The side of an equilateral triangle expands at the rate of $$\sqrt{3} \mathrm{~cm} / \mathrm{sec}$$. When the side is $$12 \mathrm{~cm}$$, the rate of increase of its area is

If $$f(x)=2 x^3+9 x^2+\lambda x+20$$ is a decreasing function of $$x$$ in the largest possible interval $$(-2,-1)$$, then $$\lambda$$ is equal to

$$ \text { The point on the curve } x^2=x y \text { which is closest to }(0,5) \text { is } $$

For a given curve $$y=2 x-x^2$$, when $$x$$ increases at the rate of 3 units/sec, then how does the slope of the curve change?