answer：Okay, so I have this problem about an ellipse and a circle. Let me try to understand it 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). One of its foci coincides with the center of the given circle. The circle's equation is x² + y² - 2√2 x = 0. Hmm, I remember that to find the center and radius of a circle from its equation, I need to complete the square.Let me rewrite the circle's equation:x² - 2√2 x + y² = 0To complete the square for the x terms, I take the coefficient of x, which is -2√2, divide by 2 to get -√2, and square it to get (√2)² = 2. So I add and subtract 2:(x² - 2√2 x + 2) + y² = 2Which simplifies to:(x - √2)² + y² = 2So the center of the circle is at (√2, 0), and the radius is √2. That means one of the foci of the ellipse is at (√2, 0). Since the ellipse is centered at the origin, the other focus should be at (-√2, 0). Now, the standard equation of an ellipse centered at the origin is (x²/a²) + (y²/b²) = 1, where a is the semi-major axis, b is the semi-minor axis, and c is the distance from the center to each focus, with c² = a² - b².We know that c = √2 because the focus is at (√2, 0). So, c² = 2. Therefore, a² = b² + 2.The ellipse also passes through the point (√2, 1). Let's plug that into the ellipse equation:( (√2)² ) / a² + (1²) / b² = 1Simplify:2 / a² + 1 / b² = 1But we know that a² = b² + 2, so let's substitute that into the equation:2 / (b² + 2) + 1 / b² = 1Let me solve this equation for b². Let me denote b² as t for simplicity.2 / (t + 2) + 1 / t = 1Multiply both sides by t(t + 2) to eliminate denominators:2t + (t + 2) = t(t + 2)Simplify left side:2t + t + 2 = 3t + 2Right side:t² + 2tSo the equation becomes:3t + 2 = t² + 2tBring all terms to one side:t² + 2t - 3t - 2 = 0Simplify:t² - t - 2 = 0Factor:(t - 2)(t + 1) = 0So t = 2 or t = -1. Since t = b² can't be negative, t = 2. Therefore, b² = 2, and a² = b² + 2 = 4.So the standard equation of the ellipse is (x²/4) + (y²/2) = 1.Alright, that seems to make sense. Let me double-check by plugging the point (√2, 1) into the ellipse equation:( (√2)² ) / 4 + (1²) / 2 = 2/4 + 1/2 = 0.5 + 0.5 = 1. Perfect, it works.Now, moving on to part (2). A line passes through point P(0,1) and intersects the ellipse at points A and B. The origin is O. The condition given is that vector AP = 2 vector PB. I need to find the area of triangle AOB.First, let me visualize this. Point P is at (0,1). The line passes through P and intersects the ellipse at A and B. The vector condition means that P divides the segment AB in the ratio AP:PB = 2:1. So, P is closer to B than to A.Let me denote the coordinates of A as (x₁, y₁) and B as (x₂, y₂). The vector