Let $$g\left( x \right) = 1 + x - \left[ x \right]$$ and \[f\left( x \right)\left\{ {\begin{array}{*{20}{c}}
{ - 1,}\\
{0,}\\
{1,}
\end{array}} \right.\,\begin{array}{*{20}{c}}
{x < 0}\\
{x = 0}\\
{x > 0}
\end{array}.\] Then for all $$x,f\left( {g\left( x \right)} \right)$$ is equal to
A.
$$x$$
B.
1
C.
$$f\left( x \right)$$
D.
$$g\left( x \right)$$
Answer :
1
Solution :
$$g\left( x \right) = 1 + x - \left[ x \right];$$
\[f\left( x \right)\left\{ {\begin{array}{*{20}{c}}
{ - 1,}\\
{0,}\\
{1,}
\end{array}} \right.\,\begin{array}{*{20}{c}}
{x < 0}\\
{x = 0}\\
{x > 0}
\end{array}\]
For integral values of $$x;g\left( x \right) = 1$$
For $$x < 0;$$ (but not integral value) $$x - \left[ x \right] > 0 \Rightarrow g\left( x \right) > 1$$
For $$x > 0;$$ (but not integral value) $$x - \left[ x \right] > 0 \Rightarrow g\left( x \right) > 1$$
$$\therefore g\left( x \right) \geqslant 1,\forall x\,\therefore f\left( {g\left( x \right)} \right) = 1\forall x$$
Releted MCQ Question on Calculus >> Function
Releted Question 1
Let $$R$$ be the set of real numbers. If $$f:R \to R$$ is a function defined by $$f\left( x \right) = {x^2},$$ then $$f$$ is: