Question

1. 函数$y = \ln|x - 1|$的图象大致形状是( )

a.
\

$$\begin{tikzpicture} \\draw-> (-1,0) -- (3,0) noderight {$x$}; \\draw-> (0,-1) -- (0,2) nodeabove {$y$}; \\drawdashed (1,0) -- (1,2); \\draw (0,1) nodeleft {1}; \\draw (1,0) nodebelow {1}; \\draw (0,0) nodebelow left {$o$}; \\drawsmooth, thick, domain=1:3 plot (\\x, {ln(\\x - 1) + 1}); \\drawsmooth, thick, domain=-1:1 plot (\\x, {ln(1 - \\x) + 1}); \\end{tikzpicture}$$

b.
\

$$\begin{tikzpicture} \\draw-> (-1,0) -- (3,0) noderight {$x$}; \\draw-> (0,-1) -- (0,2) nodeabove {$y$}; \\drawdashed (1,0) -- (1,2); \\draw (0,1) nodeleft {1}; \\draw (1,0) nodebelow {1}; \\draw (0,0) nodebelow left {$o$}; \\drawsmooth, thick, domain=1:3 plot (\\x, {1 - ln(\\x - 1)}); \\drawsmooth, thick, domain=-1:1 plot (\\x, {1 - ln(1 - \\x)}); \\end{tikzpicture}$$

c.
\

$$\begin{tikzpicture} \\draw-> (-1,0) -- (3,0) noderight {$x$}; \\draw-> (0,-2) -- (0,2) nodeabove {$y$}; \\drawdashed (1,0) -- (1,2); \\draw (0,1) nodeleft {1}; \\draw (1,0) nodebelow {1}; \\draw (0,0) nodebelow left {$o$}; \\drawsmooth, thick, domain=1:3 plot (\\x, {1 - ln(\\x - 1)}); \\drawsmooth, thick, domain=-1:1 plot (\\x, {ln(1 - \\x)}); \\end{tikzpicture}$$

d.
\

$$\begin{tikzpicture} \\draw-> (-1,0) -- (3,0) noderight {$x$}; \\draw-> (0,-2) -- (0,2) nodeabove {$y$}; \\drawdashed (1,0) -- (1,2); \\draw (0,1) nodeleft {1}; \\draw (1,0) nodebelow {1}; \\draw (0,0) nodebelow left {$o$}; \\drawsmooth, thick, domain=1:3 plot (\\x, {ln(\\x - 1) - 1}); \\drawsmooth, thick, domain=-1:1 plot (\\x, {ln(1 - \\x) - 1}); \\end{tikzpicture}$$

Explanation:

Step1: Analyze the function's domain

The function is \( y = \ln|x - 1| \). The argument of the logarithm must be positive, so \( |x - 1|>0 \), which means \( x
eq1 \). So the domain is \( (-\infty, 1)\cup(1, +\infty) \). We can eliminate options A and B because their graphs are defined at \( x = 1 \) (they have a point at \( x = 1 \)), while our function is not defined there.

Step2: Analyze the function's behavior for \( x>1 \)

When \( x>1 \), \( |x - 1|=x - 1 \), so the function becomes \( y=\ln(x - 1) \). The natural logarithm function \( \ln(u) \) is an increasing function when \( u>0 \). Here, \( u=x - 1 \) is increasing for \( x>1 \), so \( y=\ln(x - 1) \) is increasing for \( x>1 \).

Step3: Analyze the function's behavior for \( x<1 \)

When \( x<1 \), \( |x - 1|=1 - x \), so the function becomes \( y=\ln(1 - x) \). Let \( u = 1 - x \), which is a decreasing function for \( x<1 \). The natural logarithm function \( \ln(u) \) is increasing, so by the chain rule, \( y=\ln(1 - x) \) is decreasing for \( x<1 \) (since the outer function is increasing and the inner function is decreasing, the composite function is decreasing).

Now, let's check the options. Option C: For \( x>1 \), the graph is decreasing, which contradicts our analysis that \( y=\ln(x - 1) \) should be increasing for \( x>1 \). Option D: For \( x>1 \), the graph is increasing, and for \( x<1 \), the graph is decreasing, which matches our analysis.

Answer:

D