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Solution of |1+(3/x)| is A. (0,3) B. [1,0) C. (1,0)U(0,3) D. (2,0)U(0,3)

Solution of |1+(3/x)| is
A. (0,3)
B. [1,0)
C. (1,0)U(0,3)
D. (2,0)U(0,3)

Grade:12

2 Answers

Arun
25750 Points
5 years ago
Dear Anirudh
 
I think that you have not completed your question. Please check and repost it. You can attach an image too.
I will be happy to help you.
 
Regards
Arun
Samyak Jain
333 Points
5 years ago
| 1 + 3/x | > 2
  • If x \leq -3 or x > 0         …...........(1)
           | 1 + 3/x |  \geq 0
           So 1 + 3/x  >  2 ,   (x + 3 – 2x) / x  > 0
           {x(x-3)} / x2  > 0  or
           x(x – 3)  > 0,                    [x2 > 0]
           which implies x \epsilon (0,3)              …................(2)
           Taking intersection of (1) & (2), we get 
           x \epsilon (0,3)                                  …......................(3)
  • If  -3 < x < 0            ….............(4)
           | 1 + 3/x | < 0
           So 1 + 3/x < -2   or   (x + 3 + 2x) / x  <  0
           {3 x (x + 1)} / x2 < 0
           x (x + 1) < 0,     which implies
           x \epsilon (-1,0)                       …..........................(5)
           Taking intersection of (4) & (5), we get 
           x \epsilon (-1,0)                               …...................(6)
From (3) & (6),       x \epsilon (0,3)   or   x \epsilon (-1,0)
Therefore, the answer should be 
                                                  x \epsilon (-1,0)       \cup       x \epsilon (0,3)    

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