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Rajdeep's Message Board


This is my message board. I am moderator. My job is to keep the community clean and engaged. Any types of questions can be asked here. Any irrelevant comment , note or problem(report the problem and leave the link here) can be conveyed through here or slack.

\[\huge \color{red}{\text{Happy}} \quad \huge \color{blue}{\text{Brillianting !}}\]

Note by Rajdeep Dhingra
2 years, 6 months ago

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Rajdeep, it has come to my attention that you are aggressively @mentioning several members numerous times a day. Please refrain from doing so, especially if you notice that they have not responded to any of your past mentions.

In general, unless you have a strong reason, please refrain from @ mentioning any person more than once a day. If you continually bombard people with such requests, they will (similar to the boy who cried wolf) most likely start to ignore you. Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin sorry Rajdeep Dhingra · 2 years, 5 months ago

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@Vighnesh Raut The solution to your question \(\int{\frac{dx}{1 + \tan^{4}{x}}}\) is given below.

Solution : \[\int{\frac{dx}{1+ tan^4 {x}}} \\ \text{Substitute tan x = t . Then the integral will become } \\ \int{\frac{dt}{(1 + t^2)(1 + t^4 )}} \\ \text{ Separating using partial fractions we get} \\ \int{(\frac{1}{2(1 +t^2)} + \frac{2t - \sqrt{2}}{4\sqrt{2}(-t^2 + \sqrt{2}t - 1)} + \frac{2t + \sqrt{2}}{4\sqrt{2}(t^2 + \sqrt{2}t + 1)})} \\ \text{Integrating we get} \\ \frac{1}{2}\arctan{t} - \frac{1}{4\sqrt{2}}\log{(t^2 - \sqrt{2}t + 1)} + \frac{1}{4\sqrt{2}}\log{(t^2 + \sqrt{2}t + 1}) \\ \text{We can easily substitute back in t = tan x and then simplify.}\] Rajdeep Dhingra · 2 years, 5 months ago

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@Azhaghu Roopesh M Fully agreed! @Azhaghu Roopesh M . Please try my 100 followers problem! Mehul Arora · 2 years, 4 months ago

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Sorry for misprint here \(\phi = \varphi\)

Now If body is in equilibrium. Resultant = 0

So that means \(mg\sin{\theta} = T\cos{\phi}\)

Plugging in values and solving we get

\[T = \frac{10 \times 9.8 \times \sin{45^{\circ}}}{\cos{25^{\circ}}}\] If Doubt comment otherwise reply that you understood. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra ohh.. thank you.. I got it.. :D :D Rishabh Tripathi · 2 years, 4 months ago

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@Rishabh Tripathi Your welcome! Rajdeep Dhingra · 2 years, 4 months ago

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unable to download fiitjee big bang sample papers. anybody who knows where to find them? Himanshi Sethi · 10 months ago

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@Harsh Shrivastava Please post your JOMC Questions here . Rajdeep Dhingra · 1 year, 5 months ago

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In the above solution mass 1 = 1 kg , mass 2 = 2 kg , mass 3 = 3 kg. This will make calculation easier. You can try and generalize it for variables . If any doubt persists then do ask. Rajdeep Dhingra · 1 year, 7 months ago

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@Rajdeep Dhingra How do we get that block m1 moves upwards? If we did m1 downwards we get a negative value? @Rajdeep Dhingra Anik Mandal · 1 year, 7 months ago

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@Anik Mandal Yes. @Anik Mandal Rajdeep Dhingra · 1 year, 7 months ago

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Comment deleted Dec 24, 2015

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@Rajdeep Dhingra Yes thanks for your co-operation Anik Mandal · 1 year, 7 months ago

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@Rajdeep Dhingra Well I would be happy if you post problems on Newton's sums. Swapnil Das · 2 years, 1 month ago

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Try my new questions Find the Truth 7 and Dirty Algebra. Archit Boobna · 2 years, 2 months ago

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Rajdeep try my question "I think factorization may help" Archit Boobna · 2 years, 3 months ago

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@Archit Boobna I have already done it. Rajdeep Dhingra · 2 years, 3 months ago

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@Rajdeep Dhingra Do "Dirty Maxima" then Archit Boobna · 2 years, 3 months ago

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Here's the link to the staff list , in case you missed out on my comment on Calvin sir's Message Board . Azhaghu Roopesh M · 2 years, 4 months ago

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@Azhaghu Roopesh M This is not the staff list. Also I read your comment on Sir's message board.


Just help me out in this: -

AB and CD are 2 straight lines meeting at O and XY is another straight line. Show that in general 2 points can be found in XY which are equidistant from AB and CD . When is there only one such point ?


See this Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Did you check out the whole page ?

Also I see that Shashwat has explained it to you already . Azhaghu Roopesh M · 2 years, 4 months ago

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Guys I have a problem in questions like this one.


