| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | C. (-inf, -1) U (4, inf) Reasoning Skill: Identifying Pros And Cons In this question, the student needs to identify the domain of the function, which means finding the values of x for which the function is defined. The function is defined as long as the denominator is not zero and the expression inside the square root is non-negative. The student needs to analyze the expression inside the square root, x^2 - 3x - 4, and find its roots. Factoring the quadratic expression, we get (x - 4)(x + 1) = 0, which gives us x = 4 and x = -1 as the roots. Now, the student needs to consider the pros and cons of each interval: * (-inf, -1): The expression inside the square root is negative, so this interval is not part of the domain. * (-1, 4): The expression inside the square root is positive, so this interval is part of the domain. * (4, inf): The expression inside the square root is positive, so this interval is part of the domain. However, the student also needs to consider the denominator, which cannot be zero. The denominator is zero when x = -1 and x = 4. Therefore, these points must be excluded from the domain. Considering all the pros and cons, the student can conclude that the domain of the function is (-inf, -1) U (4, inf). |
C |
| deepseekr1 | The domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2 - 3x - 4}} \) requires the denominator \(\sqrt{x^2 - 3x - 4}\) to be defined and non-zero. This means the radicand \( x^2 - 3x - 4 \) must be positive. Factoring the quadratic expression gives \( (x - 4)(x + 1) \). Solving the inequality \( (x - 4)(x + 1) > 0 \) using a sign chart reveals the solution intervals \( x < -1 \) or \( x > 4 \). Since \( x = -1 \) and \( x = 4 \) make the denominator zero (invalid), they are excluded. Thus, the domain is \( (-\infty, -1) \cup (4, \infty) \). **Answer: C** |
C |
| gpto1 | To find the domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2 - 3x - 4}} \), we need to determine for which real numbers \( x \) the function is defined. 1. **Denominator Must Not Be Zero or Negative**: Since there's a square root in the denominator, the expression under the square root must be **positive** (not zero, because division by zero is undefined). So: \[ x^2 - 3x - 4 > 0 \] 2. **Find the Critical Points**: Solve the quadratic inequality: \[ x^2 - 3x - 4 = 0 \] Using the quadratic formula: \[ x = \frac{3 \pm \sqrt{( -3)^2 - 4 \cdot 1 \cdot ( -4 )}}{2} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm 5}{2} \] Thus, the critical points are \( x = -1 \) and \( x = 4 \). 3. **Test Intervals**: Determine the sign of \( x^2 - 3x - 4 \) in the intervals determined by these critical points: - **For \( x < -1 \)**: Choose \( x = -2 \): \[ (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 > 0 \] The expression is positive. - **For \( -1 < x < 4 \)**: Choose \( x = 0 \): \[ (0)^2 - 3(0) - 4 = -4 < 0 \] The expression is negative. - **For \( x > 4 \)**: Choose \( x = 5 \): \[ (5)^2 - 3(5) - 4 = 25 - 15 - 4 = 6 > 0 \] The expression is positive. 4. **Conclusion**: The function is defined for \( x < -1 \) and \( x > 4 \). **Answer:** C |
C |
| GPT-4o | To determine the domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2-3x-4}} \), we need to ensure that the expression under the square root, \( x^2 - 3x - 4 \), is positive, since the square root function is only defined for non-negative numbers, and also the denominator should not be zero. First, we find the zeros of the quadratic expression \( x^2 - 3x - 4 \): \[ x^2 - 3x - 4 = 0 \] Factoring the quadratic equation, we get: \[ (x - 4)(x + 1) = 0 \] So, the roots are \( x = 4 \) and \( x = -1 \). The quadratic \( x^2 - 3x - 4 \) changes sign at these roots. We need \( x^2 - 3x - 4 > 0 \) for the function to be defined. We can test the intervals determined by the roots: \( (-\infty, -1) \), \((-1, 4)\), and \( (4, \infty) \). 1. For \( x \in (-\infty, -1) \): Choose \( x = -2 \), \[ (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 > 0 \] 2. For \( x \in (-1, 4) \): Choose \( x = 0 \), \[ 0^2 - 3(0) - 4 = -4 < 0 \] 3. For \( x \in (4, \infty) \): Choose \( x = 5 \), \[ 5^2 - 3(5) - 4 = 25 - 15 - 4 = 6 > 0 \] Thus, the quadratic expression \( x^2 - 3x - 4 \) is positive for \( x \in (-\infty, -1) \cup (4, \infty) \). Therefore, the domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2 - 3x - 4}} \) is: \[ (-\infty, -1) \cup (4, \infty) \] So the correct answer is: C. \((-inf, -1) \cup (4, inf)\) |
