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Let A=x∣x2−4x+3<0,x∈R; B=x∣21−x+p≤0;x2−2(p+7)x+5≤0 If B⊆A, then p∈

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a

[−4,−1]

b

[−4,∞)

c

(−∞,1]

d

[0,1]

answer is A.

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Detailed Solution

A=(1,3) For B⊆A21−1+p≤0⇒p≤−121−3+p≤0⇒p≤−1/4p≤−1f(x)=x2−2(p+7)x+5 f(1)≤0 ⇒p≥−4f(3)≤0⇒P≥−1P≥−4 So p∈[−4,−1]

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