| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the value of $b$ such that the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ intersect at only one point, we need to set the two equations equal to each other and solve for $x$. This will give us the $x$-coordinate of the intersection point. Setting $y_1 = y_2$, we get: $$x^2 + 2x + 7 = 6x + b$$ Rearranging the equation, we get: $$x^2 - 4x + (7 - b) = 0$$ Since the parabola and the line intersect at only one point, this quadratic equation must have only one solution. This means that the discriminant of the quadratic equation must be equal to zero. The discriminant is given by: $$\Delta = (-4)^2 - 4(1)(7 - b) = 16 - 28 + 4b$$ Setting the discriminant equal to zero, we get: $$16 - 28 + 4b = 0$$ Solving for $b$, we get: $$4b = 12$$ $$b = 3$$ Therefore, the value of $b$ is 3. The final answer is B. |
B |
| deepseekr1 | To determine the value of \( b \) such that the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, set the equations equal to each other: \[ x^2 + 2x + 7 = 6x + b \] Rearranging terms gives the quadratic equation: \[ x^2 - 4x + (7 - b) = 0 \] For this quadratic to have exactly one solution, its discriminant must be zero. The discriminant \( D \) is calculated as: \[ D = (-4)^2 - 4 \cdot 1 \cdot (7 - b) = 16 - 4(7 - b) \] Simplify: \[ D = 16 - 28 + 4b = -12 + 4b \] Set \( D = 0 \): \[ -12 + 4b = 0 \implies 4b = 12 \implies b = 3 \] Verification: Substituting \( b = 3 \) into \( y_2 = 6x + 3 \), the equation \( x^2 - 4x + 4 = 0 \) factors to \( (x - 2)^2 = 0 \), confirming a single intersection point at \( x = 2 \). Thus, the correct answer is **B**. **Answer: B** |
B |
| gpto1 | The problem requires us to find the value of \( b \) such that the parabola \( y = x^2 + 2x + 7 \) and the line \( y = 6x + b \) intersect at exactly one point. This means the two equations have one common solution, so when we set them equal to each other, the resulting quadratic equation will have a discriminant \( D = 0 \). Set the two equations equal to each other: \[ x^2 + 2x + 7 = 6x + b \] Bring all terms to one side to form a quadratic equation: \[ x^2 - 4x + (7 - b) = 0 \] Compute the discriminant \( D \): \[ D = (-4)^2 - 4 \times 1 \times (7 - b) = 16 - 4(7 - b) \] Set \( D \) to zero for one real solution: \[ 16 - 4(7 - b) = 0 \] \[ 16 - 28 + 4b = 0 \] \[ -12 + 4b = 0 \] \[ 4b = 12 \] \[ b = 3 \] **Answer: B** |
B |
| GPT-4o | To find the value of \( b \) such that the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, we need to set the equations equal to each other and solve for \( x \): \[ x^2 + 2x + 7 = 6x + b \] Rearrange the equation to form a standard quadratic equation: \[ x^2 + 2x + 7 - 6x - b = 0 \] Simplify: \[ x^2 - 4x + (7 - b) = 0 \] For the line and the parabola to intersect at exactly one point, the quadratic equation must have exactly one solution. This occurs when the discriminant of the quadratic equation is zero. The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by: \[ \Delta = b^2 - 4ac \] In our quadratic equation \( x^2 - 4x + (7 - b) = 0 \), \( a = 1 \), \( b = -4 \), and \( c = 7 - b \). Plugging these values into the discriminant formula, we get: \[ \Delta = (-4)^2 - 4(1)(7 - b) \] \[ \Delta = 16 - 4(7 - b) \] \[ \Delta = 16 - 28 + 4b \] \[ \Delta = 4b - 12 \] Set the discriminant equal to zero for the quadratic to have exactly one solution: \[ 4b - 12 = 0 \] Solve for \( b \): \[ 4b = 12 \] \[ b = 3 \] Thus, the value of \( b \) that makes the line and the parabola intersect at exactly one point is \( 3 \). Answer: B |
B |
| llama3.1-70B-Instruct | A Explanation: The correct answer is A. To solve this problem, set the two equations equal to each other and solve for b : \( \begin{array}{l}x^{2}+2 x+7=6 x+b \\ x^{2}-4 x+(7-b)=0\end{array} \) Since the parabola and the line intersect at only one point, the discriminant of the quadratic equation must equal 0. The discriminant is the value under the radical in the quadratic formula, $b^{2}-4 a c$ . Set the discriminant equal to 0 and solve for b : \( \begin{array}{l}(-4)^{2}-4(1)(7-b)=0 \\ 16-28+4 b=0 \\ 4 b=12 \\ b=3\end{array} \) The best answer is A. |
B |
