Question

If $f:R\to R,f\left(x\right)=\{\begin{array}{c}1x>0\\ 0x=0\\ -1x<0\end{array}$ and $g:R\to R,g\left(x\right)=\left[x\right]$, then $(f\circ g)\left(\pi \right)$ is:

• A. $\pi$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

Answer: (C)

$1$

Given that $g\left(x\right)=\left[x\right]$

$\Rightarrow$ $g\left(\pi \right)=g\left(\mathrm{3.141....}\right)$

i.e $g\left(\pi \right)=3$

$(f\circ g)\left(x\right)=f\left(g\right(x\left)\right)$

$\Rightarrow$ $(f\circ g)\left(\pi \right))=f(g\left(\pi \right))$

But $g\left(\pi \right)=3$

$\Rightarrow f\left(3\right)=1$ ...... $[\because f(x)=1\text{for}x>0]$