🧠:[<think>Okay, so I have this problem about an ellipse. Let me try to figure it out step by step. First, part (1) asks for the standard equation of the ellipse. The ellipse is centered at the origin, which is (0,0). The left focus is at F1(-√3, 0), and the right vertex is at D(2, 0). Hmm. So, since the center is at the origin, the standard equation of an ellipse is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis. Now, the right vertex is at (2,0), so that means the major axis is along the x-axis because the vertex is on the x-axis. Therefore, a is the distance from the center to the vertex, so a = 2. Next, we know the coordinates of the left focus are (-√3, 0). For an ellipse, the distance from the center to each focus is c, so here c = √3. The relationship between a, b, and c in an ellipse is c² = a² - b². We have a = 2 and c = √3. Let's plug these into the equation. c² = a² - b²(√3)² = 2² - b²3 = 4 - b²So, solving for b²: b² = 4 - 3 = 1. Therefore, b = 1. So the standard equation of the ellipse should be (x²/4) + (y²/1) = 1, which simplifies to (x²/4) + y² = 1. Let me double-check. The major axis is along the x-axis with a = 2, so vertices at (±2, 0). The foci are at (±c, 0) where c² = a² - b². Given c = √3, then c² = 3, so 3 = 4 - b² gives b² = 1. Yeah, that seems right. So part (1) should be x²/4 + y² = 1. Moving on to part (2). If P is a moving point on the ellipse, find the trajectory equation of the midpoint M of segment PA. Point A is given as (1, 1/2). So, for any point P(x1, y1) on the ellipse, the midpoint M of PA would have coordinates ((x1 + 1)/2, (y1 + 1/2)/2). But since P is on the ellipse, it satisfies x1²/4 + y1² = 1. We need to express the coordinates of M in terms of x and y and eliminate x1 and y1. Let me denote the coordinates of M as (h, k). So,h = (x1 + 1)/2 => x1 = 2h - 1k = (y1 + 1/2)/2 => y1 = 2k - 1/2Since (x1, y1) lies on the ellipse, substitute x1 and y1 into the ellipse equation:( (2h - 1)² ) /4 + (2k - 1/2)² = 1So, expand this equation and then replace h and k with x and y to get the trajectory equation.Let me compute each term step by step. First term: (2h - 1)² /4Let's compute (2h - 1)²: 4h² -4h +1Divide by 4: (4h² -4h +1)/4 = h² - h + 1/4Second term: (2k - 1/2)²Compute (2k - 1/2)²: 4k² - 2k + 1/4So adding both terms:(h² - h + 1/4) + (4k² - 2k + 1/4) = 1Combine like terms:h² - h + 1/4 +4k² -2k +1/4 =1So h² +4k² -h -2k + 1/2 =1Subtract 1 from both sides:h² +4k² -h -2k -1/2=0Multiply both sides by 2 to eliminate the fraction:2h² +8k² -2h -4k -1=0But maybe instead of multiplying, I should rearrange the equation. Let's see. Alternatively, perhaps complete the square for h and k. Starting from:h² - h +4k² -2k +1/2 =1Wait, original equation after combining terms was h² +4k² -h -2k +1/2 =1. Then subtract 1 to get h² +4k² -h -2k -1/2=0.Alternatively, maybe let's complete the squares for h and k. Let's try.First, group h terms and k terms:h² - h +4k² -2k = 1/2For the h terms: h² - h. To complete the square, take coefficient of h, which is -1, divide by 2: -1/2, square it: 1/4. So add and subtract 1/4.Similarly for the k terms: 4k² -2k. Factor out 4 first: 4(k² - (2/4)k) = 4(k² - (1/2)k). Now complete the square inside: take coefficient of k, which is -1/2, divide by 2: -1/4, square: 1/16. So add and subtract 1/16.So:[h² - h +1/4 -1/4] +4[ k² - (1/2)k +1/16 -1/16 ] =1/2This becomes:[(h - 1/2)^2 -1/4] +4[(k - 1/4)^2 -1/16] =1/2Expand:(h -1/2)^2 -1/4 +4(k -1/4)^2 -4*(1/16)=1/2Simplify:(h -1/2)^2 +4(k -1/4)^2 -1/4 -1/4=1/2Combine the constants:-1/4 -1/4 = -1/2So:(h -1/2)^2 +4(k -1/4)^2 -1/2 =1/2Add 1/2 to both sides:(h -1/2)^2 +4(k -1/4)^2 =1So the equation is:(h -1/2)^2 +4(k -1/4)^2 =1Therefore, replacing h and k with x and y, the trajectory of M is:(x -1/2)^2 +4(y -1/4)^2 =1Which is the equation of an ellipse centered at (1/2, 1/4) with major and minor axes. So that's the trajectory equation. Let me check my steps again. Original midpoint coordinates:h = (x1 +1)/2 => x1 =2h -1k = (y1 +0.5)/2 => y1 =2k -0.5Then substitute into ellipse equation x1²/4 + y1²=1:( (2h -1)^2 ) /4 + (2k -0.5)^2 =1Expand:(4h² -4h +1)/4 +4k² -2k +0.25=1Which is h² - h +0.25 +4k² -2k +0.25=1Combine like terms: h² +4k² -h -2k +0.5=1So h² +4k² -h -2k =0.5Then moving to completing squares:h² -h +4k² -2k=0.5Complete square for h: h² -h + (1/4) - (1/4)Complete square for k: 4(k² -0.5k + (1/16)) -4*(1/16)So, 4(k² -0.5k +1/16)=4(k -1/4)^2, and subtract 4*(1/16)=1/4Therefore:(h -1/2)^2 -1/4 +4(k -1/4)^2 -1/4=0.5Thus:(h -1/2)^2 +4(k -1/4)^2 -0.5=0.5Adding 0.5 gives:(h -1/2)^2 +4(k -1/4)^2 =1Yes, so that's correct. Therefore, the trajectory is this ellipse equation. So the answer is (x -1/2)^2 +4(y -1/4)^2=1. Alternatively, if we need to present in standard form, maybe expand it, but the problem says trajectory equation, so this should suffice. Let me confirm with an example. Suppose point P is at (2,0), the right vertex. Then midpoint M between (2,0) and (1,1/2) would be ((2+1)/2, (0 + 0.5)/2)=(1.5, 0.25). Plug into the equation: (1.5 -0.5)^2 +4*(0.25 -0.25)^2= (1)^2 +0=1. Which satisfies the equation. Another point: if P is (-2,0), left vertex. Midpoint between (-2,0) and (1,0.5): ((-2+1)/2, (0 +0.5)/2)= (-0.5, 0.25). Plug into equation: (-0.5 -0.5)^2 +4*(0.25 -0.25)^2= (-1)^2 +0=1. Correct. Another point: if P is (0,1), which is on the ellipse since 0/4 +1=1. Midpoint between (0,1) and (1,0.5): (0.5, 0.75). Plug into equation: (0.5 -0.5)^2 +4*(0.75 -0.25)^2=0 +4*(0.5)^2=4*(0.25)=1. Correct. So all these points satisfy the equation. Therefore, it seems correct. Final Answer(1) The standard equation of the ellipse is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The trajectory equation of the midpoint ( M ) is boxed{left(x - dfrac{1}{2}right)^2 + 4left(y - dfrac{1}{4}right)^2 = 1}.</think>Given an ellipse centered at the origin with the left focus at ( F_1(-sqrt{3}, 0) ) and the right vertex at ( D(2, 0) ):(1) To find the standard equation of the ellipse:- The semi-major axis ( a ) is the distance from the center to the vertex, so ( a = 2 ).- The distance from the center to each focus ( c ) is ( sqrt{3} ).- Using the relationship ( c^2 = a^2 - b^2 ), we find ( b ): [ (sqrt{3})^2 = 2^2 - b^2 implies 3 = 4 - b^2 implies b^2 = 1 implies b = 1 ]- The standard equation of the ellipse is: [ boxed{dfrac{x^2}{4} + y^2 = 1} ](2) To find the trajectory equation of the midpoint ( M ) of segment ( PA ) where ( A ) is ( (1, frac{1}{2}) ):- Let ( P ) be a point ((x_1, y_1)) on the ellipse. The midpoint ( M ) has coordinates (left( dfrac{x_1 + 1}{2}, dfrac{y_1 + frac{1}{2}}{2} right)).- Express ( x_1 ) and ( y_1 ) in terms of ( h ) and ( k ): [ x_1 = 2h - 1 quad text{and} quad y_1 = 2k - frac{1}{2} ]- Substitute ( x_1 ) and ( y_1 ) into the ellipse equation: [ dfrac{(2h - 1)^2}{4} + (2k - frac{1}{2})^2 = 1 ]- Simplify and complete the square: [ (h - frac{1}{2})^2 + 4(k - frac{1}{4})^2 = 1 ]- The trajectory equation of the midpoint ( M ) is: [ boxed{left(x - dfrac{1}{2}right)^2 + 4left(y - dfrac{1}{4}right)^2 = 1} ],<think>Okay, let me try to solve this problem step by step. First, part (1) asks for the standard equation of the ellipse. The problem states that the ellipse is centered at the origin, has a left focus at F₁(-√3, 0), and a right vertex at D(2, 0). I remember that the standard equation of an ellipse centered at the origin with major axis along the x-axis is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis, and the distance from the center to each focus is c, with c² = a² - b².Given that the right vertex is at (2, 0), that means the semi-major axis a is 2, because the vertex is at distance a from the center. So a = 2.Also, the left focus is at (-√3, 0), so the distance from the center to the focus c is √3. Therefore, c = √3.Now, since we know c and a, we can find b using the relationship c² = a² - b². Let's compute that:c² = (√3)² = 3a² = 2² = 4So, 3 = 4 - b² => b² = 4 - 3 = 1 => b = 1.Therefore, the standard equation of the ellipse is (x²/4) + (y²/1) = 1, which simplifies to (x²)/4 + y² = 1.Wait, let me double-check. If a=2, c=√3, then b² = a² - c² = 4 - 3 = 1, yes, so b=1. So the equation is correct. Okay, that seems right for part (1).Now part (2): If P is a moving point on the ellipse, find the trajectory equation of the midpoint M of segment PA. Point A is given as (1, 1/2).Hmm, so we need to parametrize point P on the ellipse, find the midpoint between P and A, and then eliminate the parameter to get the equation in terms of x and y.Let me think. First, since P is on the ellipse (x²/4) + y² = 1, we can parametrize P using parametric equations of the ellipse. For an ellipse, the standard parametric equations are x = a cosθ, y = b sinθ, where θ is the parameter. Here, a=2, b=1, so P can be written as (2 cosθ, sinθ). Then, the coordinates of point A are (1, 1/2). The midpoint M of PA would have coordinates that are the average of the coordinates of P and A. So:x-coordinate of M: (2 cosθ + 1)/2y-coordinate of M: (sinθ + 1/2)/2Let me write that as:x = (2 cosθ + 1)/2y = (sinθ + 1/2)/2We need to eliminate θ from these equations to find the relationship between x and y.Let me solve for cosθ and sinθ in terms of x and y.From the x-coordinate:x = (2 cosθ + 1)/2Multiply both sides by 2:2x = 2 cosθ + 1Subtract 1:2x - 1 = 2 cosθDivide by 2:cosθ = (2x - 1)/2Similarly, from the y-coordinate:y = (sinθ + 1/2)/2Multiply both sides by 2:2y = sinθ + 1/2Subtract 1/2:sinθ = 2y - 1/2Now, we can use the identity cos²θ + sin²θ = 1.Substitute:[(2x - 1)/2]^2 + [2y - 1/2]^2 = 1Let me compute each term:First term: [(2x - 1)/2]^2 = (4x² - 4x + 1)/4Second term: [2y - 1/2]^2 = (4y² - 2y + 1/4)Adding them together:(4x² - 4x + 1)/4 + 4y² - 2y + 1/4 = 1Multiply the first term by 1/4:(4x² - 4x + 1)/4 = x² - x + 1/4So the equation becomes:x² - x + 1/4 + 4y² - 2y + 1/4 = 1Combine like terms:x² - x + 4y² - 2y + 1/4 + 1/4 = 1Which simplifies to:x² - x + 4y² - 2y + 1/2 = 1Subtract 1 from both sides:x² - x + 4y² - 2y + 1/2 - 1 = 0 => x² - x + 4y² - 2y - 1/2 = 0Hmm, this seems a bit messy. Maybe completing the squares for x and y would help to write it in standard form.Let's group x terms and y terms:x² - x + 4y² - 2y = 1/2For the x terms: x² - x. To complete the square, take coefficient of x, which is -1, divide by 2: (-1/2), square it: 1/4. So:x² - x + 1/4 = (x - 1/2)^2Similarly for the y terms: 4y² - 2y. Factor out the 4 first:4(y² - (1/2)y). Now, complete the square inside the parentheses. Coefficient of y is -1/2, half of that is -1/4, square is 1/16. So:4[y² - (1/2)y + 1/16 - 1/16] = 4[(y - 1/4)^2 - 1/16] = 4(y - 1/4)^2 - 4*(1/16) = 4(y - 1/4)^2 - 1/4Therefore, substituting back into the equation:(x - 1/2)^2 - 1/4 + 4(y - 1/4)^2 - 1/4 = 1/2Combine the constants on the left:(x - 1/2)^2 + 4(y - 1/4)^2 - 1/4 - 1/4 = 1/2Which is:(x - 1/2)^2 + 4(y - 1/4)^2 - 1/2 = 1/2Add 1/2 to both sides:(x - 1/2)^2 + 4(y - 1/4)^2 = 1So, the equation simplifies to:(x - 1/2)^2 + 4(y - 1/4)^2 = 1This is an ellipse centered at (1/2, 1/4) with semi-major axis 1 (since under x term, the denominator would be 1) and semi-minor axis 1/2 (since under y term, the coefficient is 4, which is 1/(1/2)^2). The major axis is along the x-axis because the larger denominator is under the x term.Wait, but the standard form of an ellipse is ((x - h)^2)/a² + ((y - k)^2)/b² = 1. In this case, it's written as (x - 1/2)^2 + 4(y - 1/4)^2 = 1. So to match the standard form, it's equivalent to:((x - 1/2)^2)/1 + ((y - 1/4)^2)/(1/4) = 1Therefore, a² = 1, so a = 1, and b² = 1/4, so b = 1/2. Hence, the major axis is along the x-direction because a > b (since a=1 and b=1/2). So the center is (1/2, 1/4), major axis length 2a=2, minor axis length 2b=1.Therefore, the trajectory of the midpoint M is an ellipse with equation (x - 1/2)^2 + 4(y - 1/4)^2 = 1.But the problem asks for the trajectory equation, so we can write it in standard form as above. Alternatively, expanding it back into general form.But let me check my steps again to make sure I didn't make any mistakes.Starting from parametric equations:x = (2 cosθ + 1)/2y = (sinθ + 1/2)/2Expressed in terms of cosθ and sinθ:cosθ = (2x - 1)/2sinθ = 2y - 1/2Then, using cos²θ + sin²θ = 1:[(2x - 1)/2]^2 + [2y - 1/2]^2 = 1Expanding:(4x² - 4x + 1)/4 + 4y² - 2y + 1/4 = 1Which becomes x² - x + 1/4 + 4y² - 2y + 1/4 = 1Then combining constants: 1/4 + 1/4 = 1/2, so x² - x + 4y² - 2y + 1/2 = 1Subtract 1: x² - x + 4y² - 2y - 1/2 = 0Then completing the squares:x² - x becomes (x - 1/2)^2 - 1/44y² - 2y becomes 4[(y - 1/4)^2 - 1/16] = 4(y - 1/4)^2 - 1/4Putting together:(x - 1/2)^2 - 1/4 + 4(y - 1/4)^2 - 1/4 = 1/2So left side: (x - 1/2)^2 + 4(y - 1/4)^2 - 1/2 = 1/2Adding 1/2 to both sides: (x - 1/2)^2 + 4(y - 1/4)^2 = 1Yes, that looks correct. So the trajectory equation is an ellipse as derived. Therefore, the answer is (x - 1/2)² + 4(y - 1/4)² = 1.Alternatively, if they want it expanded, let's do that:Expand (x - 1/2)^2: x² - x + 1/4Expand 4(y - 1/4)^2: 4(y² - (1/2)y + 1/16) = 4y² - 2y + 1/4So adding them together:x² - x + 1/4 + 4y² - 2y + 1/4 = 1Combine terms:x² + 4y² - x - 2y + 1/2 = 1Subtract 1:x² + 4y² - x - 2y - 1/2 = 0But the standard form is better, so probably (x - 1/2)^2 + 4(y - 1/4)^2 = 1 is the preferred answer. Let me check once more if this makes sense.Since the midpoint M is between point A (1, 1/2) and a point P moving on the original ellipse. So as P moves, M should trace another ellipse, scaled down and shifted towards A. The calculations seem to confirm that. So the answer for part (2) is the ellipse equation we derived.Therefore, summarizing:1) The standard equation of the given ellipse is x²/4 + y² = 1.2) The trajectory of the midpoint M is (x - 1/2)^2 + 4(y - 1/4)^2 = 1.Final Answer(1) The standard equation of the ellipse is boxed{dfrac{x^2}{4} + y^2 = 1}.(2) The trajectory equation of the midpoint ( M ) is boxed{left(x - dfrac{1}{2}right)^2 + 4left(y - dfrac{1}{4}right)^2 = 1}.</think>Given an ellipse centered at the origin with a left focus at ( F_1(-sqrt{3}, 0) ) and a right vertex at ( D(2, 0) ):1. Standard Equation of the Ellipse: - The semi-major axis ( a ) is 2 (distance from the center to the vertex). - The distance from the center to each focus ( c ) is ( sqrt{3} ). - Using the relationship ( c^2 = a^2 - b^2 ), we find ( b ): [ c^2 = 3, quad a^2 = 4 implies 3 = 4 - b^2 implies b^2 = 1 implies b = 1 ] - The standard equation of the ellipse is: [ boxed{dfrac{x^2}{4} + y^2 = 1} ]2. Trajectory Equation of the Midpoint ( M ) of Segment ( PA ): - Point ( A ) has coordinates ( (1, frac{1}{2}) ). - Parametrize point ( P ) on the ellipse as ( (2 cos theta, sin theta) ). - Midpoint ( M ) coordinates are: [ x = dfrac{2 cos theta + 1}{2}, quad y = dfrac{sin theta + dfrac{1}{2}}{2} ] - Express ( cos theta ) and ( sin theta ) in terms of ( x ) and ( y ): [ cos theta = dfrac{2x - 1}{2}, quad sin theta = 2y - dfrac{1}{2} ] - Using the identity ( cos^2 theta + sin^2 theta = 1 ): [ left( dfrac{2x - 1}{2} right)^2 + left( 2y - dfrac{1}{2} right)^2 = 1 ] - Simplify and complete the square: [ left( x - dfrac{1}{2} right)^2 + 4 left( y - dfrac{1}{4} right)^2 = 1 ] - The trajectory equation of the midpoint ( M ) is: [ boxed{left( x - dfrac{1}{2} right)^2 + 4 left( y - dfrac{1}{4} right)^2 = 1} ]]

🔑:(1) From the given information, the ellipse has its foci on the x-axis. Let the standard equation of the ellipse be frac{x^2}{a^2} + frac{y^2}{b^2} = 1 (a > b > 0).The ellipse passes through the point D(2, 0) and has its left focus at F(-sqrt{3}, 0).Thus, a = 2 and c = sqrt{3}.We can find b using the formula b = sqrt{a^2 - c^2} = 1.Therefore, the standard equation of the ellipse is frac{x^2}{4} + y^2 = 1.(2) Let point P have coordinates (x_0, y_0) and let the midpoint M of segment PA have coordinates (x, y).Using the midpoint formula, we have:begin{cases} x = frac{x_0 + 1}{2} y = frac{y_0 + frac{1}{2}}{2} end{cases}Solving for x_0 and y_0, we get:begin{cases} x_0 = 2x - 1 y_0 = 2y - frac{1}{2} end{cases}Since point P lies on the ellipse, we have:frac{(2x - 1)^2}{4} + (2y - frac{1}{2})^2 = 1Simplifying and rearranging the equation, we get:left(x - frac{1}{2}right)^2 + frac{left(y - frac{1}{4}right)^2}{frac{1}{4}} = 1Hence, the trajectory equation of the midpoint M of segment PA is boxed{left(x - frac{1}{2}right)^2 + 4left(y - frac{1}{4}right)^2 = 1}.

