (x-2)(x+1)
Always been confused with these types of questions. Cheers.
EDIT: Another question,
find the exact value
sin(cos^{-1}2/3 + tan^{-1}-3/4)
Using our general formula:Comment: Would like to mention that the "formula" is known as the "division transformation". :)
P(x) = (x-1)(x+1)q(x)+Ax+B
Substituing in the particular values for x, in this case, 2 and -1.
P(-1) = 8, therefore:
8 = -A+B
From the given information, we know that upon division of the polynomial by (x-2), the remainder is 5, therefore P(2) = 5.
Also, division with (x+1) leaves 8, therefore P(-1) = 8.
Using our general formula:
P(x) = (x-1)(x+1)q(x)+Ax+B
Substituing in the particular values for x, in this case, 2 and -1.
P(-1) = 8, therefore:
8 = -A+B
P(2) = 5
5 = 2A+B
Solving the simultaneous equations, we get:
A = -1 & B = 7
Therefore the remainder present upon the division by (x-2)(x+1) is -x + 7
Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + B
Cheers.
Ah, okay. But how do you know that the remainder for (x+1)(x-2) is in the form Ax + BFurthering onto what Jake said now.
Cheers.
How would you do the following questions attached?See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii).
See post #429 here for the first one. The method for (iii) is given and alluded to for part (ii).
I don't have the time to do lengthy questions right now.
Aha thanks Jamon :)
If cosx = -3/5 for 0<x<pi then tan(x/2) = ?
Hey! I need a hand with this binomial theorem question (its 11f from the 2012 Ext1 HSC:
I have done part i) with no issues, the answer is -1760, its just part ii) I'm really stuck on.
Thanks, this website is saving my HSC :)
At this point, you need to read the textbook and make sure you understand what inverse sine is.
I'm sorry to say this, but this is because your answer is wrong for a foundation level question.
(In particular, the domain)
Also, the derivative is also in the textbook
I got it nowTwo ways to analyse c), both of which lead to the answer "no".
Anyways the question was let f(x)=sin^-1 x
Following up on that question, the questions were
c. Is f'(x) undefined at x=1 and x=-1. Deduce the tangents to the curve at these points
d. Show that the line y=x is a tangent to the curve y=f(x) at (0,0)
Could someone here please give me hints
Thanks
Lol'ed ;DThe ∞^{th} commandment:
I was wondering is there a way to always be able to graph an exponential function and finding the exponential equation from the graph?
I am really struggling with exponential graphs in terms of drawing them gievn the equation and being unable to find the equation given the graph. The tranformations that can be possible are also being quite an obstacle for me. Please help :'(
Perhaps a set of rough rules will help!!
Consider the general form of an exponential function:
(Note that it doesn't have to be the with an 'e', any number works fine,though if you have a negative number there you'll need to adjust this a littlethe base won't be - this requires complex analysis)
- Higher values of A stretches the curve in the vertical direction. Really, the noticeable impact is that A is the y-intercept, but only if there is no constant C added to the exponent
- Higher values of B means the exponent will increase at a faster rate, the right hand side of the exponential will increase its slope noticeably
- C causes a horizontal shift to the left (for positive values) or right (for negative values)
- D causes a vertical shift upwards (positive values) or downwards (negative values)
These rules might seem very wish-washy, and that's because they are, they should only be a rough guide. You aim should be to gradually develop an intuition for what the different parts of the exponential do. Have a play with sketching different functions on Desmos or something similar, and try to notice the patterns in what you see, that is the best way to develop that intuition ;D
Thank you
I was also wondering how would you evalulate it as a single power and evaulate. Also does evaluate simply mean solve?
Thank you rui ace
I also tried these questions but i am not getting the right answers. I got 2+25 for a
For a) you should get 2 + 125. You've correctly square rooted the 4, but after you square root the 25 (to get 5) you need to put it to the power of 3 (resulting in 125).
For b) you can use indices laws to multiply the indices in brackets, then convert to 9 to 3^2. Then, you can add powers when multiplying with the same base
For the answer for a the sheet says its 2/125? Im still a bit confused for b i got (5^2 -3^4)÷ 3^1
I got 5^2 -3^2? But the answer is -2?
I also tried this question but im not sure to do next
@anotherworld2b
Just to clear things up - Your textbook is full of typos. I think the answers they provide can be reached if all those TYPOs were amended.
Hi, I'm having trouble with part b of this question. I've tried finding the cartesian equation of each projectile and equating them, but I keep getting a weird answer. (the textbook answer is x=11.5m and y=8.2m). I'm not sure what I'm doing wrong, help would be appreciated! Thanks! :)Note: Part a) looked fine.
Note: Part a) looked fine.
