Rajdeep's Message Board


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

Note by Rajdeep Dhingra
4 years, 5 months ago

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</code>...<code></code> ... <code>.">   Easy Math Editor

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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 - 4 years, 3 months ago

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sorry

Rajdeep Dhingra - 4 years, 3 months ago

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@Vighnesh Raut The solution to your question dx1+tan4x\int{\frac{dx}{1 + \tan^{4}{x}}} is given below.

Solution : dx1+tan4xSubstitute tan x = t . Then the integral will become dt(1+t2)(1+t4) Separating using partial fractions we get(12(1+t2)+2t242(t2+2t1)+2t+242(t2+2t+1))Integrating we get12arctant142log(t22t+1)+142log(t2+2t+1)We can easily substitute back in t = tan x and then simplify.\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 - 4 years, 3 months ago

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Sorry for misprint here ϕ=φ\phi = \varphi

Now If body is in equilibrium. Resultant = 0

So that means mgsinθ=Tcosϕmg\sin{\theta} = T\cos{\phi}

Plugging in values and solving we get

T=10×9.8×sin45cos25T = \frac{10 \times 9.8 \times \sin{45^{\circ}}}{\cos{25^{\circ}}} If Doubt comment otherwise reply that you understood.

Rajdeep Dhingra - 4 years, 3 months ago

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

Rishabh Tripathi - 4 years, 3 months ago

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Your welcome!

Rajdeep Dhingra - 4 years, 3 months ago

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@Archit Boobna Please tell me your FTRE marks

Rajdeep Dhingra - 4 years, 5 months ago

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333

Archit Boobna - 4 years, 5 months ago

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Archit Boobna In your new question Be Careful.

To evaluate C can the integers be same.

Like-(1)1000=1also11000=1(-1)^{1000} = 1 \\ also \quad 1^{1000} = 1.

Do we have take both of them as different cases or same.

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra Read the question carefully. The values of nn are redundant. Only the integers which can be expressed as n100 , nZn^{100}~,~n\in \mathbb{Z} are to be taken into consideration. You cannot express (1)(-1) as n100n^{100} for any nZn\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 - 4 years, 4 months ago

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

Also look at the other question . posted in this note

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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@Prasun Biswas Thanks @Prasun Biswas

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna You're welcome. :)

Prasun Biswas - 4 years, 4 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 2s2s. Prove that ss < AP+BP+CPAP + BP + CP < 2s2s

Also could you give a solution for tanveen dhingra Problem Above

Rajdeep Dhingra - 4 years, 4 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 22 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=2sAB+BC+AC=2s. Using the property specified in the beginning, we have,

AP+BP>AB,AP+CP>AC,BP+CP>BCAP+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)>(AB+BC+AC)=2s    AP+BP+CP>2s2=s2(AP+BP+CP)\gt (AB+BC+AC)=2s \implies AP+BP+CP\gt \dfrac{2s}{2}=s

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

When PP is a point inside Δ\DeltaABCABC, we always have,

AP<AB  AP<ACBP<BC  BP<ABCP<AC  CP<BCAP\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<AB+BC+ACAP+BP+CP<AC+AB+BCAP+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<AB+BC+AC=2sAP+BP+CP\lt AB+BC+AC=2s

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

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

s<AP+BP+CP<2ss\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 - 4 years, 4 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 - 4 years, 4 months ago

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@Rajdeep Dhingra I have updated the solution to include the proof of the upper bound.

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas what is \land and \lor

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra \land denotes the logical "AND" operator and \lor denotes the logical "OR" operator.

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas Could you please solve @tanveen dhingra Question.

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 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 - 4 years, 4 months ago

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@Rajdeep Dhingra Just Use Binomial

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna It won't work.Already tried.

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra x=21990mod1990x=1024199mod1990x1024=(10241991024)mod1990x1024=1024(10241981)mod1990Tofindit:1024(10241981)=1990n+a512(10241981)=995n+a2512(219801)=995n+a2x={ 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(10241981)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 - 4 years, 4 months ago

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@Archit Boobna @Archit Boobna How would factorization help?

