💡First, calculate Brendan's total earnings from his hourly wage:2 shifts * 8 hours/shift = 16 hours1 shift * 12 hours/shift = 12 hoursTotal hours = 16 + 12 = 28 hoursTotal wage income = 28 hours * 6/hour = 168Next, calculate his total tips for the week:Total tips = 28 hours * 12/hour = 336Now, determine the total income if all tips were reported:Total income = 168 (wage) + 336 (tips) = 504Calculate the taxes if all income were reported:Taxes = 20% of 504 = 0.20 * 504 = 100.80However, Brendan only pays 56 in taxes. Calculate the taxes paid on his wage:Taxes on wage = 20% of 168 = 0.20 * 168 = 33.60Determine the taxes paid on reported tips:Taxes on tips = 56 - 33.60 = 22.40Find the amount of tips reported:Reported tips = 22.40 / 0.20 = 112Finally, calculate the fraction of tips reported:Fraction = 112 / 336 = 1/3

💡First, identify the number of marbles Josh lost and the number he found.Josh lost 23 marbles and found 9 marbles.To determine how many more marbles he lost than he found, subtract the number of marbles found from the number of marbles lost.23 (lost) - 9 (found) = 14Therefore, Josh lost 14 more marbles than he found.

💡Okay, so I have this problem about an ellipse with foci at F₁ = (0,0) and F₂ = (4,0). It says the ellipse intersects the x-axis at the point (1,0) and one other point, and I need to find that other point. Hmm, okay, let me think about how to approach this.First, I remember that an ellipse is defined as the set of all points where the sum of the distances from the two foci is constant. So, for any point P on the ellipse, the sum PF₁ + PF₂ is the same. Since the ellipse intersects the x-axis at (1,0), I can use this point to find that constant sum.Let me calculate the distances from (1,0) to each focus. The distance from (1,0) to F₁ = (0,0) is just the distance between (1,0) and (0,0), which is √[(1-0)² + (0-0)²] = √1 = 1. Similarly, the distance from (1,0) to F₂ = (4,0) is √[(1-4)² + (0-0)²] = √[(-3)²] = √9 = 3. So, the sum of these distances is 1 + 3 = 4. That means for any point on the ellipse, the sum of the distances to F₁ and F₂ is 4.Now, I need to find another point where the ellipse intersects the x-axis. Let's denote this point as (x, 0). Since it's on the x-axis, the y-coordinate is 0. So, the point is (x, 0). Let's apply the same logic: the sum of the distances from (x, 0) to F₁ and F₂ should be 4.Calculating the distances, the distance from (x, 0) to F₁ = (0,0) is √[(x - 0)² + (0 - 0)²] = √x² = |x|. Similarly, the distance from (x, 0) to F₂ = (4,0) is √[(x - 4)² + (0 - 0)²] = √(x - 4)² = |x - 4|.So, the equation becomes |x| + |x - 4| = 4. Now, I need to solve this equation for x. Since absolute value functions can be tricky, I should consider different cases based on the value of x.Case 1: x ≥ 4. In this case, |x| = x and |x - 4| = x - 4. So, the equation becomes x + (x - 4) = 4. Simplifying, that's 2x - 4 = 4, so 2x = 8, which gives x = 4. So, x = 4 is a solution in this case.Case 2: 0 ≤ x ≤ 4. Here, |x| = x and |x - 4| = 4 - x. Plugging into the equation: x + (4 - x) = 4. Simplifying, that's 4 = 4, which is always true. Hmm, so this suggests that any x between 0 and 4 satisfies the equation. But wait, the ellipse can only intersect the x-axis at two points, right? We already know one of them is (1,0), so the other must be at x = 4, which we found in Case 1.But hold on, in Case 2, it seems like all x between 0 and 4 satisfy the equation. That doesn't make sense because an ellipse can't intersect the x-axis at infinitely many points. I must be missing something here. Maybe I need to consider the definition of an ellipse more carefully.Wait a second, the ellipse is determined by the sum of distances to the foci being constant, which we found to be 4. So, for points on the x-axis, the sum |x| + |x - 4| must equal 4. But when x is between 0 and 4, |x| + |x - 4| simplifies to x + (4 - x) = 4, which is always true. That suggests that the entire line segment between (0,0) and (4,0) is part of the ellipse, but that can't be right because an ellipse is a closed curve, not a line segment.I think I'm confusing the definition here. Actually, the ellipse is the set of points where the sum of the distances is constant, but not all points on the line segment between the foci satisfy this condition unless the constant is exactly equal to the distance between the foci. In this case, the distance between F₁ and F₂ is 4 units, which is the same as the constant sum we found. So, when the constant sum equals the distance between the foci, the ellipse collapses to the line segment between the two foci. But in our problem, the ellipse intersects the x-axis at (1,0) and another point, implying it's a proper ellipse, not a line segment.This means there must be a mistake in my earlier reasoning. Let me double-check. The sum of distances from (1,0) is 4, which is equal to the distance between the foci. That suggests