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Time for gaokao proffecor

Sir simply We can use ramanujan formula ∞ ∫((dx)/(xⁿ+1))=(π/n)/(sin(π/n)) 0

You are the king of integral professor

91181 by picking top and bottom sticks of the second digit and putting the first stick with the first digit to make it 9, and add the second stick after the last digit, this will make it 91181

This is such a nice video

大a

Great work as always professor🎉

You can also solve it by working backwards

Hello! I'd like to ask you to solve the question 3 of this paper (Suneung test - South Korea - I don't know what year). I don't know why, but RUclips removes my comment whenever I try to link the document, so I will type the problem: "There are 2 students from each of these countries: Korea, China and Japan. These 6 students will each randomly choose to sit in one out of the six seats numbered as in the diagram below. What is the probability that 2 students from the same country sit such that the difference between their seat numbers will be either 1 or 10" DIAGRAM (2x3 table): 11 ------ 12 ------- 13 21 ------ 22 ------- 23 Alternatives: a) 1/20 b) 1/10 c) 3/20 d) 1/5 e) 1/4 First off, I thought the exercise was asking about the cases where AT LEAST two students were sitting in such a way that the condition was satisfied, but I got to 13/15 (through combinatorics), which isn't even an option in the alternatives. Then, I shifted into thinking that maybe I should calculate only the cases where ALL OF THE 6 students were sitting in compliance with the condition and I finally arrived at 1/5 (letter D). Nonetheless, I couldn't find the official answer. So, I'd highly appreciate some help! Thank you!

To the people who are saying its "easy" try doing this in 2 mins + relatively same level of physics and chemistry

Hey there, I am from India, I don't know about you, but that's how I did it, (I am just a high-school student so if I did something wrong forgive me) ●divided top multiplication by n^n and multiplied by n! ●multiplied bottom multiplication by (n!)^2 and divided by n^2n ●then got that multiplication you got ~reimann uhh multiplication ●raised top and bottom term separately to the power of x/n like you write (a/b)^n =a^n/b^n, then raised the top and bottom half separately to the power of e by taking natural log on them ●next by the property of log and taking reimann sum, the top expression simplified to e^({integral}(1 to x) [ln(k)dk), here I figured that x is any contant integer, so I took x/n as my dk ●repeating above the bottom expression simplified to e^({integral}(0 to x) [ln(1+k^2)]dk), here also I took x/n as my dk as x is just any constant (but infinity as it would cause problems) ●The final expression becomes (e^(xln(x)-x+1))/(e^(xln(1+x^2)+2tan^-1(x)-2x)) ●I analysed the function for which I concluded that it's a decreasing function ●Further analysis showed me that the derivative is a increasing function ●So I got option a,c and d as correct options Please correct me if I got anything wrong 😅😅 Love from India ❤❤

Graphing the given equation will give you a line m+n-33=0 and a point (-33, -33). This is a very nice problem!

This one was indeed tough. Anyone who said this was easy without showing their work is a liar

You are the best integration master in RUclips. I know people on X know you are the best

This is so beautifully done professor, you are the best

The method you used to solve the first one was time efficient

Very super sir Still you are uploading such nice integrals It can be highly useful for me And provide more your techniques I wanted to learn!!👍🏻

Gorgeous

Very nice!

I am innnnn loooove with your integrals, and your integration skills

Nice one by perfect professor

What a brilliant video

This is just the perfect video

Brilliant! I solved it very differently: First i set a=33 for convenience and so the equation became: A^3 = m^3+n^3+3amn. I substituted a = m+n+delta on both sides, so i got: (m+n+delta)^3 = m^3 + n^3 + 3mn(m+n+delta). Opening the brackets and subtracting both sides by the lhs and rhs of the original equation, all terms left had delta in them so delta=0 derives the set of m+n=33 solutions. Dividing by delta we're left with a quadratic equation whose determinant is -3(m-n)^2, which eliminates all solutions where m is not equal to n. Now substituting m=n in our original equation, we are left with: A^3=2n^3+3an. Now using the same trick: setting this time a=delta-n, simplifying we get: Delta^2*(delta-3n) = 0. So either delta is 0 which implies a=-n=-m which gives the last solution, or a=2n=2m which means m+n=a which is already covered in previous solutions. Overall, for a general non zero value of a: Either m=n=-a - one solution, or m+n=a, which implies all solutions (i, a-i) where min(a,0)<=i<=max(a,0) - a+1 solutions. A+2 solutions total. My choice of delta in each step was such that delta=0 was a relatively easy to see solution.

You are the best professor

Or just use from the start one of euler's identity a^3+b^3+c^3-3abc= {(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]}/2 For a=m b=n c=-33

Fantastic video professor

What a brilliant and nice video professor

you have a talent to explain complex sums so easily…gives me a lot of motiv:)

Can you do ploynomial divison after you find the factor m+n-33 or no?

you help me get a motivation for learning for my a-levels that are next year, im 13 and i love ur videos

I got lost at the 3rd step.

Why not the other arguments for each that differ by 2pi?

Nice and great. love this video a lot

Fantastic video professor. I know this hard limit question this is a famous one. But no one solved it like you. You are the best🎉

That was a tough question, but your clear and thorough explanation made it easy to grasp. Fantastic work, sir!

After watching ur usual videos, this does NOT match up to the level of "The hardest"

Hey in which uni/college do u work in Btw well explained

You are the best professor

What a fantastic video

Sooooo handsome

This is really a nice video

Another nice video professor🎉

What a Clever method

Make a series preparing for korean sat calculus questions

What I would have done is used the fact that sin(x) = Im(e^ix) to deduce that I=integral from -inf to inf of sin(x)/(x^2-2x+5) wrt x= Im(J) where J= integral from -inf to inf of e^ix/(x^2-2x+5) wrt x, this way I only have to deal with one exponential

Contour method is really cool, and smart in my opinion. But I wonder what are its limits.

Lol=8

X, 2x-5=9

This is obviously very well-explained video