2198012^{1980} - 1 = (29901)(2990+1)(2^{990}-1)(2^{990} + 1) = (24951)(2495+1)(2990+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 - 4 years, 4 months ago

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@Rajdeep Dhingra But my approach is halfway to the solution

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna Yes , But how to prove 2198012^{1980} - 1 mod995\mod{995} = 0\boxed{0}

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra http://www.wolframalpha.com/input/?i=Factorize+x%5E1980-1

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna But during the exam we can't use wolfram alpha

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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@Rajdeep Dhingra 219801≢0(mod1990)2^{1980}-1\not \equiv 0 \pmod{1990}

I checked using the Chinese Remainder Theorem. The result comes out to be 21980996(mod1990)2^{1980}\equiv 996 \pmod{1990}

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas but the answer is coming right using this assumption

Rajdeep Dhingra - 4 years, 4 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 219801(mod995)2^{1980}\equiv 1\pmod{995}. That doesn't necessarily mean that we'll also have 219801(mod1990)2^{1980}\equiv 1\pmod{1990}.

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas I said 2^1980-1 mod 995 =0

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna Oh sorry. That was a typo on my part. :P

Prasun Biswas - 4 years, 4 months ago

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@Rajdeep Dhingra By the way, 219801(mod995)2^{1980}\equiv 1\pmod{995}, so @Archit Boobna's method is correct (as far as I can see).

Prasun Biswas - 4 years, 4 months ago

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@Prasun Biswas So you mean 219802^{1980} is a factor of 995

Rajdeep Dhingra - 4 years, 4 months ago

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@Prasun Biswas http://www.wolframalpha.com/input/?i=2%5E1980-1+mod+995&dataset= @Prasun Biswas

Archit Boobna - 4 years, 4 months ago

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@Rajdeep Dhingra you are online

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna I am trying to solve it

Archit Boobna - 4 years, 4 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 - 4 years, 4 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 - 4 years, 4 months ago

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@Rajdeep Dhingra Done

Archit Boobna - 4 years, 4 months ago

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@Archit Boobna I am just writing it in daum

Archit Boobna - 4 years, 4 months ago

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@Prasun Biswas Prasun Biswas Try my new integration problem here

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months ago

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do you like S.S.T?

Adarsh Kumar - 4 years, 5 months ago

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Archit Boobna Calvin lin liked and reshared my integration cauchy

a a b b

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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Well a neat diagram and simple geometry would do it.

Look at this diagram.

Yo Yo Yo Yo

Here APAP is the crease along which paper is to be folded.Hence MM becomes the reflection of BB

Since MM is the reflection of BB hence BN=NMBN=NM.

Also let side of square be xx then MC=x2MC=\frac{x}{2}

Also tan(θ)=12,sec(θ)=52tan(\theta)=\frac{1}{2},\Rightarrow sec(\theta)=\frac{\sqrt{5}}{2}

Now BM=5x2BM=\frac{\sqrt{5}x}{2} (By applying pythagorean theorom in BMCBMC)

Since NN is the mid-point of BMBM hence BN=5x4BN=\frac{\sqrt{5}x}{4}

Finally in triangle BNPBNP by simple trignometry we have :

BNsec(θ)=BP=5x4×52=5x8BNsec(\theta)=BP=\frac{\sqrt{5}x}{4} \times \frac{\sqrt{5}}{2} = \frac{5x}{8}

PC=BCBP=3x8PC=BC-BP=\frac{3x}{8}

Finally we are getting PC:PB::3:5PC : PB :: 3:5

Hence proved.

Ronak Agarwal - 4 years, 4 months ago

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Thanks!!

Rajdeep Dhingra - 4 years, 4 months ago

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And by the way how did you make the figure.

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra Paint.

Ronak Agarwal - 4 years, 4 months ago

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@Ronak Agarwal And how did you mark the θ\theta and the perpendicular sign

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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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.

