Question

Let $S = \\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\\}$. Define $f : S → S$ as
$f(n) = \begin{cases} 2n, & \text{if } n \in \\{1,2,3,4,5\\} \\\ 2n-11, & \text{if } n \in \\{6,7,8,9,10\\} \end{cases}$
Let $g : S → S$ be a function such that.
$(f \circ g)(n) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\\ n - 1, & \text{if } n \text{ is even} \end{cases}$
Then $g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5))$ is equal to __________.

Answer 1

∵$$f(n) = \begin{cases} 2n, & \text{if } n \in \\{1,2,3,4,5\\} \\\ 2n-11, & \text{if } n \in \\{6,7,8,9,10\\} \end{cases}
$∴ f(1) = 2, f(2) = 4, … , f(5) = 10$
and $f(6) = 1, f(7) = 3, f(8) = 5, … , f(10) = 9$
Now,
$f(g(n)) = \begin{cases} n + 1, & \text{if } n \text{ is odd} \\\ n - 1, & \text{if } n \text{ is even} \end{cases}$
$∴ f(g(10)) = 9 \implies g(10) = 10$
$f(g(1)) = 2 \implies g(1) = 1$
$f(g(2)) = 1 \implies g(2) = 6$
$f(g(3)) = 4 \implies g(3) = 2$
$f(g(4)) = 3 \implies g(4) = 7$
$f(g(5)) = 6 \implies g(5) = 3$
$∴$ $g(10) \cdot (g(1) + g(2) + g(3) + g(4) + g(5)) = 190$