When I simulated the relevant Cartesian equations of motion on GeoGebra, I found the diagram that they provided was highly deceptive. Unless my equation was wrong, particle B literally soars right over particle A.
When I plugged the values for the answer in, I found that particle A travels through the given coordinates, but not particle B.
These are the equations I had for particle B. Tell me what you think.
Yeah, I had similar equations but since the velocity is in the opposite direction [relative to the other projectile], acceleration is -10, velocity is -10 - 20sin(45) and therefore displacement is -5t^2 - 20 sin(45) t. (Just a sign difference).The direction of the velocity is flipped for the horizontal component but not the vertical.
The direction of the velocity is flipped for the horizontal component but not the vertical.
If when t=0, y'=-20sin(45), then we'd be implying that the projectile was being fired into the ground, not upwards.
Bear in mind maths in focus is full of mistakes. I might be wrong but I'm pretty sure I didn't do anything stupid here (if anyone would like to check, please do).
Hi,
I've only just recently started doing some binomial questions. In a textbook, I got through 8.1 fine, but got stuck on a few 8.2 questions. (The first two questions are from a textbook). Also, I found that I could do some of the past paper questions on binomial theorem, but the last two pics are 2 questions which I didn't how to approach.
Thanks.
For the last one ii) the binomial identity, i've tried integrating, factoring out the x then subbing in x equals -1 and it gives you the lhs but not the rhs :/
Will work on this further
But wait! After integrating and subbing in x=0 you get the entire lhs equals to 0 and the rhs equalling 1/n +1
Summing the two would give the final identity. I think I might have broken a maths law haha could someone check?
For the last one ii) the binomial identity, i've tried integrating, factoring out the x then subbing in x equals -1 and it gives you the lhs but not the rhs :/
Will work on this further
But wait! After integrating and subbing in x=0 you get the entire lhs equals to 0 and the rhs equalling 1/n +1
Summing the two would give the final identity. I think I might have broken a maths law haha could someone check?
Thanks Jamon, Spencerr and Rui for you help!Just quickly regarding the definite integral question (I'm busy right now)
For the 1992 q, for the definite integral, how would you know to use 0 and -1 or does any combo of 1/-1/0 still work fine?
For the 1996 q, is the pic below the kind of working out required for the 2 marks of the questions?
Thanks.
So from the textbook questions, Q9 can be solved in a similar way to the 1996 question, but how would you solve Q7?
Also, again in the pic is the working to Q9. Is this how you would set out the working to 'show' the given equation?
Thanks Jamon, Spencerr and Rui for you help!For your equating coefficients thing, you can just say that the coefficient of x^{something} in the expansion of ... is ...
For the 1992 q, for the definite integral, how would you know to use 0 and -1 or does any combo of 1/-1/0 still work fine?
For the 1996 q, is the pic below the kind of working out required for the 2 marks of the questions?
Thanks.
Hi Rui, thanks for answering. I understand why we use x=0 and x=-1 for the indefinite method, but previously, I couldn't see why you had chosen these from a quick glance at the question with the definite method, because I'm used to using to definite method with resisted motion questions where the limits of the integration are given with the initial and final conditions.Just pick 0. Usually if your thing is tidy then 0 is involved.
I really like this method because it's really quick :D Thanks for your time.
(http://uploads.tapatalk-cdn.com/20160830/f7ab2bd6fd855e696d342cbb6152427e.jpg)
how to do part b ii)
Hey! My solution is below :)Just check your working out again. Your answer gets A > (3√3) / 2
(http://i.imgur.com/K2pjoxE.png)
Jake
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpsco1hzpaf.png)
If you play around with GeoGebra you'll see what we're trying to do. Notice how as A gets larger, the stationary points seemingly poke outwards more. We are basically telling it to not poke over the x-axis.
I was wondering how do you do these two questions?
Hey! Just for everyone's else sake,
THIS IS NOT HSC!
Whilst I have done matrices, I don't really understand the terminology used, so won't be able to help you out here. Sorry!
Jake
I was wondering how do you do these two questions?
For this question im not sure how to start it
For this question im not sure how to start itMatrices are in no way a part of the HSC course.
I can handle this one ;DHow did you even know this anyway :o
First, a definition a shear, in case you need it. A shear parallel to a given line through the origin is a transformation in which the component of a vector perpendicular to the line is unchanged, and the component parallel to the line is increased by an amount proportional to the perpendicular component. If the proportionality constant is k, we call it a k–shear.
For this question im not sure how to start it
Sorry for the inconvenience caused.i wasnt wure where to post my questionsAll good
How did you even know this anyway :o
(http://uploads.tapatalk-cdn.com/20160907/ae4126f0bb6b882c43323b42292c0630.jpg)
how to do part b ii) a and b
For a), we know that the amount of oil being poured into the hemisphere is
Therefore, after 8 minutes, there will be a volume of
Setting this as V,
Just looking at the equation, it's pretty clear that the solution is h=1m
For b), we need to start by finding the rate of change equation.