Calvin Lin Staff - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months ago

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

Satvik Golechha - 4 years, 4 months ago

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Its a RMO 1990 problem

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

Rajdeep Dhingra - 4 years, 4 months ago

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Could You Guys help me?

Find the remainder when 219902^{1990} is divided by 1990.

@Calvin Lin @Satvik Golechha @Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Sreejato Bhattacharya

tanveen dhingra - 4 years, 4 months ago

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You can see the solution here

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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@Calvin Lin Please help

Rajdeep Dhingra - 4 years, 4 months ago

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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.

Calvin Lin Staff - 4 years, 4 months ago

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Thank you very much

Rajdeep Dhingra - 4 years, 4 months ago

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

tanveen dhingra - 4 years, 4 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 - 4 years, 4 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 - 4 years, 4 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 - 4 years, 4 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 - 4 years, 4 months ago

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

Tanveen Dhingra - 4 years, 4 months ago

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

Sorry for the trouble

tanveen dhingra - 4 years, 4 months ago

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If a,b,c and d are any 4 positive real numbers, then prove that ab+bc+cd+da4\frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \ge 4

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

Rajdeep Dhingra - 4 years, 4 months ago

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Simply apply A.MG.M.A.M \ge G.M.

Ronak Agarwal - 4 years, 4 months ago

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@Ronak Agarwal Have you tried my problem

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra Yep, that was too easy.

Ronak Agarwal - 4 years, 4 months ago

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@Rajdeep Dhingra You should try this Easy Integral

@Rajdeep Dhingra

Ronak Agarwal - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months ago

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

Rajdeep Dhingra - 4 years, 4 months ago

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@Rajdeep Dhingra No,No apply A.M.H.M.A.M. \ge H.M. to quantities ab,bc,cd,da\frac{a}{b},\frac{b}{c},\frac{c}{d},\frac{d}{a} to directly get the result.

Ronak Agarwal - 4 years, 4 months ago

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@Ronak Agarwal Thank you

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 3 months ago

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@Rajdeep Dhingra Here's a hint consider : F(n,m)=0π/2sinnxcosmxdxF(n,m) = \displaystyle \int _{ 0 }^{ \pi /2 }{ \sin ^{ n }{ x } \cos ^{ m }{ x } dx }

Ronak Agarwal - 4 years, 3 months ago

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@Ronak Agarwal @Ronak Agarwal How is this related to your question ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra No further hints.

Ronak Agarwal - 4 years, 3 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 - 4 years, 3 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 - 4 years, 3 months ago

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@Ronak Agarwal Could you help me with this limni=0n1F(i) \displaystyle \lim _{ n\rightarrow \infty }{ \sum _{ i = 0 }^{ n }{ \frac { 1 }{ F(i) } } } .

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

F(0)=0,F(1)=1,F(2)=1,F(3)=2,,F(n)=F(n1)+F(n2)F(0) = 0 , F(1) = 1 , F(2) = 1 , F(3) = 2 , \dots \dots , F(n) = F(n-1) + F(n-2)

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra I am working on it.

Ronak Agarwal - 4 years, 3 months ago

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@Ronak Agarwal Are you a moderator? Have you added the solution to your problem ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra Yes.

Ronak Agarwal - 4 years, 3 months ago

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@Ronak Agarwal Yes to both ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra To only first question.

Ronak Agarwal - 4 years, 3 months ago

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@Rajdeep Dhingra I have added solution to my problem. You can check it out @Rajdeep Dhingra

Ronak Agarwal - 4 years, 3 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 - 4 years, 3 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 , ζ(3) \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 - 4 years, 3 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 2.2.212.213.2152.2.2^{\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- ζ(3)\zeta{(3)} was just a joke. Please don't take it seriously and please keep helping me Ronak agarwal. I won't reapeat the ζ(3)\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 - 4 years, 3 months ago

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@Ronak Agarwal So that means F(n,m)=12B[12(n+1),12(m+1)]F(n,m) = \frac{1}{2}B[\frac{1}{2}(n+1),\frac{1}{2}(m+1)]. Now what ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra Differentiate both sides of the expression first with respect to mm and then with respect to nn and then put m=n=0m=n=0

Ronak Agarwal - 4 years, 3 months ago

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@Ronak Agarwal How to differentiate Beta function

Rajdeep Dhingra - 4 years, 3 months ago

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You can also use the Rearrangement Inequality. That trivializes the problem.

