💡Okay, so I have this problem about an ellipse and a parabola sharing the same foci. The ellipse is given by the equation ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) where ( a > b > 0 ). The parabola is ( y^2 = 4x ). They share the same foci, which are ( F_1 ) and ( F_2 ). Point ( P ) is where the parabola and ellipse intersect in the first quadrant, and the line ( PF_1 ) is tangent to the parabola. I need to find the length of the major axis of the ellipse.First, I remember that for an ellipse, the foci are located at ( (pm c, 0) ) where ( c = sqrt{a^2 - b^2} ). For a parabola ( y^2 = 4px ), the focus is at ( (p, 0) ). In this case, the parabola is ( y^2 = 4x ), so ( p = 1 ), meaning the focus is at ( (1, 0) ). Since the ellipse shares the same foci, that means the foci of the ellipse are at ( (-1, 0) ) and ( (1, 0) ). So, ( c = 1 ).Therefore, for the ellipse, ( c = sqrt{a^2 - b^2} = 1 ), which implies ( a^2 - b^2 = 1 ). So, that's one equation relating ( a ) and ( b ).Now, point ( P ) is the intersection of the ellipse and the parabola in the first quadrant. So, ( P ) must satisfy both equations. Let me denote ( P ) as ( (x, y) ). Then, ( y^2 = 4x ) and ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ).Also, the line ( PF_1 ) is tangent to the parabola at point ( P ). So, the line connecting ( P ) and ( F_1 ) is tangent to the parabola. ( F_1 ) is at ( (-1, 0) ), so the line ( PF_1 ) goes from ( (-1, 0) ) to ( (x, y) ).I need to find the equation of this line and use the condition that it's tangent to the parabola. For a line to be tangent to a parabola, it must satisfy the condition that the system of equations (the line and the parabola) has exactly one solution.Let me find the equation of the line ( PF_1 ). The slope ( m ) of the line connecting ( (-1, 0) ) and ( (x, y) ) is ( m = frac{y - 0}{x - (-1)} = frac{y}{x + 1} ). So, the equation of the line is ( y = frac{y}{x + 1}(x + 1) ). Wait, that's just ( y = frac{y}{x + 1}(x + 1) ), which simplifies to ( y = y ). Hmm, that doesn't help. Maybe I should write it in terms of a variable.Let me denote the slope as ( m ), so ( m = frac{y}{x + 1} ). Then, the equation of the line is ( y = m(x + 1) ).Since this line is tangent to the parabola ( y^2 = 4x ), substituting ( y = m(x + 1) ) into the parabola equation gives:( [m(x + 1)]^2 = 4x )Expanding this:( m^2(x^2 + 2x + 1) = 4x )( m^2x^2 + 2m^2x + m^2 - 4x = 0 )This is a quadratic equation in ( x ). For the line to be tangent to the parabola, this quadratic must have exactly one solution, so the discriminant must be zero.The discriminant ( D ) of ( ax^2 + bx + c = 0 ) is ( D = b^2 - 4ac ). Here, ( a = m^2 ), ( b = 2m^2 - 4 ), and ( c = m^2 ).So,( D = (2m^2 - 4)^2 - 4(m^2)(m^2) = 0 )Let me compute this:First, expand ( (2m^2 - 4)^2 ):( (2m^2 - 4)^2 = 4m^4 - 16m^2 + 16 )Then, compute ( 4(m^2)(m^2) = 4m^4 )So,( D = (4m^4 - 16m^2 + 16) - 4m^4 = -16m^2 + 16 = 0 )So,( -16m^2 + 16 = 0 )Divide both sides by 16:( -m^2 + 1 = 0 )( m^2 = 1 )Thus, ( m = pm 