🧠:[<think>Alright, so I have this problem here about finding the number of zeros of an even function f(x) defined on all real numbers. When x is greater than or equal to 0, f(x) is given as the logarithm of (x squared minus 3x plus 3). Hmm, okay. Let me try to unpack this step by step.First off, since the function is even, that means f(-x) = f(x) for all x in R. So, the graph of the function is symmetric about the y-axis. Therefore, whatever happens on the right side of the y-axis (x ≥ 0) is mirrored on the left side (x ≤ 0). That's a key point here. So, if I can figure out the number of zeros for x ≥ 0, then I can just double that number (if there are zeros only on one side) or adjust accordingly. But wait, actually, if there's a zero at x=0, it's the same on both sides since 0 is its own mirror image. So, I need to check if x=0 is a zero as well. Let me check that later.But first, let's focus on the given part: for x ≥ 0, f(x) = lg(x² - 3x + 3). The logarithm here is base 10, I assume, since it's written as lg. Wait, actually, in some contexts, "lg" can mean log base 2, but in others, it's log base 10. Hmm. Wait, actually, in Chinese math notation, "lg" is often used for log base 10, while "ln" is natural log. But since the problem is written in English here, maybe "lg" is log base 10. But maybe the base isn't too important here because the logarithm of 1 is 0 regardless of the base. So, when does f(x) equal zero? That would be when the argument of the logarithm is 1, because log_b(1) = 0 for any base b. So, solving x² - 3x + 3 = 1 would give me the zeros for x ≥ 0. Then, because the function is even, any positive zero would correspond to a negative zero, and if x=0 is a zero, it's only counted once.So, let's start by solving x² - 3x + 3 = 1. Subtract 1 from both sides: x² - 3x + 2 = 0. Then factor that quadratic: x² - 3x + 2 factors into (x - 1)(x - 2) = 0. So the solutions are x = 1 and x = 2. Since we're considering x ≥ 0, both of these are valid. Therefore, on the right side of the y-axis, the function f(x) has zeros at x = 1 and x = 2.Now, since the function is even, the zeros on the left side (x ≤ 0) would be at x = -1 and x = -2. So, that's two zeros on the right and two on the left. But wait, we also need to check if x = 0 is a zero. Let's plug x = 0 into the expression for x ≥ 0: f(0) = lg(0² - 3*0 + 3) = lg(3). Since lg(3) is approximately 0.477, which is not zero. Therefore, x = 0 is not a zero of the function. Therefore, the zeros are at x = ±1 and x = ±2, totaling four zeros. So, the number of zeros of f(x) on R is 4.Wait a second, but before I jump to conclusions, I need to make sure that the logarithm is defined for all x ≥ 0. The argument of the logarithm is x² - 3x + 3. For x ≥ 0, we need to ensure that x² - 3x + 3 > 0. Let me check if this quadratic is always positive.The quadratic x² - 3x + 3. The discriminant is (-3)^2 - 4*1*3 = 9 - 12 = -3. Since the discriminant is negative, the quadratic does not cross the x-axis and is always positive (since the coefficient of x² is positive). Therefore, for all real x, x² - 3x + 3 > 0, so the logarithm is defined for all x ≥ 0. Therefore, the function f(x) is defined for all x ≥ 0, and since it's even, also for x ≤ 0 by symmetry.Therefore, there are no issues with the domain. So, the zeros occur where x² - 3x + 3 = 1, which we found at x = 1 and x = 2 on the right side, mirrored to x = -1 and x = -2 on the left side. Thus, four zeros in total.Wait, but hold on, let me visualize the graph. Since f(x) is even, the graph is symmetric. The original function for x ≥ 0 is lg(x² - 3x + 3). Let's think about the behavior of x² - 3x + 3. Since it's a quadratic opening upwards, with vertex at x = -b/(2a) = 3/(2) = 1.5. Plugging x = 1.5 into the quadratic gives (1.5)^2 - 3*(1.5) + 3 = 2.25 - 4.5 + 3 = 0.75. So the minimum value is 0.75, which is positive. Therefore, the quadratic is always positive, as we concluded earlier.Therefore, the logarithm of this quadratic, which is f(x) for x ≥ 0, will have its minimum at x = 1.5, where the quadratic is minimized. The value there is lg(0.75), which is negative because 0.75 is less than 1. So the graph of f(x) for x ≥ 0 is a curve that starts at x=0 with lg(3) ≈ 0.477, decreases to a minimum at x=1.5 with lg(0.75) ≈ -0.1249, and then increases towards infinity as x increases, since x² -3x +3 grows without bound. But wait, logarithm of a quadratic that goes to infinity would go to infinity, but logarithm is a slow function. So as x increases, lg(x² -3x +3) behaves like lg(x²) = 2 lg|x|, so it goes to infinity as x approaches infinity.But importantly, since the quadratic is minimized at 0.75, which is still positive, the logarithm is defined everywhere. So, the function f(x) for x ≥ 0 crosses the x-axis when the quadratic equals 1, which happens at x=1 and x=2, as we found. Then, since it's even, it also crosses at x=-1 and x=-2. So, four zeros in total.Therefore, the answer should be 4. But let me double-check.Wait, another way to think about this is that even functions have symmetry about the y-axis. So, each positive zero has a corresponding negative zero, except possibly at zero itself. But in this case, there is no zero at x=0, so all zeros come in pairs. Since we found two zeros on the positive side, their negatives make two more, totaling four. That makes sense.Therefore, the number of zeros of f(x) on R is 4. So, the final answer is boxed{4}.Final Answerboxed{4}</think>The function ( y = f(x) ) is an even function defined on (mathbb{R}). For ( x geq 0 ), ( f(x) = lg(x^2 - 3x + 3) ). To determine the number of zeros of ( f(x) ), we need to solve ( f(x) = 0 ).Since ( f(x) ) is even, any zero at ( x = a ) implies a zero at ( x = -a ). For ( x geq 0 ), ( f(x) = 0 ) when ( lg(x^2 - 3x + 3) = 0 ). This occurs when the argument of the logarithm is 1:[x^2 - 3x + 3 = 1]Solving this equation:[x^2 - 3x + 2 = 0 implies (x - 1)(x - 2) = 0 implies x = 1 text{ or } x = 2]Thus, there are zeros at ( x = 1 ) and ( x = 2 ) for ( x geq 0 ). Due to the evenness of ( f(x) ), there are corresponding zeros at ( x = -1 ) and ( x = -2 ).Checking if ( x = 0 ) is a zero:[f(0) = lg(0^2 - 3 cdot 0 + 3) = lg(3) neq 0]Thus, ( x = 0 ) is not a zero. The quadratic ( x^2 - 3x + 3 ) has a negative discriminant (( -3 )), confirming it is always positive, so the logarithm is defined for all ( x ).Therefore, the zeros are at ( x = pm 1 ) and ( x = pm 2 ), resulting in 4 zeros in total.[boxed{4}],<think>Okay, let's tackle this problem step by step. So, we have an even function f(x) defined on all real numbers. When x is greater than or equal to 0, f(x) is given as the logarithm of (x squared minus 3x plus 3). We need to find the number of zeros of f(x) on the real line. Hmm, zeros of the function mean the values of x where f(x) = 0. Since the function is even, whatever happens on the right side (x ≥ 0) is mirrored on the left side (x ≤ 0). So, if I can figure out how many zeros there are for x ≥ 0, I can just double that number (accounting for symmetry) to get the total number of zeros on the entire real line. But wait, if there's a zero at x = 0, it would only count once because it's its own mirror image. So I need to check if x = 0 is a zero or not as well.First, let me recall that an even function satisfies f(-x) = f(x) for all x. So, the graph is symmetric about the y-axis. Therefore, if there's a zero at some positive x = a, there must be a corresponding zero at x = -a. Similarly, if x = 0 is a zero, it remains a single zero.Alright, so first step: analyze f(x) for x ≥ 0, which is given as f(x) = lg(x² - 3x + 3). The logarithm here is base 10, I assume, since it's denoted as lg. Wait, actually, sometimes in different contexts, lg can mean log base 2, but in many math problems, especially in Chinese textbooks, lg is log base 10. Hmm, the problem is written in Chinese maybe? Wait, the original problem is in Chinese? No, the user input is in English. Wait, no, the user just says "lg". In some contexts, "lg" is log base 10, while "ln" is natural log. So, I think here, it's safe to assume that lg is log base 10. So, f(x) = log10(x² - 3x + 3) for x ≥ 0.So, to find the zeros of f(x), we need to solve f(x) = 0. Since log10(z) = 0 when z = 1, because 10^0 = 1. So, the equation becomes x² - 3x + 3 = 1. So, x² - 3x + 3 = 1 ⇒ x² - 3x + 2 = 0. Let's factor this quadratic: x² - 3x + 2 factors into (x - 1)(x - 2) = 0. So the solutions are x = 1 and x = 2. Therefore, for x ≥ 0, the zeros are at x = 1 and x = 2. Since these are both positive numbers, their mirror images will be at x = -1 and x = -2. Therefore, in total, there are 4 zeros? Wait, hold on. Wait, but we need to make sure that these x values are within the domain where the original function is defined. Wait, but the function f(x) is defined on all real numbers because it's an even function. However, when x ≥ 0, f(x) is defined as log(x² - 3x + 3). But we have to check whether x² - 3x + 3 is positive for x ≥ 0 because the logarithm is only defined for positive arguments. So, even though the problem says "the function y = f(x) is defined as an even function on ℝ", we need to ensure that when x ≥ 0, the expression inside the log is positive. So, let's check if x² - 3x + 3 > 0 for all x ≥ 0. Let's see: x² - 3x + 3. The discriminant of this quadratic is (-3)² - 4*1*3 = 9 - 12 = -3, which is negative. Therefore, the quadratic x² - 3x + 3 is always positive for all real x. Therefore, the logarithm is defined for all x ≥ 0, and since the function is even, for x ≤ 0, f(x) = f(-x) = log((-x)^2 - 3*(-x) + 3). Wait, no, hold on. Wait, actually, when x ≤ 0, since the function is even, f(x) = f(-x). But f(-x) is defined as log((-x)^2 - 3*(-x) + 3) when -x ≥ 0, i.e., when x ≤ 0. Wait, but (-x)^2 is x², and -3*(-x) is +3x. So, f(-x) = log(x² + 3x + 3). Therefore, for x ≤ 0, f(x) = log(x² + 3x + 3). But wait, x is negative here, but x² is positive, and 3x is negative. Wait, but x² + 3x + 3. Let's check if that is positive for all x ≤ 0. Let's take x ≤ 0. Let me compute x² + 3x + 3. Let's see, the quadratic x² + 3x + 3. The discriminant is 9 - 12 = -3, which is also negative, so this quadratic is always positive as well. Therefore, f(x) is defined for all real numbers, as the logarithm arguments are always positive. So, f(x) is indeed defined on all ℝ.Therefore, going back, for x ≥ 0, f(x) = log(x² - 3x + 3). The zeros occur when log(x² - 3x + 3) = 0, which gives x² - 3x + 3 = 1, leading to x = 1 and x = 2. Therefore, on the non-negative side, there are two zeros. Since the function is even, these zeros are mirrored at x = -1 and x = -2. Therefore, total zeros are at x = -2, -1, 1, 2. So four zeros in total. But wait, before finalizing this answer, I need to check if x = 0 is a zero. Let's compute f(0). Since f is even, f(0) = f(-0) = f(0), so f(0) = log(0² - 3*0 + 3) = log(3). log(3) is approximately 0.477, which is not zero. Therefore, x = 0 is not a zero. Therefore, the zeros are indeed at x = ±1 and ±2, so four zeros in total.Wait, but let me double-check the equations. For x ≥ 0, solving log(x² - 3x + 3) = 0 gives x² - 3x + 3 = 10^0 = 1. So x² - 3x + 2 = 0, which factors as (x -1)(x -2) = 0. So x = 1 and 2. Correct.Now, since the function is even, the graph for x ≤ 0 is a mirror image of x ≥ 0. Therefore, the zeros at x = 1 and 2 are mirrored to x = -1 and -2. Therefore, total zeros are four. So, the answer should be 4.But let me just verify if there are any other zeros. Since the function is even, the only possible zeros are symmetric. Since for x ≥ 0, there are two zeros, then the total number should be four. Wait, but let me confirm the definition again. The function is even, so f(x) for x < 0 is equal to f(-x). Therefore, if x is negative, f(x) = f(-x) = log((-x)^2 - 3*(-x) + 3) = log(x² + 3x + 3). Wait, but when x < 0, the expression inside the log is x² + 3x + 3, but x is negative. For example, take x = -1: f(-1) = log(1 + (-3) + 3) = log(1) = 0. Wait, that's interesting. Wait, hold on, let me compute f(-1). Since x = -1 is negative, f(-1) = f(1) = log(1² - 3*1 + 3) = log(1 - 3 + 3) = log(1) = 0. So f(-1) = 0, which is correct. Similarly, f(-2) = f(2) = log(4 - 6 + 3) = log(1) = 0. So indeed, x = -1 and -2 are zeros. So that's consistent. So, the zeros at x = 1 and 2 lead to zeros at x = -1 and -2. Therefore, total four zeros.But let me check if there are any other zeros. For x ≥ 0, f(x) = log(x² - 3x + 3). When does this equal zero? Only when x² - 3x + 3 = 1, which gives two solutions. For x < 0, f(x) = log(x² + 3x + 3). Let's see if this can ever be zero. For x < 0, set log(x² + 3x + 3) = 0 ⇒ x² + 3x + 3 = 1 ⇒ x² + 3x + 2 = 0. Factor: (x + 1)(x + 2) = 0. Solutions x = -1 and x = -2. But x < 0, so these are valid. Therefore, the zeros for x < 0 are x = -1 and -2. Therefore, total zeros at ±1 and ±2, so four in total.Therefore, the number of zeros of f(x) on ℝ is 4.Wait, but hold on. Let me check the quadratics again. For x ≥ 0, the quadratic is x² - 3x + 3. For x < 0, f(x) = log(x² + 3x + 3). Wait, but how do we get that? Let me verify. Since the function is even, f(x) = f(-x). For x < 0, -x > 0, so f(-x) = log((-x)^2 - 3*(-x) + 3) = log(x² + 3x + 3). Therefore, for x < 0, f(x) = log(x² + 3x + 3). Therefore, to find zeros for x < 0, solve log(x² + 3x + 3) = 0 ⇒ x² + 3x + 3 = 1 ⇒ x² + 3x + 2 = 0 ⇒ (x + 1)(x + 2) = 0 ⇒ x = -1, x = -2. Since x < 0, these solutions are valid. Therefore, on the negative side, x = -1 and -2. Therefore, total zeros at x = -2, -1, 1, 2. So four zeros. Hence, the answer is 4.But let me think if there's any possibility of overlapping zeros or something. For example, if x = 0 were a zero, but we saw that f(0) = log(3) ≈ 0.477 ≠ 0, so no. Also, the quadratics x² - 3x + 3 and x² + 3x + 3 both have discriminant negative, so they never cross zero; they are always positive, so the log is always defined. Therefore, the only zeros are the four points we found. Therefore, the number of zeros is 4.Final AnswerThe number of zeros of the function ( f(x) ) on ( mathbb{R} ) is boxed{4}.</think>The function ( y = f(x) ) is an even function defined on (mathbb{R}). For ( x geq 0 ), ( f(x) = lg(x^2 - 3x + 3) ). We need to determine the number of zeros of ( f(x) ) on (mathbb{R}).1. Finding zeros for ( x geq 0 ): - The function ( f(x) = lg(x^2 - 3x + 3) ) is zero when the argument of the logarithm is 1. - Solve ( x^2 - 3x + 3 = 1 ): [ x^2 - 3x + 2 = 0 implies (x - 1)(x - 2) = 0 implies x = 1 text{ or } x = 2 ] - Thus, there are zeros at ( x = 1 ) and ( x = 2 ) for ( x geq 0 ).2. Considering the even nature of the function: - Since ( f(x) ) is even, zeros at ( x = 1 ) and ( x = 2 ) imply corresponding zeros at ( x = -1 ) and ( x = -2 ).3. Checking if ( x = 0 ) is a zero: - Evaluate ( f(0) ): [ f(0) = lg(0^2 - 3 cdot 0 + 3) = lg(3) neq 0 ] - Therefore, ( x = 0 ) is not a zero.4. Verifying the domain of the logarithm: - The quadratic ( x^2 - 3x + 3 ) has a negative discriminant (( -3 )), so it is always positive for all real ( x ). - Similarly, the quadratic ( x^2 + 3x + 3 ) (for ( x < 0 )) also has a negative discriminant (( -3 )), ensuring the logarithm is defined for all real ( x ).Thus, the zeros of ( f(x) ) are at ( x = pm 1 ) and ( x = pm 2 ), resulting in a total of four zeros.[boxed{4}]]

🔑:Given that f(x) is an even function, we know that f(-x)=f(x).For xgeqslant 0, f(x)=lg (x^{2}-3x+3).To find the zeros of the function, we solve the equation:lg (x^{2}-3x+3)=0This simplifies to x^{2}-3x+3=1, which gives us (x-1)(x-2)=0.Solving for x, we get x=1 or x=2.Since f(x) is an even function, these zeros are symmetric about the y-axis, giving us two more zeros at x=-1 and x=-2.Thus, the function f(x) has boxed{4} zeros on mathbb{R}.

🧠:[<think>Okay, so I need to solve this geometry problem. Let me try to visualize and understand the given information step by step. We have triangle ABC with N being the midpoint of side AB. It's given that angle A is greater than angle B. Then there's a point D on ray AC such that CD equals BC. So, D is extending beyond point C on the ray AC, right? Because CD is equal to BC, so if I imagine triangle ABC, point D would be such that from C, moving a distance equal to BC along AC's extension. Next, point P is on ray DN and is on the same side of line BC as point A. So, DN is a ray starting from D through N, and P is somewhere on that ray but staying on the same side of BC as A. Also, angle PBC is equal to angle A. Hmm, that seems important. So angle PBC is equal to angle BAC. Then, PC intersects AB at point E, and BC intersects DP at point T. The task is to find the value of the expression (BC/TC) - (EA/EB). First, I need to draw a diagram to get a clear picture. Let me sketch triangle ABC, mark N as the midpoint of AB. Then extend AC beyond C to point D such that CD = BC. Then draw ray DN and place point P on this ray such that angle PBC equals angle A. Then connect PC to find intersection E on AB, and find T where DP intersects BC. Once the diagram is set, maybe I can use coordinate geometry, but that might get messy. Alternatively, look for similar triangles, use Ceva's theorem, Menelaus' theorem, or trigonometric identities. Since angles are involved, maybe trigonometric Ceva?Wait, let me note down the given conditions:1. N is midpoint of AB: so AN = NB.2. Angle A > angle B: so in triangle ABC, angle at A is larger than at B.3. D is on ray AC with CD = BC: so triangle BCD is isosceles with BC = CD.4. P is on ray DN, same side of BC as A, with angle PBC = angle A.5. PC intersects AB at E.6. DP intersects BC at T.7. Find BC/TC - EA/EB.Hmm. Let's see if we can find relations between these segments.First, since CD = BC, triangle BCD is isosceles with base BD. Wait, no. CD = BC, so sides BC and CD are equal, so triangle BCD is isosceles with base BD? Wait, BD is not necessarily a base. Wait, points B, C, D. If CD = BC, then triangle BCD is isosceles with BC = CD, so angles at B and D are equal. Wait, but BD is the base here. Wait, but CD is equal to BC, so the triangle has sides BC and CD equal. So angle at C is between BC and CD, so the two equal sides are BC and CD, so the base angles at B and D are equal. Therefore, angle CBD = angle CDB. But maybe that's not directly useful yet. Let's think about point P. Since angle PBC = angle A, maybe we can use some angle chasing here. Let's denote angle BAC as α, which is equal to angle PBC. Then angle ABC is β, with α > β. Since angle PBC = α, which is given. Since P is on ray DN, which starts at D and goes through N. N is the midpoint of AB, so DN is a median-like line, but starting from D. Perhaps we can use Ceva's theorem on triangle ABC with point E? Wait, E is the intersection of PC and AB. So Ceva's theorem states that (AF/FB) * (BD/DC) * (CE/EA) = 1 for concurrent lines, but here, PC intersects AB at E, but I don't know if the other lines are concurrent. Alternatively, Menelaus' theorem for transversal cutting the sides of the triangle. Alternatively, use coordinates. Let's try coordinate geometry.Let me set coordinate system. Let me place point B at origin (0,0), point C on the x-axis at (c, 0), point A somewhere in the plane. Since angle A > angle B, so point A should be positioned such that angle at A is larger than at B. Let me assume coordinates for simplicity. Let me choose coordinates such that B is at (0,0), C is at (1,0), and AB is such that N is the midpoint. Let me define point A as (a, b), so midpoint N would be ((a/2), (b/2)). But maybe it's better to set BC as a specific length. Let's say BC = 1, so CD = 1 as well. Then coordinate system: B(0,0), C(1,0), D is on the extension of AC beyond C, such that CD = BC = 1. So if point C is at (1,0), then D is located such that vector from C to D is in the direction opposite to B. Wait, no. AC is a ray from A to C, so extending beyond C would be beyond point C. So if AC is from A to C, then D is on that ray beyond C, with CD = BC = 1. So coordinate of D depends on coordinate of A. Alternatively, set coordinate system with point B at (0,0), point C at (1,0). Let me let point A be at (0,2), so that AB is from (0,0) to (0,2), midpoint N is at (0,1). Then AC is from (0,2) to (1,0). Extending beyond C, point D should be such that CD = BC. BC is from (0,0) to (1,0), so BC length is 1. Therefore, CD should be 1. Since AC is the line from (0,2) to (1,0), parametric equation of AC is (t, 2 - 2t) where t from 0 to 1. Extending beyond C (t >1), point D is at (1 + t, 0 - 2t) such that the distance from C (1,0) to D (1 + t, -2t) is 1. So compute distance CD: sqrt(t^2 + ( -2t)^2 ) = sqrt(5t^2) = t*sqrt(5) = 1. Therefore, t = 1/sqrt(5). Therefore, coordinates of D would be (1 + 1/sqrt(5), -2/sqrt(5)). Wait, but this might get complicated. Maybe choosing coordinates where computations are easier. Let me try a different approach. Let me set point B at (0,0), point C at (1,0), and point A at (0,1). Then AB is from (0,0) to (0,1), midpoint N is at (0, 0.5). Then AC is from (0,1) to (1,0). Extending AC beyond C, point D is such that CD = BC = 1. The direction from C to D is same as from A to C. Wait, AC is from (0,1) to (1,0), so direction vector is (1, -1). So to go from C (1,0) to D, moving in direction (1, -1) for length CD=1. The length of vector (1,-1) is sqrt(2), so unit vector is (1/sqrt(2), -1/sqrt(2)). Therefore, moving 1 unit from C, D is (1 + 1/sqrt(2), 0 - 1/sqrt(2)) = (1 + 1/√2, -1/√2). Then, DN is the ray from D to N (0, 0.5). Let me parametrize DN. The parametric equations for DN: starting at D (1 + 1/√2, -1/√2) and going towards N (0, 0.5). The direction vector is (-1 - 1/√2, 0.5 + 1/√2). So parametric form: x = 1 + 1/√2 - t(1 + 1/√2), y = -1/√2 + t(0.5 + 1/√2), where t ≥ 0.Point P is on this ray DN, on the same side of BC as A. Since BC is the x-axis from (0,0) to (1,0), the line BC is the x-axis. The "same side as point A" would be the upper half-plane (since A is at (0,1)). But point D is below the x-axis at (1 + 1/√2, -1/√2), and N is at (0, 0.5). The ray DN goes from D upwards towards N. So points on DN beyond N would be on the other side, but since P is on ray DN starting at D, passing through N, but P is on the same side of BC as A. Wait, since DN starts at D (below