AP is from A to P, so it's (0 - x₁, 1 - y₁). The vector PB is from P to B, so it's (x₂ - 0, y₂ - 1). The condition is that AP = 2 PB, which gives:( -x₁, 1 - y₁ ) = 2( x₂, y₂ - 1 )This gives two equations:-x₁ = 2x₂ --> x₁ = -2x₂1 - y₁ = 2(y₂ - 1) --> 1 - y₁ = 2y₂ - 2 --> -y₁ = 2y₂ - 3 --> y₁ = -2y₂ + 3So, we have relationships between x₁ and x₂, and y₁ and y₂.Now, since points A and B lie on the ellipse, they satisfy the ellipse equation:x₁² / 4 + y₁² / 2 = 1x₂² / 4 + y₂² / 2 = 1Also, since A and B lie on the line passing through P(0,1), let me find the equation of the line. Let me assume the line has a slope m, so its equation is y = m x + 1.So, points A and B lie on this line, so:y₁ = m x₁ + 1y₂ = m x₂ + 1But from earlier, we have y₁ = -2y₂ + 3. Let me substitute y₁ and y₂ from the line equation into this:m x₁ + 1 = -2(m x₂ + 1) + 3Simplify:m x₁ + 1 = -2m x₂ - 2 + 3m x₁ + 1 = -2m x₂ + 1Subtract 1 from both sides:m x₁ = -2m x₂Divide both sides by m (assuming m ≠ 0, which I think is safe here because if m=0, the line is horizontal, but let me check later):x₁ = -2x₂Which is consistent with our earlier result x₁ = -2x₂.So, that's consistent. Now, let me use the ellipse equation for points A and B.From point A:x₁² / 4 + y₁² / 2 = 1But y₁ = m x₁ + 1, so:x₁² / 4 + (m x₁ + 1)² / 2 = 1Similarly for point B:x₂² / 4 + (m x₂ + 1)² / 2 = 1But since x₁ = -2x₂, let me substitute x₁ = -2x₂ into the equation for point A:(-2x₂)² / 4 + (m*(-2x₂) + 1)² / 2 = 1Simplify:(4x₂²)/4 + (-2m x₂ + 1)² / 2 = 1Which simplifies to:x₂² + [ ( -2m x₂ + 1 )² ] / 2 = 1Let me expand ( -2m x₂ + 1 )²:= (4m² x₂² - 4m x₂ + 1 )So, substituting back:x₂² + (4m² x₂² - 4m x₂ + 1 ) / 2 = 1Multiply through by 2 to eliminate the denominator:2x₂² + 4m² x₂² - 4m x₂ + 1 = 2Combine like terms:(2 + 4m²) x₂² - 4m x₂ + 1 - 2 = 0Simplify:(2 + 4m²) x₂² - 4m x₂ - 1 = 0Now, this is a quadratic equation in x₂. Let me write it as:(4m² + 2) x₂² - 4m x₂ - 1 = 0Let me denote this as:A x₂² + B x₂ + C = 0Where A = 4m² + 2, B = -4m, C = -1Now, the solutions to this quadratic will give me x₂. But since we have a line intersecting the ellipse, there should be two points, A and B, so two solutions. However, we already have a relationship between x₁ and x₂, so maybe we can find m.Alternatively, let's consider that the line intersects the ellipse at points A and B, so the quadratic equation should have two solutions. But since we have the vector condition, maybe we can find m such that the ratio is satisfied.Alternatively, maybe I can use the fact that x₁ = -2x₂ and substitute into the ellipse equation.Wait, let me think differently. Let me parametrize the line as y = m x + 1 