Решебник по алгебре и начала математического анализа 11 класс Алимов Задание 896

\[\boxed{\mathbf{896}\mathbf{.}}\]

\[y = ax - 2f(x) = 1 + \ln x\]

\[f^{'}\left( x_{0} \right) = (1)^{'} + \left( \ln x \right)^{'} =\]

\[= 0 + \frac{1}{x} = \frac{1}{x_{0}}\]

\[f\left( x_{0} \right) = 1 + \ln x_{0}\]

\[y = 1 + \ln x_{0} + \frac{1}{x_{0}} \bullet \left( x - x_{0} \right) =\]

\[= 1 + \ln x_{0} + \frac{x}{x_{0}} - 1 =\]

\[= \frac{x}{x_{0}} + \ln x_{0}.\]

\[x_{0}:\]

\[\ln x_{0} = - 2\]

\[\ln x_{0} = \ln e^{- 2}\]

\[x_{0} = \frac{1}{e^{2}}.\]

\[a = \frac{1}{x_{0}} = 1\ :\frac{1}{e^{2}} = e^{2}.\]

\[Ответ:\ \ a = e^{2}.\]