BC) and goes through N (above BC), so the ray DN starts at D, goes through N, and continues beyond. The part from D to N crosses BC, but beyond N, it's on the same side as A. Wait, but the problem says P is on ray DN and is on the same side of line BC as point A. So P must be on the part of DN that is above BC, which is from N onwards. Therefore, P is on the extension of DN beyond N. But angle PBC = angle A. Let me compute angle A. In triangle ABC, with A at (0,1), B at (0,0), C at (1,0). So angle at A is angle BAC. Let's compute it. The coordinates are A(0,1), B(0,0), C(1,0). So vectors AB is (0,-1), AC is (1,-1). The angle at A is between AB and AC. The cosine of angle A is (AB · AC)/( |AB||AC| ) = (0*1 + (-1)(-1))/(1 * sqrt(1 + 1)) = 1 / sqrt(2). Therefore, angle A is 45 degrees. So angle PBC is 45 degrees. So angle between PB and BC is 45 degrees. Point P is on DN beyond N. So we need to find point P on DN such that angle PBC is 45 degrees. Let's parametrize point P. Let me consider parametric coordinates of DN beyond N. Let me parameterize DN as starting at D and going towards N. Wait, but since P is on the same side as A, which is above BC, so P is on DN beyond N. So from N, moving in the direction opposite to D. Wait, DN is from D to N, so beyond N would be the ray starting at N and going in the direction from D to N. Wait, but D is below BC, N is above. So direction from D to N is upwards. So beyond N, the direction is the same as from D to N. Wait, but actually, parametrizing DN: if we think of DN as a line, starting at D, passing through N, then beyond N is continuing in the same direction. So parametrize P as N + k*(vector from D to N), where k > 1. Vector from D to N is N - D = (0 - (1 + 1/√2), 0.5 - (-1/√2)) = (-1 - 1/√2, 0.5 + 1/√2). So parametric coordinates for P would be N + k*(vector DN). But maybe it's better to use a parameter t such that P is expressed as N + t*(DN direction). Alternatively, find coordinates of P such that angle PBC is 45 degrees. Let me denote coordinates of P as (x,y). Since P is on DN beyond N, which we can parametrize. Let me compute DN's equation. Coordinates of D: (1 + 1/√2, -1/√2), N: (0, 0.5). The slope of DN is (0.5 - (-1/√2)) / (0 - (1 + 1/√2)) = (0.5 + 1/√2) / (-1 - 1/√2). Let me compute that:Numerator: 0.5 + 1/√2 ≈ 0.5 + 0.707 ≈ 1.207Denominator: -1 - 1/√2 ≈ -1 - 0.707 ≈ -1.707So slope ≈ 1.207 / (-1.707) ≈ -0.707, which is -1/√2. Interesting. So the slope is -1/√2.Therefore, equation of DN: starting at D (1 + 1/√2, -1/√2), with slope -1/√2. Let me verify with point N: when x = 0, does y = 0.5?Starting from D, moving left by (1 + 1/√2) units along x-axis, with slope -1/√2. So the equation is y - (-1/√2) = -1/√2 (x - (1 + 1/√2)). Simplify:y + 1/√2 = -1/√2 x + (1 + 1/√2)(-1/√2)Compute the constant term: (1 + 1/√2)(-1/√2) = -1/√2 - 1/2Therefore, equation becomes:y = -1/√2 x -1/√2 -1/2 -1/√2 = -1/√2 x - (2/√2 + 1/2)Wait, this seems messy. Maybe better to parametrize DN with a parameter t.Let me set t = 0 at D, t = 1 at N. So parametric equations:x(t) = (1 + 1/√2) + t*(0 - (1 + 1/√2)) = (1 + 1/√2)(1 - t)y(t) = (-1/√2) + t*(0.5 + 1/√2)So for t=0: x = 1 + 1/√2, y = -1/√2For t=1: x=0, y=0.5For t >1, we go beyond N towards negative x direction. Wait, but when t >1, x becomes negative? Wait, if t=2, x = (1 + 1/√2)(1 - 2) = -(1 + 1/√2), which is negative. But since P is on the same side of BC as A, which is the upper half-plane (y >0). So when t >1, let's check y-coordinate:y(t) = -1/√2 + t*(0.5 + 1/√2)At t=1: y=0.5At t=2: y= -1/√2 + 2*(0.5 + 1/√2) = -1/√2 +1 + 2/√2 = 1 + (1/√2)Which is positive. So for t >1, y(t) is positive. So P is at some t >1.We need to find t such that angle PBC = 45 degrees. Let's compute coordinates of P as (x(t), y(t)), then compute angle PBC.Point B is at (0,0), point C at (1,0), and P is at (x(t), y(t)). So angle PBC is the angle between vectors BP and BC.Vector BP is (x(t), y(t)), and vector BC is (1,0). The angle between them is 45 degrees. The tangent of angle PBC is y(t)/x(t) = tan(45) = 1. Therefore, y(t)/x(t) = 1 => y(t) = x(t)So we set y(t) = x(t):-1/√2 + t*(0.5 + 1/√2) = (1 + 1/√2)(1 - t)Let me solve for t:Left side: -1/√2 + t*(0.5 + 1/√2)Right side: (1 + 1/√2) - (1 + 1/√2)tBring all terms to left:-1/√2 + t*(0.5 + 1/√2) - (1 + 1/√2) + (1 + 1/√2)t =0Combine like terms:[-1/√2 -1 -1/√2] + t*(0.5 + 1/√2 +1 +1/√2) =0Simplify constants:-1 - 2/√2Coefficient of t:(0.5 +1) + (1/√2 +1/√2) = 1.5 + 2/√2So equation:-1 - 2/√2 + t*(1.5 + 2/√2) =0Solve for t:t = (1 + 2/√2)/(1.5 + 2/√2)Multiply numerator and denominator by √2 to rationalize:Numerator: √2 + 2Denominator: 1.5√2 + 2But let's compute:Numerator: 1 + 2/√2 = 1 + √2Wait, 2/√2 = √2, so numerator is 1 + √2Denominator: 1.5 + 2/√2 = 3/2 + √2Therefore, t = (1 + √2)/(3/2 + √2) = multiply numerator and denominator by 2:(2 + 2√2)/(3 + 2√2)Multiply numerator and denominator by (3 - 2√2):[(2 + 2√2)(3 - 2√2)] / [(3 + 2√2)(3 - 2√2)]Compute denominator: 9 - (2√2)^2 = 9 - 8 =1Numerator: 2*3 + 2*(-2√2) + 2√2*3 + 2√2*(-2√2) =6 -4√2 +6√2 -8Combine terms: (6 -8) + (-4√2 +6√2) = -2 + 2√2Therefore, t = (-2 + 2√2)/1 = 2(√2 -1)Therefore, t = 2(√2 -1)Therefore, coordinates of P:x(t) = (1 + 1/√2)(1 - t) = (1 + 1/√2)(1 - 2√2 + 2)Wait, t = 2(√2 -1), so 1 - t =1 -2√2 +2 = 3 -2√2Wait, wait, compute x(t):x(t) = (1 + 1/√2)(1 - t) where t=2(√2 -1)1 - t =1 -2√2 +2 = 3 -2√2So x(t) = (1 + 1/√2)(3 -2√2)Similarly, y(t) = -1/√2 + t*(0.5 +1/√2)Compute y(t):t=2(√2 -1), so:y(t) = -1/√2 +2(√2 -1)*(0.5 +1/√2)First compute (0.5 +1/√2):0.5 =1/2, so 1/2 +1/√2 = (√2 +2)/2√2Wait, maybe better to compute numerically:0.5 +1/√2 ≈0.5 +0.707 ≈1.207But let's keep it exact:0.5 +1/√2 = (√2/2 +1)/√2 = Not sure. Let's compute 2(√2 -1)*(0.5 +1/√2):First, multiply terms:2(√2 -1)*(1/2 +1/√2) = 2*(√2 -1)*(1/2 +1/√2)Multiply out the terms inside:(√2 -1)*(1/2 +1/√2) = √2*(1/2) + √2*(1/√2) -1*(1/2) -1*(1/√2)= (√2/2 +1 -1/2 -1/√2)Combine like terms:(√2/2 -1/√2) + (1 -1/2)= ( (√2/2 - √2/2) ) +1/2Wait, √2/2 -1/√2 = (√2/2 - √2/2) =0? Wait, no:Wait, 1/√2 = √2/2, so √2/2 -1/√2 = √2/2 - √2/2 =0. Then, 1 -1/2 =1/2. So total is 0 +1/2=1/2Therefore, 2*(√2 -1)*(1/2 +1/√2)=2*(1/2)=1Therefore, y(t)= -1/√2 +1=1 -1/√2Similarly, x(t)= (1 +1/√2)(3 -2√2). Let me compute this:First, multiply (1)(3 -2√2) + (1/√2)(3 -2√2)=3 -2√2 +3/√2 -2*(√2)/√2=3 -2√2 +3/√2 -2= (3 -2) + (-2√2 +3/√2)=1 -2√2 +3/√2Convert 3/√2 to 3√2/2:=1 -2√2 + (3√2)/2=1 - (4√2)/2 + (3√2)/2=1 - (√2)/2Therefore, x(t)=1 - (√2)/2, and y(t)=1 -1/√2So coordinates of P are (1 - √2/2, 1 -1/√2)Now, we need to find PC intersects AB at E.Point C is at (1,0), point P is at (1 - √2/2, 1 -1/√2). So equation of PC:From C(1,0) to P(1 - √2/2, 1 -1/√2). Let's find parametric equations.Parametric equations for PC: x =1 - t*(√2/2), y =0 + t*(1 -1/√2), where t ranges from 0 to1 gives PC. But since E is the intersection with AB, which is the line from A(0,1) to B(0,0), which is the y-axis x=0.So to find E, set x=0 and find t:0 =1 - t*(√2/2) => t= 2/√2 = √2But t=√2 would give y-coordinate:y=0 + √2*(1 -1/√2)= √2 -1Therefore, point E is at (0, √2 -1)Wait, but AB is from (0,0) to (0,1). Wait, in our coordinate system, A was at (0,1), B at (0,0), so AB is the line x=0 from (0,0) to (0,1). However, in our calculation, E is at (0, √2 -1). Since √2 ≈1.414, so √2 -1 ≈0.414, which is between 0 and1, so E is on AB between B and A. So coordinates of E are (0, √2 -1). Therefore, EA = distance from E to A: |1 - (√2 -1)| = |2 -√2|. EB = distance from E to B: |√2 -1 -0| = √2 -1. Therefore, EA/EB = (2 -√2)/(√2 -1)Simplify (2 -√2)/(√2 -1): Multiply numerator and denominator by (√2 +1):[(2 -√2)(√2 +1)] / [(√2 -1)(√2 +1)] = [2√2 +2 -√2*√2 -√2*1]/[2 -1] = [2√2 +2 -2 -√2]/1 = (√2)/1 = √2Therefore, EA/EB = √2Now, need to find BC/TC - EA/EB. BC is the length from B(0,0) to C(1,0), which is1. So BC=1.Now, find TC. Point T is the intersection of DP and BC.Point D is at (1 +1/√2, -1/√2). Point P is at (1 -√2/2,1 -1/√2). So DP is the line from D to P. Let me find equation of DP.Coordinates:D: (1 +1/√2, -1/√2)P: (1 -√2/2,1 -1/√2)Compute the direction vector:P - D = (1 -√2/2 -1 -1/√2, 1 -1/√2 - (-1/√2)) = (-√2/2 -1/√2, 1 -1/√2 +1/√2) = (- (√2/2 +1/√2),1)Simplify √2/2 =1/√2, so √2/2 +1/√2= 2/√2= √2. Therefore, direction vector is (-√2,1). So the direction vector is (-√2,1). Therefore, parametric equations for DP can be written as:x =1 +1/√2 -√2 * ty= -1/√2 + tWe need to find intersection T with BC, which is the x-axis (y=0). So set y=0:-1/√2 + t =0 => t=1/√2Then x-coordinate:x=1 +1/√2 -√2*(1/√2)=1 +1/√2 -1=1/√2Therefore, point T is at (1/√2,0)Thus, TC is the distance from T(1/√2,0) to C(1,0):TC=1 -1/√2Therefore, BC/TC =1 / (1 -1/√2)= [1]/[(√2 -1)/√2]= √2/(√2 -1)Multiply numerator and denominator by (√2 +1):√2(√2 +1)/[(√2)^2 -1]= (2 +√2)/(2 -1)=2 +√2Therefore, BC/TC=2 +√2Therefore, the expression BC/TC - EA/EB= (2 +√2) -√2=2Therefore, the value is2.But let me verify calculations again to be sure.First, EA/EB was found as √2. BC/TC=2 +√2. Therefore, subtracting gives 2. So the answer is 2.Final Answerboxed{2}</think>Given triangle ( triangle ABC ) with ( N ) as the midpoint of ( AB ), ( angle A > angle B ), and point ( D ) on ray ( AC ) such that ( CD = BC ). Point ( P ) is on ray ( DN ) and satisfies ( angle PBC = angle A ). ( PC ) intersects ( AB ) at ( E ), and ( DP ) intersects ( BC ) at ( T ). We need to find the value of ( frac{BC}{TC} - frac{EA}{EB} ).1. Coordinates and Setup: - Set ( B ) at ( (0, 0) ), ( C ) at ( (1, 0) ), and ( A ) at ( (0, 1) ). - ( N ) is the midpoint of ( AB ), so ( N ) is at ( (0, 0.5) ). - ( D ) is on ray ( AC ) such that ( CD = BC = 1 ). Using parametric equations, ( D ) is found to be ( (1 + frac{1}{sqrt{2}}, -frac{1}{sqrt{2}}) ).2. Finding Point ( P ): - Parametrize ray ( DN ) and find ( P ) such that ( angle PBC = 45^circ ). Solving for coordinates, ( P ) is found to be ( (1 - frac{sqrt{2}}{2}, 1 - frac{1}{sqrt{2}}) ).3. Intersection Points: - ( E ) is the intersection of ( PC ) with ( AB ). Solving parametric equations, ( E ) is found to be ( (0, sqrt{2} - 1) ). - ( T ) is the intersection of ( DP ) with ( BC ). Solving parametric equations, ( T ) is found to be ( (frac{1}{sqrt{2}}, 0) ).4. Calculating Ratios: - ( EA ) and ( EB ) are calculated as ( 2 - sqrt{2} ) and ( sqrt{2} - 1 ) respectively, giving ( frac{EA}{EB} = sqrt{2} ). - ( BC = 1 ) and ( TC = 1 - frac{1}{sqrt{2}} ), giving ( frac{BC}{TC} = 2 + sqrt{2} ).5. Final Calculation: - Subtracting the ratios: ( frac{BC}{TC} - frac{EA}{EB} = (2 + sqrt{2}) - sqrt{2} = 2 ).Thus, the value of the expression is (boxed{2}).,<think>Okay, let's try to tackle this geometry problem. It seems a bit complex with all these points and conditions, but maybe if I break it down step by step, it will make more sense.First, let's parse the problem again to make sure I understand all the parts:We have triangle ABC with N being the midpoint of AB. It's given that angle A is greater than angle B. Then there's a point D on the extension of ray AC beyond C such that CD = BC. Point P is on ray DN (so starting at D and going through N) and is on the same side of line BC as point A. They also specify that angle PBC equals angle A. PC intersects AB at point E, and BC intersects DP at point T. We need to find the value of the expression (BC/TC) - (EA/EB).Alright, so first, let me try to draw a rough sketch of this triangle ABC. Let me label the points: A, B, C. N is the midpoint of AB. Since angle A is greater than angle B, side BC must be longer than side AC? Wait, no, in a triangle, the larger angle is opposite the longer side. So if angle A > angle B, then BC > AC. Hmm, okay.Point D is on ray AC, so that could be on the segment AC or beyond C. But since CD = BC, and BC is a side of the triangle, if D were between A and C, CD would be shorter than AC, but BC could be longer or shorter. Wait, but maybe D is beyond C? Because CD has to be equal to BC. If BC is longer than AC, then D must be beyond C. But if BC is shorter, maybe not. Wait, but angle A is larger than angle B, so BC should be longer than AC. So perhaps D is beyond C on the extension of AC.So, CD = BC. So starting from C, moving a distance BC along the ray AC, which would be beyond C if BC is longer than AC. So D is such that CD = BC. So that's a key point.Then P is on ray DN (so starting at D and going through N) and is on the same side of line BC as point A. So line BC divides the plane into two halves; point A is on one side, and we need to place P on that same side. Then angle PBC equals angle A. So angle at B, between PB and BC, is equal to angle at A of triangle ABC.PC intersects AB at E, and BC intersects DP at T. Then compute BC/TC - EA/EB.Hmm. So maybe I need to use some properties of similar triangles, Ceva's theorem, Menelaus' theorem, mass point geometry, or coordinate geometry? Let's see.Since N is the midpoint of AB, that might be a clue. Also, since CD = BC, triangle BCD might be isoceles? Wait, if CD = BC, then triangle BCD is isoceles with base BD? Wait, no, CD is along AC. Hmm.Alternatively, since CD = BC, maybe triangle BCD is congruent or similar to another triangle? Not sure yet.Angle PBC = angle A. So angle at B is equal to angle at A. Maybe that gives a similarity between triangles? Let's see. If angle PBC = angle BAC, then maybe triangles ABC and PBC are similar? Wait, not necessarily. Let's check:In triangle ABC, angle at A is angle BAC. In triangle PBC, angle at B is angle PBC. If angle PBC = angle BAC, then maybe there is some similarity. Let's see. If we have two angles equal, then triangles are similar. Let's suppose angle PBC = angle BAC. Also, angle B is common? Wait, no. If angle PBC is equal to angle BAC, and if there is another angle that's common, maybe. Alternatively, maybe triangle PBC is similar to another triangle. Hmm.Alternatively, maybe using trigonometric Ceva's theorem? Since we have cevians in the triangle.Alternatively, coordinate geometry. Assign coordinates to points and compute the required ratios. Let me try that approach. Let's place the triangle ABC on coordinate plane.Let me let point B be at (0,0), point C at (c,0), and point A somewhere in the plane. Since N is the midpoint of AB, coordinates of N would be ((Ax + Bx)/2, (Ay + By)/2) = (Ax/2, Ay/2) if B is at (0,0). But maybe it's better to set up coordinates so that calculations are easier.Alternatively, place point B at the origin, point C at (c,0), point A at (a,b). Then N is the midpoint of AB, so coordinates of N would be (a/2, b/2). Then point D is on ray AC such that CD = BC. Let's compute coordinates of D.First, vector AC is from A(a,b) to C(c,0). The vector AC is (c - a, -b). To get point D on ray AC beyond C such that CD = BC. The length BC is the distance from B(0,0) to C(c,0), which is c. So CD should be equal to BC, so CD = c. The direction from C to D is the same as from A to C. So the vector CD is a scalar multiple of vector AC. Wait, but AC is from A to C, so if we go beyond C, the direction is the same as AC. Wait, actually, if we are moving along ray AC starting at A, then beyond C, the direction is from A to C. So vector AC is (c - a, -b), so direction from C is same as vector AC. Wait, no. If we start at C and move in the direction of AC, which is from A to C, but if we are moving along the ray AC, which starts at A and goes through C, so beyond C, the direction is still along AC. So to get from C to D, we need to move in the direction from A to C. So vector AC is (c - a, -b), so unit vector in direction AC is (c - a, -b)/|AC|. Then CD is length BC = c. So coordinates of D would be C plus vector CD in direction AC. Wait, but |AC| is sqrt((c - a)^2 + b^2), and CD = BC = c. So D is C + (c/|AC|)*(c - a, -b). Hmm, this might get complicated. Maybe coordinate geometry is not the best approach here.Alternatively, let's consider using Ceva's theorem. Ceva's theorem states that for concurrent cevians, the product of certain ratios equals 1. But here, we might have multiple intersections. Let's see.Alternatively, mass point geometry. Since N is the midpoint of AB, masses at A and B would be equal. Then perhaps we can assign masses and use that to find ratios.Alternatively, use Menelaus' theorem for transversal lines cutting the sides of the triangle. For example, line DP intersects BC at T, so maybe applying Menelaus' theorem to triangle BCD with transversal DP? Not sure.Wait, let me try to outline the key elements:1. N is the midpoint of AB. So AN = NB.2. D is on ray AC such that CD = BC. So triangle BCD is isoceles with CD = BC. Therefore, angle CBD = angle CDB? Wait, in triangle BCD, sides BC and CD are equal, so yes, angles opposite them are equal. So angle CBD = angle CDB.3. P is on ray DN such that angle PBC = angle A. So angle PBC is equal to angle BAC. This seems like a key condition. Maybe we can use some angle chasing here.4. PC intersects AB at E, and BC intersects DP at T. We need to compute BC/TC - EA/EB.Hmm. Maybe there is a relation between these ratios that simplifies to a constant.Alternatively, could the expression BC/TC - EA/EB evaluate to 1? Let me check with a specific case.Alternatively, let me consider an example triangle where calculations are easier.Suppose triangle ABC is such that angle A is 60 degrees, angle B is 30 degrees. Then BC is longer than AC, since angle A > angle B. Let's assign coordinates:Let’s place point B at (0,0), point C at (2,0). Since angle at A is 60 degrees, angle at B is 30 degrees. Then side BC is 2 units. Then using the Law of Sines:In triangle ABC,a / sin A = b / sin B = c / sin CHere, side BC is opposite angle A, so BC = a = 2.Angle A = 60°, angle B = 30°, so angle C = 90°.Therefore, sides:a = BC = 2, opposite angle A (60°)b = AC, opposite angle B (30°): b = (2 * sin 30°)/ sin 60° = (2 * 0.5) / (√3/2) = 1 / (√3/2) = 2/√3 ≈ 1.1547c = AB, opposite angle C (90°): c = (2 * sin 90°)/ sin 60° = (2 * 1)/ (√3/2) = 4/√3 ≈ 2.3094So coordinates:Point B: (0,0)Point C: (2,0)Point A: Let's compute coordinates. Since angle at C is 90°, triangle is right-angled at C. Wait, but angle C is 90°, so sides: Wait, no, angle C is 90°, then sides:Wait, no, according to Law of Sines, if angle A is 60°, angle B is 30°, angle C is 90°, then sides:a = BC = opposite angle A (60°): 2b = AC = opposite angle B (30°): 1c = AB = opposite angle C (90°): √3Wait, that can't be. Wait, if angle C is 90°, then AB would be the hypotenuse, so AB = 2 (BC) is opposite angle A (60°). Wait, no. Wait, confusion here.Wait, if angle A is 60°, angle B is 30°, angle C is 90°, then side BC is opposite angle A (60°), side AC is opposite angle B (30°), and side AB is opposite angle C (90°). So in a right-angled triangle at C, sides:AC = adjacent to angle B (30°), so AC = AB * cos(30°)BC = opposite to angle B (30°), so BC = AB * sin(30°)But BC is given as 2. Then AB = BC / sin(30°) = 2 / 0.5 = 4. Then AC = AB * cos(30°) = 4 * (√3/2) = 2√3 ≈ 3.464.Wait, but according to the problem, angle A is greater than angle B, which is satisfied here (60° > 30°). Then coordinates:Point C is at (0,0) if it's the right angle? Wait, maybe I need to adjust coordinates.Wait, if it's right-angled at C, then coordinates can be:C: (0,0)B: (BC, 0) = (2, 0)A: (0, AC) = (0, 2√3)Then AB is the hypotenuse from (0, 2√3) to (2, 0), which has length sqrt((2)^2 + (2√3)^2) = sqrt(4 + 12) = sqrt(16) = 4, which matches.So coordinates:A: (0, 2√3)B: (2, 0)C: (0, 0)Wait, but then midpoint N of AB would be ((0 + 2)/2, (2√3 + 0)/2) = (1, √3)Point D is on ray AC beyond C such that CD = BC = 2. Since AC is from A(0, 2√3) to C(0,0). So ray AC is the line x=0 from A downwards to C and beyond. So moving from C(0,0) along AC (which is the same line) beyond C would be in the negative y-direction. But CD = BC = 2. BC is the distance from B(2,0) to C(0,0), which is 2 units. So CD is 2 units from C(0,0) along AC. But AC is vertical. Wait, AC is from A(0, 2√3) to C(0,0). So direction is along the negative y-axis. So beyond C, along AC (which is the same line), point D would be at (0, -2). Because CD = 2, moving down from C(0,0) to (0, -2).Therefore, coordinates of D: (0, -2)Then ray DN: from D(0, -2) through N(1, √3). Let's find the equation of line DN.The direction vector from D to N is (1 - 0, √3 - (-2)) = (1, √3 + 2). So parametric equations for DN: x = 0 + t*1, y = -2 + t*(√3 + 2), where t ≥ 0.Point P is on this ray DN on the same side of BC as A. Line BC is from B(2,0) to C(0,0), so it's the x-axis. The "same side of line BC as point A" would be the upper half-plane (since A is at (0, 2√3), above the x-axis). But point D is at (0, -2), below the x-axis, and N is at (1, √3), above the x-axis. So the ray DN goes from below BC to above BC. So the part of DN beyond N is in the upper half-plane. Since P is on the same side as A, which is above BC, so P is on the part of DN beyond N.Additionally, angle PBC = angle A = 60°. So angle between PB and BC at point B is 60°. Let's find coordinates of P such that this angle condition is satisfied.Point B is at (2, 0). Let’s denote point P as (t, -2 + t*(√3 + 2)), where t ≥ 1 (since beyond N, which is at t=1, x=1, y=√3). Wait, no: parametric equations for DN are x = t, y = -2 + t*(√3 + 2). Wait, no, direction vector is (1, √3 + 2), so starting at D(0, -2), so parametric equations:x = 0 + t*1 = ty = -2 + t*(√3 + 2)So when t=0, it's D(0, -2); when t=1, it's N(1, √3). So for P beyond N, t > 1.So point P is (t, -2 + t*(√3 + 2)) for t > 1.We need angle PBC = 60°. Let's compute angle PBC.Point B is (2,0). The angle at B between points P, B, and C.Vector BP is from B(2,0) to P(t, -2 + t*(√3 + 2)): (t - 2, -2 + t*(√3 + 2) - 0) = (t - 2, -2 + t*(√3 + 2))Vector BC is from B(2,0) to C(0,0): (-2, 0)The angle between vectors BP and BC is 60°. Let's use the dot product formula:cos(theta) = (BP . BC) / (|BP| |BC|)Given theta = 60°, so cos(theta) = 0.5Compute BP . BC = (t - 2)(-2) + (-2 + t*(√3 + 2))(0) = -2(t - 2) + 0 = -2t + 4|BP| = sqrt((t - 2)^2 + (-2 + t*(√3 + 2))^2)|BC| = sqrt((-2)^2 + 0^2) = 2So:(-2t + 4) / (|BP| * 2) = 0.5Multiply both sides by 2|BP|:-2t + 4 = |BP|But |BP| = sqrt((t - 2)^2 + (-2 + t*(√3 + 2))^2)So:-2t + 4 = sqrt((t - 2)^2 + (-2 + t*(√3 + 2))^2)Let me square both sides to eliminate the square root:(-2t + 4)^2 = (t - 2)^2 + (-2 + t*(√3 + 2))^2Expand left side: (4t^2 - 16t + 16)Right side: (t^2 - 4t + 4) + [(-2 + t*(√3 + 2))^2]First compute (-2 + t*(√3 + 2))^2:= [t*(√3 + 2) - 2]^2= t^2*(√3 + 2)^2 - 4t*(√3 + 2) + 4So expanding:= t^2*(3 + 4√3 + 4) - 4t*(√3 + 2) + 4= t^2*(7 + 4√3) - 4t*(√3 + 2) + 4Therefore, right side:(t^2 - 4t + 4) + [t^2*(7 + 4√3) - 4t*(√3 + 2) + 4]= t^2 - 4t + 4 + t^2*(7 + 4√3) - 4t*(√3 + 2) + 4= t^2*(1 + 7 + 4√3) + t*(-4 -4*(√3 + 2)) + (4 + 4)= t^2*(8 + 4√3) + t*(-4 -4√3 -8) + 8= t^2*(8 + 4√3) + t*(-12 -4√3) + 8Left side: 4t^2 -16t +16Set equal:4t^2 -16t +16 = t^2*(8 + 4√3) + t*(-12 -4√3) +8Bring all terms to left:4t^2 -16t +16 - t^2*(8 + 4√3) - t*(-12 -4√3) -8 = 0Simplify:(4 -8 -4√3)t^2 + (-16 +12 +4√3)t + (16 -8) =0= (-4 -4√3)t^2 + (-4 +4√3)t +8 =0Divide both sides by -4 to simplify:(1 + √3)t^2 + (1 - √3)t -2 =0Now, we have a quadratic equation:(1 + √3)t^2 + (1 - √3)t -2 =0Let’s solve for t using quadratic formula:t = [ - (1 - √3) ± sqrt( (1 - √3)^2 -4*(1 + √3)*(-2) ) ] / [2*(1 + √3)]First compute discriminant:D = (1 - √3)^2 -4*(1 + √3)*(-2)= (1 - 2√3 +3) +8*(1 + √3)= (4 -2√3) +8 +8√3= 12 +6√3Therefore,t = [ - (1 - √3) ± sqrt(12 +6√3) ] / [2*(1 + √3)]Simplify sqrt(12 +6√3). Let’s see if this can be expressed as sqrt(a) + sqrt(b):Suppose sqrt(12 +6√3) = sqrt(x) + sqrt(y). Then squaring:12 +6√3 = x + y + 2√(xy)Equate rational and irrational parts:x + y =122√(xy)=6√3 → √(xy)=3√3 → xy=27Thus, solving x + y =12 and xy=27. The solutions are roots of t^2 -12t +27=0. Discriminant: 144 -108=36. So t=(12±6)/2=9 or 3. So sqrt(12 +6√3)=sqrt(9)+sqrt(3)=3+√3.Therefore, sqrt(12 +6√3)=3 +√3.Therefore, t = [ - (1 - √3) ± (3 +√3) ] / [2*(1 + √3)]Compute numerator for both signs:First with +:- (1 - √3) + (3 +√3) = -1 + √3 +3 +√3 = 2 + 2√3With -:- (1 - √3) - (3 +√3) = -1 +√3 -3 -√3 = -4Therefore, t=(2 + 2√3)/[2*(1 + √3)] or t=(-4)/[2*(1 + √3)]Simplify t= (2 + 2√3)/[2*(1 + √3)] = [2(1 + √3)]/[2(1 + √3)] =1The other solution is t= -4/[2*(1 + √3)] = -2/(1 + √3). Since t>1 for point P beyond N, we discard the negative solution. So t=1. But t=1 is point N. However, P is on ray DN beyond N, so t>1. Wait, but here, the solution gives t=1, which is N. But angle PBC is supposed to be angle A=60°, but at t=1, point P is N. Let's check angle at B if P is N.Wait, point N is the midpoint of AB. Coordinates of N are (1, √3). So vector BP would be from B(2,0) to N(1, √3): (-1, √3). The vector BC is (-2,0). The angle between vectors (-1, √3) and (-2,0).Compute the angle:cos(theta) = [(-1)(-2) + (√3)(0)] / [sqrt(1 + 3) * sqrt(4)]= [2]/[2 * 2] = 2/4 = 0.5Thus, theta=60°, which matches angle A=60°. Wait, so point P is N? But according to the problem statement, P is on ray DN and is on the same side of line BC as A. N is on AB, which is on the same side as A. So in this specific case, P=N.But then PC intersects AB at E. If P=N, then line PC is line NC. But N is the midpoint of AB. So line NC connects N(1, √3) to C(0,0). Let's find where this line intersects AB. Wait, but AB is from A(0, 2√3) to B(2,0). The line NC is already passing through N, which is on AB. Therefore, the intersection point E is N itself. But the problem states that E is the intersection of PC and AB. If P=N, then PC is NC, which intersects AB at N. So E=N. Therefore, EA/EB = AN/NB =1, since N is the midpoint. Therefore, EA/EB=1.Then BC=2, TC is the intersection of DP and BC. But DP is DN, since P=N. DN is the line from D(0,-2) to N(1,√3). Let's find where DN intersects BC (the x-axis). Parametrize DN:x= t, y= -2 + t*(√3 + 2). We need to find T where y=0.Set y=0:-2 + t*(√3 + 2)=0 => t= 2/(√3 + 2)Therefore, coordinates of T are ( t, 0 ) where t=2/(√3 + 2)But t is the x-coordinate, which is also the parameter from before. Wait, in parametrization of DN, x=t, so x=2/(√3 + 2). So point T is (2/(√3 + 2), 0). Therefore, TC is the distance from T to C(0,0), which is 2/(√3 + 2). BC is 2, so BC/TC= 2 / [2/(√3 + 2)] = √3 +2.Then BC/TC - EA/EB = (√3 + 2) -1 = √3 +1. But the problem asks for the value of this expression, which in this specific case is √3 +1. However, the problem probably expects a constant value, not depending on the specific triangle. But √3 +1 is approximately 2.732, which is not an integer. But in the problem statement, they might have constructed a general case where this expression equals 1, but in my specific case, it's √3 +1. So either my approach is wrong, or the specific triangle I chose isn't general.Wait, this is a problem. The answer should be a constant, regardless of the triangle. Therefore, my assumption that P=N in this specific case might be incorrect. Wait, but according to the calculation, when angle PBC=60°, which is angle A, and in this case, point P=N satisfies that. But then the result is not a constant. Therefore, there must be a mistake.Wait, perhaps in this specific case, the construction leads to P=N, but in the general case, P is different. Wait, but in this case, with the specific angles, it turned out that P=N, but maybe in other triangles, P is different. However, the problem states "Let the midpoint of side AB...", so the configuration is general. Therefore, perhaps my coordinate approach is not the right way, or I made a miscalculation.Alternatively, maybe the answer is always 1. Let me check with another triangle.Suppose triangle ABC is isoceles with AB=AC, but angle A > angle B. Wait, no, if AB=AC, then angle B=angle C, which contradicts angle A>angle B. So perhaps take another triangle.Alternatively, take a triangle where angle A is 90°, angle B is 45°, so angle C is 45°, but this would make BC=AC. But angle A> angle B, so BC>AC. But in this case, angle A=90°, angle B=45°, so BC is the hypotenuse, which is longer. Let me try this.Let’s set up coordinates:Let’s place point B at (0,0), point C at (2,0), and point A at (0,2). Then AB is from (0,2) to (0,0), AC is from (0,2) to (2,0). Then midpoint N of AB is (0,1).Point D is on ray AC such that CD=BC. BC is distance from B(0,0) to C(2,0) is 2. So CD=2. AC is from A(0,2) to C(2,0). Parametrize AC: any point on AC can be written as (2t, 2 - 2t) for t ≥0. C is at t=1: (2,0). To find D beyond C such that CD=2. The direction from C is along AC, which is the same as the vector from A to C: (2, -2). Unit vector in direction AC: (2, -2)/sqrt(8) = (1, -1)/√2. Then moving from C(2,0) in this direction for length 2: D = C + 2*(1, -1)/√2 * sqrt(8)/sqrt(8)? Wait, maybe better to parametrize.The vector AC is (2, -2). To go beyond C by length BC=2 in the direction of AC. Since AC has length sqrt(2^2 + (-2)^2)=sqrt(8)=2√2. So unit vector in direction AC is (2, -2)/(2√2)= (1, -1)/√2. Then CD vector is 2*(1, -1)/√2 = (√2, -√2). Therefore, D = C + (√2, -√2) = (2 + √2, 0 - √2) = (2 + √2, -√2)Then ray DN is from D(2 + √2, -√2) through N(0,1). Let's parametrize DN.Direction vector from D to N: (0 - (2 + √2), 1 - (-√2)) = (-2 - √2, 1 + √2)Parametric equations:x=2 + √2 + t*(-2 - √2)y= -√2 + t*(1 + √2)We need to find point P on this ray (t ≥0) such that angle PBC= angle A=90°.Angle PBC is the angle at B between points P, B, and C. Since angle A is 90°, we need angle PBC=90°. So PB should be perpendicular to BC.Vector BC is from B(0,0) to C(2,0), which is (2,0). Vector BP is from B(0,0) to P(x,y), which is (x,y). To have angle PBC=90°, the vectors BP and BC must be perpendicular. Wait, no. Angle PBC is the angle at B between PB and BC. If angle PBC=90°, then PB is perpendicular to BC.Vector BC is (2,0). Vector PB is (x -0, y -0)=(x,y). The angle between PB and BC is 90°, so their dot product is zero:(x)(2) + (y)(0)=0 => 2x=0 => x=0.Therefore, point P must lie on the y-axis. But P is on ray DN. So find t such that x-coordinate of P is 0.From parametric equation:x=2 + √2 + t*(-2 - √2)=0Solve for t:(2 + √2) + t*(-2 - √2)=0 => t= (2 + √2)/(2 + √2)=1Therefore, when t=1, x=0. Then y-coordinate:y= -√2 +1*(1 + √2)= -√2 +1 + √2=1Therefore, point P is (0,1). Wait, point N is (0,1), which is the midpoint of AB. So P=N again. Then PC is the line from P(0,1) to C(2,0). Let's find E, the intersection of PC with AB.AB is from A(0,2) to B(0,0). Line PC is from (0,1) to (2,0). Parametrize PC: x=0 + 2s, y=1 - s, where s ∈ [0,1]. AB is x=0. So intersection E is at x=0. Solve for s when x=0=2s => s=0. Then y=1 -0=1. So E is (0,1), which is point N. Therefore, EA/EB = distance from E to A over distance from E to B. Since E is (0,1), A is (0,2), B is (0,0). EA=1, EB=1, so EA/EB=1.Now, BC=2. TC is the intersection of DP and BC. But DP is from D(2 + √2, -√2) to P(0,1). Let's find where DP intersects BC (the x-axis, y=0).Parametrize DP: from D(2 + √2, -√2) to P(0,1). Direction vector is (-2 - √2,1 + √2). Parametric equations:x=2 + √2 + t*(-2 - √2)y= -√2 + t*(1 + √2)Find t when y=0:-√2 + t*(1 + √2)=0 => t= √2 / (1 + √2)Then x-coordinate:x=2 + √2 + (√2 / (1 + √2))*(-2 - √2)Simplify denominator: multiply numerator and denominator by (1 - √2):t= √2*(1 - √2)/[(1 + √2)(1 - √2)] = √2*(1 - √2)/(1 -2)= -√2*(1 - √2)/1= -√2 + 2Wait, this is getting complicated. Alternatively, compute x:x=2 + √2 - (√2*(2 + √2))/(1 + √2)Factor numerator:√2*(2 + √2) = √2*2 + (√2)^2 = 2√2 +2Denominator:1 + √2Thus,x=2 + √2 - (2√2 +2)/(1 + √2)Multiply numerator and denominator by (1 - √2):(2√2 +2)(1 - √2)/(1 -2)= (2√2 +2 -2*2 -2√2)/(-1)= (2√2 +2 -4 -2√2)/(-1)= (-2)/(-1)=2Therefore, x=2 + √2 -2=√2Wait, this seems incorrect. Let me check again.Wait, original expression:x=2 + √2 + t*(-2 - √2)We found t= √2 / (1 + √2)Therefore,x=2 + √2 + [√2 / (1 + √2)]*(-2 - √2)Let me compute:[√2 / (1 + √2)]*(-2 - √2)= √2*(-2 - √2)/(1 + √2)Factor numerator: -√2(2 + √2)Denominator:1 + √2Note that 2 + √2= (1 + √2) +1. Not helpful. Wait, let's multiply numerator and denominator by (1 - √2):-√2(2 + √2)(1 - √2)/[(1 + √2)(1 - √2)] = -√2(2(1 - √2) + √2(1 - √2))/(-1)= -√2[2 -2√2 + √2 -2]/(-1)= -√2[ (2 -2) + (-2√2 +√2) ]/(-1)= -√2[ -√2 ]/(-1)= -√2*(-√2)/(-1)= (2)/(-1)= -2Therefore, x=2 + √2 -2=√2Therefore, point T is (√2,0). Therefore, TC is distance from T(√2,0) to C(2,0) is |2 - √2|. Since 2 >√2, TC=2 -√2.Therefore, BC=2, so BC/TC=2/(2 -√2). Multiply numerator and denominator by (2 +√2):2*(2 +√2)/[(2 -√2)(2 +√2)] = (4 + 2√2)/(4 -2)= (4 +2√2)/2=2 +√2Then EA/EB=1, so BC/TC - EA/EB= (2 +√2) -1=1 +√2≈2.414. Again, not a constant. So in this case, the value is 1 +√2, in the previous case it was √3 +1. These are different, implying that the expression is not a constant. But the problem states "Find the value of the expression", implying that it is a constant. So my approach must be wrong.Alternatively, maybe I misapplied the problem's conditions. Let me double-check.Wait, in the first example, when I took angle A=60°, angle B=30°, and the triangle was right-angled at C, then P turned out to be N, leading to BC/TC - EA/EB=√3 +1 -1=√3. Wait, no, in that case, BC=2, TC=2/(√3 +2), so BC/TC= (2)/(2/(√3 +2))= √3 +2. Then EA/EB=1, so BC/TC - EA/EB=√3 +2 -1=√3 +1. But in the second example, angle A=90°, angle B=45°, it's 1 +√2. These are different. Therefore, unless there's a mistake in my calculations, which is possible, the answer isn't a constant, which contradicts the problem's implication. Therefore, I must have made a mistake.Wait, perhaps the problem has a different configuration. Let me re-examine the problem statement."D is a point on ray AC such that CD = BC". So ray AC starts at A and goes through C. So D is on the extension beyond C of AC. So in my first example, where AC is from A(0, 2√3) to C(0,0), then D would be beyond C, so at (0, -2). But in my second example, AC is from A(0,2) to C(2,0). So ray AC beyond C is the line from A through C to infinity. So parametric equations beyond C would be x=2 + t*2, y=0 + t*(-2) for t>0. Wait, no, the direction from A to C is (2, -2). So beyond C, moving in the same direction. So D is located at C + k*(2, -2), where k>0. Then CD length is |k*(2, -2)|=k*sqrt(8)=2k√2. But CD=BC=2 (in second example). Therefore, 2k√2=2 => k=1/√2. Therefore, coordinates of D are C + (2/√2, -2/√2)= (2 + √2, -√2). Which matches my previous calculation.But in that case, point T is (√2,0), and TC=2 -√2. Then BC/TC=2/(2 -√2)=2*(2 +√2)/ (4 -2)= (4 + 2√2)/2=2 +√2. Then EA/EB=1. So 2 +√2 -1=1 +√2.But the problem says "Find the value of the expression". So the answer must be a constant. Hence, my coordinate approach might not be suitable, or I may have misapplied the conditions.Alternatively, maybe there is a synthetic solution.Given that N is the midpoint of AB, and D is such that CD=BC. Let's consider triangles.Since CD=BC, triangle BCD is isoceles with BC=CD. Therefore, angles at B and D are equal: angle CBD=angle CDB.Given that angle PBC=angle A.Also, since N is the midpoint of AB, and D is such that CD=BC, maybe there's some reflection or rotation symmetry.Alternatively, use Ceva's theorem in triangle ABC with point E.But E is the intersection of PC and AB. To apply Ceva's theorem, we need three cevians. However, maybe not directly.Alternatively, consider triangle BTC and some relations.Wait, let's try using Ceva's theorem in triangle BCD with cevians BT, CP, and DN. But not sure.Alternatively, let's think about the given angles. Angle PBC=angle A. Since angle A + angle B + angle C=180°, angle A > angle B.Given that, and since CD=BC, which gives triangle BCD isoceles, as mentioned.Perhaps, construct point P such that angle PBC=angle A, then PC meets AB at E. Then we have to relate the ratios BC/TC and EA/EB.Wait, maybe using the Law of Sines in some triangles.In triangle PBC, angle PBC=angle A, angle PCB=?Wait, angle PBC=angle A= angle BAC. Let me consider triangle ABC and triangle PBC.In triangle ABC, angle at A is angle BAC. In triangle PBC, angle at B is angle PBC= angle BAC. If these angles are equal, perhaps there is some similarity.But unless another angle is equal, we can't establish similarity. However, if we can find another angle equality.Alternatively, let's consider triangle ABC and triangle PBC. If angle PBC=angle BAC, and if angle BPC=angle ABC, then triangles would be similar. Let's check:If angle PBC=angle BAC (given), angle BPC=angle ABC (unknown), then triangles ABC and PBC would be similar. If this is the case, then ratios of sides would be proportional.But unless we know that, we can't confirm. Alternatively, maybe using Law of Sines in triangle PBC:PB/sin(angle PCB)= BC/sin(angle BPC)= PC/sin(angle PBC)Similarly, in triangle ABC:AB/sin(angle ACB)= BC/sin(angle BAC)= AC/sin(angle ABC)But not sure.Alternatively, consider the point P such that angle PBC=angle A. This is similar to constructing a point P such that PB is a certain proportion. Maybe using the Law of Sines in triangle PBC and ABC.Alternatively, since angle PBC=angle A, and in triangle ABC, angle A + angle B + angle C=180°, so angle PBC=angle A, which is given.Alternatively, let's consider the circumcircle of triangle ABC. Maybe point P lies somewhere on it. Not sure.Alternatively, use coordinate geometry but in a general triangle.Let’s assume general triangle ABC with coordinates:Let’s place point B at (0,0), point C at (c,0), and point A at (a,b). N is the midpoint of AB, so coordinates of N are (a/2, b/2).Point D is on ray AC beyond C such that CD=BC. BC has length c (distance from (0,0) to (c,0)). Vector AC is (c -a, -b). To find point D beyond C such that CD=BC=c. Then D can be expressed as C + k*(vector AC), where k is a scalar such that |CD|=c.The vector CD is k*(c -a, -b). So |CD|=k*sqrt((c -a)^2 + b^2)=c. Thus, k= c / sqrt((c -a)^2 + b^2).Therefore, coordinates of D are:D = (c,0) + k*(c -a, -b) = (c + k*(c -a), 0 + k*(-b)).Now, ray DN is from D to N. Parametrize DN as:x = c + k*(c -a) + t*(a/2 - [c + k*(c -a)])y = -k*b + t*(b/2 - (-k*b))But this is getting too messy. Maybe there's a better way.Alternatively, use vectors.Let’s represent points as vectors. Let’s denote vectors with bold letters.Let’s set point B as the origin. Vector B = 0. Let’s denote vector C as c, and vector A as a. Then midpoint N is (a + 0)/2 = a/2.Point D is on ray AC beyond C such that CD=BC= |C - B|= |c -0|=|c|.Vector AC is C - A = c - a. So point D can be written as C + t*(C - A) where t>0 such that |D - C|=BC=|c|. Thus,D = C + t*(C - A)|D - C|= |t*(C - A)|= t*|C - A|= |c|Therefore, t= |c| / |C - A|Thus, D= C + (|c| / |C - A|)*(C - A)But in coordinate terms, this is complex.Alternatively, let’s use mass point geometry.Given that N is the midpoint of AB, so mass at A and B is 1 each, mass at N is 2.Point D is such that CD=BC. If we can relate the masses or ratios in triangle BCD.Alternatively, consider the intersection points and apply Ceva's theorem.In triangle BCD, cevians are CP, DN, and BT. But not sure.Alternatively, in triangle ABC, point E is on AB, point T is on BC, and we need to relate BC/TC - EA/EB.Let me think about the expression BC/TC - EA/EB. If we can express both terms in terms of other ratios and find that their difference is a constant.Alternatively, since BC/TC is a ratio along BC, and EA/EB is a ratio along AB, maybe there's a relation through the lines PC and DP.Wait, since PC intersects AB at E and DP intersects BC at T, maybe use Menelaus' theorem on triangle ABC with transversal EPC? Not sure.Alternatively, use Menelaus on triangle BCD with transversal TDP.Wait, Menelaus' theorem states that for a triangle and a transversal line, the product of the segment ratios is -1. But signs have to be considered.Alternatively, let's apply Menelaus' theorem to triangle BCD with transversal T-P-E? Not sure.Alternatively, consider triangle DBC. Line PC intersects DB at E and BC at T. Hmm, not sure.Alternatively, think of the problem in terms of projective geometry, cross ratios, but that might be too advanced.Alternatively, use Ceva's theorem in triangle ABC. For Ceva's theorem, the cevians are AD, BE, and CF, and their concurrency implies (AF/FB)(BD/DC)(CE/EA)=1. But here, we have point E and T, but not sure.Alternatively, note that the expression BC/TC - EA/EB can be rewritten as (BC/TC) + ( - EA/EB ). If we can relate these ratios through some other theorem.Wait, suppose we apply Ceva's theorem to triangle BCP with cevians BT, CD, and some other line. Not sure.Alternatively, since angle PBC=angle A, and considering triangle ABC and triangle PBC. Maybe there's a spiral similarity.If angle PBC=angle BAC, and if there's a similarity transformation that maps one triangle to the other.Spiral similarity involves rotation and scaling. Suppose we rotate triangle ABC around point B by some angle and scale it to get triangle PBC. Since angle PBC=angle BAC, maybe this is the case.Let’s consider a spiral similarity centered at B that maps angle BAC to angle PBC. So rotate triangle ABC around B by angle theta and scale by factor k to get triangle PBC.The rotation angle would be the difference between angle ABC and angle PBC. Wait, but angle PBC=angle A. So maybe the rotation angle is angle A - angle ABC?Alternatively, the rotation would map BA to BP, and BC to BC. Not sure.Alternatively, since angle PBC=angle BAC, and if we can rotate triangle BAC around B to align with PBC.But this is getting vague.Alternatively, use trigonometric Ceva's theorem in triangle ABC for point P.Wait, point P is outside the triangle, but maybe not.Alternatively, consider the trigonometric form of Ceva's theorem: if concurrency of cevians, then (sin(angle1)/sin(angle2))...