We are to find the range of the values of $k$ such that the inequality

$$\frac { \log 3 x } { 4 x + 1 } \leqq \log \left( \frac { 2 k x } { 4 x + 1 } \right) \tag{1}$$

holds for all positive real numbers $x$, where $\log$ is the natural logarithm.\\
(1) For $\mathbf{A}$ and $\mathbf{B}$ in the following sentences, choose the correct answer from among (0) $\sim$ (8) below.

By transforming inequality (1) we obtain

$$\log k \geqq \mathbf { A } . \tag{2}$$

Here, when the right side of (2) is denoted by $g ( x )$ and this $g ( x )$ is differentiated with respect to $x$, we have

$$g ^ { \prime } ( x ) = \mathbf { B } .$$

(0) $\frac { \log 3 x } { 4 x + 1 } - \log ( 4 x + 1 ) - \log 2 x$\\
(1) $\frac { \log 3 x } { 4 x + 1 } - \log ( 4 x + 1 ) + \log 2 x$\\
(2) $\frac { \log 3 x } { 4 x + 1 } + \log ( 4 x + 1 ) + \log 2 x$\\
(3) $\frac { \log 3 x } { 4 x + 1 } + \log ( 4 x + 1 ) - \log 2 x$\\
(4) $\frac { 4 \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$\\
(5) $\frac { 3 x + 2 + \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$\\
(6) $- \frac { 4 \log 3 x } { ( 4 x + 1 ) ^ { 2 } }$\\
(7) $\frac { 3 x - 2 - \log 2 x } { ( 4 x + 1 ) ^ { 2 } }$\\
(8) $- \frac { 3 \log 2 x } { ( 4 x + 1 ) ^ { 2 } }$

(2) In the following sentences, for $\mathbf { E } , \mathbf { F }$ and $\mathbf { G }$, choose the correct answer from among (0) $\sim$ (3) below. For the other blanks, enter the correct number.

Over the interval $0 < x < \frac { \mathbf { C } } { \mathbf{D} }$, $g ( x )$ is $\mathbf { E }$ and over the interval $\frac { \mathbf{C} } { \mathbf{D} } < x$, $g ( x )$ is $\mathbf { F }$. Hence at $x = \frac { \mathbf { C } } { \mathbf{D} }$, $g ( x )$ is $\mathbf { G }$.

From the above, the range of the value of $k$ such that inequality (1) holds for all positive real numbers $x$ is

$$k \geqq \frac { \mathbf { H } } { \mathbf { I } }$$

(0) increasing\\
(1) decreasing\\
(2) maximized\\
(3) minimized