JEE 2014 का ग़ज़ब का सवाल || V.K. Bansal Sir ही सही solve कर पाये थे || Bhannat Maths | Aman Sir

[@dhruvgupta9530]

sochne wali baat hai, jin students ne us question ko khud se solve kiya hoga, vo kitna brilliant mindset rkhte honge in maths

[@shreyassharma8277]

Han bro,aise aise questions bane hi in bacchon ke liye hote hain jinhone apna bachpan bigad kar is exam ko diya hai

[@kattaratheist6617]

@@shreyassharma8277 depend karta hai kuch bacchon ke liye padhai hi unka bachpan hota hai.woh padhai ko enjoyment ki tarah lete hain na ki burden ki tarah.

[@jeeaspirant3550]

@Secret Diary bhai us samay ho sakta hai ho aur uske baad se ab Sare teachers is type ke concepts batate ho Question to brilliant hai mai matlab shock me raha tha tha jab khud se solve kiya tha pehle ki aisa kaise ho gya

[@jeeaspirant3550]

@Secret Diary bhai tum brilliant ho ham jaise average bacchon ke liye shock ki hi baat hai Bhai kuch to baat hogi ki bahuton se nahi bana

[@jeeaspirant3550]

@Secret Diary I agree but bhia dhyan nahi rehta sab ko

[@prabhatk]

If we put a=2, numerator will become -1. Keeping in mind that power is not exactly 2 but it's limiting value is 2. And -1 raised to the power 1.9999999 or 2.0000000001 will become imaginary number. Brilliant question!

[@Dream_Eater_Darkrai]

I will agree with this solution 👌

[@harshharsh3488]

😊

[@TitanGaming-ps5fw]

Naa bro -1^1.9999999999 = -1 you can use calculator

[@ThatGuyInIIIT]

Coaching classes needs 2-3 days to solve this question while students are expected to sove in few minutes

[@Biome_AZ]

Edit:

[@ramesh-vm6kr]

@@Hetubanna vhai ye question mene 30 second me kr liya kya baat kr rhe ho kitna easy question h

[@indianengineer5802]

@@Hetubanna 😂😂kuch bhi ....this was one of the easy and doable question

[@sneaky6969]

Students unhi coaching walo k padhaye hue hote hain

[@sneaky6969]

@@Hetubanna kuch bhi

[@MohitKumar-ls6bb]

I was his student in 2015 in Bansal classes He was a real legend🙏

[@phonexlegend69420]

IIT nikla?

[@suyashdixit8896]

​@@phonexlegend69420 Sabka thode na bro IIT selection ho jata hai 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂

[@harshkumar0101]

​@@phonexlegend69420sabhi iit nikalne ke liye nhi padhte

[@narutouzamaki53]

​@@phonexlegend69420bhai iit nikal neh ke baad log yt pehle bas lecture dekhte hai

[@navrez3100]

@@phonexlegend69420 maybe wo MIT Massachusetts chl ho? Jruri thodi na h ki sb iit hi choose kre

[@anubhavchandra6222]

Teachers taking 1 day for the question paper and students were supposed to do it within 3 hours..

[@maddy8229]

Bilkul sahi sir

[@varadkhairnar7752]

@Prîñcêss behen apne ko bas ek sal Milta hai teachers logo ke pass toh 7 8 sal ka experience rehta hai,hum unse better nahi rehte fir aise task jo woh kar nahi skte hum ein karna padta hai

[@siddhantrohila4889]

Such questions are supposed to be left. They are even made to be left. Otherwise Jee Adv toppers would have scored full marks till now.

[@dontwastetimeyouarelosingr8172]

All questions are not meant to be solved during exam it's just a trap

[@jitugangwar721]

@Prîñcêss 0 kaise hoga

[@SarthakKapat]

2014 Rank 1.Chitang Murdia Isliye 117 out of 120 score kiya tha😅?

[@vishishjaiswal4589]

The most important thing in calculus is domain which most of students don't check. IIT had given the clue to check domain by writing non negative integar. Answer is by putting a=2 base becomes -ve and base is +ve on putting a=0 do like if you agree

[@soumya_who]

Maan gye bhaiya Op thinking 🔥

[@utkarshkatiyar]

Last line ✍️ kr lo..

