MCQ
Let $g(x) = 1 + x - [x]$ and $f(x) = \left\{ \begin{array}{l} - 1,\;x < 0\\0,\;\;x = 0,\;\\{\rm{1,}}\;\;\;{\rm{x}} > {\rm{0}}\end{array} \right.$then for all $x,\;f(g(x))$ is equal to
  • A
    $x$
  • $1$
  • C
    $f(x)$
  • D
    $g(x)$

Answer

Correct option: B.
$1$
Here $g(x) = 1 + n - n = 1,\,x = n \in Z$
$1 + n + k - n = 1 + k , x = n + k ($where $n \in Z,\,0 < k < 1)$
Clearly, $g(x) > 0$ for all $x$.
So, $f(g(x)) = 1$ for all $x.$

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