Thread to share Good Questions from Calculus which requires deep thinking

#1

Please share some good questions of Calculus for JEE Advanced.

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#2

Try this, maybe not requires deep thinking

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#3

I'm getting 2e-2

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#4

Recheck calculations.
Maybe the definite integral you did wrong

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#5

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#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.

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#7

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


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

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#8

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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.

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#11

@Jagdish_Singh sir how you solved limit step?

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#14

RESONANCE STAR BATCH (DROPPER)

Pg #1

Pg#2

Pg#3

Pg#4

Pg#5

Pg#6

Pg#7

Pg#8

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#15

Please share some nice integrals ..

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#16


Level->easy to medium.

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#17

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

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#18

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

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#19

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

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#20

Whats ATP? @Anindya_Sikdar

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#21

any time padhai-- Utube channel

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#22

Ohok.

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#23

can anyone prove this identity ? Written below

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