💡Okay, so I have this problem with three parts, all related to the quadratic function f(x) = ax² + (b - 1)x + 1, where a and b are real numbers and a is positive. Let me try to tackle each part step by step.**Part I: Finding f(x) given x₁ = 1 and symmetry about x = 2**First, the problem states that x₁ = 1 is a root of f(x). So, plugging x = 1 into f(x) should give zero.f(1) = a(1)² + (b - 1)(1) + 1 = a + b - 1 + 1 = a + b = 0.So, from this, I get the equation: a + b = 0. That's equation (1).Next, it says that for any x ∈ ℝ, f(2 - x) = f(2 + x). This means the function is symmetric about the line x = 2. For a quadratic function, the axis of symmetry is given by x = -B/(2A), where the quadratic is in the form Ax² + Bx + C.In our case, the quadratic is f(x) = ax² + (b - 1)x + 1, so the axis of symmetry is x = -(b - 1)/(2a). Since the function is symmetric about x = 2, this must equal 2.So, -(b - 1)/(2a) = 2.Let me write that as equation (2): -(b - 1) = 4a => -b + 1 = 4a => -b = 4a - 1 => b = -4a + 1.Now, from equation (1): a + b = 0. Substitute b from equation (2) into this.a + (-4a + 1) = 0 => a - 4a + 1 = 0 => -3a + 1 = 0 => -3a = -1 => a = 1/3.Then, b = -4*(1/3) + 1 = -4/3 + 3/3 = (-4 + 3)/3 = -1/3.So, a = 1/3 and b = -1/3. Therefore, the function f(x) is:f(x) = (1/3)x² + (-1/3 - 1)x + 1 = (1/3)x² - (4/3)x + 1.Wait, let me double-check that:f(x) = ax² + (b - 1)x + 1.a = 1/3, b = -1/3.So, f(x) = (1/3)x² + (-1/3 - 1)x + 1 = (1/3)x² - (4/3)x + 1. Yeah, that seems correct.**Part II: Determining if f(x) = |2x - a| + 2 has negative real roots when b = 2a - 3**Given that b = 2a - 3, we need to see if the equation f(x) = |2x - a| + 2 has negative real roots. If it does, find the range of these negative roots.First, let's write f(x) with b = 2a - 3:f(x) = ax² + (2a - 3 - 1)x + 1 = ax² + (2a - 4)x + 1.So, f(x) = ax² + (2a - 4)x + 1.We need to solve the equation:ax² + (2a - 4)x + 1 = |2x - a| + 2.Let me denote the right-hand side as |2x - a| + 2. Since absolute value functions can be split into cases, I'll consider two cases based on the expression inside the absolute value.**Case 1: 2x - a ≥ 0 ⇒ x ≥ a/2**In this case, |2x - a| = 2x - a. So, the equation becomes:ax² + (2a - 4)x + 1 = 2x - a + 2.Simplify the right-hand side: 2x - a + 2.So, moving all terms to the left:ax² + (2a - 4)x + 1 - 2x + a - 2 = 0.Combine like terms:ax² + (2a - 4 - 2)x + (1 + a - 2) = 0.Simplify:ax² + (2a - 6)x + (a - 1) = 0.So, quadratic equation: ax² + (2a - 6)x + (a - 1) = 0.Let me compute the discriminant to see if real roots exist.Discriminant D = (2a - 6)² - 4*a*(a - 1).Compute:(2a - 6)² = 4a² - 24a + 36.4*a*(a - 1) = 4a² - 4a.So, D = (4a² - 24a + 36) - (4a² - 4a) = -24a + 36 + 4a = (-20a) + 36.So, D = -20a + 36.For real roots, D ≥ 0: -20a + 36 ≥ 0 ⇒ 20a ≤ 36 ⇒ a ≤ 36/20 = 9/5 = 1.8.But since a > 0, a ∈ (0, 1.8].But in this case, x ≥ a/2, so if a is positive, x is positive. So, in this case, any roots would be positive, which we are not interested in for this part. So, negative roots can't come from this case.**Case 2: 2x - a < 0 ⇒ x < a/2**Here, |2x - a| = -(2x - a) = -2x + a.So, the equation becomes:ax² + (2a - 4)x + 1 = -2x + a + 2.Move all terms to the left:ax² + (2a - 4)x + 1 + 2x - a - 2 = 0.Combine like terms:ax² + (2a - 4 + 2)x + (1 - a - 2) = 0.Simplify:ax² + (2a - 2)x + (-a - 1) = 0.So, quadratic equation: ax² + (2a - 2)x - (a + 1) = 0.Compute discriminant D:D = (2a - 2)² - 4*a*(-a - 1).Compute:(2a - 2)² = 4a² - 8a + 4.-4*a*(-a - 1) = 4a² + 4a.So, D = (4a² - 8a + 4) + (4a² + 4a) = 8a² - 4a + 4.Since D = 8a² - 4a + 4, let's check if it's always positive.Compute discriminant of D as a quadratic in a: D' = (-4)² - 4*8*4 = 16 - 128 = -112 < 0.Since the discriminant is negative, D is always positive for all real a. So, the quadratic equation has two real roots.Now, we need to check if these roots are negative.Given that x < a/2, and a > 0, so a/2 is positive. So, if the roots are less than a/2, but we need to see if they are negative.Let me denote the roots as x₁ and x₂.Using quadratic formula:x = [-(2a - 2) ± sqrt(8a² - 4a + 4)] / (2a).Simplify numerator:-(2a - 2) = -2a + 2.So, x = (-2a + 2 ± sqrt(8a² - 4a + 4)) / (2a).Let me factor numerator:= [ -2(a - 1) ± sqrt(8a² - 4a + 4) ] / (2a).Let me write sqrt(8a² - 4a + 4) as sqrt(4*(2a² - a + 1)) = 2*sqrt(2a² - a + 1).So, x = [ -2(a - 1) ± 2*sqrt(2a² - a + 1) ] / (2a) = [ - (a - 1) ± sqrt(2a² - a + 1) ] / a.So, two roots:x₁ = [ - (a - 1) + sqrt(2a² - a + 1) ] / a,x₂ = [ - (a - 1) - sqrt(2a² - a + 1) ] / a.We need to check if these roots are negative.First, let's analyze x₁:x₁ = [ - (a - 1) + sqrt(2a² - a + 1) ] / a.Let me compute numerator:- (a - 1) + sqrt(2a² - a + 1) = -a + 1 + sqrt(2a² - a + 1).Is this positive or negative?Let me denote N = -a + 1 + sqrt(2a² - a + 1).We can check if N is positive or negative.Compute N:sqrt(2a² - a + 1) is always positive, since 2a² - a + 1 is always positive (discriminant: (-1)^2 - 8 = -7 < 0).So, sqrt(2a² - a + 1) > |a - 1|?Wait, let's compute N:N = sqrt(2a² - a + 1) - (a - 1).Let me square both sides to compare:[sqrt(2a² - a + 1)]² = 2a² - a + 1,(a - 1)² = a² - 2a + 1.So, 2a² - a + 1 vs a² - 2a + 1.Compute difference:2a² - a + 1 - (a² - 2a + 1) = a² + a.Since a > 0, a² + a > 0. So, sqrt(2a² - a + 1) > |a - 1|.But since sqrt(2a² - a + 1) is always positive, and (a - 1) can be positive or negative depending on a.Wait, when a > 1, (a - 1) is positive, so N = sqrt(2a² - a + 1) - (a - 1).Since sqrt(2a² - a + 1) > sqrt(a² - 2a + 1) = |a - 1|, so sqrt(2a² - a + 1) > a - 1 when a > 1.So, N = sqrt(...) - (a - 1) > 0.When a < 1, (a - 1) is negative, so N = sqrt(...) - (negative) = sqrt(...) + positive, which is definitely positive.When a = 1, N = sqrt(2 - 1 + 1) - 0 = sqrt(2) > 0.So, in all cases, N > 0. Therefore, x₁ = N / a > 0, since a > 0.Therefore, x₁ is positive.Now, x₂ = [ - (a - 1) - sqrt(2a² - a + 1) ] / a.Compute numerator:- (a - 1) - sqrt(2a² - a + 1) = -a + 1 - sqrt(2a² - a + 1).Let me denote M = -a + 1 - sqrt(2a² - a + 1).Is M positive or negative?Compute M:sqrt(2a² - a + 1) > 0, so M = (-a + 1) - sqrt(2a² - a + 1).Let me see if (-a + 1) can be positive or negative.If a < 1, then (-a + 1) > 0.If a ≥ 1, then (-a + 1) ≤ 0.But sqrt(2a² - a + 1) is always positive.So, when a < 1:M = positive - positive. Which could be positive or negative.When a ≥ 1:M = non-positive - positive = negative.So, let's check when a < 1:Is (-a + 1) > sqrt(2a² - a + 1)?Let me square both sides:(-a + 1)² = a² - 2a + 1,[ sqrt(2a² - a + 1) ]² = 2a² - a + 1.Compare a² - 2a + 1 vs 2a² - a + 1.Difference: (a² - 2a + 1) - (2a² - a + 1) = -a² - a.Since a > 0, -a² - a < 0. So, a² - 2a + 1 < 2a² - a + 1.Therefore, (-a + 1) < sqrt(2a² - a + 1).So, M = (-a + 1) - sqrt(...) < 0.Therefore, in all cases, M < 0.Thus, x₂ = M / a < 0, since a > 0.Therefore, x₂ is negative.So, in Case 2, we have one negative root x₂ and one positive root x₁.But in Case 1, we had roots only when a ≤ 1.8, but those roots are positive.So, overall, the equation f(x) = |2x - a| + 2 has exactly one negative real root, which is x₂.Now, we need to find the range of this negative root.Let me express x₂ in terms of a:x₂ = [ - (a - 1) - sqrt(2a² - a + 1) ] / a.Let me write it as:x₂ = [ -a + 1 - sqrt(2a² - a + 1) ] / a.Let me factor out a negative sign:x₂ = - [ (a - 1) + sqrt(2a² - a + 1) ] / a.Let me denote t = 1/a, since a > 0, t > 0.But maybe it's better to express in terms of t = something else.Alternatively, let me make substitution:Let me set t = a - 1/2, but not sure.Alternatively, let me consider t = sqrt(2a² - a + 1).Wait, perhaps it's better to analyze the expression.Let me denote S = sqrt(2a² - a + 1).So, x₂ = [ -a + 1 - S ] / a = (-a + 1)/a - S/a = (-1 + 1/a) - S/a.Let me write it as:x₂ = (-1 + 1/a) - sqrt(2a² - a + 1)/a.Let me factor 1/a inside the sqrt:sqrt(2a² - a + 1)/a = sqrt( (2a² - a + 1)/a² ) = sqrt(2 - 1/a + 1/a²).So, x₂ = (-1 + 1/a) - sqrt(2 - 1/a + 1/a²).Let me set t = 1/a. Since a > 0, t > 0.So, x₂ = (-1 + t) - sqrt(2 - t + t²).So, x₂ = t - 1 - sqrt(t² - t + 2).Now, we need to find the range of x₂ as t varies over (0, ∞).Let me denote x₂(t) = t - 1 - sqrt(t² - t + 2).We need to find the range of x₂(t) for t > 0.Let me analyze the function x₂(t):x₂(t) = t - 1 - sqrt(t² - t + 2).Let me compute its derivative to see if it's increasing or decreasing.Compute dx₂/dt:d/dt [t - 1 - sqrt(t² - t + 2)] = 1 - (1/(2*sqrt(t² - t + 2)))*(2t - 1).Simplify:= 1 - (2t - 1)/(2*sqrt(t² - t + 2)).Let me denote D = 2*sqrt(t² - t + 2).So, derivative is:1 - (2t - 1)/D.We can analyze the sign of the derivative.Let me compute:1 - (2t - 1)/D.Let me set this equal to zero to find critical points:1 = (2t - 1)/D ⇒ D = 2t - 1.But D = 2*sqrt(t² - t + 2).So, 2*sqrt(t² - t + 2) = 2t - 1.Square both sides:4*(t² - t + 2) = (2t - 1)².Compute:Left: 4t² - 4t + 8.Right: 4t² - 4t + 1.Set equal:4t² - 4t + 8 = 4t² - 4t + 1 ⇒ 8 = 1, which is impossible.So, no critical points. Therefore, the derivative does not cross zero.Let me check the sign of the derivative.Compute derivative at t = 1:D = 2*sqrt(1 - 1 + 2) = 2*sqrt(2).Derivative = 1 - (2*1 - 1)/(2*sqrt(2)) = 1 - (1)/(2*sqrt(2)) ≈ 1 - 0.3535 ≈ 0.6465 > 0.So, derivative is positive at t = 1.Since there are no critical points, and derivative is positive at t = 1, and as t approaches 0 and infinity, let's see:As t → 0+:x₂(t) = t - 1 - sqrt(t² - t + 2) ≈ 0 - 1 - sqrt(0 - 0 + 2) = -1 - sqrt(2) ≈ -2.414.Derivative as t → 0+:D = 2*sqrt(0 - 0 + 2) = 2*sqrt(2).Derivative = 1 - (0 - 1)/(2*sqrt(2)) = 1 + 1/(2*sqrt(2)) > 0.So, derivative is positive as t approaches 0.As t → ∞:x₂(t) = t - 1 - sqrt(t² - t + 2).Approximate sqrt(t² - t + 2) ≈ t - 1/2 + ... (using binomial expansion).So, x₂(t) ≈ t - 1 - (t - 1/2) = -1 + 1/2 = -1/2.But let's compute more accurately:sqrt(t² - t + 2) = t*sqrt(1 - 1/t + 2/t²) ≈ t*(1 - 1/(2t) + (2)/(2t²) + ... ) = t - 1/2 + 1/t + ... .So, x₂(t) ≈ t - 1 - (t - 1/2 + 1/t) = t -1 - t + 1/2 - 1/t = -1 + 1/2 - 1/t = -1/2 - 1/t.As t → ∞, x₂(t) approaches -1/2.So, as t increases, x₂(t) approaches -1/2 from below.Since the derivative is always positive, x₂(t) is increasing from -1 - sqrt(2) ≈ -2.414 towards -1/2.Therefore, the range of x₂ is (-1 - sqrt(2), -1/2).So, the negative real root x₂ lies in the interval (-1 - sqrt(2), -1/2).