[@amritkumar5953]

Great job

[@yogeshkushwaha9174]

Pls tell why base can't be negative

[@yashsharma9607]

Bhai non negative wali condition base ke liye nhi 'a' ke liye hai

[@iitjeemathspro]

Amazing personality VK sir, the legend 🔥🔥🔥. His impact on education shall be felt forever!

[@harshsahu1966]

@@kingdomalmighty119 why are you watching this?

[@adarshhoizal1164]

Bansal sir had contacts in all IITs. He used to directly obtain solutions from the IIT professors who had set the actual JEE question papers in that particular year. Hence it was technically not possible for the answers published on the website of his coaching institute to be wrong.

[@bikramdehuri7624]

Ye andar ki baat hai. Kam logon ko yahi sachhai k baare me pata hota hai..

[@informatives7035]

ghanta ..professer sre like judge ...ias upscthey are so age ,nothing about their future and famioy tension . anyone cant threat,bribe them .

[@parijatsutradhar]

I would have guaranteed marked 0 in the exam (especiallywhen the timer is running), and after seeing the key, i would have cried for making a silly mistake of not considering the negetive part...

[@BHUMIHAR.BRAHAMAN]

@@parijatsutradhar where are u today do you qualify for this exam or not

[@parijatsutradhar]

@@BHUMIHAR.BRAHAMAN results are not yet out bro

[@ashishkumarsingh1659]

Bansal sir was king of kota at that time..he was the true inspiration for us at that time🙏❣️

[@srii.abhigyan2938]

Correct but . Too much superstition is not good at all, talent and inventiveness are more or less all, Depends on how much enthusiasm and persistence one is proving.

[@alien3200]

​@@srii.abhigyan2938yeah, talent is relative to a person

[@YashMRSawant]

One thing to always remember when dealing with limits. If the raise to factor is function of limiting variable then it's effect should be invariant whether that raise to factor is expressed as odd/even, even/odd, even/even or odd/odd. In our example, the limiting value of raise to factor is 2, so RHS should be 1/4 when left hand side is evaluated with the value of raise to factor is expressed as 199999/99999~2.

[@Yashkarale120]

No thats not the case u can evaluate easily power can be irrational for example u can compute 2½,2^2½ because Let 2^2½=t logt to base 2 =2½, which is defined U can further it by antilog concepts solve .....thats not the reason,,..... true reason is that when a=2, It is in form of (-ve)^something =t Logt to the base (-ve)=something This is not possible in this universe

[@shilpirai6537]

​@@Yashkarale120ghvh

[@umamishra956]

Actually if -ve^ eventh root is not defined and here by putting a=2 we are getting f(x) = -ve and that's why a.=0 is answer

[@YashMRSawant]

@@Yashkarale120 please read what i described carefully before commenting.

[@animationworld4997]

❤

[@kunaljaju646]

Looking at this question the function is defined at 1 as the ans for the limit is 1/4,so this limit is going to be continuous for 1 and LHL=RHL=f(1),when you solve for LHL you get (0/1-a)^2 now if a is anything except 2 the ans for LHL will be approaching 0 and if a=0 then it will be in the form 0/0 which means it could be further simplified,also RHL gives 1/4 when a=0[RHL=(a+1/2)^2]

[@AkashRaj-vj6br]

I checked it in Arihant 40 years iit jee main and advanced books ..i bought it in 2019 (edition) and it's written a=2

[@Seriouslyfunny1]

If possible, use brilliant tutorial problems instead of arihant. They are much better quality. After that, use Arihant only to get an idea of PYQ.

[@shaktisinghbhati9473]

The point here is that you can't replace (1-x)/1-√x by (1+√x) because at x=1 first expression is not defined but second one is.If you solve this question without substitution of 1+√x in place of (1-x)/(1-√x), by taking log both sides then using multiplication of functions property of limit and solving log part in the same way sir has done here and apply L'hospital rule in (1-x)/(1-√x). you will get only ans= 0

[@dbjindian]

yes....

[@rajendersony6301]

absolutely right

[@mitsu7june]

Why can't we replace?? Since it's tends to 1 not absolute 1

[@simransharma5352]

Agar unka sahi tha toh aap ka bhi toh sahi hi tha Hats off for u sir😀😀😊

[@liberalacedemi200]

I am also agree with you sir because (1- a)^2 =1 When a=0 Sq of (1) or |1| is always 1 But When a= 2 Then sq of (-1) is 1 & |-1| is (1, -1) Hence last answer is a=0 .......... Ans.