1 ). Since point ( P ) is in the first quadrant, the slope should be positive, so ( m = 1 ).Therefore, the equation of the tangent line is ( y = 1 cdot (x + 1) ), so ( y = x + 1 ).Now, since point ( P ) lies on both the parabola and the ellipse, let's find its coordinates.From the parabola equation ( y^2 = 4x ), and the line ( y = x + 1 ), substitute ( y ) into the parabola equation:( (x + 1)^2 = 4x )Expand:( x^2 + 2x + 1 = 4x )Bring all terms to one side:( x^2 + 2x + 1 - 4x = 0 )Simplify:( x^2 - 2x + 1 = 0 )Factor:( (x - 1)^2 = 0 )So, ( x = 1 ). Then, ( y = 1 + 1 = 2 ). So, point ( P ) is ( (1, 2) ).Now, since ( P ) is also on the ellipse, substitute ( x = 1 ) and ( y = 2 ) into the ellipse equation:( frac{1^2}{a^2} + frac{2^2}{b^2} = 1 )Simplify:( frac{1}{a^2} + frac{4}{b^2} = 1 )We already know that ( a^2 - b^2 = 1 ) from earlier. So, we have two equations:1. ( frac{1}{a^2} + frac{4}{b^2} = 1 )2. ( a^2 - b^2 = 1 )Let me denote ( a^2 = A ) and ( b^2 = B ) for simplicity. Then, the equations become:1. ( frac{1}{A} + frac{4}{B} = 1 )2. ( A - B = 1 )From equation 2, ( A = B + 1 ). Substitute this into equation 1:( frac{1}{B + 1} + frac{4}{B} = 1 )Let me find a common denominator for the left side. The common denominator is ( B(B + 1) ):( frac{B}{B(B + 1)} + frac{4(B + 1)}{B(B + 1)} = 1 )Combine the numerators:( frac{B + 4(B + 1)}{B(B + 1)} = 1 )Simplify the numerator:( B + 4B + 4 = 5B + 4 )So,( frac{5B + 4}{B(B + 1)} = 1 )Multiply both sides by ( B(B + 1) ):( 5B + 4 = B(B + 1) )Expand the right side:( 5B + 4 = B^2 + B )Bring all terms to one side:( B^2 + B - 5B - 4 = 0 )Simplify:( B^2 - 4B - 4 = 0 )Now, solve this quadratic equation for ( B ):Using the quadratic formula, ( B = frac{4 pm sqrt{16 + 16}}{2} = frac{4 pm sqrt{32}}{2} = frac{4 pm 4sqrt{2}}{2} = 2 pm 2sqrt{2} )Since ( B = b^2 ) must be positive, both solutions are positive, but we need to check which one makes sense in the context.From equation 2, ( A = B + 1 ). So, if ( B = 2 + 2sqrt{2} ), then ( A = 3 + 2sqrt{2} ). If ( B = 2 - 2sqrt{2} ), then ( A = 3 - 2sqrt{2} ). But ( A = a^2 ) must be greater than ( B = b^2 ), since ( a > b ). Let's compute both:1. If ( B = 2 + 2sqrt{2} ), then ( A = 3 + 2sqrt{2} ). Since ( 3 + 2sqrt{2} > 2 + 2sqrt{2} ), this is valid.2. If ( B = 2 - 2sqrt{2} ), then ( A = 3 - 2sqrt{2} ). But ( 3 - 2sqrt{2} ) is approximately ( 3 - 2.828 = 0.172 ), which is positive, but we need to check if ( A > B ). ( 3 - 2sqrt{2} ) vs ( 2 - 2sqrt{2} ). ( 3 - 2sqrt{2} approx 0.172 ) and ( 2 - 2sqrt{2} approx -0.828 ). So, ( A = 0.172 ) and ( B = -0.828 ). But ( B = b^2 ) can't be negative. So, this solution is invalid.Therefore, the valid solution is ( B = 2 + 2sqrt{2} ) and ( A = 3 + 2sqrt{2} ).So, ( a^2 = 3 + 2sqrt{2} ). Therefore, ( a = sqrt{3 + 2sqrt{2}} ).Wait, can I simplify ( sqrt{3 + 2sqrt{2}} )?Yes, because ( sqrt{3 + 2sqrt{2}} = sqrt{2} + 1 ). Let me check:( (sqrt{2} + 1)^2 = 2 + 2sqrt{2} + 1 = 3 + 2sqrt{2} ). Yes, that's correct.So, ( a = sqrt{2} + 1 ). Therefore, the major axis length is ( 2a = 2(sqrt{2} + 1) = 2sqrt{2} + 2 ).So, the length of the major axis is ( 2sqrt{2} + 2 ), which is option B.**Final Answer**boxed{B}

