💡Okay, so I have this problem about complex numbers p, q, and r being zeros of a polynomial Q(z) = z³ + sz + t. The sum of their magnitudes squared is 300, and they form a right triangle with hypotenuse k. I need to find k².First, I remember that for a cubic polynomial like Q(z) = z³ + sz + t, the sum of the roots is equal to the coefficient of z², which is zero in this case. So, p + q + r = 0. That means the centroid of the triangle formed by p, q, and r is at the origin. That might be useful later.Now, the problem mentions that the points p, q, and r form a right triangle. Without loss of generality, I can assume that the right angle is at one of the vertices, say q. So, the triangle has a right angle at q, making the sides pq and qr the legs, and pr the hypotenuse. Wait, no, actually, if the right angle is at q, then the hypotenuse should be the side opposite to q, which would be pr. So, pr is the hypotenuse. Therefore, the length of the hypotenuse k is |p - r|.But I need to find k², which is |p - r|². Hmm, okay.I also know that |p|² + |q|² + |r|² = 300. Maybe I can relate this to the sides of the triangle.Since p + q + r = 0, I can express one of the roots in terms of the other two. Let me express r as r = -p - q. Then, I can write |r|² as | -p - q |², which is |p + q|².So, |p|² + |q|² + |r|² = |p|² + |q|² + |p + q|² = 300.Let me expand |p + q|². That's equal to |p|² + |q|² + 2 Re(p overline{q}). So, substituting back, we have:|p|² + |q|² + (|p|² + |q|² + 2 Re(p overline{q})) = 300Simplify that:2|p|² + 2|q|² + 2 Re(p overline{q}) = 300Divide both sides by 2:|p|² + |q|² + Re(p overline{q}) = 150Hmm, not sure if that's helpful yet. Maybe I need another approach.Since p, q, r form a right triangle with the right angle at q, the vectors pq and qr are perpendicular. In complex numbers, if two vectors are perpendicular, their dot product is zero. But in complex plane terms, that would mean that (p - q) is perpendicular to (r - q). So, the inner product of (p - q) and (r - q) should be zero.In complex numbers, the inner product can be represented as Re[(p - q) overline{(r - q)}] = 0. Alternatively, the imaginary part of (p - q) overline{(r - q)} should be zero because the real part is the dot product and the imaginary part is the cross product. Wait, no, actually, for perpendicular vectors, the inner product is zero, so Re[(p - q) overline{(r - q)}] = 0.But maybe another way is to use the property that if two complex numbers are perpendicular, then one is a purely imaginary multiple of the other. So, (p - q) = i * λ * (r - q) for some real λ. Hmm, but I'm not sure if that's the easiest way.Alternatively, since the triangle is right-angled at q, by the Pythagorean theorem, |p - q|² + |r - q|² = |p - r|². So, that gives me |p - q|² + |r - q|² = |p - r|².But I need to relate this to |p|² + |q|² + |r|² = 300.Wait, let me think about the centroid. Since the centroid is at the origin, the centroid formula is (p + q + r)/3 = 0, so p + q + r = 0. That might help in expressing some relations between p, q, and r.Let me denote vectors from the origin to p, q, r as vectors OP, OQ, OR. Then, the centroid being at the origin means that OP + OQ + OR = 0.Now, in the triangle PQR, the centroid is the average of the three vertices, which is the origin.I also know that in a triangle, the sum of the squares of the sides is equal to three times the sum of the squares of the medians. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use the formula for the sum of the squares of the distances from the centroid.Wait, the formula is that for any point O, the sum |OP|² + |OQ|² + |OR|² = 3|OG|² + |PG|² + |QG|² + |RG|², where G is the centroid. But since G is the origin, |OG|² = 0, so |OP|² + |OQ|² + |OR|² = |PG|² + |QG|² + |RG|².But I'm not sure if that helps here.Alternatively, maybe I can express |p|² + |q|² + |r|² in terms of the sides of the triangle.Wait, let's recall that in any triangle, the sum of the squares of the sides is equal to 2 times the sum of the squares of the medians. But since the centroid is at the origin, maybe the medians relate to the vectors from the origin.Wait, the medians of the triangle PQR are the lines from each vertex to the midpoint of the opposite side. Since the centroid is the origin, each median passes through the origin.So, for example, the median from p goes to the midpoint of qr, which is (q + r)/2. But since p + q + r = 0, (q + r)/2 = (-p)/2. So, the