[@satyenjha8699]

This was a simple question,but students tend to do mistake in such questions as they apply limits partially. Partial limits can be used only in product form.

[@siddhantgureja7803]

Partial limits apply Kari mtlb.??

[@satyenjha8699]

@@siddhantgureja7803 means say for example you have lim x->a (f(x))^g(x) then you can't do it like lim x->a (f(x))^lim x->a g(x) unless it is not an indeterminate form. Here it is an indeterminate form so you can't seperately apply limit to base and exponent

[@siddhantgureja7803]

@@satyenjha8699 Bhai yaha pe kaunsi indeterminate form hai.?? Power wali toh shyd bas 1 raise to the power infinity hoti hai.. Jisme base 1 ko tend karta hai aur power infinity ko.

[@satyenjha8699]

@@siddhantgureja7803 power aur base dono seperately 0/0 wala indeterminate form hai, isliye dono pe seperately limits use nahi kar sakte

[@siddhantgureja7803]

@@satyenjha8699 Haa 1 put karne pe 0^0 wali indeterminate form ban toh rahi hai.

[@naveenkhare1979]

I have great interest in Maths as on today at the age of 63. Still watching match problems,Olympiad question and try to solve the problem first then tally. You are genius.I admir your way of presenting explanation .

[@masroormalik7964]

I agree. I also watch such videos.

[@YogeshKumar-sq7qe]

On solving we get a=0,2[For a=2, base of given limit approaches -1/2 (clearly base of given limit is -ve) and exponent approaches to 2. Since, base of limit cannot be -ve (kyo ki agar limit ka base -ve ho gya to limit inside ln will not be defined, to define limit inside ln, (1-a)/2 must be greater than o, from this we get, a

[@uthrisar3874]

Base of limit kyu negative aa nahi sakta bro???

[@s_iiitg]

Kyu bhai base - ve kyu nhi ho sakh ta

[@anshumanagrawal346]

@@uthrisar3874 Bhai tum khud hi socho (-0.501)^2.001 hota h kya real (just an example)

[@uthrisar3874]

@@anshumanagrawal346 thanks a lot Bhai 👍👍👍

[@anshumanagrawal346]

@@uthrisar3874 👍👍

[@ajaychaudharycom.chaudhary6181]

(1-a)^2=1 1-a=+_1 Sice1-a= positive value So, 1-a=1hoga Then a=0

[@soumyepratapsingh6090]

My maths teacher taught us this question normally in class and we didn't feel any difficulty didn't knew it was that important at that time!

[@AltafHussain-zv4lh]

Thoda cross check karne se correct option mil jata

[@adityamaurya2177]

@@AltafHussain-zv4lh Yaarr iit ka hi question paper aisa h jiske answer mein clash hota h even india ke top teachers ka..

[@chemifi1686]

Your teacher did the right thing. This man is just creating hype with the name of Bansal sir.

[@SandeepKumar-nm5wz]

These kind of teacher is appropriate for our society.

[@chandankumarchand6686]

@@chemifi1686 exactly.....he is talking like it's end of the world...

[@arish8598]

Sir you are 100 percent correct 💯 Online graphing calculator also shows At a=0, lim x >>1 is 1/4 As rhl is y=0.2491 at x=1.01

[@shubhamtripathi5199]

@@SahajOp Ye kaha likha hai ki non-positive aur non-negative dono nhi ho skta? Waha likha hai non-negative nahi ho skta bas. Non-negative matlab less than 0. To 0 to ho hi skta hai answer.

[@SahajOp]

@@shubhamtripathi5199 ok , non negative is > equal to 0

[@rishavraj6027]

Because sir for a =2 base of above limit approaches -1/2 and exponent approach to 2 and since base cannot be negative Hence limit doesn't exist 🙏

[@rhlsx]

@G.U.D. Temper Very well explained! 🙌🏻✨💥 Wish more emphasis is given on understanding the meaning of the concept rather than just solving questions! 🙌🏻 Thank you so much for such a detailed and awesome explanation! ✨

[@dawnstudios7813]

If we allow complex exponentiation, then a=2 should be the correct answer, one can even find lim (-1)^x as x tends to 2 on Wolfram alpha, and look at the series expansion near 2, it includes imaginary terms as well.

