按照笔记内容，制作成上海高三生的数学第一轮复习教学视频，简单易懂---Here is the complete and accurate extraction of the content from the image in a structured plain text format: --- **Title:** 直线中的数形结合思想 (Mathematical-Graphical Combined Thinking in Straight Lines) **Section: 类型一: 斜率 (Type One: Slope)** * **Explanation:** * `y-b / x-a` is considered the slope of two points `(x,y)` and `(a,b)`. * In particular, `y/x` is considered the slope of two points `(x,y)` and `(0,0)`. **Question 1** * **Question Stem:** 已知实数x, y满足方程`x+2y = 6`, 当`1 ≤ x ≤ 3`时, `(y-1) / (x-2)` 的取值范围为______。 (Given real numbers x, y satisfying the equation `x+2y = 6`, when `1 ≤ x ≤ 3`, the range of `(y-1) / (x-2)` is ______) * **Chart Description:** * **Type:** Cartesian Coordinate System with a line segment. * **Axes:** X-axis and Y-axis are labeled. * **Points:** * `A (1, 5/2)` * `B (3, 3/2)` * `P (x, y)` on the line segment AB. * A distinct point `(2, 1)`. * **Lines:** * A straight line segment connecting `A` and `B`. * Dashed vertical lines from `A` and `B` to the X-axis. * Dashed horizontal lines from `A` and `B`. * **Annotations:** * `P(x,y) 在 AB 上` (P(x,y) is on AB) * `kPA = (5/2 - 1) / (1 - 2) = (3/2) / (-1) = -3/2` * `kPB = (3/2 - 1) / (3 - 2) = (1/2) / 1 = 1/2` * `=> k ∈ [-3/2, 1/2]` **Question 2** * **Question Stem:** 若函数 `f(x) = log₂(x+1)`, 且 `c > b > a > 0`, 且 `abc ≠ 0`, 则 `f(a)/a`, `f(b)/b`, `f(c)/c` 的大小关系是 ( ). (If function `f(x) = log₂(x+1)`, and `c > b > a > 0`, and `abc ≠ 0`, then the relationship between `f(a)/a`, `f(b)/b`, `f(c)/c` is ( ).) * **Options:** * A. `f(a)/a > f(b)/b > f(c)/c` * B. `f(c)/c > f(b)/b > f(a)/a` * C. `f(a)/b > f(b)/a > f(c)/c` * D. `f(a)/a > f(b)/b > f(c)/b` * **Chart Description:** * **Type:** Graph of a logarithmic function. * **Axes:** X-axis and Y-axis. * **Function:** A curved line representing `f(x) = log₂(x+1)`. * **Points:** * Origin `(0,0)`. * Points `(a, f(a))`, `(b, f(b))`, `(c, f(c))` on the curve. * **Lines:** Straight line segments connecting `(0,0)` to each of `(a, f(a))`, `(b, f(b))`, `(c, f(c))`. These segments are labeled `k₁`, `k₂`, `k₃` respectively. * **Annotations:** * `f(a)/a, f(b)/b, f(c)/c` are considered as the slopes of `(a,f(a))` and `(0,0)`, `(b,f(b))` and `(0,0)`, `(c,f(c))` and `(0,0)` respectively. * `由图像知 k₁ > k₂ > k₃, 故选 A` (From the image, `k₁ > k₂ > k₃`, so choose A) **Question 3** * **Question Stem:** 求函数 `f(x) = (sqrt(x+1) - 2) / (x+4)`, `x ∈ [0,3]` 的值域. (Find the range of the function `f(x) = (sqrt(x+1) - 2) / (x+4)`, `x ∈ [0,3]`.) * **Solution Hint/Transformation:** * `= (y-2) / (x+4)` * Let `y = sqrt(x+1)`, for `x ∈ [0,3]`. * `->` This represents the slope of `(x,y)` and `(-4,2)`. * **Chart Description:** * **Type:** Cartesian Coordinate System with a curve and points. * **Axes:** X-axis and Y-axis. * **Function:** A curved line segment representing `y = sqrt(x+1)` for `x ∈ [0,3]`. * **Points:** * `A (0,1)` (start of curve) * `B (3,2)` (end of curve) * `P (-4,2)` (a fixed point) * **Lines:** Straight line segments connecting `P` to `A` and `P` to `B`. * **Annotations:** * `kPB = 0` (Slope of PB) * `kPA = (1-2) / (0 - (-4)) = -1/4` (Slope of PA) * `=> k ∈ [-1/4, 0]` **Section: 暑假讲义第12讲思考4——旧题新做 (Summer Lecture Notes, Lecture 12, Reflection 4 — Old problem, new approach)** **Reflection 4** * **Question Stem:** 函数 `f(x) = (cos x + 1) / (sin x - 2)` 的值域为_______. (The range of function `f(x) = (cos x + 1) / (sin x - 2)` is _______.) * **Answer/Range:** `[-4/3, 0]` * **Solution Hint/Transformation:** * `1°` If `cos x = -1`, then `f(x) = 0`. * `2°` If `cos x ≠ -1`, then `1/f(x) = (sin x - 2) / (cos x + 1)`. * This is interpreted as the slope of `(cos x, sin x)` and `(-1, 2)`. * **Chart Description:** * **Type:** Cartesian Coordinate System with a unit circle and an external point. * **Axes:** X-axis and Y-axis. * **Shapes:** A circle centered at `(0,0)` with radius `1` (representing `(cos x, sin x)` points). * **Points:** * Point `P(-1,2)`. * A point on the circle `(x,y)` representing `(cos x, sin x)`. * **Lines:** * A line connecting `P(-1,2)` to a point on the circle, representing a slope. * Tangent lines from `P(-1,2)` to the circle. * A radius to the tangent point, forming a right angle with the tangent. * **Annotations:** * `y - 2 = k(x + 1)` (Equation of a line through `P(-1,2)`) * `d = |k+2| / sqrt(k^2+1) = 1` (Distance from origin `(0,0)` to the line `kx - y + k + 2 = 0` is `1`) * `=> k^2 + 4k + 4 = k^2 + 1` * `=> 4k = -3` * `=> k = -3/4` * `k ∈ [-3/4, ∞)` * `=> 1/f(x) ∈ [-3/4, ∞)` * `=> f(x) ∈ [-4/3, 0)` (This represents the range when `cos x ≠ -1`. Combined with `f(x)=0` from the first case, the total range is `[-4/3, 0]`) * **Problem Explanation:** 虽然圆还没复习到，但是基本的圆心和半径各位同学还是没问题的。这道题试试看？ (Although circles haven't been reviewed yet, the basic center and radius should be fine for all students. Try this problem?) **Question 5** * **Question Stem:** 在直角坐标平面上, 点 `P(x,y)` 的坐标满足 `x² - 2x + y² = 0`, 点 `Q(a,b)` 的坐标满足 `a² + b² + 6a - 8b + 24 = 0`, 则 `(y-b) / (x-a)` 的取值范围是 ( ). (In the Cartesian coordinate plane, point `P(x,y)` satisfies `x² - 2x + y² = 0`, point `Q(a,b)` satisfies `a² + b² + 6a - 8b + 24 = 0`, then the range of `(y-b) / (x-a)` is ( ).) * **Options:** * A. `[-2, 2]` * B. `[-4-sqrt(7)/3, -4+sqrt(7)/3]` * C. `[-1, 3]` * D. `[-(6+sqrt(7))/3, (6+sqrt(7))/3]` (Corrected likely typo from image) **Section: 类型二: 点到点的距离 (Type Two: Point to Point Distance)** * **Explanation:** * The formula `sqrt((x-x₁)² + (y-y₁)²)` can be considered as the distance from `(x,y)` to `(x₁,y₁)`. * **Example:** `y = sqrt(x²-2x+2) = sqrt((x-1)²+1²)`, which can be considered as the distance from `(x,0)` to `(1,1)`. * **Lecture Note:** 暑假讲义第10讲思考2 问题翻译过来就是两点距离的平方 (暑假已经讲过，这里不再赘述) (Summer Lecture Notes, Lecture 10, Reflection 2. The problem translates to the square of the distance between two points (already covered in summer, no further details here).) **Question 6** * **Question Stem:** 若x, a, b均为实数，且 `(a+2)² + (b-3)² = 1`, 则 `(x-a)² + (ln x - b)²` 的最小值为 ( ). (If x, a, b are real numbers, and `(a+2)² + (b-3)² = 1`, then the minimum value of `(x-a)² + (ln x - b)²` is ( ).) * **Options:** * A. `3√2` * B. `18` * C. `3√2 - 1` * D. `19 - 6√2` **Question 7** * **Question Stem:** 求函数 `y = sqrt(x²+1) + sqrt(x²-4x+8)` 的值域. (Find the range of the function `y = sqrt(x²+1) + sqrt(x²-4x+8)`.) * **Answer/Range:** `[sqrt(13), +∞)` * **Solution Hint/Transformation:** * `= sqrt(x²+1²) + sqrt((x-2)²+2²)` * This represents the sum of distances from `(x,0)` to `(0,1)` and from `(x,0)` to `(2,2)`. * **Chart Description:** * **Type:** Cartesian Coordinate System with a triangle formation. * **Axes:** X-axis is labeled. Y-axis is implied. * **Points:** * `A (0,1)` * `B (2,2)` * `P (x,0)` on the X-axis. * `A' (0,-1)` (reflection of