median from p is the line from p to (-p)/2, which is a vector of length (3/2)|p|.Similarly, the length of the median from p is (3/2)|p|, and similarly for the other medians.Wait, so the length of each median is (3/2) times the distance from the origin to the respective vertex.Therefore, the length of the median from p is (3/2)|p|, from q is (3/2)|q|, and from r is (3/2)|r|.But the formula for the sum of the squares of the sides in terms of the medians is:|pq|² + |qr|² + |rp|² = (4/3)(|m_p|² + |m_q|² + |m_r|²)Where m_p, m_q, m_r are the medians from p, q, r respectively.But since |m_p| = (3/2)|p|, then |m_p|² = (9/4)|p|², and similarly for the others.So, substituting:|pq|² + |qr|² + |rp|² = (4/3)( (9/4)|p|² + (9/4)|q|² + (9/4)|r|² ) = (4/3)(9/4)(|p|² + |q|² + |r|² ) = 3(|p|² + |q|² + |r|² )Given that |p|² + |q|² + |r|² = 300, so:|pq|² + |qr|² + |rp|² = 3 * 300 = 900.But in our case, the triangle is right-angled, so by Pythagoras, |pq|² + |qr|² = |pr|².Therefore, |pr|² = |pq|² + |qr|² = (|pq|² + |qr|² + |pr|²) - |pr|² = 900 - |pr|².Wait, that seems a bit circular. Let me think again.Wait, in a right-angled triangle, the sum of the squares of the two legs equals the square of the hypotenuse. So, if the right angle is at q, then |pq|² + |qr|² = |pr|².But we also have that |pq|² + |qr|² + |pr|² = 900.Substituting |pr|² = |pq|² + |qr|² into the sum, we get:(|pq|² + |qr|²) + |pr|² = 900But |pr|² = |pq|² + |qr|², so:(|pq|² + |qr|²) + (|pq|² + |qr|²) = 900Which simplifies to 2(|pq|² + |qr|²) = 900Therefore, |pq|² + |qr|² = 450But since |pr|² = |pq|² + |qr|², then |pr|² = 450.Therefore, k² = |pr|² = 450.Wait, that seems straightforward. Let me verify.We started with |p|² + |q|² + |r|² = 300.Using the centroid property and the relation between the sum of the squares of the sides and the sum of the squares of the medians, we found that the sum of the squares of the sides is 900.Since the triangle is right-angled, one of the sides squared is equal to the sum of the squares of the other two sides. Therefore, the hypotenuse squared is equal to the sum of the squares of the other two sides, which is 450.Therefore, k² = 450.I think that makes sense. Let me see if there's another way to approach it to confirm.Alternatively, since p + q + r = 0, we can write r = -p - q.Then, |p|² + |q|² + |r|² = |p|² + |q|² + |p + q|² = 300.Expanding |p + q|², we get |p|² + |q|² + 2 Re(p overline{q}) = 300.So, |p|² + |q|² + |p|² + |q|² + 2 Re(p overline{q}) = 300Which simplifies to 2|p|² + 2|q|² + 2 Re(p overline{q}) = 300Divide both sides by 2:|p|² + |q|² + Re(p overline{q}) = 150Now, in the triangle, since it's right-angled at q, the vectors pq and qr are perpendicular. So, (p - q) and (r - q) are perpendicular.Expressing r as -p - q, then r - q = -p - 2q.So, (p - q) and (-p - 2q) are perpendicular.Their dot product should be zero. In complex numbers, the dot product is Re[(p - q) overline{(r - q)}] = 0.So, Re[(p - q) overline{(-p - 2q)}] = 0.Let me compute that:(p - q) overline{(-p - 2q)} = (p - q)(- overline{p} - 2 overline{q}) = -p overline{p} - 2p overline{q} + q overline{p} + 2q overline{q}Taking the real part:Re[-|p|² - 2 Re(p overline{q}) + Re(q overline{p}) + 2|q|²] = 0But Re(q overline{p}) = Re(p overline{q}), so:Re[-|p|² - 2 Re(p overline{q}) + Re(p overline{q}) + 2|q|²] = 0Simplify:Re[-|p|² - Re(p overline{q}) + 2|q|²] = 0Which gives:-|p|² - Re(p overline{q}) + 2|q|² = 0From earlier, we had |p|² + |q|² + Re(p overline{q}) = 150Let me denote Re(p overline{q}) as x for simplicity.Then, we have:-|p|² - x + 2|q|² = 0 ...(1)and|p|² + |q|² + x = 150 ...(2)Let me solve these equations.From equation (1):-|p|² - x + 2|q|² = 0 => |p|² = 2|q|² - xSubstitute into equation (2):(2|q|² - x) + |q|² + x = 150Simplify:3|q|² = 150 => |q|² = 50Then, from equation (2):|p|² + 50 + x = 150 => |p|² + x = 100From equation (1):|p|² = 2*50 - x = 100 - xSo, |p|² = 100 - xSubstitute into |p|² + x = 100:(100 - x) + x = 100 => 100 = 100Which is always true, so it doesn't give new information.So, we have |q|² = 50, and |p|² = 100 - x, where x = Re(p overline{q}).But we need to find k² = |p - r|².Since r = -p - q, then p - r = p - (-p - q) = 2p + q.So, |p - r|² = |2p + q|² = 4|p|² + 4 Re(p overline{q}) + |q|²We know |q|² = 50, and |p|² = 100 - x, where x = Re(p overline{q}).So, substitute:|p - r|² = 4(100 - x) + 4x + 50 = 400 - 4x + 4x + 50 = 450So, k² = 450.That confirms the earlier result. Therefore, the answer is 450.

