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

If $$\mathrm{U}_{\mathrm{n}}(\mathrm{n}=1,2)$$ denotes the $$\mathrm{n}^{\text {th }}$$ derivative $$(\mathrm{n}=1,2)$$ of $$\mathrm{U}(x)=\frac{\mathrm{L} x+\mathrm{M}}{x^2-2 \mathrm{~B} x+\mathrm{C}}$$ (L, M, B, C are constants), then $$\mathrm{PU}_2+\mathrm{QU}_1+\mathrm{RU}=0$$, holds for

$$ \text { If } y=\tan ^{-1}\left[\frac{\log _e\left(\frac{e}{x^2}\right)}{\log _e\left(e x^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log _e x}{1-6 \cdot \log _e x}\right] \text {, then } \frac{d^2 y}{d x^2}= $$

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