and substitute into the ellipse equation:x² / 4 + (m x + 1)² / 2 = 1Multiply through by 4 to eliminate denominators:x² + 2(m x + 1)² = 4Expand:x² + 2(m² x² + 2m x + 1) = 4Simplify:x² + 2m² x² + 4m x + 2 = 4Combine like terms:(1 + 2m²) x² + 4m x + (2 - 4) = 0Which simplifies to:(1 + 2m²) x² + 4m x - 2 = 0So, the quadratic equation in x is:(1 + 2m²) x² + 4m x - 2 = 0Let me denote this as:A x² + B x + C = 0Where A = 1 + 2m², B = 4m, C = -2Now, the solutions to this equation are the x-coordinates of points A and B. Let me denote them as x₁ and x₂.From quadratic formula:x = [ -B ± sqrt(B² - 4AC) ] / (2A)So,x = [ -4m ± sqrt( (4m)^2 - 4*(1 + 2m²)*(-2) ) ] / (2*(1 + 2m²))Simplify discriminant:(16m²) - 4*(1 + 2m²)*(-2) = 16m² + 8*(1 + 2m²) = 16m² + 8 + 16m² = 32m² + 8So,x = [ -4m ± sqrt(32m² + 8) ] / (2*(1 + 2m²))Factor sqrt(8) from sqrt(32m² + 8):sqrt(8*(4m² + 1)) = 2√2 * sqrt(4m² + 1)So,x = [ -4m ± 2√2 sqrt(4m² + 1) ] / (2*(1 + 2m²)) = [ -2m ± √2 sqrt(4m² + 1) ] / (1 + 2m²)So, the two solutions are:x₁ = [ -2m + √2 sqrt(4m² + 1) ] / (1 + 2m²)x₂ = [ -2m - √2 sqrt(4m² + 1) ] / (1 + 2m²)But from earlier, we have x₁ = -2x₂. Let me set up the equation:x₁ = -2x₂Substitute the expressions:[ -2m + √2 sqrt(4m² + 1) ] / (1 + 2m²) = -2 * [ (-2m - √2 sqrt(4m² + 1)) / (1 + 2m²) ]Simplify the right side:-2 * [ (-2m - √2 sqrt(4m² + 1)) / (1 + 2m²) ] = [4m + 2√2 sqrt(4m² + 1)] / (1 + 2m²)So, the equation becomes:[ -2m + √2 sqrt(4m² + 1) ] = [4m + 2√2 sqrt(4m² + 1)]Multiply both sides by (1 + 2m²) to eliminate denominators, but since they are already multiplied, I can equate numerators:-2m + √2 sqrt(4m² + 1) = 4m + 2√2 sqrt(4m² + 1)Bring all terms to one side:-2m - 4m + √2 sqrt(4m² + 1) - 2√2 sqrt(4m² + 1) = 0Simplify:-6m - √2 sqrt(4m² + 1) = 0Factor out -√2:-√2 ( (6m)/√2 + sqrt(4m² + 1) ) = 0Wait, that might not be the best approach. Let me instead move terms:-6m = √2 sqrt(4m² + 1)Square both sides to eliminate the square root:( -6m )² = ( √2 sqrt(4m² + 1) )²36m² = 2*(4m² + 1)Simplify:36m² = 8m² + 236m² - 8m² = 228m² = 2m² = 2 / 28 = 1/14So, m = ±1/√14But let me check if squaring introduced extraneous solutions. Let's take m = 1/√14 and see if it satisfies the original equation:-6m = √2 sqrt(4m² + 1)Plug m = 1/√14:-6*(1/√14) = √2 sqrt(4*(1/14) + 1)Simplify left side:-6/√14Right side:√2 sqrt( (4/14) + 1 ) = √2 sqrt( (2/7) + 1 ) = √2 sqrt(9/7) = √2*(3/√7) = 3√2 / √7But left side is negative, right side is positive. So, m = 1/√14 doesn't satisfy the equation.Now, try m = -1/√14:-6*(-1/√14) = √2 sqrt(4*(1/14) + 1)Left side:6/√14Right side:Same as before: 3√2 / √7Simplify 6/√14:6/√14 = (6√14)/14 = (3√14)/73√2 / √7 = 3√2 / √7 = (3√14)/7So, both