Amazing Solution of Difficult JEE Main Problem Based on Integration
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- เผยแพร่เมื่อ 29 ก.ย. 2024
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In this video we will be solving a difficult integration problem from JEE Main 2022
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#bhannatmaths #jeepyq #integration #jee2024 #jee2023
Being a JEE aspirant ,in Aman Sir's comment's section lot of students talk about concepts which we have not even heard, like Lambert W function etc.
If u have any source or information about it, can u share it. I would like to learn too
Inverse of x*e^x@@vibhashrivastava8846
@@vibhashrivastava8846 inverse of x*e^x
@@vibhashrivastava8846 i talked about it once. here is a small explanation
set f(x)=xe^x
then f^-1(x)=W(x)
so basically foW(x)=Wof(x)=x
thus w(x)e^w(x)=x and W(xe^x)=x
it is used to solve equations where the variable is exponentiated as well linear. like 2^a + a=5
inf
series expansion is W(x)= ∑(-n)^(n-1)x^n/n!
n=1
Yeah
The Way you teach Maths, wish you 1M Subs Fam soon!! Sir
Sir yaha pe sirf 3 inetegration karna hai
Sqrt(2x)
Sqrt(2x-x²)
Sqrt(1-y²) and
y²/2
Fir inki values ko har jagah daal dena hai.
And not 7 integrals to be solved
amazing method sir ❤❤
Why u use only half curve can anyone explain
Sir kya aap IIT jam ki preparation krate hai
Sir linear /linear ka graph ma ek video la aya.A sab student ka biniti ha please 🙏😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭
Bhai asan hota hai Aod padhoge to khud banaloge
Bhai simply itna yaad rakho range uska all real except coff of x in Nr / coff of x in Dr
Tough question
I am a 31 year old doctor who has not solved mathematics since 2008.
But this is putting fun in my otherwise boring professional life!
This can also be done with simple integration by observing that the messiest one to integrate is sqrt(1-y^2) which luckily is present in all options and in LHS and RHS. It is the same trick that led you to get a simple solution using geometry! Had that part not been common it would have been messy using geometry or otherwise!
Left hand side:
First part:
integral(sqrt(2x)) dx|0 to 2 = 2sqrt(2)/3*x(3/2)|0 to 2 = 8/3.
Second one:
integral (sqrt(2x-x^2)) dx|0 to 2 = integral(sqrt(1-(x-1)^2)) dx|0 to 2 = integral(sqrt(1-y^2))dy|-1 to 1
We notice that the same integral is present on right side with limits 0 to 1. Add it to both sides makes LHS:
8/3 - integral(sqrt(1-y^2))dy|-1 to 0
= 8/3 - integral(sqrt(1-y^2)) dy | 0 to 1 (y = -x substitution followed by flipping limits and substitute x = y)
Right hand side becomes:
integral(1)dy|0 to 1 + integral(2)dy|1 to 2 - integral(y^2/2)dy|0 to 2 + I
= 1 + 2 - 4/3 + I = 5/3 + I
=> I = 8/3 - 5/3 - integral(sqrt(1-y^2)) dy | 0 to 1
=> I = integral(1 - sqrt(1-y^2)) dy | 0 to 1
Option C.
Yeah thats something that will click in exam rather than going for graphs for me. Nice solution bro. Keep it up 👍👍👍
Hello...Your method is amazing..
I have a small doubt .. You mentioned (in the 3rd paragraph )that integration -1to 0 √(1-y²)dy can be converted to integration 0 to 1√(1-y²)dy ...
Can u please explain this step ..I mean how the limits are flipped without changing the expression...If it's some rule or property, please tell me it's name so that I can learn about it.
im not reading all that but congratulations
beautiful observations sir, but its very difficult to think all this in exam
Tabhi toh maths ka difficulty sabse high rehta hai mere bhai
Very very true
The problem is based on an old Russian book of Engineering Maths and the topic is double integral..... The same graph is shown.....
Problem:
lim n infty 1 2^ n ( 1 sqrt(1 - 1/(2 ^ n)) + 1/(sqrt(1 - 2/(2 ^ n))) + 1/(sqrt(1 - 3/(2 ^ n))) +......+ 1 sqrt 1- 2^ n -1 2^ n ) is equal to -
Put 2^n=t and proceed
As 2^n->inf
t->inf
@@syed3344 thank you for suggesting a approach 🤗
The toughest part in this is to figure out areas along the y- axis. We usually don't do it along Y-axis, so understanding is difficult.