This should be intuitive by now; set what you want to find on the left hand side, then use what you can find out to produce a right hand side. We know the change in volume over time; we just need to change h over V.
I'll leave you to sub the relevant numbers int
(http://uploads.tapatalk-cdn.com/20160908/db3542b435c3758342f6b2bfde7636c6.jpg)That question doesn't make sense to any of us. Mind telling us where you got it from?
how to do part a iii)
In each game of chess that Bobby plays against Boris there is a probability of 1/3 that Bobby wins, a prob. of 1/6 that Boris wins, and prob. of 1/2 that the game is drawn. They play 4 games.
i) find prob. that Bobby wins 2 games and Boris wins 2 games
ii) find prob. that Bobby wins 1 game, Boris wins 1 game and the other 2 games are drawn.
answer:
i) 4C2(1/3)^2(1/6)^2
ii)4C2(1/2)^2(1/3)(1/6) *2
i get the answer for i) but i dont understand why there is a need to multiply 2 for part ii)
ques2 :
Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral
i) normal at P is x+py=2ap+ap^3
ii) R(0, a(2+p^2)
iii) prove SP is equal in length to latus rectum, that is 4a units
not sure how to do part iii)
ques2 :
Consider x^2=4ay with focus S. The normal at P(2ap, ap^2) meets y axis at R and triangle SPR is equilateral
i) normal at P is x+py=2ap+ap^3
ii) R(0, a(2+p^2)
iii) prove SP is equal in length to latus rectum, that is 4a units
not sure how to do part iii)
Can anyone give me a rundown on parametrics or tell me where i can find something that will explain it easily?
Can anyone give me a rundown on parametrics or tell me where i can find something that will explain it easily?Look for where it says parametrics
Look for where it says parametrics
Watch eddie woo videos on Youtube
What should you do in the 5 mins reading time?
Thank you!!!
Edit: Posts merged, avoid double posting, use the Modify and Insert Quote functions instead!
What should you do in the 5 mins reading time?
Thank you!!!
Edit: Posts merged, avoid double posting, use the Modify and Insert Quote functions instead!
In the five minutes, I usually try to answer as many multiple choice as I can after I quickly flick through the paper. In Ext 1, you're not going to really get far on any major question, so I think multis is the best way to go!On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those.
On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those.
On the other hand (but still overlapping) I only do about 1-3 multiple choice questions. I take enough time to look at the paper and actually know what's going on. I look out for questions that will throw me off and brace myself for those.
Hey guys how do you do part ii and iii for the q attached??Part (iii) is easy. Clearly r=6 and n=12 so just sub it into the formula for (ii).
Hi! This is a silly question but I never understood how to do these types of polynomials :/The answer is just (-2x+6)
P(x)=(x+1)(x-3)Q(x)+(-2x+6)
Where Q(x) is a polynomial.
Find the remainder when P(x) is divided by (x+1)(x-3)
Sorry if this is dumb D: Thank you!
Neutron
Hi! This is a silly question but I never understood how to do these types of polynomials :/
P(x)=(x+1)(x-3)Q(x)+(-2x+6)
Where Q(x) is a polynomial.
Find the remainder when P(x) is divided by (x+1)(x-3)
Sorry if this is dumb D: Thank you!
Neutron
And finally, 5 minutes is enough time to say 'I think I can' about 150 times. At the end of the day, no matter what you do in the 5 minutes reading time, the important thing is to be in the right mindset. If finding tricky questions, and feeling safe in knowing you can figure out how to do them, is the way to make you feel comfortable, go for that. If reading the formula sheet, or listing pneumonic devices, is your thing; do that. Reading time isn't a huge factor in exam success (although, if you use it in a way best suited to you, it can be very effective)
hi! can someone please explain this question for me :)
Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.That question is kinda mean.
another one, how do you do part ii guysss??
Guys how on earth would you do the q attached in an exam?? The answer is C but I don't get how you can get without using something like wolfram alpha. Any tips would be greeeatly appreciated.
Hi can anyone help me with question 12b)i) and ii) in the 2015 hsc 3u exam? Its not letting me attach a pic of the question so if someone could look it up that would be great! I have no idea how to do these types of questions so any help is extremely appreciated!These are one mark circle geometry questions. Tips:
These are one mark circle geometry questions. Tips:
(i) <ACD is right next to it and labelled 30 degrees. That's something I immediately keep in mind. And then just read the question. They explicitly tell you BD is a diameter, so what does that say about <BCD?