sides are equal. Therefore, m = -1/√14 is the valid solution.So, the slope m is -1/√14. Let me rationalize that: m = -√14 / 14.Now, with m known, I can find the coordinates of points A and B.From earlier, the x-coordinates are:x₁ = [ -2m + √2 sqrt(4m² + 1) ] / (1 + 2m²)x₂ = [ -2m - √2 sqrt(4m² + 1) ] / (1 + 2m²)Let me compute 4m² + 1:m² = 1/14, so 4m² = 4/14 = 2/7Thus, 4m² + 1 = 2/7 + 1 = 9/7So, sqrt(4m² + 1) = 3/√7Now, compute x₁:x₁ = [ -2*(-1/√14) + √2*(3/√7) ] / (1 + 2*(1/14))Simplify numerator:2/√14 + 3√2 / √7Note that √14 = √2 * √7, so 2/√14 = 2/(√2 √7) = √2 / √7Similarly, 3√2 / √7 is as is.So, numerator:√2 / √7 + 3√2 / √7 = 4√2 / √7Denominator:1 + 2/14 = 1 + 1/7 = 8/7Thus, x₁ = (4√2 / √7) / (8/7) = (4√2 / √7) * (7/8) = (4√2 * 7) / (8√7) = (28√2) / (8√7) = (7√2)/(2√7)Rationalize denominator:(7√2)/(2√7) * (√7/√7) = 7√14 / 14 = √14 / 2Similarly, x₂:x₂ = [ -2*(-1/√14) - √2*(3/√7) ] / (8/7)Numerator:2/√14 - 3√2 / √7 = √2 / √7 - 3√2 / √7 = (-2√2)/√7Denominator: 8/7So, x₂ = (-2√2 / √7) / (8/7) = (-2√2 / √7) * (7/8) = (-14√2)/(8√7) = (-7√2)/(4√7)Rationalize:(-7√2)/(4√7) * (√7/√7) = (-7√14)/28 = (-√14)/4So, x₁ = √14 / 2, x₂ = -√14 / 4Now, find y₁ and y₂ using y = m x + 1, where m = -1/√14.For x₁ = √14 / 2:y₁ = (-1/√14)*(√14 / 2) + 1 = (-1/2) + 1 = 1/2For x₂ = -√14 / 4:y₂ = (-1/√14)*(-√14 / 4) + 1 = (1/4) + 1 = 5/4So, points A and B are:A: (√14 / 2, 1/2)B: (-√14 / 4, 5/4)Now, I need to find the area of triangle AOB, where O is the origin (0,0).The area of a triangle given three points can be found using the determinant formula:Area = (1/2) | x₁ y₂ - x₂ y₁ |Since O is the origin, the vectors OA and OB are just the coordinates of A and B.So, vectors OA = (√14 / 2, 1/2) and OB = (-√14 / 4, 5/4)The area is (1/2) | (√14 / 2)(5/4) - (-√14 / 4)(1/2) | = (1/2) | (5√14 / 8) + (√14 / 8) | = (1/2) | 6√14 / 8 | = (1/2)(6√14 / 8) = (3√14)/8So, the area of triangle AOB is 3√14 / 8.Let me double-check the calculations to make sure I didn't make any errors.First, the slope m was found to be -1/√14, which seems correct.Then, computing x₁ and x₂, I used the quadratic solutions and substituted m correctly. The simplifications seem okay.For y₁ and y₂, using y = m x + 1, I plugged in the x-values and got 1/2 and 5/4, which seems correct.Then, using the determinant formula for the area, I think that's the right approach since O is the origin. The formula is (1/2)|x₁ y₂ - x₂ y₁|.Plugging in the values:x₁ y₂ = (√14 / 2)(5/4) = 5√14 / 8x₂ y₁ = (-√14 / 4)(1/2) = -√14 / 8So, subtracting: 5√14 / 8 - (-√14 / 8) = 5√14 / 8 + √14 / 8 = 6√14 / 8Multiply by 1/2: (1/2)(6√14 / 8) = 3√14 / 8Yes, that seems correct.So, the area is 3√14 / 8.

