Question

Let $g\left(x\right)=1+x-\left[x\right]$ (where [] represents GIF) and $f\left(x\right)=\{\begin{array}{ccc}-1& :& x<0\\ 0& :& x=0\\ 1& :& x>0\end{array}$
Then ${\int }_0^{4}f\left(g\right(x\left)\right)$ is

• A. 1
• B. 0
• C. 2
• D. 4
  

Answer 1

The correct option is D 4

$g\left(x\right)=1+x-\left[x\right]=1+\left\{x\right\}$, where $\left\{\right\}$ represents fractional part function.
Now we know that, $\left\{x\right\}\in [0,1)$
$\Rightarrow g\left(x\right)>0\forall x\in R\therefore f\left(g\right(x\left)\right)=1,\forall x$
$\therefore {\int }_0^{4}f\left(g\right(x\left)\right)dx={\int }_0^{4}1dx=[x{]}_0^4=4$