# 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.

$\huge \color{#D61F06}{\text{Happy}} \quad \huge \color{#3D99F6}{\text{Brillianting !}}$

Note by Rajdeep Dhingra
4 years, 9 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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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.

Staff - 4 years, 7 months ago

sorry

- 4 years, 7 months ago

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

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

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

- 4 years, 7 months ago

@Archit Boobna Please tell me your FTRE marks

- 4 years, 9 months ago

333

- 4 years, 9 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

@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

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

Thanks @Prasun Biswas

- 4 years, 8 months ago

You're welcome. :)

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

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.

- 4 years, 8 months ago

@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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

what is $\land$ and $\lor$

- 4 years, 8 months ago

$\land$ denotes the logical "AND" operator and $\lor$ denotes the logical "OR" operator.

- 4 years, 8 months ago

Could you please solve @tanveen dhingra Question.

- 4 years, 8 months ago

Well, you already gave him the link to the solution of his problem, didn't you?

- 4 years, 8 months ago

Is that solution the best and easiest.I don't think so.Could you please make it shoter

- 4 years, 8 months ago

Just Use Binomial

- 4 years, 8 months ago

It won't work.Already tried.

- 4 years, 8 months ago

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

(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

- 4 years, 8 months ago

@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

- 4 years, 8 months ago

But my approach is halfway to the solution

- 4 years, 8 months ago

Yes , But how to prove $2^{1980} - 1$ $\mod{995}$ = $\boxed{0}$

- 4 years, 8 months ago

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

- 4 years, 8 months ago

But during the exam we can't use wolfram alpha

- 4 years, 8 months ago

No! I mean to say that 2^1980-1 can be factorised... but we are just not getting how

- 4 years, 8 months ago

$2^{1980}-1\not \equiv 0 \pmod{1990}$

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

- 4 years, 8 months ago

but the answer is coming right using this assumption

- 4 years, 8 months ago

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 $2^{1980}\equiv 1\pmod{1990}$.

- 4 years, 8 months ago

I said 2^1980-1 mod 995 =0

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

So you mean $2^{1980}$ is a factor of 995

- 4 years, 8 months ago

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

- 4 years, 8 months ago

you are online

- 4 years, 8 months ago

I am trying to solve it

- 4 years, 8 months ago

By the way, i have another proof. If you will help me prove that 2^1980 mod 1990 =1 , then I can prove it

- 4 years, 8 months ago

Let me think about it.

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

- 4 years, 8 months ago

Done

- 4 years, 8 months ago

I am just writing it in daum

- 4 years, 8 months ago

Prasun Biswas Try my new integration problem here

- 4 years, 8 months ago

Rajdeep if you have any problems with my question, discuss it on the solutions page or disputes page of the problem

- 4 years, 8 months ago

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

- 4 years, 8 months ago

do you like S.S.T?

- 4 years, 9 months ago

Archit Boobna Calvin lin liked and reshared my integration cauchy

a b

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

Thanks!!

- 4 years, 8 months ago

And by the way how did you make the figure.

- 4 years, 8 months ago

Paint.

- 4 years, 8 months ago

And how did you mark the $\theta$ and the perpendicular sign

- 4 years, 8 months ago

By drawing curve feature and oval feature of paint. You know you can draw curves in paint.

- 4 years, 8 months ago

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

Staff - 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Its a RMO 1990 problem

- 4 years, 8 months ago

Could You Guys help me?

Find the remainder when $2^{1990}$ is divided by 1990.

- 4 years, 8 months ago

You can see the solution here

- 4 years, 8 months ago

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

Is there a wiki.

- 4 years, 8 months ago

- 4 years, 8 months ago

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.

Staff - 4 years, 8 months ago

Thank you very much

- 4 years, 8 months ago

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

- 4 years, 8 months ago

@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.

- 4 years, 8 months ago

I sent you an email to the email address that you used on the account. Please respond accordingly.

Staff - 4 years, 8 months ago

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.

- 4 years, 8 months ago

@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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

Sorry for the trouble

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

@Ronak Agarwal Have you tried my problem

- 4 years, 8 months ago

Yep, that was too easy.

- 4 years, 8 months ago

You should try this Easy Integral

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

Thank you

- 4 years, 8 months ago

@Ronak Agarwal Its not working.

- 4 years, 8 months ago

@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.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

@Ronak Agarwal How is this related to your question ?

- 4 years, 8 months ago

No further hints.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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 :

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

- 4 years, 8 months ago

I am working on it.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Yes.

- 4 years, 8 months ago

Yes to both ?

- 4 years, 8 months ago

To only first question.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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.

- 4 years, 7 months ago

@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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Differentiate both sides of the expression first with respect to $m$ and then with respect to $n$ and then put $m=n=0$

- 4 years, 8 months ago

How to differentiate Beta function

- 4 years, 8 months ago

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 $a=b=c=d$

- 4 years, 8 months ago

@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 $Directed Sum \ge \frac{a}{c} + \frac{b}{a} + \frac{d}{b} + \frac{c}{d} \ge Reverse Sum$

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

Simple use of Rearrangemet inequality.

- 4 years, 7 months ago

Which question ?

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

- 4 years, 8 months ago

- 4 years, 8 months ago

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! :)

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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.

- 4 years, 8 months ago

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 $n$ $\gt$ $1$

- 4 years, 8 months ago

Interesting Question. On it.

- 4 years, 8 months ago

@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.

- 4 years, 8 months ago

Here's the solution to your problem.

- 4 years, 8 months ago

Prasun Have you checked it out !

Will you participate ? Check the edit in it.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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 :

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Nishi Sofat Arushi Sofat

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

- 4 years, 7 months ago

Will you please tell me how did you get thiscorrect?

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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

here is the solution:

- 4 years, 7 months ago

Yeah i did the same but i didn't want to waste my chance and points thats why i asked.Thanks.

- 4 years, 7 months ago

@Rajdeep Dhingra Can You please help me with this ?

- 4 years, 7 months ago

Have you tried solving it ?

- 4 years, 7 months ago

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....

- 4 years, 7 months ago

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

- 4 years, 7 months ago

- 4 years, 7 months ago

Should I tell you the next step ?

- 4 years, 7 months ago

yeah.. plzzz

- 4 years, 7 months ago

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)

- 4 years, 7 months ago

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

- 4 years, 7 months ago

Because the constant is in the form of $\dfrac{a}{b}$

- 4 years, 7 months ago

not getting it :( maybe I will be able to solve more of such problems after this one

- 4 years, 7 months ago

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

- 4 years, 7 months ago

how should I start on this one?

- 4 years, 7 months ago

Done it ?

- 4 years, 7 months ago

on it .wait pls :)

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

Hint: Use the Remainder-Factor Theorem for polynomials.

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

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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.

- 4 years, 7 months ago

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

- 4 years, 7 months ago

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:
$\ln L=\int\limits_0^1\ln x\,dx$