💡Okay, so I need to figure out how many oranges were picked in total. Let me read the problem again to make sure I understand it correctly.Mary picked 14 oranges and Jason picked 41 oranges. Keith picked 38 apples. How many oranges were picked in all?Alright, so Mary and Jason picked oranges, and Keith picked apples. The question is asking about the total number of oranges picked. That means I should focus on the oranges and ignore the apples that Keith picked.First, I'll note down the number of oranges each person picked:- Mary picked 14 oranges.- Jason picked 41 oranges.Now, to find the total number of oranges picked by both Mary and Jason, I need to add these two numbers together.So, 14 (Mary's oranges) plus 41 (Jason's oranges) equals... let me do the math.14 + 41 = 55.Wait, is that right? Let me double-check my addition to be sure.Starting with 14, if I add 40, that would be 54, and then adding the remaining 1 makes it 55. Yep, that seems correct.Just to make sure I didn't miss anything, I'll look back at the problem. It mentions Keith picking 38 apples, but since the question is only about oranges, I don't need to include Keith's apples in my calculation. So, I can safely ignore that part.Therefore, the total number of oranges picked by Mary and Jason together is 55.

💡First, I need to determine how many additional cans Jennifer bought after meeting Mark. She initially bought 40 cans and ended up with 100 cans, so she bought an additional 60 cans.Next, I'll set up the ratio of cans Jennifer bought to the cans Mark bought. For every 5 cans Mark bought, Jennifer bought 6 additional cans. This means Jennifer bought 11 cans for every 5 cans Mark purchased.Using the ratio, I can set up the equation 60 (Jennifer's additional cans) divided by 11 equals Mark's cans divided by 5.Solving for Mark's cans, I multiply both sides by 5 to find that Mark bought 27.27 cans. Since Mark cannot buy a fraction of a can, I round down to the nearest whole number.Therefore, Mark purchased 27 cans of milk.

💡Okay, so I've got this function f(x) = x² - a x ln x - 1, and I need to do two things. First, prove that f(x) + x² f(1/x) = 0. Second, if f(x) has three distinct zeros x₁, x₂, x₃ with x₁ < x₂ < x₃, I need to find the range of a and prove that x₁ + x₃ > 2a - 2.Starting with part (1). I need to show that f(x) + x² f(1/x) equals zero. Let me compute f(1/x) first. So, f(1/x) would be (1/x)² - a*(1/x)*ln(1/x) - 1. Simplify that: (1/x²) - a*(1/x)*(-ln x) - 1, because ln(1/x) is -ln x. So that becomes 1/x² + a ln x / x - 1.Now, multiply f(1/x) by x²: x²*(1/x² + a ln x / x - 1) = 1 + a x ln x - x². So, x² f(1/x) = 1 + a x ln x - x².Now, f(x) is x² - a x ln x - 1. So, f(x) + x² f(1/x) is (x² - a x ln x - 1) + (1 + a x ln x - x²). Let's see: x² - x² cancels, -a x ln x + a x ln x cancels, and -1 +1 cancels. So, indeed, f(x) + x² f(1/x) = 0. That was straightforward.Moving on to part (2). The function f(x) has three distinct zeros. So, I need to find the range of a for which this is true. Also, I have to prove that x₁ + x₃ > 2a - 2.First, let's analyze the function f(x). It's defined for x > 0 since ln x is involved. Let's see its behavior as x approaches 0 and as x approaches infinity.As x approaches 0 from the right, ln x approaches negative infinity. So, the term -a x ln x: if a is positive, then -a x ln x approaches positive infinity because ln x is negative. If a is negative, it approaches negative infinity. The x² term approaches 0, and the constant term is -1. So, overall, as x approaches 0+, f(x) approaches positive infinity if a > 0, negative infinity if a < 0, and -1 if a = 0.As x approaches infinity, x² dominates, so f(x) approaches positive infinity regardless of a.At x = 1, f(1) = 1 - a*1*ln 1 - 1 = 1 - 0 -1 = 0. So, x=1 is always a zero of f(x).So, f(1) = 0. So, x=1 is one of the zeros, say x₂=1. So, the other two zeros are x₁ <1 and x₃ >1.To have three distinct zeros, the function must cross the x-axis three times. So, it must have a local maximum above the x-axis and a local minimum below the x-axis, or vice versa, but given that as x approaches 0, f(x) approaches positive infinity if a>0, and as x approaches infinity, it approaches positive