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Question

# Match the elements of the list I, which contains inequalities, with the elements of list II, which has solution sets.List-I(Inequality)List-II (Solution set)A) x2−4x+3>01) (3,4)B) x2−5x+6≤02) (−1,1)∪(2,4)C) x2+6x−27>0,−x2+3x+4>03) (−∞,1)∪(3,∞)D) x2−3x−4<0,x2−3x+2>04) [3,4]5) [2,3]Then the correct order for A B C D, is

A
1 2 3 4
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B
3 5 1 2
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C
1 3 5 4
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D
3 2 4 1
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Solution

## The correct option is C 3 5 1 2A: x2−4x+3>0⇒(x−1)(x−3)>0So, x∈(−∞,1)∪(3,∞) →(3)B: x2−5x+6≤0⇒(x−2)(x−3)≤0So, x∈[2,3] →(5)C: x2+6x−27>0 and −x2+3x+4>0 ⇒(x+9)(x−3)>0 and (x−4)(x+1)<0So, x∈(−∞,−9)∪(3,∞) and x∈(−1,4) Taking intersection, we getx∈(3,4) →(1)D: x2−3x−4<0 and x2−3x+2>0 ⇒(x−4)(x+1)<0 and (x−1)(x−2)>0So, x∈(−1,4) and x∈(−∞,1)∪(2,∞) Taking intersection, we getx∈(−1,1)∪(2,4) →(2)∴ The final order for A B C D is 3 5 1 2.Hence, option B.

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