| llama3.1-405B-Instruct | To find the value of $b$ such that the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ intersect at only one point, we need to set the two equations equal to each other and solve for $x$. This will give us the $x$-coordinate of the intersection point. Setting $y_1 = y_2$, we get: $$x^2 + 2x + 7 = 6x + b$$ Rearranging the equation to get a quadratic equation in terms of $x$, we get: $$x^2 - 4x + (7 - b) = 0$$ For the parabola and the line to intersect at only one point, the discriminant of the quadratic equation must be equal to zero. The discriminant is given by: $$D = (-4)^2 - 4(1)(7 - b)$$ Setting $D = 0$, we get: $$16 - 4(7 - b) = 0$$ Simplifying the equation, we get: $$16 - 28 + 4b = 0$$ Solving for $b$, we get: $$4b = 12$$ $$b = 3$$ Therefore, the value of $b$ is 3. The final answer is B. |
B |
| DeepSeek-Coder-V2 | To determine the value of \( b \) such that the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, we need to set the equations equal to each other and solve for \( x \). The condition for having only one intersection point is that the resulting quadratic equation has exactly one solution, which means its discriminant must be zero. First, set the equations equal to each other: \[ x^2 + 2x + 7 = 6x + b \] Rearrange the equation to standard quadratic form: \[ x^2 + 2x + 7 - 6x - b = 0 \] \[ x^2 - 4x + 7 - b = 0 \] This is a quadratic equation in the form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -4 \), and \( c = 7 - b \). For the quadratic equation to have exactly one solution, the discriminant must be zero. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \] Substitute \( a = 1 \), \( b = -4 \), and \( c = 7 - b \) into the discriminant formula: \[ \Delta = (-4)^2 - 4(1)(7 - b) \] \[ \Delta = 16 - 4(7 - b) \] \[ \Delta = 16 - 28 + 4b \] \[ \Delta = 4b - 12 \] Set the discriminant equal to zero for the equation to have exactly one solution: \[ 4b - 12 = 0 \] \[ 4b = 12 \] \[ b = 3 \] Thus, the value of \( b \) is \( 3 \). Therefore, the correct answer is: \[ \boxed{B} \] |
B |
| Qwen2-72B-Instruct | For the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ to intersect at only one point, they must be tangent to each other. This means the quadratic equation formed by setting $y_1 = y_2$ must have only one solution, implying the discriminant must be zero. Set $y_1 = y_2$: \[x^2 + 2x + 7 = 6x + b\] Rearrange the equation: \[x^2 - 4x + (7 - b) = 0\] The discriminant $\Delta$ of a quadratic equation $ax^2 + bx + c = 0$ is given by $\Delta = b^2 - 4ac$. For the equation to have only one solution, $\Delta = 0$. For our equation, $a = 1$, $b = -4$, and $c = 7 - b$: \[\Delta = (-4)^2 - 4(1)(7 - b) = 0\] \[16 - 28 + 4b = 0\] \[4b = 12\] \[b = 3\] Thus, the answer is B. 3. |
B |
| baichuan4 | To find the value of $b$, we need to solve the equation $y_1 = y_2$ for $x$ and $b$. Substituting $y_1$ and $y_2$ gives us: $x^2 + 2x + 7 = 6x + b$ Rearranging the terms, we get: $x^2 - 4x + 7 - b = 0$ Since the parabola and the line intersect at only one point, the discriminant of the quadratic equation must be zero. Therefore, we have: $(-4)^2 - 4(1)(7 - b) = 0$ Solving for $b$, we get: $16 - 28 + 4b = 0$ $4b = 12$ $b = 3$ So, the correct answer is B. 3. |
B |
| doubao | D. 4. First find the intersection point by setting the two equations equal to each other: $x^2 + 2x + 7 = 6x + b$. Rearranging gives $x^2 - 4x + 7 - b = 0$. Since they intersect at only one point, the discriminant of this quadratic equation is 0. The discriminant is $(-4)^2 - 4(1)(7-b) = 0$. Solving for b gives $b = 4$. |
B |
| minimax | For the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ to intersect at only one point, they must be tangent to each other. This means that the quadratic equation formed by setting $y_1$ equal to $y_2$ has exactly one solution. So, we set the two equations equal to each other: $x^2 + 2x + 7 = 6x + b$ Now, rearrange the equation to bring all terms to one side: $x^2 + 2x - 6x + 7 - b = 0$ $x^2 - 4x + (7 - b) = 0$ For this quadratic equation to have exactly one solution, its discriminant must be zero. The discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by $b^2 - 4ac$. In our case, $a = 1$, $b = -4$, and $c = 7 - b$. So, the discriminant is: $(-4)^2 - 4(1)(7 - b) = 0$ $16 - 28 + 4b = 0$ $4b = 12$ $b = 3$ Therefore, the value of $b$ for which the parabola and the line intersect at only one point is 3. Answer: B. 3 |
