`|(x-3)/(x+1)| le 1` or `-1 le (x-3)/(x+1) le 1` `rArr (x-3)/(x+1)-1 le 0 " and " 0 le (x-3)/(x+1)+1` `rArr (-4)/(x+1) le 0 " and " 0 le (2x -2)/(x+1)` `rArr x lt -1 " and " { x lt - 1 or x ge 1}` `rArr x ge 1`
Topper's Solved these Questions
SET THEORY AND REAL NUMBER SYSTEM
CENGAGE|Exercise Exercise 1.1|12 Videos
SET THEORY AND REAL NUMBER SYSTEM
CENGAGE|Exercise Exercise 1.2|8 Videos
SEQUENCE AND SERIES
CENGAGE|Exercise Question Bank|36 Videos
SETS AND RELATIONS
CENGAGE|Exercise Question Bank|15 Videos
Similar Questions
Explore conceptually related problems
Solve : (2x -1)/(x +2) le 3
Solve : |(1)/(2x-1)| le 3, x ne 1//2
Solve the following : (a) 1 le |x-2| le 3 " (b) "0 lt |x-3|le 5 (c ) |x-2|+|2x-3|=|x-1| " (d) " |(x-3)/(x+1)| le1
Solve (x)/(x+2) le (1)/(|x|)
Solve : (5x)/(4)-1 le (4x-1)/(3)
Solve : (1)/(x^(2) +x) le(1)/(2x^(2) + 2x+3)
Solve : (2x-1)/(2) le 2x + (1)/(2) le (11)/(2) + x
Solve log_(x)(x^(2)-1) le 0 .
Solve: (1)/(x-1)le2,x in R
CENGAGE-SET THEORY AND REAL NUMBER SYSTEM -Archieves