| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The expressions inside each square root must be non-negative. Thus, $25-x^2\ge0$, so $-5\le x\le 5$, and $-(x-2)\ge0$, so $x\le 2$. Therefore, the domain of $h$ is $[-5,2]$, which has width $\boxed{7}$. The best answer is A. |
A |
| deepseekr1 | The function \( h(x) = \sqrt{25 - x^2} + \sqrt{-(x - 2)} \) requires both square roots to be non-negative. For \( \sqrt{25 - x^2} \), solving \( 25 - x^2 \geq 0 \) gives \( x \in [-5, 5] \). For \( \sqrt{-(x - 2)} \), simplifying \( -(x - 2) \geq 0 \) leads to \( x \leq 2 \). The intersection of \( [-5, 5] \) and \( (-\infty, 2] \) is \( [-5, 2] \). The width of this interval is \( 2 - (-5) = 7 \). **Answer: A** |
A |
| gpto1 | A | A |
| GPT-4o | To determine the domain of the function \( h(x) = \sqrt{25 - x^2} + \sqrt{-(x-2)} \), we need to analyze the conditions under which the expressions inside the square roots are non-negative, as the square root function is only defined for non-negative values. 1. For \( \sqrt{25 - x^2} \): \[ 25 - x^2 \geq 0 \implies x^2 \leq 25 \implies -5 \leq x \leq 5 \] 2. For \( \sqrt{-(x-2)} \): \[ -(x-2) \geq 0 \implies x - 2 \leq 0 \implies x \leq 2 \] The domain of \( h(x) \) is the intersection of these two intervals: \[ -5 \leq x \leq 5 \quad \text{and} \quad x \leq 2 \] Thus, the combined interval is: \[ -5 \leq x \leq 2 \] The width of this interval is: \[ 2 - (-5) = 2 + 5 = 7 \] Therefore, the width of the interval is \( 7 \). Answer: A |
A |
| llama3.1-70B-Instruct | A The answer is A. |
A |
| llama3.1-405B-Instruct | The expressions inside each square root must be non-negative. Thus, $25-x^2\ge0$, so $-5\le x\le 5$, and $-(x-2)\ge0$, so $x\le 2$. Therefore, the domain of $h$ is $[-5,2]$, which has width $\boxed{7}$. The best answer is A. |
A |
| DeepSeek-Coder-V2 | To determine the domain of the function \( h(x) = \sqrt{25 - x^2} + \sqrt{-(x - 2)} \), we need to ensure that both square root terms are defined. This requires the expressions inside the square roots to be non-negative. 1. For the term \(\sqrt{25 - x^2}\): \[ 25 - x^2 \geq 0 \] Solving this inequality: \[ 25 \geq x^2 \implies -5 \leq x \leq 5 \] 2. For the term \(\sqrt{-(x - 2)}\): \[ -(x - 2) \geq 0 \implies x - 2 \leq 0 \implies x \leq 2 \] Combining these two conditions, we get: \[ -5 \leq x \leq 5 \quad \text{and} \quad x \leq 2 \] The intersection of these intervals is: \[ -5 \leq x \leq 2 \] The width of this interval is: \[ 2 - (-5) = 2 + 5 = 7 \] Thus, the width of the domain is 7. Therefore, the correct answer is: \[ \boxed{A} \] |
A |
| Qwen2-72B-Instruct | To find the domain of $h(x) = \sqrt{25-x^2}+\sqrt{-(x-2)}$, we need to find the values of $x$ for which both square roots are defined. For $\sqrt{25-x^2}$ to be defined, we need $25-x^2 \geq 0$. This implies $-5 \leq x \leq 5$. For $\sqrt{-(x-2)}$ to be defined, we need $-(x-2) \geq 0$. This implies $x \leq 2$. The intersection of these two conditions is $-5 \leq x \leq 2$. The width of this interval is $2 - (-5) = 7$. Answer: A. 7 |
A |
| baichuan4 | B | A |
| doubao | For the first square root to be defined, $25-x^2 \geq 0$, which gives $-5 \leq x \leq 5$. For the second square root to be defined, $-(x-2) \geq 0$, that is $x \leq 2$. The common interval is $-5 \leq x \leq 2$. The width of this interval is $2-(-5)=7$. So the answer is A. | A |
