In∫ππ11+2sin(x/2)(sin(nx/2)sin(x/2))2 dx \large \displaystyle I_n \int_{\pi}^\pi \frac1{1 + 2^{\sin(x/2)} } \left( \frac{\sin(nx/2)}{\sin(x/2)} \right)^2 \, dx In∫ππ1+2sin(x/2)1(sin(x/2)sin(nx/2))2dx
Hi brilliant! I've a doubt in calculus! The question is above.
The options are:
A) In=In+1∀n≥1{ I }_{ n }={ I }_{ n+1 }\forall n\ge 1In=In+1∀n≥1
B) I0,I1,I2,...,InareinAP{ I }_{ 0 },{ I }_{ 1 },{ I }_{ 2 },{ ...,I }_{ n }\quad are\quad in\quad API0,I1,I2,...,InareinAP
C) ∑m=09I2m=90π\sum _{ m=0 }^{ 9 }{ { I }_{ 2m }=90\pi } ∑m=09I2m=90π
D) ∑m=010Im=55π\sum _{ m=0 }^{ 10 }{ { I }_{ m }=55\pi } ∑m=010Im=55π
More than one options are right
Please provide detail solutions!
Please don't fluke answers.
Note by Aditya Kumar 5 years, 6 months ago
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First substitute x by -x and add to get rid of the denominator. Then use the steps in the solution of this problem. This should help you.
@Pi Han Goh @Kartik Sharma @Surya Prakash @Sudeep Salgia please help!
First substitute x by -x and add to get rid of the denominator.And then evaluate I0I_0I0,I1I_1I1,I2I_2I2.... which will come out to be 0,π,2π.....0,\pi ,2\pi.....0,π,2π..... hence they will be in AP .So B,C,D.
And that problem by Sudeep can also be evaluated like that.
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Ooh nice :)
i think you have written the statement wrong it should be -pi to pi , correct it then only answers can be given @Aditya Kumar if it's -pi to pi then the answer would be B,C,D .....just correct it !
I'm sure that it's one of a, b, c or d.
Thanks for enlightening!
There is no one :(
What????
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
*italics*
or_italics_
**bold**
or__bold__
paragraph 1
paragraph 2
[example link](https://brilliant.org)
> This is a quote
\(
...\)
or\[
...\]
to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
Comments
First substitute x by -x and add to get rid of the denominator. Then use the steps in the solution of this problem. This should help you.
@Pi Han Goh @Kartik Sharma @Surya Prakash @Sudeep Salgia please help!
First substitute x by -x and add to get rid of the denominator.And then evaluate I0,I1,I2.... which will come out to be 0,π,2π..... hence they will be in AP .So B,C,D.
And that problem by Sudeep can also be evaluated like that.
Log in to reply
Ooh nice :)
i think you have written the statement wrong it should be -pi to pi , correct it then only answers can be given @Aditya Kumar if it's -pi to pi then the answer would be B,C,D .....just correct it !
I'm sure that it's one of a, b, c or d.
Log in to reply
Thanks for enlightening!
There is no one :(
Log in to reply
What????