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-- Mathematics - Section B1
Relations and Functions
Let $g(x) =1+ x −[x]$ and $f (x) = sgn(x)$. Then for all x, $f\{g(x)\}$ is equal t
$x$
1
$f(x)$
$g(x)$
$∴g(x)=1+x-[x]=1+\{x\}≥1\,∀\,x∈R$
by definition of $f(x)=1,x>0$
$f(g(x)≥1)=1$