infinity. So, to have three zeros, f(x) must dip below the x-axis somewhere between 0 and 1, and then rise again above the x-axis beyond 1.Therefore, the function must have a local maximum at some point between 0 and 1, and a local minimum at some point beyond 1. For this to happen, the derivative f’(x) must have two critical points: one in (0,1) and one in (1, ∞).Let's compute the derivative f’(x). f(x) = x² - a x ln x -1. So, f’(x) = 2x - a (ln x + 1). Because derivative of x² is 2x, derivative of -a x ln x is -a (ln x +1) by product rule, and derivative of -1 is 0.So, f’(x) = 2x - a (ln x +1). To find critical points, set f’(x) = 0: 2x - a (ln x +1) =0.So, 2x = a (ln x +1). So, x = [a (ln x +1)] / 2.We need to find when this equation has two solutions: one in (0,1) and one in (1, ∞). Because for f(x) to have three zeros, f’(x) must have two critical points.Let me define g(x) = f’(x) = 2x - a (ln x +1). So, g(x) = 2x - a ln x -a.We need to analyze g(x). Let's compute its derivative: g’(x) = 2 - a /x.So, g’(x) = 2 - a/x. Setting this equal to zero: 2 - a/x =0 => x = a/2.So, the function g(x) has a critical point at x = a/2. Now, depending on the value of a, this critical point is in different intervals.If a >0, then x = a/2 is positive. If a <0, x = a/2 is negative, which is outside our domain x>0.So, for a >0, g(x) has a critical point at x = a/2. Let's see the behavior of g(x):As x approaches 0+, ln x approaches -infty, so -a ln x approaches +infty if a>0, so g(x) approaches +infty.As x approaches infinity, 2x dominates, so g(x) approaches +infty.At x = a/2, g(x) has a critical point. Let's compute g(a/2):g(a/2) = 2*(a/2) - a ln(a/2) -a = a - a ln(a/2) -a = -a ln(a/2).So, g(a/2) = -a ln(a/2).For g(x) to have two zeros (i.e., for f’(x)=0 to have two solutions), the function g(x) must dip below zero somewhere. So, the minimum of g(x) must be negative.So, the minimum of g(x) is at x = a/2, which is g(a/2) = -a ln(a/2). For this to be negative, we need -a ln(a/2) <0.Since a >0, because otherwise, as we saw, the critical point is negative, which is outside our domain.So, for a >0, -a ln(a/2) <0 implies that ln(a/2) >0, because -a is negative, so multiplying both sides by -1 reverses the inequality: ln(a/2) >0 => a/2 >1 => a >2.So, if a >2, then g(a/2) <0, which means that g(x) has two zeros: one in (0, a/2) and one in (a/2, ∞). But wait, we need the critical points of f(x) to be in (0,1) and (1, ∞). So, we need x = a/2 to be in (0,1) or in (1, ∞).Wait, if a >2, then a/2 >1. So, the critical point of g(x) is at x = a/2 >1. So, the function g(x) is decreasing on (0, a/2) and increasing on (a/2, ∞). At x = a/2, it reaches a minimum.Since g(a/2) <0, and as x approaches 0+, g(x) approaches +infty, and as x approaches infinity, g(x) approaches +infty, so g(x) must cross zero once in (0, a/2) and once in (a/2, ∞). But since a/2 >1, the second zero is in (a/2, ∞), which is beyond 1.But we need the critical points of f(x) to be in (0,1) and (1, ∞). So, the first zero of g(x) is in (0, a/2). But if a/2 >1, then (0, a/2) includes (0,1). So, the first zero is in (0,1), and the second zero is in (a/2, ∞), which is beyond 1.Therefore, for a >2, f’(x)=0 has two solutions: one in (0,1) and one in (1, ∞). Therefore, f(x) has a local maximum at x=m in (0,1) and a local minimum at x=n in (1, ∞). For f(x) to have three zeros, the local maximum must be above zero and the local minimum must be below zero.We already know that f(1)=0. So, if the local minimum at x=n is below zero, then f(x) will cross the x-axis again beyond x=n. Similarly, if the local maximum at x=m is above zero, f(x) will cross the x-axis again before x=m.But wait, f(1)=0, so if the local minimum at x=n is below zero, then f(x) must cross the x-axis at x=1 and then again at x=n, but since f(x) approaches infinity as x approaches infinity, it must cross again at some x>1. Similarly, if the local maximum at x=m is above zero, f(x) must cross the x-axis at x=m and then again at x=1.Wait, but f(1)=0, so if f(x) has a local maximum above zero at x=m <1, then f(x) must cross the x-axis at x=m and then again at x=1. Similarly, if f(x) has a local minimum below zero at x=n >1, then f(x) must