Problem: Which expression is equal to
β£aβ2β(aβ1)2β£\left|a-2-\sqrt{(a-1)^{2}}\right| β£β£β£β£βaβ2β(aβ1)2ββ£β£β£β£β
for a<0a<0a<0?
Answer Choices:
A. 3β2a3-2 a3β2a B. 1βa1-a1βa C. 111 D. a+1a+1a+1 E. 333
Solution:
Because aaa is negative, aβ1<0a-1<0aβ1<0, so (aβ1)2=β£aβ1β£=1βa\sqrt{(a-1)^{2}}=|a-1|=1-a(aβ1)2β=β£aβ1β£=1βa. Therefore distributing gives aβ2β(1βa)=2aβ3a-2-(1-a)=2 a-3aβ2β(1βa)=2aβ3. Because aaa is negative, 2aβ3<02 a-3<02aβ3<0, so β£2aβ3β£=(A)3β2a|2 a-3|=(\text{A})\boxed{3-2a}β£2aβ3β£=(A)3β2aβ.
The problems on this page are the property of the MAA's American Mathematics Competitions