that (1,0) is actually on the line segment between F₁ and F₂, making the ellipse degenerate. But the problem states it's an ellipse, so it must be non-degenerate. Therefore, my initial assumption that the sum is 4 might be incorrect.Wait, no, the sum of distances for an ellipse must be greater than the distance between the foci. If it's equal, it's a line segment. So, if the sum is equal to 4, which is the distance between F₁ and F₂, then the ellipse is degenerate. But the problem says it's an ellipse, so perhaps the sum is greater than 4. Maybe I made a mistake in calculating the sum.Let me recalculate the sum for the point (1,0). The distance from (1,0) to (0,0) is 1, and to (4,0) is 3. So, 1 + 3 = 4. That's correct. So, if the sum is 4, which is equal to the distance between the foci, the ellipse is degenerate. But the problem states it's an ellipse, so perhaps the given point (1,0) is not on the major axis? Wait, no, (1,0) is on the x-axis, which is the major axis since the foci are on the x-axis.Hmm, this is confusing. Maybe the problem is misstated, or perhaps I'm misunderstanding something. Let me think again. If the sum of distances is equal to the distance between the foci, it's a line segment. If it's greater, it's an ellipse. Since the problem says it's an ellipse, the sum must be greater than 4. But according to the point (1,0), the sum is 4. That seems contradictory.Wait, maybe the point (1,0) is not on the ellipse? But the problem says it is. So, perhaps the ellipse is degenerate, but the problem refers to it as an ellipse. Maybe in this context, they consider the degenerate case as an ellipse. If that's the case, then the ellipse would be the line segment between (0,0) and (4,0), and it would intersect the x-axis at all points between 0 and 4, including (1,0) and (4,0). But the problem mentions another point, so maybe (4,0) is the other intersection point.But earlier, when I considered x ≥ 4, I found x = 4 as a solution. So, perhaps the other intersection point is (4,0). Let me verify. If I plug x = 4 into the equation |x| + |x - 4|, I get 4 + 0 = 4, which satisfies the equation. So, (4,0) is indeed another intersection point.But wait, if the ellipse is degenerate, then it's just the line segment from (0,0) to (4,0), and it intersects the x-axis at every point between 0 and 4, not just at two points. But the problem says it intersects at (1,0) and one other point, implying only two intersection points. This is conflicting.Maybe I need to reconsider. Perhaps the ellipse is not degenerate, and the sum of distances is greater than 4. But then, how does the point (1,0) lie on the ellipse? Because for (1,0), the sum is 4, which would mean it's on the degenerate ellipse. This is confusing.Alternatively, maybe the problem is correct, and the ellipse is non-degenerate, but the sum of distances is greater than 4, and (1,0) is just one of the intersection points. But then, how do we find the other point?Wait, perhaps I made a mistake in assuming that the sum is 4. Let me think again. If the ellipse is non-degenerate, the sum of distances must be greater than 4. So, maybe the point (1,0) is not on the x-axis? No, it is on the x-axis. Hmm.Wait, perhaps the ellipse is not centered at the midpoint of the foci. The midpoint of F₁ and F₂ is (2,0). So, the center of the ellipse is at (2,0). The major axis length is 2a, and the distance between foci is 2c, so 2c = 4, so c = 2. For an ellipse, a > c, so a must be greater than 2. The sum of distances is 2a, so 2a > 4, so a > 2.Given that, the ellipse equation can be written as [(x - 2)²]/a² + y²/b² = 1, where b² = a² - c² = a² - 4.Now, since the ellipse passes through (1,0), let's plug that into the equation:[(1 - 2)²]/a² + 0²/b² = 1(1)/a² = 1So, a² = 1, which implies a = 1. But wait, a must be greater than c = 2, so this is impossible. Therefore, there must be a mistake in my approach.Wait, maybe I shouldn't assume the standard form of the ellipse. Alternatively, perhaps the ellipse is not aligned with the coordinate axes? But since the foci are on the x-axis, the major axis is along the x-axis, so the standard form should apply.But according to the calculation, plugging (1,0) into the standard form gives a² = 1, which contradicts a > 2. Therefore, perhaps the ellipse is degenerate, and the sum of distances is 4, making it the line segment between (0,0) and (4,0). In that case, the other intersection point is (4,0).But the problem refers to it as an ellipse, which is typically considered non-degenerate. Maybe the problem allows for the degenerate case. If so, then the other intersection point is (4,0).Alternatively, perhaps the problem has a typo, and the point is not (1,0), but another point. But assuming the problem is correct, I think the other intersection point is (4,0).So, after considering all this, I think the other point of intersection on the x-axis is (4,0).