=1.But not sure.Alternatively, let's get back to the coordinate examples. Since in both cases, the expression evaluated to 1 + something, but different each time. However, the problem states to find the value, implying it's a constant. Therefore, my coordinate examples must have errors.Wait, in the first example, when angle A=60°, angle B=30°, right-angled at C. Point P=N. Then EA/EB=1. BC/TC=√3 +2. Then BC/TC - EA/EB=√3 +1. But the problem might have the answer 1, which is not the case here.Wait, but maybe there's a mistake in assuming P=N. Let me double-check.In the first example, angle PBC=60°, which is equal to angle A. If P=N, then vector BP is from B(2,0) to N(1, √3). The angle between BP and BC is computed as 60°, which matches angle A. So that's correct. Then in this case, the expression evaluates to √3 +1. But according to the problem's expectation, the answer should be a nice integer, likely 1 or 2. So there's a contradiction.Alternatively, maybe my coordinate choice is incorrect. Let's consider another coordinate system where calculations are easier.Let’s take triangle ABC with AB=2, angle at A=60°, angle at B=30°, so angle at C=90°.Wait, earlier attempt may have mixed up the coordinates.Alternatively, let's use the following coordinates:Let’s set point B at (0,0), point C at (1,0), and point A somewhere such that angle A=60°, angle B=30°. Then BC=1, AC=√3, AB=2.Wait, Law of Sines:BC / sin(A) = AC / sin(B) = AB / sin(C)Given angle A=60°, angle B=30°, angle C=90°.Thus,BC / sin(60°) = AC / sin(30°) = AB / sin(90°)So BC / (√3/2) = AC / (1/2) = AB /1Let’s set AB=2, then BC= (√3/2)*2=√3, and AC= (1/2)*2=1. But in this case, BC=√3, AC=1, AB=2.Coordinates: Place B at (0,0), point C at (√3,0). Since angle at C is 90°, point A must be at (√3,1). Wait, no, because if angle at C is 90°, then AC and BC are perpendicular. If B is at (0,0), C at (√3,0), then A would be at (√3, y). Wait, let's compute.Wait, in a triangle with AB=2, BC=√3, AC=1, right-angled at C.Coordinates: C is at (0,0), B at (√3,0), A at (0,1). Then AB= sqrt((√3)^2 +1^2)=sqrt(4)=2. BC=√3, AC=1. Angle at C is 90°. Angles:Angle at A: arctan(√3/1)=60°, angle at B: arctan(1/√3)=30°. Yes, this works.So coordinates:A: (0,1)B: (√3,0)C: (0,0)Midpoint N of AB: ((√3 +0)/2, (0 +1)/2)= (√3/2, 0.5)Point D is on ray AC beyond C such that CD=BC=√3. AC is from A(0,1) to C(0,0), so ray AC beyond C is the negative y-axis. CD=√3, so D is at (0,0 -√3)= (0, -√3)Ray DN: from D(0, -√3) to N(√3/2, 0.5). Parametric equations:x=0 + t*(√3/2 -0)= t*√3/2y= -√3 + t*(0.5 + √3)Find point P on this ray (t ≥0) such that angle PBC=60°.Point B is at (√3,0). We need angle PBC=60°, which is the angle between PB and BC.Vector BC is from B(√3,0) to C(0,0): (-√3,0)Vector BP is from B(√3,0) to P(x,y): (x -√3, y -0)= (x -√3, y)The angle between vectors BP and BC is 60°. Using dot product:cos(theta)= (BP . BC) / (|BP| |BC|)BP . BC= (x -√3)(-√3) + y*0= -√3(x -√3)|BP|= sqrt((x -√3)^2 + y^2)|BC|= sqrt( (√3)^2 +0^2 )=√3Thus,cos(60°)= [ -√3(x -√3) ] / [ sqrt((x -√3)^2 + y^2) * √3 ]Simplify:0.5= [ - (x -√3) ] / sqrt((x -√3)^2 + y^2 )Multiply both sides by denominator:0.5 * sqrt((x -√3)^2 + y^2 ) = - (x -√3)Square both sides:0.25 * ((x -√3)^2 + y^2 ) = (x -√3)^2Multiply both sides by 4:(x -√3)^2 + y^2 =4(x -√3)^2Thus,y^2 =3(x -√3)^2Thus, y=±√3(x -√3)But since P is on the same side of BC as A, which is above the x-axis (since A is at (0,1)), so y>0. Thus, y=√3(x -√3)Therefore, the coordinates of P satisfy y=√3(x -√3). But P is also on ray DN.From parametric equations of DN:x= t*√3/2y= -√3 + t*(0.5 +√3)Set y=√3(x -√3)=√3(t*√3/2 -√3)=√3*( (t*3/2 ) -√3 )Wait, wait:Wait, x= t*√3/2, so √3(x -√3)=√3(t*√3/2 -√3)=√3*( (t*√3/2 -√3) )=√3*(√3/2 t -√3)= (3/2 t -3 )Thus, y=3/2 t -3.But from DN parametrization:y= -√3 + t*(0.5 +√3)So set equal:3/2 t -3 = -√3 + t*(0.5 +√3)Solve for t:3/2 t -3 = -√3 +0.5 t +√3 tBring all terms to left:3/2 t -3 +√3 -0.5 t -√3 t=0Combine like terms:(3/2 t -0.5 t) + (-√3 t) + (-3 +√3)=0(1 t) -√3 t + (-3 +√3)=0t(1 -√3) + (-3 +√3)=0Thus,t= (3 -√3)/(1 -√3)Multiply numerator and denominator by (1 +√3):t= (3 -√3)(1 +√3)/[(1 -√3)(1 +√3)]= (3*1 +3*√3 -√3*1 -√3*√3)/(1 -3)= (3 +2√3 -3)/(-2)= (2√3)/(-2)= -√3But t ≥0 since P is on ray DN. Negative t contradicts. So no solution? This is a problem.Wait, this suggests that there's no such point P on ray DN which satisfies angle PBC=60°, but the problem states that such a point P exists. Therefore, there must be an error in calculations.Wait, let's recast the equations.We have two expressions for y:From angle condition: y=√3(x -√3)From parametrization: x= t*√3/2, y= -√3 + t*(0.5 +√3)Substitute x into the angle condition:y=√3(t*√3/2 -√3)=√3*( (t*√3/2 -√3) )=√3*( (t*√3/2 - (2√3)/2 ) )=√3*( (t√3 -2√3)/2 )=√3*(√3(t -2)/2 )= (3(t -2))/2Thus, y= (3(t -2))/2From parametrization:y= -√3 + t*(0.5 +√3)Set equal:(3(t -2))/2 = -√3 + t*(0.5 +√3)Multiply both sides by 2:3(t -2) = -2√3 + t*(1 +2√3)Expand left side:3t -6Right side: -2√3 + t*(1 +2√3)Bring all terms to left:3t -6 +2√3 -t*(1 +2√3)=0Factor t:t*(3 -1 -2√3) + (-6 +2√3)=0t*(2 -2√3) + (-6 +2√3)=0Factor 2 from first term:2t*(1 -√3) + (-6 +2√3)=0Divide both sides by 2:t*(1 -√3) + (-3 +√3)=0Thus,t= (3 -√3)/(1 -√3)Which is same as before. Multiply numerator and denominator by (1 +√3):t= (3 -√3)(1 +√3)/ (1 -3)= [3(1) +3√3 -√3 -3]/ (-2)= [ (3 -3) + (3√3 -√3) ] / (-2)= (2√3)/ (-2)= -√3Negative t, which is invalid. So in this coordinate system, there is no such point P on ray DN. Contradicts problem statement. So either my coordinates are wrong, or the problem has specific conditions where P exists.Alternatively, maybe in this configuration, such a point P doesn't exist, implying that my coordinate choice is invalid.But the problem states that such a point P exists. Therefore, I must have made an error in coordinate setup.Perhaps in some triangles, P exists on ray DN, and in others, it doesn't, but the problem implies that it does. Therefore, my coordinate examples might not satisfy all conditions. So maybe a better approach is needed.Let me try to solve the problem using Ceva's theorem and Menelaus' theorem synthetically.Given triangle ABC, midpoint N of AB, D on AC extended beyond C such that CD=BC, P on ray DN beyond N such that angle PBC=angle A. PC intersects AB at E, and DP intersects BC at T. Find (BC/TC) - (EA/EB).First, since CD=BC, triangle BCD is isoceles with CB=CD. Therefore, angle CBD= angle CDB.Given angle PBC= angle A. Let's denote angle A=α, angle B=β, so α>β.In triangle ABC, angles:α + β + γ=180°, where γ=angle C.In triangle BCD, angles at B and D are equal: angle CBD=angle CDB=θ.Therefore, angle BCD=180° - 2θ.But angle BCD is also equal to angle BCA + angle ACD. Since D is on extension of AC beyond C, angle ACD=180° - angle BCA=180° - γ.Wait, no. Point D is on ray AC beyond C, so angle BCD is the angle at C between BC and CD. Since CD is colinear with AC, angle BCD= angle BCA. Wait, no. If you extend AC beyond C to D, then at point C, angle between BC and CD is equal to angle BCA.Wait, in triangle BCD, angle at C is angle BCD= angle BCA. Then angles in triangle BCD:angle CBD=angle CDB=θ, angle BCD=γ.Thus, γ + 2θ=180°, so θ=(180° -γ)/2.But in triangle ABC, γ=180° -α -β. Therefore, θ=(α +β)/2.Given that angle PBC=α.In triangle PBC, angle PBC=α, angle BPC=?In triangle ABC, angle at B is β.If we can relate angles in triangle PBC to those in triangle ABC.Alternatively, consider triangle PBC and triangle ABC.Angle PBC=angle A=α. If we can find another angle relation.Alternatively, use Ceva's theorem in triangle ABC with point E, but E is the intersection of PC and AB.But Ceva's theorem would involve three cevians. Alternatively, use Menelaus' theorem on triangle ABC with transversal E-P-C.Wait, Menelaus' theorem states that for a triangle and a transversal line, the product of the segment ratios is -1.In triangle ABC, if line EPC intersects AB at E, BC at T, and ... Wait, no, Menelaus' theorem requires the transversal to intersect all three sides (or their extensions). Here, PC intersects AB at E and BC at T. But where does it intersect the third side?Alternatively, consider Menelaus' theorem for triangle BCD with transversal T-P-E.Hmm.Alternatively, apply Ceva's theorem in triangle BCD. Cevians are BT, CP, and DN. If they are concurrent, then:(BT/TD)*(DP/PN)*(NC/CB)=1But not sure.Alternatively, since N is midpoint of AB, and D is defined such that CD=BC, maybe there's a homothety or reflection that maps certain points.Alternatively, since CD=BC and N is the midpoint, perhaps consider vectors or coordinate transformations.Alternatively, use the Law of Sines in triangles PBC and ABC.In triangle PBC, angle PBC=α, angle BPC=?, angle BCP=?In triangle ABC, angle at C is γ, angle at A is α, angle at B is β.In triangle PBC, angles sum to 180°, so α + angle BPC + angle BCP=180°.But angle BCP is part of angle BCA=γ. So angle BCP=γ - angle PCT, where T is the intersection of DP and BC.But this is getting too vague.Alternatively, let's consider the ratios we need: BC/TC and EA/EB. The expression is their difference.Suppose we can express both ratios in terms of a common variable and find their difference.Let’s denote TC=x, so BC/TC=BC/x.EA/EB= let's denote EB=y, EA= AB - EB= AB - y.But AB is a fixed length.Alternatively, use mass point geometry for ratios.In mass point, masses are assigned to vertices such that the ratios of lengths are inversely proportional to the masses.For example, if TC=x and TB=BC -x= BC -x, then mass at C is m_C= TB=BC -x, mass at B is m_B=TC=x. Then the mass at T is m_T= m_B + m_C= BC -x +x=BC.Similarly, for point E on AB, masses at A and B are determined by the ratio EA/EB.But connecting these two might be challenging.Alternatively, since E is the intersection of PC and AB, and T is the intersection of DP and BC, maybe there's a relation between the two ratios.Alternatively, use the concept of duality or reciprocal figures.Alternatively, consider the following:Since CD=BC, and N is the midpoint of AB, perhaps quadrilateral BNCD is a parallelogram or has some symmetry. But not necessarily.Alternatively, consider that since CD=BC, then BD=BC + CD=2BC if D is beyond C. Wait, no, BD is not necessarily twice BC. Because CD=BC, but BD depends on the angle at C.Alternatively, if we consider triangle BCD isoceles with BC=CD, then BD=2BC*cos(theta), where theta is angle at B. But not sure.Alternatively, recall that in the problem statement, angle PBC=angle A. So maybe using trigonometric Ceva.In triangle ABC, if there exists a point P such that angle PBC=angle A, then the cevians might satisfy certain conditions.Alternatively, use the trigonometric form of Ceva's theorem for concurrency:[sin(angle PBA)/sin(angle PBC)] * [sin(angle QCB)/sin(angle QCA)] * [sin(angle RAC)/sin(angle RAB)] =1But not sure.Alternatively, since angle PBC=angle A, and P is on ray DN, we might need to construct P such that this angle condition holds.Alternatively, use the Law of Sines in triangle PBC and triangle ABC.In triangle PBC:PB / sin(angle BCP) = BC / sin(angle BPC)In triangle ABC:AB / sin(angle ACB) = BC / sin(angle BAC)= BC / sin(angle PBC)Thus, BC / sin(angle PBC)=AB / sin(angle ACB)Thus, BC / sin(alpha)= AB / sin(gamma)Therefore, AB= BC * sin(gamma)/ sin(alpha)Similarly, in triangle PBC:PB / sin(angle BCP)= BC / sin(angle BPC)But angle BCP= angle BCA - angle PCT= gamma - angle PCT.Not sure.Alternatively, since angle PBC=alpha and angle BAC=alpha, perhaps triangles PBC and BAC are similar.If angle PBC=angle BAC=alpha, and angle BPC=angle ABC=beta, then triangles PBC and BAC would be similar.If that's the case, then PB/BA=BC/AC=PC/BC.But this requires angle BPC=beta.But angle BPC=180° - alpha - angle BCP.Not sure.Assuming similarity, then PB/BA=BC/AC => PB= (BA * BC)/ACSimilarly, PC/BC=BC/AC => PC= (BC^2)/ACThis might help in finding ratios.But need to verify if triangles are indeed similar.Alternatively, since angle PBC=angle BAC=alpha, and if angle BPC=angle ABC=beta, then similarity follows.Assuming this similarity, then ratios hold.Given that, then PC= (BC^2)/ACThus, PC/BC= BC/AC => PC= (BC^2)/ACThus, if we can express PC in terms of BC and AC, then we can find the ratio EC/EA.Wait, E is the intersection of PC and AB.Using Menelaus' theorem on triangle ABC with transversal P-E-C:(AP/PB)*(BE/EA)*(AC/CB)=1But not sure.Alternatively, use Menelaus on triangle ABC with transversal E-P-C:No, Menelaus requires a line cutting through all three sides.Alternatively, use Menelaus on triangle ABP with transversal E-C-.This is getting too convoluted.Perhaps the key insight is to recognize that the expression BC/TC - EA/EB equals 1.But how?Alternatively, consider that T is the midpoint of BC. Then TC=BC/2, so BC/TC=2. If EA/EB=1, then the expression is 2 -1=1. But in previous examples, this wasn't the case. Unless in some configurations T is the midpoint.Alternatively, use harmonic division or projective geometry.Alternatively, consider that the problem is designed such that the value is 1, regardless of the triangle. Therefore, the answer is boxed{1}.But in the coordinate examples I tried, the answer wasn't 1. Unless I made a mistake.Alternatively, check with an equilateral triangle where angles are equal, but the problem states angle A>angle B. So not applicable.Alternatively, check with isoceles triangle where AB=AC, but angle A>angle B.Wait, if AB=AC, then angles at B and C are equal, contradicting angle A> angle B. So cannot.Therefore, the answer is likely 1, and my coordinate examples had errors.Alternatively, let's consider that in the expression BC/TC - EA/EB, the terms BC/TC and EA/EB are related through Ceva or Menelaus, and their difference is 1.Alternatively, since D is such that CD=BC, and N is the midpoint, then by constructing P such that angle PBC=angle A, there is a homothety that centers at B, mapping C to D and A to P. Not sure.Alternatively, use the fact that N is the midpoint and CD=BC to apply midline theorem.Alternatively, the problem might be designed with the answer 1, and through the problem's conditions, the two ratios cancel each other's difference to 1.Given that in both coordinate examples, the expression was greater than 1, but the problem likely expects 1, I might have made a mistake.Alternatively, reconsider the first example:Triangle ABC right-angled at C, angle A=60°, angle B=30°, BC=2, AC=2√3/√3=2, no. Wait, earlier confusion. Let me recast.Wait, if angle A=60°, angle B=30°, angle C=90°, then sides:BC=2 (opposite angle A=60°)AC=1 (opposite angle B=30°)AB=√3 (hypotenuse)Wait, no, in a 30-60-90 triangle, sides are 1, √3, 2. The side opposite 30° is 1, opposite 60° is √3, hypotenuse 2.Thus, if angle A=60°, angle B=30°, then BC=1 (opposite angle A), AC=√3 (opposite angle B=30°), AB=2.Coordinates:B at (0,0), C at (1,0), A at (0,√3)Midpoint N of AB: ((0 +0)/2, (√3 +0)/2)= (0, √3/2)Point D is on ray AC beyond C such that CD=BC=1. AC is from A(0,√3) to C(1,0). So direction vector AC is (1, -√3). Unit vector in direction AC: (1, -√3)/2. Then D=C + 1*(1, -√3)/2 *2= C + (1, -√3)= (1 +1, 0 -√3)= (2, -√3)Ray DN is from D(2, -√3) to N(0, √3/2). Direction vector: (-2, (√3/2 +√3))= (-2, (3√3/2))Parametric equations: x=2 -2t, y= -√3 + (3√3/2)t, t≥0Find point P on this ray such that angle PBC=60°. Point B is at (0,0).Angle PBC=60°, so angle between PB and BC is 60°.Vector BC is from B(0,0) to C(1,0): (1,0)Vector BP is from B(0,0) to P(x,y): (x,y)Dot product: x*1 + y*0= x|BP|=sqrt(x^2 + y^2)|BC|=1cos(theta)=x / sqrt(x^2 + y^2)=cos(60°)=0.5Thus,x / sqrt(x^2 + y^2) =0.5 => x=0.5 sqrt(x^2 + y^2)Square both sides:x^2=0.25(x^2 + y^2) => 4x^2=x^2 + y^2 => 3x^2=y^2 => y=±√3 xSince P is on the same side of BC as A (upper half-plane), y>0, so y=√3 xBut P is also on ray DN: x=2 -2t, y= -√3 + (3√3/2)tSet y=√3 x:-√3 + (3√3/2)t=√3(2 -2t)Divide both sides by √3:-1 + (3/2)t=2 -2tMultiply both sides by 2:-2 +3t=4 -4t7t=6 => t=6/7Thus, point P:x=2 -2*(6/7)=2 -12/7=2/7y=√3*(2/7)=2√3/7Check if this is on ray DN:x=2 -2t=2 -12/7=2/7, y= -√3 + (3√3/2)*(6/7)= -√3 + (18√3)/14= -√3 + (9√3)/7= (-7√3 +9√3)/7=2√3/7. Correct.Thus, coordinates of P are (2/7, 2√3/7)Line PC connects P(2/7, 2√3/7) to C(1,0). Find E, the intersection with AB.AB is from A(0,√3) to B(0,0). Parametric equations for AB: x=0, y=√3 -√3 s, s∈[0,1]Parametric equations for PC: x=2/7 + (1 -2/7)k=2/7 +5k/7, y=2√3/7 -2√3/7 k, k≥0Intersection E is where x=0. Set 2/7 +5k/7=0 => k=]

🔑:Solution: As shown in the figure, extend BP to intersect line AC at F. Thus, triangle ACB sim triangle BCF, which implies AC cdot CF = BC^2, so CF = frac{BC^2}{AC}.Notice that line DTN intersects triangle ABC, applying Menelaus' theorem we get frac{AN}{NB} cdot frac{BT}{TC} cdot frac{CD}{DA} = 1.Thus, frac{BT}{TC} = frac{DA}{CD} cdot frac{BN}{AN} = frac{DA}{CD} = frac{AC}{BC} + 1,so frac{BC}{TC} = frac{BT}{TC} + 1 = frac{AC}{BC} + 2.Similarly, by line DNP intersecting triangle ABF,we get frac{FP}{PB} cdot frac{BN}{NA} cdot frac{AD}{DF} = 1.By line CEP intersecting triangle ABF, we get frac{FP}{PB} cdot frac{BE}{EA} cdot frac{AC}{CF} = 1, sofrac{EA}{BE} = frac{FP}{PB} cdot frac{AC}{CF} = frac{AN}{BN} cdot frac{FD}{AD} cdot frac{AC}{CF} = frac{FD}{AD} cdot frac{AC}{CF}= frac{frac{BC^2}{AC} + BC}{AC + BC} cdot frac{AC}{frac{BC^2}{AC}} = frac{BC}{AC} cdot frac{AC^2}{BC^2} = frac{AC}{BC}, therefore,frac{BC}{TC} - frac{EA}{BE} = 2.

🧠:[<think>Okay, let's tackle this geometry problem step by step. So, we have four statements (A to D) about lines and planes in space, and we need to figure out which one is false. The question gives that m and n are two different lines, and α and β are two different planes. Let's go through each option one by one.Starting with Option A: If m is perpendicular to n, m is perpendicular to plane α, and n is not a subset of α (meaning n isn't lying entirely within α), then n is parallel to α. Hmm. Let me visualize this. If m is perpendicular to plane α, then m is a line that's standing straight up relative to α. Now, if m is also perpendicular to another line n, which isn't in α, where does that leave n? Well, if m is perpendicular to α, then m is orthogonal to every line in α. But n is not in α and is perpendicular to m. So, maybe n has to lie in a plane that's parallel to α? Because if n is perpendicular to m, and m is perpendicular to α, then n must be parallel to α. Wait, that seems right. Because if you imagine m as a vertical line, then n being perpendicular to m would have to be horizontal. If it's not in α, which is a horizontal plane (if m is vertical), then n would have to be in another horizontal plane, hence parallel. So, I think A is true. So maybe A isn't the false statement.Moving on to Option B: If m is perpendicular to β, and α is perpendicular to β, then m is parallel to α or m is a subset of α. Let me think. If α and β are perpendicular planes, their dihedral angle is 90 degrees. Now, if m is perpendicular to β, then m is a line that's orthogonal to β. If α is perpendicular to β, then α could be considered as a plane that's intersecting β along some line. Now, if m is perpendicular to β, where is m in relation to α? If m is perpendicular to β, then m must lie in a plane that's parallel to α or is part of α. Wait, let me try to imagine this. Let’s say β is the xy-plane. Then α, being perpendicular to β, would be a vertical plane, like the xz-plane. If m is perpendicular to β (the xy-plane), then m is a vertical line. Now, is this vertical line either parallel to α (the xz-plane) or lying within α? Well, if m is the z-axis, then it's lying in the xz-plane (so m is a subset of α). If m is some other vertical line not in the xz-plane, like the line parallel to the z-axis but passing through (0,1,0), then would that line be parallel to α (xz-plane)? Wait, lines can't really be parallel to planes unless they lie in a parallel plane or are parallel in direction. Wait, actually, a line is parallel to a plane if it's either lying in the plane or doesn't intersect the plane and is directionally parallel. But in three dimensions, a line not in a plane can still be parallel if it doesn't intersect and has direction vectors lying within the plane's direction. Hmm. Wait, but in this case, if α is the xz-plane and m is a vertical line (along the z-axis direction) not in α, then it's actually parallel to α because it's going in the same direction as the z-axis, which is part of α. Wait, but the xz-plane has directions along x and z. So a vertical line (along z) not in α would still have a direction that's within the plane's direction. Wait, maybe not. If α is the xz-plane, then any line parallel to the z-axis is in α or parallel? Wait, actually, no. If you have a line that's parallel to the z-axis but shifted along the y-axis, it's not in the xz-plane, but it's still parallel because it doesn't intersect α and its direction is within α's direction. So, yes, such a line would be parallel to α. Similarly, if m is in α, then it's a subset. So in either case, m is either parallel to α or in α. Therefore, B seems true. So maybe B is not the false one.Option C: If m is parallel to α, and α is perpendicular to β, then m is perpendicular to β. Hmm. Let me parse this. So, m is a line parallel to plane α. Plane α is perpendicular to plane β. Does that mean m is perpendicular to β? Let's take an example. Let α be the xy-plane and β be the xz-plane (so they are perpendicular). Now, a line m parallel to α (the xy-plane) would be a horizontal line, maybe along the x-axis or y-axis. If m is along the x-axis, then m is actually lying in both α and β (since x-axis is in both xy and xz planes). But wait, m is parallel to α, so it can be in α or parallel. If m is parallel to α but not in α, then m would be a horizontal line shifted upwards. But if α is the xy-plane and β is the xz-plane, then β is vertical. Is a horizontal line (parallel to α) perpendicular to β? Well, the xz-plane has normal vector along the y-axis. A horizontal line (say, along the x-direction) would not be perpendicular to β. Because the direction of the line is along x, and the normal of β is along y. The line would actually lie within β if it's along the x-axis, but if it's shifted along y, it's parallel. Wait, but in either case, the line isn't necessarily perpendicular to β. For example, if m is parallel to the x-axis but above the xy-plane, then it's parallel to α (xy-plane), but the line's direction is along x, which is in β (xz-plane). So the angle between m and β would be zero, not 90 degrees. Wait, but how do we define a line being perpendicular to a plane? A line is perpendicular to a plane if it's perpendicular to every line in the plane, which is equivalent to being parallel to the plane's normal vector. So, if β is the xz-plane, its normal is along the y-axis. So a line perpendicular to β would be along the y-axis. But our m is along the x-axis or some horizontal direction, which is not along y. Therefore, m is not perpendicular to β. Hence, the conclusion in C is false. Therefore, C is the false statement. Wait, but let me check again. If α is perpendicular to β, and m is parallel to α, does that imply m is perpendicular to β? No, because being parallel to a plane just means the line is either in the plane or parallel in direction, but not necessarily related to the other plane. So, for example, if two planes are perpendicular, a line parallel to one plane isn't necessarily perpendicular to the other. So, yes, C seems false. Let me hold that thought and check D to be sure.Option D: If m is perpendicular to n, m is perpendicular to α, and n is perpendicular to β, then α is perpendicular to β. Let's see. So m ⊥ n, m ⊥ α, n ⊥ β. Then α ⊥ β? Let's try to visualize. If m is perpendicular to α, then m is a normal vector to α. Similarly, n is a normal vector to β (since n is perpendicular to β). Now, m and n are perpendicular lines. So, the normals to α and β are perpendicular. If the normals of two planes are perpendicular, then the planes themselves are perpendicular. Yes, because the angle between two planes is equal to the angle between their normals. So if normals are perpendicular, the planes are perpendicular. Therefore, D is true. So D is not the false one.Therefore, the false statement is C. Let me just confirm once again for C. Suppose α and β are two perpendicular planes. Take a line m that's parallel to α. Does that line have to be perpendicular to β? Let's take α as the xy-plane and β as the xz-plane. A line parallel to α could be the line along the x-axis (which is in both α and β), or a line parallel to the x-axis but shifted up along z. Wait, that line would actually be in β (the xz-plane) if it's along the x-axis. But if it's shifted along y, then it's parallel to α. Wait, maybe I need a different example. Suppose α is the xy-plane, β is the yz-plane (so they are perpendicular). Take a line m that's parallel to α; for example, a horizontal line in the x-direction but at some height z = h. This line is parallel to α (xy-plane), but is it perpendicular to β (yz-plane)? The line's direction is along the x-axis. The normal vector to β (yz-plane) is along the x-axis. So, the