Sirji itna analysis me 5min se jyada hi lagega pakka
Sahi bole bhai ☺️☺️
Sir honestly bata ra hu mene bhi area ke concept se socha tha but me kuch miss kar raha tha mere soution me so answer match nahi kr raha tha aur sabhi books me standard meathod diya thai jisme bas integrate karke compare kiya hai, aaj finally jake mughe meri mistake ka pata chala jo me kr raha tha so THANK YOU SO MUCH SIR for this amazing solution😍
Phir bhi sir 4-5 min to lg hi jayenge scratch se aise soch kar solve krne mein kha 2.5 min mein hoga ye
That is why skip such insane questions and attempt them at the end if any time is left.
Sir is explaining thats why for sir it took so much time. In exam hall you dont have to explain everything you are doing, so your speed will be much faster.
gajab sir aapratim
Wow this is a very beautifully constructed Problem 🗿
Why u use only half curve can anyone explain
@@Harshitkumar-cq1vb you mean y=√(2π) right ?
If this is your problem then I will just say that since the values inside the sqrt should not be -ve that is 2π y also be +ve . The other half curve is y=-ve so that is why it doesn't exist in the coordinate plane it only will only exit in Argand Plane .
@@omkumarsingh7 y2= 4ax graph to negative mebhi jata hai coordinate plane pr
@@Harshitkumar-cq1vb yes bro y² = 4ax jata hai negative but bro sawaal me hamme y = √(4ax) ke format me tha naaki y² = 4ax
Agar abhi bhi clear na ho toh aap Google pe jaake ek yeh dono graph banana .
@@omkumarsingh7 sorry to disturb yoy but what is difference between y, 2 4ax and y =underroot 4ax
Bsc 1st year math mathod book integral calculus
Sir please please make a video of this solving all these integral
i was one of them who got right ans.
sir we can also transform root 2x-x2 to root 1-(x-1)^2 and by putting x-1= z it will be 2 * intreagtion 0 to 1 root 1-z^2 . now we can balance this above integral and match the constants on both side. I am getting C too
Extremely complex question and tough solution.
U r bsc student
@@monujhembrom9279 I am a graduate engineer.
Gajab 🛐🛐🛐
Beautiful solution Aman Sir ! 👏👏👏
Like for this LEGEND MATHEMAGICIAN ❤😊
Are bhai definite integration solve karo fir ans match karo
Simple
Ha do fir 10 min
@@AvikGupta-ch7oh Bhai pagal hai kya, ye integration solve karne me 2 min bhi ni lagte. Agar tujhe 10 min lag raha hai, to sorry u need more practice
@@Raj-xt4fk timer lagake 7 integrals solve kariye aap, 2 min se neeche ya uske aas paas bhi nahi ayega
@@shivanshnigam4015 bhai tu meri baat sun. Wo jo root 2x - x2 wala integral hai, aur wo jo root 1- y2 wala hai, wo dono ekdam same hai. Rhs wale ko lhs mei bhej aur uska limit 0 se 1 ho jayga. Fir mentally trig sub karke kar le. Aur wo wala y2/2 wala to lulla hai. Ye sab to ho hi jaata hai.
@@Raj-xt4fk trig sub ki jaruat nahi hai vo sqrt(1-y^2) wala sab jagah hai to usse side me rakh ke bhi kiya jaa sakta hai, lekin poora poora solve karne me toh thoda time lagega hi
Sir you are the best in the world
Good explanation
Bhai dekh toh le
Itne hard integral nhi hai agar dekha jae tho √1-y² ka integral y/2√1-y² + a²/2sin-1(y/a) se likha ja skta hai and jo limits di hai since wo kafi asan hai tho jab 0 and 1 put kroge tho simpy π/4 aajaega orr 2x-x2 wali ko bhi king rule lga kr 2×int(√1-x²)0to1 likha ja skta hai.
Ya phir ek orr Kam kr skte hai 2x-x² wali term ko 2×int(√1-x²)0to1 likh kr isse L bol do orr uss trf wali 1-y² wali bhi L ho jaegi orr ab I ko L ke terms mein solve krke L mein hi answer compare krlo
Graphs wala method unique hai and most probably or kisi easy ko bohot aasan bna skta hai but uss wale question ke liye yeh second wali approach (L assume krne walo) better hai
Sir mujhe lagta hai ye solution integrals solve karne se zyada complex way hai k
Absolutely genius! Only problem is, one mostly can't think in this way while giving the paper.
Amazing explanation ❤❤
8th audience 😂
Sir, amazing question and way to crack. i could never ever solve it this way. AMAZING
Sir pace series is best . I hope app shabi log ache hoge m bhi hu ekdham bhannat
Why u use only half curve can anyone explain
Indeed a good question ❤ , but had a different approach.
Sir may you create a video on maclaurin series expansion ?