[@anshumanagrawal346]

It's still wrong, (-1)^(irrational) is always undefined

[@varun8762]

Complex exponentiation not in jee advance syllabus

[@priyam6496]

On putting a=2....(-1/2)^2 comes out but rhl and lhl is not defined becoz base is negative for example (-1/2)^1.99999999 is undefined nd (-1/2)^2.00000001 is also undefined therefore limit doesn exist.😇Always check domain in limit is must after getting ans.

[@masroormalik7964]

Zabardast Kahaani banayi hai aap ne. I am not a student but a retired engineer and I do love watching math videos. Wonderful.

[@adhritimandal948]

If we put a=2,then it is approaching -1/2,which is negative....but we know that for a limit to exist in f(x)^g(x) form, f(x) must be greater than 0....so we reject a=2.....

[@cgnytro5460]

Oo tai naki

[@thegreatindia8013]

Oo tai naki means that's why

[@aneekmandal5430]

@@thegreatindia8013 No,it means 'oh,is it?' I am also a bengali from w.b ....

[@pokhiti]

Aise tough questions ko easy banane ke liye hi hain hamare Aman sir

[@anandsnambiar3147]

we cannot put a=2, because when we substitute a=2 to the function we would get a negative value. but since f(x)^g(x) is only defined when f(x) is positive, and here since f(x) is coming as negative when a=2, we cannot give value of a as 2, so the ans is 0. Pls pin this comment sir if i am right :)

[@Mk-hf6dv]

Bhai tum to jee me top karoge

[@sanayamitall2470]

Ge

[@sinistershivank8010]

But don't you think when you put a=2 base will be negative but exponent is 2 which is possible for negative bases Bcoz as much i can remember negative bases are not defined for a^x functions not for x^2 function So for negative base reason a=0 or 2 both correct

[@prajojeet]

@National Thinking Framework (India Chapter) what is your source?

[@prajojeet]

@National Thinking Framework (India Chapter) means how do you come across all such awesome content ?? From where??

[@2ndraistar415]

Negative start from -1 and before ot 0 comes first therefore it is the largest integer

[@thunderbolt6276]

Reason: on solving we get two values of a=0 and a=2. On putting a=0 we get the base as tending to 1/2 and exponent as tending to 2, so limit=1/4. But on putting a=2 we get base tending to -1/2 and exponent tending to 2, now one would say (-1/2)²=1/4, so what's the problem but Emphasis has to be paid to the word exponent tending to 2 but not 2, so it wouldn't be defined.

[@krishnamohapatra2432]

Ekdum sahi hai...yeahi batt hai bhai 🙏

[@Cm-zc2zx]

@@krishnamohapatra2432 yarr 😂😂🤣 Main smjh gyaa chodo

[@purvanshsharma9705]

sir meine ek method se kiya jisme direct a = 0 aaraha tha, Meine pure expression ko exp(limx-->1ln(expresion)^power) mein convert kardiya aur phir solve kiya

[@prateeksingh432]

The expression after taking common and dividing become f(x)^g(x) form and we know in this form f(x) cannot be -ve. so If we put a =2 the f(x) becomes -ve . Hence ans will be zero.

[@samadfaridi1309]

Ashish sir is legend

[@acos126]

Lakshay se ho na ?

[@nitinverma.6236]

Note -Always take log on both side when such type of questions asked.It will always give you correct answer.

[@MathXCompt]

Yes ...Nitin you are right 24 min ki video bna di iss question pr aur baccho ko paka dala

[@Sankalp-sd6fm]

@@MathXCompt ye bolte h vk bansal sir se solve nahi ho paya tha us time Aur ye khud 7 saal baad aaye h solve karne

[@Rexghh]

agreed

[@nehakushwah7158]

@@kingdomalmighty119 chotu 2 v rkhena tab v 1/4 hi ayega

[@kingdomalmighty119]

@@nehakushwah7158 kuchh bhi

[@JagdishTandi-zs3jo]

14:04 at the moment he knew who is king of maths.

[@VinaySingh-bb8mz]

Sir , We have (-a+1) as a base so if we 1)put a=2 then we get result-1 2)put a=0 then we get result 1 Hence, it is clear that a=0 is the largest one

[@archanabansal688]

In my opinion , they tell non negative neither negative nor positive , this reason is applicable in my thought.