answer：First, I need to establish the relationship between the length and the width of the rectangular field. The problem states that the length is double the width, so I can express the length as ( l = 2w ).Next, I'll calculate the area of the rectangular field using the formula for the area of a rectangle, which is length multiplied by width:[text{Area of field} = l times w]The problem also mentions a square-shaped pond with a side length of 8 meters. The area of the pond can be calculated as:[text{Area of pond} = 8 times 8 = 64 text{ m}^2]According to the problem, the area of the pond is ( frac{1}{8} ) of the area of the field. This gives me the equation:[64 = frac{1}{8} times (l times w)]Substituting ( l = 2w ) into the equation:[64 = frac{1}{8} times (2w times w)]Simplifying the equation:[64 = frac{2w^2}{8}][64 = frac{w^2}{4}]To solve for ( w^2 ), I'll multiply both sides by 4:[256 = w^2]Taking the square root of both sides to find ( w ):[w = sqrt{256} = 16 text{ m}]Finally, using the relationship ( l = 2w ) to find the length:[l = 2 times 16 = 32 text{ m}]

answer：First, I need to determine the circumference of the circular field to find out the total length of the fence required.The area of a circle is given by the formula:[ A = pi r^2 ]where ( A ) is the area and ( r ) is the radius of the circle.Given that the area is 13.86 hectares, I should convert this to square meters since the cost of fencing is provided per meter. There are 10,000 square meters in one hectare, so:[ 13.86 text{ hectares} times 10,000 text{ square meters/hectare} = 138,600 text{ square meters} ]Next, I'll solve for the radius ( r ) using the area formula:[ 138,600 = pi r^2 ][ r^2 = frac{138,600}{pi} ][ r^2 approx frac{138,600}{3.14159} ][ r^2 approx 44,127.294 ][ r approx sqrt{44,127.294} ][ r approx 210.065 text{ meters} ]Now that I have the radius, I can calculate the circumference ( C ) of the circle:[ C = 2pi r ][ C = 2 times 3.14159 times 210.065 ][ C approx 1319.468 text{ meters} ]With the total cost of fencing being Rs. 6070.778380479544 and the length of the fence being approximately 1319.468 meters, I can find the rate per meter by dividing the total cost by the length:[ text{Rate per meter} = frac{text{Total cost}}{text{Length of fence}} ][ text{Rate per meter} = frac{6070.778380479544}{1319.468} ][ text{Rate per meter} approx 4.60 ]Therefore, the rate of fencing per meter is approximately Rs. 4.60.

answer：First, I need to determine the total number of televisions produced by the factory in the first year. The factory produces 10 televisions each day. Assuming there are 365 days in a year, the total production for the first year is calculated by multiplying the daily production rate by the number of days in a year.Next, the factory reduces its total production by 10% in the second year. To find the reduced production, I will calculate 10% of the first year's total production and then subtract this amount from the first year's total to obtain the second year's production.Finally, I will present the calculated total production for the second year as the final answer.

answer：Alright, so I have this problem with two vectors, a and b. Both are unit vectors because their components are cosine and sine of some angles, right? So, vector a is (cosα, sinα) and vector b is (cosβ, sinβ). The problem gives me a relation involving these vectors: the magnitude of k times vector a plus vector b equals the square root of 3 times the magnitude of vector a minus k times vector b. I need to find the dot product of a and b in terms of k, and then determine if a can be perpendicular or parallel to b.Starting with part (1). I remember that the magnitude squared of a vector is equal