a. B ∈ R, then f (x) = x | SiNx + a | + B is an odd function if and only if () A. a2+b2=0B. ab=0C. ba=0D. a2-b2=0

a. B ∈ R, then f (x) = x | SiNx + a | + B is an odd function if and only if () A. a2+b2=0B. ab=0C. ba=0D. a2-b2=0

Because the domain of function is r, f (0) = 0. So B = 0. So f (x) = x | SiNx + a |. Because f (x) is odd, f (- x) = - f (x), that is - x | - SiNx + a | = - x | - SiNx + a |, so | - SiNx + a | = - SiNx + a |, so a = 0