**Part III: Finding the minimum value of h(a) when a ≥ 2, x₂ - x₁ = 2, and g(x) = -f(x) + 2(x₂ - x) has maximum h(a) on (x₁, x₂)**Given:- a ≥ 2- x₂ - x₁ = 2- g(x) = -f(x) + 2(x₂ - x)- Find the minimum value of h(a), where h(a) is the maximum of g(x) on (x₁, x₂)First, let's recall that f(x) = ax² + (b - 1)x + 1.Given that x₁ and x₂ are roots, so f(x) = a(x - x₁)(x - x₂).Given that x₂ - x₁ = 2, so x₂ = x₁ + 2.Let me denote x₁ = p, so x₂ = p + 2.So, f(x) = a(x - p)(x - p - 2).Expanding:f(x) = a[(x - p)(x - p - 2)] = a[(x - p)^2 - 2(x - p)] = a(x² - 2px + p² - 2x + 2p).But maybe it's better to keep it as f(x) = a(x - p)(x - p - 2).Now, g(x) = -f(x) + 2(x₂ - x) = -a(x - p)(x - p - 2) + 2((p + 2) - x).Simplify:g(x) = -a(x - p)(x - p - 2) + 2(p + 2 - x).Let me expand the first term:-a(x - p)(x - p - 2) = -a[(x - p)(x - p - 2)] = -a[(x - p)^2 - 2(x - p)].So, g(x) = -a(x - p)^2 + 2a(x - p) + 2(p + 2 - x).Let me simplify term by term:First term: -a(x - p)^2.Second term: +2a(x - p).Third term: +2(p + 2 - x) = 2p + 4 - 2x.So, combine all terms:g(x) = -a(x - p)^2 + 2a(x - p) + 2p + 4 - 2x.Let me collect like terms:Let me expand -a(x - p)^2:= -a(x² - 2px + p²) = -a x² + 2a p x - a p².Then, +2a(x - p) = +2a x - 2a p.Then, +2p + 4 - 2x.So, combining all:g(x) = (-a x² + 2a p x - a p²) + (2a x - 2a p) + (2p + 4 - 2x).Combine like terms:- x² term: -a x².x terms: 2a p x + 2a x - 2x = (2a p + 2a - 2) x.Constants: -a p² - 2a p + 2p + 4.So, g(x) = -a x² + (2a p + 2a - 2) x + (-a p² - 2a p + 2p + 4).But this seems complicated. Maybe there's a better way.Alternatively, let me write g(x) as:g(x) = -a(x - p)(x - p - 2) + 2(p + 2 - x).Let me factor out terms:= -a(x - p)(x - p - 2) - 2(x - p - 2).Wait, notice that 2(p + 2 - x) = -2(x - p - 2).So, g(x) = -a(x - p)(x - p - 2) - 2(x - p - 2).Factor out (x - p - 2):g(x) = (x - p - 2)(-a(x - p) - 2).So, g(x) = (x - p - 2)(-a(x - p) - 2).Let me write it as:g(x) = (x - p - 2)(-a(x - p) - 2) = (x - p - 2)(-a x + a p - 2).Alternatively, factor out negative sign:= -(x - p - 2)(a x - a p + 2).But maybe it's better to write it as:g(x) = (x - p - 2)(-a(x - p) - 2).Let me denote u = x - p.Then, g(x) = (u - 2)(-a u - 2).So, g(u) = (u - 2)(-a u - 2).Expand this:= -a u(u - 2) - 2(u - 2) = -a u² + 2a u - 2u + 4.So, g(u) = -a u² + (2a - 2)u + 4.Now, since u = x - p, and x ∈ (x₁, x₂) = (p, p + 2), so u ∈ (0, 2).So, we need to find the maximum of g(u) = -a u² + (2a - 2)u + 4 on u ∈ (0, 2).This is a quadratic function in u, opening downward (since coefficient of u² is -a < 0).Therefore, the maximum occurs at the vertex.The vertex of a quadratic au² + bu + c is at u = -b/(2a). Wait, but here the quadratic is -a u² + (2a - 2)u + 4.So, the vertex is at u = -(2a - 2)/(2*(-a)) = (2a - 2)/(2a) = (a - 1)/a.So, u = (a - 1)/a.We need to check if this u is within (0, 2).Since a ≥ 2, (a - 1)/a = 1 - 1/a.Since a ≥ 2, 1/a ≤ 1/2, so 1 - 1/a ≥ 1 - 1/2 = 1/2.Also, 1 - 1/a < 1, since 1/a > 0.So, u = (a - 1)/a ∈ (1/2, 1) ⊂ (0, 2). Therefore, the maximum occurs at u = (a - 1)/a.Compute g(u) at this point:g((a - 1)/a) = -a*((a - 1)/a)^2 + (2a - 2)*((a - 1)/a) + 4.Let me compute each term:First term: -a*((a - 1)^2 / a²) = - (a - 1)^2 / a.Second term: (2a - 2)*(a - 1)/a = [2(a - 1)]*(a - 1)/a = 2(a - 1)^2 / a.Third term: +4.So, combining:g = - (a - 1)^2 / a + 2(a - 1)^2 / a + 4 = [ - (a - 1)^2 + 2(a - 1)^2 ] / a + 4 = (a - 1)^2 / a + 4.So, h(a) = (a - 1)^2 / a + 4.Simplify:= (a² - 2a + 1)/a + 4 = a - 2 + 1/a + 4 = a + 1/a + 2.So, h(a) = a + 1/a + 2.Now, we need to find the minimum value of h(a) for a ≥ 2.Compute derivative of h(a):h'(a) = 1 - 1/a².Set derivative equal to zero to find critical points:1 - 1/a² = 0 ⇒ 1 = 1/a² ⇒ a² = 1 ⇒ a = ±1.But since a ≥ 2, the critical point a = 1 is not in our domain.Therefore, h(a) is increasing for a ≥ 2, because h'(a) = 1 - 1/a² > 0 for a > 1.Since a ≥ 2, h(a) is increasing, so the minimum occurs at a = 2.Compute h(2):h(2) = 2 + 1/2 + 2 = 4 + 0.5 = 4.5 = 9/2.Therefore, the minimum value of h(a) is 9/2.**Final Answer**(I) boxed{dfrac{1}{3}x^2 - dfrac{4}{3}x + 1}(II) The equation has negative real roots in the range boxed{(-1 - sqrt{2}, -dfrac{1}{2})}(III) The minimum value of ( h(a) ) is boxed{dfrac{9}{2}}

💡Okay, so I've got this equation to solve: (x^2 - 4lfloor x rfloor + 3 = 0), where (x) is a real number. I need to figure out how many solutions this equation has. The options are 2, 3, 1, or 4 solutions. Hmm, let's break this down step by step.First, I remember that (lfloor x rfloor) is the floor function, which means it gives the greatest integer less than or equal to (x). So, for example, if (x = 2.3), then (lfloor x rfloor = 2). This is important because it means (lfloor x rfloor) is always an integer, and it's related to the interval in which (x) lies.Let me denote (lfloor x rfloor) as (n), where (n) is an integer. So, (n leq x < n + 1). Substituting this into the equation, I get:(x^2 - 4n + 3 = 0)Which simplifies to:(x^2 = 4n - 3)Now, since (x^2) is always non-negative, the right side of the equation must also be non-negative. So, (4n - 3 geq 0), which implies:(4n geq 3)(n geq frac{3}{4})But (n) is an integer, so the smallest possible value for (n) is 1. That means (n) can be 1, 2, 3, and so on. However, I should also consider the upper limit because (x^2) can't be infinitely large, but since (x) is real, we'll see how far this goes.Now, for each integer (n), I can solve (x^2 = 4n - 3) and check if the solutions lie within the interval ([n, n + 1)). Let's start with (n = 1):1. For (n = 1): (x^2 = 4(1) - 3 = 1) So, (x = pm 1) But since (x) must be in ([1, 2)), only (x = 1) is valid.2. For (n = 2): (x^2 = 4(2) - 3 = 5) So, (x = pm sqrt{5}) (sqrt{5}) is approximately 2.236, which is in ([2, 3)), so (x = sqrt{5}) is valid. The negative root is -2.236, which is not in ([2, 3)), so it's invalid.3. For (n = 3): (x^2 = 4(3) - 3 = 9) So, (x = pm 3) (x = 3) is in ([3, 4)), so it's valid. The negative root is -3, which is not in ([3, 4)), so it's invalid.4. For (n = 4): (x^2 = 4(4) - 3 = 13) So, (x = pm sqrt{13}) (sqrt{13}) is approximately 3.606, which is in ([4, 5))? Wait, no, 3.606 is actually in ([3, 4)), but (n = 4) implies (x) should be in ([4, 5)). So, 3.606 is not in ([4, 5)), so this solution is invalid. The negative root is -3.606, which is not in ([4, 5)), so it's also invalid.Wait, that seems confusing. Let me double-check. If (n = 4), then (x) should be in ([4, 5)), but (sqrt{13}) is approximately 3.606, which is actually less than 4. So, it doesn't fall into the interval for (n = 4). Therefore, there's no solution for (n = 4).Let me try (n = 5):5. For (n = 5): (x^2 = 4(5) - 3 = 17) So, (x = pm sqrt{17}) (sqrt{17}) is approximately 4.123, which is in ([5, 6))? No, it's actually in ([4, 5)), so it doesn't fall into the interval for (n = 5). The negative root is -4.123, which is not in ([5, 6)), so it's invalid.Hmm, it seems like as (n) increases, the positive root (sqrt{4n - 3}) is always less than (n + 1), so it doesn't fall into the interval ([n, n + 1)). Let me verify this.Suppose (x = sqrt{4n - 3}). We need (x geq n) for it to be in the interval ([n, n + 1)). So:(sqrt{4n - 3} geq n)Squaring both sides:(4n - 3 geq n^2)(n^2 - 4n + 3 leq 0)Solving the quadratic inequality:(n^2 - 4n + 3 = 0)Using the quadratic formula:(n = frac{4 pm sqrt{16 - 12}}{2} = frac{4 pm 2}{2})So, (n = 3) or (n = 1). Therefore, the inequality (n^2 - 4n + 3 leq 0) holds for (1 leq n leq 3).This means that for (n = 1, 2, 3), the positive root (sqrt{4n - 3}) is greater than or equal to (n), so it falls into the interval ([n, n + 1)). For (n > 3), (sqrt{4n - 3}) is less than (n), so it doesn't fall into the interval ([n, n + 1)).Therefore, we only need to consider (n = 1, 2, 3).Let's summarize:- For (n = 1): (x = 1) is valid.- For (n = 2): (x = sqrt{5}) is valid.- For (n = 3): (x = 3) is valid.So, we have three solutions: (x = 1), (x = sqrt{5}), and (x = 3).Wait, but earlier when I tried (n = 4), I thought (sqrt{13}) was approximately 3.606, which is actually in ([3, 4)), but (n = 4) requires (x) to be in ([4, 5)). So, (sqrt{13}) is not in ([4, 5)), hence invalid. Similarly, for higher (n), the roots don't fall into the required intervals.Therefore, the only valid solutions are for (n = 1, 2, 3), giving us three real solutions.I think that's it. So, the equation has three solutions.

💡Alright, so I've got this problem here about n elements forming n pairs, and there's this condition about when two elements form a pair. It says that if and only if two elements a_i and a_j form a pair, then their corresponding pairs p_i and p_j have a common element. And I need to prove that each element belongs to exactly two pairs. Hmm, okay, let me try to unpack this.First, let me make sure I understand the setup correctly. We have n elements, let's call them a1, a2, ..., an. These elements form n pairs, which are p1, p2, ..., pn. So each pair pi is a pair of elements from the set {a1, a2, ..., an}. Now, the condition is that if and only if two elements ai and aj form a pair, then their corresponding pairs pi and pj share a common element.Wait, that seems a bit circular. Let me rephrase it: For any two elements ai and aj, they form a pair (i.e., they are paired together in one of the pairs p1 to pn) if and only if their corresponding pairs pi and pj have a common element. So, if ai and aj are paired together, then pi and pj share an element. Conversely, if pi and pj share an element, then ai and aj must be paired together.Okay, so this is a kind of bi-conditional relationship between the pairing of elements and the sharing of elements in their corresponding pairs. Interesting. So, the structure here is such that the pairing of elements directly relates to the intersection of their pairs.Now, I need to prove that each element belongs to exactly two pairs. That is, for each ai, there are exactly two pairs among p1 to pn that contain ai. So, each element is paired with exactly two other elements. Hmm, that sounds like it's forming some kind of regular graph, where each vertex has degree two. In graph theory, a regular graph of degree two is a collection of cycles. So, maybe