💡Okay, so I have this inequality to solve: [frac{5x^2 + 20x - 56}{(3x - 4)(x + 5)} < 2.] Hmm, I remember that when dealing with inequalities involving rational expressions, I need to be careful about the denominator and the sign changes. Let me try to break this down step by step.First, I think I should get all terms on one side to make the inequality easier to handle. So, I'll subtract 2 from both sides:[frac{5x^2 + 20x - 56}{(3x - 4)(x + 5)} - 2 < 0.]Now, to combine these terms into a single fraction, I need a common denominator. The denominator is already given as ((3x - 4)(x + 5)), so I'll rewrite 2 as (frac{2(3x - 4)(x + 5)}{(3x - 4)(x + 5)}). That way, both terms have the same denominator:[frac{5x^2 + 20x - 56 - 2(3x - 4)(x + 5)}{(3x - 4)(x + 5)} < 0.]Okay, now I need to simplify the numerator. Let me expand the term (2(3x - 4)(x + 5)). First, multiply out ((3x - 4)(x + 5)):[(3x - 4)(x + 5) = 3x cdot x + 3x cdot 5 - 4 cdot x - 4 cdot 5 = 3x^2 + 15x - 4x - 20 = 3x^2 + 11x - 20.]So, multiplying by 2 gives:[2(3x^2 + 11x - 20) = 6x^2 + 22x - 40.]Now, subtract this from the original numerator:[5x^2 + 20x - 56 - (6x^2 + 22x - 40) = 5x^2 + 20x - 56 - 6x^2 - 22x + 40.]Let me combine like terms:- (5x^2 - 6x^2 = -x^2)- (20x - 22x = -2x)- (-56 + 40 = -16)So, the numerator simplifies to:[-x^2 - 2x - 16.]Wait, that doesn't look right. Let me double-check my subtraction:Original numerator: (5x^2 + 20x - 56)Subtracting (6x^2 + 22x - 40):(5x^2 - 6x^2 = -x^2)(20x - 22x = -2x)(-56 - (-40) = -56 + 40 = -16)Hmm, so it's correct. So the numerator is (-x^2 - 2x - 16). Maybe I can factor out a negative sign to make it easier:[-(x^2 + 2x + 16).]Wait, can I factor (x^2 + 2x + 16)? Let me check the discriminant: (b^2 - 4ac = 4 - 64 = -60). Since it's negative, this quadratic doesn't factor nicely over the real numbers. Hmm, that complicates things a bit.But wait, maybe I made a mistake earlier. Let me go back to the numerator:Original numerator after subtraction: (-x^2 - 2x - 16). Wait, that seems off because when I subtracted, I should have:(5x^2 + 20x - 56 - 6x^2 - 22x + 40)Which is:(5x^2 - 6x^2 = -x^2)(20x - 22x = -2x)(-56 + 40 = -16)So that's correct. So the numerator is indeed (-x^2 - 2x - 16), which is the same as (-(x^2 + 2x + 16)). Since this quadratic doesn't factor, maybe I should consider completing the square or using the quadratic formula, but I'm not sure if that's necessary here.Wait, maybe I made a mistake in expanding (2(3x - 4)(x + 5)). Let me check that again:((3x - 4)(x + 5)) is (3x^2 + 15x - 4x - 20 = 3x^2 + 11x - 20). Then multiplying by 2 gives (6x^2 + 22x - 40). That seems correct.So, the numerator is indeed (-x^2 - 2x - 16). Hmm, maybe I should factor out a negative sign to make it easier:[-(x^2 + 2x + 16).]Since (x^2 + 2x + 16) doesn't factor, I might need to consider the sign of this expression. Since the coefficient of (x^2) is positive, the parabola opens upwards, and since the discriminant is negative, it never crosses the x-axis. So, (x^2 + 2x + 16) is always positive. Therefore, (-(x^2 + 2x + 16)) is always negative.Wait, so the numerator is always negative. That might simplify things. So, the inequality becomes:[frac{-(x^2 + 2x + 16)}{(3x - 4)(x + 5)} < 0.]Since the numerator is always negative, the sign of the entire expression depends on the denominator. So, the inequality is equivalent to:[frac{-text{(always positive)}}{(3x - 4)(x + 5)} < 0.]Which simplifies to:[frac{-1}{(3x - 4)(x + 5)} < 0.]Wait, but that might not be the case. Let me think again. The numerator is (- (x^2 + 2x + 16)), which is always negative because (x^2 + 2x + 16) is always positive. So, the numerator is negative, and the denominator is ((3x - 4)(x + 5)). So, the entire expression is negative divided by something. So, the sign of the entire expression depends on the denominator.So, the inequality is:[frac{text{Negative}}{(3x - 4)(x + 5)} < 0.]Which is equivalent to:[frac{text{Negative}}{text{Denominator}} < 0.]So, for the entire expression to be less than zero, the denominator must be positive because a negative divided by a positive is negative. So, we need:[(3x - 4)(x + 5) > 0.]So, now I need to solve ((3x - 4)(x + 5) > 0). To do this, I'll find the critical points where each factor is zero:- (3x - 4 = 0 Rightarrow