if $A = {x : x^{2} = 1}$ and $B = {x : x^{4} = 1}$ , then $A Δ B$ is equal to

{i, - i}

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{- 1 - 1}

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{1 - 1, i, - i}

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{1, i}

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Solution

The correct option is A ${i, - i}$
Given: $A = {x : x^{2} = 1}$ and $B = {x : x^{4} = 1}$
Consider, $A = {x : x^{2} = 1}$
$\Rightarrow x^{2} = 1$
$\Rightarrow x = \pm 1$
So, $A = {- 1, 1}$
and $B = {x : x^{4} = 1}$
$\Rightarrow x^{2} = \pm 1$
$\Rightarrow x^{2} = 1$ or $x^{2} = - 1$
$\Rightarrow x = \pm 1$ or $x = \pm - i$
So, $B = {- 1, 1, - i, i}$
Now $A Δ B = (A - B) \cup (B - A) = ϕ \cup {i, - i}$
So, $A$ is correct option.

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Similar questions

A = {x : x^{2} = 1}

B = {x : x^{4} = 1}

A △ B

A = (\begin{matrix} i & - i - i & i \end{matrix})

B = (\begin{matrix} 1 & - 1 - 1 & 1 \end{matrix})

A^{8}

A_{i} = [\begin{matrix} a^{i} & b^{i} b^{i} & a^{i} \end{matrix}]

| a | < 1, | b | < 1

\sum_{i = 1}^{\infty} | A_{i} |

$If a = 1 + i, then a^{2} equals$

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