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Brilliant Sub Junior Calculus Contest (Season-1) Note-3

This is the third note to the season of Brilliant Sub Junior Calculus Contest (Season-1).

Click here for First Note (if you want to visit it)

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Hi everyone! This is the first season of Brilliant Sub Junior Calculus Contest. This contest is for beginners or intermediate ones who want to sharpen their skills of problem-solving in overall calculus.

The aim of this contest is to improve the skill of in the computation in all sorts of problem (of basic level) in calculus like integrals (both definite and indefinite), differentiation, limits, ecetra by learning from each other and of course to have fun!

Eligibility:- People should fulfill either of the 2 following

  • 17 years or below

  • Level 4 or below in Calculus

    Eligible people here may participate in this contest.

The rules are as follows:

  • I will start by posting the first problem. If there is a user solves it, then they must post a new one.

  • You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

  • Only make substantial comment that will contribute to the discussion.

  • Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

  • If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

  • The scope of questions is only computation of basic level problems in calculus.

  • It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

  • You are also NOT allowed to post a solution using a contour integration or residue method.

Answer shouldn't contain any Special Function.

Please post your solution and your proposed problem in a single new thread.

Format your post is as follows:

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2
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5
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**SOLUTION OF PROBLEM xxx (number of problem) :**

**[Post your solution here]**

**PROBLEM xxx (number of problem) :**

**[Post your problem here]**

The comments will be easiest to follow if you sort by "Newest":

Note by Akshay Yadav
1 year, 7 months ago

No vote yet
1 vote

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Comments

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Problem 40 :

Evaluate \[\int_{0}^{1} \dfrac{5x^3+3x-1}{(x^3+3x+1)^3} dx\]

Samuel Jones - 1 year, 7 months ago

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\(I=\displaystyle \int { \frac { -({ x }^{ 3 }+3x+1)+6{ x }^{ 3 }+6x }{ { ({ x }^{ 3 }+3x+1) }^{ 3 } } dx } \)

This can be further written as,

\(I=\displaystyle \int { \frac { -{ ({ x }^{ 3 }+3x+1) }^{ 2 }+{ 2x({ x }^{ 3 }+3x+1) }({ 3x }^{ 2 }+3) }{ { ({ x }^{ 3 }+3x+1) }^{ 4 } } dx }\)

This is like the derivative form of \( \displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 }} \)

The integral simplifies to,

\(I=\displaystyle \frac { -x }{ { ({ x }^{ 3 } }+3x+1)^{ 2 } } +constant\)

Now putting the limits \(0\) to \(1\) we get the value as \(-1/25\)

Saarthak Marathe - 1 year, 7 months ago

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Just for enlightening, I'm posting my method too.

\[ \begin{aligned}\int\frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3}\, dx &=\int\frac{5x^3 + 3x-1}{\left(\sqrt{x}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)\right)^3}\,dx\\&=\int\frac{5x^3 + 3x-1}{x^{3/2}\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{5x^{3/2}+3x^{-1/2}-x^{-3/2}}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\, dx\\&=\int\frac{2\,d(x^{5/2}+3\sqrt{x}+x^{-1/2})}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^3}\\& -\frac{1}{\left(x^{5/2}+3\sqrt{x}+x^{-1/2}\right)^2}+C\\& \end{aligned} \]

\[ \implies \int_{0}^{1} \frac{5x^3 + 3x - 1}{\left(x^3 + 3x + 1\right)^3} dx = \boxed{-\dfrac{1}{25}} \]

Samuel Jones - 1 year, 7 months ago

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Nice method.

Saarthak Marathe - 1 year, 7 months ago

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Problem 41:

Evaluate

\[\displaystyle \int _{0}^{\pi} \frac{1}{ (5+\cos (x))^{\frac{1}{3}}} dx\]

Saarthak Marathe - 1 year, 7 months ago

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@Saarthak Marathe Can you post your solution to this question? Thanks.

Samuel Jones - 1 year, 7 months ago

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To be honest,I don't know how to solve this one. When I first saw it,I thought it would be easy. But found out later that it is quite tough. It uses some kind of special function to solve. I managed to bring it to an algebraic form,

\(I=\displaystyle \int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { x } \sqrt { 1-x } { (2+x) }^{ 1/3 } } } \)

I have posted this question on Brilliant hoping to get a less complicated solution to this question.

