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Question: If\[F(n) = {( - 1)^{k - 1}} \cdot (n - 1)\], \[G(n) = n - F(n)\] then what is the value of \[\left( ...

IfF(n)=(1)k1(n1)F(n) = {( - 1)^{k - 1}} \cdot (n - 1), G(n)=nF(n)G(n) = n - F(n) then what is the value of (GoG)(n)\left( {GoG} \right)(n), where kk is odd?

1)1 2)n 3)2 4)n1  1)1 \\\ 2)n \\\ 3)2 \\\ 4)n - 1 \\\
Explanation

Solution

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

Complete step-by-step solution:
Here we are given the function F(n)F(n) as,
F(n)=(1)k1(n1)F(n) = {( - 1)^{k - 1}} \cdot (n - 1)
And we are also given another function G(n)G(n) as, G(n)=nF(n)G(n) = n - F(n). We will put the value of F(n)F(n) in the previous equation to get the value of G(n)G(n) as,
G(n)=n((1)k1(n1))G(n) = n - \left( {{{( - 1)}^{k - 1}} \cdot (n - 1)} \right)
We know that (GoG)(n)\left( {GoG} \right)(n) can be written as,
(GoG)(n)=G(G(n))\left( {GoG} \right)(n) = G\left( {G\left( n \right)} \right), we will now put the value of G(n)G(n) obtained from above step as,
(GoG)(n)=G(n((1)k1(n1)))\Rightarrow \left( {GoG} \right)(n) = G\left( {n - \left( {{{( - 1)}^{k - 1}} \cdot (n - 1)} \right)} \right)
We further solve it as,
(GoG)(n)=n(1)k1(n1)(1)k1((n1)(1)k1(n1))\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)k1=1{( - 1)^{k - 1}} = 1 as kk is odd and 11 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$$’.