| minimax | To find the domain of the function $h(x) = \sqrt{25-x^2}+\sqrt{-(x-2)}$, we need to consider the domain of each square root separately, because the domain of a function is the set of all possible input values (x-values) for which the function is defined. For the first square root, $\sqrt{25-x^2}$, the expression inside the square root must be greater than or equal to zero for the square root to be defined. This gives us the inequality: $25 - x^2 \geq 0$ Solving this inequality, we get: $x^2 \leq 25$ $-5 \leq x \leq 5$ So the domain of the first square root is $[-5, 5]$. For the second square root, $\sqrt{-(x-2)}$, the expression inside the square root must also be greater than or equal to zero: $-(x-2) \geq 0$ $x - 2 \leq 0$ $x \leq 2$ So the domain of the second square root is $(-\infty, 2]$. The domain of the entire function $h(x)$ is the intersection of the domains of the two square roots, which is the set of x-values that satisfy both conditions: $[-5, 5] \cap (-\infty, 2] = [-5, 2]$ The width of this interval is the difference between the upper and lower bounds: Width = $2 - (-5) = 2 + 5 = 7$ Therefore, the width of the domain of the function $h(x)$ is 7. Answer: A. 7 |
A |
| kimi | To find the domain of the function \( h(x) = \sqrt{25-x^2} + \sqrt{-(x-2)} \), we need to determine the values of \( x \) for which the function is defined. This means both the expressions inside the square roots must be non-negative. 1. For \( \sqrt{25-x^2} \) to be defined, the expression inside the square root must be non-negative: \[ 25 - x^2 \geq 0 \] \[ x^2 \leq 25 \] This implies that \( -5 \leq x \leq 5 \). 2. For \( \sqrt{-(x-2)} \) to be defined, the expression inside the square root must also be non-negative: \[ -(x-2) \geq 0 \] \[ x - 2 \leq 0 \] This implies that \( x \leq 2 \). The domain of \( h(x) \) is the intersection of these two conditions, which is \( -5 \leq x \leq 2 \). The width of this interval is \( 2 - (-5) = 7 \). Answer: A. 7 |
A |
| wenxinyiyan | A Explanation: 1. For the square root function $\sqrt{25-x^2}$ to be defined, its argument must be non-negative: $25 - x^2 \geq 0$ This gives: $x^2 \leq 25$ Or: $-5 \leq x \leq 5$ 2. For the square root function $\sqrt{-(x-2)}$ to be defined, its argument must be non-negative: $-(x-2) \geq 0$ This gives: $x-2 \leq 0$ Or: $x \leq 2$ 3. Combining the two conditions, we get the domain of $h(x)$ to be: $-5 \leq x \leq 2$ 4. The width of this interval is: $2 - (-5) = 7$ Hence, the answer is A. |
A |
| zhipu | To determine the domain of the function \( h(x) = \sqrt{25 - x^2} + \sqrt{-(x - 2)} \), we need to ensure that both square root expressions are defined. 1. **First term**: \( \sqrt{25 - x^2} \) - This term is defined when \( 25 - x^2 \geq 0 \). - Solving \( 25 - x^2 \geq 0 \): \[ 25 \geq x^2 \implies -5 \leq x \leq 5 \] 2. **Second term**: \( \sqrt{-(x - 2)} \) - This term is defined when \( -(x - 2) \geq 0 \). - Solving \( -(x - 2) \geq 0 \): \[ -x + 2 \geq 0 \implies x \leq 2 \] Combining both conditions: - From the first term: \( -5 \leq x \leq 5 \) - From the second term: \( x \leq 2 \) The intersection of these intervals is \( -5 \leq x \leq 2 \). Now, we calculate the width of this interval: - The width is given by the difference between the upper and lower bounds: \[ 2 - (-5) = 2 + 5 = 7 \] Therefore, the width of the domain interval is 7. The correct answer is: A. 7 |
A |
| gpto1 | A | A |
| deepseekv3 | To determine the domain of the function \( h(x) = \sqrt{25 - x^2} + \sqrt{-(x - 2)} \), we need to find the values of \( x \) for which both square root expressions are defined (i.e., their arguments are non-negative). 1. **First Square Root:** \( \sqrt{25 - x^2} \) is defined when \( 25 - x^2 \geq 0 \). This implies: \( x^2 \leq 25 \) So: \( -5 \leq x \leq 5 \) 2. **Second Square Root:** \( \sqrt{-(x - 2)} \) is defined when \( -(x - 2) \geq 0 \). This implies: \( x - 2 \leq 0 \) So: \( x \leq 2 \) 3. **Combining the Conditions:** The domain of \( h(x) \) is the intersection of the two intervals: \( -5 \leq x \leq 2 \) 4. **Width of the Domain:** The width of the interval \( [-5, 2] \) is: \( 2 - (-5) = 7 \) Thus, the width of the domain is **7**. **Answer:** A. 7 |
A |