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Genuinely, hats off to you for ploughing on with the question. You are at a disadvantage with these questions being advanced and specialised in your field sometimes, as oppposed to, as you mention at the start of video knowing some basic high school equations, that students sitting this exam will have known. The standard projectile SUVAT formulas and equation for the trajectory of a projectile rather than deriving them from first principles like you did. Then later on they will have known the various formulas for cos2x etc. This is classic STEP mechanics question, fairly basic mechanics, then proceeds to include calculus, trig and some messy algebra. Hope it does not put you off doing other mechanics questions!! Thanks again.

I submitted it, from exam papers called STEP maths (the entrance exam taken for Cambridge Uiniversity Maths degree), this is taken from STEP Maths 1 2016 Q11

Do you know where they got this question from? love ur vids btw

loved that question, took me ages to get the diagram correct thou as you had it @4.55. After that I found the height of lines from centre to the horizontal line of the spheres, as sqrt(r^2-1) and sqrt (r^2-4), then created another right angled triangle which has height as the difference of the above, base 7 and hypotenuse 2r. Then solved

Hey ! I think you have the right answer, not sure though as i don't have a correction... You probably already know it but the Gk = {exp((2*i*n*pi)/k) | n in [0, k-1]} are pretty common groups called the k-th roots of the unity as they satisfy the equation z^k = 1 and are usually denoted Uk, k being a non-zero natural number, here in France. I have the confirmation that these groups are a part of the solution. For the infinite groups, i agree that U (the unity circle or trigonometric circle) is a compact subgroup though i don't have a justification for it being the only one. Your argument with the rational / irrationnal numbers might be enough as if you have a rationnal number in a subgroup, you will end up generating one of the Uk and if you have an irrationnal in the subgroup, you will end up with the whole circle, therefore showing that you cannot generate something else than the Uk or U (as the irrationnal and rationnal numbers form a partition of the real numbers) I also want to apologize for taking so long to answer i just really wanted to find an answer to the infinite group problem but ended up finding none, so yeah, i'm really sorry D:

6/14 = 1/7? I'm not entirely sure that's correct. 😛

Another way to see this is that Ax=x is the same thing as (A-I)x=0. But A-I is going to have a zero on the first diagonal element (and indeed the first column is all zeros) so it's not invertible (you can also see this as the determinant, here the product of the diagonals, will obviously be zero). Thus the statement that "(A-I)x=0 implies x=0" must be false, since that's the same as saying A-I is invertible.

Option B) can also be falsified directly: In a triangular matrix, the eigenvalues are directly on the main diagonal. So, among other things, 1 is an eigenvalue. This means that for A*x = 1*x, x is an eigenvector for the eigenvalue 1. And since an eigenvector is not the zero vector, it cannot follow from A*x = 1*x that x = 0. More precisely: A*x = 1*x COULD theoretically mean that x = 0, but since 1 is an eigenvalue, there must exist an vector, that is not the zero vector, and fulfills A*x = 1*x, so someone cannot conclude from A*x = x, that x must necessary be the zero vector.

The Oxford exam assumes no knowledge of complex numbers, even thou most people sitting it know them. Your solution at the end was the intended answer. In the most most basic sense of factor theorem if (x²+1) is a factor it means x^2= -1 leads to the function being zero, put that into the function, you end up with 4ⁿ - (-4)ⁿ needing to be zero, hence n needs to be even.

ooh this is a great question, it entirely threw me at first, but it really kind of tests you for your understanding of why we do what we do when we find roots of a polynomial. If something factors something else then that means when that thing that factors out is 0, the whole thing is 0. So if x^2+1=0, and it factors the equation, then the equation should equal 0 when x^2=-1. So just sub in x^2=-1 and equate it to 0. (3+1)^n-(-1+3)^n(-1-1)^n=0 4^n-(2)^n(-2)^n=0 4^n-(-1)^n4^n=0 So for this to hold n must be even.

