Let f(x) = x3 + ax2 + bx + 5 sin2x be an increasing function on the set R. Then, a and b satisfy.

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a2 − 3b + 15 < 0 Explanation: f\left( x ight) = x^3 + a x^2 + bx + 5 \sin^2 x f'\left( x ight) = 3 x^2 + 2ax + \left( b + 5 \sin 2x ight) ext {Given}:f\left( x ight) ext { is increasing on R }. \Rightarrow f'\left( x ight) > 0, \forall x \in R \Rightarrow 3 x^2 + 2ax + \left( b + 5 \sin 2x ight) > 0, \forall x \in R ext { Since this quadratic function is >0, its discriminant is } <0. \Rightarrow \left( 2a ight)^2 - 4\left( 3 ight)\left( b + 5 \sin 2x ight) < 0 \Rightarrow 4 a^2 - 12b - 60 \sin 2x < 0 \Rightarrow a^2 - 3b - 15 \sin 2x < 0 ext { We know that the minimum value of sin 2x is−1}. herefore a^2 - 3b + 15 < 0

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