The solution set for the
inequality |x+2| < |4x+1|
(a) -3/5 < x
< 1/3
(b) -1/3 < x
< 3/5
(c) 0 < x <
1/3
(d) -1/3 < x
< 1/3
(e) None of these
Solution: (E)
Explanation:
Since the expressions on
both sides of the inequality are under modulus, you can square on both sides
without disturbing the inequality.
(x+2)2 <
(4x+1)2
x2 + 4x + 4
< 16x2 + 8x + 1
15x2 + 4x – 3 >
0
(5x + 3)(3x – 1) > 0
Hence, x € (-¥, -3/5) U (1/3, +¥)