CAT 2003 Slot 2 — QA Question 12

Consider the case when b is negative:

i.e. say b = –k, where k ≥ 1

Then, x = –|a|b = –|a| × (–k) = |a|k

∴ xb = –|a|k²

∴ a – xb = a + |a|k²

Now,

If a > 0, then a – xb = a + |a|k² > 0 since all the terms will be positive

If a < 0 (say a = –2), then a – xb = –2 + 2k² ≥ 0, since 2k² ≥ 2 as k ≥ 1

However, if a = 0, then a – xb = 0 + 0 = 0

Hence, when b is negative, a – xb ≥ 0

Now, consider the case when b is positive:

i.e. say b = +k, where k ≥ 1

Then, x = –|a|b = –|a| × (k) = –|a|k

∴ xb = –|a|k²

This is the same value of xb as we got in the previous case. Hence, the same conclusions will hold.

∴ For all cases, a − xb ≥ 0

Hence, option (b).