If A - x -1- x -1 - x - R - B - x - x - 0 or x - 2 - x - R and A - B - R - C- then set C isA- x - 1- x -2 - x - R B- x - 1 - x -2 - x - R C- x - 1- x - 2 - x - R D- None of these

The correct option is B {x: 1 ≤ x 2; x ∈ℝ}A∪ nbsp;B = (−1, 1) ∪ ( ∞, 0] ∪ [2, ∞) = ( ∞, 1) ∪ [2, ∞) = ℝ [1,2) ...

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If A = x :-1<...

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If $A = {x : - 1 < x < 1; x \in R}$ , $B = {x : x \leq 0 or x \geq 2; x \in R}$ and $A \cup B = R - C$ , then set $C$ is

{x : 1 < x < 2; x \in R}

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{x : 1 < x \leq 2; x \in R}

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{x : 1 \leq x < 2; x \in R}

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None of these

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A = [x ∶ x \in R, | x | < 1];

B = [x ∶ x \in R, | x - 1 | \geq 1]

A \cup B = R - D

D

The range of the function $f (x) = \frac{x^{2} - x}{x^{2} + 2 x}$ is

Let A = { x : x ∈ R, |x| < 1}; B = {x : x ∈ R, |x-1| ≥ 1} and

A ∪ B = R - D, then the set D is

a, b, c, a_{1}, b_{1}, c_{1}, \in R

a x^{2} + b x + c > 0 \forall x \in R

a_{1} x^{2} + b_{1} x + c_{1} > 0 \forall x \in R .

f (x) = {\begin{matrix} x^{3} + 1, & x < 0 x^{2} + 1, & x \geq 0 \end{matrix}, g (x) = ⎧ ⎨ ⎩ \begin{matrix} (x - 1)^{\frac{1}{3}}, & x < 1 (x - 1)^{\frac{1}{2}}, & x \geq 1 \end{matrix}

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