If \(\displaystyle f \left( \dfrac{x+y}{2} \right) = \dfrac{f(x) + f(y)}{2}. \)
\(\text{Find } f(2) \text{ given } f(0) = 1 \text{ and } f'(0) = -1. \)


Also give me links to many such problems so I can practice them. Any book or wiki or PDF also will work.


Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra You might want to see this . Azhaghu Roopesh M · 2 years, 4 months ago

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Also why do you want to become a mod ? Just curious . Azhaghu Roopesh M · 2 years, 4 months ago

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@Azhaghu Roopesh M @Azhaghu Roopesh M , You can Solve the problem now... I have reposted it Mehul Arora · 2 years, 4 months ago

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@Mehul Arora It was easy but good. When I tried the same problem with the wrong answer I wondered where had I went wrong.I thought of disputing it but didn't do it as the changed problem had come. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Yeah! sorry for the inconvenience caused. Mehul Arora · 2 years, 4 months ago

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@Azhaghu Roopesh M Well I wanted to become one since I came to know about it. I thought it was cool. Well know I have changed my mind(Looking at people like you and megh choksi).
P.S - If I get a request to become one well I won't deny it. \(\ddot\smile\) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Frankly speaking , Megh has a better chance of becoming a mod . Look at the number of his followers ! Azhaghu Roopesh M · 2 years, 4 months ago

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@Azhaghu Roopesh M Yes, But I think you'll also be chosen if megh is. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Really ? We'll see . Azhaghu Roopesh M · 2 years, 4 months ago

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@Azhaghu Roopesh M Yes


Could you help the guy out who posted a question on the RMO/INMO board. The \(\omega\) one. Rajdeep Dhingra · 2 years, 4 months ago

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@Prasun Biswas @Rajdeep Dhingra help me in starting that question . Please help. Mehul Arora · 2 years, 4 months ago

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@Mehul Arora @Prasun Biswas Help him. I am exhausted. Not because of you but other things. I typed almost 50 comments each no less than 200 - 300 words. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra rajdeep, I got my answer as 2013/2008 why did I go wrong? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora From the given fact that:

\(P(1)=P(2)=P(3)=P(4)=P(5)=1\)

\(\Rightarrow P(x) = A(x-1) (x-2) (x-3) (x-4) (x-5)+1\)

Therefore,

\(\dfrac {P(2015)-1}{P(2014)-1} \)

\(= \dfrac {A(2015-1) (2015-2) (2015-3) (2015-4) (2015-5)+1-1}{A(2014-1) (2014-2) (2014-3) (2014-4) (2014-5)+1-1} \)

\(= \dfrac {A(2014) (2013) (2012) (2011) (2010)}{A(2013) (2012) (2011) (2010) (2009)} = \dfrac {2014}{2009} \)

\(\Rightarrow a = 2014\), \(b = 2009\) and \(a+b = 2014+2009 = \boxed{4023}\) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Hahaha.. I truly understood how exhausted u are... This is @Chew-Seong Cheong Sir's solution... XD Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Well yes, I didn't want to type this whole stuff out. So I did copy his. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra One last question, Is the polnomial always having +1 in the end? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora No

P.S - I will post 3 problems tomorrow on this. Then you can check your newly learnt skill. \(\ddot\smile\) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Thanks!! In those problems can You add a solution to 1 problem which helps me to understand how the polynomial expansion will end i.e. with which constant? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Well your concepts about Polynomial are scattered. No problem, Well that happens when attain higher knowledge in an unappropiate order.Well there is no particular constant, it may be 0 also.
See , A polynomial is anything in the form \(a_0 x^n + a_1 x^{n-1} + ......... + a_{n-1} x + a_n\).Provided that only \(a_0 \neq 0\). Rest anything is possible.There no particular constant. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Where can I read more about this?? Is there a brilliant wiki? or something like aops or wolfram|alpha? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Well the brilliant wiki is empty. I learnt it from class 9 NCERT. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra k.. I have that!! Which Chapter? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Sorry class 10th. Here's the link.

offline Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Hell man.... I got that as 4022 !!! Though I finally understood all this.... Thank you so much for your help @Rajdeep Dhingra :* Mehul Arora · 2 years, 4 months ago

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@Rajdeep Dhingra Can You please help me with this ? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Hint: Use the Remainder-Factor Theorem for polynomials.

