] Given: Function f(x) is increasing on the set R Therefore, f′(x)>0 3x2+2ax+b−5sin2x>0 Since, f′(x)>0;D<0, Here D is discriminant (2a)2−(4×3)×(b−5sin2x)<0 4a2−12b+60sin2x<0 a2−3b+15sin2x<0 Since, x∈[0,
π
2
] Therefore, sin2x>0 sin2x is always positive Hence, a2−3b+15<0