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B
{−2,1}
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C
{−3,2}
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D
{−4,−2}
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Solution
The correct option is B{−2,1} |x||x+1|=2 ∵[|a||b|=|ab|] ⇒|x(x+1)|=2 ⇒x2+x=±2 ⇒x2+x−2=0 or x2+x+2=0 ⇒(x−1)(x+2)=0 or x=−1±√1−4(1)22 ⇒x=1,−2 or
no real value of x(as D<0)
Hence, required solution is x={1,−2}