# CONStant

Let the function f(x) be positive on the interval [a,b] . Prove that the expression
$\int_{a}^{b} f(x) dx \cdot$$\int_{a}^{b}\frac{dx}{f(x)}$ reaches the least value when f(x) is constant on the interval

# Calculas – periodic Function

let f : $\mathbb{R}\implies\mathbb{R}$ be a monotonic function for which there exist a,b , c, d $\in$ with a $\neq$ 0 , c $\neq$ 0 such that for all x the following equalities hold :
$\int_{x}^{x+\sqrt{3}}$f(t)dt = ax+b and $\int_{x}^{x+\sqrt{2}}$f(t)dt =cx+d.
Prove that f is a linear function , i.e, f(x)=mx+n for some m,n$\in$$\mathbb{R}$

Solution (AOPS) :

#6VPM
neelanjan wrote:
let f : $\mathbb{R}\implies\mathbb{R}$ be a monotonic function for which there exist a,b , c, d $\in$ with a $\neq$ 0 , c $\neq$ 0 such that for all x the following equalities hold :
$\int_{x}^{x+\sqrt{3}}$f(t)dt = ax+b and $\int_{x}^{x+\sqrt{2}}$f(t)dt =cx+d.
Prove that f is a linear function , i.e, f(x)=mx+n for some m,n$\in$$\mathbb{R}$

$\int_{x}^{x+\sqrt{3}}f(t)dt=ax+b;\int_{x}^{x+\sqrt{2}}f(t)dt =cx+d$

$\\ \Rightarrow \left ( \int_{x}^{x+\sqrt{3}}f(t)dt \right )'=(ax+b)' \Leftrightarrow f(x+\sqrt{3})-f(x)=a;f(x+\sqrt{2})-f(x)=c$

$f(x+\sqrt{3})-f(x)=a$ g(x)=a $\Rightarrow g(x)$ is diffrenable$\Rightarrow \exists$diffrenable $(f(x+\sqrt{3})-f(x))$

Diffrenable is lim ;$\lim {(f(x+\sqrt{3})-f(x))} =\lim{f(x+\sqrt{3})}+\lim{f(x)} \Rightarrow \exists f'(x)$

$\\ \Leftrightarrow f'(x+\sqrt{3})=f'(x);f'(x+\sqrt{2})=f'(x)$

$\\ g(x)=f'(x) \Rightarrow g(x+\sqrt{2})=g(x);g(x+\sqrt{3})=g(x) \forall x \in R$

$\\ \sqrt{3};\sqrt{2}$ are period of $g(x) \Rightarrow \sqrt{3}=n\sqrt{2} (n\in Z)$

It is nonsense $\Rightarrow g(x)=const \Rightarrow f'(x)=c \Rightarrow f(x)=mx+n$

# Euler Constant

Prove that the sequence $\displaystyle{\sum_{i=1}^{n}}$ $\frac{1}{n}$ – log n is convergert . [And the constant is denoted by $\gamma$and it is still unknown wheather it is irrational or not ].

Solution :Monotony: It is easy to prove that the sequence is decreasing. Indeed:

$\gamma_{n+1} - \gamma_n = \left ( \mathcal{H}_{n+1}- \ln (n+1) \right ) - \left ( \mathcal{H}_n - \ln n \right ) = \frac{1}{n}+ \ln \left ( 1- \frac{1}{n} \right )<0$
The inequality is valid since $\ln (1-x)$ is a concave function hence lies beneath the line $y=-x$ that is its tangent to its graph at $x_0=0$. Plugging $x=1/n$ yields the result.

Boundness: Using the fact that:

$\mathcal{H}_n = \sum_{k=1}^{n}\frac{1}{k}> \int_{1}^{n}\frac{{\rm d}t}{t}= \ln (n+1)> \ln n$
we have that $\gamma_n = \mathcal{H}_n -\ln n >0$. Hence $\gamma_n$ is bounded from below.

Therefore the sequence converges.

The limit of this sequence is a very well known constant, namely Euler – Mascheroni constant.