(ii) I see the angle formed by a chord (AD) and a tangent (XD). I immediately think about alternate segment theorem. There are two triangles to use alternate segment theorem on, so just look at both of them and find a useful once.
Remember - You're only looking for like one thing for a one marker. Don't overanalyse in these scenarios.
(Exclusion: That 1 marker is the last question of the paper. Maybe think a bit there)
That makes sense, thank you!!
Also, could you please explain how to do 13a)iii) from the 2015 hsc paper?
Mod edit: You're most welcome, but please stick to editing prior posts in these situations over unnecessarily posting
I only understand how you got c=5 but not the rest. How did you find a=2 and what n equals to?There's a modify button in the top right corner of every post you make. Just use that and you're all good.
Also im not sure how to edit my previous posts
Hi guys! Can you help me with this question? Thanks!
Thanks! I thought I needed to expand the cos(x+h) so I got confused. :-X
What about this question? I'm a bit confused on how to take the theta out. Thanks!
Hi Jake,
I am doing Extension 1 mathematics next year, as I did really well in prelim Ext 1. The thing is, I have heard many rumours from students in the year above me that it is really hard to do Extension 1 by itself (without Ext 2). However they say, it's really easy for people doing Extension 2. I just wanted to know what your take on that is (how true is the statement?) and whether extension 1 is really worth doing as it only counts for a single unit in my case.
Thanks heaps,
Samuel
Hi Jake,Those rumours are ridiculous. Seconding everything Jake said.
I am doing Extension 1 mathematics next year, as I did really well in prelim Ext 1. The thing is, I have heard many rumours from students in the year above me that it is really hard to do Extension 1 by itself (without Ext 2). However they say, it's really easy for people doing Extension 2. I just wanted to know what your take on that is (how true is the statement?) and whether extension 1 is really worth doing as it only counts for a single unit in my case.
Thanks heaps,
Samuel
Thanks very much Jake.You do practice questions for every single topic. Singling out perms and combs doesn't make a difference.
I also have another question regarding Permutations and Combinations. I personally found it the hardest maths topics I have ever done in the sense that, 1. at times I didn't really know what the question was actually asking and 2. there was no way to check my method/ answer.
I just wanted some advice on how to study the topic: Tonnes of Practice? What textbook maybe (Cambridge/ Maths in Focus)? or Maybe just past papers?
Thanks
Samuel
(http://uploads.tapatalk-cdn.com/20161001/ec185216686fdc6cb969583e461bf6a7.jpg)Read the question again. b and a are actually y-coordinates, not x-coordinates
(Or technically I should say they are x-coordinates and not t-coordinates)
They can definitely both be stationary points because SHM is described by a sine curve, which has infintely many stationary points
OHH but when it says "whose distances from the same side of the origin..." isn't the origin referring to the centre of motion?Recall: The origin is x=0 for our purposes of analysing motion
Hi, could someone please explain how to work through this question (both parts) ? I've looked through the solutions and it still doesn't make much sense
hiya folk! This is from the 2012 HSC exam (Question 11). I understand how to do part i of the question, but part ii is like...waht?
I'm mainly confused about what a 'Non-zero constant term' is.
Thanks for the help!
how do we answer iii? (2014 MX1)
Could someone please help me with question 14b)iii) from the 2013 hsc paper?
If you could post the actual question in the future that'd be awesome.
I think the Board of Studies does a good job with this answer, so I've just posted it below. Let me know if you need any clarification on any particular line or working; I feel like it is probably clear, and to be honest I'm not good enough at LaTeX to type out a proper answer.
(http://i.imgur.com/RrSouq1.png)
Jake
No this is fine, thank you!
No problem! I always hated these kinds of questions (still do), so if you find that you struggle with them, don't worry! In my opinion, they are the most difficult maths you do in the 3U course (ie, really hard binom questions) just because of the terminology, logic and guesswork that is required.2013 binomial theorem question was insane though to be fair.
2013 binomial theorem question was insane though to be fair.
Doing it for the MX1 Revision lecture brought back nightmares ;)You just purposely chose to bring back many baaaad memories
Hi, can someone please explain how to do this question?
As an exercise, you may wish to expand (2x^{3}-1/x)^{n} on WolframAlpha. Use n=1,2,3,5,6,7,9,10,11 and you will find there is no constant term (or in other words, the constant term is 0). When n=4,8,12,... however, you'll find you get a constant
Edit: Actually Jake answered this one already. Refer to post 601.
how do you do this guys?Not 100% confident on this one but if the answer is A I can share my working. ::)
sorry i have another question. How do you do part iv, it's one mark but wth :S
On the reference sheet there is two general expansions for binomial theorem. However, if you used the other expansion you get a different answer? I get n = (4/3)k
how do you do this guys?
i) for an inverse to exist, f(x) must be a one to one function (there must be a unique y-value for every x-value). By restricting the domain to one side of the turning point at x =2 (after which y-values will begin to repeat) we turn f(x) into a one-to-one function, thus an inverse exists for values including and on one side of the point (2,0), in this case x >= 2.Love the answer. Comments though.
