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## Comments

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TopNewestRajdeep, 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.

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sorry

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

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

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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.

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ohh.. thank you.. I got it.. :D :D

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Your welcome!

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@Archit Boobna Please tell me your FTRE marks

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333

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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.

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@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

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@Prasun Biswas

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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

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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

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.

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@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

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@tanveen dhingra Question.

Could you please solveLog in to reply

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(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

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@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

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I checked using the Chinese Remainder Theorem. The result comes out to be \(2^{1980}\equiv 996 \pmod{1990}\)

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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 \(2^{1980}\equiv 1\pmod{1990}\).

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@Archit Boobna's method is correct (as far as I can see).

By the way, \(2^{1980}\equiv 1\pmod{995}\), soLog in to reply

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@Prasun Biswas

http://www.wolframalpha.com/input/?i=2%5E1980-1+mod+995&dataset=Log in to reply

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The question is not of proof it is of finding the remainder

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here

Prasun Biswas Try my new integration problemLog in to reply

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@Archit Boobna Is class there tomorrow ? Tomorrow's date is 14/2/2015

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do you like S.S.T?

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Archit Boobna Calvin lin liked and reshared my integration cauchy

a

b

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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.

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Well a neat diagram and simple geometry would do it.

Look at this diagram.

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.

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Thanks!!

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And by the way how did you make the figure.

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You are much more likely to get a response by posting that as a note instead of just as a comment on your messageboard.

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Thanks for the suggestion. Anyways, do you know how to solve it.

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I think this is a previous RMO (maybe 1991) problem, you can find a solution online.

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Its a RMO 1990 problem

Tried but couldn't find a solution.Please help @Satvik Golechha

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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

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You can see the solution here

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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.

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@Calvin Lin Please help

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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.

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Thank you very much

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@Calvin Lin Sir please change my age. It is 12 years.My DOB is 14th August 2002. Thanks! In advance

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@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.

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note earlier. I wanted someone to verify whether my method is correct or not. I even tagged you there, but no response.

Can you please help me with this? I posted aLog in to reply

@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

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@Calvin Lin Sir My parents have replied you back. Sir , Please restore my account.

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

Sorry for the trouble

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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

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Simply apply \(A.M \ge G.M. \)

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@Ronak Agarwal Have you tried my problem

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Easy Integral

You should try this@Rajdeep Dhingra

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@Ronak Agarwal I solved it using hyperbolic tangent integral. How did you solve it?

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

Please Contribute a Solution.@Prasun Biswas @Calvin Lin

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How did you expand (a+b+c+d)^4

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@Ronak Agarwal Its not working.

Please help @Calvin Lin @Satvik Golechha @Sreejato Bhattacharya

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@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.

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@Ronak Agarwal How is this related to your question ?

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

\(F(0) = 0 , F(1) = 1 , F(2) = 1 , F(3) = 2 , \dots \dots , F(n) = F(n-1) + F(n-2)\)

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@Rajdeep Dhingra

I have added solution to my problem. You can check it outLog in to reply

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.

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@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.

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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.

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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 \(a=b=c=d\)

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@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

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Simple use of Rearrangemet inequality.

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Which question ?

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Could you give a algebraic solution to this \(\text{ Harmonic }\) at its best

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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

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@Prasun Biswas @Ronak Agarwal @Satvik Golechha @satvik pandey @Calvin Lin @Sreejato Bhattacharya Please help

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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! :)

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Just read the theorem and went on to try the example.

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@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.

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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

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Interesting Question. On it.

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@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.

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.

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Here's the solution to your problem.

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Prasun Have you checked it out !

Will you participate ? Check the edit in it.

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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

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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 :

\( I = \dfrac { x }{ \sqrt { 1-{ x }^{ 4 } } } +C \)

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

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Nishi Sofat Arushi Sofat

Please ask Rahul Sir to solve this.\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\]

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Will you please tell me how did you get thiscorrect?

