1.

The solution set of the equation `|(x+1)/(x)|+|x+1|=((x+1)^(2))/(|x|)`, isA. `{x:xge0}`B. `{x:x gt 0} cup{-1}`C. `{-1,1}`D. `{x:x ge1" or ",x le-1}`

Answer» Correct Answer - B
We have,
`|(x+1)/(x)|+|x+1|=((x+1)^(2))/(|x|)`
`implies(|x+1|)/(|x|)+|x+1|=(|x+1|^(2))/(|x|)`
`implies|x+1|{(1)/(|x|)+1-(|x+1|)/(|x|)}=0`
`implies |x+1|=0" or ",(1)/(|x|)+1=(|x+1|)/(|x|)`
`implies |x+1|=0" or ",1+|x|=|x+1|`
`implies x=-1" or ",1+|x|=|x+1|`
In order to solve the equation `1+|x|=|x+1|`, we consider the following cases :
In this case, we have
|x|=-x and |x+1|=-(x+1)
`:. 1+|x|=|x+1|`
`implies 1-x=-(x+1)`, which is absurd.
CASE II When `-1 lex lt 0`
In this case, we have
|x|=-a and |x+1|=x+1
`:. 1+|x|=|x+1|implies1-x=x+1 implies x=0`
But, `-1 le x lt 0`
So, there is no solution in this case.
CASE III When `x ge0`
In this case, we have
|x|=x and |x+1|=x+1
`:. 1+|x|=|x+1|`
`implies 1+x=x+1`, which is true for all x.
Clearly, the given equation is meaningful for `x ne 0`
Hence, the solution set of the given equation is
`{x:x gt 0}cup{-1}`


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