<s>[INST] find the value of ''a'' and ''b'' for which function is always increasing 
Question Let f(x)= x$^{3}+ax^{2}+bx+5sin^{2}x$ be an increasing
function$\forall$ x $\in$$\mathbb{R}$$ Then$

(a) $a^2 -3b-15$<0
(b) $a^2 -3b-15$>0
(c)$a^2 -3b+15<$0
(d)a>0,b>0
My Approach i tried by making perfect square of the derivative
$\left\{ x+\frac{a}{3}\right\} ^{2}$+$\frac{b}{3}$-$\frac{a}{9}-5\geq$0
$\left\{ sin2x\geq-1\right\} $ $\Longrightarrow$a$^{2}$-3b+45$\leq$0
but it doesn't gives the required result
 [/INST] When we differentiate the function we get
\begin{align*}
3x^2+2ax+b+10\sin(x)\cos(x)=3x^2+2ax+b+5\sin(2x)
\end{align*}
Now as the function should be increasing, its differential should be everywhere nonnegative. But if it is everywhere nonnegative, so will
\begin{align*}
3x^2+2ax+b-5
\end{align*}
as $\sin(2x)$ is between -1 and 1. So if we now solve
\begin{align*}
3x^2+2ax+b-5=0
\end{align*}
we get
\begin{align*}
x=\frac{-2a\pm \sqrt{4a^2-4\cdot 3(b-5)}}{6}
\end{align*}
As it should be everywhere nonnegative, we are only allowed to have at max 1 solution or no solutions, so the argument squareroot should be negative:
\begin{align*}
4a^2-12b+60\leq 0
\end{align*}
which is equivalent to
\begin{align*}
a^2-3b+15\leq 0
\end{align*}
I'm missing a minussign, but I think this method should work, hopefully it is clear.
</s>