# b.stat amd b.math 2012

1.Let n$\geq$1,and S = $\{1,2,3,,,,,,,,,,n\}$. For a function f:$S\implies S$ , a subset D, $D \subseteq S$ is said t be invarient under f , if $f(x)\in D$ . Note that the empty set and S are invariant for all f . Let deg(f) be the number of subsets of S invariant under f .
(1)Show that there is a function f :$S\implies S$ such that deg(f) = 2.
(2)Further show that for any k such that 1 $\leq$ k $\leq$ n ther is a function f:$S\implies S$ that is deg(f)=$2^k$.

Solution :Consider the function defined piecewise as f(x) = x – 1 is x ($\neq1$) and f(x) = n if x = 1

Of course null set and the entire sets are invariant subsets. We prove that there are no other invariant subsets.

Suppose D = {$(a_1 , a_2 , … , a_k$)} be an invariant subset with at least one element.

Since we are working with natural numbers only, it is possible to arrange the elements in ascending order (there is a least element by well ordering principle).

Suppose after rearrangement D = {$(b_1 , b_2 , … , b_k)$} where $(b_1$) is the least element of the set

If $(b_1$$\neq$ 1) then $(f(b_1) = b_1$ -1) is not inside D as $(b_1$) is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.

This $(b_1$) must equal to 1.

As D is invariant subset $(f(b_1) = n$) must belong to D. Again f(n) = n-1 is also in D and so on. Thus all the elements from 1 to n are in D making D=S.

Hence we have proved that degree of this function is 2.

(b)

For a natural number ‘k’ to find a function with deg(f) = $(2^k$) define the function piecewise as

f(x) = x for (1$\leq x$ $\leq(k-1)$)
= n for x=k
= x-1 for the rest of elements in ‘n’

To construct an invariant subset the ‘k-1’ elements which are identically mapped, and the entirety of the ‘k to n’ elements considered as a unit must be considered. Thus there are total k-1 + 1 elements with which subsets are to be constructed. There are $(2^k$) subsets possible.

# a inequality posted on aops

Let $a,\,b,\,c$ are non-nagetive numbers such that $ab+bc+ca>0.$ Prove that
$$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \geqslant \frac{3}{2} + \frac{9k(a^3+b^3+c^3)}{(a+b+c)^3},$$where $k = \frac{13\sqrt{13}-35}{32}.$

Solution :at first use Nesbitt inequality whih follows that
(a/b+c)+(b/c+a)+(c/a+b)$\geq$ 3/2.
Then use AM of 3-rd power $\geq$ 3 rd power of A-M which follows that

$a^3+b^3+c^3$/$(a+b+c)^3$ $\geq$ 3/27 = 1/9 ,multypling into the both sides by 9 we get whole result $\geq$ 1.And multypling by k into the both sides we get the whole result $\geq$ k. then add two inequality and we will get the nesbitt inequality as:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+k \geqslant \frac{3}{2} + \frac{9k(a^3+b^3+c^3)}{(a+b+c)^3},$$\geq$ 3/2 + k( because we have prove the second inequality , that is $\geq$ 1.
Eliminate k from the both sides , and we will obtained the nesbitt inequality(john nesbitt , 1902 , England). so it is not true not only k = $k = \frac{13\sqrt{13}-35}{32}.$ , but also any positive values of k . So , we are done

# inequality

Let $a, b, c$ be positive real numbers satisfying $a^3+b^3+c^3=a^4+b^4+c^4$. Show that
$$\frac{a}{a^2+b^3+c^3}+\frac{b}{a^3+b^2+c^3}+\frac{c}{a^3+b^3+c^2} \geq 1$$

Solution :$\sum \frac{a}{a^2+b^3+c^3}=\sum \frac{a^2}{a^3+ab^3+ac^3}$$\geq \frac{(a+b+c)^2}{a^3+b^3+c^3+\sum_{sym}a^3b}$$= \frac{(a+b+c)^2}{a^4+b^4+c^4+\sum_{sym}a^3b}$$= \frac{(a+b+c)^2}{(a+b+c)(a^3+b^3+c^3)}$$= \frac{(a+b+c)(a^4+b^4+c^4)^2}{(a^3+b^3+c^3)^3}$

But the last expression is obviously $\geq 1$ by Holder inequality . so , we are done

