The maximum area of a triangle whose one vertex is at $$(0,0)$$ and the other two vertices lie on the curve $$y=-2 x^2+54$$ at points $$(x, y)$$ and $$(-x, y)$$, where $$y>0$$, is :

The function $$f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$$

The function $$f(x)=2 x+3(x)^{\frac{2}{3}}, x \in \mathbb{R}$$, has

Consider the function $$f:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}$$ defined by $$f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$$. Consider the statements

(I) The curve $$y=f(x)$$ intersects the $$x$$-axis exactly at one point.

(II) The curve $$y=f(x)$$ intersects the $$x$$-axis at $$x=\cos \frac{\pi}{12}$$.

Then