C |
| llama3.1-70B-Instruct | C |
C |
| llama3.1-405B-Instruct | C. (-inf, -1) U (4, inf) |
C |
| DeepSeek-Coder-V2 | C. (-inf, -1) U (4, inf) | C |
| Qwen2-72B-Instruct | C. (-inf, -1) U (4, inf) | C |
| baichuan4 | To find the domain of the function $f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$, we need to determine the values of $x$ for which the function is defined. The function is undefined when the denominator is 0 or when the expression inside the square root is negative. The denominator is $\sqrt{x^2-3x-4}$. Let's set the expression inside the square root equal to 0 and solve for $x$: $$x^2 - 3x - 4 = 0$$ This is a quadratic equation that can be factored as: $$(x - 4)(x + 1) = 0$$ So, $x = 4$ or $x = -1$. These are the values that make the denominator 0, and thus the function is undefined at these points. Next, we need to determine for which values of $x$ the expression $x^2 - 3x - 4$ is positive because the square root is only defined for non-negative numbers. We can do this by analyzing the factors we found: - When $x > 4$ or $x < -1$, the expression $x^2 - 3x - 4$ is positive. - When $-1 < x < 4$, the expression $x^2 - 3x - 4$ is negative. Therefore, the domain of the function is all real numbers except for $x = 4$ and $x = -1$, and excluding the interval between -1 and 4 where the expression under the square root is negative. This means the correct answer is: C. (-inf, -1) U (4, inf) |
C |
| doubao | D. For the function to be defined, the denominator cannot be zero and the radicand in the square root must be non-negative. The denominator is $\sqrt{x^2-3x-4}$, so $x^2-3x-4>0$. Factoring gives $(x-4)(x+1)>0$. This is true when $x<-1$ or $x>4$. Also, $x$ cannot be equal to -1 because the denominator would be zero. So the domain is $(-\infty,-1)\cup(-1,4)\cup(4,\infty)$. |
C |
| minimax | To find the domain of the function $f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$, we need to determine the set of all real numbers x for which the function is defined. The function will be undefined in two cases: 1. When the denominator is zero, because division by zero is undefined. 2. When the expression under the square root is negative, because the square root of a negative number is not a real number. Let's first find the values of x for which the denominator is zero: $\sqrt{x^2-3x-4} = 0$ Squaring both sides to eliminate the square root gives: $x^2 - 3x - 4 = 0$ Factoring the quadratic expression: $(x - 4)(x + 1) = 0$ This gives us two solutions: $x = 4$ and $x = -1$ These are the points where the denominator is zero, so the function is undefined at x = 4 and x = -1. Next, we need to ensure that the expression under the square root is non-negative: $x^2 - 3x - 4 \geq 0$ We already factored this expression above, and it factors into $(x - 4)(x + 1)$. The expression is non-negative when either both factors are non-negative or both are non-positive: 1. Both non-negative: $x \geq 4$ or $x \geq -1$ 2. Both non-positive: $x \leq 4$ or $x \leq -1$ However, since we are looking for non-negative values, ***bining the conditions, we get: $x \leq -1$ or $x \geq 4$ Therefore, the domain of the function is the union of two intervals: $(-\infty, -1)$ and $(4, \infty)$. The correct answer is: C. (-inf, -1) U (4, inf) |
C |