Consider the following sequences: {a,b,c,d}\{a,b,c,d\} and {1d,1c,1b,1a}\left \{\dfrac{1}{d},\dfrac{1}{c},\dfrac{1}{b},\dfrac{1}{a}\right \} of positive reals and without loss of generality, assume abcda\geq b\geq c\geq d. Then, we also have 1d1c1b1a\dfrac{1}{d}\geq \dfrac{1}{c}\geq \dfrac{1}{b}\geq \dfrac{1}{a}

Using the Rearrangement Inequality, we have,

Directed SumRandom SumReverse Sum    ad+bc+cb+daab+bc+cd+daaa+bb+cc+dd=1+1+1+1=4\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

ab+bc+cd+da4\therefore \frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq 4

and equality holds iff a=b=c=da=b=c=d

Prasun Biswas - 4 years, 4 months ago

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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 DirectedSumac+ba+db+cdReverseSumDirected Sum \ge \frac{a}{c} + \frac{b}{a} + \frac{d}{b} + \frac{c}{d} \ge Reverse Sum

Help @Ronak Agarwal @Calvin Lin

Rajdeep Dhingra - 4 years, 4 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 - 4 years, 4 months ago

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Simple use of Rearrangemet inequality.

siddharth bhatt - 4 years, 3 months ago

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Which question ?

Rajdeep Dhingra - 4 years, 3 months ago

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

Rajdeep Dhingra - 4 years, 3 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 - 4 years, 4 months ago

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The sum is, xiaiyizix\equiv \displaystyle \sum_i a_iy_iz_i where ziyi1(modni)z_i\equiv y_i^{-1}\pmod{n_i} for the ithi^{\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 - 4 years, 4 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 - 4 years, 4 months ago

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@Rajdeep Dhingra Determine the largest 3-digit prime factor of the integer (20001000){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 - 4 years, 4 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 n4n^{4} + 4n4^{n} is composite for all integer value of nn >\gt 11

Please help.@Prasun Biswas @Ronak Agarwal @Rajdeep Dhingra @Satvik Golechha

Micheal Faraday - 4 years, 4 months ago

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Interesting Question. On it.

Rajdeep Dhingra - 4 years, 4 months ago

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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 22 ). 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+42n+1(2n + 1)^{4} \quad + \quad 4^{2n + 1} \Rightarrow 16n4+32n3+24n2+8n+1+42n+116n^4 + 32n^3 + 24n^2 + 8n + 1 + 4^{2n+1}

4(4n4+8n3+6n2+2n+42n4n^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,

(4n4+8n3+6n2+2n+42n4n^4 + 8n^3 + 6n^2 + 2n + 4^{2n}) mod10\mod{ 10 } .

now the cyclicity is 10\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 - 4 years, 4 months ago

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Here's the solution to your problem.

Prasun Biswas - 4 years, 4 months ago

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Prasun Have you checked it out !

Will you participate ? Check the edit in it.