[@ajaymokta3258]

For a= 2 base of above limit approaches to -1/2 and exponent approaches to 2 and since base cannot negative hence for a= 2 limit does not exist

[@Phoenix-rg5kt]

3:59 😂😂 I can't stop my laugh 😂legendery backbencher... BTW VK Sir is Undescribable🔥🔥

[@crickethungama-ho5dp]

Sir a=0 hoga kyoki a=2 par f(x)=[{-ax+sin(x-1)+a} / {x+sin(x-1)-1}] ka value negative hai . g(x)=1+√x , f(x) ke power me hai . Yah defined hoga jab f(x) positive ho.

[@kshitizmangalbajracharya5152]

I am not really aware about the exam this video refers to, but I believe it's not so big of an exam that you'd need to study complex analysis. So, that makes this question a simple calculus question - the ones dealt in early undergraduate level or intermediate level. And, the approach taught in calculus to evaluate the limit of the form f(x)^g(x) is to take its logarithm and find its limit. Then, exponentiating that limit shall give the required limit. To do so, you will need f(x) to be positive as it will appear in the operation of logarithm. This is the underlying assumption of such problems in the basic calculus course. You can check it by taking logarithm of the given function and proceeding with the evaluation of its limit. Since the function inside log needs to be positive, its limit will consquently be positive (by using monotonicity of limits). Taking a=2, the limit will no longer be positive. So, the only option left is to take a=0. That's what is happening. This is indeed a really good question. Now on, I will always discuss this question whenever I teach calculus.

[@Gopal_kg]

Great, a wonderful explanation!

[@AliRaza-cv9ug]

Me too.... Sir

[@AliRaza-cv9ug]

You know mathematics brother..

[@ritviksharma5949]

Not so big of an exam? Just look it up once. It is given by 12 appearing students but the questions are of much higher level. This exam is one of the toughest exams on this planet.

[@sarvendrashukla8051]

Satisfactory words!

[@VSTMathsClasses]

VK Bansal sir was best....and will remain best in our hearts ever...😪

[@srii.abhigyan2938]

Correct but . Too much superstition is not good at all, talent and inventiveness are more or less all, Depends on how much enthusiasm and persistence one is proving.

[@shubham_bandhavakar8826]

Agar even root lenge toh andar negative nhi hona chahiye aur kyuki power approaching hai toh denominator bohot bada number hoga jiske last ke digits 00000 hone chahiye which is even aur agar a 2 hogya toh under root mai negative number which is not possible for even root (power is approaching to 2)

[@madhavgarg8405]

As final ans is 1/4 , which is only possible when (k)^2 ., if 0

[@AnishKumar-kp9kt]

Ashish sir solved this question and we also don't find it difficult

[@harshpatil924]

sir aaaap bohat hi jyada struggle kr rhe ho hmare liye thank u sir for doing sooooooooo😍😍😍😍😍😍😍😍😍

[@thelyceum6093]

At very first place it is obvious that a=0 . Because if you don't put a=0 limit will not be indeterminate form (say if you put a=2 then numerator/denominator will be -2+1 /1+1 ) So to limit to be exist at first place a=0 .

[@shoryaprakash8945]

In exponent raised to irrational power is defined in terms of logarithm function. Like if power is rational like a^(p/q) by definition is root of equation x^q=a^p but for irrational power it's defined as follows a^b= e^(b.ln(a) ) Here power is irrational so we need to work limit by converting it to log and since we are speaking about real function when a>1 for x aprroches1+ limit num sin(x-1)-a(x-1) is 0 hence it is out of domain

[@Chandankumar96245]

0 is currect ans because if we put 0, in place of (a) then we get right hand side equal to left hand side.

