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Question

The domain of 1xx2+3x12x2 is:

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Solution

Consider the problem

xx2=0x(1x)=0x=0,1

And

3x12x2=02x2+3x1=02x23x+1=02x22xx+1=02x(x1)1(x1)=0(x1)(2x1)=0x=1,x=12

3x12x2<0whenx(,2)(1,)
and xx20whenx(,0][1,)

Therefore, Domain of given function is
R{{(,2)(1,)}{(,0][1,)}}

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