Saarthak Marathe - 1 year, 7 months ago

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@Saarthak Marathe @Saarthak Marathe That goes against the rules. Please post questions that you know the solution to from next time.

Samuel Jones - 1 year, 7 months ago

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@Saarthak Marathe Why did you change the problem? I solved the original problem and came here to post the solution, only to find that it has been changed.

Samuel Jones - 1 year, 7 months ago

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Hmm was it changed? I mean he gave a problem which was there in brilliant and he posted another one which is this one.

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma @Aditya Sharma @Saarthak Marathe Earlier problem was to evaluate \(\displaystyle \int_{0}^{\frac{\pi}{2}} \ln (\sin^2 x +k^2 \cos^2 x) \ dx\)

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones Ya. This was pointed out to be a problem on brilliant with a name "the one with log and trigonometry" (maybe). So he changed the problem

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Oh ok. Btw, do you know why alternate solutions have been deleted in JEE contest?

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones The contest which sandeep sir started? I thnik only particular solutions are kept for each problem.

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Why is that so? I think if a solution is entirely different from the accepted one, it shouldn't be deleted.

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones I do agree with u personally in that case, maybe the moderator wasn't too much careful while removing the solution. If sandeep sir or Calvin sir is informed they will rectify it

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma I have informed them. Hope they'll resolve the issue.

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones Yeah and @Saarthak Marathe should post the solution as the allotted time is over.

Aditya Narayan Sharma - 1 year, 7 months ago

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I think the solution should be posted

Aditya Narayan Sharma - 1 year, 7 months ago

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@Akshay Yadav @Nihar Mahajan @Samuel Jones what you guyz suggest about continuing the contest as @Saarthak Marathe hasn't posted the solution yet . I think someone should post a new problem and keep it going !

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Since Sarthak Marathe has exceeded the time limit, from this moment onwards anyone can post the new question.

Akshay Yadav - 1 year, 7 months ago

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Can you verify the problem?

Aditya Kumar - 1 year, 7 months ago

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@Aditya Kumar The problem is correct.

Saarthak Marathe - 1 year, 7 months ago

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Hey, Can you lift the bar on the level. I would also like to solve a few good problems

Vignesh S - 1 year, 7 months ago

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Problem 46:

\(\displaystyle \int_{-\pi}^{\pi} \dfrac{2x(1+sinx)}{1+{cos}^{2}x}dx \)

Saarthak Marathe - 1 year, 7 months ago

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Problem 43:

Evaluate

\[ \huge \int_{0}^{\frac{\pi}{2}} \dfrac{\ln (\cos x)}{\sin x} dx\]

Samuel Jones - 1 year, 7 months ago

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Substitute \(\cos x = t\), the integral becomes

\(\displaystyle \text{I} = \int_{0}^{1} \dfrac{\log t}{1-t^2} \ dt \)

\(\displaystyle = \sum_{n=0}^{\infty} \int_{0}^{1} t^{2n} \log t \ dt \quad \left(\text{Using infinite G.P.} \right)\)

\(\displaystyle \int_{0}^{1} t^{2n} \log t \ dt = - \dfrac{1}{(2n+1)^2} \) which can be shown using integration by parts many times.

\(\displaystyle \implies \text{I} = - \sum_{n=0}^{\infty} \dfrac{1}{(2n+1)^2} \)

\(\displaystyle = - \left(\sum_{n=1}^{\infty} \dfrac{1}{n^2} - \sum_{n=1}^{\infty} \dfrac{1}{(2n)^2} \right) \)

\(\displaystyle = - \dfrac{3}{4} \sum_{n=1}^{\infty} \dfrac{1}{n^2}\)

\(\displaystyle = \boxed{-\dfrac{\pi^2}{8}}\)

Samuel Jones - 1 year, 7 months ago

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Nice observation samuels (+1)!!

Samarth Agarwal - 1 year, 7 months ago

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You should post the solution now.

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Sharma I've posted the solution, but you post the next problem, since you had solved it first.

