JEE 2014 का ग़ज़ब का सवाल || V.K. Bansal Sir ही सही solve कर पाये थे || Bhannat Maths | Aman Sir
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- Опубликовано: 10 сен 2024
- JEE 2014 का ग़ज़ब का सवाल || V.K. Bansal Sir ही सही solve कर पाये थे || Bhannat Maths | Aman Sir
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sochne wali baat hai, jin students ne us question ko khud se solve kiya hoga, vo kitna brilliant mindset rkhte honge in maths
Han bro,aise aise questions bane hi in bacchon ke liye hote hain jinhone apna bachpan bigad kar is exam ko diya hai
@@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.
@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
@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
@Secret Diary I agree but bhia dhyan nahi rehta sab ko
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!
I will agree with this solution 👌
😊
Coaching classes needs 2-3 days to solve this question while students are expected to sove in few minutes
Edit:
@@Hetubanna vhai ye question mene 30 second me kr liya kya baat kr rhe ho kitna easy question h
@@Hetubanna 😂😂kuch bhi ....this was one of the easy and doable question
Students unhi coaching walo k padhaye hue hote hain
@@Hetubanna kuch bhi
Amazing personality VK sir, the legend 🔥🔥🔥.
His impact on education shall be felt forever!
@@kingdomalmighty119 why are you watching this?
I was his student in 2015 in Bansal classes
He was a real legend🙏
IIT nikla?
@@phonexlegend69420 Sabka thode na bro IIT selection ho jata hai 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂
@@phonexlegend69420sabhi iit nikalne ke liye nhi padhte
@@phonexlegend69420bhai iit nikal neh ke baad log yt pehle bas lecture dekhte hai
@@phonexlegend69420 maybe wo MIT Massachusetts chl ho? Jruri thodi na h ki sb iit hi choose kre
2014 Rank 1.Chitang Murdia Isliye 117 out of 120 score kiya tha😅?
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
Maan gye bhaiya
Op thinking 🔥
Last line ✍️ kr lo..
Great job
Pls tell why base can't be negative
Bhai non negative wali condition base ke liye nhi 'a' ke liye hai
Bansal sir was king of kota at that time..he was the true inspiration for us at that time🙏❣️
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.
@@srii.abhigyan2938yeah, talent is relative to a person
Teachers taking 1 day for the question paper and students were supposed to do it within 3 hours..
Bilkul sahi sir
@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
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.
All questions are not meant to be solved during exam it's just a trap
@Prîñcêss 0 kaise hoga
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.
Ye andar ki baat hai. Kam logon ko yahi sachhai k baare me pata hota hai..
ghanta ..professer sre like judge ...ias upscthey are so age ,nothing about their future and famioy tension .
anyone cant threat,bribe them .
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...
@@parijatsutradhar where are u today do you qualify for this exam or not
@@BHUMIHAR.BRAHAMAN results are not yet out bro
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.
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
@@Yashkarale120ghvh
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
@@Yashkarale120 please read what i described carefully before commenting.
❤
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
@@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.
@@shubhamtripathi5199 ok , non negative is > equal to 0
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.
Partial limits apply Kari mtlb.??
@@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
@@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.
@@siddhantgureja7803 power aur base dono seperately 0/0 wala indeterminate form hai, isliye dono pe seperately limits use nahi kar sakte
@@satyenjha8699
Haa 1 put karne pe 0^0 wali indeterminate form ban toh rahi hai.
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]
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!
Thoda cross check karne se correct option mil jata
Your teacher did the right thing.
This man is just creating hype with the name of Bansal sir.
Same mera tution me bhi krwaha tha
These kind of teacher is appropriate for our society.
@@chemifi1686 exactly.....he is talking like it's end of the world...
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 .
I agree. I also watch such videos.
Agar unka sahi tha toh aap ka bhi toh sahi hi tha
Hats off for u sir😀😀😊
3:59 😂😂 I can't stop my laugh 😂legendery backbencher... BTW VK Sir is Undescribable🔥🔥
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
Base of limit kyu negative aa nahi sakta bro???
Kyu bhai base - ve kyu nhi ho sakh ta
@@uthrisar3874 Bhai tum khud hi socho (-0.501)^2.001 hota h kya real (just an example)
@@anshumanagrawal346 thanks a lot Bhai 👍👍👍
@@uthrisar3874 👍👍
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
yes....
absolutely right
Why can't we replace?? Since it's tends to 1 not absolute 1
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 🙏
@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! ✨
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.
It's still wrong, (-1)^(irrational) is always undefined
Complex exponentiation not in jee advance syllabus
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
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.
It was an Easy Question...
You just exaggerated it too much dk why??
Yeah Exactly 💯
@@poplu8422 yess why should a student waste around 30 min to a single question of limit which is also not very difficult???🤔
Bhai is bande ne zindagi ke 10 saal barbaad kr diye ise abhi tk question hi samjh nhi aaya h 😂
VK Bansal sir was best....and will remain best in our hearts ever...😪
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.
