Question

If $A=\{y:y=\frac{a+1}{2},aϵW\text{and}a\le 5\},B\{y:y=\frac{2n-1}{2},nϵW\text{and}n<5\}\text{and}C=\{-1,\frac{-1}{2},1,\frac{3}{2},2\}$ then $A-(B\cup C)=$

• A. $\left\{1\right\}$
• B. $\left\{2\right\}$
• C. $\left\{3\right\}$
• D. $\{1,2,3\}$

Answer 1

The correct option is C $\left\{3\right\}$

Given that,
$A=\{y:y=\frac{(a+1)}{2},a\in W\text{and}a\le 5\}$
$a=\left\{0,1,2,3,\dots \right\}$

Now,
$y=\frac{(0+1)}{2}=\frac{1}{2}$ for $a=0$
$y=1$ for $a=1$
$y=\frac{3}{2}$ for $a=2$
$y=2$ for $a=3$
$y=\frac{5}{2}$ for $a=4$
$y=3$ for $a=5$

$\therefore A=\{\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},3\}$

$B=\left\{y:y=\frac{(2n-1)}{2},n\in Wandn<5\right\}$
$n=0,1,2,3,4\Rightarrow Forn=0,y=-\frac{1}{2}$
For $n=1,y=\frac{(2x1-1)}{2}=\frac{1}{2}$
For $n=2,y=\frac{(2x2-1)}{2}=\frac{3}{2}$
For $n=3,y=\frac{(2x3-1)}{2}=\frac{5}{2}$
For $n=4,y=\frac{(2x4-1)}{2}=\frac{7}{2}$

$\therefore B=\left\{-\frac{1}{2},\frac{1}{2},\frac{3}{2},\frac{5}{2},\frac{7}{2}\right\}$

$C=-1,-\frac{1}{2},1,\frac{3}{2},2$

Hence,
$B\cup C=\left\{-1,-\frac{1}{2},\frac{1}{2},1,\frac{3}{2},2,\frac{5}{2},\frac{7}{2}\right\}$

$A-(B\cup C)=3$