Embibe Experts Solutions for Chapter: Relations & Functions, Exercise 2: Exercise 2

The function $f:R\to R$ defined by $f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)$ is

If $R$ be a relation from $A=\{1,2,3,4\}$ to $B=\{1,3,5\}$ such that $\left(a,b\right)\in R\iff a<b$, then $RO{R}^{-1}$ is

The least value of $x$ for which $f\left(x\right)=\frac{\sqrt{x^2-4}}{{\sin }^{-1}(2-x)}$ is defined is

$f\left(x\right)=\frac{\cos x}{\left[{\displaystyle \frac{x}{\pi }}\right]+{\displaystyle \frac{1}{2}}},$ where $x$ is not an integral multiple of $\pi$ and $\left[x\right]$ denote the greatest integer function, is

Let $f\left(x\right)=\frac{\sin x}{\sqrt{1+{\tan }^{2}x}}-\frac{\cos x}{\sqrt{1+{\cot }^{2}x}}$. Then range of $f\left(x\right)$ is

If $f\left(x\right)=\frac{1-x}{1+x};x\ne -1, 0$, and $x>0$ then $f\left[f\right(x\left)\right]+f\left[f\left(\frac{1}{x}\right)\right]$

If $3f\left(x\right)-f\left(\frac{1}{x}\right)={\log }_e{x}^4,$ the $f\left(e^{-x}\right)$ is

The function $f\left(x\right)=\frac{x}{e^x-1}+\frac{x}{2}+1$ is