If $f=\{(1,2),(3,5),(4,1)\}$ and $g=\{(2,3),(5,1),(1,3)\}$, then $g \circ f=\ldots \ldots \ldots$ and $f \circ g=\ldots \ldots \ldots$.
Given that,
$$\begin{aligned} f & =\{(1,2),(3,5),(4,1)\} \text { and } g=\{(2,3),(5,1),(1,3)\} \\ g \circ f(1) & =g\{f(1)\}=g(2)=3 \\ g \circ f(3) & =g\{f(3)\}=g(5)=1 \\ g \circ f(4) & =g\{f(4)\}=g(1)=3 \\ g \circ f & =\{(1,3),(3,1),(4,3)\} \\ \text{Now,}\quad f \circ g(2) & =f\{g(2)\}=f(3)=5 \\ f \circ g(5) & =f\{g(5)\}=f(1)=2 \\ f \circ g(1) & =f\{g(1)\}=f(3)=5 \\ f \circ g & =\{(2,5),(5,2),(1,5)\} \end{aligned}$$
If $f: R \rightarrow R$ be defined by $f(x)=\frac{x}{\sqrt{1+x^2}}$, then $(f \circ f \circ f)(x)=$ .............. .
$$\begin{aligned} \text{Given that,}\quad f(x) & =\frac{x}{\sqrt{1+x^2}} \\ (f \circ f \circ f)(x) & =f[f\{f(x)\}] \\ & =f\left[f\left(\frac{x}{\sqrt{1+x^2}}\right)\right]=f\left(\frac{\sqrt{1+x^2}}{\sqrt{1+\frac{x^2}{1+x^2}}}\right) \\ & =f\left[\frac{x \sqrt{1+x^2}}{\sqrt{1+x^2\left(\sqrt{2 x^2}+1\right)}}\right]=f\left(\frac{x}{\sqrt{1+2 x^2}}\right) \\ & =\frac{\sqrt{\sqrt{1+2 x^2}}}{\sqrt{1+\frac{x^2}{1+2 x^2}}}=\frac{x \sqrt{1+2 x^2}}{\sqrt{1+2 x^2} \sqrt{1+3 x^2}} \\ & =\frac{x}{\sqrt{1+3 x^2}}=\frac{x}{\sqrt{3 x^2+1}} \end{aligned}$$
If $f(x)=\left[4-(x-7)^3\right]$, then $f^{-1}(x)=$ ............ .
Given that,
$$\begin{aligned} f(x) & =\left\{4-(x-7)^3\right\} \\ \text{Let}\quad y & =\left[4-(x-7)^3\right] \end{aligned}$$
$(x-7)^3=4-y$
$(x-7)=(4-y)^{1 / 3}$
$\Rightarrow\quad x=7+(4-y)^{1 / 3}$
$f^{-1}(x)=7+(4-x)^{1 / 3}$
Let $R=\{(3,1),(1,3),(3,3)\}$ be a relation defined on the set $A=\{1,2,3\}$. Then, $R$ is symmetric, transitive but not reflexive.
If $f: R \rightarrow R$ be the function defined by $f(x)=\sin (3 x+2) \forall x \in R$. Then, $f$ is invertible.