Thread to share Good Questions from Calculus which requires deep thinking

#473

68dfbbec114b876974dd4642e3dac095ed2d32ff

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#474

@Azimuddin_Sheikh @Hari_Shankar @Viraj @Viraam_Rao @Amay_2018 help

#475

I am getting answer as \frac {sqrt(\pi)}{2}

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#476

@Azimuddin_Sheikh

Bro, what was your method? Only thing I could think of trying was reduction formula but it isnt leading me to any answer.

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#477

Did you use differentiating under integral?

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#478

I tried to bound the integrand with the Gaussian function

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#479

How bro ? Pls show

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#480

@pratyaksh_tyagi

Is this from any Allen test?

If it is, then there should be some solution with techniques within JEE syllabus right.
Gaussian function JEE syllabus mein Hain @Azimuddin_Sheikh ?

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#481

Kvpy 2019 as I rmmbr

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#482

A friend sent me this.
No gaussian function and integral out of syllabus. Maybe there is some other method

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#483

I remember this problem was in kvpy and they ask some problems of the level of isi , so we can expect to use some out of syllabus theorems for this problem

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#484

One thing we can also notice is that : Using binomial approximation, we have (for n>1),
(1+x^2)^n > 1+nx^2
Hence,
I_n=\int_0^1 \frac{dx}{(1+x^2)^n} < \int_0^1 \frac{dx}{1+nx^2} = \frac{1}{\sqrt{n}} \int_0^1 \frac{dp}{1+p^2}=\frac{\pi }{4\sqrt{n}} where p=\sqrt{n} x

Hence, \sqrt{n} I_n < \frac{\pi }{4} < 2 Hence, L<2

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Doubt from Limits
#485

Nice.
Thanks bro

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#486

Bro you didn't change the limits?
But the conclusion will be same nevertheless

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#488

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#489

IMG_20190515_204952

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#490

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#491

Thanks sir

#492

Nice problem IMG_20190516_122734

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#493

IMG_20190516_124204

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