Extra hint: Try forming a new polynomial whose roots are the given \(x\)'s. Prasun Biswas · 2 years, 4 months ago

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@Prasun Biswas \[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\] Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra That's an easy one. You simply need to take the limit as \(L\) and then apply natural logarithm on both sides to get,

\[\ln L=\lim_{x\to\infty}\frac{1}{x}\cdot \sum_{i=1}^x f\left(\frac{i}{x}\right)\]

This is a Riemann sum and can be evaluated by using limit sum definition of definite integral with \(f(x)=\ln x\). We proceed as follows:

\[\ln L=\int\limits_0^1\ln x\,dx\]

Using IBP, we have,

\[\ln L=\left[x\ln x-x\right]_0^1\implies \ln L=(1\times \ln 1-1)-\lim_{x\to 0^+}(x\ln x-x)\]

The limit \(\displaystyle\lim_{x\to 0^+}(x\ln x)\) can be evaluated to be \(0\) using L'Hopital's rule. Thus,

\[\ln L=(0-1)-(0-0)=(-1)\implies \boxed{L=e^{-1}}\] Prasun Biswas · 2 years, 4 months ago

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@Prasun Biswas Thanks a lot ! How about this one

You are watching four people play bridge, where a hand begins by dealing each player 13 cards.

a. After a hand is dealt, you ask a player, “Do you have at least one Ace?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

b. On a later hand, you ask another player, “Do you have the Ace of Spades?” She says, “Yes.” What is the probability that she’s holding more than one Ace?

See when we apply bayes theorem I can't understand how to evaluate P(A) and P(B).The probabilities I asked you.So, instead just solve the whole question please. Rajdeep Dhingra · 2 years, 4 months ago

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@Prasun Biswas Hey bro . Did you solve that probability problem I had asked ? Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Nope, I got busy with some NT and inequality problems. I'll look into it later. I was actually gonna solve it in the afternoon but then I had to sleep since I stayed up late last night. Prasun Biswas · 2 years, 4 months ago

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@Mehul Arora Have you tried solving it ? Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra I have a really low level of knowledge in Functions and all so can you please help me out? with this and the basics and all.... Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Ok fine.......
Its gonna take time to type but if you have patience I will help you. Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Sure.. Im ready Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Should I tell you the next step ? Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra yeah.. plzzz Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Ok Now the question says \[\large f(1)=2 \\ \large f(2)=3 \\ \large f(3)=4 \\ \large f(4)=5 \\\large f(5)=6 \]
So for values up to 5 the function is just \(\large f(n) = n + 1\)
Now we can set up are polynomial
\[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1\]
Now try the question. I am sure you'll be able to do it. If not comment.(I am going offline will be online in half an hour) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Ohk... then why is the answer asked in the form of m/n? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Because the constant is in the form of \(\dfrac{a}{b}\) Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra not getting it :( maybe I will be able to solve more of such problems after this one Mehul Arora · 2 years, 4 months ago

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@Mehul Arora ok
Now we have to find \(C \text{ from } f(x)\).
They have given us the value of the function at \(f(8)\) which is \(f(8) = 8-1 = x-1\).
So now we have \[\large f(x) = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x + 1 \\ \large f(8) = C(8-1)(8-2)(8-3)(8-4)(8-5) + 8 + 1 \\ \text{Plugging in value of the function at 8 , we get} \\ 7 = C\dfrac{7!}{2!} + 9 \\ \text{Finding value of C we get} \\ C = (-2)\dfrac{2!}{7!} \\ \text{Now our polynomial changes to n} \\ \large f(x) = (-2)(\dfrac{2!}{7!})(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) + x +1 \\ \text{Plugging 9 we get} \\ \large f(9) = \dfrac{14}{3} \]
Comment if doubt, otherwise reply that you understood. Here is a problem same as this . Try this to clear concepts Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra how should I start on this one? Mehul Arora · 2 years, 4 months ago

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@Mehul Arora Done it ? Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra on it .wait pls :) Mehul Arora · 2 years, 4 months ago

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@Mehul Arora See any n degree polynomial can be expressed in the form \[P(x) = C(x - a_1)(x - a_2)(x - a_3).........(x - a_n) + k\].
Try the problem now. If unable to solve I will help you. Rajdeep Dhingra · 2 years, 4 months ago

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Will you please tell me how did you get thiscorrect? Gautam Sharma · 2 years, 4 months ago

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@Gautam Sharma But the person who has posted it has written himself that this question is incorrect .Is the question correct? Gautam Sharma · 2 years, 4 months ago

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@Gautam Sharma Yes it is but you need to assume it in complex plane.

here is the solution:

Rajdeep Dhingra · 2 years, 4 months ago

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@Rajdeep Dhingra Yeah i did the same but i didn't want to waste my chance and points thats why i asked.Thanks. Gautam Sharma · 2 years, 4 months ago

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@Rajdeep Dhingra A square always ends in 21 41 61 01 or 81 ok Any no is in form of 10k+1....................................10k+9 Hope u can square them now Kaustubh Miglani · 1 year, 4 months ago

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Nishi Sofat Arushi Sofat

Please ask Rahul Sir to solve this.\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\] Rajdeep Dhingra · 2 years, 5 months ago

