Let $g(x) =1+ x −[x]$ and $f (x) = sgn(x)$. Then for all x, $f\{g(x)\}$ is equal t
Answer & explanation
Correct answer: option 2
$∴g(x)=1+x-[x]=1+\{x\}≥1\,∀\,x∈R$
by definition of $f(x)=1,x>0$
$f(g(x)≥1)=1$
Let $g(x) =1+ x −[x]$ and $f (x) = sgn(x)$. Then for all x, $f\{g(x)\}$ is equal t
Correct answer: option 2
$∴g(x)=1+x-[x]=1+\{x\}≥1\,∀\,x∈R$
by definition of $f(x)=1,x>0$
$f(g(x)≥1)=1$