I = 5 sin (100Q) satisfies:- (A) 0

C = 5sin2(100Q) satisfies: (A) 0 S T < 1. (B) 1ST < 2. 8sinc+ 17 sin 14T for $ < 0 18. Let QI be a constant so that the function for 2 0 Qi'.r is differentiable at x 0. Let Q = In(3 + QI l). Then T 5sin2(100Q) satisfies: (A) (B) 1ST < 2. (C) T < 3. (D) 3ST < 4.