Problem: If x is a real number and ∣x−4∣+∣x−3∣<a where a >0, then:
Answer Choices:
A. 0<a<.01
B. .01< a <1
C. 0< a <1
D. 0<a≦1
E. a>1
Solution:
When x≧4,∣x−4∣+∣x−3∣=x−4+x−3≧1
When x≦3,∣x−4∣+∣x−3∣=4−x+3−x≧1
When 3<x<4,∣x−4∣+∣x−3∣=4−x+x−3=1
Since a>∣x−4∣+∣x−3∣, then a>1.