MCQ
Let g(x) = 1 + x - [x] and $\text{f(x)}=\begin{cases}-1,&\text{x}<0\\0,&\text{x}=0\\1,&\text{x}>0\end{cases}$ where [x] denotes the greatest integer less than or equal to x. Then for all x, f(g(x)) is equal to:
  • A
    x
  • 1
  • C
    f(x)
  • D
    g(x)

Answer

Correct option: B.
1
When, -1 < x < 0
Then, g(x) = 1 + x - [x]

= 1 + x - (-1) = 2 + x

$\therefore$ f(g(x)) = 1

When, x = 0

Then, g(x) = 1 + x - [x]

= 1 + x - 0 = 1 + x

$\therefore$ f(g(x)) = 1

When, x > 1

Then, g(x) = 1 + x - [x]

= 1 + x - 1 = x

$\therefore$ f(g(x)) = 1

Therefore, for each interval f(g(x)) = 1

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