line m is parallel to the normal of β. Therefore, m is perpendicular to β. Wait, hold on, that contradicts my earlier thought. If the line is parallel to the normal of β, then the line is perpendicular to β. Hmm. Wait, in this case, if β is the yz-plane, its normal is along the x-axis. So a line along the x-axis is indeed perpendicular to β. But in my previous example, where β was the xz-plane, the normal is along the y-axis. If I take a line parallel to α (xy-plane), say along the x-axis, then it's in β (xz-plane), so it's not perpendicular. Wait, but if the line is parallel to α, which is the xy-plane, then a line parallel to α would have a direction vector lying in the xy-plane. If β is the xz-plane, then the normal of β is along the y-axis. So, if a line is in the xy-plane, its direction is in xy, so the angle between the line and β would depend. Wait, maybe I confused the planes. Let me clarify.Let me set up coordinates: Let α be the xy-plane (normal along z), β be the xz-plane (normal along y). Then α and β are not perpendicular because their normals (z and y) are not perpendicular; wait, no, the dot product of y and z is zero, so they are perpendicular. Wait, the normals are y and z, which are perpendicular, so planes α (xy) and β (xz) are indeed perpendicular. Now, take a line m that's parallel to α (xy-plane). Let's say m is along the x-axis direction but shifted up along z. So, direction vector (1,0,0). Then, is this line perpendicular to β (xz-plane)? The normal of β is along y. The line's direction is along x. Since the line's direction is not along the normal of β, the line is not perpendicular to β. Therefore, the line is parallel to α but not perpendicular to β. Therefore, statement C is false. Wait, but in another configuration, if β's normal is along x, then a line along x would be perpendicular. But in this case, depending on how α and β are arranged, the result changes. Wait, but the problem says "α and β are two different planes", but doesn't specify their orientation. So in the general case, if α and β are perpendicular, and m is parallel to α, does m have to be perpendicular to β? From the example above, it's not necessarily the case. If α is the xy-plane and β is the xz-plane (perpendicular), then a line parallel to α (direction in xy) is not necessarily perpendicular to β. For example, a line along the x-axis is in both α and β, but not perpendicular. A line along the y-axis (parallel to α) is in α but is perpendicular to β (since β's normal is along z? Wait, no. Wait, β is xz-plane, normal is along y. So a line along y would be along the normal of β, hence perpendicular. Wait, confusion here. Let me clarify:If α is the xy-plane (normal along z), β is the xz-plane (normal along y). Then, a line parallel to α (xy-plane) could be in the x-direction or y-direction. If it's in the x-direction (direction vector (1,0,0)), then this line is in β (xz-plane) as well. Is it perpendicular to β? The line is in β, so it's not perpendicular. If it's in the y-direction (direction vector (0,1,0)), then since β's normal is along y, this line is along the normal of β, hence perpendicular. So, depending on the direction of m, it might or might not be perpendicular. But the statement C says "if m is parallel to α, then m is perpendicular to β". But in the case where m is along x-direction (parallel to α but also in β), it's not perpendicular. Hence, the statement C is not always true, so it's false. Therefore, C is the false statement.Just to confirm once again, let's see. If m is parallel to α, then m's direction vector lies in α's direction. Since α and β are perpendicular, the normal of β is in some direction. For m to be perpendicular to β, m's direction must be parallel to β's normal. But unless α's plane contains β's normal, this isn't necessarily the case. So unless the normal of β is in α, which isn't necessarily true just because α and β are perpendicular. Therefore, m could be in a direction within α that isn't aligned with β's normal, hence not perpendicular. So yes, C is false.Final Answerboxed{C}</think>Given the problem, we need to determine which statement among A, B, C, and D is false. Here is the detailed reasoning:Option A: If ( m perp n ), ( m perp alpha ), and ( n nsubseteq alpha ), then ( n parallel alpha ). - If ( m ) is perpendicular to plane ( alpha ), ( m ) is a normal line to ( alpha ). If ( m ) is also perpendicular to ( n ), and ( n ) is not in ( alpha ), then ( n ) must be parallel to ( alpha ). This is true.Option B: If ( m perp beta ), ( alpha perp beta ), then ( m parallel alpha ) or ( m subseteq alpha ). - If ( m ) is perpendicular to ( beta ) and ( alpha perp beta ), then ( m ) must be parallel to ( alpha ) or lie within ( alpha ). This is true because a line perpendicular to ( beta ) would be parallel to the normal of ( beta ), and since ( alpha perp beta ), the normal of ( beta ) lies in ( alpha ). Hence, ( m ) is either parallel to ( alpha ) or within ( alpha ).Option C: If ( m parallel alpha ), ( alpha perp beta ), then ( m perp beta ). - If ( m ) is parallel to ( alpha ) and ( alpha perp beta ), ( m ) does not necessarily have to be perpendicular to ( beta ). For example, if ( alpha ) is the xy-plane and ( beta ) is the xz-plane, a line ( m ) parallel to the x-axis (which is parallel to ( alpha )) is not perpendicular to ( beta ). This is false.Option D: If ( m perp n ), ( m perp alpha ), ( n perp beta ), then ( alpha perp beta ). - If ( m ) and ( n ) are perpendicular, and ( m perp alpha ), ( n perp beta ), then the normals of ( alpha ) and ( beta ) (which are ( m ) and ( n )) are perpendicular. Hence, ( alpha perp beta ). This is true.Thus, the false statement is (boxed{C}).,<think>Okay, let's see. I need to figure out which of these statements about lines and planes is false. The options are A, B, C, D. Let me go through each one step by step. Starting with statement A: If m is perpendicular to n, and m is perpendicular to plane α, and n is not a subset of α (so n isn't lying in α), then n is parallel to α. Hmm. Let me visualize. If m is perpendicular to plane α, then m is like a vertical line sticking up from α. Then n is another line that's perpendicular to m. So if m is vertical, then n would have to be horizontal? But if n isn't in α, does that mean it's parallel? Wait, if n is horizontal and not in α, then yes, it should be parallel to α. Because any line perpendicular to a line that's perpendicular to the plane would be parallel to the plane. That seems right. So statement A might be true. Not sure yet, but maybe.Moving on to B: If m is perpendicular to β, and α is perpendicular to β, then m is parallel to α or m is contained in α. Let me think. If m is perpendicular to β, that means m is a line that's orthogonal to every line in β. And α and β are perpendicular planes, so they intersect along a line. If m is perpendicular to β, could m be lying in α? Well, if α is perpendicular to β, then α contains all lines that are perpendicular to β, right? Wait, no. Actually, if two planes are perpendicular, then a line perpendicular to one plane might be parallel to the other plane or lie within it. For example, take the xy-plane and yz-plane as perpendicular planes. A line along the z-axis is perpendicular to the xy-plane. The z-axis is parallel to the yz-plane? Wait, no, the z-axis is actually part of the yz-plane. Wait, no. The yz-plane is the plane where x=0, and the z-axis is part of that. So if a line is perpendicular to the xy-plane (like the z-axis), then it's contained within the yz-plane. But if the line is perpendicular to the xy-plane but not along the z-axis, say, some other line perpendicular to the xy-plane, would that line be parallel to the yz-plane? Wait, any line perpendicular to the xy-plane is vertical, so if the yz-plane is vertical, then a vertical line not in the yz-plane would be parallel to it? No, parallel lines in space need to be in the same direction and not intersect, but lines perpendicular to the xy-plane are all vertical. So a vertical line not in the yz-plane would still be parallel to the yz-plane? Hmm. Wait, actually, in 3D space, two lines that are both perpendicular to the same plane are parallel. So all lines perpendicular to the xy-plane are parallel to each other, hence they are all parallel to the z-axis. So if α is the yz-plane, then the z-axis is in α, and any other vertical line (perpendicular to xy-plane) is parallel to α. Because the yz-plane contains the z-axis, and any parallel line to the z-axis would be parallel to the plane? Wait, no. If a line is parallel to a plane, it means that the line is either in the plane or parallel to it but not intersecting. Wait, actually, no. In 3D geometry, a line is parallel to a plane if it lies in a plane parallel to the given plane. Wait, maybe I need to recall the exact definition. A line is parallel to a plane if it doesn't intersect the plane and is not in the plane. So if a vertical line is not in the yz-plane, it would be parallel to the yz-plane? Because the yz-plane is x=0, so any vertical line with x ≠ 0 would not intersect it and isn't in it, hence is parallel. Therefore, if m is perpendicular to β (the xy-plane), and α (yz-plane) is perpendicular to β, then m is either in α (if it's the z-axis) or parallel to α (if it's another vertical line). So statement B is true. So B is not the false one.Now, statement C: If m is parallel to α, and α is perpendicular to β, then m is perpendicular to β. Wait, this seems suspect. Let's take an example. Let α be the yz-plane and β be the xy-plane. Then α is perpendicular to β. If m is a line parallel to α (the yz-plane), then m could be any line that's parallel to the yz-plane, which would mean it's a horizontal line in the x-direction or some other direction? Wait, no. If a line is parallel to the yz-plane, it means it doesn't intersect the yz-plane and is going in a direction that's parallel. Wait, actually, lines parallel to a plane can have any direction that's parallel to that plane. For example, in the yz-plane, lines can be along the y-axis, z-axis, or any direction in the yz-plane. If a line is parallel to the yz-plane, then it must be parallel to some line in the yz-plane. So, for instance, a line in the direction of the y-axis but shifted along the x-axis would be parallel to the yz-plane. Now, if α is the yz-plane and β is the xy-plane (which are perpendicular), then m is parallel to α. So m could be, say, a line along the y-axis direction but at some x ≠ 0. Then, is m perpendicular to β (the xy-plane)? The xy-plane has normal vector along the z-axis. A line along the y-axis is not perpendicular to the xy-plane. In fact, it's lying within the xy-plane. Wait, but if m is parallel to the yz-plane, it could be along the z-axis direction. Wait, if m is parallel to the yz-plane, then it's direction vector is in the y-z direction. For example, a line in the direction of the z-axis but shifted along x. Then, such a line would be perpendicular to the xy-plane? Because the xy-plane's normal is the z-axis, and a line along the z-axis is perpendicular. So if m is parallel to the yz-plane, then m's direction is a combination of y and z. If it's along z, then it's perpendicular to the xy-plane. If it's along y, it's parallel to the xy-plane. Wait, so maybe statement C is not always true. For example, if m is parallel to α (yz-plane) and α is perpendicular to β (xy-plane), then m could be in the y-direction (parallel to xy-plane) or z-direction (perpendicular to xy-plane). Therefore, m isn't necessarily perpendicular to β. So statement C is false. Wait, but let me check again. If a line is parallel to a plane, does that mean it's direction is in the plane's direction? So, if α is the yz-plane, then direction vectors in the yz-plane would be any vector with x-component zero. So if m is parallel to α, then m's direction vector has x-component zero. Then, β is a plane perpendicular to α, say, the xy-plane. Then, the normal vector of β is along the z-axis. So if m's direction is in the yz-plane (so direction vector (0, a, b)), then the dot product with the normal vector (0,0,1) of β would be b. So unless b is zero, the line isn't perpendicular. Wait, the line is perpendicular to β if it's parallel to the normal vector of β. So, if m is parallel to the normal vector of β, then it's perpendicular. So if m is parallel to α (yz-plane), then m can have direction (0, a, b). To be perpendicular to β (xy-plane), m's direction must be along (0,0,1). So unless a=0, m isn't perpendicular. Therefore, if m is parallel to α, which is the yz-plane, then m can have direction (0,a,b), which is only perpendicular to β (xy-plane) if a=0. Otherwise, it's not. So statement C is not necessarily true. Therefore, statement C is false. So maybe C is the answer. Wait, but let me check the other options to be sure.Statement D: If m is perpendicular to n, m is perpendicular to α, and n is perpendicular to β, then α is perpendicular to β. Hmm. Let me visualize. So m is perpendicular to both n and α. Since m is perpendicular to α, m is a normal vector to α. Similarly, n is perpendicular to β, so n is a normal vector to β. If m is perpendicular to n, then the normals of α and β are perpendicular, which implies that α and β are perpendicular. Yes, because the angle between planes is the angle between their normals. So if the normals are perpendicular, the planes are perpendicular. So statement D is true.Therefore, the false statement is C. Let me verify again with another example. Suppose α is the yz-plane, β is the xz-plane, which are both perpendicular to the xy-plane. Wait, no. Wait, if α is the yz-plane (normal along x-axis) and β is the xy-plane (normal along z-axis). Then, if m is a line along the x-axis (perpendicular to α, since α's normal is x-axis), and n is a line along the z-axis (perpendicular to β). Then m is along x-axis, n is along z-axis, so m is perpendicular to n. Then according to D, α (yz-plane) and β (xy-plane) are perpendicular, which they are. So that's correct. Another example for statement C: Let α be the yz-plane, β be the xy-plane. Take a line m that's parallel to the y-axis, so it's parallel to α. Then α is perpendicular to β. But m is parallel to β (since the y-axis is in β). So m is not perpendicular to β, it's lying in β. Wait, but if m is parallel to α (yz-plane), then if m is in β (xy-plane), then it's not necessarily lying in β. Wait, if m is parallel to α (yz-plane), which is perpendicular to β (xy-plane), then m could be in β or not? Wait, a line parallel to the yz-plane can be in a different plane. For example, take a line that's along the y-axis but shifted along the z-axis. Wait, but that line would still be in the yz-plane. Hmm. Maybe I need a different example. If m is parallel to the yz-plane but not lying in the yz-plane, like a line that is parallel to the y-axis but at some x ≠ 0 and z ≠ 0. Then, such a line is parallel to the y-axis, which is in both α (yz-plane) and β (xy-plane). But if the line is in 3D space, say, along the direction of the y-axis but located at (1, t, 0) for t ∈ ℝ. Then this line is parallel to the y-axis, hence parallel to α (yz-plane). But this line is also in the xy-plane (β), so it's not perpendicular to β. Therefore, statement C is false because being parallel to α and α perpendicular to β does not imply the line is perpendicular to β; it could be in β or parallel to β. Therefore, C is indeed the false statement.Final Answerboxed{C}</think>To determine the false statement among the given options, we analyze each statement step by step:Statement A: If ( m perp n ), ( m perp alpha ), and ( n nsubseteq alpha ), then ( n parallel alpha ). - If ( m ) is perpendicular to plane ( alpha ), ( m ) is a normal vector to ( alpha ). A line ( n ) perpendicular to ( m ) and not lying in ( alpha ) must be parallel to ( alpha ). This is true.Statement B: If ( m perp beta ), ( alpha perp beta ), then ( m parallel alpha ) or ( m subseteq alpha ). - If ( m ) is perpendicular to ( beta ), ( m ) is a normal vector to ( beta ). Since ( alpha perp beta ), ( m ) must either lie in ( alpha ) or be parallel to ( alpha ). This is true.Statement C: If ( m parallel alpha ), ( alpha perp beta ), then ( m perp beta ). - A line ( m ) parallel to ( alpha ) can have a direction vector within ( alpha ). However, ( alpha perp beta ) means the normals of ( alpha ) and ( beta ) are perpendicular. A line parallel to ( alpha ) is not necessarily perpendicular to ( beta ). For example, a line in the yz-plane (parallel to ( alpha )) can be along the y-axis, which is not perpendicular to the xy-plane (( beta )). This is false.Statement D: If ( m perp n ), ( m perp alpha ), ( n perp beta ), then ( alpha perp beta ). - ( m ) and ( n ) are normals to ( alpha ) and ( beta ) respectively. If ( m perp n ), the normals are perpendicular, implying ( alpha perp beta ). This is true.Thus, the false statement is (boxed{C}).]

🔑:If m perp n and m perp alpha, then the relationship between n and alpha is that n is either inside alpha or n is parallel to alpha. Since n nsubseteq alpha, it follows that n parallel alpha, hence statement A is true. If m perp beta and alpha perp beta, then m parallel alpha or m subseteq alpha, hence statement B is true. If m parallel alpha and alpha perp beta, then m could either be parallel to beta or intersect with beta, hence statement C is false. If m perp n and m perp alpha, then n nsubseteq alpha or n parallel alpha, and given n perp beta, it follows that alpha perp beta, hence statement D is true. Therefore, the correct choice is boxed{C}.

🧠:[<think>Okay, so I have this function: f(x) = 2 sin x cos x - 2 cos²x + 1. The problem has two parts. First, I need to find the maximum value of f(x) and the corresponding x value. Second, if f(θ) = 3/5, find the value of cos[2(π/4 - 2θ)]. Alright, let's tackle part (1) first.Hmm, looking at the function f(x) = 2 sin x cos x - 2 cos²x + 1. It seems like there are some trigonometric identities I can use here to simplify it. Let me recall that 2 sin x cos x is equal to sin 2x. So replacing that term, the function becomes sin 2x - 2 cos²x + 1. Now, the next term is -2 cos²x. There's another identity for cos²x: cos²x = (1 + cos 2x)/2. Let me substitute that in. So replacing cos²x, we get:sin 2x - 2*(1 + cos 2x)/2 + 1. Simplifying the terms: the 2 and denominator 2 cancel out, so that becomes sin 2x - (1 + cos 2x) + 1. Then distributing the negative sign: sin 2x - 1 - cos 2x + 1. The -1 and +1 cancel each other, so we end up with sin 2x - cos 2x.Okay, so f(x) simplifies to sin 2x - cos 2x. That's a much simpler expression. Now, to find the maximum value of this function. Since it's a combination of sine and cosine with the same argument 2x, I can write this as a single sine or cosine function using the amplitude-phase form. Remember that A sin θ + B cos θ can be written as C sin(θ + φ) or C cos(θ + φ), where C = sqrt(A² + B²). Let me apply that here.So, f(x) = sin 2x - cos 2x. Let's think of this as A sin 2x + B cos 2x, where A = 1 and B = -1. Then the amplitude C would be sqrt(1² + (-1)²) = sqrt(1 + 1) = sqrt(2). Therefore, we can write f(x) as sqrt(2) sin(2x + φ), where φ is some phase shift. Alternatively, maybe it's easier to write it as sqrt(2) cos(2x + φ'). Let me check.Alternatively, to combine sin 2x - cos 2x into a single sine function, let's recall that sin α cos β - cos α sin β = sin(α - β). Wait, but here we have sin 2x - cos 2x. Maybe another approach: factor out sqrt(2). So sqrt(2)*( (1/sqrt(2)) sin 2x - (1/sqrt(2)) cos 2x ). Then notice that 1/sqrt(2) is sin π/4 and cos π/4. Therefore, this becomes sqrt(2)*( sin 2x cos π/4 - cos 2x sin π/4 ). Using the sine subtraction formula, this is sqrt(2)*sin(2x - π/4). Yes, that works. So f(x) = sqrt(2) sin(2x - π/4). The maximum value of sin is 1, so the maximum value of f(x) is sqrt(2)*1 = sqrt(2). Therefore, the maximum value is sqrt(2), and it occurs when sin(2x - π/4) = 1. Solving for x: 2x - π/4 = π/2 + 2πk, where k is any integer. Then 2x = 3π/4 + 2πk, so x = 3π/8 + πk. So the maximum occurs at x = 3π/8 + πk. Since the question asks for "the corresponding value of x", perhaps they want the principal value, which is 3π/8. But depending on the context, they might accept all solutions, but since it's not specified, maybe just 3π/8.Wait, let me verify my steps again to make sure.Original function: 2 sin x cos x is sin 2x, correct. Then -2 cos²x +1. Using identity cos²x = (1 + cos 2x)/2, so -2*(1 + cos 2x)/2 = -1 - cos 2x. Then adding 1 gives sin 2x - cos 2x. Correct. Then rewriting as sqrt(2) sin(2x - π/4). Let me confirm that. If I have sin 2x - cos 2x, then to write it as C sin(2x - φ), let's expand C sin(2x - φ) = C sin 2x cos φ - C cos 2x sin φ. Comparing coefficients with sin 2x - cos 2x, we have:C cos φ = 1 (coefficient of sin 2x)-C sin φ = -1 (coefficient of cos 2x)Therefore, from the first equation: C cos φ = 1From the second equation: C sin φ = 1Therefore, dividing the two equations: (C sin φ)/(C cos φ) = 1/1 => tan φ = 1 => φ = π/4.Then, C is sqrt( (1)^2 + (1)^2 ) = sqrt(2). So, yes, that works. Therefore, sin 2x - cos 2x = sqrt(2) sin(2x - π/4). Therefore, the maximum value is sqrt(2). So part (1) is done.Now, part (2): if f(θ) = 3/5, find the value of cos[2(π/4 - 2θ)].First, let's write down f(θ) = sin 2θ - cos 2θ = 3/5. We need to find cos[2(π/4 - 2θ)]. Let me simplify the argument inside the cosine first.2(π/4 - 2θ) = π/2 - 4θ. So we need to find cos(π/2 - 4θ). But cos(π/2 - x) is sin x, so cos(π/2 - 4θ) = sin 4θ. Wait, but hold on, the problem says cos[2(π/4 - 2θ)], which is cos(π/2 - 4θ). But cos(π/2 - 4θ) = sin 4θ. Therefore, the expression simplifies to sin 4θ. Therefore, the problem is asking for sin 4θ. So perhaps we can find sin 4θ given that sin 2θ - cos 2θ = 3/5.Alternatively, maybe I miscalculated the argument. Let me check again. 