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If $f(x)=x^3+a x^2+b x+5 \sin ^2 x$ is an increasing function on $R$, then
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$a^2-3 b+15 < 0$
$\begin{array}{ll} & \because f^{\prime}(x)>0 \\ \Rightarrow & 3 x^2+2 a x+b+10 \sin x \cos x>0 \\ \Rightarrow & 3 x^2+2 a x+b+5 \sin 2 x>0 \\ \Rightarrow & 3 x^2+2 a x+b-5>0 \\ \Rightarrow \quad & 3 x^2+2 a x+(b-5)>0 \\ \text { Here, } & a>0, a < 0 \\ \therefore & (2 a)^2-4 \times 3 \times(b-5) < 0 \\ & 4 a^2-12(b-5) < 0 \\ & a^2-3(b-5) < 0 \\ & a^2-3 b+15 < 0 .\end{array}$
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