B |
| kimi | To find the value of \( b \) for which the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, we need to set \( y_1 \) equal to \( y_2 \) and solve for \( x \). The equation will be: \[ x^2 + 2x + 7 = 6x + b \] Rearrange the terms to form a quadratic equation: \[ x^2 - 4x + (7 - b) = 0 \] For the parabola and the line to intersect at exactly one point, this quadratic equation must have exactly one solution. This happens when the discriminant is zero. The discriminant \(\Delta\) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \] In our case, \( a = 1 \), \( b = -4 \), and \( c = 7 - b \). So the discriminant is: \[ \Delta = (-4)^2 - 4 \cdot 1 \cdot (7 - b) \] \[ \Delta = 16 - 4(7 - b) \] \[ \Delta = 16 - 28 + 4b \] \[ \Delta = 4b - 12 \] Set the discriminant equal to zero for the quadratic to have exactly one solution: \[ 4b - 12 = 0 \] \[ 4b = 12 \] \[ b = 3 \] So the value of \( b \) is \(\boxed{B. 3}\). |
B |
| wenxinyiyan | A Explanation: Given that the parabola $y_1 = x^2 + 2x + 7$ and the line $y_2 = 6x + b$ intersect at only one point, we can set them equal to each other: $x^2 + 2x + 7 = 6x + b$ Rearranging this equation gives: $x^2 - 4x + (7 - b) = 0$ Since there is only one point of intersection, the discriminant of this quadratic equation must be zero. The discriminant is given by: $\Delta = (-4)^2 - 4 \cdot 1 \cdot (7 - b)$ Setting the discriminant to zero and solving for $b$ yields: $16 - 4(7 - b) = 0$ $16 - 28 + 4b = 0$ $4b = 12$ $b = 7$ Therefore, the value of $b$ is 7, which corresponds to option A. |
B |
| zhipu | To determine the value of \( b \) for which the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, we need to set the equations equal to each other and solve for \( x \): \[ x^2 + 2x + 7 = 6x + b \] Rearrange the equation to form a standard quadratic equation: \[ x^2 + 2x + 7 - 6x - b = 0 \] Simplify: \[ x^2 - 4x + (7 - b) = 0 \] For the parabola and the line to intersect at only one point, this quadratic equation must have exactly one solution. This occurs when the discriminant of the quadratic equation is zero. The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by: \[ \Delta = b^2 - 4ac \] In our equation \( x^2 - 4x + (7 - b) = 0 \), we have \( a = 1 \), \( b = -4 \), and \( c = 7 - b \). Plugging these into the discriminant formula: \[ \Delta = (-4)^2 - 4 \cdot 1 \cdot (7 - b) \] \[ \Delta = 16 - 4(7 - b) \] \[ \Delta = 16 - 28 + 4b \] \[ \Delta = 4b - 12 \] Set the discriminant equal to zero for a single intersection point: \[ 4b - 12 = 0 \] Solve for \( b \): \[ 4b = 12 \] \[ b = 3 \] Thus, the value of \( b \) is \( 3 \). The correct answer is: B. 3 |
B |
| gpto1 | The problem requires us to find the value of \( b \) such that the parabola \( y = x^2 + 2x + 7 \) and the line \( y = 6x + b \) intersect at exactly one point. This means the two equations have one common solution, so when we set them equal to each other, the resulting quadratic equation will have a discriminant \( D = 0 \). Set the two equations equal to each other: \[ x^2 + 2x + 7 = 6x + b \] Bring all terms to one side to form a quadratic equation: \[ x^2 - 4x + (7 - b) = 0 \] Compute the discriminant \( D \): \[ D = (-4)^2 - 4 \times 1 \times (7 - b) = 16 - 4(7 - b) \] Set \( D \) to zero for one real solution: \[ 16 - 4(7 - b) = 0 \] \[ 16 - 28 + 4b = 0 \] \[ -12 + 4b = 0 \] \[ 4b = 12 \] \[ b = 3 \] **Answer: B** |
B |
| deepseekv3 | To determine the value of \( b \) such that the parabola \( y_1 = x^2 + 2x + 7 \) and the line \( y_2 = 6x + b \) intersect at only one point, we follow these steps: 1. **Set the equations equal to each other:** \[ x^2 + 2x + 7 = 6x + b \] 2. **Rearrange the equation to form a quadratic equation:** \[ x^2 + 2x + 7 - 6x - b = 0 \\ x^2 - 4x + (7 - b) = 0 \] 3. **Determine the condition for a single intersection point:** A quadratic equation \( ax^2 + bx + c = 0 \) has exactly one real solution when its discriminant \( D \) is zero. The discriminant is given by: \[ D = b^2 - 4ac \] For our equation \( x^2 - 4x + (7 - b) = 0 \): - \( a = 1 \) - \( b = -4 \) - \( c = 7 - b \) So, \[ D = (-4)^2 - 4(1)(7 - b) \\ D = 16 - 28 + 4b \\ D = 4b - 12 \] Set the discriminant to zero for a single solution: \[ 4b - 12 = 0 \\ 4b = 12 \\ b = 3 \] **Answer:** \(\boxed{B}\) |
B |