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Question

Number of integer values of x satisfying the inequality
|x3|+|2x+4|+|x|11 is

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Solution

|x3|+|2x+4|+|x|11
If x3,
x3+2x+4+x11
4x10
x2.5 No solution

If 0x<3,
x+3+2x+4+x11x2x[0,2]

If 2x<0,
x+3+2x+4x11
711 which is true
x[2,0)

If x<2,
x+32x4x11
4x111
x3x[3,2)

Hence, x[0,2][2,0)[3,2)
i.e., x[3,2]

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