this structure is a union of cycles.But let's not jump to conclusions yet. Let me try to approach this step by step.First, let's model this situation. Let's consider the elements a1, a2, ..., an as vertices in a graph. Each pair pi is an edge connecting two vertices. So, we have a graph with n vertices and n edges. Now, the condition given is that two vertices ai and aj are adjacent (i.e., connected by an edge) if and only if their corresponding pairs pi and pj share a common element.Wait, that seems a bit confusing. Let me think again. Each pair pi is a pair of elements, so pi = {a_k, a_l} for some k and l. So, pi is an edge between a_k and a_l. Now, the condition is that ai and aj form a pair (i.e., there exists some pi such that pi = {ai, aj}) if and only if pi and pj share a common element.Wait, that doesn't make sense because pi and pj are pairs, so they share a common element only if they share one of the elements a1 to an. But if pi and pj share a common element, that means that the pairs pi and pj have one element in common. So, for example, if pi = {ai, ak} and pj = {aj, ak}, then pi and pj share the element ak.But according to the condition, this sharing happens if and only if ai and aj form a pair. So, if ai and aj are paired together, then their pairs pi and pj share a common element. Conversely, if pi and pj share a common element, then ai and aj must be paired together.Wait, that seems a bit tangled. Let me try to formalize it.Let me denote the pairs as follows: For each i, pi is a pair {ai, aj} for some j. So, each pi is an unordered pair of two distinct elements from the set {a1, a2, ..., an}.Now, the condition is: For any i and j, ai and aj form a pair (i.e., there exists some k such that pk = {ai, aj}) if and only if pi and pj share a common element.Wait, that seems a bit off because pi and pj are pairs, so they share a common element only if they share one of the elements a1 to an. But if ai and aj form a pair, then pi and pj share a common element. But pi is {ai, something} and pj is {aj, something else}, so unless pi and pj share a common element, which would mean that pi = {ai, ak} and pj = {aj, ak} for some k, then pi and pj share ak.But according to the condition, this sharing happens if and only if ai and aj form a pair. So, if ai and aj are paired together, then pi and pj share a common element. Conversely, if pi and pj share a common element, then ai and aj must be paired together.Wait, that seems like it's defining a kind of incidence structure. Maybe it's a projective plane or something similar, but I'm not sure. Let me think differently.Let me consider the incidence between elements and pairs. Each element ai is in some number of pairs. Let's denote the number of pairs that ai is in as di. So, di is the degree of ai in this graph.Now, the total number of pairs is n, and each pair contains two elements, so the sum of all di is 2n. That's because each pair contributes to the degree of two elements.Now, the condition given is that two elements ai and aj form a pair if and only if their corresponding pairs pi and pj share a common element. So, if ai and aj are paired together, then pi and pj share a common element. Conversely, if pi and pj share a common element, then ai and aj must be paired together.Wait, that seems like it's saying that the adjacency of two elements is equivalent to their pairs sharing a common element. So, in graph terms, two vertices are adjacent if and only if their corresponding edges share a common vertex.But in a simple graph, edges don't share vertices unless they are incident to the same vertex. So, if two edges share a common vertex, that means they are both incident to that vertex. So, in this case, if pi and pj share a common element, say ak, then both pi and pj are incident to ak. So, pi = {ai, ak} and pj = {aj, ak}.But according to the condition, this sharing implies that ai and aj form a pair. So, ai and aj must be paired together, meaning that there exists some pair pk = {ai, aj}.Wait, that seems like it's defining a kind of triangle in the graph. If pi and pj share ak, then ai and aj are connected through ak, and also directly connected by pk.But in a simple graph, this would mean that ai, aj, and ak form a triangle. So, each time two edges share a common vertex, the two endpoints of those edges are also connected, forming a triangle.But in our case, we have n vertices and n edges, so it's a graph with n vertices and n edges. Now, in a connected graph, the number of edges is at least n-1, so n edges would imply that it's connected and has exactly one cycle, making it a unicyclic graph.But if every two edges that share a common vertex also have their endpoints connected, forming a triangle, then the graph must be such that every edge is part of a triangle. But in a unicyclic graph, there's only one cycle, so unless the cycle is a triangle, this might not hold.Wait, maybe I'm overcomplicating this. Let me try a different approach.Let me consider the incidence between elements and pairs. Each element ai is in di pairs, and each pair pi contains