Let the function f(x) be positive on the interval [a,b] . Prove that the expression

reaches the least value when f(x) is constant on the interval

# Calculas – periodic Function

let f : be a monotonic function for which there exist a,b , c, d with a 0 , c 0 such that for all x the following equalities hold :

f(t)dt = ax+b and f(t)dt =cx+d.

Prove that f is a linear function , i.e, f(x)=mx+n for some m,n

Solution (AOPS) :

f(t)dt = ax+b and f(t)dt =cx+d.

Prove that f is a linear function , i.e, f(x)=mx+n for some m,n

g(x)=a is diffrenablediffrenable

Diffrenable is lim ;

are period of

It is nonsense

# Euler Constant

Prove that the sequence – log n is convergert . [And the constant is denoted by and it is still unknown wheather it is irrational or not ].

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

The inequality is valid since is a concave function hence lies beneath the line that is its tangent to its graph at . Plugging yields the result.

**Boundness**: Using the fact that:

we have that . Hence 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 n1,and S = . For a function f: , a subset D, is said t be invarient under f , if . 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 : such that deg(f) = 2.

(2)Further show that for any k such that 1 k n ther is a function f: that is deg(f)=.

Solution :Consider the function defined piecewise as f(x) = x – 1 is x () 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 = {)} 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 = {} where ) is the least element of the set

If 1) then -1) is not inside D as ) is the smallest element in D. Hence D is no more an invariant subset which is contrary to our initial assumption.

This ) must equal to 1.

As D is invariant subset ) 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) = ) define the function piecewise as

f(x) = x for (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 ) subsets possible.

# a inequality posted on aops

Let are non-nagetive numbers such that Prove that

where

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 be positive real numbers satisfying . Show that

Solution :

But the last expression is obviously by Holder inequality . so , we are done

# IMO 1993

Problem no. 25 ( IMO 1993) Let f(x) = , where 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) = . 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) = , 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 = , which is false since 3+5 = 8.

case 2 :

if n is odd we would have 0 = , 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 rn – 2, and hence the coefficients of x, are all zero. Since the coefficient of x is zero, we have: , so is divisible by 3.

Now , we will use the induction hypothesis .

We can

now proceed by induction.

Assume are all divisible by 3. Then

consider the coefficient of . If s-1 t+1, then = linear combination of . If s-1 t+1, then = linear combination of some or all of

Either way, is divisible by 3.

Now we will consider the coefficients of x, ,

… , which gives us that all the a’s are multiples of 3. Now consider the coefficientof 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 grid divided into subgrids of size 3*3 . This grid has 81 cells , 9 in each subgrid .Now consider an grid divided into subgrids of size n*n . Find the number of ways in which you can select cells from the grid such that there is exactly one cell coming from each subgrid , one from each row and one from each column.

# Indian national mathematical olympiad

Find all functions f: such that , for all x , y , z in

Solution :Let be the assertion

and so and so

So, either , which gives the solution

Either and so and is bijective.

Then and is multiplicative.

So and so , still true when

And so 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 ( 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 (a , b).