💡Alright, so I've got this problem here about mixing two solutions, X and Y. Solution X has 20% of material A and 80% of material B. Solution Y has 30% of material A and 70% of material B. When you mix these two solutions, the final mixture has 22% of material A. The question is asking how much of solution X is present in this mixture.Okay, let's break this down. I think I need to figure out the ratio or the amount of solution X and Y that, when mixed together, give me a final concentration of 22% material A. Since both solutions have different percentages of material A, the mixture's percentage will depend on how much of each solution I use.Maybe I can set up some variables to represent the amounts of each solution. Let's say I use 'x' amount of solution X and 'y' amount of solution Y. Then, the total amount of the mixture would be x + y.Now, the amount of material A in solution X would be 20% of x, which is 0.20x. Similarly, the amount of material A in solution Y would be 30% of y, which is 0.30y. The total amount of material A in the mixture would be the sum of these two, so 0.20x + 0.30y.According to the problem, the final mixture has 22% material A. So, 22% of the total mixture (which is x + y) should be equal to the total material A from both solutions. That gives me the equation:0.20x + 0.30y = 0.22(x + y)Hmm, that looks like a good start. Now, I need to solve for x or y. But I have two variables here, so maybe I need another equation. Wait, if I assume that the total amount of the mixture is 100 units, that might simplify things. Let's try that.If I assume the total mixture is 100 units, then x + y = 100. That means y = 100 - x. Now I can substitute y in the first equation with (100 - x):0.20x + 0.30(100 - x) = 0.22 * 100Simplifying the right side, 0.22 * 100 is 22. So now the equation is:0.20x + 0.30(100 - x) = 22Let's expand the left side:0.20x + 30 - 0.30x = 22Combine like terms:-0.10x + 30 = 22Now, subtract 30 from both sides:-0.10x = -8Divide both sides by -0.10:x = (-8)/(-0.10) = 80So, x is 80 units. That means there are 80 units of solution X in the mixture. To check, if I use 80 units of X and 20 units of Y (since y = 100 - x = 20), the total material A would be:0.20 * 80 + 0.30 * 20 = 16 + 6 = 22Which matches the 22% of the total mixture (22 units out of 100). So, it seems correct.But wait, what if the total mixture isn't 100 units? Does this ratio still hold? I think it does because percentages are proportions, and the ratio of X to Y remains the same regardless of the total amount. So, even if the total mixture is different, the proportion of X would still be 80%.So, in conclusion, the amount of solution X in the mixture is 80 units, assuming the total mixture is 100 units. If the total mixture is different, the actual amount would scale accordingly, but the percentage of X remains 80%.