The answer B ( even n) is right, and I just found a proof that I really like! In fact tpwards the end you kinda got the idea, as you can think about as working modulo (x^2+1), which is what you did by setting x^2 = -1. So, for the polynomial on the left, x^4 + 3, euclidean division will show you that this is equal to (x^2+1)(x^2-1) + 4, which is 4 modulo (x^2+1), and with the power of n this becomes 4^n. for the polynomial on the right, (x^2+3)(x^2-1), add and subtract 2 to each term (or again just do euclidean disision), and you will find that it is equal to (x^2+1)-4, which is just -4 modulo x^2+1, or again (-4)^n when we include the n. Thus, the whole thing is 4^n - (-4)^n modulo x^2+1, which is 0 if and only if n is even!

But I don't understand the relationship between the fact that we are searching for this number and that it is a root. (Yes, I am a bit stupid).

Solution that is a little less casework. There is symmetry between the mono increasing and mono decreases cases, so just consider the mono increasing case Clearly, the sequence 123456 is somewhat special, so we can just get that out of the way. Consider removing 1 from a sequence. The sequence after must be 23456. If the 1 were first, we would be back to our special sequence. If 1 were second, we would be double counting since we could also remove the 2. If 1 were anywhere else (4 spots left), it would work and be unique (no double counting). The same logic follows when removing 6 from a sequence. So, in total, for sequences that we remove a 1 or a 6 from there are 8 unique sequences and 2 double-counted sequences. Consider removing 2 from a sequence. The sequence after must be 13456. If the 2 were second, we would be back to our special sequence. If 2 were first or third, we would be double counting since we could also remove the 1 or 3, respectively. If 2 were anywhere else (3 spots left), it would work and be unique (no double counting). The same logic follows when removing either 3, 4, or 5 from a sequence. So, in total, for sequences that we remove a 2, 3, 4, or 5 from there are 12 unique sequences and 8 double-counted sequences. That is all the cases. Summing up, we get 1 + 8 + 2/2 + 12 + 8/2 = 26 By symmetry, total answer for mono increasing and decreasing would be 52.

Great problem It's not just that the longest cycle for a^n mod 10 is 4, but that every cycle length is a factor of 4. You can repeat a 2 cycle twice to get a 4 cycle. If there was a 3 cycle you would have to separately show that b^c^d ≡ b^c mod 3 (which it also happens to be).

Hi, all your answers were correct.If anyone is interested the question was taken from a UK maths entrance exam for The University of Cambridge. The papers are called STEP papers, STEP 1, STEP 2 and STEP 3, each paper getting a bit harder i.e STEP 3 is the most challenging. This was Q2 from STEP 1 2016. The format of the exam is you are given 3 hours per paper to solve about 5-6 questions from each paper. I did actually mange to solve this one, but took me about 45 minutes. After the differentation the simplification you did is really important so you can easily spot the values of the constants needed. Thanks for doing, thought people would find it interesting finding integrals from a differentiated function, not seen that before.

I had no idea how to approach this at first, but as soon as you wrote down N = 2,020 + 10,000 k, it became clear and I was able to find the answer. Nice one!

That was such a really clever solution. I'm sure most people would have found centre of circle (-a,-b) and the radius, r^2=(a^2+b^2+c). Distance from origin to centre, d^2=a^2+b^2, so r^2>d^2, leads to c>0.

5:10 ah, my favourite prime test "sounds prime" haha! I have to admit that's exactly what I was thinking as you said it. I love that little simplification at the end where you recognised the 3 sets of the same digits! It makes sense given its multiplied by 101, but I am pretty sure I'd have entirely missed that.

I tripped myself up on this one for a second when I wrote 2(1-x)^2 = 2x^2 on my paper for some reason instead of 2(1-x)^2 = x^2 before I caught it. Still not sure where that extra coefficient in my head came from the first time.

Nice I have been hooked up with the channel From what you did just now I was able to understand a bit But could you please elaborate a bit more on the determinant formula? Thank you

hey ! i got a question that absolutely stumped me during an oral exam and i'd like to see you give it a shot. here's the question : Find all compact subgroups of (C*, x)

This is a very simple application of the fact that I = int [ f ( x ) , a to b satisfies (b- a) * Min (f(x) ) < I < max (f (x ) .

