Question

If$F(n) = {( - 1)^{k - 1}} \cdot (n - 1)$, $G(n) = n - F(n)$ then what is the value of $\left( {GoG} \right)(n)$, where $k$ is odd?

$$1)1 \\\ 2)n \\\ 3)2 \\\ 4)n - 1 \\\$$

Answer 1

Here we are given a function $G(n)$ and its relation with another function $F(n)$ is also given. We have to find the function of that function. We do this by simply getting the value of $G(n)$ using the relation given above. Then we find the value of $\left( {GoG} \right)(n)$ by using the obtained value of $G(n)$. Then we will solve further and make use of the fact that $k$ here is odd. In this way we can reach our result.

Complete step-by-step solution:
Here we are given the function $F(n)$ as,
$F(n) = {( - 1)^{k - 1}} \cdot (n - 1)$
And we are also given another function $G(n)$ as, $G(n) = n - F(n)$. We will put the value of $F(n)$ in the previous equation to get the value of $G(n)$ as,
$G(n) = n - \left( {{{( - 1)}^{k - 1}} \cdot (n - 1)} \right)$
We know that $\left( {GoG} \right)(n)$ can be written as,
$\left( {GoG} \right)(n) = G\left( {G\left( n \right)} \right)$, we will now put the value of $G(n)$ obtained from above step as,
$\Rightarrow \left( {GoG} \right)(n) = G\left( {n - \left( {{{( - 1)}^{k - 1}} \cdot (n - 1)} \right)} \right)$
We further solve it as,
$\Rightarrow \left( {GoG} \right)(n) = n - {( - 1)^{k - 1}}(n - 1) - {( - 1)^{k - 1}}((n - 1) - {( - 1)^{k - 1}}(n - 1))$
We know that ${( - 1)^{k - 1}} = 1$ as $k$ is odd and $1$ minus odd is even. So, we use this in above equation and move further ahead as,

\Rightarrow \left( {GoG} \right)(n) = n - (n - 1) - \left( {\left( {n - 1} \right) - \left( {n - 1} \right)} \right) \\\ \Rightarrow \left( {GoG} \right)(n) = n - (n - 1) \\\ \Rightarrow \left( {GoG} \right)(n) = 1 $$ Hence the value of $$\left( {GoG} \right)(n)$$ comes out to be $$1$$ **Hence the correct option is $$1)$$.** **Note:** This is to note that this is not the real value of the function $$\left( {GoG} \right)(n)$$. This value is true only when the value of $$k$$ is odd as is the case here. For the even value of $$k$$ we might have got another result. Here the function we have found, $$\left( {GoG} \right)(n)$$ is called as the function of a function and is read as ‘$$G$$ of $$G$$’.