[@hiteshsirclassesmathematic5093]

It's my first comment on KZbin Answer is 0 because root of √ 1-a not equal to -1 ,√1-a=1 answer will be 0 , square root of any real value is not negative , baki sir aap batiyega apke explanation ka wait rahega 🙏

[@ramanujancollegeofmathemat6087]

YOU ARE CORRECT 👍 ANSWER WILL BE 0

[@priyanshubisht3163]

Negative Non negative Positive All are different from each other

[@pratikshasharma5555]

1-a^2) ^2 = 1.. so we can write ( 1-a^2) ^2 = 1^2 ... square se square cut gya... therefore 1- a^2 =1 ... == 1 se 1 cut gya so answer will be zero :) ""a =0""

[@chanduudarapu7906]

This Q is solved by Ashish sir in Lakshya Jee batch, and sir said most people say this Q is solved only by VK bansal sir but it's not, some students have solved it

[@unknwn9274]

Bhaii yahi comment mai bhi karne wala tha😅...

[@healer1609]

Because if we a=2 in Main equation then value comes out negative or non positive so answer is 0

[@_AmanKumar2592AA]

Sir sinx ki indeterminate form ka expansion karke khol do Sinx = x - x^3/6 + x^5/120 - ........x^n/n factorial +...... Aise question hamare bsc math indeterminate form ke chapter's main hain

[@tanishqthakur2728]

I think here we can say that since 1-x/1-rootx is a seperate function and we are applying limit on it and getting the ans as 2 which means the function in approaching to 2 not exactly equal to 2 .

[@umamishra956]

Sir ye question to maine khud se hi karliya x=t+1 se🎉

[@Ayush__Prajapati]

A notice to all teachers and students----- ""Don't compare ur talent with him "" He is totally different and father of kota educational junction 🙂🙂😇

[@praveenkaintura7278]

Absolutely

[@nibbles5007]

Tu pagal hai khaha kiya compare

[@Ayush__Prajapati]

@@nibbles5007 ye sabhi k liye hai 🤣

[@srii.abhigyan2938]

Correct but . Too much superstition is not good at all, talent and inventiveness are more or less all, Depends on how much enthusiasm and persistence one is proving.

[@amlanbiswas2766]

22:18 sir this is because agr hum (1-a)^2 se power of 2 hatatey hai toh √1 aa jaata hai and we know that ki sqaure root k andr value 0 yah 0 se jyada honi chahiye naaki negative since idhr jikar hai integer ki toh complex numbers ki concept nhi lagega nahh...

[@aryangautam3575]

Let me try to explain: As we put a=2 in the function we get negative base,right? And if we put x=0 we get positive base. Now the power is tending to 2 .As 2 is an even no. it doesn't mean that the no. tending to 2 is also an even no matter how much it is close to 2 .Now,only the even power of a non negative no. is also a non negative no. But a no. tending to 2 is not an even no. As it is given the whole expression=1/4 i.e. base can never be negative as power is not an even no. So,the value a=2 is discarded. So the only possible value of a is 0.

[@iitianaakash5582]

Apke charan kidhar ha sir😐🤣

[@aryangautam3575]

@@iitianaakash5582 What happened?

[@anish8373]

Bro hatsoff

[@aryangautam3575]

@@anish8373 Thanks bhai

[@satyenjha8699]

No this is not correct logic. The thing is even if you reject 2 using this logic,you have still used wrong concept of limits and coincidentally got 0 as the answer. The only way to do this is to take log on both sides. You can't simply reduce the exponent to 2 as you can't apply limits seperately to base and exponent until and unless it's not an indeterminate form.

[@jayatemihir5390]

Hi, a=2 cannot be the solution because then the inner term (sin(x-1)/(x-1) - a)/(sin (x-1)/(x-1)) will be negative in the neighbourhood of 1 and the exponent 1 + sqrt(x) will be fractional in the neighbourhood of x=1. So, due to this in the left or right neighbourhood of the limit will be an expression (-ve)^(fractional real number) which won't be a real number at all and hence the limit won't exist in real domain that case. So, a=0 will be the only case where the limit will take the form (+ve)^real which is allowed. I am already graduated from NIT Allahabad but I still watch your videos to solve problems in my mind.

[@vishnukrishnadas9233]

Sir , FOR (a) =2 the base approaches to -1by2 and hence this value must be rejected.

[@narendramudi7984]

There is nothing about "Approach" as the limit is on "x" not on "a"

[@Shivkumar-ev9mv]

These online teachers are less teachers more actors

[@FunnyReels119]

For a=2 we get a negative number raised to the power of a rational number. This is not always defined. Hence a=0 is the right answer.