2*(π/4 - 2θ) = π/2 - 4θ, correct. So cos(π/2 - 4θ) = sin 4θ. So the value we need is sin 4θ.So now, the problem reduces to: given that sin 2θ - cos 2θ = 3/5, find sin 4θ.Alternatively, maybe I can use some trigonometric identities to relate these.First, let's let’s denote φ = 2θ. Then the given equation is sin φ - cos φ = 3/5, and we need to find sin 2φ (since 4θ = 2φ). So if we can find sin 2φ, that will be the answer.So let me rephrase: given sin φ - cos φ = 3/5, find sin 2φ.Let me square both sides of the given equation: (sin φ - cos φ)^2 = (3/5)^2.Left side: sin²φ - 2 sin φ cos φ + cos²φ = (sin²φ + cos²φ) - 2 sin φ cos φ = 1 - sin 2φ. Therefore, 1 - sin 2φ = 9/25. Solving for sin 2φ: sin 2φ = 1 - 9/25 = 16/25.Therefore, sin 2φ = 16/25. But 2φ is 4θ, so sin 4θ = 16/25. Therefore, cos[2(π/4 - 2θ)] = sin 4θ = 16/25.Wait, let me check this again step by step.Given f(θ) = sin 2θ - cos 2θ = 3/5.Let’s set φ = 2θ. Then sin φ - cos φ = 3/5.Square both sides: (sin φ - cos φ)^2 = 9/25.Expanding: sin²φ - 2 sin φ cos φ + cos²φ = 9/25.But sin²φ + cos²φ = 1, so 1 - 2 sin φ cos φ = 9/25.Therefore, -2 sin φ cos φ = 9/25 - 1 = -16/25.Multiply both sides by (-1): 2 sin φ cos φ = 16/25.But 2 sin φ cos φ = sin 2φ. Therefore, sin 2φ = 16/25.Since φ = 2θ, then 2φ = 4θ. Therefore, sin 4θ = 16/25.But earlier, we saw that cos[2(π/4 - 2θ)] = sin 4θ. Therefore, the value is 16/25.Therefore, the answer to part (2) is 16/25.Wait, seems straightforward. Let me check if there's an alternative approach to confirm.Alternatively, starting from f(θ) = sin 2θ - cos 2θ = 3/5.We can write this as sqrt(2) sin(2θ - π/4) = 3/5, as we had in part (1). Therefore, sin(2θ - π/4) = 3/(5 sqrt(2)).But maybe this approach is more complicated. Alternatively, the previous approach by squaring is straightforward.Alternatively, maybe use the expression for sin 4θ. Since sin 4θ = 2 sin 2θ cos 2θ. If we can find sin 2θ and cos 2θ, we can compute sin 4θ.From the given equation: sin 2θ - cos 2θ = 3/5.Let me call a = sin 2θ, b = cos 2θ. Then a - b = 3/5. Also, we know that a² + b² = 1.So we have:a - b = 3/5,a² + b² = 1.Let me solve for a and b. Let's express a = b + 3/5.Substitute into the second equation: (b + 3/5)^2 + b² = 1.Expanding: b² + (6/5)b + 9/25 + b² = 1.Combine like terms: 2b² + (6/5)b + 9/25 - 1 = 0.Simplify 9/25 - 1 = -16/25. So:2b² + (6/5)b - 16/25 = 0.Multiply both sides by 25 to eliminate denominators:50b² + 30b - 16 = 0.Divide equation by 2:25b² + 15b - 8 = 0.Use quadratic formula: b = [-15 ± sqrt(15² - 4*25*(-8))]/(2*25)Calculate discriminant: 225 + 800 = 1025.So sqrt(1025) = sqrt(25*41) = 5*sqrt(41).Thus, b = [-15 ± 5√41]/50 = [-3 ± √41]/10.Therefore, two possible solutions for b (cos 2θ):b = [ -3 + √41 ] /10 ≈ (-3 + 6.403)/10 ≈ 3.403/10 ≈ 0.3403orb = [ -3 - √41 ] /10 ≈ (-3 -6.403)/10 ≈ -9.403/10 ≈ -0.9403Then a = b + 3/5.If b = [ -3 + √41 ] /10, then a = [ -3 + √41 ] /10 + 6/10 = [ 3 + √41 ] /10 ≈ (3 + 6.403)/10 ≈ 9.403/10 ≈ 0.9403.If b = [ -3 - √41 ] /10, then a = [ -3 - √41 ] /10 + 6/10 = [ 3 - √41 ] /10 ≈ (3 -6.403)/10 ≈ -3.403/10 ≈ -0.3403.Now, since sin 4θ = 2ab.Compute 2ab for each case.First case: a = [3 + √41]/10, b = [ -3 + √41 ]/10.ab = [3 + √41][ -3 + √41 ] / 100.Multiply numerator: (√41 +3)(√41 -3) = (√41)^2 - 3^2 = 41 -9 =32.So ab =32/100 =8/25.Thus, 2ab =16/25.Second case: a = [3 - √41]/10, b = [ -3 - √41 ]/10.ab = [3 - √41][ -3 - √41 ] /100.Multiply numerator: (3)(-3) + 3*(-√41) - √41*(-3) - √41*(-√41)= -9 -3√41 +3√41 +41= (-9 +41) + (-3√41 +3√41)= 32 +0=32.Thus ab=32/100=8/25. Then 2ab=16/25.Therefore, in both cases, sin 4θ=16/25. So regardless of the values of a and b, we have sin 4θ=16/25. Therefore, the answer is 16/25. So part (2) is 16/25.Therefore, the final answers are:(1) Maximum value is sqrt(2) at x=3π/8 +kπ, but likely 3π/8 is the principal solution.(2) The value is 16/25.Final Answer(1) The maximum value of ( f(x) ) is (boxed{sqrt{2}}) and it occurs at ( x = boxed{dfrac{3pi}{8} + kpi} ) for any integer ( k ).(2) The value of ( cos 2left( dfrac{pi}{4} - 2theta right) ) is (boxed{dfrac{16}{25}}).</think>Given the function ( f(x) = 2sin x cos x - 2cos^2 x + 1 ):1. Finding the maximum value of ( f(x) ) and the corresponding value of ( x ): - Simplify the function using trigonometric identities: [ f(x) = 2sin x cos x - 2cos^2 x + 1 = sin 2x - (1 + cos 2x) + 1 = sin 2x - cos 2x ] - Rewrite the simplified function using the amplitude-phase form: [ f(x) = sqrt{2} sinleft(2x - frac{pi}{4}right) ] - The maximum value of ( sin theta ) is 1, so the maximum value of ( f(x) ) is ( sqrt{2} ). - The maximum occurs when ( 2x - frac{pi}{4} = frac{pi}{2} + 2kpi ), solving for ( x ) gives: [ x = frac{3pi}{8} + kpi quad text{for any integer } k ]2. Finding the value of ( cos 2left( frac{pi}{4} - 2theta right) ) given ( f(theta) = frac{3}{5} ): - Given ( f(theta) = sin 2theta - cos 2theta = frac{3}{5} ). - Square both sides: [ (sin 2theta - cos 2theta)^2 = left(frac{3}{5}right)^2 implies 1 - sin 4theta = frac{9}{25} ] - Solving for ( sin 4theta ): [ sin 4theta = 1 - frac{9}{25} = frac{16}{25} ] - Note that ( cos 2left( frac{pi}{4} - 2theta right) = cosleft( frac{pi}{2} - 4theta right) = sin 4theta ), thus: [ cos 2left( frac{pi}{4} - 2theta right) = frac{16}{25} ] Final Answer(1) The maximum value of ( f(x) ) is (boxed{sqrt{2}}) and it occurs at ( x = boxed{dfrac{3pi}{8} + kpi} ) for any integer ( k ).(2) The value of ( cos 2left( dfrac{pi}{4} - 2theta right) ) is (boxed{dfrac{16}{25}}).,<think>Alright, let's tackle this problem step by step. The function given is ( f(x) = 2sin x cos x - 2cos^2 x + 1 ). There are two parts: first, finding the maximum value of ( f(x) ) and the corresponding ( x ); second, if ( f(theta) = frac{3}{5} ), finding the value of ( cos 2left( frac{pi}{4} - 2theta right) ).Starting with part (1). The function looks like a combination of sine and cosine terms. Maybe I can simplify it using trigonometric identities to make it easier to find the maximum.First, let's recall some identities. The term ( 2sin x cos x ) is familiar—it's the double-angle identity for sine. Specifically, ( 2sin x cos x = sin 2x ). So that simplifies the first term.Next, the term ( -2cos^2 x ). There's a double-angle identity for cosine squared: ( cos^2 x = frac{1 + cos 2x}{2} ). If I use that, then ( -2cos^2 x = -2 times frac{1 + cos 2x}{2} = -1 - cos 2x ).Putting it all together, substitute these into the original function:( f(x) = sin 2x - 1 - cos 2x + 1 ).Simplify the constants: ( -1 + 1 = 0 ), so the function reduces to:( f(x) = sin 2x - cos 2x ).Hmm, that's simpler. So now ( f(x) = sin 2x - cos 2x ). To find the maximum value of this function, I can use the amplitude formula for a linear combination of sine and cosine. The general form is ( Asin theta + Bcos theta ), which has an amplitude of ( sqrt{A^2 + B^2} ). But here, we have ( sin 2x - cos 2x ), so it's similar but with coefficients 1 and -1.Wait, actually, let's write it as ( sin 2x - cos 2x = sqrt{1^2 + (-1)^2} sin(2x + phi) ), where ( phi ) is some phase shift. The amplitude would be ( sqrt{1 + 1} = sqrt{2} ). Therefore, the maximum value of ( f(x) ) is ( sqrt{2} ), and the minimum is ( -sqrt{2} ).But let me confirm that. If we have ( a sin theta + b cos theta ), then the amplitude is ( sqrt{a^2 + b^2} ). Here, ( a = 1 ), ( b = -1 ), so amplitude is ( sqrt{2} ). Therefore, the maximum value is indeed ( sqrt{2} ).But wait, sometimes the formula is written as ( R sin(theta + phi) ), where ( R = sqrt{a^2 + b^2} ). So yes, maximum is ( R ), minimum is ( -R ). Therefore, the maximum value of ( f(x) ) is ( sqrt{2} ).Now, to find the corresponding value of ( x ). Let's express ( f(x) = sin 2x - cos 2x ) as a single sine function with phase shift. Let's do that.Let me recall that ( a sin theta + b cos theta = R sin(theta + phi) ), where ( R = sqrt{a^2 + b^2} ), and ( phi = arctan left( frac{b}{a} right ) ) or something like that. Wait, no, actually, the phase shift depends on both a and b. Let me recall the exact formula.Alternatively, we can write ( a sin theta + b cos theta = R sin(theta + phi) ), where ( R = sqrt{a^2 + b^2} ), and ( phi = arctan left( frac{b}{a} right ) ). Wait, but let's verify:Expanding ( R sin(theta + phi) ):( R sin theta cos phi + R cos theta sin phi ).Comparing with ( a sin theta + b cos theta ), we have:( a = R cos phi )( b = R sin phi )Therefore, ( tan phi = frac{b}{a} ), so ( phi = arctan left( frac{b}{a} right ) ).But in our case, ( a = 1 ), ( b = -1 ). Therefore, ( R = sqrt{2} ), and ( phi = arctan(-1/1) = arctan(-1) = -pi/4 ).Therefore, the function can be written as:( f(x) = sqrt{2} sin left( 2x - frac{pi}{4} right ) ).Therefore, the maximum value is ( sqrt{2} ), which occurs when ( sin left( 2x - frac{pi}{4} right ) = 1 ), i.e., when ( 2x - frac{pi}{4} = frac{pi}{2} + 2pi n ), where ( n ) is an integer. Solving for ( x ):( 2x = frac{3pi}{4} + 2pi n )( x = frac{3pi}{8} + pi n ).So, the maximum value is ( sqrt{2} ), achieved at ( x = frac{3pi}{8} + pi n ), where ( n ) is any integer.Alternatively, if the problem expects a specific value in a certain interval, but since it's not specified, the general solution is okay. Let me check if my steps are correct.Original function: simplified to ( sin 2x - cos 2x ), then expressed as ( sqrt{2} sin(2x - pi/4) ). That seems right. Let me verify by expanding:( sqrt{2} sin(2x - pi/4) = sqrt{2} left[ sin 2x cos pi/4 - cos 2x sin pi/4 right ] = sqrt{2} left[ sin 2x cdot frac{sqrt{2}}{2} - cos 2x cdot frac{sqrt{2}}{2} right ] = sqrt{2} cdot frac{sqrt{2}}{2} (sin 2x - cos 2x) = 1 cdot (sin 2x - cos 2x) ). Perfect, that's correct. So the expression is equivalent.Therefore, the maximum value is indeed ( sqrt{2} ), achieved when ( 2x - pi/4 = pi/2 + 2pi n ), which leads to ( x = 3pi/8 + pi n ). So part (1) seems done.Moving on to part (2): Given ( f(theta) = frac{3}{5} ), find the value of ( cos 2left( frac{pi}{4} - 2theta right ) ).First, let's note that ( f(theta) = sin 2theta - cos 2theta = frac{3}{5} ). We need to find ( cos 2left( frac{pi}{4} - 2theta right ) ).Let me simplify the expression inside the cosine first. The argument is ( 2left( frac{pi}{4} - 2theta right ) = frac{pi}{2} - 4theta ). Therefore, ( cos left( frac{pi}{2} - 4theta right ) ). Wait, but cosine of ( frac{pi}{2} - 4theta ) is equal to ( sin 4theta ), since ( cos(pi/2 - x) = sin x ). Therefore, ( cos 2left( frac{pi}{4} - 2theta right ) = sin 4theta ).Wait, let me verify that again. Let's compute the argument step by step:( 2 times left( frac{pi}{4} - 2theta right ) = frac{pi}{2} - 4theta ). So the expression is ( cos left( frac{pi}{2} - 4theta right ) ). Using the co-function identity, ( cosleft( frac{pi}{2} - x right ) = sin x ), so ( cosleft( frac{pi}{2} - 4theta right ) = sin 4theta ).Therefore, the expression we need to find is ( sin 4theta ).So, the problem reduces to: given ( sin 2theta - cos 2theta = frac{3}{5} ), find ( sin 4theta ).Alright, so we need to find ( sin 4theta ) given that ( sin 2theta - cos 2theta = frac{3}{5} ).Let me denote ( phi = 2theta ). Then the given equation becomes ( sin phi - cos phi = frac{3}{5} ), and we need to find ( sin 2phi ).So, if we can find ( sin 2phi ), that will be ( sin 4theta ). Let's proceed.Given ( sin phi - cos phi = frac{3}{5} ). Let's square both sides to use the Pythagorean identity.( (sin phi - cos phi)^2 = left( frac{3}{5} right )^2 )Expanding the left side:( sin^2 phi - 2sin phi cos phi + cos^2 phi = frac{9}{25} )Simplify using ( sin^2 phi + cos^2 phi = 1 ):( 1 - 2sin phi cos phi = frac{9}{25} )Therefore,( -2sin phi cos phi = frac{9}{25} - 1 = -frac{16}{25} )Multiply both sides by (-1):( 2sin phi cos phi = frac{16}{25} )But ( 2sin phi cos phi = sin 2phi ), so:( sin 2phi = frac{16}{25} )Therefore, ( sin 4theta = frac{16}{25} ).Wait, that seems straightforward. Let me check again:We set ( phi = 2theta ), so the given equation is ( sin phi - cos phi = 3/5 ).Square both sides: ( 1 - sin 2phi = 9/25 ).Wait, hold on, actually, ( (sin phi - cos phi)^2 = sin^2 phi - 2sin phi cos phi + cos^2 phi = 1 - sin 2phi ). Because ( 2sin phi cos phi = sin 2phi ). So ( 1 - sin 2phi = 9/25 ).Therefore, ( sin 2phi = 1 - 9/25 = 16/25 ). So yes, ( sin 2phi = 16/25 ). But ( phi = 2theta ), so ( 2phi = 4theta ), so ( sin 4theta = 16/25 ). Therefore, the value we need is ( 16/25 ).So, part (2) answer is ( 16/25 ).But let me confirm once more. Starting with ( sin 2theta - cos 2theta = 3/5 ). Squared, gives ( 1 - sin 4theta = 9/25 ), so ( sin 4theta = 1 - 9/25 = 16/25 ). Therefore, ( cos 2(pi/4 - 2theta) = sin 4theta = 16/25 ). That's correct.Alternatively, to cross-verify, perhaps compute ( theta ) and substitute? But that might be more involved. Alternatively, use another method.Alternatively, express ( sin 2theta - cos 2theta = sqrt{2} sin(2theta - pi/4) = 3/5 ). Then, ( sin(2theta - pi/4) = 3/(5sqrt{2}) ). Then, to find ( sin 4theta ), note that ( 4theta = 2 times 2theta ). Let ( alpha = 2theta - pi/4 ). Then, ( 2theta = alpha + pi/4 ), so ( 4theta = 2alpha + pi/2 ). Then, ( sin 4theta = sin(2alpha + pi/2) = sin 2alpha cos pi/2 + cos 2alpha sin pi/2 = cos 2alpha ).But ( alpha = arcsin(3/(5sqrt{2})) ). Therefore, ( cos 2alpha = 1 - 2sin^2 alpha = 1 - 2 times (9/(25 times 2)) = 1 - 9/25 = 16/25 ). So that's another way to get the same result. Therefore, confirms that ( sin 4theta = 16/25 ).Therefore, the answer to part (2) is ( boxed{dfrac{16}{25}} ).But let me check if there's any possible mistake here. For example, when we squared both sides, we could introduce extraneous solutions, but since we only need the value of ( sin 4theta ), which depends on the square, maybe the sign isn't an issue. However, in this case, since ( sin 4theta ) is directly obtained from squaring, and the result is positive (16/25). Let's see: the original equation ( sin 2theta - cos 2theta = 3/5 ). The left side is ( sqrt{2} sin(2theta - pi/4) ). The maximum of that is ( sqrt{2} approx 1.414 ), and 3/5 = 0.6, which is within the range, so it's valid. Therefore, the solution is valid, and ( sin 4theta = 16/25 ).Therefore, both parts seem solved correctly.Final Answer(1) The maximum value of ( f(x) ) is ( boxed{sqrt{2}} ) at ( x = boxed{dfrac{3pi}{8} + npi} ) for any integer ( n ).(2) The value of ( cos 2left( dfrac{pi}{4} - 2theta right) ) is ( boxed{dfrac{16}{25}} ).</think>Given the function ( f(x) = 2sin x cos x - 2cos^2 x + 1 ):(1) To find the maximum value of ( f(x) ) and the corresponding value of ( x ):First, simplify the function using trigonometric identities:[f(x) = 2sin x cos x - 2cos^2 x + 1 = sin 2x - (1 + cos 2x) + 1 = sin 2x - cos 2x]This can be rewritten using the amplitude form:[f(x) = sqrt{2} sin left( 2x - frac{pi}{4} right)]The maximum value of ( sqrt{2} sin left( 2x - frac{pi}{4} right) ) is ( sqrt{2} ), which occurs when:[2x - frac{pi}{4} = frac{pi}{2} + 2pi n implies 2x = frac{3pi}{4} + 2pi n implies x = frac{3pi}{8} + pi n quad text{for any integer } n](2) Given ( f(theta) = frac{3}{5} ), find the value of ( cos 2left( frac{pi}{4} - 2theta right) ):First, note that:[f(theta) = sin 2theta - cos 2theta = frac{3}{5}]Let ( phi = 2theta ). Then:[sin phi - cos phi = frac{3}{5}]Squaring both sides:[(sin phi - cos phi)^2 = left( frac{3}{5} right)^2 implies 1 - 2sin phi cos phi = frac{9}{25} implies 2sin phi cos phi = frac{16}{25} implies sin 2phi = frac{16}{25}]Since ( phi = 2theta ), we have ( 2phi = 4theta ). Therefore:[sin 4theta = frac{16}{25}]We need to find ( cos 2left( frac{pi}{4} - 2theta right) ):[2left( frac{pi}{4} - 2theta right) = frac{pi}{2} - 4theta implies cos left( frac{pi}{2} - 4theta right) = sin 4theta = frac{16}{25}]Final answers:(1) The maximum value of ( f(x) ) is ( boxed{sqrt{2}} ) at ( x = boxed{dfrac{3pi}{8} + npi} ) for any integer ( n ).(2) The value of ( cos 2left( dfrac{pi}{4} - 2theta right) ) is ( boxed{dfrac{16}{25}} ).]

🔑:Solution: (1) We have f(x)=sin 2x-(2cos ^{2}x-1)=sin 2x-cos 2x= sqrt {2}sin (2x- dfrac {pi}{4}), Therefore, when 2x- dfrac {pi}{4}=2kpi+ dfrac {pi}{2}, that is x=kpi+ dfrac {3}{8}pi(kinmathbb{Z}), f(x) reaches its maximum value boxed{sqrt {2}}; (2) Given f(theta)=sin 2theta-cos 2theta, and f(theta)= dfrac {3}{5}, we have: sin 2theta-cos 2theta= dfrac {3}{5}, Squaring both sides, we get 1-sin 4theta= dfrac {9}{25}, thus sin 4theta= dfrac {16}{25}; Therefore, cos 2( dfrac {pi}{4}-2theta)=cos ( dfrac {pi}{2}-4theta)=sin 4theta= boxed{dfrac {16}{25}}.

🧠:[<think>Okay, let me try to figure out this problem. So, the question is about a function f(x) whose domain is all real numbers. We are told that f(1 - 2x) is an even function, and f(x - 1) is an odd function. Then we need to determine which of the given statements (A, B, C, D) is true based on that information. The options are about specific values of f at certain points: f(0), f(-2), f(-3), and f(-5). First, let me recall what even and odd functions are. An even function satisfies the property g(-x) = g(x) for all x in its domain. An odd function satisfies g(-x) = -g(x) for all x in its domain. So, the problem states that f(1 - 2x) is even. That means if we substitute x with -x in f(1 - 2x), we should get the same function. Similarly, f(x - 1) is odd, so substituting x with -x in f(x - 1) should give us the negative of the original function. Let me write down these conditions mathematically. For the even function condition:f(1 - 2(-x)) = f(1 - 2x)Simplify the left side:f(1 + 2x) = f(1 - 2x)So, for all x, f(1 + 2x) must equal f(1 - 2x). Let me try to make sense of this. If we let t = 2x, then the equation becomes f(1 + t) = f(1 - t) for all t. That suggests that f is symmetric around t = 1, or in other words, f is symmetric about the line x = 1. Because replacing t with -t gives the same value. So, for any t, f(1 + t) = f(1 - t). Therefore, f(x) is symmetric about x = 1. So, f(1 + t) = f(1 - t). So, in terms of x, if I let x = 1 + t, then f(x) = f(2 - x). So, f(x) = f(2 - x) for all x. That seems like a reflection over x = 1. Okay, so that's the first condition: f is symmetric about x = 1, so f(2 - x) = f(x).Now, the second condition is that f(x - 1) is an odd function. Let's write that out. If we let h(x) = f(x - 1), then h(-x) = -h(x). So, substituting, h(-x) = f(-x - 1) and -h(x) = -f(x - 1). Therefore, we have:f(-x - 1) = -f(x - 1) for all x.Let me make a substitution here. Let me set y = x - 1. Then x = y + 1. Substitute into the equation:f(- (y + 1) - 1) = -f(y)Simplify the left side:f(-y - 1 - 1) = f(-y - 2) = -f(y)So, f(-y - 2) = -f(y). Let's replace y with x for simplicity:f(-x - 2) = -f(x) for all x.So, this is another functional equation. So, combining both conditions: f is symmetric about x = 1, which gives f(2 - x) = f(x), and f(-x - 2) = -f(x).Now, let's see if we can use these two equations to find relationships between specific values of f. First, let's try to find f(0). Let's see. Maybe using the symmetry about x=1.If we set x = 0 in f(2 - x) = f(x), we get f(2 - 0) = f(0), so f(2) = f(0). Hmm. So, f(2) equals f(0). Not sure if that helps immediately.Alternatively, using the other equation: f(-x - 2) = -f(x). Let's set x = 0 here. Then:f(-0 - 2) = -f(0) => f(-2) = -f(0). So, f(-2) is the negative of f(0). But from the first equation, f(2) = f(0). So, f(-2) = -f(2). That might be useful.Wait, so f(-2) = -f(2), but from the first condition, since f is symmetric about x=1, then f(2) = f(0). Therefore, f(-2) = -f(0). So, combining these, we have f(-2) = -f(0) and f(0) = f(2). So, f(-2) = -f(2). That's an interesting relationship. But does that help us? Let's hold onto that thought.Now, maybe we can find other relationships. Let's try to find f(1). Let x =1 in the symmetry equation: f(2 -1) = f(1), so f(1) = f(1). That's trivial. Not helpful.How about using the second equation f(-x - 2) = -f(x). Let's set x = 1. Then f(-1 - 2) = -f(1) => f(-3) = -f(1). So, f(-3) is the negative of f(1). Also, from the symmetry equation, if we set x = 1 + t, then f(1 + t) = f(1 - t). So, setting t = -2, f(1 - 2) = f(1 + 2), so f(-1) = f(3). Similarly, if we set t = -4, f(1 - 4) = f(1 + 4) => f(-3) = f(5). Wait, but from the second equation, we had f(-3) = -f(1). So, combining these, f(5) = -f(1). Similarly, f(-3) = -f(1) and f(-3) = f(5). So, f(5) = -f(1). Is there a way to relate f(1) to something else? Maybe using the symmetry equation. Let's set x =1 in the symmetry equation: f(2 -1) = f(1) => f(1) = f(1). Again, trivial.Alternatively, use the second equation again. Let me set x = -1 in f(-x - 2) = -f(x). Then:f(-(-1) -2) = -f(-1) => f(1 -2) = -f(-1) => f(-1) = -f(-1). That implies f(-1) = 0. Because if f(-1) = -f(-1), then adding f(-1) to both sides gives 2f(-1) =0, so f(-1)=0. Ah! That's a good point. Let's do that again. If we set x = -1 in the second equation:Left side: f(-(-1) -2) = f(1 - 2) = f(-1)Right side: -f(-1)Therefore, f(-1) = -f(-1), so 2f(-1) =0 => f(-1)=0. So, f(-1)=0.Now, since f(-1)=0, from the symmetry equation (since f(2 - x)=f(x)), set x = -1:f(2 - (-1)) = f(-1) => f(3) = f(-1) =0. Therefore, f(3)=0.So, f(3)=0. Wait, but the answer options are about f(0), f(-2), f(-3), f(-5). So, f(3)=0, which is not in the options. Maybe we can use this to find other zeros.From the second equation f(-x -2)= -f(x). Let's set x=3. Then:f(-3 -2) = -f(3) => f(-5) = -f(3). But f(3)=0, so f(-5)= -0=0. Therefore, f(-5)=0. That's option D. So, is that the answer? Wait, let me check.Yes, if we set x=3 in f(-x -2)= -f(x), we get f(-5)= -f(3). Since f(3)=0, then f(-5)=0. Therefore, D is true. But let me check if other options can be confirmed or denied.Option A: f(0)=0. Let's see. From the first condition, f(2 -0)=f(0) => f(2)=f(0). From the second condition, set x=2: f(-2 -2)= -f(2) => f(-4)= -f(2). Also, from the first condition, f(-4) = f(2 - (-4)) = f(6). So, f(6)= -f(2). But unless we know something else about f(2), we can't conclude f(2)=0. So, f(0)=f(2), but unless there's a way to show f(2)=0, we can't say f(0)=0. So, A is not necessarily true.Option B: f(-2)=0. From the second equation, set x=0: f(-0 -2)= -f(0) => f(-2)= -f(0). If we can show f(0)=0, then f(-2)=0. But we don't know if f(0)=0. From the first condition, f(0)=f(2). If we had f(0)=0, then f(2)=0, but unless we can show that, it's not necessarily true. So, unless there's a relation, f(-2)= -f(0), but f(0) could be anything. So, B is not necessarily true.Option C: f(-3)=0. From the second equation, set x=1: f(-1 -2)= -f(1) => f(-3)= -f(1). Earlier, we found f(-1)=0. From the first condition, f(-1)=f(3). So, f(3)=0. Then from the second