(A,C) For option (A) & (B) Let h1(x) = eg(x) f(x) Now h1(a) = h1(b) = 0 and h1(x) is continuous also, so by Rolles theorem h1’(x) = 0 has atleast one root in (a, b) ⇒ eg(x)(f ’(x) + f(x) g’(x)) = 0 has atleast one root in (a, b) ⇒ Option (A) is correct. Similarly assume h2(x) = ekx g(x) for option (C) and (D) and apply same concept.