# Thread to share Good Questions from Calculus which requires deep thinking

#389

Now its correct !! please share ur solution

1 Like
#390

Getting 0 via approximation method

#391

Put x^{2018} =t , then it converts to beta function B(a, b)
And then I think we can use the identity B(a, b) = \frac{ \Gamma (a) \Gamma (b) }{ \Gamma (a+b) }

1 Like
#392

@arush_kumar_singh only that options are given
@Rohan_Shinde1 , @pratyaksh_tyagi bro, solution is related to set theory and function
@Rohan_Shinde1 that's the beauty of question

1 Like
#393

How is it related to set theory?

2 Likes
#394

Lol approximation method

5 Likes
#395

I can't think of a better way other than Beta function. Overkills the problem within 3 lines

1 Like
#396

@Tushar_Rathore
@Yash_Srivastava2
But can make questions cakewalk

#397

Btw beta fn is not in jee syllabus na?

1 Like
#398

Nope It isnt

1 Like
#399
3 Likes
#400

I still don't get it where is the set theory? I see function part overall but not set theory?

3 Likes
#401

@Tushar_Rathore
I m not getting
How you prove that
\int _0^1 f(x) - f^{-1}(x) = 0

1 Like
#402

i wanna say relation

2 Likes
#403

Now that does seem much better. BTW nice solution.

3 Likes
#404

Try this one. This is an superb question

Find 2I-1


[.] This is only parenthesis not gif

#405

He applied the property that
\displaystyle \int_a^b f(x)dx +\int_{f(a)}^{f(b)} f^{-1}(x ) dx=bf(b)-af(a)

2 Likes
#406

0? @Tushar_Rathore

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#407

Yup, share ur solution

#408

yes it can be solved by using stirling approximation for n! you might have read it.