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Solve the inequation log_(x^(2)+2x-3)((|x+4|-|x|)/(x-1)|gt0 |
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Answer» The given inequation is valid for `(|x+4|-|x|)/((x-1))gt0` and `x^(2)+2x-3gt0,!=1`…….i Now consider the following cases: CASE I If `0ltx^(2)+2x-3lt1` `implies4lt(x+1)^(2)lt5` `implies-sqrt(5)lt(x+1)lt-2` or `2ltx+1ltsqrt(5)` `implies-sqrt(5)-1ltxle-3` or `1ltxltsqrt(5)-1` `:.x epsilon (-sqrt(5)-1,-3)uu(1,sqrt(5)-1)`...ii Ten `(|x+4|-|x|)/((x-1))lt1` Now `xlt-4`, ten `(-(x+4)+x)/((x-1))lt1` `implies 1+4/(x-1)gt0` `implies((x+3))/((x-1))gt0` `:.x epsilon (-oo,-3)uu(1,oo)` `impliesx epsilon (-oo,-4)[:' x lt -4]`.iii `-4lexlt0`, then `(x+4+x)/((x-1))-1lt0` `implies((x+5))/((x-1))LT0` `:.x epsilon (-5,1)` `impliesx epsilon [-4,0)[:'-4lexlt0]` ..iv and `xge0` then `((x+4-x)/((x-1))lt1` `implies1=4/(x-1)gt0` `implies((x-5))/((x-1))gt0` `:.x epsilon (-oo,1)uu(5,oo)` `impliesx epsilon [0,1)uu(5,oo)[ :' x ge0]`.......v From eqs (iii), (iv) and (v) we get `x epsilon (-oo,1)uu(5,oo)` .vi Now common valuesin Eq. (ii) and (iv) is `x epsilon (-sqrt(5)-1,-3)`..vii Case II If `x^(2)+2x-3gt1` `impliesx^(2)+2x+1gt5implies(x+1)^(2)gt5` `impliesx+1lt-sqrt(5)` or `x+1gtsqrt(5)` `:.x epsilon(-oo,-1-sqrt(5))uu(sqrt(5)-1,oo)`........viii Then `(|x+4|-|x|)/((x-1))gt1` Now `xlt-4`, then `(-4)/(x-1)gt1` `implies1+4/(x-1)lt0` `implies (x+3)/(x-1)lt0` `:. x epsilon (-3,1)` Which is false `[:' x lt-4]` `-4lexlt0` then `(2x+4)/((x-1))-1gt0` ltbRgt `implies((x+5))/((x-1))gt0` `:.x epsilonn (-oo,-5)uu(1,oo)` Which is false `[:'-4lexlt0]` and `xge0` then `4/(x-1)gt1` `implies1-4/(x-1)lt0` `implies(x-5)/(x-1)lt0` `:.x epsilon (1,5)`....ix which is false `[:'xge0]` Now comon values in Eq (viii) and (ix) is `:. x epsilon (sqrt(5)-1,5)` .......x Combinin Eqs (viii) and (x) we get `x epsilon (-sqrt(5)-1,-3)uu(sqrt(5)-1,5)` |
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