# IMO 1993

Problem no. 25 ( IMO 1993) Let f(x) = $x^n+ 5x^{n-1}+3$ , where $n\geq 1$ is n integer . Prove that f(x) cannot be expressed as the product of non constant polynomial with integer coefficients .
SOLUTION:(by Neelanjan Mondal )
Suppose
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x + 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x + 1)$. We show that all the a’s are divisible by 3 and using that we will establish a contradiction .[Here , we will show using only + ve sign of 3 , the other case is
f(x) = $(x^r + a_{r-1}x^{r-1} + ... + a_1x - 3)(x^s + b_{s-1}x^{s-1} + ... + b_1x - 1)$ , in which constant terms contains only – ve terms and the casee follows only the + ve case . So , we will show here only the first case] .
r and s must be greater than 1.Because
for if r = 1, then 3 is a root . Now we will make two cases :
case 1:
n is even, then it follows that 0 = $3^n +5\cdot$$3^{n- 1} + 3 = 3^{n-1}( 3 +5)+ 3$, which is false since 3+5 = 8.
case 2 :
if n is odd we would have 0 = $3^{n-1}(3 + 5) + 3$, which is false since 3 + 5 = 8.
If s = 1, then 1 is a root and we will argue by contradiction in the same way .
So r$\leq$n – 2, and hence the coefficients of x, $x^2, ... , x^r$ are all zero. Since the coefficient of x is zero, we have: $a_1+3b_1 = 0$, so $a_1$ is divisible by 3.
Now , we will use the induction hypothesis .
We can
now proceed by induction.
Assume $a_1, ... , a_t$ are all divisible by 3. Then
consider the coefficient of $x_{t+1}$. If s-1 $\geq$t+1, then $a_{t+1}$ = linear combination of $a_1,... , a_t + 3b_{t+1}$. If s-1 $\leq$ t+1, then $a_{t+1}$ = linear combination of some or all of $a_1,... , a_t.$
Either way, $a_{t+1}$ is divisible by 3.
Now we will consider the coefficients of x, $x^2$,
… , $x^{r-1}$ which gives us that all the a’s are multiples of 3. Now consider the coefficientof $x^r$ which is also zero.
Now we will get , that is a sum of of terms which are multiples of 3 .
Then it does not becomes 0.
So , follows a contradiction !
So , the factorization is not possible .( proved ).

# isi entrance problem

The following figure shows a $3^2*3^2$ grid divided into $3^2$ subgrids of size 3*3 . This grid has 81 cells , 9 in each subgrid .Now consider an $n^2*n^2$ grid divided into $n^2$ subgrids of size n*n . Find the number of ways in which you can select $n^2$ cells from the grid such that there is exactly one cell coming from each subgrid , one from each row and one from each column.

Find all functions f: $\mathbb { R }\implies\mathbb{ R }$ such that $f(x^2+yf(z)) = x f(x)+z f(y)$ , for all x , y , z in $\mathbb { R }$

Solution :Let $P(x,y,z)$ be the assertion $f(x^2+yf(z))=xf(x)+zf(y)$

$P(0,0,0)$ $\implies$ $f(0)=0$

$P(0,1,xf(x))$ $\implies$ $f(f(xf(x)))=xf(x)f(1)$
$P(0,x,x)$ $\implies$ $f(xf(x))=xf(x)$ and so $f(f(xf(x)))=xf(x)$ and so $xf(x)(f(1)-1)=0$

So, either $f(x)=0$ $\forall x\ne 0$, which gives the solution $\boxed{\text{S1 : }f(x)=0\text{ }\forall x}$
Either $f(1)=1$ and so $P(0,1,x)$ $\implies$ $f(f(x))=x$ and $f(x)$ is bijective.

Then $P(0,x,f(y))$ $\implies$ $f(xy)=f(x)f(y)$ and $f(x)$ is multiplicative.
So $P(x,0)$ $\implies$ $f(x)^2=f(x^2)=xf(x)$ and so $f(x)=x$ $\forall x\ne 0$, still true when $x=0$

And so $\boxed{\text{S2 : }f(x)=x\text{ }\forall x}$ which indeed is a solution

# differentiable function

Let f and g be two non- decreasing twice differentiable function defined on an interval (a,b) such that for each x $\in$ ( a, b) , f”(x) = g(x) and g”(x) = f(x) . Suppose also that g(x)f(x) is linear in x on ( a ,b) . Show that we must have f(x) = g(x) = 0 for all x$\in$ (a , b).