two elements. So, the total number of incidences is 2n, as each pair contributes two elements.Now, the condition says that two elements ai and aj form a pair if and only if their corresponding pairs pi and pj share a common element. So, if ai and aj are paired together, then pi and pj share a common element. Conversely, if pi and pj share a common element, then ai and aj must be paired together.Wait, that seems like it's defining a kind of dual relationship between elements and pairs. Maybe it's a kind of incidence geometry where elements and pairs are duals.But perhaps I can model this as a bipartite graph, where one partition is the set of elements {a1, a2, ..., an} and the other partition is the set of pairs {p1, p2, ..., pn}. Then, an element ai is connected to a pair pj if ai is in pj.But according to the condition, two elements ai and aj are connected (i.e., there is a pair containing both) if and only if their corresponding pairs pi and pj share a common element. Wait, that seems a bit circular.Alternatively, maybe I can think of this as a graph where the vertices are the elements, and the edges are the pairs. Then, the condition is that two vertices are adjacent if and only if their corresponding edges share a common vertex.Wait, that's interesting. So, in graph terms, this would mean that the adjacency of two vertices is equivalent to their corresponding edges sharing a common vertex. That seems like a kind of self-dual property.But I'm not sure if that's helpful. Let me think about small cases to get some intuition.Let's take n=3. So, we have three elements a1, a2, a3, and three pairs p1, p2, p3. Each pair is a pair of elements. Now, according to the condition, two elements form a pair if and only if their corresponding pairs share a common element.Wait, but with n=3, each pair must consist of two elements, so p1, p2, p3 would be the three possible pairs: {a1,a2}, {a1,a3}, {a2,a3}. Now, according to the condition, for any two elements, say a1 and a2, they form a pair (i.e., p1 = {a1,a2}) if and only if p1 and p2 share a common element. But p1 = {a1,a2} and p2 = {a1,a3} share a1, so according to the condition, a1 and a2 form a pair, which they do. Similarly, p1 and p3 share a2, so a1 and a3 form a pair, which they do.Wait, but in this case, each element is in exactly two pairs. For example, a1 is in p1 and p2, a2 is in p1 and p3, and a3 is in p2 and p3. So, each element is in exactly two pairs, which is what we're supposed to prove.Okay, so for n=3, it works out. Let's try n=4.For n=4, we have four elements a1, a2, a3, a4, and four pairs p1, p2, p3, p4. Now, each pair is a pair of elements, so we have to choose four pairs out of the six possible pairs.But according to the condition, two elements form a pair if and only if their corresponding pairs share a common element. Hmm, let's see.Suppose we have p1 = {a1,a2}, p2 = {a1,a3}, p3 = {a1,a4}, and p4 = {a2,a3}. Now, let's check the condition.For a1 and a2, they form a pair p1. Do p1 and p2 share a common element? Yes, a1. Similarly, p1 and p3 share a1, and p1 and p4 share a2. But according to the condition, a1 and a2 form a pair if and only if p1 and p2 share a common element. Wait, but p1 and p2 share a1, which is true, and a1 and a2 do form a pair. Similarly, p1 and p4 share a2, and a1 and a2 do form a pair.But wait, p2 and p3 share a1, so according to the condition, a2 and a3 should form a pair, but in our setup, a2 and a3 are paired in p4, which is correct. Similarly, p2 and p4 share a2, so a1 and a3 should form a pair, which they do in p2.Wait, but in this setup, a1 is in three pairs: p1, p2, p3. But according to the problem statement, each element should be in exactly two pairs. So, this setup doesn't satisfy the condition because a1 is in three pairs.Hmm, so maybe my initial choice of pairs is incorrect. Let me try a different setup.Suppose we have p1 = {a1,a2}, p2 = {a2,a3}, p3 = {a3,a4}, p4 = {a4,a1}. Now, this forms a cycle: a1-a2-a3-a4-a1. Now, let's check the condition.For a1 and a2, they form a pair p1. Do p1 and p2 share a common element? Yes, a2. Similarly, p1 and p4 share a1. So, according to the condition, a1 and a2 form a pair, which they do.For a2 and a3, they form a pair p2. Do p2 and p1 share a2? Yes. Do p2 and p3 share a3? Yes. So, according to the condition, a2 and a3 form a pair, which they do.Similarly, for a3 and a4, they form a pair p3. Do p3 and p2 share a3? Yes. Do p3 and p4 share a4? Yes. So, a3 and a4 form a pair, which they do.For a4 and a1, they form a pair p4. Do p4 and p3 share a4? Yes. Do p4 and p1 share a1? Yes. So, a4 and a1 form a pair, which they do.Now, in this setup, each element is in exactly two pairs: a1 is in p1 and p4, a2 is in p1 and p2, a3 is in p2 and p3, and a4 is in p3 and p4. So, this satisfies the condition that each element is in exactly two pairs.But wait, in this case, the pairs form a cycle, which is a 4-cycle. So, each element is in exactly two pairs, and the condition is satisfied.But in my first attempt, I had a1 in three pairs, which didn't satisfy the condition. So, it seems that the