Samuel Jones - 1 year, 7 months ago

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\(\text{Let the integral be J = } \int_{0}^{\frac{\pi}{2}} \frac{log(cosx)}{sinx}dx\)

\(\text{Put cosx = t } \)

\(-sinx dx=dt\implies dx = -\frac{dt}{\sqrt{1-t^2}}\)

\(\text{We have } J = -\int_{0}^{1}\frac{ln(t)}{1-t^2}\)

\(\text{Put ln(t) = u }\implies \frac{1}{t}dt=du\implies dt=tdu=e^u du\)

\(J = -\int_{\infty}^{0} \frac{xe^x}{1-e^{2x}}dx = -\frac{1}{2}\int_{\infty}^{0} \frac{x[(e^x+1)+(e^x-1)]}{1-e^{2x}}dx\)

\(2J = -\int_{\infty}^{0} [\frac{x}{1-e^x} - \frac{x}{e^x+1}]dx\)

\(2J = -\color{red}{\underbrace{\int_{0}^{\infty} [\frac{x}{e^x-1}}} + \color{blue}{\underbrace{\frac{x}{e^x+1}]dx}}\)

\(\text{In this step we are going to implement some of the special functions , their definitions are as below : }\)

\(\text{Polygarithm Function :}\)

\(\large Li_{s}(z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}-1}dx\)

\(\large -Li_{s}(-z) = \frac{1}{\Gamma({s})}\int_{0}^{\infty} \frac{x^{s-1}}{\frac{e^x}{z}+1}dx\)

\(\text{The special values where s=2 & z=1,-1 are : } Li_{2}(1) = \frac{(\pi)^2}{6} , Li_{2}(-1) = -\frac{(\pi)^2}{12}\)

\(-2J = \color{red}{\underbrace{Li_{2}(1)}} + \color{blue}{\underbrace{-Li_{2}(-1)}}\implies -2J = \frac{(\pi)^2}{6}+\frac{(\pi)^2}{12}\implies \boxed{J=-\frac{\pi^2}{8}}\)

\(\text{There might be an method to do this by elementary functions . Please post if anyone gets this solved. }\)

Aditya Narayan Sharma - 1 year, 7 months ago

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Hey,some of my friends told me that it is either quite hard or not possible to solve this without using special functions,does your solution also use any such functions? @Samuel Jones

Adarsh Kumar - 1 year, 7 months ago

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@Aditya Sharma @Adarsh Kumar Do you want me to write the solution, or do I wait until you guys figure out your mistake ?

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones Let the alotted time be over, then post your solution. Lets see if we can make it in that time @Samuel Jones

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Sure.

Samuel Jones - 1 year, 7 months ago

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I have solved without special functions.

Samuel Jones - 1 year, 7 months ago

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@Samuel Jones Is it the answer ?

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma No, the answer is \( -\dfrac{\pi^2}{8}\)

Samuel Jones - 1 year, 7 months ago

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Yeah I tried it , maybe i m wrong but are special functions required? I too have the same question as adarsh

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma I saw your solution, but in the step \(\frac {ln(t)}{1-t^2}\), you have forgotten a negative sign when substituting. Therefore your answer is positive rather than negative. I will give a solution without special functions though so just wait for a while.

Eddy Li - 1 year, 7 months ago

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@Eddy Li Thanks! I will fix it, I wasn't sure where I have missed the sign.

Aditya Narayan Sharma - 1 year, 7 months ago

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\(\mathbf{Problem }\) \(\mathbf{ 42}\)

\(\text{Integrate : } \large \int_{0}^{1} \frac{x^4(1-x)^4}{1+x^2}dx\)

Aditya Narayan Sharma - 1 year, 7 months ago

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Using partial fractions,

\( \displaystyle \int_{0}^{1} \dfrac{x^4(1-x)^4}{1+x^2} dx = \int_{0}^{1} \left( x^6 -4x^5 +5x^4 -4x^2 - \dfrac{4}{x^2+1} +4 \right) dx\)

\( = \boxed{\dfrac{22}{7} - \pi} \)

Interesting Note : Since the integrand is positive, the integral must also be positive, giving us \(\pi < \dfrac{22}{7}\) which is the widely used approximation for \(\pi\) .

