Consider two sets A=x Z:|(|x-3|-3)| 1 and B=x R-1,2:((x-2)(x-4))/(x-1) _e(|x-2|)=0.

Question

Consider two sets $A={x\in Z:|(|x-3|-3)|\le 1}$ and

$B={x\in R-{1,2}:\frac{(x-2)(x-4)}{x-1}{\log }_{e}(|x-2|)=0}$. Then the number of

onto functions $f:A	o B$ is equal to :

Accepted Answer

$\begin{gathered} A={x\in Z:|(|x-3|-3)|\le 1} \ |(|x-3|-3)|\le 1\, {|x|\le a\Rightarrow x\in [-a,a]} \end{gathered}$

$\Rightarrow$ $(|x-3|-3)\in [-1,1]$

$\Rightarrow$ $|x-3|\in [2,4]$

$|x|\in [\alpha ,\beta ]\Rightarrow x\in [-\beta ,-\alpha ]\cup [\alpha ,\beta ]$

$x-3\in [-4,-2]\cup [2,4]$

$\Rightarrow$ $x\in [-1,1]\cup [5,7]$

Since $x\in Z$ (integers), $x=-1,0,1,5,6,7$

So set $A={-1,0,1,5,6,7}$

Set $B={x\in R-{1,2},\frac{(x-2)(x-4)}{(x-1)}{\log }_{e}(|x-2|)=0}$

$\frac{(x-2)(x-4)}{(x-1)}{\log }_{e}(|x-2|)=0$

$\Rightarrow$ $x-2=0\Rightarrow x=2$ but $x
e 2$

$\Rightarrow$ $x-4=0\Rightarrow x=4$

$\Rightarrow$ ${\log }_{e}|x-2|=0\Rightarrow |x-2|=1$

$\begin{gathered} x-2=\pm 1\Rightarrow x=3,1x
e 1 \ B={3,4} \ n(A)n(B)AB \ \ n(A)=6,n(B)=2 \end{gathered}$

Number of onto function $f:A	o B$ is

= Total function - Into function

Total function $=(n(B){)}^{n(A)}={2}^{6}$

number of Into function can be find using inclusion-exclusion principle

Number of into function $={}^{2}{C}_{1}(1{)}^{6}$

${}^{2}{C}_{1}(1{)}^{6}=$ only one image or range left from set $B$.

Number of onto function $={2}^{6}-{}^{2}{C}_{1}(1{)}^{6}$

$=64-2=62$

Solutions · 1