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Solution

For $f(x)$ increasing ⇒ $f'(x) \ge 0$ for all $x$. $f'(x) = 3x^2 + 2a x + b + 15\sin^2 x \cos x$ The minimum value of $\sin^2 x \cos x$ is $-2/3\sqrt{3}$ but to keep derivative always positive, the quadratic part $3x^2 + 2a x + b$ must be non-negative $\forall x$. Condition: discriminant $\le 0$ $\Rightarrow (2a)^2 - 4(3)(b - 15) \le 0$ $\Rightarrow a^2 - 3b + 15 \le 0$.