−1≤x≤3
x≥3 and x≤−1
x≥3 or x<−1
−1≤x<3
Correct answer is B
x2−2x≥3⟹x2−2x−3≥0
x2+x−3x−3=(x+1)(x−3)≥0
x=−1;x=3
Check: x=−1:(−1)2−2(−1)=1+2≥3 (satisfied)
−1<x<3:02−2(0)=0≥3 (not satisfied)
x<−1:(−2)2−2(−2)=4+4=8≥3 (satisfied)
x=3:32−2(3)=9−6=3≥3 (satisfied)
x>3:42−2(4)=16−8=8≥3 (satisfied)
∴x2−2x≥3 is satisfied in the region x≤-1 and x≥3
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