cross the x-axis at x=n and then approach infinity.But wait, f(x) is zero at x=1, so if the local maximum at x=m is above zero, then f(x) must cross the x-axis at x=m, then at x=1, and then again at x=n. But that would require three zeros: x₁ < m <1, x₂=1, and x₃ >n. Wait, but n is the point where f(x) has a local minimum. So, if f(n) <0, then f(x) must cross the x-axis at x=n, but since f(x) approaches infinity as x approaches infinity, it must cross again at some x> n. But we already have x=1 as a zero. So, actually, the three zeros would be x₁ < m <1, x₂=1, and x₃ >n. But wait, that would require f(x) to cross the x-axis at x=1, which is already a zero, but if f(n) <0, then f(x) must cross the x-axis at x=n, but since f(x) approaches infinity as x approaches infinity, it must cross again at some x> n. But that would make four zeros, which is not possible.Wait, maybe I'm getting confused. Let me think again.f(x) is zero at x=1. If f(x) has a local maximum at x=m <1, then f(m) >0. Since f(x) approaches infinity as x approaches 0, and f(m) >0, then f(x) must cross the x-axis at some x <m, say x₁. Then, between x₁ and m, f(x) is decreasing from +infty to f(m) >0, so it doesn't cross zero again. Then, from m to 1, f(x) decreases from f(m) >0 to f(1)=0. So, it crosses zero at x=1. Then, beyond x=1, f(x) has a local minimum at x=n >1. If f(n) <0, then f(x) must cross the x-axis again at some x >n, say x₃. So, in total, three zeros: x₁ <m <1, x₂=1, and x₃ >n.Yes, that makes sense. So, for f(x) to have three zeros, we need f(m) >0 and f(n) <0.We already have that f’(x)=0 has two solutions m in (0,1) and n in (1, ∞) when a >2.Now, we need to ensure that f(m) >0 and f(n) <0.But f(m) is the local maximum at x=m. Since f(m) >0, that's good. f(n) is the local minimum at x=n. We need f(n) <0.So, let's compute f(n). Since n is a critical point, f’(n)=0, so 2n -a (ln n +1)=0 => a = 2n / (ln n +1).So, f(n) = n² - a n ln n -1. Substitute a:f(n) = n² - [2n / (ln n +1)] * n ln n -1 = n² - [2n² ln n / (ln n +1)] -1.Simplify:f(n) = n² - 2n² ln n / (ln n +1) -1.Let me factor n²:f(n) = n² [1 - 2 ln n / (ln n +1)] -1.Simplify the expression in the brackets:1 - 2 ln n / (ln n +1) = [ (ln n +1) - 2 ln n ] / (ln n +1) = (1 - ln n) / (ln n +1).So, f(n) = n² (1 - ln n) / (ln n +1) -1.We need f(n) <0:n² (1 - ln n) / (ln n +1) -1 <0.Let me denote t = ln n. Since n >1, t >0.So, f(n) becomes:n² (1 - t) / (t +1) -1 <0.But n = e^{t}, so n² = e^{2t}.Thus, f(n) = e^{2t} (1 - t)/(t +1) -1 <0.So, e^{2t} (1 - t)/(t +1) <1.We need to find t >0 such that e^{2t} (1 - t)/(t +1) <1.Let me define h(t) = e^{2t} (1 - t)/(t +1). We need h(t) <1.Compute h(0): e^{0}*(1-0)/(0+1)=1*1/1=1.Compute h(t) as t increases:For t approaching 0+, h(t) approaches 1.Compute h(t) for t=1: e^{2}*(1-1)/(1+1)=0.For t=0.5: e^{1}*(1 -0.5)/(0.5 +1)= e*(0.5)/(1.5)= e*(1/3)≈0.906 <1.For t=0.2: e^{0.4}*(0.8)/(1.2)= e^{0.4}*(2/3)≈1.4918*(0.6667)≈1.0 >1.Wait, so at t=0.2, h(t)≈1.0, which is equal to 1. At t=0.5, h(t)≈0.906 <1. So, there must be a t between 0.2 and 0.5 where h(t)=1.Let me solve h(t)=1:e^{2t} (1 - t)/(t +1) =1.This is a transcendental equation, probably can't be solved analytically. But we can see that for t >0, h(t) starts at 1 when t=0, decreases to 0 as t approaches infinity, and has a maximum somewhere.Wait, actually, at t=0, h(t)=1. As t increases, h(t) first increases? Wait, no, at t=0.2, h(t)≈1.0, at t=0.5, h(t)≈0.906. So, it's decreasing from t=0 onwards.Wait, let me compute the derivative of h(t):h(t) = e^{2t} (1 - t)/(t +1).Let me compute h’(t):h’(t) = d/dt [e^{2t} (1 - t)/(t +1)].Use product rule:= e^{2t} * 2 * (1 - t)/(t +1) + e^{2t} * [ -1*(t +1) - (1 - t)*1 ] / (t +1)^2.Simplify:= 2 e^{2t} (1 - t)/(t +1) + e^{2t} [ - (t +1) - (1 - t) ] / (t +1)^2.Simplify the numerator in the second term:- (t +1) - (1 - t) = -t -1 -1 + t = -2.So, h’(t) = 2 e^{2t} (1 - t)/(t +1) - 2 e^{2t} / (t +1)^2.Factor out 2 e^{2t} / (t +1)^2:= 2 e^{2t} / (t +1)^2 [ (1 - t)(t +1) -1 ].Expand (1 - t)(t +1):= (1)(t +1) - t(t +1) = t +1 - t² - t = 1 - t².So, h’(t) = 2 e^{2t} / (t +1)^2 [ (1 - t²) -1 ] = 2 e^{2t} / (t +1)^2 [ -t² ].Thus, h’(t) = -2 t² e^{2t} / (t +1)^2 <0 for t >0.So, h(t) is strictly decreasing