x = frac{4}{3})- (x + 5 = 0 Rightarrow x = -5)These critical points divide the real number line into intervals. I'll test each interval to determine the sign of the product.The intervals are:1. (x < -5)2. (-5 < x < frac{4}{3})3. (x > frac{4}{3})Let me test each interval:1. For (x < -5), say (x = -6): - (3(-6) - 4 = -18 - 4 = -22) (negative) - (-6 + 5 = -1) (negative) - Product: negative * negative = positive2. For (-5 < x < frac{4}{3}), say (x = 0): - (3(0) - 4 = -4) (negative) - (0 + 5 = 5) (positive) - Product: negative * positive = negative3. For (x > frac{4}{3}), say (x = 2): - (3(2) - 4 = 6 - 4 = 2) (positive) - (2 + 5 = 7) (positive) - Product: positive * positive = positiveSo, the product ((3x - 4)(x + 5)) is positive when (x < -5) or (x > frac{4}{3}).But remember, we also need to consider the domain of the original inequality. The denominator ((3x - 4)(x + 5)) cannot be zero, so (x neq frac{4}{3}) and (x neq -5).Therefore, the solution to ((3x - 4)(x + 5) > 0) is (x in (-infty, -5) cup (frac{4}{3}, infty)).But wait, earlier I concluded that the numerator is always negative, so the entire expression (frac{text{Negative}}{(3x - 4)(x + 5)}) is less than zero when the denominator is positive, which is when (x in (-infty, -5) cup (frac{4}{3}, infty)).However, I need to make sure that the original inequality is correctly transformed. Let me recap:Original inequality:[frac{5x^2 + 20x - 56}{(3x - 4)(x + 5)} < 2.]Subtracting 2:[frac{5x^2 + 20x - 56 - 2(3x - 4)(x + 5)}{(3x - 4)(x + 5)} < 0.]Simplifying numerator:[-x^2 - 2x - 16 = -(x^2 + 2x + 16).]So, the inequality becomes:[frac{-(x^2 + 2x + 16)}{(3x - 4)(x + 5)} < 0.]Since (x^2 + 2x + 16) is always positive, the numerator is always negative. Therefore, the inequality is equivalent to:[frac{-1}{(3x - 4)(x + 5)} < 0.]Which simplifies to:[frac{1}{(3x - 4)(x + 5)} > 0.]Because multiplying both sides by -1 reverses the inequality.So, now I need to solve:[frac{1}{(3x - 4)(x + 5)} > 0.]Since the numerator is 1, which is positive, the inequality holds when the denominator is positive. So, again, we have:[(3x - 4)(x + 5) > 0.]Which, as before, gives (x in (-infty, -5) cup (frac{4}{3}, infty)).But wait, I think I might have made a mistake earlier. Let me double-check the transformation.Starting from:[frac{-(x^2 + 2x + 16)}{(3x - 4)(x + 5)} < 0.]Since the numerator is negative, the inequality is equivalent to:[frac{1}{(3x - 4)(x + 5)} > 0.]Because if (frac{-A}{B} < 0), then (frac{A}{B} > 0), since multiplying both sides by -1 reverses the inequality.So, yes, the inequality reduces to (frac{1}{(3x - 4)(x + 5)} > 0), which is true when ((3x - 4)(x + 5) > 0), as I concluded.Therefore, the solution is (x in (-infty, -5) cup (frac{4}{3}, infty)).But wait, let me check if this makes sense. Let me pick a test value in each interval and see if the original inequality holds.1. For (x < -5), say (x = -6): - Original expression: (frac{5(-6)^2 + 20(-6) - 56}{(3(-6) - 4)(-6 + 5)}) - Numerator: (5*36 - 120 -56 = 180 - 120 -56 = 4) - Denominator: ((-18 -4)(-1) = (-22)(-1) = 22) - So, (frac{4}{22} = frac{2}{11} approx 0.18), which is less than 2. So, the inequality holds.2. For (-5 < x < frac{4}{3}), say (x = 0): - Numerator: (5*0 + 0 -56 = -56) - Denominator: ((0 -4)(0 +5) = (-4)(5) = -20) - So, (frac{-56}{-20} = 2.8), which is greater than 2. So, the inequality does not hold.3. For (x > frac{4}{3}), say (x = 2): - Numerator: (5*4 + 40 -56 = 20 +40 -56 = 4) - Denominator: ((6 -4)(2 +5) = (2)(7) =14) - So, (frac{4}{14} approx 0.2857), which is less than 2. So, the inequality holds.Wait, but according to my earlier solution, the inequality holds for (x in (-infty, -5) cup (frac{4}{3}, infty)), which matches the test values. So, that seems correct.However, I'm a bit confused because when I first simplified the numerator, I thought it was (-x^2 - 2x -16), which is always negative. So, the entire expression is negative over the denominator. Therefore, the inequality (frac{text{Negative}}{text{Denominator}} < 0) is equivalent to (frac{1}{text{Denominator}} > 0), which is what I did.But just to be thorough, let me consider the critical points where the expression is zero or undefined. The expression is undefined at (x = -5) and (x = frac{4}{3}). It is zero when the numerator is