Let S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Defi | vedclass

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(D) Given $f(n) = \begin{cases} 2n, & n \in \{1, 2, 3, 4, 5\} \ 2n - 11, & n \in \{6, 7, 8, 9, 10\} \end{cases}$.We find $f^{-1}(n)$ by solving $f(x)

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$190$

Vedclass · Answer

(D) Given $f(n) = \begin{cases} 2n, & n \in \{1, 2, 3, 4, 5\} \ 2n - 11, & n \in \{6, 7, 8, 9, 10\} \end{cases}$.We find $f^{-1}(n)$ by solving $f(x) = n$:If $n \in \{2, 4, 6, 8, 10\}$,$2x = n \implies x = n/2$.If $n \in \{1, 3, 5, 7, 9\}$,$2x - 11 = n \implies x = (n + 11)/2$.Thus,$f^{-1}(n) = \begin{cases} n/2, & n \in \{2, 4, 6, 8, 10\} \ (n + 11)/2, & n \in \{1, 3, 5, 7, 9\} \end{cases}$.Given $f(g(n)) = \begin{cases} n + 1, & n 	ext{ is odd} \ n - 1, & n 	ext{ is even} \end{cases}$,we have $g(n) = f^{-1}(f(g(n)))$.For $n$ odd,$g(n) = f^{-1}(n + 1)$. Since $n+1$ is even,$g(n) = (n + 1)/2$.For $n$ even,$g(n) = f^{-1}(n - 1)$. Since $n-1$ is odd,$g(n) = (n - 1 + 11)/2 = (n + 10)/2$.Calculating values:$g(1) = (1+1)/2 = 1$,$g(2) = (2+10)/2 = 6$,$g(3) = (3+1)/2 = 2$,$g(4) = (4+10)/2 = 7$,$g(5) = (5+1)/2 = 3$,$g(10) = (10+10)/2 = 10$.Therefore,$g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5)) = 10 \cdot (1 + 6 + 2 + 7 + 3) = 10 \cdot 19 = 190$.

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