Man, for each k=1,2,..., you have log(k) < int_{k}^{k+1} log(x) dx < log(k+1) For some n, add all the inequalities for k=1,2,...,n-1, sum log(k) < int_{k=1}^{n} log(x) < sum log(k+1) log((n-1)!) < (log(x^x)-x)|_{1}^{n} < log(n!) log((n-1)!) < log(n^n)-n+1 < log(n!) Now you take the power of e, (n-1)! < n^ne^{1-n} < n! In other words, instead of opening each integral, taking the e power of each inequality, multiplying everything and using telescoping reasoning for the product, it is better to add all inequalities, so the sum of integrals is the total integral, take the e power and it is done. That's clean. And intuitive. And if I actually did this test, I would leave the following message for the person that made this question: "Really? "n^ne^{1-n}"? (n/e)^ne is way better!"

I really wanted to use modular arithmetic here, and I got close with considering both representations mod 4, this collapses the base 8 representation to just a, and it works nicely for the base 11 given that 11=-1 mod 4, and so you can eliminate a by setting the two equations equal. But there isn't an easy way to eliminate another variable modulo another number as far as I can see. You get some nice relationships between a, b, and c that can give you some good hints to start, but I always had to result to just start guessing numbers to see what worked. I was kind of hoping to get the numbers algebraically directly.

BTW that was a question from the University of Cambridge entrance exam for maths in 1957. Brutal question for 18 year olds!!

Thank you so much for doing that question, I thought it was really challenging. I got to the last inequality you got to, but tried multiplying by the expression by (n-1)^(n-1) but did not help. Thank you again.

The ones I can solve in my head are my favorites :D (There aren't a lot of them) Why didn't you just do "the sum from 1 to 14" instead of "the sum from 1 to 15, then subtract 15"? ;)

Nice question. I changed log 4 (x) into 1/2 log 2 (x) and log 8 (2x) into 1/3 log 2 (2x). Then found r by using 1/2 log 2 (x)*r=log 2(x), so r is 2. Using this in first two terms, 2*1/3 log 2 (2x)=1/2 log 2 (x), 1+log 2 (x) = 3/4 log 2 (x), leads to x.

Thanks for doing question, thought it was just a nice little question. Quite a straightforward question that looks a bit intimidating at the start, but plugging in values of n=0,1,2.. you can quite quickly see the pattern and the connection to the sum of a geometric sequence.

Before you try to include the answers that is within the interval, take the equation and solve for x as if you were to find all of the solutions. Here, we know that 2cos(2x)+2=nπ. Now, 2cos(2x)=nπ-2, so cos(2x)=(nπ-2)/2. Convert cos(2x) to either of the two forms. I'll go with 2(cos(x))^2-1=(nπ-2)/2. Add to 1 both sides and then divide by 2. What we have there is (cos(x))^2=nπ/4, so cos(x)=±√(nπ)/2, so x=2kπ±arccos(±√(nπ)/2) where k and n are integers. Looking at the interval there, if you plug in some integer values for both k and n, we only see that there are 2 solutions within that interval, which is choice A.

The answer is correct, but somewhere in the middle of the step is incorrect because the derivative of log(x) is 1/(xln(10)). From there, if you differentiate both sides of the equation, you get 1/(xln(10))=4cx^3, so 4cx^4ln(10)=1. Now substitute cx^4=log(x), which is 4log(x)ln(10)=1. Note that log(x)=ln(x)/ln(10). Basically, 4ln(x)=1, so x=e^(1/4). If you raise both sides by 4, we have x^4=e. Putting back in the equation that was differentiated, we 4ce=1. Solving for c gives us c=1/(4e). Hence, the answer must be choice A.

Derivative if log(x) is not 1/x. Graphing tech confirms that the given solution does not work for this question. If you meant ln(x) = cx^4, then it does work. The way it is wtitten, the correct answer is c = log(e)/(4e)

I numerically computed the integral for 3 different values of n to be safe and subsitituted those values in the options to find which matched lol

bro i dont even know what im looking at. this could be sanscript for all i know

Thanks for doing the question. I thought it was an interesting one. I struggled a lot with this question. I also set k's equal to each other, got nowhere. I then focused on derivatives as it said met tangentially, that did not help me either, because unlike you who realised that meant dy/dx =1 I didn't. But I think the key to the question was supposed to be the fact that they are inverse functions, hence reflected on the line y=x as you said. But if they meet, they must meet on the line y=x. Hence replace the y with x in the second equation, end up with a quadratic in x. If only one solution the discriminant is zero, leads to the answer. As a previous viewer said. Hope you found it interesting.

