Name: Number of Onto Functions
Uploaded: 2019-07-19
Duration: 9 min 3 s
Description: In this video, I'll show you how to prove the number of onto functions formula. This will prove handy for a better understanding of this concept.

Question

If

f (x)

be continuous function in

[0, 2 π]

and

f (0) = f (2 π)

, then prove that there exists a point

c \in (0, π)

such that

f (c) = f (c + π)

.

Accepted Answer

Let a function

$g (x) = f (x) - f (x + π)$

Put $x = 0$

$\Rightarrow g (0) = f (0) - f (0 + π)$

$= f (0) - f (π)$

Again put $x = π$

$\Rightarrow g (π) = f (π) - f (2 π)$

Given that $f (0) = f (2 π)$

$\Rightarrow g (π) = f (π) - f (0)$

$= - [f (0) - f (π)]$

$\Rightarrow g (0) g (π) = - [f (0) - f (π)]^{2}$

$\Rightarrow g (0) g (π) < 0$

$\Rightarrow g (0)$ and $g (π)$ are of opposite sign.

By intermediate value theorem

There is a point $c \in (a, b)$ for which

$g (c) = 0$

$\Rightarrow g (c) = f (c) - f (π + c) = 0$

$\Rightarrow f (c) - f (π + c) = 0$

Hence, there exist a point $c \in (a, b)$

for which $f (c) = f (π + c)$ .

If $f (x)$ be continuous function in $[0, 2 π]$ and $f (0) = f (2 π)$ , then prove that there exists a point $c \in (0, π)$ such that $f (c) = f (c + π)$ .

Important Questions on Functions

Learn and Prepare for any exam you want !

If fx be continuous function in 0, 2π and f0=f2π, then prove that there exists a point c∈0, π such that f(c)=fc+π.

Important Questions on Functions

Learn and Prepare for any exam you want !

If $f (x)$ be continuous function in $[0, 2 π]$ and $f (0) = f (2 π)$ , then prove that there exists a point $c \in (0, π)$ such that $f (c) = f (c + π)$ .