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Please Find \[\int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx}\] Please help @Ronak Agarwal @Prasun Biswas @megh choksi Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra That's a simple one : Put \( \displaystyle x=\sqrt{sin\theta} \) to get :

\( \displaystyle I=\frac { 1 }{ 2 } \int { (2\sec ^{ 2 }{ \theta } -1)\sqrt { \csc { \theta } } d\theta } \)

\(\displaystyle I = \frac { 1 }{ 2 } (\int { (2\sec ^{ 2 }{ \theta } \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } ) \)

Applying product rule to first integral we have :

\( \displaystyle u = \sqrt { \csc { \theta } } , v'=\sec ^{ 2 }{ \theta } \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } -\int { 2\tan { \theta } \frac { -\csc { \theta } \cot { \theta } }{ 2\sqrt { \csc { \theta } } } } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle I = \frac { 1 }{ 2 } (2\tan { \theta } \sqrt { \csc { \theta } } +\int { \sqrt { \csc { \theta } } d\theta } -\int { \sqrt { \csc { \theta } } d\theta } )+C \)

\( \displaystyle \Rightarrow I = \tan { \theta } \sqrt { \csc { \theta } } +C \)

Reverting back to our substitution we have :

\( I = \dfrac { x }{ \sqrt { 1-{ x }^{ 4 } } } +C \) Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal Today, I also found a new way. \[ \int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx} \\ \text{Dividing the inner things by (x^2) we get} \\ \int{\frac{\frac{1 + x^4}{x^3}}{(\frac{ 1 - x^4}{x^2})^{3/2}}dx} \\ \int{\frac{\frac{1}{x^3} + x}{(\frac{1}{x^2} + x^2)^{3/2}}dx} \\ \text{Now we have a derivative, So we substitute the down expression. Integrating and substituting back we get} \\ \frac{1}{\sqrt{\frac{1}{x^2} - x^2}} + C \\ \frac{x}{\sqrt{1 - x^4}} + C \] Rajdeep Dhingra · 2 years, 5 months ago

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Hi Guys, I read your conversation and was fascinated to see your knowledge?

Could you solve one of my questions? Ques1. Prove that \(n^{4}\) + \(4^{n}\) is composite for all integer value of \(n\) \(\gt\) \(1\)

Please help.@Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Satvik Golechha Micheal Faraday · 2 years, 5 months ago

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@Micheal Faraday Here's the solution to your problem. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Prasun Have you checked it out !

Will you participate ? Check the edit in it. Rajdeep Dhingra · 2 years, 5 months ago

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@Micheal Faraday @PRIYANSHU MISHRA For all n \(\in\) Even . It can be easily proved that it composite(As it is a multiple of \(2\) ). For the odd I Have an approach but is not complete. Still working on that .

Here is my approach for odd.

Let the odd numbers be of the form 2n + 1 where n \(\in\) Natural numbers. Then the question will change to

\((2n + 1)^{4} \quad + \quad 4^{2n + 1}\) \(\Rightarrow\) \(16n^4 + 32n^3 + 24n^2 + 8n + 1 + 4^{2n+1}\)

4(\(4n^4 + 8n^3 + 6n^2 + 2n + 4^{2n}\)) + 1 . Now If start inserting values into it then we'll see that for all n except 2 , it is divisible by 5 . Now we'll find last digit of this to prove it is divisible by 5. Last digit can be found out by doing modulo 10.So,

(\(4n^4 + 8n^3 + 6n^2 + 2n + 4^{2n}\)) \(\mod{ 10 }\) .

now the cyclicity is \(\boxed{10}\) . So we find remainders for n = 1 to 10.

Here I am getting problems.Solving this would take time . Could you simplify the mod expression.@Calvin Lin @Satvik Golechha @Peter Taylor @Prasun Biswas @Ronak Agarwal Please help. Rajdeep Dhingra · 2 years, 5 months ago

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@Micheal Faraday Interesting Question. On it. Rajdeep Dhingra · 2 years, 5 months ago

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@Prasun Biswas @Ronak Agarwal @Calvin Lin @Satvik Golechha

I have a doubt in the wiki on Chinese Remainder Theorem

I didn't understand how did this theorem is applied. In the example please tell this step.Which values we multipied in this?

Image

Image

Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra The sum is, \(x\equiv \displaystyle \sum_i a_iy_iz_i\) where \(z_i\equiv y_i^{-1}\pmod{n_i}\) for the \(i^{\textrm{th}}\) congruence of the system and the other symbols have their usual meanings as given in that wiki. Read it carefully and try the example problems yourself to get a better grasp on how to utilize CRT to solve a system of congruencies. A level 5 like you would've no problem in doing that. Cheers! :) Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Sorry for the trouble . Didn't read the section on how to use the theorem.

Just read the theorem and went on to try the example. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Determine the largest 3-digit prime factor of the integer \({2000\choose 1000}\).

