Let S = \ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10\ . Define f: S → S as f(n)...

Given, f\left(n\right) = \left\{\begin{cases}2n, & ifn = 1, 2, 3, 4, 5 \\ 2n - 11 & ifn = 6, 7, 8, 9, 10\end{cases}\right)</p><p>So,f\left(1\right) = 2$,$f\left(2\right) = 4$, .....$f\left(5\right) = 10$</p><p>And$f\left(6\right) = 1$,$f\left(7\right) = 3$,$f\left(8\right) = 5$......$f\left(10\right) = 9$</p><p>$f\left(g\left(n\right)\right) = \left\{$\begin{cases}n + 1 & ; & n \in odd \\ n - 1 & ; & n \in even\end{cases}$\right)

So, $f\left(g\left(10\right)\right) = 9 ⇒ g\left(10\right) = 10$

$f\left(g\left(1\right)\right) = 2 ⇒ g\left(1\right) = 1$

$f\left(g\left(2\right)\right) = 1 ⇒ g\left(2\right) = 6$

$f\left(g\left(3\right)\right) = 4 ⇒ g\left(3\right) = 2$

$f\left(g\left(4\right)\right) = 3 ⇒ g\left(4\right) = 7$

$f\left(g\left(5\right)\right) = 6 ⇒ g\left(5\right) = 3$

So, $g\left(10\right) \cdot \left[g\left(1\right) + g\left(2\right) + g\left(3\right) + g\left(4\right) + g\left(5\right)\right]$

$= 10 \cdot \left[1 + 6 + 2 + 7 + 3\right] = 190$