Samuel Jones - 1 year, 7 months ago

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It is a JEE problem yeah?(This came in the year my sister wrote IIT :P)

Adarsh Kumar - 1 year, 7 months ago

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Oh you identified :p no need to mention the year, yes it's a Jee Advanced problem

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma haha I had seen it some time before!That is why I was able to identify,i won't mention don't worry!

Adarsh Kumar - 1 year, 7 months ago

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@Adarsh Kumar Haha it is worth sharing IIT JEE problems

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Yeah.

Harsh Shrivastava - 1 year, 7 months ago

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@Harsh Shrivastava I just completed reading the solution from a book!:P So,i guess i would wait sometime,in case somebody else is posting.

Adarsh Kumar - 1 year, 7 months ago

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@Adarsh Kumar Is the solution very hard?

Harsh Shrivastava - 1 year, 7 months ago

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@Harsh Shrivastava Nah,just requires a small trick!

Adarsh Kumar - 1 year, 7 months ago

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@Adarsh Kumar Yeah basically the method is to decompose and expand smartly

Aditya Narayan Sharma - 1 year, 7 months ago

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@Aditya Narayan Sharma Exactly,I will post the solution by night,if nobody else does,then we will solve whole night!If you are free !

Adarsh Kumar - 1 year, 7 months ago

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@Adarsh Kumar Ya ofcourse !! If it's about maths and brilliant nothing comes more imp than it @mention532456:Adarsh Kumar . we will have a night maths party :-)

Aditya Narayan Sharma - 1 year, 7 months ago

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Problem 45:

\(\displaystyle \int { \frac { x }{ { (7x-10-{ x }^{ 2 }) }^{ 3/2 } } dx } \)

Saarthak Marathe - 1 year, 7 months ago

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Solution:

\(I=\displaystyle \int { \frac { x }{ ({ (2-x)(x-5) })^{ 3/2 } } dx } \)

\(I= \displaystyle \int { \frac { x }{ \frac { { (x-5) }^{ 3/2 } }{ { (2-x) }^{ 3/2 } } { (2-x) }^{ 3 } } dx } \)

Take \(t=\frac{x-5}{2-x} \) which gives \(x=\frac{2t+5}{1+t} \)

\( I=\displaystyle \frac { 1 }{ 9 } \int { \frac { \frac { (2t+5) }{ (1+t) } }{ { t }^{ 3/2 }\frac { 1 }{ (1+t) } } } dt \)

This simplifies to ,

\(I=\displaystyle \frac { 2 }{ 9 } \int { { t }^{ -1/2 }dt+\frac { 5 }{ 9 } \int { { t }^{ -3/2 }dt } } \)

This on integration gives,

\(I=\displaystyle \frac { 4 }{ 9 } { t }^{ 1/2 }-\frac { 10 }{ 9 } { t }^{ -1/2 } \)

Resubstituting \(t=\frac{x-5}{2-x}\) we get our answer.

Saarthak Marathe - 1 year, 7 months ago

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@Akshay Yadav Should I post the solution to this problem ,as no one has posted one yet ?

Saarthak Marathe - 1 year, 7 months ago

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Go ahead, post the solution. The time limit has been already exceeded for anyone to solve it.

Akshay Yadav - 1 year, 7 months ago

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@Akshay Yadav @Akshay Yadav Should I post the next question?

Saarthak Marathe - 1 year, 7 months ago

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\(\text{Problem 44 :}\)

\(\large \int \frac{dx}{\sec^2x+\tan^2x}\)

Aditya Narayan Sharma - 1 year, 7 months ago

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Take \( tanx=t\)

Then the integral becomes,

\( I=\displaystyle \int { \frac { dt }{ (1+{ t }^{ 2 })(1+{ 2t }^{ 2 }) } } \)

Now separate the terms,

\(I=\displaystyle 2\int { \frac { dt }{ (1+2{ t }^{ 2 }) } } -\int { \frac { dt }{ ({ t }^{ 2 }+1) } } \)

which on integration and resubstituting the value \(t= \tan x\) comes out as,

\(\sqrt { 2 } { \tan }^{ -1 }(\sqrt { 2 } \tan x)-x+C \)

Saarthak Marathe - 1 year, 7 months ago

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Absolutely , Post the next problem

Aditya Narayan Sharma - 1 year, 7 months ago

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