Two functions $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined as follows: $f(x) = \begin{cases} 0, & x \text{ is rational} \\ 1, & x \text{ is irrational} \end{cases}$ and $g(x) = \begin{cases} -1, & x \text{ is rational} \\ 0, & x \text{ is irrational} \end{cases}$. Then, $(f \circ g)(\pi) + (g \circ f)(e)$ is equal to:
- A$0$
- B$-1$
- C$2$
- D$1$
Explore More
Similar Questions
If $f(x) = \frac{3x+4}{5x-7}, x \neq \frac{7}{5}$ and $g(x) = \frac{7x+4}{5x-3}, x \neq \frac{3}{5}$,then $(g \circ f)(3) = $
Let $f(x) = \frac{x-1}{x+1}$,$x \in R - \{-1, 0, 1\}$. If $f^{n+1}(x) = f(f^n(x))$ for all $n \in N$,then $f^6(6) + f^7(7) = $
DifficultJEE MAIN 2022
View SolutionLet $f: \{1,3,4\} \rightarrow \{1,2,5\}$ and $g: \{1,2,5\} \rightarrow \{1,3\}$ be given by $f = \{(1,2), (3,5), (4,1)\}$ and $g = \{(1,3), (2,3), (5,1)\}$. Write down $g \circ f$.
Easy
View SolutionIf $f: R \rightarrow R$ and $g: R^{+} \rightarrow R$ are such that $g\{f(x)\}=|\sin x|$ and $f\{g(x)\}=(\sin \sqrt{x})^2$, then a possible choice for $f$ and $g$ is
Let $R$ be a relation '$ < $' from $A$ to $B$,where $A = \{1, 2, 3, 4\}$ and $B = \{1, 3, 5\}$ such that $(a, b) \in R \iff a < b$. Then $R \circ R^{-1}$ is:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo