If $f: R \rightarrow R$ be defined by $f(x)=\left\{\begin{array}{l}2 x: x>3 \\ x^2: 1< x \leq 3 \\ 3 x: x \leq 1\end{array}\right.$ Then, $f(-1)+f(2)+f(4)$ is
If $f: R \rightarrow R$ be given by $f(x)=\tan x$, then $f^{-1}(1)$ is
Let the relation $R$ be defined in $N$ by $a R b$, if $2 a+3 b=30$. Then, $R=$ ..........
$$\begin{aligned} \text{Given that,}\quad 2 a+3 b & =30 \\ 3 b & =30-2 a \\ b & =\frac{30-2 a}{3} \end{aligned}$$
$$\begin{aligned} \text{For}\quad & a=3, b=8 \\ & a=6, b=6 \\ & a=9, b=4 \\ & a=12, b=2 \\ & R=\{(3,8),(6,6),(9,4),(12,2)\} \end{aligned}$$
If the relation $R$ be defined on the set $A=\{1,2,3,4,5\}$ by $R=\left\{(a, b):\left|a^2-b^2\right|< 8\right\}$. Then, $R$ is given by ............ .
Given,
$$\begin{aligned} & A=\{1,2,3,4,5\}, \\ & R=\left\{(a, b):\left|a^2-b^2\right|<8\right\} \\ & R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3),(4,3),(3,4),(4,4),(5,5)\} \end{aligned}$$
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,
$$\begin{aligned} f & =\{(1,2),(3,5),(4,1)\} \text { and } g=\{(2,3),(5,1),(1,3)\} \\ g \circ f(1) & =g\{f(1)\}=g(2)=3 \\ g \circ f(3) & =g\{f(3)\}=g(5)=1 \\ g \circ f(4) & =g\{f(4)\}=g(1)=3 \\ g \circ f & =\{(1,3),(3,1),(4,3)\} \\ \text{Now,}\quad f \circ g(2) & =f\{g(2)\}=f(3)=5 \\ f \circ g(5) & =f\{g(5)\}=f(1)=2 \\ f \circ g(1) & =f\{g(1)\}=f(3)=5 \\ f \circ g & =\{(2,5),(5,2),(1,5)\} \end{aligned}$$