structure needs to be such that each element is in exactly two pairs, forming a cycle, to satisfy the given condition.Okay, so for n=4, it works when the pairs form a cycle, and each element is in exactly two pairs. Now, let's see if this generalizes.Suppose we have n elements, and the pairs form a single cycle, where each element is paired with two others, forming a cycle of length n. Then, each element is in exactly two pairs, and the condition is satisfied because any two elements that are paired together share a common pair, and any two pairs that share a common element correspond to two elements that are paired together.Wait, but in the n=4 case, the pairs form a 4-cycle, and each element is in two pairs. Similarly, for n=3, the pairs form a triangle, and each element is in two pairs.So, it seems that the structure is a cycle where each element is connected to two others, forming a single cycle of length n. In this case, each element is in exactly two pairs, satisfying the condition.But the problem states that there are n pairs, which in the case of a cycle of length n, we have n edges, which correspond to the n pairs. So, this seems to fit.But wait, in the n=3 case, the cycle is a triangle, which has three edges, and in the n=4 case, the cycle has four edges. So, in general, for n elements, forming a cycle of length n would require n edges, which correspond to the n pairs.But in a cycle of length n, each element is in exactly two pairs, which is what we need to prove.But I need to make sure that this is the only possibility. Is there any other way to arrange the pairs such that the condition holds without each element being in exactly two pairs?Suppose, for example, that one element is in three pairs. Then, according to the condition, the pairs that include this element must share a common element, which is the element itself. But then, any two pairs that include this element would share it, which would imply that the corresponding elements are paired together. But if the element is in three pairs, say p1, p2, p3, then p1 and p2 share the element, so the other elements in p1 and p2 must be paired together. Similarly, p1 and p3 share the element, so the other elements in p1 and p3 must be paired together, and p2 and p3 share the element, so the other elements in p2 and p3 must be paired together.But this would create a structure where the other elements are all paired together, leading to multiple pairs, which might not be possible without creating a contradiction in the number of pairs or the degrees of the elements.Alternatively, if an element is in only one pair, then according to the condition, its pair would not share any common elements with any other pair, which would mean that the element is isolated, which contradicts the fact that we have n pairs and n elements, implying that the graph is connected.Wait, but if the graph is connected, then it must have at least n-1 edges, but we have exactly n edges, so it's a connected graph with exactly one cycle, making it a unicyclic graph. In a unicyclic graph, all vertices except those on the cycle have degree one, but in our case, we need all vertices to have degree two, which would imply that the graph is a single cycle.Therefore, the only way to satisfy the condition is if the graph is a single cycle where each element is in exactly two pairs.So, putting this all together, the structure must be a single cycle of length n, where each element is paired with two others, forming exactly two pairs per element. Therefore, each element belongs to exactly two pairs, which is what we needed to prove.I think that makes sense. Let me try to summarize the steps:1. Model the problem as a graph where elements are vertices and pairs are edges.2. The condition given translates to the graph being such that two vertices are adjacent if and only if their corresponding edges share a common vertex.3. This implies that the graph must be a cycle, as any other structure would either not satisfy the condition or lead to contradictions in the degrees of the vertices.4. In a cycle of length n, each vertex has degree two, meaning each element is in exactly two pairs.5. Therefore, each element belongs to exactly two pairs, satisfying the problem's condition.I think that covers it. I should probably write this up more formally, but this is the gist of the reasoning.

💡First, I need to determine how many ways there are to select 4 boys out of the 10 available. This can be calculated using the combination formula, which is:[binom{10}{4} = frac{10!}{4!(10-4)!} = frac{10 times 9 times 8 times 7}{4 times 3 times 2 times 1} = 210]Next, I need to determine how many ways there are to select 3 girls out of the 10 available. Again, using the combination formula:[binom{10}{3} = frac{10!}{3!(10-3)!} = frac{10 times 9 times 8}{3 times 2 times 1} = 120]Finally, to find the total number of ways to form the team, I multiply the number of ways to choose the boys by the number of ways to choose the girls:[210 times 120 = 25200]Therefore, the total number of ways to select the team is 25,200.