Not 100% confident on this one but if the answer is A I can share my working. ::)
Yhep the answers A, Howdu get it :OIt is A. I just made a typo in my working.
----
And for part iv, the answer is actually 4-k :S
It is A. I just made a typo in my working.
Oh yeaah thanks mate! ;D
Also how do you do part iv for the second q attached (ans: 4-k)
Love the answer. Comments though.
1. They don't teach interval notation in the HSC. They only use x>=2 to denote [2, inf)
2. There's no definition of 'one-to-one' and 'onto' functions in the HSC either. If a function is one-to-one, it is just called 'invertible'.
No need for bijective functions in the HSC. This is because the HSC defines the range but absolutely ignores the concept of a 'codomain'
Oh yeaah thanks mate! ;DI fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.
Also how do you do part iv for the second q attached (ans: 4-k)
I fixed it up, my bad. Because k < 2 the inverse is actually 2 - sqrt(x), not 2 + sqrt(x) like I had before.
Edit: Rui beat me to it but yolo 8)
wait i still don't get why you take the negative root, even tho k<2Uhh I think that's wrong. You don't redefine the inverse function just like that. >_<
Uhh I think that's wrong. You don't redefine the inverse function just like that. >_<
Because (k-2)^{2} > 0 even though k-2 < 0
Just use my working out.
i understand all your working except; since k-2<0 why do we take the negative branch of the abs(k-2) i.e. 2-k ?
Could someone please help me with question 14b)iii) from the 2012 hsc paper? I have the solutions, i just dont understand it...Please send an image of the question for convenience in the future. Instructions on how to do so may be found here.
Please send an image of the question for convenience in the future. Instructions on how to do so may be found here.
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zpshnjrwhfj.png)
(http://i1164.photobucket.com/albums/q566/Rui_Tong/Capture_zps1aclsfju.png)
Could anyone explain the answer to question 13)d)ii) of the 2012 HSC extension 1 exam.
Is there a set list of proofs that we need to know for circle geo?
Is there a set list of proofs that we need to know for circle geo?You may or may not be surprised, but check the syllabus. Scroll to around page 28.
So much to remember :'( but thank you RuiAce and jakesilove
Hey,
I know this is probably really really simple but i wasn't in class when this was covered so i was just wondering if someone could explain how to find the value of this question to me?
There is most certainly a proof for it but memorising it is well beyond 3U.
ahh okay, that makes sense.
Is there a proof for that identity or is it just something to memorise?
Thank you!
14 past papers in and I currently have a stack of questions that I am stumped on, so i apologise in advance for the spam that will most likely occur.If they're all past HSC papers, look at the answers first in the future. You're more than free to address any confusion with the answers though.
Can someone please explain how to solve this question from the 2008 HSC paper that I have attached below.
If they're all past HSC papers, look at the answers first in the future. You're more than free to address any confusion with the answers though.
________________________________________
________________________________________
When I said stumped I mean not even the worked solutions on BOSTES makes sense to me lol.BOSTES solutions don't always make sense because they just spit it out. Hence the link - solutions written by other sources are usually clearer.
Thank you for the explanation, i definitely needed that! It took me a while to understand the second part but I think i get it now. Thanks again :)
can someone please help me out with part ii...do not like SHM at all
Thanks
Can i Get help with this question please?
Hey! So, the question is basically asking us to find the stationary point (ie. the minimum value), and then the x value for which this will occur. Naturally, we employ calculus for this!
Our expanded equation is
So, we start by finding the derivative
If we set this equal to zero, and multiply out our denominators, we can use the quadratic formula to find solutions. We find that the first value for x is
This is where the maximum sag occurs; sub it back into the original equation to get the maximum sag!
Let me know if I can clarify anything
Jake
thank you for you help
I got confused on how to multiply out the denominators and use the quadratic formula to find solutions. How do you do this? :o
Is it supposed to be like this? ???
ohh thank you!! could you please do this one as well :)
Hey! I would expand the the denominator out, and then use completing the square to get something that looks like a^2-(x-c)^2 in the square root. From there, you can easily let u=x-c so that you can integrate using inverse sin/cos! Show us some working out, and if you still can't get it we'd be happy to write up a solution.
i think i got it! :D thanksLooking good
Hi there,The first two parts are elementary formula work
I was just wondering how did they get the answer for this question?? Could someone please explain? Thanks! :)
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?
If our working is a little dodgy (I.e made some leaps in my head) but the answer is correct can you still get full marks?Yep leaps are fine unless it's one of those show that questions. If it says show that (or prove that etc.) you should only skip the trivial steps.