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.....
Oo tai naki
Oo tai naki means that's why
@@thegreatindia8013 No,it means 'oh,is it?' I am also a bengali from w.b ....
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.
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.
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
Bhaii yahi comment mai bhi karne wala tha😅...
sir aaaap bohat hi jyada struggle kr rhe ho hmare liye thank u sir for doing sooooooooo😍😍😍😍😍😍😍😍😍
90%= time waste
10%= teaching
Gajab ke teacher ho 🫡
Sahi kaha yr
Khud ka badayi bhi chodna hai
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
:)
Bhai tum to jee me top karoge
Ge
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
@National Thinking Framework (India Chapter) what is your source?
@National Thinking Framework (India Chapter) means how do you come across all such awesome content ?? From where??
Ashish sir is legend
Lakshay se ho na ?
Because if we a=2 in Main equation then value comes out negative or non positive so answer is 0
Who frame this question how intelligent he is
Je baat 😂
How intelligent r u man that u think about this question 👍
Sir , FOR (a) =2 the base approaches to -1by2 and hence this value must be rejected.
There is nothing about "Approach" as the limit is on "x" not on "a"
Sir Arihant 42 years pyq 1979-2020 me abhi bhi ans a=2 given hai 😂
Nhi bhai 0 hai
@@classyrydm33 nahi
2 hi diya hua hai
physics wallah study material m 0 h
oooo bhai maaro kya hi copy krte he 😂😂😂😂😂😂
@@kumarsoham8734 oh, nice; but u haven't studied that module as i am 12th student 😃
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.
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.
He wasted 20 minutes out of 25. I wonder how degraded people become to make their videos lengthy .
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
Ashish sir solved this question and we also don't find it difficult
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.
Ekdum sahi hai...yeahi batt hai bhai 🙏
@@krishnamohapatra2432 yarr
😂😂🤣
Main smjh gyaa chodo
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.
A notice to all teachers and students-----
""Don't compare ur talent with him ""
He is totally different and father of kota educational junction 🙂🙂😇
Absolutely
Tu pagal hai khaha kiya compare
@@nibbles5007 ye sabhi k liye hai 🤣
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.
Those golden days of IIT JEE preparation
Bhaiya aapka Hua IIT me just asking for some tips.....
I am preparing for jee mains and I don't know if I am well prepared for advanced but I will try though....
@@vladiyogiputinraj4132 Hii Buddy, I'll definitely help you if you want suggestion then connect with me !
@@gauravindoria9309 best books??
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
Zabardast Kahaani banayi hai aap ne. I am not a student but a retired engineer and I do love watching math videos.
Wonderful.
Note -Always take log on both side when such type of questions asked.It will always give you correct answer.
Yes ...Nitin you are right
24 min ki video bna di iss question pr aur baccho ko paka dala
@@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
agreed
@@kingdomalmighty119 chotu 2 v rkhena tab v 1/4 hi ayega
@@nehakushwah7158 kuchh bhi
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.
Great, a wonderful explanation!
Me too.... Sir
You know mathematics brother..
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.
Satisfactory words!
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.
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.
sin(x-1) -2(x-1)
@@fireians2809 follow the graph of y = sin x vs y = x
(1-a)^2=1
1-a=+_1
Sice1-a= positive value
So, 1-a=1hoga
Then a=0
a = 2 rakhane par base -ve hoga vahi a = 0 rakhane par base +ve hoga
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,,,,, 💪💪💪💪🦾
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 .
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)
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.
Apke charan kidhar ha sir😐🤣
@@iitianaakash5582 What happened?
Bro hatsoff
@@anish8373 Thanks bhai
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.
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
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""
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
It's my first comment on RUclips
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 🙏
YOU ARE CORRECT 👍 ANSWER WILL BE 0
Negative
Non negative
Positive
All are different from each other
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
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
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
a ki value rakhni hai naa ki, x ki
Sorry bhai a ki hi😅
Kon kon Ashish agrawal sir ki limit ki prayash 1.0 ki class attend karke ye video dekha raha😂😂😂😂
@unknown-Mbps 😂😂😂chummu muche bhi iit bana do
As final ans is 1/4 , which is only possible when (k)^2 ., if 0
0 is currect ans because if we put 0, in place of (a) then we get right hand side equal to left hand side.
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.
In my opinion , they tell non negative neither negative nor positive , this reason is applicable in my thought.
Sir a=0 because it satisfies the given condition that the equation is equal to 1/4. Am I correct sir
1/4 or 1/2 ??
