Question

Let $g(x) = 1 + x - \lbrack x\rbrack$ and $f(x) = \left{ \begin{matrix}

• 1, & x < 0 \ 0, & x = 0 \ 1, & x > 0 \end{matrix} \right.\ $, then for all x,$f(g(x))$ is equal to

• A. x
• B. 1
• C. $f(x)$
• D. $g(x)$

Answer 1

Answer: (B)

1

Answer 2

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

$x = \frac{y + 4}{3}$, $x = n + k$ (where $n \in Z,$ $0 < k < 1$)

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

Clearly, $f(x) = 3x - 4 = y$ for all x. So, $f(g(x)) = 1$ for all x.