Question

If , f(x) + f(x-1) = 10n,
Then find the value of ,

f(1/2)+f(2/10)+f(3/10)+.... ..+f(9/10) ​

Answer 1

GiveN :-

• $\textsf{If f(x) + f(1-x) = 10 (correct) }$

To FinD :-

• The value of

$\sf f\bigg( \dfrac{1}{10}\bigg) + f\bigg( \dfrac{2}{10}\bigg)+f\bigg( \dfrac{3}{10}\bigg) +... .+ f\bigg( \dfrac{9}{10}\bigg)$

SolutioN :-

Here we need to find the value of the given expression . We can firstly simply the equation. Let's simplify out .

$\sf\to f\bigg( \dfrac{1}{10}\bigg) + f\bigg( \dfrac{2}{10}\bigg)+f\bigg( \dfrac{3}{10}\bigg) +... . f\bigg( \dfrac{9}{10}\bigg) \\\\\sf\to f\bigg( \dfrac{1}{10}\bigg)+ f\bigg( \dfrac{9}{10}\bigg) +f\bigg( \dfrac{2}{10}\bigg)+ f\bigg( \dfrac{8}{10}\bigg)+ .. .. .. +\red{ f\bigg( \dfrac{5}{10}\bigg) }$

Till now , we basically grouped out the term whose the value of x inside the function is equal to 1. Note that the term highlighted with red cannot be grouped since the number of terms is odd that is 9 .

$\sf\to \underset{\blue{\sf A }}{\underbrace{\bigg[ f\bigg( \dfrac{1}{10}\bigg) +f\bigg(1- \dfrac{1}{10}\bigg) \bigg]}}+\underset{\blue{\sf B }}{\underbrace{\bigg[ f\bigg( \dfrac{2}{10}\bigg) + f\bigg( 1-\dfrac{2}{10}\bigg) \bigg] }}+ .. .. .. + f\bigg( \dfrac{1}{2}\bigg)$

Now here see that the terms A and B are in similar form as of $\textsf{f(x) + f(1-x) = 10 }$ . So, This can be written as 10 . Here there are 4 terms . So we can write is as 4×10 .

$\sf\to 10\times 4 + f\bigg( \dfrac{1}{2}\bigg) \\\\\sf\to 40 + f\bigg( \dfrac{1}{2}\bigg) \\\\\sf\to\boxed{\red{\tt 40+f\bigg( \dfrac{1}{2}\bigg)}}$

$\rule{200}2$

Now we must find the value of f(½) .So for that substitute x = ½ in the given function .

$\sf:\implies \pink{ f(x)+f(1-x)=10}\\\\\sf:\implies f\bigg(\dfrac{1}{2}\bigg) +f\bigg(1-\dfrac{1}{2}\bigg)=10\\\\\sf:\implies f\bigg(\dfrac{1}{2}\bigg) + f\bigg(\dfrac{1}{2}\bigg) = 10 \\\\\sf:\implies 2f\bigg(\dfrac{1}{2}\bigg) =10\\\\\sf:\implies f\bigg(\dfrac{1}{2}\bigg) =\dfrac{10}{2}\\\\\sf:\implies\boxed{\orange{\tt f\bigg(\dfrac{1}{2}\bigg) = 5 }}$

$\rule{200}2$

Now put on the respective value .

$\sf\to 40 + f\bigg( \dfrac{1}{2}\bigg) \\\\\sf\to 40+5 \\\\\sf\to\pink{45}$

$\rule{200}2$

$\red{\bigstar}\underline{\textsf{ Therefore our reqiured answer is :- }}$

$\large\boxed{\purple{\tt f\bigg( \dfrac{1}{10}\bigg) + f\bigg( \dfrac{2}{10}\bigg) + ... .. + f\bigg( \dfrac{9}{10}\bigg)=\boxed{\orange{45}} }}$

$\rule{200}2$