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

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

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