Correct Answer - `(1)/(3)le xle1`
`(4)/(x+1)le3rArr (4)/(x+1)-3le0rArr(1-3x)/(x+1)le0`.
`therefore (1-3x le0 and x+1gt0) or(1-3x ge0 andx+1lt0)`
`rArr (xge(1)/(3) and xgt-1)or(xle(1)/(3)and xlt-1)`
`rArr xge(1)/(3)orxlt-1rArrx in[(1)/(3),oo)uu(-oo,-1)`
Again `(6)/(x+1)ge3 rArr (6)/(x+1)-3 ge0 rArr(3-3x)/(x+1)ge0 rArr (1-x)/(1+x)ge0`.
`therefore (1-xge0 and 1+xgt0) or(1-xle0and1+xlt0)`
`rArr (xle1 andxgt-1) or(xge1 andxlt-1)`
`-1 ltxle1 rArr x in(-1,1]`.
`therefore "solution set"={(-oo,-1)uu[(1)/(3),oo)}nn(-1,1]=[(1)/(3),1]`.
