Question

| 1 |x+3210 then xe (a) (-13,7) (b) (-13, 7) (c) (-00,-13) [7,00) (d) (-0, -13] [7,)

JEE/Engineering Exams

Maths

Solution

141

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( |x+3| geq 10 ) ( 8 b quad|x| geqslant a Rightarrow x geqslant a ) or ( x leqslant-a ) ( therefore x+3 geqslant 10 ) or ( x+3 leq-10 ) ( Rightarrow x geqslant 10-3 ) of ( x leq-10-3 ) ( x geqslant 7 ) 7 ( -13 quad j quad 7 ) ( therefore x in(-infty,-13] cup[7, infty) )