Question

Given $A=\{\mathrm{x}|\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}{x}^2-1=0\},B=\{\mathrm{x}|\mathrm{x}\mathrm{is}\mathrm{a}\mathrm{root}\mathrm{of}{\mathrm{x}}^2-2\mathrm{x}+1=0\}$, then

• A. $A\cap B=1$
• B. $A\cup B=\varphi$
• C. $A\cup B=A$
• D. $A\cap B=\varphi$

Answer 1

Answer: (A), (C)

$A\cup B=A$

The explanation for the correct option is:

Find out what elements does set A and B contains:

For set A, values will be: $x^2–1=0$

$$\begin{gathered} \Rightarrow x^2=1 \\ \Rightarrow x=\pm 1 \\ \therefore A=\{-1,1\} \end{gathered}$$

Similarly for B: $x^2–2x+1=0$

$$\begin{gathered} \Rightarrow {(\mathrm{x}–1)}^2=0 \\ \Rightarrow x=1 \\ \therefore B=\left\{1\right\} \end{gathered}$$

$$\begin{gathered} \therefore A\cap B=1 \\ \therefore A\cup B=\{1,-1\}=A \end{gathered}$$

Hence, options ‘A’ and ‘C’ are Correct.