| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. (-\infty,0)\cup (0,\infty) | B |
| deepseekr1 | To determine the range of \( f(x) = \frac{1}{1-x} \), we analyze the possible output values (\( y \)) by solving \( y = \frac{1}{1-x} \) for \( x \): 1. Rearrange the equation: \[ y(1 - x) = 1 \implies y - yx = 1 \implies x = \frac{y - 1}{y}. \] This expression for \( x \) is valid for all \( y \neq 0 \). If \( y = 0 \), the equation \( \frac{1}{1-x} = 0 \) has no solution because the numerator (1) cannot equal zero. Thus, \( y \neq 0 \). 2. Analyze the behavior of \( f(x) \): - As \( x \to 1^- \), \( 1 - x \to 0^+ \), so \( f(x) \to +\infty \). - As \( x \to 1^+ \), \( 1 - x \to 0^- \), so \( f(x) \to -\infty \). - As \( x \to \pm\infty \), \( f(x) \to 0 \), but \( f(x) \) never actually equals zero. Thus, \( f(x) \) can take any real value except \( y = 0 \), making the range \( (-\infty, 0) \cup (0, \infty) \). **Answer:** B |
B |
| gpto1 | B | B |
| GPT-4o | The correct answer is B. \((- \infty, 0) \cup (0, \infty)\). The function \( f(x) = \frac{1}{1-x} \) is defined for all \( x \neq 1 \). As \( x \) approaches 1, the denominator \( 1-x \) approaches 0, causing the function to approach \(\pm \infty\). The output of the function \( f(x) \) can be any real number except for 0, since there is no \( x \) that will make \( \frac{1}{1-x} = 0 \). Therefore, the range is all real numbers except 0, which is expressed in interval notation as \( (-\infty, 0) \cup (0, \infty) \). |
B |
| llama3.1-70B-Instruct | B |
B |
| llama3.1-405B-Instruct | The range of the function $f(x) = \frac{1}{1-x}$ can be found by considering the possible values of $f(x)$ as $x$ varies over its domain. First, note that the domain of $f(x)$ is all real numbers except $x=1$, since the denominator cannot be zero. Now, let's consider the behavior of $f(x)$ as $x$ approaches positive infinity and negative infinity: * As $x$ approaches positive infinity, $1-x$ approaches negative infinity, so $\frac{1}{1-x}$ approaches zero from the left. * As $x$ approaches negative infinity, $1-x$ approaches positive infinity, so $\frac{1}{1-x}$ approaches zero from the right. This means that the range of $f(x)$ includes all real numbers except zero. Next, let's consider the behavior of $f(x)$ as $x$ approaches $1$ from the left and right: * As $x$ approaches $1$ from the left, $1-x$ approaches zero from the right, so $\frac{1}{1-x}$ approaches positive infinity. * As $x$ approaches $1$ from the right, $1-x$ approaches zero from the left, so $\frac{1}{1-x}$ approaches negative infinity. This means that the range of $f(x)$ includes all real numbers except zero, and also includes positive and negative infinity. Therefore, the range of the function $f(x) = \frac{1}{1-x}$ is $(-\infty,0)\cup (0,\infty)$. The correct answer is B. (-\infty,0)\cup (0,\infty). |
B |
| DeepSeek-Coder-V2 | B. (-\infty,0)\cup (0,\infty) | B |
| Qwen2-72B-Instruct | B. (-\infty,0)\cup (0,\infty) | B |
| baichuan4 | A | B |
| doubao | A The function $f(x)=\frac{1}{1-x}$ is undefined when $1-x=0$, that is $x=1$. As $x$ approaches $1$ from the left, $f(x)$ approaches negative infinity, and as $x$ approaches $1$ from the right, $f(x)$ approaches positive infinity. And for other values of $x$, $f(x)$ can take any value. So the range is $(-\infty,-1)\cup (1,\infty)$. |
B |
| minimax | To find the range of the function $f(x) = \frac{1}{1-x}$, we need to consider the values that $f(x)$ can take as $x$ varies over all real numbers except the value that makes the denominator zero (since division by zero is undefined). The denominator $1-x$ becomes zero when $x = 1$. Therefore, $x$ can be any real number except 1. As $x$ approaches 1 from the left (x < 1), the denominator $1-x$ approaches 0 from the positive side, and $f(x)$ approaches negative infinity. As $x$ approaches 1 from the right (x > 1), the denominator $1-x$ approaches 0 from the negative side, and $f(x)$ approaches positive infinity. For $x$ much less than 1 (either negative or positive but far from 1), $f(x)$ will be close to 0 but positive, since the denominator will be a large positive number and the numerator is 1. For $x$ much greater than 1, $f(x)$ will be close to 0 but negative, since the denominator will be a large negative number and the numerator is 1. Therefore, the function $f(x)$ can take all real values except 0, as $x$ approaches 1 from either side. The range of $f(x)$ is all real numbers except 0. Expressed in interval notation, the range is: B. (-∞, 0) ∪ (0, ∞) |