for t >0. Therefore, h(t) starts at 1 when t=0 and decreases to 0 as t approaches infinity. So, h(t)=1 only at t=0, and for t>0, h(t) <1.Wait, but earlier, when I computed h(0.2), I thought it was approximately 1.0, but actually, let me recalculate:At t=0.2:h(t)= e^{0.4}*(1 -0.2)/(0.2 +1)= e^{0.4}*(0.8)/(1.2)= e^{0.4}*(2/3).e^{0.4}≈1.4918, so 1.4918*(2/3)≈0.9945 <1. So, actually, h(t) is always less than 1 for t>0. So, h(t) <1 for all t>0.Therefore, f(n) = h(t) -1 <0 for all t>0, which means f(n) <0 for all n>1.Therefore, for a >2, f(x) has three zeros: x₁ <1, x₂=1, and x₃ >1.Wait, but what about a=2? Let's check.If a=2, then f’(x)=2x -2(ln x +1)=2x -2 ln x -2.Set f’(x)=0: 2x -2 ln x -2=0 => x - ln x -1=0.Let me solve x - ln x -1=0.At x=1: 1 -0 -1=0. So, x=1 is a critical point.Compute the second derivative f''(x)=2 - a/x. For a=2, f''(x)=2 -2/x.At x=1, f''(1)=2 -2=0. Hmm, inconclusive.Wait, let's analyze f’(x) for a=2.f’(x)=2x -2 ln x -2.Compute f’(x) at x approaching 0+: 2x approaches 0, -2 ln x approaches +infty, so f’(x) approaches +infty.At x=1: f’(1)=2 -0 -2=0.As x increases beyond 1, f’(x)=2x -2 ln x -2. Let's see its behavior.Compute f’(2)=4 -2 ln 2 -2≈4 -1.386 -2≈0.614>0.Compute f’(e)=2e -2*1 -2≈5.436 -2 -2≈1.436>0.So, f’(x) is increasing for x>1 because f''(x)=2 -2/x>0 when x>1.Since f’(1)=0 and f’(x) increases beyond x=1, f’(x) >0 for x>1.Similarly, for x <1, let's see:At x approaching 0+, f’(x) approaches +infty.At x=1, f’(1)=0.Compute f’(0.5)=1 -2 ln 0.5 -2=1 -2*(-0.693) -2≈1 +1.386 -2≈0.386>0.So, f’(x) is positive for x <1 as well, except at x=1 where it is zero.Therefore, for a=2, f’(x) ≥0 for all x>0, with f’(1)=0. So, f(x) is increasing on (0,1) and increasing on (1, ∞). Therefore, f(x) has only one zero at x=1.Thus, for a=2, f(x) has only one zero. Therefore, the range of a for which f(x) has three distinct zeros is a >2.So, part (2)(i): a ∈ (2, ∞).Now, part (2)(ii): Prove that x₁ +x₃ >2a -2.Given that x₂=1, and x₁ <1, x₃ >1.We need to relate x₁ and x₃. From part (1), we have f(x) +x² f(1/x)=0.So, f(x) = -x² f(1/x).Since x₁ is a zero of f(x), f(x₁)=0. Therefore, f(1/x₁) = - (1/x₁²) f(x₁)=0. So, 1/x₁ is also a zero of f(x). But since x₁ <1, 1/x₁ >1. So, x₃=1/x₁.Therefore, x₃=1/x₁. So, x₁ x₃=1.We need to prove that x₁ +x₃ >2a -2.Given that x₃=1/x₁, so x₁ +x₃ =x₁ +1/x₁.We need to show that x₁ +1/x₁ >2a -2.Let me denote y =x₁. Then, y +1/y >2a -2.Since y <1, because x₁ <1.We need to express a in terms of y.From f(y)=0: y² -a y ln y -1=0.So, a y ln y = y² -1.Thus, a= (y² -1)/(y ln y).Note that ln y <0 because y <1.So, a= (y² -1)/(y ln y). Since y² -1 <0 (because y <1), and ln y <0, so a= positive number.So, a= (1 - y²)/(-y ln y)= (1 - y²)/(y |ln y|).We need to express 2a -2 in terms of y:2a -2= 2*(1 - y²)/(y |ln y|) -2.We need to show that y +1/y >2*(1 - y²)/(y |ln y|) -2.Let me rearrange the inequality:y +1/y +2 > 2*(1 - y²)/(y |ln y|).Multiply both sides by y (since y>0):y² +1 +2y > 2*(1 - y²)/|ln y|.Note that |ln y|= -ln y because y <1.So, the inequality becomes:y² +2y +1 > 2*(1 - y²)/(-ln y).Simplify the right-hand side:2*(1 - y²)/(-ln y)=2*(y² -1)/ln y.But ln y <0, so y² -1 <0, so the right-hand side is positive.Wait, let me write it as:y² +2y +1 > 2*(1 - y²)/(-ln y).Let me denote t = -ln y >0 because y <1.Then, ln y = -t, so y = e^{-t}.So, y² = e^{-2t}, 1 - y²=1 - e^{-2t}.So, the inequality becomes:e^{-2t} + 2 e^{-t} +1 > 2*(1 - e^{-2t})/t.Simplify the left-hand side:e^{-2t} + 2 e^{-t} +1 = (e^{-t} +1)^2.So, (e^{-t} +1)^2 > 2*(1 - e^{-2t})/t.Note that 1 - e^{-2t}= (1 - e^{-t})(1 + e^{-t}).So, 2*(1 - e^{-2t})/t= 2*(1 - e^{-t})(1 + e^{-t})/t.Thus, the inequality is:(e^{-t} +1)^2 > 2*(1 - e^{-t})(1 + e^{-t})/t.Simplify the right-hand side:2*(1 - e^{-t})(1 + e^{-t})/t= 2*(1 - e^{-2t})/t.Wait, but we already have that. So, perhaps another approach.Let me consider the function φ(t)= (e^{-t} +1)^2 - 2*(1 - e^{-2t})/t.We need to show that φ(t) >0 for t>0.Compute φ(t):φ(t)= e^{-2t} + 2 e^{-t} +1 - 2*(1 - e^{-2t})/t.Let me rearrange:φ(t)= e^{-2t} + 2 e^{-t} +1 - 2/t + 2 e^{-2t}/t.Combine like terms:φ(t)= (1 + e^{-2t}) + 2 e^{-t} - 2/t + 2 e^{-2t}/t.Hmm, not sure if that helps. Maybe consider expanding e^{-t} and e^{-2t} as power series.Recall that e^{-t}=1 -t + t²/2 - t³/6 + t⁴/24 -...Similarly, e^{-2t}=1 -2t + 2t² - (4t³)/3 + (8t⁴)/12 -...So, let's compute φ(t):φ(t)= (1 -2t + 2t² - (4t³)/3 + ... ) + 2*(1 -t + t²/2 - t³/6 + ...) +1 - 2/t + 2*(1 -2t + 2t² - (4t³)/3 + ...)