$$\text { If } f(x)=\left|\begin{array}{ccc} x^3 & 2 x^2+1 & 1+3 x \\ 3 x^2+2 & 2 x & x^3+6 \\ x^3-x & 4 & x^2-2 \end{array}\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }$$

Let $$f(x)=(x+3)^2(x-2)^3, x \in[-4,4]$$. If $$M$$ and $$m$$ are the maximum and minimum values of $$f$$, respectively in $$[-4,4]$$, then the value of $$M-m$$ is

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