Mathematics - Functions Question with Solution | TestHub

Given, $f\left(n\right)=\left\{\begin{array}{cl}2n,& \mathrm{if} n=1,2,3,4,5\\ 2n-11& \mathrm{if} n=6,7,8,9,10\end{array}\right.$

So, $f\left(1\right)=2$, $f\left(2\right)=4$, .....$f\left(5\right)=10$

And $f\left(6\right)=1$, $f\left(7\right)=3$, $f\left(8\right)=5$ ......$f\left(10\right)=9$

$f\left(g\left(n\right)\right)=\left\{\begin{array}{lll}n+1& ;& n\in \mathrm{odd} \\ n-1& ;& n\in \mathrm{even} \end{array}\right.$

So, $f\left(g\left(10\right)\right)=9\Rightarrow g\left(10\right)=10$

$f\left(g\left(1\right)\right)=2\Rightarrow g\left(1\right)=1$

$f\left(g\left(2\right)\right)=1\Rightarrow g\left(2\right)=6$

$f\left(g\left(3\right)\right)=4\Rightarrow g\left(3\right)=2$

$f\left(g\left(4\right)\right)=3\Rightarrow g\left(4\right)=7$

$f\left(g\left(5\right)\right)=6\Rightarrow g\left(5\right)=3$

So, $g\left(10\right)·\left[g\left(1\right)+g\left(2\right)+g\left(3\right)+g\left(4\right)+g\left(5\right)\right]$

$=10·\left[1+6+2+7+3\right]=190$