Let g ( x ) = 1 + x - [ x ] and f ( x ) = { - 1 , x 0 0 , x = 0 1 , x 0 . Then for all x , f ( g ( x ) ) is equal to [2001]

Question

Let $g (x) = 1 + x - [x]$ and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$ $.$ Then for all $x, f (g (x))$ is equal to [2001]

Smart Achievers · Accepted Answer

(2)

$g (x) = 1 + x - [x]$

and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$

For integral values of $x$ , $g (x) = 1$

For $x < 0$ (but not integral value), $x - [x] > 0 \Rightarrow g (x) > 1$

For $x > 0$ (but not integral value), $x - [x] > 0 \Rightarrow g (x) > 1$

$∴$ $g (x) \geq 1, \forall x \Rightarrow f (g (x)) = 1, \forall x$

Solution

Q.
Let $g (x) = 1 + x - [x]$ and $f (x) =$ ${\begin{matrix} - 1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0 \end{matrix}$ $.$ Then for all $x, f (g (x))$ is equal to [2001]

1 $x$

2 $1$

3 $f (x)$

4 $g (x)$

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Solution

Q. Let g(x)=1+x-[x] and f(x)={-1,x<00,x=01,x>0. Then for all x, f(g(x)) is equal to [2001]

1 x

2 1

3 f(x)

4 g(x)