[@kingdomalmighty119]

It's so simple. When found a=0 & a=2. After that put the value of 'a' one by one in this equation {(-a+1)/2}^2=1/4 and you will get L.H.S = R.H.S when you put the value of 'a'=0 , you will get L.H.S=R.H.S , I mean 1/4=1/4 .thanks a lot and all the best for next exam 👍🙏🥰 I am a neet aspirant. Please blessings me..❤❤

[@harshitdhiman1120]

jis jisko smjh ni aya ki a = 2 incorrect kyun hai! Wo isko L hopital rule se solve kro khud smjh ajaega cross check krlena !

[@paramdhamduha]

If we put the value of a=2, our answer of the question will be undefined because√2is irretional number.

[@Deadpool-yc5rs]

Answer is that because in power its actually approaching to 1 not 1 so in fraction power -ve not allowed and on putting 2 its coming -ve 1/2

[@mathsdivergence67566]

Dear Sir, exponential function a^x mai base a ki value hamesha greater than zero (a>0) and does not equal to 1 hoti hai only then function is defined. And sir is question me at a=2 par base ki value -1/2 par tend kar rahi hai joki ek negative entity hai and at a=0 par base ki value +1/2 par tend kar rahi hai, therefore a=2 is rejected and correct answer is a=0.

[@dcttournaments5262]

It's reallyyyy easy...take limit=L,log on both sides...solve and get a=0,2...now check domain of log which is a

[@sks-rd2ng]

Who frame this question how intelligent he is

[@dhruvmohanshukla8933]

Je baat 😂

[@Priyanshu_Chakraborty]

How intelligent r u man that u think about this question 👍

[@gamernscholar]

Right hand limit does not exist with a=2 since sin(x-1) - 2(x-1) < 0 when x>1 with Denominator being +ve the overall fraction is negative and (-ve)^fraction does not exist , similarly LHL also do not exist , a = 2 is rejected.

[@fireians2809]

sin(x-1) -2(x-1)

[@gamernscholar]

@@fireians2809 follow the graph of y = sin x vs y = x

[@piyushrajyadav]

a = 2 rakhane par base -ve hoga vahi a = 0 rakhane par base +ve hoga

[@ujjwal9412]

The ques is easy but the two solutions make it advanced level problem. Now people who are not able to conclude why answer was not 2 cross checking is important here. When you divide by (x-1) here numerator becomes in the form of f(x) g (x) If a=2 then f(x) becomes negative which is not possible in this form hence 2 is rejected and the final answer concludes to 0⃣. I felt good solving it in 1 attempt and getting it correct . My teacher taught me to cross check in limits questions and now i get it why its important❤

[@nimishgupta5813]

Sir Arihant 42 years pyq 1979-2020 me abhi bhi ans a=2 given hai 😂

[@classyrydm33]

Nhi bhai 0 hai

[@mustafa6543]

@@classyrydm33 nahi 2 hi diya hua hai

[@kumarsoham8734]

physics wallah study material m 0 h

[@harshpatil924]

oooo bhai maaro kya hi copy krte he 😂😂😂😂😂😂

[@nimishgupta5813]

@@kumarsoham8734 oh, nice; but u haven't studied that module as i am 12th student 😃

[@inderjeetsharma6668]

Very nicely explained dear. I am a Maths teacher. Thanks

[@vinodasati2883]

Actually air 1 of 2014 chitraang murdia got 117 out of 120 in maths just got this limits problem wrong. His mentor anna sir have told this on unacademy

[@jeeadvanced6533]

😦😦😦

[@avais4428]

He is one of most brightest students of IIT

[@kingdomalmighty119]

It's so simple. When found a=0 & a=2. After that put the value of 'a' one by one in this equation {(-a+1)/2}^2=1/4 and you will get L.H.S = R.H.S when you put the value of 'a'=0 , you will get L.H.S=R.H.S , I mean 1/4=1/4 .thanks a lot and all the best for next exam 👍🙏🥰 I am a neet aspirant. Please blessings me..❤❤

[@TusharKumar-ty8kh]

@@kingdomalmighty119 how the fuck it is simple to find value 0 and 2

[@kingdomalmighty119]

@@TusharKumar-ty8kh when you're preparing IIT exams That time, these type of questions are common.

[@satyampandey890]

7:40 Yaha nahi samajh aaya 11:18 Bada Institute 14:01 Bansal Classes 17:05 Varun Sir, IIT Kanpur 20:20 IIT ne a=0 release kiya 21:42 Why a= 0?

