Let $$f(x) = {x^3}{e^{ - 3x}},\,x > 0$$. Then the maximum value of f(x) is

Suppose $$f:R \to R$$ be given by $$f(x) = \left\{ \matrix{ 1,\,\,\,\,\,\,\,\,\,\,\mathrm{if}\,x = 1 \hfill \cr {e^{({x^{10}} - 1)}} + {(x - 1)^2}\sin {1 \over {x - 1}},\,\mathrm{if}\,x \ne 1 \hfill \cr} \right.$$

then

Let $${\cos ^{ - 1}}\left( {{y \over b}} \right) = {\log _e}{\left( {{x \over n}} \right)^n}$$, then $$A{y_2} + B{y_1} + Cy = 0$$ is possible for, where $${y_2} = {{{d^2}y} \over {d{x^2}}},{y_1} = {{dy} \over {dx}}$$

The function $$y = {e^{kx}}$$ satisfies $$\left( {{{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}}} \right)\left( {{{dy} \over {dx}} - y} \right) = y{{dy} \over {dx}}$$. It is valid for