zero, but the numerator is (-x^2 - 2x -16), which is never zero because the discriminant is negative. So, there are no zeros, only vertical asymptotes at (x = -5) and (x = frac{4}{3}).Therefore, the solution is indeed (x in (-infty, -5) cup (frac{4}{3}, infty)).Wait, but in the initial problem, the inequality is (frac{5x^2 + 20x -56}{(3x -4)(x +5)} < 2). So, I think my solution is correct.But just to be absolutely sure, let me try another test value in each interval.1. (x = -6) (in (-infty, -5)): - As before, the expression is approximately 0.18 < 2: holds.2. (x = 0) (in (-5, frac{4}{3})): - Expression is 2.8 > 2: does not hold.3. (x = 2) (in (frac{4}{3}, infty)): - Expression is approximately 0.2857 < 2: holds.So, yes, the solution is correct.But wait, I just realized something. When I subtracted 2, I might have made a mistake in the algebra. Let me double-check that step.Original inequality:[frac{5x^2 + 20x -56}{(3x -4)(x +5)} < 2.]Subtract 2:[frac{5x^2 + 20x -56}{(3x -4)(x +5)} - 2 < 0.]Express 2 as (frac{2(3x -4)(x +5)}{(3x -4)(x +5)}):[frac{5x^2 + 20x -56 - 2(3x -4)(x +5)}{(3x -4)(x +5)} < 0.]Expanding (2(3x -4)(x +5)):First, expand ((3x -4)(x +5)):(3x*x = 3x^2)(3x*5 = 15x)(-4*x = -4x)(-4*5 = -20)So, (3x^2 +15x -4x -20 = 3x^2 +11x -20).Multiply by 2: (6x^2 +22x -40).Subtract this from the numerator:(5x^2 +20x -56 -6x^2 -22x +40).Combine like terms:(5x^2 -6x^2 = -x^2)(20x -22x = -2x)(-56 +40 = -16)So, numerator is (-x^2 -2x -16), which is correct.So, the transformation is correct.Therefore, the solution is (x in (-infty, -5) cup (frac{4}{3}, infty)).But wait, in the initial problem, the user wrote the solution as ((-5, -4) cup (frac{4}{3}, 4)). That seems different from my conclusion. Did I make a mistake?Wait, let me check the user's solution again:They wrote:"Solution:[x in boxed{(-5, -4) cup left(frac{4}{3}, 4right)}.]"But according to my solution, it's ((-∞, -5) ∪ (4/3, ∞)). So, why is there a discrepancy?Wait, perhaps I made a mistake in the initial steps. Let me go back.Wait, in the user's solution, they factored the numerator as (-(x -4)(x +4)), which would be (-x^2 +16), but that's not the same as (-x^2 -2x -16). So, perhaps they made a mistake in their factoring.Wait, let me see:They wrote:"Factoring the numerator:[-(x - 4)(x + 4)]Now, the inequality becomes:[frac{-(x - 4)(x + 4)}{(3x - 4)(x + 5)} < 0][frac{(x - 4)(x + 4)}{(3x - 4)(x + 5)} > 0]"But wait, if the numerator is (-x^2 -2x -16), which is (-(x^2 +2x +16)), not (-(x -4)(x +4)). Because (-(x -4)(x +4) = -x^2 +16), which is different from (-x^2 -2x -16).So, the user made a mistake in factoring the numerator. They incorrectly factored it as (-(x -4)(x +4)), which is not correct because that would give (-x^2 +16), not (-x^2 -2x -16).Therefore, their solution is incorrect because they factored the numerator incorrectly, leading them to the wrong intervals.So, my solution is correct, and the user's solution is wrong.Therefore, the correct solution is (x in (-infty, -5) cup (frac{4}{3}, infty)).But wait, let me double-check everything once more to be absolutely sure.Starting from the beginning:Original inequality:[frac{5x^2 + 20x -56}{(3x -4)(x +5)} < 2.]Subtract 2:[frac{5x^2 + 20x -56 - 2(3x -4)(x +5)}{(3x -4)(x +5)} < 0.]Expand the numerator:(5x^2 +20x -56 - [6x^2 +22x -40] = 5x^2 +20x -56 -6x^2 -22x +40 = -x^2 -2x -16).So, numerator is (-x^2 -2x -16), which is (-(x^2 +2x +16)). Since (x^2 +2x +16) has discriminant (4 -64 = -60), it doesn't factor over reals, and is always positive. So, numerator is always negative.Therefore, the inequality is:[frac{-text{(positive)}}{(3x -4)(x +5)} < 0.]Which is equivalent to:[frac{-1}{(3x -4)(x +5)} < 0.]Multiplying both sides by -1 (and reversing inequality):[frac{1}{(3x -4)(x +5)} > 0.]So, the denominator must be positive:[(3x -4)(x +5) > 0.]Critical points at (x = -5) and (x = frac{4}{3}). Testing intervals:1. (x < -5): product positive.2. (-5 < x < frac{4}{3}): product negative.3. (x > frac{4}{3}): product positive.Therefore, solution is (x in (-infty, -5) cup (frac{4}{3}, infty)).Yes, that seems correct. So, the user's solution was incorrect because they factored the numerator incorrectly, leading them to wrong intervals.Therefore, the correct solution is (x in (-infty, -5) cup (frac{4}{3}, infty)).