@Prasun Biswas @Ronak Agarwal @Calvin Lin Help. I solved it using basic prime factorization. But I bet there has got to be a shorter solution. Rajdeep Dhingra · 2 years, 5 months ago

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If a,b,c and d are any 4 positive real numbers, then prove that \[\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4\]

@Prasun Biswas @Ronak Agarwal @Satvik Golechha @Calvin Lin Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra You can also use the Rearrangement Inequality. That trivializes the problem.

Consider the following sequences: \(\{a,b,c,d\}\) and \(\left \{\dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}\right \}\) of positive reals and without loss of generality, assume \(a\geq b\geq c\geq d\). Then, we also have \(\dfrac{1}{d}\geq \dfrac{1}{c}\geq \dfrac{1}{b}\geq \dfrac{1}{a}\)

Using the Rearrangement Inequality, we have,

\[\text{Directed Sum} \ge \text{Random Sum} \ge \text{Reverse Sum}\\ \implies \frac{a}{d}+\frac{b}{c}+\frac{c}{b}+\frac{d}{a}\geq \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq \frac{a}{a}+\frac{b}{b}+\frac{c}{c}+\frac{d}{d}=1+1+1+1=4\]

\[\therefore \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4\]

and equality holds iff \(a=b=c=d\) Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas @Prasun Biswas Nice solution.

In this Rearrangement Inequality can we change the random sum any way we like ? Should it be the Product of the squence in any order?

Like in this case \(Directed Sum \ge \frac{a}{c} + \frac{b}{a} + \frac{d}{b} + \frac{c}{d} \ge Reverse Sum \)

Help @Ronak Agarwal @Calvin Lin Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Yes, for the random sum, we choose any one of the permutations of the second sequence. There's a Brilliant wiki on the topic. You can read it if you want. Prasun Biswas · 2 years, 5 months ago

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@Rajdeep Dhingra Simply apply \(A.M \ge G.M. \) Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal How did you expand (a+b+c+d)^4 Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra No,No apply \(A.M. \ge H.M.\) to quantities \(\frac{a}{b},\frac{b}{c},\frac{c}{d},\frac{d}{a} \) to directly get the result. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal Please give a solution to the problem Logarithms, Integrals,Trigonometry

First give a hint.Like what to set as a variable to use Leibniz's integral rule. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Here's a hint consider : \(F(n,m) = \displaystyle \int _{ 0 }^{ \pi /2 }{ \sin ^{ n }{ x } \cos ^{ m }{ x } dx } \) Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal So that means \(F(n,m) = \frac{1}{2}B[\frac{1}{2}(n+1),\frac{1}{2}(m+1)]\). Now what ? Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Differentiate both sides of the expression first with respect to \(m\) and then with respect to \(n\) and then put \(m=n=0\) Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal How to differentiate Beta function Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal How is this related to your question ? Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra No further hints. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal I have tried the problem and got it incorrect and all 3 tries gone. Please help now.After i have read it you can delete it.Please give more hints. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra I am putting the whole solution in my solutions discussion, I believe that you will not like my solution since it involves gamma function, anyway you see the solution to any extent you want. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal Could you help me with this \[ \displaystyle \lim _{ n\rightarrow \infty }{ \sum _{ i = 0 }^{ n }{ \frac { 1 }{ F(i) } } } \].

\(F(n)\) is the nth Fibonacci number.Here are some values :

\(F(0) = 0 , F(1) = 1 , F(2) = 1 , F(3) = 2 , \dots \dots , F(n) = F(n-1) + F(n-2)\) Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra @Rajdeep Dhingra , Its not a good idea to ask problems for which no close form exist. This is quite famous, and called "Reciprocal Fibonacci Constant" . There is NO CLOSE form to this summation. You earlier asked , \( \zeta(3) \), which also has no close form, inspite of knowing that. Trying to waste anyone time who is helping you is not a good thing. Shivang Jindal · 2 years, 5 months ago

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@Shivang Jindal But earlier I had seen this in a problem which used this.

I don't remember the problems name (as i did it a month ago) but I remember the problem . Also there was no solution to it but had quite a many solvers.

The problem was \(2.2.2^{\frac{1}{2}}.2^{\frac{1}{3}}.2^{\frac{1}{5}}\dots \dots\) here the power is fibonacci numbers. hence to evaluate this we needed "Reciprocal Fibonacci Constant".

Sorry @Ronak Agarwal I didn't know it.

P.S- \(\zeta{(3)}\) was just a joke. Please don't take it seriously and please keep helping me Ronak agarwal. I won't reapeat the \(\zeta(3)\) thing. Thank you for helping me now in the future.