to the dot product of the vector with itself. So, if I square both sides of the given equation, I can relate the dot products. Let me write that down:|k a + b|² = 3 |a - k b|²Expanding both sides using the dot product formula. For any vectors u and v, |u + v|² = |u|² + 2 u·v + |v|². So, applying that:Left side: |k a + b|² = (k a + b)·(k a + b) = k² |a|² + 2 k a·b + |b|²Right side: 3 |a - k b|² = 3 (|a|² - 2 k a·b + k² |b|²)Since a and b are unit vectors, |a|² = |b|² = 1. So substituting that in:Left side: k² * 1 + 2 k a·b + 1 = k² + 2 k a·b + 1Right side: 3 (1 - 2 k a·b + k² * 1) = 3 (1 - 2 k a·b + k²) = 3 - 6 k a·b + 3 k²Now, setting left side equal to right side:k² + 2 k a·b + 1 = 3 - 6 k a·b + 3 k²Let me bring all terms to one side to solve for a·b:k² + 2 k a·b + 1 - 3 + 6 k a·b - 3 k² = 0Simplify the terms:(k² - 3 k²) + (2 k a·b + 6 k a·b) + (1 - 3) = 0Which is:-2 k² + 8 k a·b - 2 = 0Let me rewrite this:8 k a·b = 2 k² + 2Divide both sides by 8 k:a·b = (2 k² + 2) / (8 k) = (k² + 1) / (4 k)So, the dot product of a and b is (k² + 1)/(4k). That's part (1) done.Moving on to part (2). The first question is whether a can be perpendicular to b. If a and b are perpendicular, their dot product is zero. So, can (k² + 1)/(4k) be zero?Looking at the expression (k² + 1)/(4k), the numerator is k² + 1, which is always positive since k is a positive number. The denominator is 4k, which is also positive. So, the entire expression is positive. Therefore, the dot product can never be zero. So, a cannot be perpendicular to b.Next, can a be parallel to b? If a and b are parallel, then one is a scalar multiple of the other. Since both are unit vectors, this scalar must be either 1 or -1. So, either a = b or a = -b.If a = b, then the dot product a·b = 1. If a = -b, then a·b = -1. But from part (1), we have a·b = (k² + 1)/(4k). So, can this equal 1 or -1?First, let's check if it can be 1:(k² + 1)/(4k) = 1Multiply both sides by 4k:k² + 1 = 4kBring all terms to one side:k² - 4k + 1 = 0This is a quadratic equation. Let's solve for k:k = [4 ± sqrt(16 - 4)] / 2 = [4 ± sqrt(12)] / 2 = [4 ± 2*sqrt(3)] / 2 = 2 ± sqrt(3)Since k is positive, both solutions are valid because 2 + sqrt(3) is about 3.732 and 2 - sqrt(3) is about 0.267, both positive.Now, checking if a·b can be -1:(k² + 1)/(4k) = -1Multiply both sides by 4k:k² + 1 = -4kBring all terms to one side:k² + 4k + 1 = 0Solving this quadratic:k = [-4 ± sqrt(16 - 4)] / 2 = [-4 ± sqrt(12)] / 2 = [-4 ± 2*sqrt(3)] / 2 = -2 ± sqrt(3)Since k is positive, -2 - sqrt(3) is negative, which we discard. The other solution is -2 + sqrt(3), which is approximately -2 + 1.732 = -0.267, still negative. So, no positive solution here. Therefore, a·b cannot be -1.Thus, a can be parallel to b only when a·b = 1, which happens when k = 2 + sqrt(3) or k = 2 - sqrt(3).Wait, hold on. If a and b are parallel, they could be in the same direction or opposite directions. But since a·b is positive, it can only be 1, not -1. So, only when k = 2 + sqrt(3) or k = 2 - sqrt(3), a and b are parallel in the same direction.So, summarizing:1. The dot product is (k² + 1)/(4k).2. a cannot be perpendicular to b because the dot product is always positive. a can be parallel to b when k is 2 + sqrt(3) or 2 - sqrt(3).I think that's it. Let me double-check my steps.For part (1), expanding both sides and simplifying led me to the expression for the dot product. That seems correct.For part (2), checking for perpendicularity, since the dot product is positive, they can't be perpendicular. For parallelism, setting the dot product to 1 gives the quadratic equation, which has two positive solutions. So, yes, a can be parallel to b for those k values.Yeah, I think that's solid.

