Embibe Experts Solutions for Chapter: Functions, Exercise 3: Exercise-3

The domain of the function $f\left(x\right)=\frac{1}{\sqrt{\left|x\right|-x}}$ is

Let $f$, be a function defined by $f\left(x\right)={\left(x-1\right)}^2+1, \left(x\ge 1\right)$

Statement -$1:$ The set $\left\{x: f\left(x\right)=f^{-1}\left(x\right)\right\}=\left\{1, 2\right\}$
Statement - $2: f$, is a bijection and $f^{-1}\left(x\right)=1+\sqrt{x-1}, x\ge 1$

If $f\left(x\right)+2f\left(\frac{1}{x}\right)=3x, x\ne 0$, and $S=\{x\in R :f(x)=f(-x\left)\right\}$; then $S$,

For $\mathrm{x}\in \mathrm{R}-\{0,1\},$ let ${\mathrm{f}}_1\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}},{\mathrm{f}}_2\left(\mathrm{x}\right)=1-\mathrm{x}$ and ${\mathrm{f}}_3\left(\mathrm{x}\right)=\frac{1}{1-\mathrm{x}}$ be three given functions. If a function, $\mathrm{J}\left(\mathrm{x}\right)$ satisfies $\left({\mathrm{f}}_2\circ \mathrm{J}\circ {\mathrm{f}}_1\right)\left(\mathrm{x}\right)={\mathrm{f}}_3\left(\mathrm{x}\right)$, then $\mathrm{J}\left(\mathrm{x}\right)$ is equal to

The number of functions $\mathrm{f}$ from $\left\{1, 2, 3, ..\dots , 20\right\}$, onto $\left\{1, 2, 3, \dots \dots \dots \dots , 20\right\}$ such that $\mathrm{f}\left(\mathrm{k}\right)$ is a multiple of $3$, wherever $\mathrm{k}$ is a multiple of $4$, is

Let $f:\left(0,1\right)\to R$ be defined by $f\left(x\right)=\frac{b-x}{1-bx}$, where $b$ is a constant such that $0<b<1$. Then

For $x\in R, f\left(x\right)=\left|\log 2-\sin x\right|$ and $g\left(x\right)=f\left(f\left(x\right)\right),$ then

Let $f\left(x\right)=\left\{\begin{array}{ll}-1,& -2\le x<0\\ x^2-1,& 0\le x\le 2\end{array}\right.$ and $g\left(x\right)=\left|f\right(x\left)\right|+f\left(\right|x\left|\right).$ Then, in the interval $(-2, 2), g$ is