Which of the following is a singleton set :

{x : x^{2} + 1 = 0; x \in R}

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{x : x^{2} - 1 = 0; x \in R}

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{x : x^{3} + 2 x^{2} + 2 x + 1 = 0; x \in R}

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{x : x^{3} - 6 x^{2} + 11 x - 6 = 0; x \in R}

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Solution

The correct option is B ${x : x^{3} + 2 x^{2} + 2 x + 1 = 0; x \in R}$
$\begin{matrix} x^{2} + 1 = 0 \Rightarrow x = \pm i x^{2} - 1 = 0 \Rightarrow x = \pm 1 x^{3} + 2 x^{2} + 2 x + 1 = 0 (x + 1) (x^{2} + x + 1) = 0 x = - 1 x^{2} + x + 1 = 0 d o e s n^{'} t h a v e a n y r e a l r o o t x^{3} - 6 x^{2} + 11 x - 6 = 0 (x - 1) (x^{2} - 5 x + 6) = 0 (x - 1) (x - 2) (x - 3) = 0 H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

A = {x : x \in R; (x - 1)^{2} = 0}

Which of the following sets are equal ?

(i) A = {1, 2, 3}

(ii) B = { $x ϵ R : x^{2} - 2 x + 1 = 0$ }

(iii) C = {1, 2, 2, 3}

(iv) D = { $x ϵ R : x^{3} - 6 x^{2} + 11 x - 6 = 0$ }.

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

A = {x : x^{2} - 5 x + 6 = 0}

B = {x : x^{2} - 3 x + 2 = 0}

(A \cap B)^{'}

A = \{1, 2, 3\};

B = \{x \in R : x^{2} - 2 x + 1 = 0\};

C = \{1, 2, 2, 3\};

D = \{x \in R : x^{3} - 6 x^{2} + 11 x - 6 = 0\}

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\frac{1}{\sqrt{x + | x |}}

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