💡Okay, so I need to compare the sizes of two expressions: (2^{0.3}) and (0.3^{0.2}). Hmm, exponents can sometimes be tricky, especially when dealing with decimals. Let me think about how to approach this.First, I remember that when the base is greater than 1, like 2 in this case, the function (2^x) is an increasing function. That means as (x) increases, (2^x) also increases. Since 0.3 is greater than 0, (2^{0.3}) should be greater than (2^0), which is 1. So, (2^{0.3} > 1).On the other hand, the second expression is (0.3^{0.2}). Here, the base is between 0 and 1, which means the function (0.3^x) is a decreasing function. So, as (x) increases, (0.3^x) decreases. Since 0.2 is less than 1, (0.3^{0.2}) should be greater than (0.3^1), which is 0.3. But wait, does that mean it's greater than 0.3? Let me double-check.Actually, since the function is decreasing, a smaller exponent should result in a larger value. So, (0.3^{0.2}) is greater than (0.3^1), which is 0.3, but how does it compare to 1? Well, any number between 0 and 1 raised to a positive power will still be less than 1. So, (0.3^{0.2}) is less than 1.Putting it all together, (2^{0.3}) is greater than 1, and (0.3^{0.2}) is less than 1. Therefore, (2^{0.3}) must be greater than (0.3^{0.2}).Wait, let me make sure I didn't make a mistake. Maybe I should calculate approximate values to confirm. Let's see, (2^{0.3}) can be calculated using logarithms or a calculator. If I approximate, (2^{0.3}) is roughly (e^{0.3 ln 2}). Since (ln 2) is approximately 0.693, so (0.3 times 0.693) is about 0.2079. Then, (e^{0.2079}) is roughly 1.23.For (0.3^{0.2}), similarly, it's (e^{0.2 ln 0.3}). (ln 0.3) is approximately -1.2039, so (0.2 times (-1.2039)) is about -0.2408. Then, (e^{-0.2408}) is roughly 0.786.So, (2^{0.3} approx 1.23) and (0.3^{0.2} approx 0.786). Comparing these, 1.23 is definitely greater than 0.786. Therefore, my initial conclusion seems correct.I think I've got it. (2^{0.3}) is larger than (0.3^{0.2}).

💡Okay, so I have this problem where I need to show that two specific integrals of a characteristic function φ(t) are also characteristic functions. The integrals are:1. ∫₀¹ φ(ut) du2. ∫_{ℝ₊} e^{-u} φ(ut) duHmm, characteristic functions are Fourier transforms of probability measures, right? So, if I can express these integrals as expectations of exponentials, then they would be characteristic functions of some random variables.Let me start with the first integral: ∫₀¹ φ(ut) du.Since φ(t) is a characteristic function, I can write φ(t) = E[e^{itξ}] for some random variable ξ. So, φ(ut) would be E[e^{iutξ}]. Therefore, the integral becomes:∫₀¹ E[e^{iutξ}] duI think I can interchange the integral and the expectation because of Fubini's theorem, assuming everything is nice and measurable. So, this becomes:E[∫₀¹ e^{iutξ} du]Now, let's compute the integral inside the expectation. The integral of e^{iutξ} with respect to u from 0 to 1. Let me make a substitution: let v = ut, so dv = t du, but maybe it's simpler to just integrate directly.∫₀¹ e^{iutξ} du = [ (e^{iutξ}) / (i t ξ) ) ] from 0 to 1.So, that would be (e^{itξ} - 1) / (i t ξ).Therefore, the expectation becomes:E[ (e^{itξ} - 1) / (i t ξ) ]Hmm, this looks familiar. I think this is the expectation of (e^{itξ} - 1)/(i t ξ). Maybe this can be interpreted as the characteristic function of another random variable.Wait, if I have another random variable, say η, which is uniformly distributed on [0,1], independent of ξ, then the expectation E[e^{itξη}] would be equal to E[ E[e^{itξη} | ξ] ] = E[ ∫₀¹ e^{itξu} du ] which is exactly what I have here.So, E[e^{itξη}] = ∫₀¹ φ(ut) du. Therefore, this integral is the characteristic function of the random variable ξη, where η is uniform on [0,1]. So, that's the first part done.Now, moving on to the second integral: ∫_{ℝ₊} e^{-u} φ(ut) du.Again, φ(ut) = E[e^{iutξ}], so substituting that in, we have:∫_{ℝ₊} e^{-u} E[e^{iutξ}] duInterchanging the integral and expectation, assuming Fubini's theorem applies, we get:E[ ∫_{ℝ₊} e^{-u} e^{iutξ} du ]Simplify the integrand:∫_{ℝ₊} e^{-(1 - itξ)u} duThis is the integral of an exponential function. The integral of e^{-au} du from 0 to ∞ is 1/a, provided that Re(a) > 0.Here, a = 1 - itξ. The real part of a is 1, which is positive, so the integral converges.Thus, the integral becomes:1 / (1 - itξ)So, the expectation is E[1 / (1 - itξ)].Hmm, what does this expectation represent? If I think of another random variable, say ζ, which is exponentially distributed with parameter 1, independent of ξ, then E[e^{itξζ}] would be equal to E[ E[e^{itξζ} | ξ] ] = E[ ∫_{ℝ₊} e^{-u} e^{iutξ} du ] which is exactly what I have here.Therefore, E[e^{itξζ}] = ∫_{ℝ₊} e^{-u} φ(ut) du. So, this integral is the characteristic function of the random variable ξζ, where ζ is exponentially distributed with parameter 1.So, both integrals are indeed characteristic functions of some random variables, which are products of ξ with independent uniform or exponential random variables.I think that makes sense. I used the fact that integrating a characteristic function against another measure (uniform or exponential) corresponds to taking the expectation with respect to that measure, effectively creating a new random variable as a product. Since characteristic functions are closed under such operations, the results are also characteristic functions.**Final Answer**Both integrals are characteristic functions, so the final answers are:boxed{int_{0}^{1} varphi(ut) , du}andboxed{int_{mathbb{R}_{+}} e^{-u} varphi(ut) , du}