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Lots of fun cancellations here. The key is that (log(c/b))^2 = (log(c) - log(b))^2.

Thank you for always keeping in the mistakes, we all make them and we feel ashamed when we shouldn't! We are human, even the best of us make silly mistakes. As for the solution, I used log laws to turn all the logs into logs of single unknowns (all logs are log base a): (log(c) - log(b))^2 + 4log(b)log(c) = 0 log(c)^2 - 2log(b)log(c) + 2log(b)^2 + 4log(b)log(c) = 0 (you get a nice cancel here) log(c)^2 + 2log(b)log(c) + 2log(b)^2 = 0 (this is a perfect square) (log(c) + log(b))^2 = 0 so we must have log(c) = -log(b) in other words c=1/b

Your expansion of first term @1.24 is incorrect, it should be x² not 2x, would then have led to quadratic to then use the discriminant

Convert division into subtraction using using law of logs

Hey there! Playing around a bit more with the logarithm laws, I got that (log_a(c/b))^2 = 4log_a(1/b)log_a(c) is equivalent to (log_a(c)-log_a(b))^2 = -4log_a(b)log_a(c). Denote x=log_a(b) and y=log_a(c). Then we have (y-x)^2 = -4yx, which is equivalent to (y+x)^2=0 and so to y+x=0. Substitute back to 0 = log_a(c) + log_a(b) = log_a(bc) and exponentiate to obtain 1=bc, which yields answer D).

Gonzalez Scott Thompson Anthony Martinez Cynthia

Ive never done directional gradients before but I could intuitively see why it is as it is and also why you'd make that mistake (i initially thought it would be done how you did it).

It was much easier to use the first equation to give b= -a²-a, sub this into the other equation leads to 2a^4-a²-1=0. Which is a hidden quadratic that leads to a=1 or -1, which leads to b =0 or -2.

Thumbnail problem is different! I solved that problem and got the limit as 0, but the approach was pretty much the same as it was a nice separable ODE.

How I solved it was I made 1-6 have one complete oscillation So 1-8 would have an oscillation of 1+2(1/6) So 2017=x oscillation and 8=1+1/3 oscillation X=(2017*(1+1/3))/8 X=336+1/3 oscillation And since it completed 336 oscillations we only consider the 1/3 which should be 2 🙂

It took me a couple of tries to get the right answer, because I kept trying to make the initial state have h=0, instead of h=9, which of course doesn't satisfy x^2 + h^2 = 81.

Nice stuff

Hi, thanks for doing video. It was Q4 STEP II 2016. You will kick yourself what the intended solutions were, as you showed if you did not spot two different things the question becomes really hard. The given equation can be written as a quadratic in x by cross multiplying and gathering like terms. Then use the discriminant to get the required answer. This is the same technique used in an earlier question I sent you a while ago about showing that the function did not lie between α and ß. For the next part need to use the harmonic trig identity by changing ycosΘ - sinΘ in the form Rcos(Θ+α), where R becomes √y²+1, and maximum value of the cos term is 1, hence the result. The rest was as you did. Hope viewers found it interesting, something a bit different. Thanks again for doing the question.

If you actually solve y' + xy = y, the answer turns out to be -1. Either way, the DE is separable... so I found y like this: y' + xy = x dy/dx = x - xy dy/dx = x(1 - y) dy/(1-y) = x dx // integrate ln(1 - y) = 1/2 x^2 + C 1 - y = C e^(1/2 x^2) y = 1 - C e^(1/2 x^2) Since y(0) = -1, C=2 Therefore y = 1 - 2 e^(1/2 x^2)

I think there is a mistake. In the specification the ODE is y' + xy = y but when solving the problem you use the ODE y' +xy = x (which of course leads to a different result)

When you are looking for minimum and maximum you have to either: 1) check corner points (1 and 19 in this case) Or 2) Show that the point that you have picked is in fact maximum type of extremum, and that function is continuous

Sorry, but 13² × 7 = 1183 > 1176 = 14² × 6... 😣

"spits out math homework" ask chat gpt 200004

I immediately tried to apply the given relation recursively, which I often try in such problems, and most of the time it does not work out to be useful xD. But in this problem we get a_{n+1} = a_n/a_{n-1}=a_{n-1}/(a_{n-1}a_{n-2})=1/a_{n-2}, and so every sequence item is just the inverse of the item 3 places before in the sequence. This of course implies what you found by inspection: The sequence items repeat after 6 places.