/t.Wait, this might get messy, but let's try up to t² terms.Compute each term:e^{-2t}=1 -2t + 2t² + O(t³).2 e^{-t}=2*(1 -t + t²/2)=2 -2t + t².So, e^{-2t} +2 e^{-t}= (1 -2t +2t²) + (2 -2t +t²)=3 -4t +3t².Then, add 1: 3 -4t +3t² +1=4 -4t +3t².Now, compute -2/t + 2 e^{-2t}/t:-2/t + 2*(1 -2t +2t²)/t= -2/t + 2/t -4 +4t= (-2 +2)/t + (-4 +4t)=0 + (-4 +4t)=4t -4.So, φ(t)= (4 -4t +3t²) + (4t -4)=4 -4t +3t² +4t -4=3t².So, up to t² terms, φ(t)=3t²>0 for t≠0.Therefore, for small t>0, φ(t)>0.But we need to ensure that φ(t)>0 for all t>0.Wait, but when t increases, does φ(t) remain positive?Let me check for t=1:φ(1)= (e^{-1} +1)^2 - 2*(1 - e^{-2})/1≈(0.3679 +1)^2 -2*(1 -0.1353)= (1.3679)^2 -2*(0.8647)≈1.871 -1.729≈0.142>0.For t=2:φ(2)= (e^{-2} +1)^2 -2*(1 - e^{-4})/2≈(0.1353 +1)^2 - (1 -0.0183)= (1.1353)^2 -0.9817≈1.289 -0.9817≈0.307>0.For t approaching infinity:e^{-t} approaches 0, so φ(t)≈(0 +1)^2 -2*(1 -0)/t=1 -2/t→1>0.So, φ(t) approaches 1 as t→∞.Therefore, φ(t) >0 for all t>0.Thus, (e^{-t} +1)^2 > 2*(1 - e^{-2t})/t for all t>0.Which implies that y +1/y +2 > 2*(1 - y²)/(-ln y).Which is equivalent to y +1/y >2a -2.Therefore, x₁ +x₃ >2a -2.Hence, proved.

💡Okay, so I need to figure out how many natural numbers greater than 9 but less than 100 are relatively prime to 30. Hmm, let's break this down step by step.First, what does it mean for two numbers to be relatively prime? I remember that two numbers are relatively prime if their greatest common divisor (GCD) is 1. So, in this case, I need to find numbers between 10 and 99 that have a GCD of 1 with 30.Alright, let's think about the number 30. Its prime factors are 2, 3, and 5. So, any number that doesn't share these prime factors with 30 will be relatively prime to it. That means I need to exclude numbers that are multiples of 2, 3, or 5 from the range 10 to 99.But wait, just excluding multiples of 2, 3, and 5 isn't enough because some numbers might be multiples of more than one of these primes. For example, 6 is a multiple of both 2 and 3. If I just subtract the counts of multiples of 2, 3, and 5, I might be double-counting some numbers. So, I need a better approach to avoid that.I think the Inclusion-Exclusion Principle is what I need here. This principle helps in counting the number of elements in the union of multiple sets by including and excluding the intersections appropriately. Let me recall how it works.For three sets A, B, and C, the number of elements in their union is:|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|In this context, set A can be the set of multiples of 2, set B the set of multiples of 3, and set C the set of multiples of 5. So, I need to calculate the sizes of these sets and their intersections within the range 10 to 99.First, let's determine how many numbers are in the range 10 to 99. Since 99 - 10 + 1 = 90, there are 90 numbers in total.Now, let's find the number of multiples of 2, 3, and 5 in this range.Starting with multiples of 2:The smallest multiple of 2 greater than 9 is 10, and the largest multiple of 2 less than 100 is 98. To find how many multiples of 2 there are, I can use the formula:Number of multiples of 2 = floor(99/2) - floor(9/2)Calculating that:floor(99/2) = 49 (since 49*2=98)floor(9/2) = 4 (since 4*2=8)So, number of multiples of 2 = 49 - 4 = 45Next, multiples of 3:The smallest multiple of 3 greater than 9 is 12, and the largest multiple of 3 less than 100 is 99. Using the same formula:Number of multiples of 3 = floor(99/3) - floor(9/3)Calculating:floor(99/3) = 33 (since 33*3=99)floor(9/3) = 3 (since 3*3=9)So, number of multiples of 3 = 33 - 3 = 30Now, multiples of 5:The smallest multiple of 5 greater than 9 is 10, and the largest multiple of 5 less than 100 is 95. Again, using the formula:Number of multiples of 5 = floor(99/5) - floor(9/5)Calculating:floor(99/5) = 19 (since 19*5=95)floor(9/5) = 1 (since 1*5=5)So, number of multiples of 5 = 19 - 1 = 18Alright, so we have 45 multiples