I have a doubt that I will ask you later Ronak Agarwal. I have proceeded with the solution till some extent but could you help me out with it. Hopefully will post in 3-4 hours. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra I am working on it. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal Could you solve this :

Two boxes contain between them 65 balls of several different sizes. Each ball is white, black, red or yellow. If you take any five balls of the same colour, at least 2 of them will always be of the same size. Prove that there are at least 3 balls which lie in the same box, have the same colour and the same size. Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal Are you a moderator? Have you added the solution to your problem ? Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra I have added solution to my problem. You can check it out @Rajdeep Dhingra Ronak Agarwal · 2 years, 5 months ago

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@Rajdeep Dhingra Yes. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal Yes to both ? Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra To only first question. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal Its not working.

Please help @Calvin Lin @Satvik Golechha @Sreejato Bhattacharya Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal Thank you Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal Have you tried my problem Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra You should try this Easy Integral

@Rajdeep Dhingra Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal Try this. \(\text{ Harmonic }\) at its best

Please Contribute a Solution.@Prasun Biswas @Calvin Lin Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal @Ronak Agarwal I solved it using hyperbolic tangent integral. How did you solve it? Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Yep, that was too easy. Ronak Agarwal · 2 years, 5 months ago

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@Rajdeep Dhingra Simple use of Rearrangemet inequality. Siddharth Bhatt · 2 years, 5 months ago

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@Siddharth Bhatt Could you give a algebraic solution to this \(\text{ Harmonic }\) at its best Rajdeep Dhingra · 2 years, 5 months ago

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@Siddharth Bhatt Which question ? Rajdeep Dhingra · 2 years, 5 months ago

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@Ronak Agarwal @Satvik Golechha @Prasun Biswas @megh choksi @Calvin Lin

Do you know from where I can learn about 'Mod' ,the remainder operator, and it's properties.

Is there a wiki. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Yes, there are lots of wikis on the Modular Arithmetic. These are all listed under Number Theory - Modular Arithmetic.

You can also do a search and filter by wiki, to see the relevant pages that appear. Often, the suggested page will be the most relevant to you, or it will contain the most information that will be useful to you.

Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin Thank you very much Rajdeep Dhingra · 2 years, 5 months ago

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@Calvin Lin @Calvin Lin Sir please change my age. It is 12 years.My DOB is 14th August 2002. Thanks! In advance Tanveen Dhingra · 2 years, 5 months ago

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@Tanveen Dhingra @Calvin Lin There is a mistake in the typing of my date of birth. It is actually 14th august 2001. Please restore my streak, account and points. Or transfer those things in this account. Tanveen Dhingra · 2 years, 5 months ago

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@Tanveen Dhingra I sent you an email to the email address that you used on the account. Please respond accordingly. Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin @Calvin Lin My DOB is 14th August 2002 not 2001

Sorry for the trouble Tanveen Dhingra · 2 years, 5 months ago

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@Calvin Lin @Calvin Lin Sir My parents have replied you back. Sir , Please restore my account. Tanveen Dhingra · 2 years, 5 months ago

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@Calvin Lin @Calvin Lin Sir i have sent you there Email IDs Please reach out to them Fast as i don't want to lose my streak Tanveen Dhingra · 2 years, 5 months ago

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@Calvin Lin Can you please help me with this? I posted a note earlier. I wanted someone to verify whether my method is correct or not. I even tagged you there, but no response. Prasun Biswas · 2 years, 5 months ago

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@Rajdeep Dhingra @Calvin Lin Please help Rajdeep Dhingra · 2 years, 5 months ago

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Could You Guys help me?

Find the remainder when \(2^{1990}\) is divided by 1990.

@Calvin Lin @Satvik Golechha @Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Sreejato Bhattacharya Tanveen Dhingra · 2 years, 5 months ago

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@Tanveen Dhingra You can see the solution here Rajdeep Dhingra · 2 years, 5 months ago

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A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD.Prove that the crease will divide BC in the ratio 3:5.

@Calvin Lin @Sandeep Bhardwaj @jaikirat sandhu @Prasun Biswas @Ronak Agarwal @Satvik Golechha @megh choksi @Peter Taylor @Sreejato Bhattacharya @Suyeon Khim @Agnishom Chattopadhyay @Mursalin Habib

Please help. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Well a neat diagram and simple geometry would do it.

Look at this diagram.

Yo Yo

Yo Yo

Here \(AP\) is the crease along which paper is to be folded.Hence \(M\) becomes the reflection of \(B\)

Since \(M\) is the reflection of \(B\) hence \(BN=NM\).