I did it the same way. With 2017 I saw that it's a “complicated” number that you can't immediately see what it's divisible by (2017 is actually a prime number). Because that was too complicated for me ;-) I then came up with 2016 and fortunately you can see straight away that this number is divisible by 2 and 3, i.e. divisible by 6.

In I. you show there exists an open subset of of Z+ so not every subset of Z+ is closed.

I didn't remember the proper methodology for taking a second derivative of a parametric equation, so I did it the "hard" way, and solved for t in terms of x and substituted that into y to get a y as a function of x, then just found the second derivative the old fashioned way. It was messy, but I got the right answer in the end. Once you explained the actual method (which was correct), I solved it again the "easy" way, and spent the rest of the video screaming at the screen about your algebra! Oh my god! (6 t^3 + 6 t^2)/(t + 1) is just 6 t^2 !!!! dx/dt = 2t + 2 dy/dt = 12 t^3 + 12 t^2 y' = dy/dx = 6 t^2 y'/dt = 12 t y'' = y'/dt / dx/dt = 12t / (2t + 2) y''(2) = 4

I got confused because none of the answers matched the answer I got. Thats because answer E is written wrong. There is no 4x present in the expression. Thank you for your videos.

GRE questions are normally way above my head. But in this one I think using what I know it was OK. Using parametric differentation dy/dx is what you had, but factor out 12t² in numerator you end up with 6t². All I know, not really knowing how it so, what you proved, is the second derivative is found by differentating dy/dx and dividing by dx/dt, that leads to 6t/t+1. If the point is at x=8, setting to t² +2t =8 leads to t=2 (other solution is t= -4, not valid as t>o). Put that in to second derivative gives 4. I love watching these videos where you literally solve everything from your mathematical knowledge and prowess, things most probably, including me, just say "well that is what I have been taught to do", not knowing why and how it was derived.

The question was Q9 STEP 2 2016 if anybody interested and looking up the mark scheme. I loved your absolute from first principles approach to the question using calculus. Purity of mathematical knowledge. Most students sitting the exam would have had the benefit of having done mechanics for the previous 2 years and not have to work out the formulas from first priciples, but just know them, such as SUVAT, work done, conservation of momentum and kinetic energy. Your answer for first part is correct, but could have used loss of kinetic energy of bullet (1/2mu²) equals the work done against resistance by block (Ra). Leads to same answer you got for a. But your answers to the second part are not completely correct. In the second part the mark scheme suggested using conservation of momentum as you mentioned in video but did not use, giving velocity v of bullet/block system as v=mu/(m+M). Then using initial KE of bullet and final KE energy bullet/block system and setting difference equal to work done by reistance force inside the block, leads to the answers. Which are b=aM/(m+M) and c= amM/(m+M)². STEP question love using letters like a for distance to discombobulate you when used to having a as acceleration, s is distance etc. Thanks again for doing question, I thought it was really interesting.

Hi, the intended method is the mark scheme is even simpler. As you said this is a hidden quadratic, let y=cos²x, you end up with 9y² - 12y +7, the complete square, 9(y-2/3)²+3. Minimum is 3, when y=2/3 like you had.

Your approach is so elegant! I went through the process of finding all the inflection points of f(x) by setting f'(x) = 0, which came up with cos2(x) = 0, cos2(x) = 1, or cos2(x) = 2/3, and then I had to plug them all back in to find which one was the minimum.

Thank you for the video. Why didn't you check whether the first derivative was a maxima or minima?