Hi again,
Could someone please explain these questions? Thanks in advanced! :)
In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?
Answer: 2/11
Hi Neutron,Please quote the post you are referring to. This is definitely not a recent post you are alluding to, and potentially very outdated.
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)
Hey! Okay so i don't seem to be able to find an easy way to do this (like I got an answer but the method was waaay too complicated I reckon but I got -8 degrees and 48 seconds and -24 degrees and 54 seconds??) D: I was wondering whether you guys could help me! thanks :D
Solve the following equation:
2cos2ϴ=1-3sin2ϴ 0≤ϴ≤360
Thank you!
Neutron
Hi again,
Could someone please explain these questions? Thanks in advanced! :)
In a game, two dice are rolled and the score given is the maximum of the two numbers on the uppermost faces. For example, if the dice shows a three and five, the score is a five.
d) Given that one of the dice shows a three, what is the probability of getting a score greater than five?
Answer: 2/11
Hi Neutron,
I think, I found an easy answer for your question, 2cos(2x) = 1-3sin(2x).Before I prove it, I need to point out that cos(x) = 0 is not a solution of the equation since that forces LHS=-2(because cos(2x) =2cos^2(x) -1 ) and RHS = 1(since sin(2x) = 2 sin(x) cos(x))
Solution:
2cos(2x) = 1-3sin(2x)
3(cos(2x) + sin(2x)) = 1+cos(2x) = 2cos^2(x)
3(cos^2(x) -sin^2(x) +2sin(x)cos(x) ) = 2cos^2(x) (since cos(2x) = cos^2(x) -sin^2(x) )
collecting cos^2(x) terms yields
cos^2(x) = sin(x) (3sin(x) - 6cos(x))
1 = (sin(x) / cos(x)) ( 3sin(x) -6cos(x) )/cos(x) (dividing by cos^2(x), it is possible since cos(x) is not zero )
1 = tan(x)(3tan(x) - 6)
3tan^2(x) -6tan(x) -1 = 0
tan(x) = 1+2sqrt(3)/3 or 1-2sqrt(3)/3
From there you can find the angles.I do agree with -8 degree but I don't think -24 degree is correct, I put it into the calculator and it gave me
RHS= 1.33 , LHS =3.23
A general tip to tackle to these type of questions where you get ' sin(x)cos(x)' somewhere when you solve it is to represented in terms of tan(x), (If you get lucky! )
I hope that was helpful. (Y)
And now for your Calculus questions!
So the first one starts with some calculus:
Now in the range given, \(\sin{\theta}\) is always positive (and so is \(\theta\), clearly), so our condition is fulfilled!
Unfortunately for the rest of your questions I don't have the expression! Mind uploading what you have so far? I'll take it from there (or maybe this well help enough) ;D
Hi Jamon,
Thanks for the help so far! This was the first part of the question I forgot to upload. I couldn't get Part D, and I checked some online answers which I still didn't understand (also attached below). Could you please explain what it all means? :'D Thanks heaps!! :)
hi! can i please have some help on finding the latest time in part ii?
Have any tips into showing the working for this question?
2. b) "Given that PQ passes through (0, 6a) find the equation of locus of T"
I worked out the first part which was (a) Find out the point of intersection of T of the tangents (2ap, ap^2) and Q (2aq, aq^2) on the parabola x^2 = 4ay" in which the answer was T ( a(p+q), apq) – found by finding the normal and using simultaneous equations. Would I use T to help solve question 2b?
Hi for this question part ii, how do you solve for the constant B?One way of approaching this question is to work with negative time.
Thanks!
One way of approaching this question is to work with negative time.
Let t=0 be when the object was found, which was at 30^{o}C.
A = 22 because that's the ambient temperature (of the park).
Then using t=0, T=30 we can get B=8.
Then, try to work backwards in time to find when T=37.
Yeah, I got B=8, although the answers say B=15?>They didn't do negative time. They did positive time.
They didn't do negative time. They did positive time.
If we use positive time, initially the object's temperature is 37, not 30.
With this question part (iii), how come you can't solve by mathematical induction? (or can you?)(http://uploads.tapatalk-cdn.com/20161021/f98c2efe31bd67a85090d2e36b052389.jpg)
Hey, this may be a stupid question but when they tell us to simplify in a binomial question what do they mean?
I have attached a photo below:
/(http://uploads.tapatalk-cdn.com/20161021/33d57de474d9d2bd45ceecd44ea19771.jpg)
Hiya, how do you do d (ii) of this question? :P
Hey! For Newton's method, we need to set up an equation such that 0.3 is a zero. Recall that Newton's method is a way of finding x-intercepts! Therefore, we create the equation
You can then use Newton's method in the regular way, starting from t=20s! Let me know if you want an actual worked answer for this
and how would you solve Q6/10?
