Let g(x) = 1 + x - [x] and f(x) = l - 1, ;x < 0 0, ; ;x = 0, ; 1,… - Vidyadip

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.

$1$

b
(b) Here $g(x) = 1 + n - n = 1,\,x = n \in Z$

$1 + n + k - n = 1 + k$, $x = n + k$ (where $n \in Z,\,0 < k < 1$)

Now $f(g(x)) = \left\{ \begin{array}{l} - 1,\,\,\,\,\,g(x) < 0\\\,\,\,0,\,\,\,\,g(x) = 0\\\,\,\,1,\,\,\,\,\,g(x) > 0\end{array} \right.$

Clearly, $g(x) > 0$ for all $x$. So, $f(g(x)) = 1$ for all $x.$

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