💡Okay, so I have this function f(x) defined piecewise. It's -x³ + x² when x is less than e, and a ln x when x is less than or equal to e. Wait, hold on, that seems a bit confusing. Let me read it again.It says f(x) is equal to -x³ + x² when x < e, and a ln x when x ≤ e. Hmm, so for x less than e, it's -x³ + x², and for x less than or equal to e, it's a ln x. That seems overlapping because x < e is included in x ≤ e. Maybe it's a typo? Or maybe it's supposed to be x ≥ e? Because otherwise, the function is defined as two different expressions for x < e and x ≤ e, which is a bit redundant. Maybe the second case is supposed to be x ≥ e? That would make more sense because otherwise, the first case is x < e, and the second case is x ≤ e, so for x < e, both cases apply, which is conflicting.Wait, the original problem says: f(x) is defined as -x³ + x² when x < e, and a ln x when x ≤ e. So, actually, for x < e, it's both -x³ + x² and a ln x? That can't be right because a function can't have two different expressions for the same domain. So, maybe it's a misprint. Perhaps the second case is x ≥ e? That would make sense because then for x < e, it's one expression, and for x ≥ e, it's another. Otherwise, the function is undefined or conflicting for x < e.Alternatively, maybe it's a typo, and the second case is x > e? Hmm, but the user wrote x ≤ e. Hmm, maybe I should proceed assuming that it's supposed to be x ≥ e, otherwise, the function is not properly defined. So, I think perhaps it's a typo, and the function is f(x) = -x³ + x² when x < e, and a ln x when x ≥ e. That makes more sense because then the function is defined for all real numbers, with a piecewise definition at x = e.So, moving forward with that assumption, f(x) is:f(x) = -x³ + x², when x < e,f(x) = a ln x, when x ≥ e.Alright, so for part (1), we have to discuss the number of zeros of f(x). So, zeros are the points where f(x) = 0.So, first, let's consider the case when x < e. Then f(x) = -x³ + x². Let's set that equal to zero:-x³ + x² = 0Factor out x²:x²(-x + 1) = 0So, the solutions are x² = 0 => x = 0, and -x + 1 = 0 => x = 1.So, for x < e, the zeros are at x = 0 and x = 1.Now, for x ≥ e, f(x) = a ln x. Let's set that equal to zero:a ln x = 0Since a ≠ 0 (given), we can divide both sides by a:ln x = 0 => x = e⁰ = 1.But wait, x = 1 is less than e, so in the domain x ≥ e, x = 1 is not included. Therefore, for x ≥ e, f(x) = a ln x, which is a ln x. Since ln x is positive for x > 1 and negative for x < 1, but in our case, x ≥ e, which is greater than 1, so ln x is positive. Therefore, a ln x is positive when a > 0 and negative when a < 0.But since a ≠ 0, we have to consider both cases.Case 1: a > 0.Then, f(x) = a ln x is positive for x ≥ e, so f(x) = 0 only at x = 1, which is not in x ≥ e. Therefore, for a > 0, f(x) does not have any zeros in x ≥ e.Case 2: a < 0.Similarly, f(x) = a ln x is negative for x ≥ e, because a is negative and ln x is positive. So, f(x) = a ln x is negative for x ≥ e, so f(x) = 0 only at x = 1, which is not in x ≥ e. Therefore, for a < 0, f(x) does not have any zeros in x ≥ e.Therefore, regardless of the value of a (as long as a ≠ 0), the zeros of f(x) are only at x = 0 and x = 1, both in the domain x < e.So, the function f(x) has two zeros: x = 0 and x = 1.Wait, but hold on, x = 0 is included in x < e, so that's fine. So, the number of zeros is two, regardless of a, as long as a ≠ 0.But wait, let me double-check. If a = 0, then f(x) would be defined as -x³ + x² for x < e, and 0 for x ≥ e. But since a ≠ 0, we don't have to consider that case.So, summarizing part (1): The function f(x) has two zeros at x = 0 and x = 1, regardless of the value of a (as long as a ≠ 0).Now, moving on to part (2). It says: If there exist two points M, N on the graph of the function, such that triangle MON is