How does that equation make 0.3 a zero/root?
We want to find where C(t)=0.3, right? Like, for what value of t will C(t)=0.3. If I SHIFT the graph down 0.3, then what used to be 0.3 is now going to be zero. Thus, it will be an x-intercept. But, just logically, if I want C(t) to equal 0.3, then in that case C(t)-0.3=0.3-0.3=0. Does that make sense? Definitely a difficult concept, but once you've seen it once you'll be able to answer any similar question :)
and how would you solve Q6/10?
For 6, you can go straight to your reference sheet. Instead of a theta, you have 2x, so just let 2x equal the general solution for arcsin(a) (again, this is on your formula sheet). Divide through by two, and you're done!Question 10 was addressed in post #721
10 is an interesting one. I can't really think of a better method than guess and check to be honest. Nah, scratch that. We can see clearly that x=3 is a solution to our given equation. However, x=3 is NOT a solution to a) or b), and if it were we would be dividing by zero. So, c) and d) are left. Unfortunately, x=3 is a solution to both of them, so that's not very helpful. x=2 is also a solution to our given equation, but NOT for d), leaving us with c) as the answer :)
Jake
I've got another 3 questions ;D
Question f [attachment 1]:
Question f ii [attachment 2]:
Question 12 d i [attachment 3]:
P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!! :)
I've got another 3 questions ;DNext one addressed in post #710
Question f [attachment 1]:
Question f ii [attachment 2]:
Question 12 d i [attachment 3]:
P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!! :)
Also, for this question. How come the answers don't adjust the domain so that x+pi/6 = 7pi/3?
I've got another 3 questions ;D
Question f [attachment 1]:
Question f ii [attachment 2]:
Question 12 d i [attachment 3]:
P.s sorry for the spam of questions; and thank you for answering all of my questions - much appreciated!!! :)
Hi there!! I've seen multiple answers to this long binomial question and... nope still don't really get it :'D Could you pleaseee explain this question ('cause you guys have really good explanations)! Thanks heaps!! :)Addressed in post #253
Hey! wondering if someone can help me out with part (iv)Your photo's a bit small...
Tried to find the cartesian equation but finding a value for t and substituting that into y and as you can see (I've attached my working), it feels like I'm on a road to nowhere ahaha. So is this the right way to attack this question or am i missing something?
This is usually how I would do these but seems not to be working
Your photo's a bit small...
yeah sorry, it was the only size that fit. Woah that is so much easier...how do we know what method to use?t and 1/t, compared to t^{2} and 1/t^{2}, is immediately suggestive of squaring.
Thanks also :)
Let me lend a hand! So you are correct, T is essential, since it has the coordinates we need. But we want more information to help eliminate our parameters.
Let's try and use the new info we were given. Let's look at the gradient of PQ:
But it passes through (0,6a), so let's get another expression for that:
Equating these:
So this is a new bit of information for us to use to find our locus! Remember, the idea is to tie x and y back in to the T coordinates. Let me show you; start by considering the y-coordinate:
And actually, believe it or not, that's it. We have an equation that defines a set of points that obey the given conditions! Any point on the line \(y=-4a\) will satisfy our requirements, and so THAT is the locus of the Point T :)
Ah, I see – so you find the gradients and then simplify your way through. Although...I'm little confused as how (p^2 – 6)/2p = (p+q)/2 simplified into -2/p = q/2 . But other than that I understand how to do it now. Thanks so much for the explanation ;DI think that should be -3/p there, not -2/p
I think that should be -3/p there, not -2/p
That...makes more sense, actually. Although I'm still not sure how it simplifies into -3/p = q/2...
I did, however, get the answer:
By the messy working out below. Would it be okay to write it like that?
Hi there! Could someone please explain part e)? Would be much appreciated! :)Hints have been offered up by Jamon in post #434. Try those first.
I see. To be honest, I tried figuring out how to unpack Jamon's method but ended up getting frustrated because some figures wouldn't cancel out (cumulated in a page of crossed out working out). It genuinely slipped my mind that you could separate them into separate terms...something that tends to happens a lot. I probably should go do more practice then, huh...All good. If you tend to run into little mistakes/mishaps though, you should make a list of them. And then look at those before you walk into the exam room.
Thanks for the help ^^
Multiple choice question:
Which expression is equal to cos(x) - sin(x)?
A) (root2) cos(x+pi/4)
B) (root2) cos(x-pi/4)
C) (2) cos(x+pi/4)
D) (2) cos(x-pi/4)
And please explain how you can work it out/derive it?
