Question

Let A = {1, 2, 3, 4, 5, ..., 10} and f : A → A be an invertible function. Then, $\sum _{r=1}^{10}\left(f^{-1}of\right)\left(r\right)=$___________.

Answer 1

Given: f : A → A is an invertible function, where A = {1, 2, 3, 4, 5, ..., 10}

$$\begin{gathered} \mathrm{Since},f\mathrm{is}\mathrm{invertible} \\ \mathrm{Therefore},f^{-1}of\left(x\right)=x...\left(1\right) \\ \mathrm{Now}, \\ \sum _{r=1}^{10}\left(f^{-1}of\right)\left(r\right)=f^{-1}of\left(1\right)+f^{-1}of\left(2\right)+f^{-1}of\left(3\right)+....+f^{-1}of\left(10\right) \\ =1+2+3+....+10\left(\mathrm{From}\left(1\right)\right) \\ =\frac{10\left(10+1\right)}{2}\left(\because 1+2+3+...+n=\frac{n\left(n+1\right)}{2}\right) \\ =5\left(11\right) \\ =55 \end{gathered}$$


Hence, $\sum _{r=1}^{10}\left(f^{-1}of\right)\left(r\right)=$ 55.