Amit M Agarwal Solutions for Chapter: Fundamentals of Relation and Function, Exercise 3: Target Exercises

If $f\left(x\right)=x^2+1$, then the value of $\left(fof\right)\left(x\right)$ is equal to

If $f\left(x\right)$ is defined on $\left[0,1\right]$ by the rule $f\left(x\right)=\left\{\begin{array}{cl}x,& \mathrm{if} x\in Q\\ 1-x,& \mathrm{if} x\notin Q\end{array}\right.$, then for $x\in \left[0,1\right]$

If $f\left(x\right)=\left\{\begin{array}{ll}{x}^3+1,& x<0\\ x^2+1,& x\ge 0\end{array}, g\left(x\right)=\left\{\begin{array}{ll}{\left(x-1\right)}^{1/3},& x<1\\ {\left(x-1\right)}^{1/2},& x\ge 1\end{array}\right.\right.$, then $\left(gof\right)\left(x\right)$ is equal to

If $f\left(x\right)=\left[x\right]$ and $g\left(x\right)=x-\left[x\right]$, where$\left[·\right]$, denotes greatest integer function, then which of the following is the zero function?

If $f\left(x\right)=\frac{x}{x-1}, x\ne 1$, then $\underset{19 \mathrm{times}}{\underbrace{\left(f\circ f\circ \dots \circ f\right)}}\left(x\right)$ is equal to

Let, $f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}, x\in R-\left\{0\right\}$, then $f\left(x\right)$ is equal to

Let, $f:\left[0,1\right]\to \left[0,1\right]$ and $g:\left[0,1\right]\to \left[0,1\right]$ be two functions defined by $f\left(x\right)=\frac{1-x}{1+x}$ and $g\left(x\right)=4 x\left(1-x\right)$, then $\left(fog\right)\left(x\right)$ is equal to

Let, $f\left(x\right)=|x-1|, x\in R$, then