Let $N$ be the set of natural numbers and two functions $f$ and $g$ be defined as $f, g : N \to N$ such that $f(n) = \begin{cases} \frac{n+1}{2} & \text{if } n \text{ is odd} \\ \frac{n}{2} & \text{if } n \text{ is even} \end{cases}$ and $g(n) = n - (-1)^n$. Then $fog$ is
- Aonto but not one-one.
- Bone-one but not onto.
- Cboth one-one and onto.
- Dneither one-one nor onto.
Explore More
Similar Questions
Consider functions $f$ and $g$ such that the composite function $g \circ f$ is defined and is one-one. Are $f$ and $g$ both necessarily one-one?
Easy
View SolutionIf $g(x)=3x^{2}+2x-3,$ $f(0)=-3$ and $4g(f(x))=3x^{2}-32x+72,$ then $f(g(2))$ is equal to:
DifficultJEE MAIN 2026
View SolutionIf $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2 \sqrt{x}+x$,then $f(f(x))$ is
DifficultMHT CET 2023
View SolutionIf $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x) = x^{2} - 3x + 4$ and $g(x) = 2x + 1$,then the value of $x$ for which $f(x) = (f \circ g)(x)$ is
If $f$ and $g$ are increasing and decreasing functions respectively from $[0, \infty)$ to $[0, \infty)$,and $h(x) = f(g(x))$ with $h(0) = 0$,then $h(x) - h(1)$ 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