We could do it by one more way , I.e , pehle hamm numerator ko o to pie ka interval Kar lete , or phir numerator Mei baki jo bachta woh pie to n pie interval rehta jo ki ek increasing interval hye or f(x) bhi positive hi hye . Toh yeh saari type (t+y)/t bann jati yahan par t and y both greater then zero.. So limit Hamari greater then 1 , a jaati.. Just 15 seconds mein ho Jaata..
The term you are integrating in the numerator is always positive, so the integral from 0 to n* pi or the area under the curve for this positive function will always increase with n. So the area under the curve i.e. integral for a larger n will be more than for a smaller n regardless of what the a is. So if n=> infinity then area is greater than n=1 (the denominator), ratio is >1 for any a.
Amazing question Nd explanations
We could do it by one more way , I.e , pehle hamm numerator ko o to pie ka interval Kar lete , or phir numerator Mei baki jo bachta woh pie to n pie interval rehta jo ki ek increasing interval hye or f(x) bhi positive hi hye .
Toh yeh saari type (t+y)/t bann jati yahan par t and y both greater then zero..
So limit Hamari greater then 1 , a jaati..
Just 15 seconds mein ho Jaata..
Yes, You are right!. I appreciate your effort 😊.
@JEEBeyondNumbers 👍
The term you are integrating in the numerator is always positive, so the integral from 0 to n* pi or the area under the curve for this positive function will always increase with n. So the area under the curve i.e. integral for a larger n will be more than for a smaller n regardless of what the a is. So if n=> infinity then area is greater than n=1 (the denominator), ratio is >1 for any a.
Yes it's a much simpler way to solve the problem.
thank you please keep uploading more helps a lot in revision :)
Which city do you teach
I don't teach for now. I only make these UA-cam videos to help JEE aspirants.