equation, if we set x=3, we get f(-5)= -f(3)=0. But for x=1, f(-3)= -f(1). But is f(1)=0? Let's check. Wait, maybe we can find f(1). Let me try. From the first condition, f(2 - x)=f(x). Let x=1: f(2 -1)=f(1) => f(1)=f(1). Not helpful. From the second condition, set x=1: f(-3)= -f(1). But we know f(-3)= -f(1) and we also know from the first condition that f(-3)=f(5). So, f(5)= -f(1). But unless f(1)=0, we can't say f(-3)=0. But is there a way to show f(1)=0?Let's see. Let me use the second equation. Let x= -1. Wait, we did that already. When x=-1, we found f(-1)=0. Then, from the first condition, f(-1)=f(3)=0. So, f(3)=0. From the second equation, set x=3: f(-5)= -f(3)=0. So, f(-5)=0. But for f(1), can we find another equation?Alternatively, let's set x=2 in the second equation: f(-2 -2)= -f(2) => f(-4)= -f(2). From the first condition, f(-4)=f(6). So, f(6)= -f(2). Then, from the first condition again, f(6)=f(2 -6)=f(-4). Wait, that's the same as before. Not helpful. Maybe set x= -3 in the first equation. f(2 - (-3))=f(-3) => f(5)=f(-3). But from the second equation, f(-3)= -f(1). So, f(5)= -f(1). But f(5) is also equal to f(-3). So, this is just the same as before. Alternatively, set x=5 in the first equation: f(2 -5)=f(5) => f(-3)=f(5). But again, this is the same as before. So, unless we can find another relation involving f(1), we can't determine if f(1)=0. So, we can't conclude that f(-3)=0 unless f(1)=0, but we don't know that. So, C is not necessarily true. But we did find that f(-5)=0. So, D is true. Let me just check again. From x=3, f(-5)= -f(3)=0. And f(3)=0 because f(3)=f(-1)=0. So yes, that's solid. Therefore, D is correct. Therefore, the answer is D.Final Answerboxed{D}</think>Given that the domain of the function ( f(x) ) is ( mathbb{R} ), ( f(1 - 2x) ) is an even function, and ( f(x - 1) ) is an odd function, we need to determine which of the given statements is true.1. Even Function Condition: Since ( f(1 - 2x) ) is even, we have: [ f(1 - 2(-x)) = f(1 - 2x) implies f(1 + 2x) = f(1 - 2x) ] Letting ( t = 2x ), this implies ( f(1 + t) = f(1 - t) ) for all ( t ). Therefore, ( f(x) ) is symmetric about ( x = 1 ), giving: [ f(2 - x) = f(x) ]2. Odd Function Condition: Since ( f(x - 1) ) is odd, we have: [ f(-x - 1) = -f(x - 1) ] Letting ( y = x - 1 ), this implies: [ f(-y - 2) = -f(y) ] Replacing ( y ) with ( x ), we get: [ f(-x - 2) = -f(x) ]3. Key Deductions: - Setting ( x = -1 ) in the odd function condition: [ f(-(-1) - 2) = -f(-1) implies f(-1) = -f(-1) implies 2f(-1) = 0 implies f(-1) = 0 ] - Using symmetry about ( x = 1 ): [ f(2 - (-1)) = f(-1) implies f(3) = f(-1) = 0 implies f(3) = 0 ] - Setting ( x = 3 ) in the odd function condition: [ f(-3 - 2) = -f(3) implies f(-5) = -f(3) implies f(-5) = -0 = 0 ]Thus, the true statement is (boxed{D}).,<think>Okay, let's tackle this problem step by step. The question says that the function f(x) has a domain of all real numbers, R. Then it tells us two things: first, that f(1 - 2x) is an even function, and second, that f(x - 1) is an odd function. We need to determine which of the given statements (A to D) is true based on these conditions. Let's break down what each part means and how we can use them to find the correct answer.First, let's recall the definitions of even and odd functions. An even function satisfies the property g(-x) = g(x) for all x in its domain. An odd function satisfies the property h(-x) = -h(x) for all x in its domain. So, if we can translate the given conditions into equations involving f, we might be able to find some relationships or specific values of f.Starting with the first condition: f(1 - 2x) is an even function. Let's denote this function as g(x) = f(1 - 2x). Since g is even, we have g(-x) = g(x). Therefore:f(1 - 2(-x)) = f(1 - 2x)Simplifying the left side:f(1 + 2x) = f(1 - 2x)So, for all x, f(1 + 2x) must equal f(1 - 2x). Let's make a substitution here to make this equation more manageable. Let's set y = 2x. Then, when x varies over all real numbers, so does y. Substituting y into the equation:f(1 + y) = f(1 - y)This holds for all real numbers y. Therefore, the function f is symmetric around x = 1. In other words, for any point y away from 1, the value of f at 1 + y is the same as at 1 - y. So, this tells us that the graph of f(x) is symmetric with respect to the line x = 1. That's an important property.Moving on to the second condition: f(x - 1) is an odd function. Let's denote this function as h(x) = f(x - 1). Since h is odd, we have h(-x) = -h(x). Therefore:f(-x - 1) = -f(x - 1)Again, let's make a substitution here to make this more clear. Let's set z = x - 1. Then, x = z + 1. Substituting into the equation:f(-(z + 1) - 1) = -f((z + 1) - 1)Simplifying both sides:Left side: f(-z - 1 - 1) = f(-z - 2)Right side: -f(z + 1 - 1) = -f(z)Therefore, we have:f(-z - 2) = -f(z)Since this holds for all real numbers z, we can replace z with x (since z is just a dummy variable), giving:f(-x - 2) = -f(x) for all x.So, this is another functional equation that f must satisfy. Let's note this down as equation (1):f(-x - 2) = -f(x) for all x. --- (1)Now, let's recall the first condition that we translated earlier:f(1 + y) = f(1 - y) for all y. Let's write that in terms of x. Let me substitute y with x here:f(1 + x) = f(1 - x) for all x. --- (2)So now we have two equations (1) and (2) that the function f must satisfy. Let's see if we can combine these two to find specific values of f at certain points.First, let's explore equation (1): f(-x - 2) = -f(x). Let's see if we can find a relationship between f at some points. For example, if we set x = a in equation (1), then f(-a - 2) = -f(a). This relates the value of the function at -a - 2 to its value at a.Equation (2) says that f is symmetric around x = 1. So, f(1 + x) = f(1 - x). Let's see if we can link these two equations.Let's take equation (1) and see if we can express the argument -x - 2 in terms of the symmetry around 1. Let me think. Let's set 1 + y = -x - 2. Then, y = -x - 3. But from equation (2), f(1 + y) = f(1 - y). So substituting y = -x -3:f(1 + (-x -3)) = f(1 - (-x -3))f(1 - x -3) = f(1 + x +3)f(-x -2) = f(x +4)But from equation (1), we have f(-x -2) = -f(x). Therefore, combining these two results:f(x +4) = -f(x)So, we get another equation:f(x +4) = -f(x) --- (3)This is a functional equation that relates the value of f at x +4 to the value at x with a negative sign. Let's see if we can use this to find more relations.If we replace x with x +4 in equation (3), we get:f(x +4 +4) = -f(x +4)f(x +8) = -f(x +4)But from equation (3), f(x +4) = -f(x), so substituting into the above:f(x +8) = -(-f(x)) = f(x)Therefore, we have:f(x +8) = f(x) --- (4)This tells us that the function f is periodic with period 8. So, every 8 units, the function repeats its values. However, the problem didn't state that f is periodic, but given the conditions, we derived that it must be. So, this is a useful property.Now, let's see if we can find specific values. The question gives us options about f(0), f(-2), f(-3), and f(-5). Let's check each of these.Starting with option A: f(0) = 0.To check this, let's see if we can find f(0) using the equations we have.From equation (1): f(-x -2) = -f(x). Let's set x = -2 in this equation:f(-(-2) -2) = -f(-2)f(2 -2) = -f(-2)f(0) = -f(-2)So, f(0) = -f(-2). Therefore, if we can find f(-2), we can relate it to f(0). Alternatively, if we can find another equation involving f(0), we can solve for it.Alternatively, using equation (3): f(x +4) = -f(x). Let's set x = -4 in equation (3):f(-4 +4) = -f(-4)f(0) = -f(-4)But we don't know f(-4) yet.Alternatively, let's use equation (2): f(1 + x) = f(1 - x). Let's set x = -1 in this equation:f(1 + (-1)) = f(1 - (-1))f(0) = f(2)So, f(0) = f(2). Hmm, but how does this help?We have f(0) = -f(-2) from earlier, and f(0) = f(2). So, combining these:f(2) = -f(-2)Thus, f(2) + f(-2) = 0. But we need more information to find the actual value.Alternatively, let's try to use equation (3) again. Let's set x = 0 in equation (3):f(0 +4) = -f(0)f(4) = -f(0)Similarly, if we can relate f(4) to another value, maybe via equation (2) or (1).From equation (2), set x =3:f(1 +3) = f(1 -3)f(4) = f(-2)But from above, f(4) = -f(0) and f(0) = -f(-2), so:f(4) = -f(0) = -(-f(-2)) = f(-2)But we also have f(4) = f(-2) from equation (2). So, this is consistent, but doesn't give us new information.Hmm, perhaps we need to look for another approach.Let me recall that f is periodic with period 8 (from equation (4)), so f(x +8) = f(x). So, for example, f(-5) = f(-5 +8) = f(3). Similarly, f(-3) = f(5), and so on. Maybe this can help us relate some of these values.But let's see if we can find a fixed point where we can set up equations.Wait, perhaps we can use equation (1) and equation (3) together.From equation (1): f(-x -2) = -f(x). Let's set x =1 here:f(-1 -2) = -f(1)f(-3) = -f(1)From equation (3): f(x +4) = -f(x). Let's set x = -3:f(-3 +4) = -f(-3)f(1) = -f(-3)But from above, f(-3) = -f(1). So, substituting into this:f(1) = -(-f(1)) = f(1)Which is a tautology, so it just confirms consistency.Alternatively, let's try to express f(0) in terms of other values.From equation (3): f(0 +4) = -f(0) => f(4) = -f(0)From equation (2): f(1 + x) = f(1 - x). Let x =3: f(4) = f(-2)But from above, f(4) = -f(0), so -f(0) = f(-2). Therefore, f(-2) = -f(0). But earlier, from equation (1) with x=0:f(-0 -2) = -f(0) => f(-2) = -f(0). So, same result. So, again, no new info.But let's check equation (3) with x =2:f(2 +4) = -f(2)f(6) = -f(2)But from equation (2), f(1 +5) = f(1 -5) => f(6) = f(-4). Therefore, f(-4) = -f(2)But from equation (3), x=-4:f(-4 +4) = -f(-4)f(0) = -f(-4) => f(-4) = -f(0)So, f(-4) = -f(0) = -f(2) (since f(0) = f(2)). Therefore:-f(2) = -f(2), which is again a tautology.This seems to be going in circles. Maybe we need a different approach.Let's consider that the function is periodic with period 8, so perhaps there's a relation that after 4 shifts, you get the negative. From equation (3): f(x +4) = -f(x). Then, applying equation (3) again, f(x +8) = -f(x +4) = -(-f(x)) = f(x). So, that's how we got the periodicity.Given that, perhaps the function has some symmetry around certain points. Let's think about combining the two conditions.We know that f(1 +x) = f(1 -x) from equation (2), which is symmetry about x=1.We also know that f(x +4) = -f(x) from equation (3).Let's try to shift x by 4 in equation (2). Replace x with x +4 in equation (2):f(1 + (x +4)) = f(1 - (x +4))f(x +5) = f(-x -3)But from equation (3), f(x +5) = f((x +1) +4) = -f(x +1)Therefore, we have:-f(x +1) = f(-x -3)But from equation (1): f(-x -2) = -f(x). Let's see if we can relate these.Let me write equation (1) again: f(-x -2) = -f(x). Let's manipulate this equation.Set y = x +1. Then, x = y -1.Substituting into equation (1):f(-(y -1) -2) = -f(y -1)f(-y +1 -2) = -f(y -1)f(-y -1) = -f(y -1)Which gives:f(-y -1) = -f(y -1)But from above, we also had:-f(x +1) = f(-x -3)Wait, let's check if these can be related. Let me see:If we have f(-y -1) = -f(y -1), then replacing y with x:f(-x -1) = -f(x -1)But in the second condition given in the problem, we were told that f(x -1) is an odd function. Let's recall that h(x) = f(x -1) is odd, so h(-x) = -h(x), which translates to f(-x -1) = -f(x -1). Which is exactly the equation we just derived. So, that's consistent.Therefore, all these equations are interconnected and consistent, but perhaps not giving us direct values. Let's think differently.Let me consider specific values given in the options: f(0), f(-2), f(-3), f(-5). Let's see if we can use the equations to relate these to each other or to other known points.Starting with A: f(0) =0.From equation (1): f(-x -2) = -f(x). Let x=0:f(-0 -2) = -f(0)f(-2) = -f(0)So, f(-2) = -f(0). Therefore, if f(0)=0, then f(-2)=0. But the options are separate: A says f(0)=0, B says f(-2)=0. So, if A is true, then B must also be true, but since the options are separate, maybe both A and B are true? But the question says "which of the following statements is true", so there might be multiple, but looking at the answer choices, only one of them is correct. Wait, the problem is likely structured so that only one is correct. Let me check the original problem again.Wait, the problem says "which of the following statements is true?" and the options are A: f(0)=0, B: f(-2)=0, C: f(-3)=0, D: f(-5)=0. So, only one of them is correct. Therefore, even if A and B are linked, maybe only one is true. Let's check.Wait, if A: f(0)=0, then from f(-2) = -f(0), we get f(-2)=0. Therefore, A implies B, and vice versa: if B is true, then f(-2)=0 implies f(0)=0. So, A and B are equivalent? Therefore, if either is true, the other must be true as well. But since the answer options are separate, perhaps this suggests that neither is true, unless there's a miscalculation here.Wait, but maybe there's another way to find f(0). Let's see. From equation (3): f(x +4) = -f(x). Let's set x= -4: f(-4 +4) = -f(-4) => f(0) = -f(-4). But from equation (2): f(1 + (-5)) = f(1 - (-5)) => f(-4) = f(6). And from equation (3) with x=2: f(6) = -f(2). And from equation (2) with x=1: f(2) = f(0). Therefore, f(-4) = f(6) = -f(2) = -f(0). Therefore, f(0) = -f(-4) = -(-f(0)) = f(0). Which again is a tautology.So, this seems not helpful. Maybe we need to use the periodicity. From equation (4): f(x +8) = f(x). So, for example:f(-5) = f(-5 +8) = f(3). Then, using equation (2): f(1 +2) = f(1 -2) => f(3) = f(-1). Therefore, f(-5) = f(-1). Similarly, from equation (1): f(-x -2) = -f(x). Let x=1: f(-1 -2) = -f(1) => f(-3) = -f(1). But f(-3) is an option (C). If we can relate f(1) to something else.Alternatively, let's consider equation (3): f(x +4) = -f(x). Let's set x=1: f(5) = -f(1). Then, using equation (2): f(1 +4) = f(1 -4) => f(5) = f(-3). So, f(-3) = -f(1). But we already have f(-3) = -f(1) from equation (1). So, again, consistent.Wait, maybe if we can find f(1) in terms of itself?Alternatively, from equation (3): f(x +4) = -f(x), so f(x +8) = f(x). Then, using equation (1): f(-x -2) = -f(x). Let's set x= -5: f(-(-5) -2) = -f(-5) => f(5 -2) = -f(-5) => f(3) = -f(-5). But from periodicity, f(-5) = f(3). Therefore, f(3) = -f(3), which implies 2f(3) =0 => f(3)=0. Therefore, f(3)=0, which would imply f(-5)=0, since f(-5)=f(3). Therefore, f(-5)=0. So, that's option D. Therefore, D is true.Wait, let's verify this step by step because this seems promising.Set x = -5 in equation (1): f(-(-5) -2) = -f(-5)Simplify:f(5 -2) = -f(-5)f(3) = -f(-5)But from periodicity (equation (4)), f(-5) = f(-5 +8) = f(3). Therefore:f(3) = -f(3)Adding f(3) to both sides:2f(3) =0 => f(3)=0Therefore, f(3)=0. Hence, f(-5)=f(3)=0, so f(-5)=0. Therefore, option D is correct.Similarly, let's check if any other options can be confirmed or denied.We found that f(3)=0. Let's see if we can relate this to other points. From equation (3): f(3) = f(-1 +4) = -f(-1). Therefore, -f(-1)=0 => f(-1)=0.From equation (2): f(1 +x)=f(1 -x). Let x= -1: f(0)=f(2). So, f(0)=f(2). If we can relate f(0) to another value.From equation (1): f(-x -2) = -f(x). Let x=2: f(-2 -2)= -f(2) => f(-4) = -f(2). From equation (3): f(-4 +4)= -f(-4) => f(0) = -f(-4). But f(-4) = -f(2), so:f(0) = -(-f(2)) = f(2)Which is consistent with f(0)=f(2). But we don't know if f(0) is 0.Is there a way to show that f(0)=0?From equation (3): f(0 +4) = -f(0) => f(4) = -f(0)From equation (2): f(1 +3) = f(1 -3) => f(4) = f(-2)Therefore, f(-2) = -f(0). From equation (1): f(-0 -2)= -f(0) => f(-2) = -f(0), which is the same as above.But unless we have a specific value, we can't determine f(0). However, we found that f(-5)=0. Let's check the other options.Option C: f(-3)=0. From equation (1): f(-x -2) = -f(x). Let x=1: f(-1 -2)= -f(1) => f(-3)= -f(1). If f(1)=0, then f(-3)=0. But do we know if f(1)=0?From equation (3): f(1 +4)= -f(1) => f(5) = -f(1). From equation (2): f(1 +4)=f(1 -4) => f(5)=f(-3). Therefore, f(-3)= -f(1). But we already have that from equation (1). So, f(-3) = -f(1). So, unless f(1)=0, f(-3) isn't necessarily 0. However, do we have any information about f(1)?Wait, from equation (3): f(1 +4) = -f(1) => f(5) = -f(1). But we also found that f(5)=f(-3). And since f(-3) = -f(1), we have f(5)= -f(1). But f(-3) = -f(1) as well, so this is consistent.But if we use periodicity on f(1): f(1 +8)=f(1). Not helpful. Alternatively, from equation (2): f(1 +0)=f(1 -0) => f(1)=f(1), which is trivial.Is there a way to relate f(1) to another point where we know the value? For example, we know f(3)=0. From equation (3): f(3)= -f(-1). Therefore, since f(3)=0, then f(-1)=0. Then, using equation (1): f(-(-1) -2) = -f(-1) => f(1 -2) = -f(-1) => f(-1)= -f(-1). Therefore, f(-1)=0. Which we already knew.But how does that help with f(1)?From equation (1): f(-1 -2) = -f(1) => f(-3) = -f(1). But we found that f(-3)= -f(1) and f(-3)=0 (since f(3)=0 and f(-5)=f(3)=0, but wait, f(-3) was shown to be -f(1), but we haven't established f(-3)=0. Wait, actually, earlier we found f(3)=0, which is different from f(-3). Wait, this is where I might have made a mistake.Wait, let's clarify:From setting x=-5 in equation (1):f(-(-5) -2) = -f(-5) => f(5 -2)=f(3) = -f(-5)But from periodicity, f(-5)=f(3), so substituting:f(3) = -f(3) => 2f(3)=0 => f(3)=0.Thus, f(3)=0, and since f(-5)=f(3) (from periodicity), f(-5)=0. So, D is correct.But what about f(-3)? From equation (1): f(-x -2) = -f(x). Let x=1:f(-1 -2) = f(-3) = -f(1). So, f(-3) = -f(1). But unless we know f(1), we can't say f(-3) is zero. However, if we can find f(1), perhaps through another equation.From equation (3): f(1 +4) = -f(1) => f(5)= -f(1). From equation (2): f(1 +4)=f(1 -4) => f(5)=f(-3). Therefore, f(-3)= -f(1). But we also have f(-3)= -f(1) from equation (1). So, no new info. However, since we know f(3)=0, and f(3) = -f(-1) from equation (3): f(3) = f(-1 +4) = -f(-1) => 0 = -f(-1) => f(-1)=0. Therefore, f(-1)=0.Then, from equation (1): f(-(-1) -2) = -f(-1) => f(1 -2) = -0 => f(-1) =0, which matches.But how to relate this to f(1)?Wait, from equation (2): f(1 +x) = f(1 -x). Let x=2:f(1 +2) = f(1 -2) => f(3)=f(-1). But we know f(3)=0 and f(-1)=0, so this holds.Alternatively, x= -2 in equation (2):f(1 +(-2)) = f(1 -(-2)) => f(-1) = f(3). Which again gives f(-1)=f(3)=0.But still, not helping with f(1). So, unless there's another equation, f(1) remains unknown. Therefore, we cannot conclude that f(-3)=0 unless f(1)=0, but we don't have evidence for that.Therefore, among the options, D: f(-5)=0 is correct because we showed that f(-5)=0. The other options (A, B, C) depend on other values which we cannot confirm are zero with the given information. Even though f(-2) = -f(0), unless one of them is proven to be zero, we can't say for sure. However, wait, if we consider the periodicity:From equation (3): f(x +4) = -f(x). Let's set x= -2:f(-2 +4) = -f(-2) => f(2) = -f(-2). But from equation (2): f(0)=f(2). So, f(0) = -f(-2). Therefore, f(0) + f(-2) =0.But we don't know if they are individually zero. Similarly, if we set x=0 in equation (3): f(4)= -f(0). From equation (2): f(4)=f(-2). Therefore, f(-2)= -f(0). So, f(-2) and f(0) are negatives of each other, but without further information, we can't conclude they are zero. So, options A and B cannot be confirmed.Option C is f(-3)=0. As we saw, f(-3) = -f(1), and unless f(1)=0, which we don't know, f(-3) could be non-zero. However, if we can relate f(1) to something else:From equation (3): f(1 +4) = -f(1) => f(5) = -f(1). From periodicity, f(5) = f(-3) (since f(5 -8)=f(-3)). So, f(-3)= -f(1). But we already have f(-3) = -f(1) from equation (1). So, again, same equation. Therefore, we can't determine f(1) or f(-3).Therefore, the only value we can confirm is f(-5)=0, which is option D. Hence, the correct answer is D.Final Answerboxed{D}</think>Given that the domain of the function ( f(x) ) is ( mathbb{R} ), ( f(1 - 2x) ) is an even function, and ( f(x - 1) ) is an odd function, we need to determine which of the given statements is true.1. Even Function Condition: ( f(1 - 2x) ) is even, so ( f(1 + 2x) = f(1 - 2x) ). This implies ( f(1 + y) = f(1 - y) ) for all ( y ), meaning ( f ) is symmetric about ( x = 1 ).2. Odd Function Condition: ( f(x - 1) ) is odd, so ( f(-x - 1) = -f(x - 1) ). This implies ( f(-x - 2) = -f(x) ) for all ( x ).3. Periodicity: From the odd function condition and the even function condition, we derived that ( f(x + 4) = -f(x) ). Applying this twice, we get ( f(x + 8) = f(x) ), indicating ( f ) is periodic with period 8.4. Key Deductions: - Using the periodicity and the odd function condition, we set ( x = -5 ) in ( f(-x - 2) = -f(x) ): [ f(-(-5) - 2) = -f(-5) implies f(3) = -f(-5) ] - By periodicity, ( f(-5) = f(3) ). Substituting this back, we get: [ f(3) = -f(3) implies 2f(3) = 0 implies f(3) = 0 ] - Therefore, ( f(-5) = f(3) = 0 ).Thus, the correct answer is (boxed{D}).]

🔑:Given that the domain of the function f(x) is mathbb{R}, and we have two conditions regarding the function f(x):1. f(1-2x) is an even function. This implies that f(1-2(-x)) = f(1+2x) = f(1-2x). By substituting x with frac{x+1}{2}, we get f(x) = f(2-x). This is our first key equation.2. f(x-1) is an odd function. This implies that f(-(x-1)) = -f(x-1), which simplifies to f(-x+1) = -f(x-1). By substituting x with -x-2, we get f(-2-x) = -f(x). This is our second key equation.From the first condition, we have established that f(x) = f(2-x). From the second condition, we have f(-2-x) = -f(x). Combining these, we can infer that f(-1) = 0 because substituting x = -1 in the second equation gives us f(-1) = -f(-1), which implies f(-1) = 0.Further, using the equation f(x) = f(2-x) and setting x = -2-x, we get f(-2-x) = f(4+x). Since we also have f(-2-x) = -f(x), it follows that f(4+x) = -f(x).Now, let's evaluate the options given:- For f(0), we have f(0) = f(2) = -f(-2), which does not necessarily imply f(0) = 0 without knowing f(-2).- For f(-3), we have f(-3) = -f(1), which does not necessarily imply f(-3) = 0 without knowing f(1).- For f(-5), substituting x = -5 into f(4+x) = -f(x) gives f(-1) = -f(-5). Since we know f(-1) = 0, it follows that f(-5) = 0.Therefore, the correct answer is D: f(-5) = 0, encapsulated as boxed{D}.