Bro u need to do a whole MAT paper id love it

really like how you included your restart as well; it's making me feel less bad when I have to restart from the beginning 😅

I solved it without using the cross product. First I set up the equation of the straight line L that results from the intersection of the two given planes. This means solving the system of equations resulting from the two plane equations.This results in the following equation for the straight line L: (0,-1,4) + t*(1,0,-1). So that means, a plane that is perpendicular to this straight line L must have the coefficients a=1, b=0, c=-1 and this is option A).

Nice problem

Nice problem

Hi, I saw this video last night, I did it the same way you did. Thinking about it again today, I solved it another way, but it is not any quicker at all, and maybe uses more unfamiliar concepts than the cosine rule. Just more coordinate geometry. If you set the ▲ABC up with A at the origin, then B has coordinates (6√2,0) and C is (0,6√2). If you draw horizontal lines from points D and E to the line AB, say intersecting at E and G respectively, you end up with a series of similar ▲'s. You can get work out the base lengths and heights of these triangles using ratios. That allows you to work out the coordinates of points D and E and hence the gradients of the lines AD and AE, which end up as 3 and 1/2. Finally work out angle between the lines using the formula. As I said no quicker at all but just a bit more interesting!! Why do something in a quick and straightforward way in maths when a long convuluted method also works😃😃

The answer is correct. You can proof this by using an equivalent characterization of projections: Let P be a projection (P°P = P) then it holds that V is the direct sum of the image of P (imP) and the kernel of P (kerP) with P restricted to imP being the identity on imP. KerP is the eigenspace to the eigenvalue 0 (follows immediately from the definition) while imP is the eigenspace to the eigenvalue 1 (since P restricted on imP is the identity). Since V is the direct sum of imP and kerP it's a direct sum of eigenspaces of P and thus P is diagonalizable.

Good solution! Just two little things i noticed: In your answer to b), n=3^2 is already sufficient, no need to go for n=3^6. And in your answer to e), your example for x(n)=7, 57, of course works, if you take the correct factorization 57=3*19, instead of 6*19. Anyways, in both cases you came up with other valid examples, so no big deal.

Hi, loved your solution using vectors. I saw this question a few years ago and eventually worked out it was essentially the line ax+by-c=0 has to be a tangent to the circle with radius 1 centred on the origin. I could not proceed thou. Most elegant solution I saw online was finding the distance from the point (0,0) to the line ax+by-c=0 using the distance from a point to a line formula, d=|a(0)+b(0)-c|/√(a²+b²), and setting this equal to 1.

This is asking about the dot product of the vector u = (a,b) with a unit vector v = (x,y), which is maximized when u and v are colinear, in which case you get the norm of u.

well done. Maybe for those who couldn't solve it geometrically, like in this video: You can also get the result by differentiating the function ax+by with the constraint that x² + y² = 1. Then you have to calculate a little, but in the end you also get the result.

Hi, my very basic way to try and solve was using the fact that the sum of exterior angles is 360°. If you had 4 acute angles in a polygon, that means 4 exterior angles > 90°, but that adds up to more than 360°, hence maximum number of acute angles in any convex polygon is 3. Is this not correct?

Hi, sorry question caused you pain. I have sent you an email with intended solution. Respectfully I think you went down the wrong route using vectors rather than parametric differentiation, then finding the equation of tangents at P and at Q, if let t=p at point P and t=q at point Q, find dy/dx, find equations of tangents at these points, use given fact that pq=-1, find coordinates of intersection, prove this point lies on the given curve. Sorry again for question, intended solution was not the algebra tedium you had to endure.

The question itself was OK once worked out the limits of integration. Working with the exponents was easier when you know both functions are even, so simply going from 0 to upper limit and doubling. I always kick myself for those simple errors that mess you up.

Is the matrix (0, 1; 0, 1) a counterexample to being diagonalizable?

In option D), you can take a single-element set for the set S, for example, say S = {3}. Then the set as described in option D) corresponds to the set R\{3}. Since {3} is a closed set, its complement is open and this complement is exactly R\{3}.

the option D is really close to the definition of the closure, the only difference being the W intersect S = 0 instead of W intersect S != 0

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: