Question

Let $A=\{x\in Z:3^{(x+1)(x^2-7x+12)}=1\}\text{and}$ $B=\{x\in Z:-5<2x-1\le 7\},$ where $Z$ is the set of integers. If the number of relations from $A$ to $B$ is $2^k$ then the value of $k$ is

Answer 1

$3^{(x+1)(x^2-7x+12)}=1=3^0$
$\Rightarrow (x+1)(x-3)(x-4)=0\Rightarrow x=-1,3,4$
$\therefore A=\{-1,3,4\}$

$-5<2x-1\le 7\Rightarrow -4<2x\le 8\Rightarrow -2<x\le 4$
$\therefore x=-1,0,1,2,3,4$
$\therefore B=\{-1,0,1,2,3,4\}$

No. of relations from $A$ to $B=2^{n\left(A\right)\cdot n\left(B\right)}=2^{18}$
Hence, $k=18$