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 ❤❤
@ruslan8820 thanks, but I later realised that I took limit on the top part wrong, the reason my answer was wrong, It should have been 0 to x and the integral should be ln|1+k|dk
I know. Those peope who say it is easy are saying that because their low self-esteem. They could not solve this question and were so mad at themselves. So, that's the reason why they said it was easy because they could not solve it and hurt their ego
@@MD-kv9zo He factors out n/1 from the first factor (x+n), which gives n/1 (1/n x + 1), then factors out n/2 from the second factor (x + n/2), which gives n/2 (2/n x + 1), and so on. In the end, this gives the product n/1 times n/2 times ... times (1/n x + 1) times (2/n x + 1) times ... And n/1 times n/2 times ... can be written as n^n / product of all k, with k going from 1 to n. And (1/n x + 1) times (2/n x + 1) times ... can be written es the product of all (k/n x + 1), with k going from 1 to n. In the denominator, he uses essentially the same steps.
@@bjornfeuerbacher5514 I got a bit confused at what was dividing what. Like (1/nx+1), does that mean ((1/(nx)) +) or ((1/n)x + 1), but I think I get it now.
This is an exam given by usual high school students... And also there r many more questions like this in paper which the student has to solve within time bound of 4-5 minutes which makes it the hardest ever
@@SamarthPatil-my5mh yes I know, I'm preparing for JEE advanced myself lol. Don't keep the exam on a pedestal, there really are harder high school exams, but they don't boast about it. Heard of gaokao? Equally hard science questions and then you got Shakespearean literature (figuratively speaking)
@ I am surprised how there are so many ppl who dont know how to read. Title is "hardest" question from Jee Advanced maths, not the hardest "exam." Gosh. This question is one of the hardest questions in Jee advanced certainly. Whats wrong with you people
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 ❤❤
good work
@ruslan8820 thanks, but I later realised that I took limit on the top part wrong, the reason my answer was wrong,
It should have been 0 to x and the integral should be ln|1+k|dk
@@lazyobject5797 yeay , I agree , gj
That was a tough question, but your clear and thorough explanation made it easy to grasp. Fantastic work, sir!
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🎉
Fantastic video professor
This is just the perfect video
Very nice!
Nice one by perfect professor
Nice and great. love this video a lot
Hey in which uni/college do u work in
Btw well explained
To the people who are saying its "easy" try doing this in 2 mins + relatively same level of physics and chemistry
I know. Those peope who say it is easy are saying that because their low self-esteem. They could not solve this question and were so mad at themselves. So, that's the reason why they said it was easy because they could not solve it and hurt their ego
This one was indeed tough. Anyone who said this was easy without showing their work is a liar
I got lost at the 3rd step.
Do you mean the step starting at 2:30?
@@bjornfeuerbacher5514 1:43
@@MD-kv9zo He factors out n/1 from the first factor (x+n), which gives n/1 (1/n x + 1), then factors out n/2 from the second factor (x + n/2), which gives n/2 (2/n x + 1), and so on.
In the end, this gives the product n/1 times n/2 times ... times (1/n x + 1) times (2/n x + 1) times ...
And n/1 times n/2 times ... can be written as n^n / product of all k, with k going from 1 to n. And (1/n x + 1) times (2/n x + 1) times ... can be written es the product of all (k/n x + 1), with k going from 1 to n.
In the denominator, he uses essentially the same steps.
@@bjornfeuerbacher5514 I got a bit confused at what was dividing what. Like (1/nx+1), does that mean ((1/(nx)) +) or ((1/n)x + 1), but I think I get it now.
@@MD-kv9zo Sorry, I should have written it clearer... I meant ( (1/n) x + 1) etc.
After watching ur usual videos, this does NOT match up to the level of "The hardest"
This question is not as hard as the other questions Dr PK usually solves?
This is an exam given by usual high school students... And also there r many more questions like this in paper which the student has to solve within time bound of 4-5 minutes which makes it the hardest ever
@@SamarthPatil-my5mh even so its not the "hardest", even for high schools, there much much more harder exams
@@SamarthPatil-my5mh yes I know, I'm preparing for JEE advanced myself lol. Don't keep the exam on a pedestal, there really are harder high school exams, but they don't boast about it. Heard of gaokao? Equally hard science questions and then you got Shakespearean literature (figuratively speaking)
@ I am surprised how there are so many ppl who dont know how to read. Title is "hardest" question from Jee Advanced maths, not the hardest "exam." Gosh. This question is one of the hardest questions in Jee advanced certainly. Whats wrong with you people
大a