Can anyone help me with this?Addressed in post #372 a while back
Help with part (iii) please..I got an expression in terms of n, n+1, n+2 and (1/2) but can't figure out how to manipulate it into what the answers wantsAlready addressed a while back in post #199
and part iii of this q ~ thanks :)Answered further back in post #184
Thanks for that Rui!!
Another one that makes no sense... I got an answers but it was wrong and I don't quite understand what they did (part ii)
Thanks again
Sure! So let's start by doing what we always do to find tangents to curves, differentiate. I'll call the two graphs \(y_1\) and \(y_2\):
So we'll come back to that later; we also need the point of intersection, and at that point of intersection:
But remember the gradients from above will need to be equal too (common tangent), so we also have \(re^{rx}=\frac{1}{x}\).
What we have here are two equations linking \(r\) and \(x\); we're going to solve these simultaneously! The goal will be to use the result we were just given somewhere too. With that in mind, let's rearrange one of those equations:
Aha, theres the result from Part (i); we can then say that \(rx\approx0.56\)!
Therefore back in \(*\), we have: \(\log_e{x}=e^{0.56}\), and that's an equation you can solve for \(x\), and then substitute back to find \(r\); does that make sense? :)
Heyyo again, How do you do both parts of this question? Thanks! :)
Yes that makes sense thank you! Except one part, how can we assume that rx = 0.56 (I know we found it as an approximate about but howdy you know that it equals rx as there was not mention of that above??)
Hey! First bit:
Clearly this last part is less than zero; what is inside the bracket is always negative as long as \(k\) is a positive integer (which it is) ;D
For your next question, I'll refer you to the HSC solutions and give you a bit of an explanation, since I'd type the same thing:
(http://i.imgur.com/Ns5Y5CT.png)
So we start with the proof for the lowest case and an assumption, as usual. Basically what they do then is consider one additional term on the LHS, and then they tie in the induction assumption where indicated! Remember, if we've assumed that the previous LHS is less than zero, it's kind of like just adding the \(\frac{1}{(k+1)^2}\) term onto both sides; the inequality is maintained. Then some algebra using Part (i) is applied ;D
Sorry, would you be able to elaborate on the second last line? - I don't really understand the application of part i in the induction.
Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha :P)Question 5 is best done by guess and check.
Hiya, how do you solve these 2 questions? (I'm assuming there's a more efficient way of solving Q9 than expanding it all out haha :P)
(http://uploads.tapatalk-cdn.com/20161022/6d003ddda7afa38a4b61670771d69a36.jpg)
I can't solve for k+1
Can someone please help! Thank you!
Hey guys, just got a few questions which I'm not too confident with.
1. |x+1| +x = |x-1|
2. When you are finding the volume of a curve that has been rotated about lets say the x axis, and the question specifically asks to use simpsons rule. Do you just find y^{2}, input the values into the simpsons formula and then multiply it by pi?
3. What is the value of integral of [cos^{-1}xdx]^{a}_{-a} where -1<=a<=1
4. How many times must a die be rolled so that the probability of rolling at least one size is greater than 95%?
Oh thanks, and yeah for the third one it asks to integrate cos^{-1}x dx and evaluate it with the bounds given. Its a multiple choice question where the options are
a) 0
b) a x (pi)/2
c) a x (pi)
d) 2 x a x (pi)
Cheers guys, I'm gonna be spamming this thread with a lot of questions :P
Is there a shorter method to finding the general solution to cos2x = cosx rather than using double angle formula and solving for the quadratic?
Cheers guys, I'm gonna be spamming this thread with a lot of questions :PNot really. Other methods risk losing solutions. Better off just to pull it safe with the quadratic here
Is there a shorter method to finding the general solution to cos2x = cosx rather than using double angle formula and solving for the quadratic?
Potentially, you can sketch the two graphs and see if it's obvious where the intercepts lie. However, I would always suggest using double angle formula etc. etc. for a question like that :)But sketching is tedious pls.
hi for Q5 a) ii) of 1997 HSC Maths ext http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/97MAT3U.PDF
how are you supposed to know that the fastest speed occurs at a=0 if the particle isnt moving in SHM?
Thanks in advance :)
Heyy, how do you solve this graph question? I always get confused with graphs :P
Thanks!
Heyy, how do you solve this graph question? I always get confused with graphs :P
Thanks!
P is the point of intersection at the x-axis.
Therefore C is the correct answer, as it is the second intercept.
Personally, for questions like this, I would just sub values in for t and see which one gives you zero. You'll find that only A and C do, however A would be the first intercept (as it is smaller) making C the second intercept, and our answer. Otherwise, use the general solution on your formula sheet to do this formally
Hey:) Just revising the inverse functions topic - what would cos^-1 (-x) derive to? (its the fact that x is negative that' s got me confused)
This is why I prefer the relationship
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