Question

।x-1|<1
X + 2
Solve it for X​

Answer 1

Answer:

........................

Answer 2

Answer:

A2A

(X-1)=0

X=1

(X+2)=0

X=-2

Case 1:

X<-2

(X+2)<0

Then X-1<0

|X-1|=1-X

|X-1|/(X+2)<1

(1-X)>X+2 (Since (X+2)<0)

2X<-1

X<-0.5

Solution set1=(-infinity,-2)

Case 2:

-2<X<1

(X+2)>0

X-1<0

|X-1|=1-X

|X-1|/(X+2)<1

1-X<X+2

2X>-1

X>-0.5

Solution set2=(-0.5,1)

Case 3:

(X-1)>=0

X>=1

It is obvious that (X+2)>0

|X-1|=X-1

|X-1|/(X+2)<1

(X-1)/(X+2)<1

X-1<X+2

-1<2

It is always true.

Solution set3=[1,infinity)

So solution set for this question will be union of solution set of three cases.

Answer:

(-infinity,-2)union(-0.5,1)union[1,infinity)

=(-infinity,-2)union(-0.5,infinity)

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