Question

Gn. ( (|x-1|-2 mid=1 )
First two possible cases
( ; quad-(|x-1|-2)=1 )
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( begin{array}{ll}+(x-1)-2=1 & -(x-1)-2=1 quad-left[begin{array}{ll}(x-1)-2]=1 & -[-(x-1)-2]=1 x-1=-3 & -(x-1)=-3end{array}right. & begin{array}{l}-(x-1)-2=- (x-1)-2=-1end{array}end{array} )
( x=44 quad x-1=4-3 quad x-1=1 quad-(x-1)=1 )
( x=-2 )
( x=2 )
( x=0 )
( x=4,-2,2,0 )
( x={-2,0,2,4} )
Ans
D) ( alpha-2,0,2,4} )

# 2.8 Solution of || x – 1|-2| = 1 is (A) {-1, 0, 1, 4} (B) (-2, 0, 3, 4 (C){–2, 0, 2, 3} (D) {-2,0, 2, 4}

Solution