If 3sinx + 4cosx + r is always greater than or equal to 0, what is the smallest value that r can take?
Options:
A . 5
B . -5
C . 4
D . 3
E . Can't be determined
Answer: Option A : A 3 sinx +4 cosx + r ≥ 0 3 sinx + 4 cosx ≥ -r 5×(35sinx+45cosx)≥−r Let 35 = cos A ⇒ sin A ⇒45 5(sin x cos A +sin Acos X)≥ -r 5(sin(X+A))≥ -r We have -1 ≤ sin (angle) ≤ 1 5 sin(X+A)≥ -5 rmin =5.
