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[.] represents greatest integer function. Let $g(x)=1+x-[x]$ and $f(x)=\left\{\begin{array}{cl}-3, & x < 0 \\ 0, & x=0, \text { then } \\ 5, & x>0\end{array}\right.$ $f(g(x))$ is
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The correct answer is:
$5$
We have,
$g(x)=1+x-[x]$
$g(x)=1+\{x\} \quad[\because x-[x]=\{x\}]$
$\Rightarrow g(x) \in[1,2) \quad[\because\{x\} \in[0,1)]$
Now, $f(x)=\left\{\begin{array}{cc}-3, & x < 0 \\ 0, & x=0 \\ 5, & x>0\end{array}\right.$
$\therefore \quad f(g(x))=5 ; g(x) \geq 1$
$g(x)=1+x-[x]$
$g(x)=1+\{x\} \quad[\because x-[x]=\{x\}]$
$\Rightarrow g(x) \in[1,2) \quad[\because\{x\} \in[0,1)]$
Now, $f(x)=\left\{\begin{array}{cc}-3, & x < 0 \\ 0, & x=0 \\ 5, & x>0\end{array}\right.$
$\therefore \quad f(g(x))=5 ; g(x) \geq 1$
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