For (D) can we not just say 2^5 = 32 so log2 (30) < 5 and 3^4 = 81 so log3(85) > 4 so smaller/bigger means overall it's definitely less than 5/4. For (C) you forgot the squared on the 6!, but with all the cancelling what you get is comparing 7/4 to 7/(very big)

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Differentiating does not always work

you missed that the 6! was squared in option C, although it didn't matter because squaring it would make the number 6! times smaller

I always have a lot of fun with problems like this. Here was my approach: I started by comparing the first two (since 5/4 is nice and easy)... sqrt(7)/2 is sqrt(28)/4. sqrt(28) is a little bit more than 5, so (A) must be more than (B). Looking at (C), I canceled a bunch of terms and got sqrt(7)/2160, which is a VERY small number, so obviously that's not it. For (D), I observed that log2(30) is slightly less than 5, and log3(85) is slightly more than 4, which means it is less than 5/4. (E) was the trickiest, but I wanted to compare it to (A)... so I set up the question sqrt(7)/2 <?> (1 + sqrt(6))/3. I cross multiplied to get 3 sqrt(7) <?> 2 + 2 sqrt(6). Then I moved the coefficients inside the square roots to get sqrt(63) <?> 2 + sqrt(24). The left side is slightly less than 8 (but greater than 7), the right side is 2 + slightly less than 5. So that means the left side is greater. So the answer is (A).

yea in the one which has square root of 10!, you forgot to write the squared, which was already given in the denominator, so squaring it again would make it ^4

What program are you using?

What a nice problem! Really puts into question if the examinee understands the meaning of a derivative!

Please mention the base of the log in the question

I did this in a slightly different (worse?) way. I solved for x using the Lambert W function and got x = e^(-W(-4 c)/4). The W function isn't defined for arguments < -1/e. It is equal to -1 at -1/e, has two values between -1/e and 0, and has a single value when the argument is greater than zero. Since all of the answer options are positive, the argument to W will be negative - and so it must be -1/e, the only negative argument that yields a single real value, and therefore c must be 1/(4e).

Why does anyone need a 4 min video for this when it's a multiple choice and you can easily visualize it in your head 🤔

Here's my guess, with non-elementary functions. a^b=b^a b ln(a)=a ln(b) b ln(1/a)=a ln(1/b) 1/a ln(1/a)=1/b ln(1/b) 1/a e^(1/a)=1/b e^(1/b) W(1/a e^(1/a))=W(1/b e^(1/b)) If we take the W0 branch, I think that'll return 1/a=1/b. Then I was thinking there's a second real branch, W-1, but from the graph of x=ye^y it looks like only negative x values could have two real values for different branches. My plan was to take W-1 of one side and W0 of the other, then we could say, for example, b=1/W-1(1/a e^(1/a)), but not only am I not sure that's a valid multivalued function operation, I also think that will enter the complex plane.

I did this by finding the isomers of pentane (from chemistry) which I found to be a weird parallel between trees and hydrocarbons. It's doing exactly the same thing but because it's a common chemistry question, it was easier for me to do it that way.

This actually counts the number of trees from the Cayley's theorem that are non-isomorphic. I don't think there is a specific formula but here is a numerical approach. A tree with 5 vertices has 4 edges and hence a degree of 8. Tree’s are connected. Therefore each vertex has at least a degree of 1, adding up to 5. We are left with 8 - 5 = 3 vertices to distribute. Write as a + b + c = 3 (in the interval [0, 3]) It can be seen that the only solutions are (1, 1, 1), (2, 1, 0), (3, 0, 0).

Loved your solution. I did it the other way you said, centre of circle is (-a,-b) and r^2.=a^2+b^2+c Said distance from centre to point (1,1) had to be less than radius, led to same answer.