| kimi | The domain of a function is the set of all possible input values (x-values) for which the function is defined. For the function $f(x)=\frac{x+6}{\sqrt{x^2-3x-4}}$, we need to ensure two things: 1. The denominator cannot be zero, because division by zero is undefined. 2. The expression under the square root must be positive, because the square root of a negative number is not defined in the set of real numbers. Let's find the values of x that satisfy these conditions: 1. The denominator is $\sqrt{x^2-3x-4}$. For this to be defined, $x^2-3x-4$ must be greater than zero. 2. The numerator is $x+6$, which is defined for all real numbers. Let's solve the inequality $x^2-3x-4 > 0$: $(x-4)(x+1) > 0$ This inequality is true when either both factors are positive or both are negative: - Both factors are positive: $x > 4$ and $x > -1$, which is only true for $x > 4$. - Both factors are negative: $x < 4$ and $x < -1$, which is true for $x < -1$. So the domain is all x such that $x < -1$ or $x > 4$. Therefore, the correct answer is: C. $(-inf, -1) U (4, inf)$ |
C |
| wenxinyiyan | C | C |
| zhipu | To determine the domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2-3x-4}} \), we need to ensure that the expression under the square root is non-negative and that the denominator is not zero. 1. **Find the expression under the square root:** \[ x^2 - 3x - 4 \] 2. **Factor the quadratic expression:** \[ x^2 - 3x - 4 = (x - 4)(x + 1) \] 3. **Determine where the quadratic expression is non-negative:** The expression \((x - 4)(x + 1)\) is non-negative when both factors have the same sign: - Both factors are positive: \( x - 4 > 0 \) and \( x + 1 > 0 \), which simplifies to \( x > 4 \). - Both factors are negative: \( x - 4 < 0 \) and \( x + 1 < 0 \), which simplifies to \( x < -1 \). Therefore, the expression \((x - 4)(x + 1) \geq 0\) when \( x \in (-\infty, -1] \cup [4, \infty) \). 4. **Ensure the denominator is not zero:** The denominator is zero when \((x - 4)(x + 1) = 0\), which occurs at \( x = -1 \) and \( x = 4 \). Thus, we must exclude \( x = -1 \) and \( x = 4 \) from the domain. Combining these conditions, the domain of \( f(x) \) is: \[ (-\infty, -1) \cup (4, \infty) \] However, we must check if there are any additional restrictions from the numerator. The numerator \( x + 6 \) does not impose any additional restrictions since it is a linear expression and is defined for all real numbers. Therefore, the correct domain is: \[ (-\infty, -1) \cup (-1, 4) \cup (4, \infty) \] This matches option D. Answer: D |
C |
| gpto1 | To find the domain of the function \( f(x) = \frac{x+6}{\sqrt{x^2 - 3x - 4}} \), we need to determine for which real numbers \( x \) the function is defined. 1. **Denominator Must Not Be Zero or Negative**: Since there's a square root in the denominator, the expression under the square root must be **positive** (not zero, because division by zero is undefined). So: \[ x^2 - 3x - 4 > 0 \] 2. **Find the Critical Points**: Solve the quadratic inequality: \[ x^2 - 3x - 4 = 0 \] Using the quadratic formula: \[ x = \frac{3 \pm \sqrt{( -3)^2 - 4 \cdot 1 \cdot ( -4 )}}{2} \] \[ x = \frac{3 \pm \sqrt{9 + 16}}{2} \] \[ x = \frac{3 \pm 5}{2} \] Thus, the critical points are \( x = -1 \) and \( x = 4 \). 3. **Test Intervals**: Determine the sign of \( x^2 - 3x - 4 \) in the intervals determined by these critical points: - **For \( x < -1 \)**: Choose \( x = -2 \): \[ (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 > 0 \] The expression is positive. - **For \( -1 < x < 4 \)**: Choose \( x = 0 \): \[ (0)^2 - 3(0) - 4 = -4 < 0 \] The expression is negative. - **For \( x > 4 \)**: Choose \( x = 5 \): \[ (5)^2 - 3(5) - 4 = 25 - 15 - 4 = 6 > 0 \] The expression is positive. 4. **Conclusion**: The function is defined for \( x < -1 \) and \( x > 4 \). **Answer:** C |
C |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-8f0f3168-a847-9e32-96e4-89789138581f', 'request_id': '8f0f3168-a847-9e32-96e4-89789138581f'} | C |