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

#1

Please share some good questions of Calculus for JEE Advanced.

8 Likes
#2

Try this, maybe not requires deep thinking

10 Likes
#3

I'm getting 2e-2

1 Like
#4

Recheck calculations.
Maybe the definite integral you did wrong

1 Like
#5
5 Likes
#6

Havent solved it...but i think substituting x=1/t can help.
@ananda_krishnan

Will look at it afterwards @ananda_krishnan
After i try it out.

2 Likes
#7

@Unik-Wadhwani It is from brilliant.Take a look at solution:

I loved it ..so shared it!
As yu wish @Unik-Wadhwani

5 Likes
#8
3 Likes
Doubt from limits
#9

Let \displaystyle \mathcal{S} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{\binom{n}{k}}{n^k(k+3)} = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{n!}{n^kk!(n-k)!(k+3)}

\displaystyle = \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{n(n-1)(n-2)\cdots (n-(k-1))}{n^k k!(k+3)}

\displaystyle =\sum^{\infty}_{k=0}\frac{1}{k!(k+3)} = \sum^{n}_{k=0}\frac{(k+1)(k+2)}{(k+3)!}

\displaystyle = \sum^{\infty}_{k=0}\frac{k(k+3)+2}{(k+3)!} = \sum^{\infty}_{k=0}\frac{(k+2)-2}{(k+2)!}+2\sum^{\infty}_{k=0}\frac{1}{(k+3)!}

\displaystyle =\sum^{\infty}_{k=0}\frac{1}{(k+1)!}-2\sum^{\infty}_{k=0}\frac{1}{(k+2)!}+2\sum^{\infty}_{k=0}\frac{1}{(k+3)!}

\displaystyle = (e-1)-2(e-1-1)+2\bigg(e-1-\frac{1}{2}\bigg)

So we get \displaystyle \mathcal{S} = e-2.

12 Likes
#11

@Jagdish_Singh sir how you solved limit step?

2 Likes
#15

Please share some nice integrals ..

1 Like
#16
4 Likes
#17

is it sin100x(sinx)^100/100+c?

4 Likes
#18

Yep its correct.Gud.Doing it for the 1st time or have yu done it b4?

1 Like
#19

I saw this ques at ATP 1st. But that time also I did it on my own.

2 Likes
#20

Whats ATP? @Anindya_Sikdar

1 Like
#21

any time padhai-- Utube channel

2 Likes
#22

Ohok.

1 Like
#23
4 Likes