Let A - x - R - x -1- x -1 and B - x - R - ln x 2-4 x -4 - 0- then A - B equalsA- - 0B- 0- - -3-0D- 1- -

The correct option is A ( ∞ , 0)x 1/x gt;1 ⇒ 1/x gt;0⇒ x ∈ ( ∞ , 0) ⇒ A=( ∞ , 0) ln(x^2 4x+4)≥ 0 ⇒ x^2 4x+4≥ 1 ⇒ x^2 4x+3≥ 0 ⇒ x ∈ ( ∞ , 1]∪ [3,∞ ) ∴ B=( ...

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Byju's Answer

Standard XII

Mathematics

Special Integrals - 3

Let A = x ∈ R...

Question

Let $A = {x \in R : \frac{x - 1}{x} > 1}$ and $B = {x \in R : l n (x^{2} - 4 x + 4) \geq 0}$ , then $A \cap B$ equals

(- \infty, 0)

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(0, \infty)

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(- 3, 0)

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(1, \infty)

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Solution

The correct option is A $(- \infty, 0)$
$\frac{x - 1}{x} > 1 \Rightarrow - \frac{1}{x} > 0 \Rightarrow x \in (- \infty, 0) \Rightarrow A = (- \infty, 0) l n (x^{2} - 4 x + 4) \geq 0 \Rightarrow x^{2} - 4 x + 4 \geq 1 \Rightarrow x^{2} - 4 x + 3 \geq 0 \Rightarrow x \in (- \infty, 1] \cup [3, \infty) ∴ B = (- \infty, 1] \cup [3, \infty) ∴ A \cap B = (- \infty, 0)$

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Let A={x∈R:x−1x>1} and B={x∈R:ln(x2−4x+4)≥0}, then A∩B equals

The correct option is A (−∞,0)x−1x>1⇒−1x>0⇒x∈(−∞,0)⇒A=(−∞,0)ln(x2−4x+4)≥0⇒x2−4x+4≥1⇒x2−4x+3≥0⇒x∈(−∞,1]∪[3,∞)∴ B=(−∞,1]∪[3,∞)∴ A∩B=(−∞,0)

Let $A = {x \in R : \frac{x - 1}{x} > 1}$ and $B = {x \in R : l n (x^{2} - 4 x + 4) \geq 0}$ , then $A \cap B$ equals