Thread to share Good Questions from Calculus which requires deep thinking

#347


Aise to sarre integral zero hone chahiye na but yeh ni hai

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#348

@Yash_Srivastava2 Very well done bro, I was also searching for some integral which would oppose the conclusion arrived from improper substitution

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#349

Got it why to check that , u r right @Rohan_Shinde1 bro

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#350

Jisse tum numerator me x ko put kr sako

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#351

Without the inverse how will we be able to substitute the x in integrand? Moreover the inverse must also be differentiable within confined limits so that you replace the differential dx with appropriate substituting differential dt in our case.

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#352

Put x-1 =t
Then use Taylor series.

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#354

@Yash_Srivastava2 @Azimuddin_Sheikh

Precisely this is what we were missing. So we in fact have to find Cauchy Principle Value of the integral given.

#355

But I'm not getting how to evaluate the series.
\sum_{r=1}{\infty} \frac{1}{2^{r} r^2} plz help
I remember I did this for limit 0 to 1 and got \zeta( 2)

#356

Hey @Rohan_Shinde1 are these questions solvable by methods in jee syllabus or some out of jee course.
Asking because solving out of syllabus question won't help when 3 months is left for JEE😀

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#357

Bro in this ques i m converting limit to 0 to 1
Then getting infinite HP with x-1/2 in numerator but unable to solve further
I m doing correctly na??
@Rohan_Shinde1

#358

@pratyaksh_tyagi
The series is perfectly correct. In fact it converges to \frac {\pi^2}{12}-\frac {\ln^2 2}{2}

Well done.

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#359

Don't worry some of them, approx. 2 to 3 questions will require a little higher advanced methods but others should do fine

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#360

Bro as far as I know
That is the polylogrithmic function ,right?
Post#340
So \displaystyle P_4(1)=\sum_{k=1}^{\infty} \left(\frac{1}{k^4} \right)

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#361

How you found the sum of that series...????
@Rohan_Shinde1

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#362

@Satyendra_2019

Stupendous... Marvelous.....

Guessed it right, they are called the \color{red}{\text{Polylogarithms}}


About P_4(1)=\zeta(4)=\frac {\pi^4}{90}
So that too is correct

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#363

What about this bro ques 2 @Rohan_Shinde1

#364

Bro are these questions given to you by somebody or these are your extraworks.
Too :open_mouth:heartbeating questions
This is the link very useful

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#366

@Yash Shrivastava I didn't quite understand what you meant so I would pen down my solution. If you still get doubts feel free to ask.

\displaystyle I=\int_1^{\infty} \frac {\{x\}-\frac 12}{x}dx
\displaystyle =\lim_{n\to\infty} \left(\int_1^n dx -\int_1^n \frac {\lfloor x\rfloor +\frac 12}{x}dx\right)
\displaystyle =\lim_{n\to\infty} \left( (n-1)-\frac 12\left(\ln\left(\frac {2^3}{1^3}\cdot \frac{3^5}{2^5}\cdot \frac {4^7}{3^7}.....\frac{n^{2n+1}}{(n-1)^{2n+1}}\right)\right) \right)
\displaystyle =-1+\lim_{n\to\infty} \left(n-\frac 12\ln\left(\frac {n^{2n+1}}{(n!)^2}\right) \right)

Now just Stirling's approximation for factorials to reach the answer which is -1+\ln (\sqrt {2\pi})


@Satyendra_2019

I already have read quite a much about it. Coming to the questions, some of them are original while some are from internet or some random book.

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#367

@Rohan_Shinde1 i was doing via another method but was getting same result as in second last step but was stuck there

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#368

Those worrying about only JEE scope should then skip \color{red}{\text{Q. 6, Q.7, Q.8}} ..... I guess Q.7 is also doable

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