a right-angled triangle with O as the right-angle vertex (where O is the origin), and the midpoint of the hypotenuse MN is exactly on the y-axis, find the range of the real number a.Alright, let's parse this.We have two points M and N on the graph of f(x). So, M and N are points (x, f(x)) for some x in the domain of f.Triangle MON is right-angled at O, meaning that vectors OM and ON are perpendicular. So, the dot product of vectors OM and ON should be zero.Additionally, the midpoint of the hypotenuse MN is on the y-axis. So, the midpoint has an x-coordinate of zero.So, let's denote M as (x1, f(x1)) and N as (x2, f(x2)). Since the midpoint of MN is on the y-axis, the average of x1 and x2 is zero. Therefore:(x1 + x2)/2 = 0 => x1 + x2 = 0 => x2 = -x1.So, N is the reflection of M across the y-axis. Therefore, if M is (t, f(t)), then N is (-t, f(-t)).But wait, f(x) is defined as -x³ + x² for x < e, and a ln x for x ≥ e.So, if t is positive, then f(t) is either -t³ + t² (if t < e) or a ln t (if t ≥ e). Similarly, for x2 = -t, which is negative, f(-t) is -(-t)³ + (-t)², since x2 = -t < e (since t is positive, and e is approximately 2.718, so if t is less than e, then -t is greater than -e, but still negative). Wait, actually, for x < e, f(x) is -x³ + x², regardless of whether x is positive or negative.Wait, hold on. The function is defined as:f(x) = -x³ + x², when x < e,f(x) = a ln x, when x ≥ e.So, for x < e, regardless of whether x is positive or negative, f(x) is -x³ + x². For x ≥ e, f(x) is a ln x.Therefore, if t is positive, then f(t) is either -t³ + t² (if t < e) or a ln t (if t ≥ e). Similarly, for x2 = -t, which is negative, f(-t) is -(-t)³ + (-t)² = t³ + t².Therefore, N is (-t, t³ + t²).So, M is (t, f(t)), N is (-t, t³ + t²).Now, triangle MON is right-angled at O, so vectors OM and ON are perpendicular. Therefore, their dot product is zero.Vectors OM and ON are (t, f(t)) and (-t, t³ + t²), respectively.Dot product = t*(-t) + f(t)*(t³ + t²) = -t² + f(t)*(t³ + t²) = 0.So, the equation is:-t² + f(t)*(t³ + t²) = 0.So, f(t)*(t³ + t²) = t².Therefore, f(t) = t² / (t³ + t²) = t² / [t²(t + 1)] = 1 / (t + 1).So, f(t) = 1 / (t + 1).But f(t) is either -t³ + t² (if t < e) or a ln t (if t ≥ e).So, we have two cases:Case 1: t < e.Then f(t) = -t³ + t² = 1 / (t + 1).So, equation:-t³ + t² = 1 / (t + 1).Multiply both sides by (t + 1):(-t³ + t²)(t + 1) = 1.Let's compute the left-hand side:(-t³)(t + 1) + t²(t + 1) = -t⁴ - t³ + t³ + t² = -t⁴ + t².So, equation becomes:-t⁴ + t² = 1.Bring 1 to the left:-t⁴ + t² - 1 = 0.Multiply both sides by -1:t⁴ - t² + 1 = 0.Now, let's analyze this quartic equation: t⁴ - t² + 1 = 0.Let me make a substitution: let y = t².Then, equation becomes y² - y + 1 = 0.Discriminant D = (-1)² - 4*1*1 = 1 - 4 = -3 < 0.Therefore, no real solutions for y, hence no real solutions for t.Therefore, in Case 1 (t < e), there are no solutions.Case 2: t ≥ e.Then f(t) = a ln t = 1 / (t + 1).So, equation:a ln t = 1 / (t + 1).Therefore, a = 1 / [ (t + 1) ln t ].So, a is expressed in terms of t, where t ≥ e.We need to find the range of a such that this equation has a solution t ≥ e.Therefore, we can define a function h(t) = 1 / [ (t + 1) ln t ] for t ≥ e, and find the range of h(t).So, let's analyze h(t):h(t) = 1 / [ (t + 1) ln t ].We can note that for t ≥ e, ln t ≥ 1, and t + 1 ≥ e + 1.Therefore, h(t) is positive because both numerator and denominator are positive.Moreover, as t increases, (t + 1) increases and ln t increases, so h(t) decreases.Therefore, h(t) is a decreasing function for t ≥ e.So, let's compute the limit as t approaches e from the right:lim_{t→e+} h(t) = 1 / [ (e + 1) ln e ] = 1 / [ (e + 1)*1 ] = 1 / (e + 1).And as t approaches infinity:lim_{t→∞} h(t) = lim_{t→∞} 1 / [ (t + 1) ln t ] = 0.Therefore, h(t) is decreasing from 1/(e + 1) to 0 as t increases from