Q1 Let $f(x)=\log(4x-\log x)$, where $\log$ is the natural logarithm. We are to find a local extremum of $f(x)$ by using $f''(x)$. For $\mathbf{K}$ and $\mathbf{L}$, choose the most appropriate answer from among the choices (0)$\sim$(6) below. First of all, we have $$\begin{aligned}
f'(x) &= \frac{\mathbf{A}-\frac{\mathbf{B}}{x}}{4x-\log x} \\
f''(x) &= \frac{\frac{1}{x^{\mathbf{C}}}(4x-\log x)-\left(\mathbf{A}-\frac{\square}{x}\right)^{\mathbf{D}}}{(4x-\log x)^2}
\end{aligned}$$ which give $$\begin{aligned}
f'\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= 0 \\
f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= \frac{\mathbf{GH}}{\mathbf{I}+\log\mathbf{J}}.
\end{aligned}$$ Since $$f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) \mathbf{K} \, 0,$$ $f(x)$ has a $\mathbf{L}$ at $x=\frac{\mathbf{E}}{\mathbf{F}}$, and this value is $\log(\mathbf{M}+\log\mathbf{N})$. (0) $=$ (1) $>$ (2) $\geqq$ (3) $<$ (4) $\leqq$ (5) local maximum (6) local minimum
Q1 Let $f(x)=\log(4x-\log x)$, where $\log$ is the natural logarithm. We are to find a local extremum of $f(x)$ by using $f''(x)$.
For $\mathbf{K}$ and $\mathbf{L}$, choose the most appropriate answer from among the choices (0)$\sim$(6) below.
First of all, we have
$$\begin{aligned}
f'(x) &= \frac{\mathbf{A}-\frac{\mathbf{B}}{x}}{4x-\log x} \\
f''(x) &= \frac{\frac{1}{x^{\mathbf{C}}}(4x-\log x)-\left(\mathbf{A}-\frac{\square}{x}\right)^{\mathbf{D}}}{(4x-\log x)^2}
\end{aligned}$$
which give
$$\begin{aligned}
f'\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= 0 \\
f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) &= \frac{\mathbf{GH}}{\mathbf{I}+\log\mathbf{J}}.
\end{aligned}$$
Since
$$f''\left(\frac{\mathbf{E}}{\mathbf{F}}\right) \mathbf{K} \, 0,$$
$f(x)$ has a $\mathbf{L}$ at $x=\frac{\mathbf{E}}{\mathbf{F}}$, and this value is $\log(\mathbf{M}+\log\mathbf{N})$.
(0) $=$
(1) $>$
(2) $\geqq$
(3) $<$
(4) $\leqq$
(5) local maximum
(6) local minimum