[@mrperfect2997]

Sir answer 0 hi hoga kyuki power me 1-root(x) hai 2 lene per power irrational ho jata or irrational power ka value exact nahi aata hai that's why we reject 2 or x=0 lene per power rational hai toh answer exact value aayega

[@AnkitYadav-eh3uu]

a ki value rakhni hai naa ki, x ki

[@mrperfect2997]

Sorry bhai a ki hi😅

[@nipungrover229]

simple technique was that it was becoming 1 ki power infinite form so we shoud take log and check when a is 2 the value inside log is coming negative which is not possible so answer is 0

[@anshumanarya9215]

I wouldn’t like to take the name but our maths teacher told us this same story of 0 and 2 Now I can connect it to your one ❤️

[@Memebyadit]

But for a = 2 we get a negative number raised to the power of a rational number. and logarithm of a negative number is not define. [From (3)] This is not always defined. Hence a = 0 is the right answer

[@lg.gaming3673]

a 2 or 0. but for a= 2 base of above limit approaches be negative hence limit does not exist. Therefore a=0

[@anshpal5266]

After passing non negative numbers Just Start 0. And 1 to infinity so nearest is 0 not 2

[@pwop3879]

By taking log(natural log or simple log too) both side after step2 we will get the exact value of a as in Ln x ; x>0

[@sabyasachibiswal6277]

Exactly

[@ajeetkumar-ky5ek]

See this is the problem of limit to exponents. We solve it by taking logarithm, by this method I am getting only one answer a=0. Previous method is wrong because base and exponents are inderterminant form and limit cannot be apply on both things simultaneously.

[@sk_0321]

Vk bansal sir is known as teacher of the teacher

[@shreeomdubeydubey9803]

Bcz after putting a=2 andr ka sb -ve ho jayega and power 2+ ko tend krega jo not possible hai,,,, ashish agarwal sir student,,,,, 💪💪💪💪🦾

[@pawanjaat7]

Mujhe math nhi pr kahani sunne ke lie 24 mint ka video dekha

[@Vinay-gy2gk]

Sir a=0 because it satisfies the given condition that the equation is equal to 1/4. Am I correct sir

[@Short_1sk]

1/4 or 1/2 ??

[@samyakgupta5773]

😂😂😂😂 wo to a=2 bhi karta hai🤣🤣🤣

[@pavanmishra9232]

Agar a=2 hoga tho a ki value ko jab limit main put karega tho (-1/2)^ something 2+ aayega aur ex 9/4 ya koi aur bhi no aayega tho negative ka root ban jayega tho ye imaginary no ho jayega esliye a=0 hoga

[@shivanshbaranwal6178]

Sir i really want to study from you sir where do you teach in offline sir please reply sir

[@mkmkmk845]

He was not only only good mathematecian, but a unique personality, very bold and excellent businessman.

[@Sorya-gf7qw]

I think the reason lies within domain of function . Since we got base of Power function approaching (1-a) raise to power some function and as a approaches 2 it's negative . I.e. function is undefined so answer is 0

[@deepankartripathi9911]

The square of any no. Cannot be negative which means ..(1-a)^2 can not be ( -2 )..Square is always positive.

[@vedantjadhao8117]

Those golden days of IIT JEE preparation

[@vladiyogiputinraj4132]

Bhaiya aapka Hua IIT me just asking for some tips.....

[@vladiyogiputinraj4132]

I am preparing for jee mains and I don't know if I am well prepared for advanced but I will try though....

[@gauravindoria9309]

@@vladiyogiputinraj4132 Hii Buddy, I'll definitely help you if you want suggestion then connect with me !

[@ravipratap7530]

@@gauravindoria9309 best books??

[@ayaansrivastava01]

Where did he used to teach

[@sshortschannel_YT]

1 me se jab kuch tum nikalo phir uska square karo is it equals to 1.

[@thanav3930]

If we take log on both sides and solve We get log(2/1-a)=log2 Now only a=0 is the solution

[@ramsaini3745]

90%= time waste 10%= teaching Gajab ke teacher ho 🫡

[@rakeshsonwani6212]

Sahi kaha yr

[@adityk8138]

Khud ka badayi bhi chodna hai