Also let side of square be \(x\) then \(MC=\frac{x}{2}\)

Also \(tan(\theta)=\frac{1}{2},\Rightarrow sec(\theta)=\frac{\sqrt{5}}{2}\)

Now \(BM=\frac{\sqrt{5}x}{2}\) (By applying pythagorean theorom in \(BMC\))

Since \(N\) is the mid-point of \(BM\) hence \(BN=\frac{\sqrt{5}x}{4}\)

Finally in triangle \(BNP\) by simple trignometry we have :

\(BNsec(\theta)=BP=\frac{\sqrt{5}x}{4} \times \frac{\sqrt{5}}{2} = \frac{5x}{8}\)

\(PC=BC-BP=\frac{3x}{8}\)

Finally we are getting \(PC : PB :: 3:5\)

Hence proved. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal And by the way how did you make the figure. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Paint. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal And how did you mark the \(\theta\) and the perpendicular sign Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra By drawing curve feature and oval feature of paint. You know you can draw curves in paint. Ronak Agarwal · 2 years, 5 months ago

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@Ronak Agarwal Thanks!! Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra I think this is a previous RMO (maybe 1991) problem, you can find a solution online. Satvik Golechha · 2 years, 5 months ago

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@Satvik Golechha Its a RMO 1990 problem

Tried but couldn't find a solution.Please help @Satvik Golechha Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra You are much more likely to get a response by posting that as a note instead of just as a comment on your messageboard. Calvin Lin Staff · 2 years, 5 months ago

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@Calvin Lin Thanks for the suggestion. Anyways, do you know how to solve it. Rajdeep Dhingra · 2 years, 5 months ago

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Archit Boobna Calvin lin liked and reshared my integration cauchy

a

a

b

b

Rajdeep Dhingra · 2 years, 5 months ago

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do you like S.S.T? Adarsh Kumar · 2 years, 6 months ago

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Comment deleted Jan 23, 2015

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@Rajdeep Dhingra well,same here!just like that! Adarsh Kumar · 2 years, 6 months ago

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@Archit Boobna Please tell me your FTRE marks Rajdeep Dhingra · 2 years, 6 months ago

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@Rajdeep Dhingra 333 Archit Boobna · 2 years, 6 months ago

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@Archit Boobna @Archit Boobna Is class there tomorrow ? Tomorrow's date is 14/2/2015 Rajdeep Dhingra · 2 years, 5 months ago

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@Archit Boobna Archit Boobna In your new question Be Careful.

To evaluate C can the integers be same.

Like-\((-1)^{1000} = 1 \\ also \quad 1^{1000} = 1\).

Do we have take both of them as different cases or same. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Read the question carefully. The values of \(n\) are redundant. Only the integers which can be expressed as \(n^{100}~,~n\in \mathbb{Z}\) are to be taken into consideration. You cannot express \((-1)\) as \(n^{100}\) for any \(n\in \mathbb{Z}\). Also, you count a single number once, so there are no different cases to begin with. You just consider the number of positive integers that satisfy the given conditions of C. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas @Prasun Biswas So you mean the value of C is \(\boxed{1}\) I really do feel that the answer is wrong.

Also look at the other question . posted in this note Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Nope, there is nothing wrong with that problem. I can assure you that. It's a simple one though but it made me think a bit. It was a good problem. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Thanks @Prasun Biswas Archit Boobna · 2 years, 5 months ago

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@Archit Boobna You're welcome. :) Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Prasun Biswas Could you solve the following question:

P is any point inside a triangle ABC. The perimeter of the triangle ABC is \(2s\). Prove that \(s\) < \(AP + BP + CP\) < \(2s\)

Also could you give a solution for tanveen dhingra Problem Above Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra The lower bound of the inequality can be proved easily by using the basic property of triangles that the sum of any \(2\) sides is always greater than the third side and the upper bound is to proven using the following lemma.

Lemma: Whenever we have a point inside a triangle, the distance of it from a vertex is always less than any one of the sides of the triangle originating from that vertex.

Image

Image

Given that \(AB+BC+AC=2s\). Using the property specified in the beginning, we have,

\[AP+BP\gt AB\quad,\quad AP+CP\gt AC\quad,\quad BP+CP\gt BC\]

Adding the three inequalities, we get,

\[2(AP+BP+CP)\gt (AB+BC+AC)=2s \implies AP+BP+CP\gt \dfrac{2s}{2}=s\]

\[\therefore AP+BP+CP\gt s\ldots (i)\]

When \(P\) is a point inside \(\Delta\)\(ABC\), we always have,

\[AP\lt AB~\lor~AP\lt AC \quad \land \quad BP\lt BC~\lor~BP\lt AB\quad \land \quad CP\lt AC~\lor~CP\lt BC\]

Adding the three inequalities casewise, we have,

\[AP+BP+CP\lt AB+BC+AC\quad \lor \quad AP+BP+CP\lt AC+AB+BC\]

Note that both the cases imply the same thing which is,

\[AP+BP+CP\lt AB+BC+AC=2s\]

\[\therefore AP+BP+CP\lt 2s\ldots (ii)\]

From \((i)\) and \((ii)\), we have,

\[s\lt AP+BP+CP\lt 2s\]


P.S - I still cannot think of a rigorous proof of the lemma I stated though other than analytically verifying the validity of that lemma. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas @Prasun Biswas The Lower bound is easy.

