Find the largest nonnegative a

[@PrimeNewtons]

I copied the original question incorrectly. Replace sin(1-x) with sin (x-1) in the numerator. Your final answers will be 0 and 2 instead of 0 and -2. In that case, choose 2.

[@Maths786]

No the answer is 0. But why not 2. That's the original question "Why we reject 2 and accept 0?"

[@Maths786]

I want to contact you, but I don't have anything. How can I

[@smitpatel5107]

@@Maths786 Yes, correct answer is a=0, as it was provided in the answer key..... The reason is that at a=2, and x TENDING TO 1, the exponent TENDS TO 2, not EXACTLY 2....So, the exponent is a fraction, but not 2, very close to 2.... And the thing inside the bracket, is a negative fraction, thereby making the whole number undefined..... So, a=2 is rejected, and a=0 will be accepted.....(According to Original Question)

[@abboss3456]

You might want to look at the actual Advance question again as what made that question much or less iconic was the reason behind the elimination of 2 and actual answer being 0.

[@jennymarx9228]

I'm proud to be you fan over more than almost 1.5 yr! ❤ Such a great solution!

[@user-ox4ii2bw6x]

If i was a teacher id put this on the last question on a trig final, i think it was clever!

[@dougaugustine4075]

Really great. It looked way too intimidating to think about at first. Also, I didn't know how the limit of sinA/A as A goes to 0 equaled 1.

[@fahadbinferoz6758]

10:56 shouldn't the dominator be |-2|??

[@Juman-oi5hw]

In BHANNAT MATHS channel aman sir solved it 3 years ago and he found that a=2 and a=0 ...,,,but sir take a=0 and asked a question about why we don't take a=2 ?.. and After 3 years Now in this video you got a=-2 and a=0 and we saw clearly that a=-2 ( not required) and hence a=0 is the correct answer,, what about the answer we got in bhannat maths channel ,,, which value is a=2 or a=-2

[@PrimeNewtons]

I don't know about that video you just mentioned.

[@HrishikeshRaj-qv6lr]

@@Juman-oi5hw pw wale bhannat sir

[@Juman-oi5hw]

​@HrishikeshRaj-qv6lr yess

[@Juman-oi5hw]

​@HrishikeshRaj-qv6lr yes

[@Juman-oi5hw]

​@HrishikeshRaj-qv6lr yes

[@nikhilkumar-ic9qh]

Take sin(x-1) instead of sin(1-x) in numerator only...... And 💥🤯 Jee advanced 2014 (P1)

[@Blingsss]

This problem was a great refresher on how substitutions and trig identities come together to simplify limits. The process was straightforward, but seeing how it ties into solving for the largest 𝑎 was quite satisfying. It reminded me that even small tweaks in questions, like modifying the numerator, can completely change the solution math always keeps you on your toes. By the way, when I face tricky math problems, SolutionInn’s step by step tools have been a lifesaver for understanding similar concepts.

[@rinkukoshle6537]

Sir you are great you solved it in first try correctly. Most of the teachers in our country mada mistakes while solving, there answer was 2, but the correct answer was zero by the national answer paper of this question paper. ❤❤

[@smitpatel5107]

Yes, bhannat maths channel has also posted a video on it.... Actually two..... That VK bansal sir was the only person, who gave final answer as 0, in his solution of the paper.... Also, see the pinned comment.... Prime newtons has said that answer will be 2, he copied wrong question.... So prime newtons is wrong that way.....

[@robertveith6383]

OP -- * their answer

[@AmilQarayev41]

WHERE IS THE 3RD METHOD FOR THE INTEGRALLL???

[@PrimeNewtons]

It's coming soon

[@robertveith6383]

Stop yelling your post in all caps. You are rude and childish.

[@krisbrandenberger544]

Great video 👍!

[@SriChethanVemula]

Wow, Great explanation 👏👏... Huge fan of JEE questions and this channel

[@Maths786]

No, your question is wrong. The actual question is JEE Advanced 2014 shift-1 question number 57 lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢

[@Maths786]

Actually on numerator in Sin(1-x). It should be Sin(x-1)!

[@ThenSaidHeUntoThem]

@@Maths786 I see!. If I switch it , then the answer is 2

[@Maths786]

​@@ThenSaidHeUntoThemno the answer is 0. But give reason that why we reject 2 and accept 0?

[@thebeast4216]

The thing is in the bracket we have a negative number alright and (-ve)^non integer is kind of a problem the problem is in the exponent 1+√x it is not exactly going towards 2 but towards 1.999999999 ...or 2.0000000000....1 and the thing is (-ve)^1.999999.... is actually not a real number that's why we have taken a=0 instead of 2 in short the limit does not exist as a=2 hope it helps thank you ❤❤

[@thebeast4216]

Actually answer is 0 as if we take a=2 then we are in the situation where We have (some negative no)^1.99999999.... which is not real so that's why we have to take a=0 as limit doest not exist asy a=2 at a=0 we will have (+ve no)^decimal value which is not a issue but for negative no it is hope it helps thanks

[@ElVerdaderoAbejorro]

Mate, I love your channel but you need to be more careful with your arguments as others pointed out. I've spotted dozens of small mistakes and things that are just plain wrong in your videos. The guys pointing out the fact you reach a = 0 or a = 2 but the answer is actually a = 0 is the whole reason this question became famous but you completely ignored this crucial part (I know that is because you copied the question wrong, but still, it should be rectified).

[@HrishikeshRaj-qv6lr]

Indians assemble here ❤

[@Kakarot-kr]

❤️

[@thebeast4216]

hi

[@maths01n]

Learnt ❤❤

[@giuseppemalaguti435]

1/2

[@HrishikeshRaj-qv6lr]

That's great explanation 🙌