Animated Solution for Mathematics - Functions: Let $S=\{1,2,3,4,5,6,7,8,9,10\}$. Define $f:S\to S$ as $f(n)=\{\begin{array}{ll}2n,& \text{if }n=1,2,3,4,5\\ 2n-11,& \text{if }n=6,7,8,9,10\end{array}$. Let $g:S\to S$ be a function such that $fog(n)=\{\begin{array}{ll}n+1,& \text{if }n\text{ is odd}\\ n-1,& \text{if }n\text{ is even}\end{array}$, then $g(10)((g(1)+g(2)+g(3)+g(4)+g(5))$ is equal to:

Welcome, future engineer. Today, we are not just solving a function problem; we are decoding a secret language. When you look at a problem involving composite functions like $f(g(n))$, it is easy to feel overwhelmed.

Take a deep breath. We are going to break this down into a simple, elegant dance of numbers.

Imagine $f(n)$ as a machine with two distinct personalities. It is a piecewise function defined on the set $S=\{1,2,3,4,5,6,7,8,9,10\}$.

For the first half of our set, $n\in \{1,2,3,4,5\}$, the machine simply doubles the input:

For the second half, $n\in \{6,7,8,9,10\}$, it doubles the input and subtracts eleven:

Before we do anything else, let's map this out. If you input $1$, you get $2$. If you input $6$, you get $2(6)-11=1$.

This mapping is your anchor. It is the physical reality of our function. Never underestimate the power of writing down these mappings; it prevents the most common errors in JEE Advanced problems.

We are given $f(g(n))=h(n)$, where $h(n)$ is defined by the parity of $n$:

This looks intimidating, but remember: $f$ is a bijection. This means it is a perfect one-to-one correspondence.

Because it is a bijection, it has an inverse, $f^{-1}$. We can peel away the layers of this composite function by applying the inverse:

This is the key to the kingdom. We don't need to find a general formula for $g(n)$. We just need to find the inverse of $f$ for specific values of $h(n)$.

Let's calculate the values we need: $g(1),g(2),g(3),g(4),g(5),$ and $g(10)$.

For $n=1$ (odd), $h(1)=1+1=2$. We need $f^{-1}(2)$. Since $f(1)=2$, we have $f^{-1}(2)=1$. Thus, $g(1)=1$.

For $n=2$ (even), $h(2)=2-1=1$. We need $f^{-1}(1)$. Since $f(6)=1$, we have $f^{-1}(1)=6$. Thus, $g(2)=6$.

For $n=3$ (odd), $h(3)=3+1=4$. We need $f^{-1}(4)$. Since $f(2)=4$, we have $f^{-1}(4)=2$. Thus, $g(3)=2$.

For $n=4$ (even), $h(4)=4-1=3$. We need $f^{-1}(3)$. Since $f(7)=3$, we have $f^{-1}(3)=7$. Thus, $g(4)=7$.

For $n=5$ (odd), $h(5)=5+1=6$. We need $f^{-1}(6)$. Since $f(3)=6$, we have $f^{-1}(6)=3$. Thus, $g(5)=3$.

Finally, for $n=10$ (even), $h(10)=10-1=9$. We need $f^{-1}(9)$. Since $f(10)=9$, we have $f^{-1}(9)=10$. Thus, $g(10)=10$.

We have all our pieces. The problem asks for the value of:

Substituting our values:

Summing the terms inside the bracket:

Finally, we compute the product:

See how the complexity melted away? By systematically mapping the function and using the inverse property, we turned a daunting composite function problem into a simple arithmetic exercise. Keep this systematic approach in your toolkit, and no function will ever intimidate you again.