I believe that the upper bound can only be prove by drawing a mirror image

Please also solve @tanveen dhingra Question Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra I have updated the solution to include the proof of the upper bound. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas what is \(\land\) and \(\lor\) Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra \(\land\) denotes the logical "AND" operator and \(\lor\) denotes the logical "OR" operator. Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Could you please solve @tanveen dhingra Question. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Well, you already gave him the link to the solution of his problem, didn't you? Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas Prasun Biswas Try my new integration problem here Rajdeep Dhingra · 2 years, 5 months ago

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@Prasun Biswas Is that solution the best and easiest.I don't think so.Could you please make it shoter Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Just Use Binomial Archit Boobna · 2 years, 5 months ago

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@Archit Boobna By the way, i have another proof. If you will help me prove that 2^1980 mod 1990 =1 , then I can prove it Archit Boobna · 2 years, 5 months ago

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@Archit Boobna Let me think about it.

The question is not of proof it is of finding the remainder Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Done Archit Boobna · 2 years, 5 months ago

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@Archit Boobna I am just writing it in daum Archit Boobna · 2 years, 5 months ago

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@Archit Boobna It won't work.Already tried. Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra \(x={ 2 }^{ 1990 }mod\quad 1990\\ x={ 1024 }^{ 199 }mod\quad 1990\\ x-1024=({ 1024 }^{ 199 }-1024)\quad mod\quad 1990\\ x-1024=1024({ 1024 }^{ 198 }-1)\quad mod\quad 1990\\ \\ To\quad find\quad it:-\\ 1024({ 1024 }^{ 198 }-1)=1990n+a\\ 512({ 1024 }^{ 198 }-1)=995n+\frac { a }{ 2 } \\ 512({ 2 }^{ 1980 }-1)=995n+\frac { a }{ 2 } \)

(Here "a" is \(1024({ 1024 }^{ 198 }-1)\) mod 1990

Now we can easily factorize 2^1980-1 and we will see that it is a multiple of 995=5x199, so a=0.

So x-1024 is a multiple of 1990. So x mod 1990 is 1024 Archit Boobna · 2 years, 5 months ago

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@Archit Boobna @Archit Boobna How would factorization help?

\(2^{1980} - 1\) = \((2^{990}-1)(2^{990} + 1)\) = \((2^{495} - 1)(2^{495}+1)(2^{990} + 1)\)

How did this help? Even further cubic factorization isn't working. @Prasun Biswas @Calvin Lin Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra But my approach is halfway to the solution Archit Boobna · 2 years, 5 months ago

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@Archit Boobna Yes , But how to prove \(2^{1980} - 1\) \(\mod{995}\) = \(\boxed{0}\) Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra \[2^{1980}-1\not \equiv 0 \pmod{1990}\]

I checked using the Chinese Remainder Theorem. The result comes out to be \(2^{1980}\equiv 996 \pmod{1990}\) Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas http://www.wolframalpha.com/input/?i=2%5E1980-1+mod+995&dataset= @Prasun Biswas Archit Boobna · 2 years, 5 months ago

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@Prasun Biswas but the answer is coming right using this assumption Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra By the way, \(2^{1980}\equiv 1\pmod{995}\), so @Archit Boobna's method is correct (as far as I can see). Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas So you mean \(2^{1980}\) is a factor of 995 Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra Well, if the answer comes out correct, that doesn't necessarily mean that the process is correct too.

P.S - Check again. In his solution, he said that \(2^{1980}\equiv 1\pmod{995}\). That doesn't necessarily mean that we'll also have \(2^{1980}\equiv 1\pmod{1990}\). Prasun Biswas · 2 years, 5 months ago

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@Prasun Biswas I said 2^1980-1 mod 995 =0 Archit Boobna · 2 years, 5 months ago

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@Archit Boobna Oh sorry. That was a typo on my part. :P Prasun Biswas · 2 years, 5 months ago

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@Rajdeep Dhingra http://www.wolframalpha.com/input/?i=Factorize+x%5E1980-1 Archit Boobna · 2 years, 5 months ago

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@Archit Boobna But during the exam we can't use wolfram alpha Rajdeep Dhingra · 2 years, 5 months ago

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@Rajdeep Dhingra No! I mean to say that 2^1980-1 can be factorised... but we are just not getting how Archit Boobna · 2 years, 5 months ago

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@Rajdeep Dhingra you are online Archit Boobna · 2 years, 5 months ago

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@Archit Boobna I am trying to solve it Archit Boobna · 2 years, 5 months ago

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@Rajdeep Dhingra Rajdeep if you have any problems with my question, discuss it on the solutions page or disputes page of the problem Archit Boobna · 2 years, 5 months ago

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