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But the person who has posted it has written himself that this question is incorrect .Is the question correct?

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

here is the solution:

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@Rajdeep Dhingra Can You please help me with this ?

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Have you tried solving it ?

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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....

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Its gonna take time to type but if you have patience I will help you.

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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)

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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

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ndegree 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.

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Hint: Use the Remainder-Factor Theorem for polynomials.

Extra hint: Try forming a new polynomial whose roots are the given \(x\)'s.

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Hey bro . Did you solve that probability problem I had asked ?

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\[\lim _{x \rightarrow \infty}{\left( \frac{x!}{x^x} \right) ^{1/x}}\]

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

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,

\[\ln L=(0-1)-(0-0)=(-1)\implies \boxed{L=e^{-1}}\]

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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.

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@Prasun Biswas @Rajdeep Dhingra help me in starting that question . Please help.

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@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.

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rajdeep, I got my answer as 2013/2008 why did I go wrong?

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

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

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 \(a+b = 2014+2009 = \boxed{4023}\)

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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 helpLog in to reply

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P.S - I will post 3 problems tomorrow on this. Then you can check your newly learnt skill. \(\ddot\smile\)

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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.

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link.

Sorry class 10th. Here's theoffline

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@Chew-Seong Cheong Sir's solution... XD

Hahaha.. I truly understood how exhausted u are... This isLog in to reply

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Also why do you want to become a mod ? Just curious .

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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. \(\ddot\smile\)

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Frankly speaking , Megh has a better chance of becoming a mod . Look at the number of his followers !

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Could you help the guy out who posted a question on the RMO/INMO board. The \(\omega\) one.

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@Azhaghu Roopesh M , You can Solve the problem now... I have reposted it

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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.

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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.

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You might want to see this .

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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 .

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

Just help me out in this: -

ABandCDare 2 straight lines meeting atOandXYis another straight line. Show that in general 2 points can be found inXYwhich are equidistant fromABandCD. When is there only one such point ?See this

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Did you check out the whole page ?

Also I see that Shashwat has explained it to you already .

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Rajdeep try my question "I think factorization may help"

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I have already done it.

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Do "Dirty Maxima" then

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Try my new questions Find the Truth 7 and Dirty Algebra.

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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.

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How do we get that block m1 moves upwards? If we did m1 downwards we get a negative value? @Rajdeep Dhingra

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Yes. @Anik Mandal

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@Harsh Shrivastava Please post your JOMC Questions here .

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@Rajdeep Dhingra Hi Rajdeep Bhaiya !

This year I am bracing up for my NSEJS preparation, and recently I had a look over your personal tips for the exam. I saw some pics that you uploaded, wherein you said they contained the Physics IJSO syllabus and were taken from a "blog". If possible, can you please provide the link for the blog ? Also, are there some other blogs/websites related to IJSO preparation ?

DevanshiLog in to reply

All of them are removed now. They used to be maintained by Resonance.

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Oh ok, thank you so much !

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Here. Many people preparing for IJSO are there.

You can go toLog in to reply

@Rajdeep Dhingra

Hi Rajdeep Bhaiya !

I had a question - is learning calculus preferable and advantageous for NSEJS and its further stages ? If yes, then how and to what extent ?

DevanshiLog in to reply

Just very basic differentiation and integration in class 11th NCERT of Physics.

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Shall surely check it out. Thank you so much !

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@Rajdeep Dhingra

I had a doubt regarding chemistry.

What, according to your personal opinion, is something better to do - Practicing whole of chemistry from Raymond Chang + Coursera,

orstudying Class 8,9,10,11,12 NCERT, and solve BMA and Pearson Trishna Chemistry for NSEJS and further stages ?Log in to reply

11th + 12th NCERT has inorganic chemistry and organic chemistry also which don't come.(except their introductory chapter).I suggest you do Chang+Coursera and do the questions of NCERT of that chapter.

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Oh ok, thank you so much for your guidance bhaiya !

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Thank you so much !

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