answer：Okay, so I have this function f(x) = |x + a| + |x - 2|, and there are two parts to the problem. Let me tackle them one by one.Starting with part (1): When a = -3, find the solution set of the inequality f(x) ≥ 3.Alright, so substituting a = -3 into the function, we get f(x) = |x - 3| + |x - 2|. I need to solve |x - 3| + |x - 2| ≥ 3.Hmm, absolute value functions can be tricky because they change their behavior based on the value of x. I remember that to solve inequalities involving absolute values, it's helpful to consider different cases based on the critical points where the expressions inside the absolute values change sign. In this case, the expressions inside the absolute values are x - 3 and x - 2, so the critical points are at x = 3 and x = 2.So, I should break down the real number line into intervals based on these critical points: (-∞, 2), [2, 3), and [3, ∞). Then, I'll analyze each interval separately.First interval: x < 2.In this interval, both x - 3 and x - 2 are negative. So, |x - 3| = -(x - 3) = 3 - x, and |x - 2| = -(x - 2) = 2 - x. Therefore, f(x) = (3 - x) + (2 - x) = 5 - 2x.So, the inequality becomes 5 - 2x ≥ 3. Let's solve this:5 - 2x ≥ 3 Subtract 5 from both sides: -2x ≥ -2 Divide both sides by -2, remembering to reverse the inequality sign: x ≤ 1.But wait, in this interval, x < 2. So, the solution here is x ≤ 1.Second interval: 2 ≤ x < 3.In this interval, x - 2 is non-negative, so |x - 2| = x - 2. However, x - 3 is still negative, so |x - 3| = 3 - x. Therefore, f(x) = (3 - x) + (x - 2) = 1.So, the inequality becomes 1 ≥ 3. Hmm, that's not true. So, there are no solutions in this interval.Third interval: x ≥ 3.Here, both x - 3 and x - 2 are non-negative. So, |x - 3| = x - 3 and |x - 2| = x - 2. Therefore, f(x) = (x - 3) + (x - 2) = 2x - 5.The inequality becomes 2x - 5 ≥ 3. Let's solve this:2x - 5 ≥ 3 Add 5 to both sides: 2x ≥ 8 Divide by 2: x ≥ 4.Since we're in the interval x ≥ 3, the solution here is x ≥ 4.Putting it all together, the solution set is x ≤ 1 or x ≥ 4. So, in interval notation, that's (-∞, 1] ∪ [4, ∞).Alright, that seems solid. Let me just double-check by testing a value in each interval.For x < 2, let's pick x = 0: f(0) = |0 - 3| + |0 - 2| = 3 + 2 = 5, which is indeed ≥ 3.For 2 ≤ x < 3, let's pick x = 2.5: f(2.5) = |2.5 - 3| + |2.5 - 2| = 0.5 + 0.5 = 1, which is < 3. So, no solutions here, as expected.For x ≥ 3, let's pick x = 5: f(5) = |5 - 3| + |5 - 2| = 2 + 3 = 5, which is ≥ 3. Good.Okay, part (1) seems done. Now, moving on to part (2): If the solution set of f(x) ≤ |x - 4| includes [1, 2], find the range of values for a.Hmm, so we need to find all a such that for every x in [1, 2], f(x) ≤ |x - 4|.So, f(x) = |x + a| + |x - 2|, and we need |x + a| + |x - 2| ≤ |x - 4| for all x in [1, 2].Let me write that inequality down:|x + a| + |x - 2| ≤ |x - 4| for all x ∈ [1, 2].I need to find the range of a such that this inequality holds for all x in [1, 2].First, let's analyze the right-hand side, |x - 4|. For x in [1, 2], x - 4 is negative, so |x - 4| = 4 - x.So, the inequality becomes:|x + a| + |x - 2| ≤ 4 - x.Let me rewrite this:|x + a| + |x - 2| + x ≤ 4.Hmm, maybe that's not the most helpful. Alternatively, let's consider the left-hand side:|x + a| + |x - 2|.Since x is in [1, 2], x - 2 is negative or zero, so |x - 2| = 2 - x.So, the left-hand side becomes |x + a| + (2 - x).Therefore, the