Question

Let g(x) = 1 + x − [x] and $f(x)\left{ \begin{matrix}

• 1, & x & \begin{matrix} < & 0 \end{matrix} \ 0, & x & \begin{matrix} = & 0 \end{matrix} \ 1, & x & \begin{matrix}

& 0 \end{matrix} \end{matrix} \right.\ $= . Then for all x, f(g(x)) is equal to (2001 S)

• A. x
• B. 1
• C. f(x)
• D. g(x)

Answer 1

Answer: (B)

1

Answer 2

g(x) = 1 + x − [x]; f(x) $\left{ \begin{matrix}

• 1, \ 0, \ 1, \end{matrix}\begin{matrix} x < 0 \ x = 0 \ x > 0 \end{matrix} \right.\ $

For integral values of x; g(x) = 1 > 0

For x < 0; x – [x] > 0 ⇒ g(x) > 0

(but not integral value)

For x > 0; x −[x] > 0 ⇒ g(x) > 0

∴ g(x) > 0, \overset{̶}{V} x

∴ f(g(x)) = 1, \overset{̶}{V}x

∴ ‘’(2) is the correct alternative.