😂😂😂😂 wo to a=2 bhi karta hai🤣🤣🤣
Aise tough questions ko easy banane ke liye hi hain hamare Aman sir
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
😦😦😦
He is one of most brightest students of IIT
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..❤❤
@@kingdomalmighty119 how the fuck it is simple to find value 0 and 2
@@TusharKumar-ty8kh when you're preparing IIT exams That time, these type of questions are common.
Sir put the a=2 in the previous equation and get .
Exponential ka base negative kaise hoga that's why we reject 2
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 ❤️
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 .
Sir after getting a=2or0 then
(1-a)^2=1 for a=2 (-1)^2=1
for a=0 (1)^2=1
In the limits of form f(x)^g(x) mey f(x) -ve nahi hona hai a=2 sey -ve aaya hai tho a=0 sayi hota hai sir
If we take log on both sides and solve
We get log(2/1-a)=log2
Now only a=0 is the solution
If we put the value of a=2, our answer of the question will be undefined because√2is irretional number.
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
sir if we put a=2 in the given sum it comes (3/2)^2 which is equal to 9/4 but not equal to 1/4.while a=0 satisfy Lim x tends 1 [sin (x-1)/(x-1) +a]/ [sin(x-1)/(x-1) +1] whole raise to the power 1+√x comes equal to 1/4 .Hence right answer is a=0.Here we are being unable to apply L- Hospital rule.Perhaps i may be wrong through this interpetation.Kindly reply
Sir there might be an error in the approach. After applying L-hospitals rule on the power term, we arrive at (.....)² = 1/4
Now, the problem mentioned actually has "a" in a linear form, which means there is only one correct value for it. Therefore the right way would be to take square root on both sides, which would give us the answer.
Squaring the equation would turn a linear equation to a quadratic one, adding another solution, which is incorrect.
@@Seriouslyfunny1 Exactly right beta.Thanks.
Sahi me?? But asish sir ne kal karwaya ye question or jaida hard bhi nhi laga 🙄😅
Lakshya wallah
@@uttarabisoi3232 hm offcourse
Sir what I think is that a=2 pe jo base he woh negative ho jaega. Though power even he but still variable power variable me base positive hota he joki a=2 pe nahi balki a=0 pe hi hota he. Also in exponential chapter when we study a power x we study that a should be positive for a power x to be defined so I think that's why answer is a=0
Good explanation, In a function: y=a^x, we know a>0, & a not =1. So, here, on putting a=2, the limit becomes (-ve number)^(1+root x) which is not a defined function coz base isn't defined.
But for a=2, it becomes (+ve number)^(1+root x) which is a defined function becoz base is +ve . So obviously a=0 because a=2 doesn't belong to domain of base of the given function
@@shubhajyotidebnath5651 thanks dude. You are a jee aspirant or what
@@harshkumargupta8831 I am not jee aspi... just as normal as everyone
@@harshkumargupta8831 BTW welcome.
@@shubhajyotidebnath5651 okay
14:04 at the moment he knew who is king of maths.
a 2 or 0.
but for a= 2 base of above limit approaches be negative hence limit does not exist.
Therefore a=0
Ashish aggarwaal sir bol rhe the aisa kuch nhi hai ki sirf bansal sir hi solve kr paye the
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.
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..❤❤
Mujhe math nhi pr kahani sunne ke lie 24 mint ka video dekha
If I put a=2 in equation I get (-1)^√x which is not defined so a=0 is correct to satisfied the equation. I hope this is correct 🙏 Mathematics lover
Ha vo jee advance 2014 wala limits k que na Ashis aggarwal sir ne bhi kar ya hai class me lakshya jee❤aur aisi koi baat nhi hai ki sirf bansal sir hi kr par th
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
Exactly
Mr. K.S. SINGH, JRS Tutorials, Varanasi. I bet there is no one today like him so versatile so experienced and so savage.
Bhai..why savage
sir a=2 defined nahi haii kyuki vo [f(x)]^g(x) ki form me haii aur ye tabhi defined haii agar f(x)≥0 pr a=2 dalne pr f(x) negative ho jaega
Ashish sir op
Sir ye question f(×)^g(×) ka concept use ho sakta hai for this f(×)must be greater than zero then this will be defined but a=2 it is -1/2 which is neagtive so it ia not defined so rejected .. i think
Sir ji,
After taking the limit, we get
(1--a)power 2/2 ki pwer =1/4
Here only a=0 satisfied the equation only,
Sir a^x ma a>0 hona chahiye aur a=2 rakhana pr value -1/2 aata hai aur a=0 rakhane pr value 1/2 aata hai
Sir if a=2 then inner function becomes {1-x+sin(x-1)/sin(x-1)+(x-1)}^1+root x which does not satisfy condition that LHL=RHL
sir 2 we will not considered because for a =2 base of abve limit approches to -1/2 and sir exponent approches to 2 and we know that base cannot be negative hence limit doesn't exist
hence correct is 0
bhai base negative kyu nahi ho skta
plzz batao