The maximum value of k for which $\sin x+2x≥kx(x+1)$ for all $x∈\left[0,\frac{π}{2}\right]$, equals

$\frac{4(π+1)}{π(π+2)}$

Let $f (x)= \sin x + 2x − kx^2 − kx$

$f'(x)= \cos x + 2 − 2kx − k$

$f''(x)= −\sin x − 2k$

The inequality is trivially true for k < 0

For k ≥ 0, we have $f''(x)<0$

The curve y = f (x) is concave downwards.

If A(0, 0) and $B\left(\frac{π}{2},f(\frac{π}{2})\right)$ be two points through which curve passes then for f(x) ≥ 0 point B must lie on the dotted line and the curve y = f(x) will lie above the line joining A and B.

i.e., $f(\frac{π}{2})≥ 0$

$⇒1+π-\frac{π}{2}k(\frac{π}{2}+1)≥ 0⇒\frac{π}{4}(π+2)k≤1+π⇒k≤\frac{4(π+1)}{π(π+2)}$