A across X-axis) * **Lines:** * Line segments `AP`, `BP`, `A'P`, `A'B`. * **Annotations:** * `AP+BP min = |A'B|` * `= sqrt(2² + 3²)` * `= sqrt(4+9)` * `= sqrt(13)` **Question 8** * **Source:** 2023年上海徐汇区高三一模第12题 5分 (2023 Shanghai Xuhui District Grade 12 Mock Exam, Question 12, 5 points) * **Question Stem:** 已知正实数 a, b 满足 `3a + 2b = 6`, 则 `b + sqrt(a² + b² - 2b + 1)` 的最小值为 _______. (Given positive real numbers a, b satisfy `3a + 2b = 6`, then the minimum value of `b + sqrt(a² + b² - 2b + 1)` is _______.) * **Solution Hint/Transformation:** * `= b + sqrt(a² + (b-1)²) ` * This represents `b` plus the distance `d` from `(a,b)` to `(0,1)`. * `=> |PA| + |PH| min,` (where `P` is `(a,b)`) * **Chart Description:** * **Type:** Cartesian Coordinate System with a line and points. * **Axes:** X-axis and Y-axis are labeled. * **Lines:** * A straight line `3x + 2y = 6`. * A line segment connecting `A(0,1)` to `P(a,b)`. * A segment representing `b` (y-coordinate of P). * **Points:** * `A (0,1)` * `P (a,b)` on the line `3a + 2b = 6`. * `A' (74/13, 2/13)` (reflection of A across the line `3a+2b=6`). * **Annotations:** * `A 关于直线 3x+2y=6 的对称点 A' (74/13, 2/13) ` (A' is the reflection of A across the line `3x+2y=6`) * `min = 2/sqrt(13)` (The value 2/13 is written, but the radical sign underneath suggests it should be `2/√13` or `2sqrt(13)/13`). **Question 9 (Advanced)** * **Question Stem:** 已知 `(x-1)² + y² = 4`, 则 `(1/2) * sqrt(4+2x-2y) + sqrt(7-2x)` 的最小值为 _______. (Given `(x-1)² + y² = 4`, then the minimum value of `(1/2) * sqrt(4+2x-2y) + sqrt(7-2x)` is _______. ) * **Annotation:** 再次进阶，但试试看？ (Advanced again, but try it?) **Section: 类型三: 点到直线的距离 (Type Three: Point to Line Distance)** * **Explanation:** * Formula `|Ax+By+C| / sqrt(A²+B²)`. * `主动联想点到直线距离公式，注意看是否带绝对值，从而翻译为距离 or 有向距离` (Actively recall the point-to-line distance formula, pay attention to whether there is an absolute value, thus translating to distance or signed distance). **Question 10** * **Question Stem:** 已知实数x, y满足`x²+y²-4x+6y+12 = 0`, 则`2x-y-2`的最小值是 ( ). (Given real numbers x, y satisfy `x²+y²-4x+6y+12 = 0`, then the minimum value of `2x-y-2` is ( ).) * **Options:** * A. `-1 + √5` * B. `4 - √5` * C. `5 - √5` * D. `5 + √5` **Question 11** * **Question Stem:** 已知实数 x,y 满足 `x|x| + y|y|/3 = 1`, 则 `|sqrt(3)x + y - 6|` 的取值范围是 _______. (Given real numbers x,y satisfy `x|x| + y|y|/3 = 1`, then the range of `|sqrt(3)x + y - 6|` is _______. ) * **Problem Explanation:** 这部分的题目基本都涉及圆锥曲线 具体的过程不再展示 重点是要有意识 能看出来是一个类似于距离的公式 同时，二维函数值域的方法又可以 总结一个了——数形结合 (Most of these problems involve conic sections. The specific process will not be shown. The key is to be aware. To be able to recognize a formula similar to distance. At the same time, the method for finding the range of a two-variable function can also be summarized as—mathematical-graphical combined thinking.) **Question 12** * **Source:** (2018年高考真题上海卷第12题 5分) 自动判卷 较难 填空 ((2018 Gaokao Real Exam Shanghai Volume Question 12, 5 points) Automatic grading, Difficult, Fill-in-the-blank) * **Question Stem:** 已知实数 `x₁, x₂, y₁, y₂` 满足 `x₁² + y₁² = 1`, `x₂² + y₂² = 1`, `x₁x₂ + y₁y₂ = 1/2`, 则 `|x₁ + y₁ - 1| / √2 + |x₂ + y₂ - 1| / √2` 的最大值为 _______. (Given real numbers `x₁, x₂, y₁, y₂` satisfy `x₁² + y₁² = 1`, `x₂² + y₂² = 1`, `x₁x₂ + y₁y₂ = 1/2`, then the maximum value of `|x₁ + y₁ - 1| / √2 + |x₂ + y₂ - 1| / √2` is _______. ) ---