For a positive number $t$, the function $f ( x )$ defined on the interval $[ 1 , \infty )$ is $$f ( x ) = \begin{cases} \ln x & ( 1 \leq x < e ) \\ - t + \ln x & ( x \geq e ) \end{cases}$$ Among linear functions $g ( x )$ satisfying the following condition, let $h ( t )$ be the minimum value of the slope of the line $y = g ( x )$. For all real numbers $x \geq 1$, $( x - e ) \{ g ( x ) - f ( x ) \} \geq 0$. For a differentiable function $h ( t )$, a positive number $a$ satisfies $h ( a ) = \frac { 1 } { e + 2 }$. What is the value of $h ^ { \prime } \left( \frac { 1 } { 2 e } \right) \times h ^ { \prime } ( a )$? [4 points] (1) $\frac { 1 } { ( e + 1 ) ^ { 2 } }$ (2) $\frac { 1 } { e ( e + 1 ) }$ (3) $\frac { 1 } { e ^ { 2 } }$ (4) $\frac { 1 } { ( e - 1 ) ( e + 1 ) }$ (5) $\frac { 1 } { e ( e - 1 ) }$
For a positive number $t$, the function $f ( x )$ defined on the interval $[ 1 , \infty )$ is
$$f ( x ) = \begin{cases} \ln x & ( 1 \leq x < e ) \\ - t + \ln x & ( x \geq e ) \end{cases}$$
Among linear functions $g ( x )$ satisfying the following condition, let $h ( t )$ be the minimum value of the slope of the line $y = g ( x )$.
For all real numbers $x \geq 1$, $( x - e ) \{ g ( x ) - f ( x ) \} \geq 0$.
For a differentiable function $h ( t )$, a positive number $a$ satisfies $h ( a ) = \frac { 1 } { e + 2 }$. What is the value of $h ^ { \prime } \left( \frac { 1 } { 2 e } \right) \times h ^ { \prime } ( a )$? [4 points]\\
(1) $\frac { 1 } { ( e + 1 ) ^ { 2 } }$\\
(2) $\frac { 1 } { e ( e + 1 ) }$\\
(3) $\frac { 1 } { e ^ { 2 } }$\\
(4) $\frac { 1 } { ( e - 1 ) ( e + 1 ) }$\\
(5) $\frac { 1 } { e ( e - 1 ) }$