If A=(x ∈R/ x^2-8x+15 ∈R) and B=(x ∈R/ x-32x-5

Question

If A=(x ∈R/ x^2-8x+15 ∈R) and B=(x ∈R/ x-32x-5< x-62x-11), then A ∩B=

Accepted Answer

Correct answer: B. (b) We have A=(x ∈R/ x^2-8x+15 ∈R)B=(x ∈R/ x-32x-5< x-62x-11) On solving, we get A ≡x ∈ (- ,3 ] ∪ [5 , )( ∵x^2-8x+15 0) and B ≡x ∈ ( 52 , 112 )( ∵ 1 (2x-5 ) (2x-11 )<0) Now A ∩B=x ∈ ( 52 ,3 ) ∪ (5 , 112 )

$If A = (x \in R / x^{2} - 8 x + 15 \in R) and B = (x \in R / \frac{x - 3}{2 x - 5} < \frac{x - 6}{2 x - 11}), then A \cap B =$

Solution

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