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Which one of the following is a null set?
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Verified Answer
The correct answer is:
$\left\{x \mid x^{2}+1=0, x \in R\right\}$
Consider the set given in option 'd' $\left\{\mathrm{x} \mid \mathrm{x}^{2}+1=0, \mathrm{x} \in \mathrm{R}\right\}$
Let $\mathrm{x}^{2}+1=0 \Rightarrow \mathrm{x}^{2}=-1 \Rightarrow \mathrm{x}=\pm \mathrm{i}$ which iscomplex
But $\mathrm{x} \in \mathrm{R}$. Hence for, any $\mathrm{x} \in \mathrm{R}, \mathrm{x}^{2}+1$ can not be zero
Let $\mathrm{x}^{2}+1=0 \Rightarrow \mathrm{x}^{2}=-1 \Rightarrow \mathrm{x}=\pm \mathrm{i}$ which iscomplex
But $\mathrm{x} \in \mathrm{R}$. Hence for, any $\mathrm{x} \in \mathrm{R}, \mathrm{x}^{2}+1$ can not be zero
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