IIT JEE Main PYQ Continuity 5 April 2024 Shift 1

[@shamanjitsingh7267]

Sir, you overcomplicated it. Use this formula instead Sin3x = 3Sinx -4Sin³x then conclude alpha = -3, then limit x->0 4(sinx/x)³ = -4×1=-4

[@Tanvir2025]

Bhai lakin expansion maximum time kaam ata h so use use karna better h...

[@priyanshusodhan1382]

Thank u so much sir apki vedio helps alot

[@bhuneshwar646]

Good afternoon sir

[@surendrakverma555]

Thanks Sir

[@PS-mh8ts]

i studied math a number of years ago but i'm quite interested in math. when i was studying calculus, i was told that for a function to be continuous at x=a, it must be defined at x=a. how can the function in this given question ever be defined at x=0 because its denominator becomes 0 when x=0? my guess is something is amiss in this question.

[@eshaanuddinkhan6634]

same doubt

[@abhinavanand9032]

If left right limit exist at a then fis continuous

[@ronin632]

yes you are right about it. The function will never be defined at x=0. For example take the case of another function G(X)=(x-2)(x-3)/(x-2). This function would also not be defined at x=2 because then the denominator would become 0. although you can take out its limit which is basically the value which you get when a approach a number from both sides( left hand side(2-) and right hand side(2+)) then you would get the answer as -1 by cancelling out the x-2 term. now this question above in the video asks that for the function to be defined at x=0, what value you should assign to f(0) which would basically be its limit when x tends to 0

[@eshaanuddinkhan6634]

@@ronin632 thanks for reply

[@PS-mh8ts]

@@ronin632 😎The question is alright then & it was my own misunderstanding. Thank you for your clarification. 💯

[@vinodjoc1752]

Sir aap best ho kaha tha aabtak? ❤

[@rishinamdev9558]

thank u sir