Question

Universal set, $U=\left\{x:x^5-6x^4+11x^3-6x^2=0\right\}$, $A=\left\{x:x^2-5x+6=0\right\}$ and $B=\left\{x:x^2-3x+2=0\right\}$ .Then, $\left(A\cap B\right)\text{'}$ is equal to

• A. $\left\{1,3\right\}$
• B. $\left\{1,2,3\right\}$
• C. $\left\{0,1,3\right\}$
• D. $\left\{0,1,2,3\right\}$

Answer 1

The correct option is C

$\left\{0,1,3\right\}$

Explanation for the correct answer:

$U=\left\{x:x^5-6x^4+11x^3-6x^2=0\right\}$

This implies that $U$ is the set of all solutions satisfying the equation $x^5-6x^4+11x^3-6x^2=0$

Factorizing the equation $x^5-6x^4+11x^3-6x^2=0$ we get

$\Rightarrow$$x^2\left(x^3-6x^2+11x-6\right)=0$

$\Rightarrow$ $x^2\left(x-1\right)\left(x^2-5x+6\right)=0$

$\Rightarrow$ $x^2\left(x-1\right)\left(x-2\right)\left(x-3\right)=0$

$\Rightarrow$ $x=0,0,1,2,3$

Using Roster form $U$ can be written as

$U=\left\{0,1,2,3\right\}$

$A=\left\{x:x^2-5x+6=0\right\}$

This implies that $A$ is the set of all solutions satisfying the equation $x^2-5x+6=0$

Factorizing the equation $x^2-5x+6=0$ we get

$\Rightarrow$$\left(x-2\right)\left(x-3\right)=0$

$\Rightarrow$ $x=2,3$

Using Roster form $A$ can be written as

$A=\left\{2,3\right\}$

$B=\left\{x:x^2-3x+2=0\right\}$

This implies that $B$ is the set of all solutions satisfying the equation $x^2-3x+2=0$

Factorizing the equation $x^2-3x+2=0$ we get

$\Rightarrow$$\left(x-2\right)\left(x-1\right)=0$

$\Rightarrow$ $x=1,2$

Using Roster form $B$ can be written as

$B=\left\{1,2\right\}$

$\Rightarrow$$\left(A\cap B\right)=\left\{2\right\}$

Hence, $\left(A\cap B\right)\text{'}=U-\left(A\cap B\right)$

$\Rightarrow$ $\left(A\cap B\right)\text{'}=\left\{0,1,2,3\right\}-\left\{2\right\}$

$\Rightarrow$ $\left(A\cap B\right)\text{'}=\left\{0,1,3\right\}$

Hence, option (C) is the correct answer.