As an alternative method, use optimization:
A. if x > -3, then x(100). If x(100) then b is not less than 2. Not true.
C. if x = 3, then x(3). If x(3), then b is 2. Not true.
E. if x > 3, then x(100). If x(100) then b is not less than 2. Not true.
Thus, both x<2 and x<3 are left.
D. If x<3, then x(2.5). If x(2.5), still b<2. Must be true.
(B. if x<2, then 2.5 missing from solution set. Thus, x<2 is not true.)
+1 Kudos if my comment was helpful. Thanks!
Failure forges confidence, and confidence cultivates success.
Proving the answer choices wrong is almost better than calculating what is right.