B |
| kimi | To determine the range of the function \( f(x) = \frac{1}{1-x} \), we need to consider the values that \( f(x) \) can take as \( x \) varies over all real numbers except for \( x = 1 \), because the function is undefined at \( x = 1 \). Let's analyze the function: 1. As \( x \) approaches 1 from the left (\( x \to 1^- \)), \( f(x) \) approaches \(-\infty\). 2. As \( x \) approaches 1 from the right (\( x \to 1^+ \)), \( f(x) \) approaches \(+\infty\). 3. As \( x \) approaches \(-\infty\), \( f(x) \) approaches 0 from the negative side. 4. As \( x \) approaches \(+\infty\), \( f(x) \) approaches 0 from the positive side. The function \( f(x) \) will take on all real values except for 0 because if \( f(x) = 0 \), then \( \frac{1}{1-x} = 0 \), which is impossible. Therefore, the range of the function is all real numbers except for 0, which can be expressed in interval notation as: \( (-\infty, 0) \cup (0, \infty) \) The correct answer is: B. \( (-\infty,0)\cup (0,\infty) \) |
B |
| wenxinyiyan | B. $(-\infty,0)\cup (0,\infty)$ Explanation: The function $f(x) = \frac{1}{1-x}$ is not defined when $1-x = 0$ or $x = 1$. For all other values of $x$, $f(x)$ is defined. To find the range, we can observe that as $x$ approaches 1 from the left (i.e., $x < 1$), the denominator $1-x$ becomes very small but positive, making $f(x)$ very large. Similarly, as $x$ approaches 1 from the right (i.e., $x > 1$), the denominator $1-x$ becomes very small but negative, making $f(x)$ a very large negative number. Therefore, as $x$ approaches 1 from either side, $f(x)$ approaches either positive or negative infinity. For $x < 1$, $f(x)$ is always positive, and for $x > 1$, $f(x)$ is always negative. Combined with the fact that $f(x)$ is never zero (because $1-x$ is never zero), we can conclude that the range of $f(x)$ is $(-\infty,0)\cup (0,\infty)$. |
B |
| zhipu | To determine the range of the function \( f(x) = \frac{1}{1-x} \), let's analyze its behavior. First, consider the domain of \( f(x) \): - The function \( f(x) = \frac{1}{1-x} \) is undefined when the denominator is zero, i.e., when \( 1 - x = 0 \) or \( x = 1 \). - Thus, the domain of \( f(x) \) is all real numbers except \( x = 1 \). Now, let's determine the possible values that \( f(x) \) can take: - As \( x \) approaches \( 1 \) from the left (i.e., \( x \to 1^- \)), \( f(x) \) approaches \( -\infty \). - As \( x \) approaches \( 1 \) from the right (i.e., \( x \to 1^+ \)), \( f(x) \) approaches \( +\infty \). - As \( x \) approaches \( -\infty \), \( f(x) \) approaches \( 0^- \). - As \( x \) approaches \( +\infty \), \( f(x) \) approaches \( 0^+ \). From these observations, it is clear that \( f(x) \) can take any real value except \( 0 \). Therefore, the range of \( f(x) \) is all real numbers except \( 0 \). In interval notation, this is expressed as \( (-\infty, 0) \cup (0, \infty) \). Thus, the correct answer is: B. \((- \infty, 0) \cup (0, \infty)\) Answer: B |
B |
| gpto1 | B | B |
| deepseekv3 | To determine the range of the function \( f(x) = \frac{1}{1 - x} \), follow these steps: 1. **Identify the domain of \( f(x) \):** The function is undefined when the denominator is zero. \( 1 - x \neq 0 \) implies \( x \neq 1 \). So, the domain of \( f(x) \) is \( (-\infty, 1) \cup (1, \infty) \). 2. **Express \( y \) in terms of \( x \):** Let \( y = \frac{1}{1 - x} \). 3. **Solve for \( x \) in terms of \( y \):** \( y = \frac{1}{1 - x} \) \( y(1 - x) = 1 \) \( y - yx = 1 \) \( yx = y - 1 \) \( x = \frac{y - 1}{y} \). 4. **Determine the possible values of \( y \):** The expression \( x = \frac{y - 1}{y} \) must be valid for all \( x \) in the domain of \( f(x) \). The denominator \( y \) cannot be zero, so \( y \neq 0 \). Also, \( x \neq 1 \) implies: \( \frac{y - 1}{y} \neq 1 \) \( y - 1 \neq y \) \( -1 \neq 0 \), which is always true. Therefore, the only restriction is \( y \neq 0 \). 5. **Conclusion:** The range of \( f(x) \) is all real numbers except \( 0 \). In interval notation, this is \( (-\infty, 0) \cup (0, \infty) \). **Answer:** \boxed{B} |
B |