e to infinity.Therefore, the range of h(t) is (0, 1/(e + 1)].Therefore, for a to be in the range (0, 1/(e + 1)], there exists a t ≥ e such that a = h(t).Therefore, the range of a is 0 < a ≤ 1/(e + 1).But wait, the problem says "find the range of the real number a". So, a must be in (0, 1/(e + 1)].But let me check if a can be negative.Wait, in Case 2, t ≥ e, so f(t) = a ln t. Since t ≥ e > 1, ln t is positive. Therefore, f(t) is positive if a > 0, and negative if a < 0.But in the equation f(t) = 1/(t + 1), the right-hand side is positive because t + 1 > 0 and 1 is positive. Therefore, f(t) must be positive, which implies that a must be positive because ln t is positive.Therefore, a must be positive. So, a > 0.Therefore, the range of a is 0 < a ≤ 1/(e + 1).Hence, the range of a is (0, 1/(e + 1)].Wait, but let me double-check.If a is negative, then f(t) = a ln t would be negative, but 1/(t + 1) is positive, so equation f(t) = 1/(t + 1) would have no solution because a negative number can't equal a positive number. Therefore, a must be positive.Therefore, the range is indeed 0 < a ≤ 1/(e + 1).So, summarizing part (2): The range of a is (0, 1/(e + 1)].**Final Answer**(1) The function has two zeros: boxed{2}.(2) The range of the real number ( a ) is boxed{left(0, dfrac{1}{e+1}right]}.

💡Alright, so I have this inequality to solve: ( x < frac{1}{x} ). Hmm, okay. I need to find all the values of ( x ) that make this inequality true. Let me think about how to approach this.First off, I know that when dealing with inequalities involving fractions, I have to be careful about the sign of the denominator because multiplying both sides by a negative number reverses the inequality sign. So, I guess I should consider two cases: one where ( x ) is positive and another where ( x ) is negative. That way, I can handle the multiplication correctly without messing up the inequality direction.**Case 1: ( x > 0 )**If ( x ) is positive, then multiplying both sides of the inequality by ( x ) won't change the direction of the inequality. Let's do that:( x times x < frac{1}{x} times x )Simplifying both sides:( x^2 < 1 )Okay, so now I have a quadratic inequality. To solve ( x^2 < 1 ), I can rewrite it as:( x^2 - 1 < 0 )Factoring the left side:( (x - 1)(x + 1) < 0 )Now, to find where this product is less than zero, I can analyze the intervals determined by the roots ( x = -1 ) and ( x = 1 ). But wait, in this case, we're only considering ( x > 0 ), so the relevant interval is between ( 0 ) and ( 1 ).Testing a value in this interval, say ( x = frac{1}{2} ):( left( frac{1}{2} - 1 right)left( frac{1}{2} + 1 right) = left( -frac{1}{2} right)left( frac{3}{2} right) = -frac{3}{4} < 0 )So, the inequality holds in this interval. Therefore, for ( x > 0 ), the solution is ( 0 < x < 1 ).**Case 2: ( x < 0 )**Now, if ( x ) is negative, multiplying both sides of the inequality by ( x ) will reverse the inequality sign. Let's proceed:( x times x > frac{1}{x} times x ) (Note the inequality sign flipped)Simplifying both sides:( x^2 > 1 )Again, I have a quadratic inequality. Let's rewrite it as:( x^2 - 1 > 0 )Factoring:( (x - 1)(x + 1) > 0 )Now, analyzing the intervals determined by ( x = -1 ) and ( x = 1 ). Since we're considering ( x < 0 ), the relevant interval is ( x < -1 ).Testing a value in this interval, say ( x = -2 ):( (-2 - 1)(-2 + 1) = (-3)(-1) = 3 > 0 )So, the inequality holds in this interval. Therefore, for ( x < 0 ), the solution is ( x < -1 ).**Combining Both Cases**From Case 1, we have ( 0 < x < 1 ), and from Case 2, we have ( x < -1 ). Combining these two intervals gives the complete solution set.So, the solution set is all real numbers less than -1 and all real numbers between 0 and 1.**Final Answer**The solution set of the inequality ( x < frac{1}{x} ) is (boxed{(-infty, -1) cup (0, 1)}).