inequality is:|x + a| + (2 - x) ≤ 4 - x.Simplify both sides:Left side: |x + a| + 2 - x Right side: 4 - xSubtract (2 - x) from both sides:|x + a| ≤ 4 - x - (2 - x) |x + a| ≤ 4 - x - 2 + x |x + a| ≤ 2.So, the inequality reduces to |x + a| ≤ 2 for all x in [1, 2].Therefore, for all x in [1, 2], |x + a| ≤ 2.Which means, for all x in [1, 2], -2 ≤ x + a ≤ 2.Let me write that as two inequalities:1) x + a ≤ 2 2) x + a ≥ -2So, for all x in [1, 2], these must hold.Let me rearrange both inequalities to solve for a.From the first inequality: a ≤ 2 - x From the second inequality: a ≥ -2 - xSo, for all x in [1, 2], a must be ≤ 2 - x and a must be ≥ -2 - x.But since this must hold for all x in [1, 2], we need to find the range of a that satisfies both inequalities for the entire interval.Let me analyze each inequality separately.First inequality: a ≤ 2 - x for all x in [1, 2].The right-hand side, 2 - x, is a decreasing function in x. So, its minimum occurs at x = 2, which is 0, and its maximum occurs at x = 1, which is 1.Therefore, to satisfy a ≤ 2 - x for all x in [1, 2], a must be ≤ the minimum of 2 - x on [1, 2], which is 0.So, a ≤ 0.Second inequality: a ≥ -2 - x for all x in [1, 2].The right-hand side, -2 - x, is a decreasing function in x. Its maximum occurs at x = 1, which is -3, and its minimum occurs at x = 2, which is -4.Therefore, to satisfy a ≥ -2 - x for all x in [1, 2], a must be ≥ the maximum of -2 - x on [1, 2], which is -3.So, a ≥ -3.Putting both inequalities together, we have -3 ≤ a ≤ 0.Therefore, the range of values for a is [-3, 0].Let me just verify this result.Suppose a = -3. Then, for x in [1, 2], |x - 3| + |x - 2| ≤ |x - 4|.Wait, actually, when a = -3, f(x) = |x - 3| + |x - 2|.But in part (2), the inequality is f(x) ≤ |x - 4|.Wait, let me check for a = -3 and x = 1:f(1) = |1 - 3| + |1 - 2| = 2 + 1 = 3 |x - 4| = |1 - 4| = 3 So, 3 ≤ 3, which holds.For x = 2:f(2) = |2 - 3| + |2 - 2| = 1 + 0 = 1 |x - 4| = |2 - 4| = 2 So, 1 ≤ 2, which holds.What about a = 0? Then, f(x) = |x + 0| + |x - 2| = |x| + |x - 2|.For x in [1, 2], |x| = x, and |x - 2| = 2 - x.So, f(x) = x + (2 - x) = 2.And |x - 4| = 4 - x.So, 2 ≤ 4 - x? Let's see for x in [1, 2]:At x = 1: 2 ≤ 3, which is true. At x = 2: 2 ≤ 2, which is true.So, yes, a = 0 works.What about a value inside the interval, say a = -1.Then, f(x) = |x - 1| + |x - 2|.For x in [1, 2], |x - 1| = x - 1, and |x - 2| = 2 - x.So, f(x) = (x - 1) + (2 - x) = 1.And |x - 4| = 4 - x.So, 1 ≤ 4 - x for x in [1, 2]. At x = 1: 1 ≤ 3, true. At x = 2: 1 ≤ 2, true. So, that works.What if a is outside the range, say a = -4.Then, f(x) = |x - 4| + |x - 2|.Wait, but the inequality is f(x) ≤ |x - 4|, so |x - 4| + |x - 2| ≤ |x - 4|.Subtract |x - 4| from both sides: |x - 2| ≤ 0.But |x - 2| is zero only when x = 2. So, for x in [1, 2], except x = 2, this inequality doesn't hold. Therefore, a = -4 doesn't work, which is consistent with our result.Similarly, if a = 1, then f(x) = |x + 1| + |x - 2|.For x in [1, 2], |x + 1| = x + 1, and |x - 2| = 2 - x.So, f(x) = (x + 1) + (2 - x) = 3.And |x - 4| = 4 - x.So, 3 ≤ 4 - x for x in [1, 2].At x = 1: 3 ≤ 3, which holds. At x = 2: 3 ≤ 2, which does not hold. So, a = 1 doesn't work, which is also consistent with our result.Therefore, the range of a is indeed [-3, 0].**Final Answer**1) The solution set is boxed{{x | x leq 1 text{ or } x geq 4}}; 2) The range of values for a is boxed{[-3, 0]}.