Sachin Sir ne kraya tha isse class me, Area se hi 😃
4:34
Sir yhi to problem hai
Jab ham graph bnate hai to ye darr rheta hai kitni bar intersect karega aur kha par karega
Aur aap phele se hi soch kar 1 sec mein bol gye
Sir muje sabhi equations ke graph pta hai par aise questions par lga nhi pata
Pls make a video on this problem🙏🙏🙏
Aur sir video mein ache se explain karna bhale hi 3 hr ki ho.😊
Ek book aati play with graphs..... arihant ki shayad karlo tagdi hai
Really amazing question and solution
Why u use only half curve can anyone explain
@@Harshitkumar-cq1vb bhai x ki limits toh dekh
@@ShivamBhalerao-i9j bhai sun or reply jarur karna y2= 4ax ka graph x +ve axis ke upar bhi hota h Or niche bhi sir ne only upar wala curve liya or aap isla explanation dere ho ki limit dekho to bhai agar x + axis ke niche wale graph ko bhi liya jata tab limit kya hoti?
Bro my bad, original equation is y=root2x toh Y ka range toh 0 to infinity he Hoga na
Jab hum y^2=4ax me y=root4ax X-aixs ke upar wale portion ka equation Hota Hai and y=-root4ax X-axis ke niche wale portion ka equation
Omg!
Amazing problem and it's solution
🎉❤
😮😮
Well anlyllisis sr ji
(SIR PLEASE READ THIS BY THIS WE COULD SOLVE IT IN 2MIN 40SEC) SIMPLEST WAY U WILL FIND IN THIS COMMENT SECTION
Apply Queens property in LHS 2nd integral(splitting integral) in root 2x - x square
Apply Kings propery in second integral of RHS ( splitting integral) in root 1-y square
This can also be done with simple integration by observing that the messiest one to integrate is sqrt(1-y^2) which luckily is present in all options and in LHS and RHS. It is the same trick that led you to get a simple solution using geometry! Had that part not been common it would have been messy using geometry or otherwise!
Left hand side: First part: integral (sqrt(2x)) dx|0 to 2=2sqrt(2)/3^ * * (3/2) / 0 to 2 = 8/3 .
Second one: integral (sqrt(2x - x ^ 2)) dx|0 to 2 = integrate (sqrt(1 - (x - 1) ^ 2)) dx d * 10 to 2 = integrate (sqrt(1 - y ^ 2)) dy |-1 to 1 We notice that the same integral is present on right side with limits 0 to 1. Add it to both sides makes LHS: 8/3 - integrate (sqrt(1 - y ^ 2)) dy |-1 to c = 8/3 - integrate (sqrt(1 - y ^ 2)) dy dy | 0 to 1 ( y = - x substitution followed by flipping limits and substitute x = y)
Right hand side becomes: integral(1)dy|0 to 1 + integrate (2) dy / 1 to 2- integral (y ^ 2 / 2) dy10 to 2 + 1 = 1 + 2 - 4/3 + 1 = 5/3 + 1
=> I = 8/3 - 5/3 - integral (sqrt(1-y^2)) dy | 0 to 1 => 1 = integrate (1 - sqrt(1 - y ^ 2)) dy dy | 0 to 1 Option C.
Paper khatam ,tan tan.
Gajab Bhai, maine kiya tumhare method se 4 min se andar me hogaya 👌👌👍👍
Did the same method, got the ans in 1min
Hello...Your method is amazing..
I have a small doubt .. You mentioned (in the last 3rd paragraph )that integration -1to 0 √(1-y²)dy can be converted to integration 0 to 1√(1-y²)dy ...
Can u please explain this step ..I mean how the limits are flipped without changing the expression...If it's some rule or property, please tell me it's name so that I can learn about it.
@@aimanfatima6292being an even function it gives same area from -a to 0 as 0 to a
Eighth
Amazing explanation sir
Wow sir, what an outstanding process of thinking !
Why u use only half curve can anyone explain
@@Harshitkumar-cq1vbbecause limit is from 1 to 2 😊 not from 0
😶🫡🔥
Sir. e^x² ka integration batana
This is non integrable in indefinite
But most of definite integrals can be solved by laplace transformation
@@j.u.4.n620 can be solved only using polar coordinates
Jaldi se comment kar deta hu first😂
Differentiate karke check kar sakte kya
U can't differentiate becoz both are constants (area)
Definite integral hai bhai differentiate karne pe zero aaega
Matlab khud solve 11 min me karo aur bate 2-3 min me solve karne ki 😅
Samjha bhi toh rahe hai hai bhai😅 aur starting ka samay bhi toh gaya tha iss question ke discussion mai related to how many students solved it and all. Tu karde isko part by part solve krke exam mai
Aagar Bina intro aur Bina samjhaya hua solve karange to sir isko bus 2 minutes me hi kar denga
Samjha bhi to rahe hai sir, waise sir ko ye question solve karne me in reality ek minute bhi nahi lagega
He's also explaining. He could've solved it in less than 2 minutes since he knows the logic.
@@navyaaru62950:11