Question

Universal set, $U=\left\{x:x^5-6x^4+11x^3-6x^2=0\right\},A=\left\{x:x^2-5x+6=0\right\}\mathrm{and}B=\left\{x:x^2-3x+2=0\right\}$. 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:

Find the value of $\left(A\cap B\right)\text{'}$ :

Given,

$$\begin{gathered} U=\{x:x^5–6x^4+11x^3–6x^2=0\} \\ \Rightarrow U=\{x:x^2(x^3-6x^2+11x-6)=0\} \\ \Rightarrow U=\{x:x^2(x-1)(x-2)(x-3)=0\} \\ A=\{x:(x-2)(x-3)=0\} \\ B=\{x:(x-1)(x-2)=0\} \end{gathered}$$

Now,

$$\begin{gathered} (A\cap B)\text{'}=\{x:x^2(x-1\left)\right(x-3)=0\} \\ \Rightarrow (A\cap B)\text{'}=\left\{0,1,3\right\}. \end{gathered}$$

Hence, the correct option is C.