💡Alright, so I have these two inequalities to solve. Let me take them one at a time.Starting with the first one: (frac{x - 3}{x + 7} < 0). Hmm, okay, so this is a rational inequality. I remember that to solve such inequalities, I need to find the values of (x) where the expression is negative. That means the numerator and denominator have opposite signs.First, I should find the critical points where the expression equals zero or is undefined. The numerator is (x - 3), which equals zero when (x = 3). The denominator is (x + 7), which equals zero when (x = -7). So, these points divide the number line into intervals that I can test.So, the critical points are at (x = -7) and (x = 3). These divide the number line into three intervals:1. (x < -7)2. (-7 < x < 3)3. (x > 3)Now, I need to test each interval to see where the expression (frac{x - 3}{x + 7}) is negative.Starting with the first interval, (x < -7). Let's pick a test point, say (x = -8).Plugging into the numerator: (-8 - 3 = -11) (negative).Plugging into the denominator: (-8 + 7 = -1) (negative).So, (frac{-11}{-1} = 11), which is positive. So, this interval doesn't satisfy the inequality.Next interval: (-7 < x < 3). Let's choose (x = 0).Numerator: (0 - 3 = -3) (negative).Denominator: (0 + 7 = 7) (positive).So, (frac{-3}{7} = -frac{3}{7}), which is negative. This interval satisfies the inequality.Last interval: (x > 3). Let's pick (x = 4).Numerator: (4 - 3 = 1) (positive).Denominator: (4 + 7 = 11) (positive).So, (frac{1}{11}) is positive. Doesn't satisfy the inequality.Also, I should check the critical points:At (x = -7), the expression is undefined, so it's not included.At (x = 3), the expression equals zero, which doesn't satisfy the strict inequality.So, the solution is (-7 < x < 3).Okay, that seems straightforward. Now, moving on to the second problem.The inequality is (x^2 - 4ax + 4a^2 + a leq 0), and we're told that the solution set is empty. We need to find the range of values for (a).Hmm, if the solution set is empty, that means there are no real numbers (x) that satisfy the inequality. So, the quadratic expression (x^2 - 4ax + 4a^2 + a) is always positive for all real (x).For a quadratic (Ax^2 + Bx + C), it is always positive if the leading coefficient (A > 0) and the discriminant (B^2 - 4AC < 0).In this case, the quadratic is (x^2 - 4ax + 4a^2 + a). So, (A = 1), which is positive, so that condition is satisfied.Now, let's compute the discriminant:(B^2 - 4AC = (-4a)^2 - 4(1)(4a^2 + a))Calculating step by step:((-4a)^2 = 16a^2)(4AC = 4 * 1 * (4a^2 + a) = 16a^2 + 4a)So, discriminant (= 16a^2 - (16a^2 + 4a) = 16a^2 - 16a^2 - 4a = -4a)For the quadratic to have no real roots, the discriminant must be less than zero:(-4a < 0)Dividing both sides by -4 (remembering to flip the inequality sign when multiplying or dividing by a negative):(a > 0)So, the range of values for (a) is all positive real numbers.Wait, let me double-check that.Starting with discriminant:(B^2 - 4AC = (-4a)^2 - 4(1)(4a^2 + a) = 16a^2 - 16a^2 - 4a = -4a)Set discriminant < 0:(-4a < 0)Divide both sides by -4 (inequality flips):(a > 0)Yes, that seems correct.So, the range of (a) is (a > 0).Alright, I think I got both problems figured out.**Final Answer**1. The solution to the inequality is (boxed{{x mid -7 < x < 3}}).2. The range of values for (a) is (boxed{(0, +infty)}).

💡First, I recognize that angle ( AOD ) is a straight angle, which means it measures ( 180^circ ).Given that ( angle AOB = 50^circ ), I can find ( angle BOD ) by subtracting ( angle AOB ) from ( angle AOD ):[angle BOD = 180^circ - 50^circ = 130^circ]Since ( OC ) is the bisector of ( angle BOD ), it divides ( angle BOD ) into two equal parts. Therefore, each of the angles ( angle BOC ) and ( angle COD ) is half of ( angle BOD ):[angle COD = frac{1}{2} times 130^circ = 65^circ]

💡First, determine the number of large cakes Gary can make with 5 pounds of flour:5 pounds ÷ 0.75 pounds per cake = 6.666... cakesSince partial cakes aren't possible, Gary can make 6 large cakes.Next, calculate the number of cupcakes using the remaining 3 pounds of flour:3 pounds ÷ 0.25 pounds per cupcake = 12 cupcakesNow, calculate the cost of additional ingredients:- Large cakes: 6 cakes × 1.50 = 9.00- Cupcakes: 12 cupcakes × 0.75 = 9.00Total additional ingredients cost = 9.00 + 9.00 = 18.00Add the cost of flour and baking equipment:Total cost = 4.00 (flour) + 10.00 (equipment) + 18.00 (ingredients) = 32.00Calculate the revenue from sales:- Large cakes: 6 cakes × 6.50 = 39.00- Cupcakes: 12 cupcakes × 2.50 = 30.00Total revenue = 39.00 + 30.00 = 69.00Finally, determine the profit:Profit = Total revenue - Total cost = 69.00 - 32.00 = 37.00

💡Okay, so I have this function f(x) = a^x + x² - x ln a, where a is greater than 0 and not equal to 1. I need to do two things: first, discuss the monotonicity of f(x), and second, determine the range of a such that there exist x₁ and x₂ in the interval [-1, 1] where the absolute difference |f(x₁) - f(x₂)| is at least e - 1.Starting with part (1): Discussing the monotonicity of f(x). Monotonicity usually refers to whether the function is increasing or decreasing over certain intervals. To find this out, I should compute the derivative of f(x) with respect to x because the sign of the derivative will tell me if the function is increasing or decreasing.So, let's compute f'(x). The function f(x) has three terms: a^x, x², and -x ln a. Taking the derivative term by term:- The derivative of a^x with respect to x is a^x ln a.- The derivative of x² is 2x.- The derivative of -x ln a is -ln a.Putting it all together, f'(x) = a^x ln a + 2x - ln a.Hmm, I can factor out ln a from the first and third terms: f'(x) = (a^x - 1) ln a + 2x.Now, to analyze the sign of f'(x), which will tell me about the monotonicity. The expression (a^x - 1) ln a is interesting because it depends on the value of a and x. Let's consider two cases: when a > 1 and when 0 < a < 1.Case 1: a > 1. In this case, ln a is positive. Now, a^x is an exponential function. For x > 0, a^x > 1, so (a^x - 1) is positive. Therefore, (a^x - 1) ln a is positive. Also, 2x is positive when x > 0. So, both terms are positive, meaning f'(x) > 0 for x > 0. Thus, f(x) is increasing on (0, ∞).For x < 0, a^x < 1 because a > 1 and x is negative, so (a^x - 1) is negative. Since ln a is positive, (a^x - 1) ln a is negative. Also, 2x is negative because x < 0. So, both terms are negative, meaning f'(x) < 0 for x < 0. Thus, f(x) is decreasing on (-∞, 0).Case 2: 0 < a < 1. Here, ln a is negative. For x > 0, a^x < 1 because a is between 0 and 1, so (a^x - 1) is negative. Multiplying by ln a (which is also negative) gives a positive term. Additionally, 