Two functions $f: R \rightarrow R, g: R \rightarrow R$ are defined as follows $f(x)= \begin{cases}0, & x…

Two functions $f: R \rightarrow R, g: R \rightarrow R$ are defined as follows $f(x)= \begin{cases}0, & x \text { is rational } \\ 1, & x \text { is irrational }\end{cases}$ $g(x)=\left\{\begin{array}{cc}-1, & x \text { is rational } \\ 0, & x \text { is irrational }\end{array}\right.$ Then, $(f \circ g)(\pi)+(g \circ f)(e)$ is equal to
  1. $0$
  2. $-1$
  3. $2$
  4. $1$

Solution

$\begin{aligned} & (g \circ f)(e)=g(f(e))=g(1)=-1 \\ & (f \circ g)(\pi)=f(g(\pi))=f(-1)=0 \\ & \therefore \quad(f \circ g)(\pi)+(g \circ f)(e)=0-1=-1 \\ & \end{aligned}$

Asked in: AP EAMCET 2001

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