Rajdeep Dhingra - 4 years, 3 months ago

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Please Find 1+x4(1x4)3/2dx\int{\frac{1+ x^4}{( 1 -x^4)^{3/2}}dx} Please help @Ronak Agarwal @Prasun Biswas @megh choksi

Rajdeep Dhingra - 4 years, 3 months ago

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That's a simple one : Put x=sinθ \displaystyle x=\sqrt{sin\theta} to get :

I=12(2sec2θ1)cscθdθ \displaystyle I=\frac { 1 }{ 2 } \int { (2\sec ^{ 2 }{ \theta } -1)\sqrt { \csc { \theta } } d\theta }

I=12((2sec2θcscθdθcscθdθ)\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 :

u=cscθ,v=sec2θ \displaystyle u = \sqrt { \csc { \theta } } , v'=\sec ^{ 2 }{ \theta }

I=12(2tanθcscθ2tanθcscθcotθ2cscθcscθdθ)+C \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

I=12(2tanθcscθ+cscθdθcscθdθ)+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

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

Reverting back to our substitution we have :

I=x1x4+C I = \dfrac { x }{ \sqrt { 1-{ x }^{ 4 } } } +C

Ronak Agarwal - 4 years, 3 months ago

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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

Rajdeep Dhingra - 4 years, 3 months ago

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Nishi Sofat Arushi Sofat

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

Rajdeep Dhingra - 4 years, 3 months ago

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Will you please tell me how did you get thiscorrect?

Gautam Sharma - 4 years, 3 months ago

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

Gautam Sharma - 4 years, 3 months ago

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

here is the solution:

Rajdeep Dhingra - 4 years, 3 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 - 4 years, 3 months ago

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@Rajdeep Dhingra Can You please help me with this ?

Mehul Arora - 4 years, 3 months ago

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Have you tried solving it ?

Rajdeep Dhingra - 4 years, 3 months ago

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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....

Mehul Arora - 4 years, 3 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 - 4 years, 3 months ago

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@Rajdeep Dhingra Sure.. Im ready

Mehul Arora - 4 years, 3 months ago

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@Mehul Arora Should I tell you the next step ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra yeah.. plzzz

Mehul Arora - 4 years, 3 months ago

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@Mehul Arora Ok Now the question says f(1)=2f(2)=3f(3)=4f(4)=5f(5)=6\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 f(n)=n+1\large f(n) = n + 1
Now we can set up are polynomial
f(x)=C(x1)(x2)(x3)(x4)(x5)+x+1\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 - 4 years, 3 months ago

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@Rajdeep Dhingra Ohk... then why is the answer asked in the form of m/n?

Mehul Arora - 4 years, 3 months ago

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@Mehul Arora Because the constant is in the form of ab\dfrac{a}{b}

Rajdeep Dhingra - 4 years, 3 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 - 4 years, 3 months ago

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@Mehul Arora ok
Now we have to find C from f(x)C \text{ from } f(x).
They have given us the value of the function at f(8)f(8) which is f(8)=81=x1f(8) = 8-1 = x-1.
So now we have f(x)=C(x1)(x2)(x3)(x4)(x5)+x+1f(8)=C(81)(82)(83)(84)(85)+8+1Plugging in value of the function at 8 , we get7=C7!2!+9Finding value of C we getC=(2)2!7!Now our polynomial changes to nf(x)=(2)(2!7!)(x1)(x2)(x3)(x4)(x5)+x+1Plugging 9 we getf(9)=143\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 - 4 years, 3 months ago

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@Rajdeep Dhingra how should I start on this one?

Mehul Arora - 4 years, 3 months ago

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@Mehul Arora Done it ?

Rajdeep Dhingra - 4 years, 3 months ago

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@Rajdeep Dhingra on it .wait pls :)

Mehul Arora - 4 years, 3 months ago

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@Mehul Arora See any n degree polynomial can be expressed in the form P(x)=C(xa1)(xa2)(xa3).........(xan)+kP(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 - 4 years, 3 months ago

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Hint: Use the Remainder-Factor Theorem for polynomials.

Extra hint: Try forming a new polynomial whose roots are the given xx's.

Prasun Biswas - 4 years, 3 months ago

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

Rajdeep Dhingra - 4 years, 3 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 - 4 years, 3 months ago

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

Rajdeep Dhingra - 4 years, 3 months ago

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

lnL=limx1xi=1xf(ix)\ln L=\lim_{x\to\infty}\frac{1}{x}\cdot \sum_{i=1}^x f\left(\frac{i}{x}\right)