Question

If $f\left(x\right)+f(1-x)=10$, then the value of $f\left(\frac{1}{10}\right)+f\left(\frac{2}{10}\right)+.....+f\left(\frac{9}{10}\right)$

Answer 1

$f\left(x\right)+f(1-x)=10$

$f\left(\frac{1}{10}\right)+f\left(\frac{2}{10}\right)+........+f\left(\frac{9}{10}\right)=f\left(\frac{1}{10}\right)+f\left(\frac{9}{10}\right)+f\left(\frac{2}{10}\right)+f\left(\frac{8}{10}\right)+......+f\left(\frac{5}{10}\right)$

$=\left[f\right(\frac{1}{10})+f(1-\frac{1}{10}\left)\right]+\left[f\right(\frac{2}{10})+f(1-\frac{2}{10}\left)\right]+.....+f\left(\frac{5}{10}\right)$

$=4\left(10\right)+f\left(\frac{1}{2}\right)$

Now, $f\left(\frac{1}{2}\right)+f(1-\frac{1}{2})=10$

$2\left[f\right(\frac{1}{2}\left)\right]=10$

$f\left(\frac{1}{2}\right)=5$

Hence, required value is $40+5=45$.