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.

@vibhashrivastava8846

6 ай бұрын

If u have any source or information about it, can u share it. I would like to learn too

@syed3344

6 ай бұрын

Inverse of x*e^x​@@vibhashrivastava8846

@syed3344

6 ай бұрын

@@vibhashrivastava8846 inverse of x*e^x

@TheHellBoy05

6 ай бұрын

@@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

@adityajha2889

6 ай бұрын

Yeah

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.

@shailnair2243

6 ай бұрын

Yeah thats something that will click in exam rather than going for graphs for me. Nice solution bro. Keep it up 👍👍👍

@aimanfatima6292

3 ай бұрын

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.

@JEE-oq1me

2 ай бұрын

im not reading all that but congratulations

Beautiful solution Aman Sir ! 👏👏👏

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.

beautiful observations sir, but its very difficult to think all this in exam

@danishhsable

6 ай бұрын

Tabhi toh maths ka difficulty sabse high rehta hai mere bhai

@shailnair2243

6 ай бұрын

Very very true

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😍

The Way you teach Maths, wish you 1M Subs Fam soon!! Sir

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

Amazing explanation sir

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 -

@syed3344

6 ай бұрын

Put 2^n=t and proceed As 2^n->inf t->inf

@LDR_Crafts

6 ай бұрын

@@syed3344 thank you for suggesting a approach 🤗

Amazing explanation ❤❤

Sir, amazing question and way to crack. i could never ever solve it this way. AMAZING

Omg! Amazing problem and it's solution

Gajab 🛐🛐🛐

The problem is based on an old Russian book of Engineering Maths and the topic is double integral..... The same graph is shown.....

Indeed a good question ❤ , but had a different approach.

Sir you are the best in the world

Well anlyllisis sr ji

Absolutely genius! Only problem is, one mostly can't think in this way while giving the paper.

Good explanation

@omkamble5366

6 ай бұрын

Bhai dekh toh le

Sir may you create a video on maclaurin series expansion ?

Wow sir, what an outstanding process of thinking !

@Harshitkumar-cq1vb

6 ай бұрын

Why u use only half curve can anyone explain

@smarter4163

4 ай бұрын

​@@Harshitkumar-cq1vbbecause limit is from 1 to 2 😊 not from 0

Sir pace series is best . I hope app shabi log ache hoge m bhi hu ekdham bhannat

Sir kya aap IIT jam ki preparation krate hai

amazing method sir ❤❤

@Harshitkumar-cq1vb

6 ай бұрын

Why u use only half curve can anyone explain

Wow this is a very beautifully constructed Problem 🗿

@Harshitkumar-cq1vb

6 ай бұрын

Why u use only half curve can anyone explain

@omkumarsingh7

6 ай бұрын

​​@@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 .

@Harshitkumar-cq1vb

6 ай бұрын

@@omkumarsingh7 y2= 4ax graph to negative mebhi jata hai coordinate plane pr

@omkumarsingh7

6 ай бұрын

@@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 .

@Harshitkumar-cq1vb

6 ай бұрын

@@omkumarsingh7 sorry to disturb yoy but what is difference between y, 2 4ax and y =underroot 4ax

🎉❤

Like for this LEGEND MATHEMAGICIAN ❤😊

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

Extremely complex question and tough solution.

@monujhembrom9279

6 ай бұрын

U r bsc student

@mdasifeqbal2323

6 ай бұрын

@@monujhembrom9279 I am a graduate engineer.

Sachin Sir ne kraya tha isse class me, Area se hi 😃

Why u use only half curve can anyone explain

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

@abhirupkundu2778

6 ай бұрын

That is why skip such insane questions and attempt them at the end if any time is left.

Really amazing question and solution

@Harshitkumar-cq1vb

6 ай бұрын

Why u use only half curve can anyone explain

@user-he8rc1ro9e

6 ай бұрын

@@Harshitkumar-cq1vb bhai x ki limits toh dekh

@Harshitkumar-cq1vb

6 ай бұрын

@@user-he8rc1ro9e 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?

@user-he8rc1ro9e

6 ай бұрын

Bro my bad, original equation is y=root2x toh Y ka range toh 0 to infinity he Hoga na

@user-he8rc1ro9e

6 ай бұрын

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

😮😮

Sirji itna analysis me 5min se jyada hi lagega pakka

@shailnair2243

6 ай бұрын

Sahi bole bhai ☺️☺️

Tough question

Sir mujhe lagta hai ye solution integrals solve karne se zyada complex way hai k

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.😊

@noname-sl3to

5 ай бұрын

Ek book aati play with graphs..... arihant ki shayad karlo tagdi hai

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

😶🫡🔥

Differentiate karke check kar sakte kya

@SIVA_GANESH.B

6 ай бұрын

U can't differentiate becoz both are constants (area)

@DistinguishedGentleman1554

6 ай бұрын

Definite integral hai bhai differentiate karne pe zero aaega

Sir. e^x² ka integration batana

@j.u.4.n620

6 ай бұрын

This is non integrable in indefinite But most of definite integrals can be solved by laplace transformation

@syed3344

6 ай бұрын

@@j.u.4.n620 can be solved only using polar coordinates

Eighth

Sir linear /linear ka graph ma ek video la aya.A sab student ka biniti ha please 🙏😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭😭

@debmalyadas7675

6 ай бұрын

Bhai asan hota hai Aod padhoge to khud banaloge

@ABH656

6 ай бұрын

Bhai simply itna yaad rakho range uska all real except coff of x in Nr / coff of x in Dr

Are bhai definite integration solve karo fir ans match karo Simple

@AvikGupta-ch7oh

6 ай бұрын

Ha do fir 10 min

@Raj-xt4fk

6 ай бұрын

@@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

@shivanshnigam4015

6 ай бұрын

@@Raj-xt4fk timer lagake 7 integrals solve kariye aap, 2 min se neeche ya uske aas paas bhi nahi ayega

@Raj-xt4fk

6 ай бұрын

​@@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.

@shivanshnigam4015

6 ай бұрын

@@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

8th audience 😂

Jaldi se comment kar deta hu first😂

(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.

@syed3344

6 ай бұрын

Paper khatam ,tan tan.

@shivanshnigam4015

6 ай бұрын

Gajab Bhai, maine kiya tumhare method se 4 min se andar me hogaya 👌👌👍👍

@annapurnasarangi9999

6 ай бұрын

Did the same method, got the ans in 1min

@aimanfatima6292

3 ай бұрын

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.

@abhinavrawat6632

3 ай бұрын

​​@@aimanfatima6292being an even function it gives same area from -a to 0 as 0 to a

Matlab khud solve 11 min me karo aur bate 2-3 min me solve karne ki 😅

@navyaaru6295

6 ай бұрын

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

@liopard8357

6 ай бұрын

Aagar Bina intro aur Bina samjhaya hua solve karange to sir isko bus 2 minutes me hi kar denga

@iconicsolos1234

5 ай бұрын

Samjha bhi to rahe hai sir, waise sir ko ye question solve karne me in reality ek minute bhi nahi lagega

@Shogun507

4 ай бұрын

He's also explaining. He could've solved it in less than 2 minutes since he knows the logic.

@Daksh-ek5ne

3 ай бұрын

​@@navyaaru62950:11