of 2, 30 multiples of 3, and 18 multiples of 5.But as I thought earlier, some numbers are multiples of more than one of these, so I need to subtract the overlaps.First, let's find the number of multiples of both 2 and 3, which is multiples of 6.Number of multiples of 6 = floor(99/6) - floor(9/6)Calculating:floor(99/6) = 16 (since 16*6=96)floor(9/6) = 1 (since 1*6=6)So, number of multiples of 6 = 16 - 1 = 15Next, multiples of both 2 and 5, which is multiples of 10.Number of multiples of 10 = floor(99/10) - floor(9/10)Calculating:floor(99/10) = 9 (since 9*10=90)floor(9/10) = 0 (since 0*10=0)So, number of multiples of 10 = 9 - 0 = 9Then, multiples of both 3 and 5, which is multiples of 15.Number of multiples of 15 = floor(99/15) - floor(9/15)Calculating:floor(99/15) = 6 (since 6*15=90)floor(9/15) = 0 (since 0*15=0)So, number of multiples of 15 = 6 - 0 = 6Now, I need to consider the overlap where numbers are multiples of all three, which is multiples of 30 (since 30 is the least common multiple of 2, 3, and 5).Number of multiples of 30 = floor(99/30) - floor(9/30)Calculating:floor(99/30) = 3 (since 3*30=90)floor(9/30) = 0 (since 0*30=0)So, number of multiples of 30 = 3 - 0 = 3Now, applying the Inclusion-Exclusion Principle:Total numbers not relatively prime to 30 = (multiples of 2 + multiples of 3 + multiples of 5) - (multiples of 6 + multiples of 10 + multiples of 15) + multiples of 30Plugging in the numbers:Total = (45 + 30 + 18) - (15 + 9 + 6) + 3Total = (93) - (30) + 3Total = 93 - 30 + 3Total = 66So, there are 66 numbers between 10 and 99 that are not relatively prime to 30.But we need the numbers that are relatively prime to 30. Since there are 90 numbers in total, we subtract the 66 that are not relatively prime:Relatively prime numbers = Total numbers - Numbers not relatively primeRelatively prime numbers = 90 - 66Relatively prime numbers = 24Wait, that seems a bit low. Let me double-check my calculations.First, the total numbers: 90. That's correct because 99 - 10 + 1 = 90.Multiples of 2: 45, multiples of 3: 30, multiples of 5: 18. That adds up to 93.Subtracting the overlaps: multiples of 6:15, multiples of 10:9, multiples of 15:6. That's 30.Adding back the multiples of 30:3.So, 93 - 30 + 3 = 66. That seems right.Therefore, 90 - 66 = 24. Hmm, okay, maybe it is correct.But I feel like 24 is a small number. Let me think differently. Maybe I can use Euler's Totient Function to verify.Euler's Totient Function φ(n) gives the number of integers less than n that are relatively prime to n. But in this case, our range is from 10 to 99, which is 90 numbers. However, φ(30) gives the count of numbers less than 30 that are relatively prime to 30.φ(30) = 30 * (1 - 1/2) * (1 - 1/3) * (1 - 1/5) = 30 * 1/2 * 2/3 * 4/5 = 8Wait, that's 8 numbers less than 30 that are relatively prime to 30. But our range is from 10 to 99, which is more than 30. Maybe I can use φ(30) to find the density and then apply it to the range.Density of numbers relatively prime to 30 is φ(30)/30 = 8/30 = 4/15.So, in 90 numbers, the expected count would be 90 * (4/15) = 24. That matches my previous result.Okay, so that seems to confirm that 24 is the correct number.Therefore, there are 24 natural numbers greater than 9 but less than 100 that are relatively prime to 30.

💡First, I need to determine the monthly incomes of M, N, and O based on the given information.I know that the average income of M and N is Rs. 5050. This means that the total income of M and N together is 2 times 5050, which is 10100.Next, the average monthly income of M and O is Rs. 5200. Therefore, the total income of M and O together is 2 times 5200, which is 10400.I also know that the income of M is Rs. 4000. Using this information, I can find the income of N by subtracting M's income from the total income of M and N: 10100 minus 4000 equals 6100.Similarly, I can find the income of O by subtracting M's income from the total income of M and O: 10400 minus 4000 equals 6400.Now that I have the incomes of N and O, I can calculate their average monthly income by adding their incomes together and dividing by 2: (6100 + 6400) divided by 2 equals 6250.Therefore, the average monthly income of N and O is Rs. 6250.