2x is positive for x > 0. So, both terms are positive, meaning f'(x) > 0 for x > 0. Therefore, f(x) is increasing on (0, ∞).For x < 0, a^x > 1 because a is less than 1 and x is negative, so (a^x - 1) is positive. Multiplying by ln a (negative) gives a negative term. Also, 2x is negative for x < 0. So, both terms are negative, meaning f'(x) < 0 for x < 0. Thus, f(x) is decreasing on (-∞, 0).So, summarizing part (1): Regardless of whether a is greater than 1 or between 0 and 1, f(x) is decreasing on (-∞, 0) and increasing on (0, ∞). So, the function has a minimum at x = 0.Moving on to part (2): We need to find the range of a such that there exist x₁, x₂ in [-1, 1] with |f(x₁) - f(x₂)| ≥ e - 1.Given that f(x) is decreasing on [-1, 0] and increasing on [0, 1], the minimum value of f(x) on [-1, 1] occurs at x = 0. The maximum value will occur at one of the endpoints, either x = -1 or x = 1. So, the maximum of f(x) on [-1, 1] is the larger of f(-1) and f(1).Therefore, the maximum difference |f(x₁) - f(x₂)| will be the difference between the maximum value at the endpoints and the minimum at x = 0. So, we need either f(1) - f(0) ≥ e - 1 or f(-1) - f(0) ≥ e - 1.Let's compute f(0): f(0) = a^0 + 0² - 0 * ln a = 1 + 0 - 0 = 1.Now, compute f(1): f(1) = a^1 + 1² - 1 * ln a = a + 1 - ln a.Similarly, f(-1) = a^{-1} + (-1)^2 - (-1) * ln a = 1/a + 1 + ln a.So, we have two expressions:1. f(1) - f(0) = (a + 1 - ln a) - 1 = a - ln a.2. f(-1) - f(0) = (1/a + 1 + ln a) - 1 = 1/a + ln a.We need either a - ln a ≥ e - 1 or 1/a + ln a ≥ e - 1.Let's analyze these inequalities.First, consider the case when a > 1. In this case, since a > 1, ln a is positive. Also, from part (1), f(x) is decreasing on [-1, 0] and increasing on [0, 1], so f(1) is the maximum on [0, 1], and f(-1) is the maximum on [-1, 0]. But since a > 1, let's see which of f(1) or f(-1) is larger.Compute f(1) - f(-1): (a + 1 - ln a) - (1/a + 1 + ln a) = a - 1/a - 2 ln a.We can analyze this expression. Let me define a function g(a) = a - 1/a - 2 ln a for a > 1.Compute g'(a): 1 + 1/a² - 2/a.Set g'(a) = 0: 1 + 1/a² - 2/a = 0.Multiply both sides by a²: a² + 1 - 2a = 0 ⇒ (a - 1)^2 = 0 ⇒ a = 1.But a > 1, so g'(a) is positive for a > 1 because:g'(a) = 1 + 1/a² - 2/a. Let's see for a > 1:1 is positive, 1/a² is positive, and -2/a is negative. Let's plug in a = 2: 1 + 1/4 - 1 = 1/4 > 0. Similarly, for a approaching 1 from the right: g'(1) = 1 + 1 - 2 = 0. For a approaching infinity: g'(a) approaches 1 + 0 - 0 = 1 > 0. So, g'(a) > 0 for a > 1, meaning g(a) is increasing on (1, ∞). Since g(1) = 1 - 1 - 0 = 0, and g(a) is increasing, g(a) > 0 for a > 1. Therefore, f(1) - f(-1) > 0, so f(1) > f(-1) when a > 1.Therefore, for a > 1, the maximum of f(x) on [-1, 1] is f(1). So, the maximum difference is f(1) - f(0) = a - ln a. We need this to be at least e - 1.So, a - ln a ≥ e - 1.We need to solve this inequality for a > 1.Let me define h(a) = a - ln a. We need h(a) ≥ e - 1.Compute h(e): h(e) = e - ln e = e - 1, which is exactly the value we need. Now, let's see the behavior of h(a):h'(a) = 1 - 1/a. For a > 1, h'(a) = 1 - 1/a > 0 because 1/a < 1. So, h(a) is increasing for a > 1. Since h(e) = e - 1, and h(a) is increasing, for all a ≥ e, h(a) ≥ e - 1. Therefore, a ≥ e.So, for a > 1, the condition a - ln a ≥ e - 1 is satisfied when a ≥ e.Now, consider the case when 0 < a < 1. In this case, ln a is negative. From part (1), f(x) is decreasing on [-1, 0] and increasing on [0, 1]. So, f(-1) is the maximum on [-1, 0], and f(1) is the maximum on [0, 1]. Let's see which of f(-1) or f(1) is larger.Compute f(-1) - f(1): (1/a + 1 + ln a) - (a + 1 - ln a) = 1/a - a + 2 ln a.Let me define k(a) = 1/a - a + 2 ln a for 0 < a < 1.Compute k'(a): -1/a² - 1 + 2/a.Set k'(a) = 0: -1/a² - 1 + 2/a = 0.Multiply both sides by a²: -1 - a² + 2a = 0 ⇒ -a² + 2a - 1 = 0 ⇒ a² - 2a + 1 = 0 ⇒ (a - 1)^2 = 0 ⇒ a = 1.But a < 1, so let's analyze the sign of k'(a):k'(a) = -1/a² - 1 + 2/a.Let me rewrite it: k'(a) = (-1 - a² + 2a)/a².The denominator a² is positive. The numerator is -1 - a² + 2a.Let me define m(a) = -1 - a² + 2a.Compute m(a): m(a) = -a² + 2a - 1 = -(a² - 2a + 1) = -(a - 1)^2 ≤ 0.So, m(a) ≤ 0 for all a, and m(a) = 0 only when a = 1. Therefore, k'(a) ≤ 0 for all a in (0, 1). So, k(a) is decreasing on (0, 1).Compute the limit as a approaches 1 from the left: k(1) = 1 - 1 + 2*0 = 0.Compute the limit as a approaches 0 from the right: 1/a approaches infinity, -a approaches 0, and 2 ln a approaches -infinity. So, 1/a - a + 2 ln a: 1/a dominates, so it approaches infinity.Wait, that seems conflicting because k(a) is decreasing on (0,1), starting from infinity at a approaches 0 and approaching 0 at a approaches 1. So, k(a) is positive on (0,1). Therefore, f(-1) - f(1) = k(a) > 0 for 0 < a < 1. So, f(-1) > f(1) when 0 < a < 1.Therefore, for 0 < a < 1, the maximum of f(x) on [-1,1] is f(-1). So, the maximum difference is f(-1) - f(0) = 1/a + ln a. We need this to be at least e - 1.So, 1/a + ln a ≥ e - 1.Let me define p(a) = 1/a + ln a for 0 < a < 1.We need p(a) ≥ e - 1.Compute p(1/e): 1/(1/e) + ln(1/e) = e + (-1) = e - 1. So, p(1/e) = e - 1.Now, let's analyze p(a):Compute p'(a): -1/a² + 1/a.Set p'(a) = 0: -1/a² + 1/a = 0 ⇒ (-1 + a)/a² = 0 ⇒ a = 1.But a < 1, so let's see the behavior of p(a):For a approaching 0 from the right: 1/a approaches infinity, ln a approaches -infinity, but 1/a dominates, so p(a) approaches infinity.For a approaching 1 from the left: 1/a approaches 1, ln a approaches 0, so p(a) approaches 1.Compute p'(a): -1/a² + 1/a. Let's see the sign:p'(a) = (-1 + a)/a². For 0 < a < 1, the numerator is (-1 + a) < 0, denominator is positive, so p'(a) < 0. Therefore, p(a) is decreasing on (0,1).Since p(a) is decreasing on (0,1) and p(1/e) = e - 1, then for a ≤ 1/e, p(a) ≥ e - 1.Wait, let's verify:Since p(a) is decreasing, as a increases, p(a) decreases. So, when a = 1/e, p(a) = e - 1. For a < 1/e, p(a) > e - 1, and for a > 1/e, p(a) < e - 1.But since we are considering 0 < a < 1, and we need p(a) ≥ e - 1, this occurs when a ≤ 1/e.Therefore, for 0 < a ≤ 1/e, p(a) ≥ e - 1.So, combining both cases:- For a > 1, we need a ≥ e.- For 0 < a < 1, we need a ≤ 1/e.Therefore, the range of a is (0, 1/e] ∪ [e, ∞).Wait, but the original condition is that a > 0 and a ≠ 1, so this range is valid.Let me just double-check:For a = e, f(1) - f(0) = e - ln e = e - 1, which satisfies the condition.For a = 1/e, f(-1) - f(0) = e + ln(1/e) = e - 1, which also satisfies the condition.For a > e, f(1) - f(0) = a - ln a > e - 1 because h(a) = a - ln a is increasing for a > 1.For 0 < a < 1/e, p(a) = 1/a + ln a > e - 1 because p(a) is decreasing and p(1/e) = e - 1.Thus, the range of a is indeed (0, 1/e] ∪ [e, ∞).