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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?
\[
  1)1 \\
  2)n \\
  3)2 \\
  4)n - 1 \\
 \]

seo-qna
Last updated date: 22nd Jul 2024
Total views: 350.7k
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Answer
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Hint: 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 \[G\]’.