🧠:[<think>Okay, let me try to solve these two problems step by step. Starting with part (1): We need to find the adjoint vector OM of the function g(x) = cos(x + π/3) + cos x and then compute the magnitude of this vector. First, let's recall that the adjoint vector of a function a sin x + b cos x is the vector (a, b). So, if we can express g(x) in the form a sin x + b cos x, then the coefficients of sin x and cos x will give us the components of the vector OM.Given g(x) = cos(x + π/3) + cos x. Let's expand cos(x + π/3) using the cosine addition formula: cos(A + B) = cos A cos B - sin A sin B. So, cos(x + π/3) = cos x cos(π/3) - sin x sin(π/3). We know that cos(π/3) = 1/2 and sin(π/3) = √3/2. Plugging these values in:cos(x + π/3) = cos x * (1/2) - sin x * (√3/2) = (1/2) cos x - (√3/2) sin x.Therefore, g(x) = [ (1/2) cos x - (√3/2) sin x ] + cos x. Let's combine like terms:The cos x terms: (1/2 cos x + cos x) = (3/2 cos x)The sin x term: - (√3/2 sin x)So, g(x) can be written as - (√3/2) sin x + (3/2) cos x. Comparing this with the standard form a sin x + b cos x, we have a = -√3/2 and b = 3/2. Wait, hold on. The standard form given is a sin x + b cos x. So, in g(x), the coefficient of sin x is -√3/2 and the coefficient of cos x is 3/2. Therefore, the adjoint vector OM is (a, b) = (-√3/2, 3/2). Now, we need to find the magnitude of this vector. The magnitude |OM| is sqrt(a² + b²). Let's compute that:a² = (-√3/2)² = (3/4)b² = (3/2)² = 9/4Therefore, |OM| = sqrt(3/4 + 9/4) = sqrt(12/4) = sqrt(3). Wait, 3/4 + 9/4 = 12/4 = 3, so sqrt(3). That's straightforward. So, part (1) seems done. The adjoint vector is (-√3/2, 3/2) and its magnitude is sqrt(3). Let me check if I did the expansion correctly. Starting with cos(x + π/3) + cos x.Yes, expanding cos(x + π/3) gives (1/2 cos x - √3/2 sin x) + cos x. Then combining cos x terms: 1/2 + 1 = 3/2. The sin x term is -√3/2. So, coefficients are a = -√3/2 (sin x coefficient) and b = 3/2 (cos x coefficient). Then magnitude is sqrt( (√3/2)^2 + (3/2)^2 ) which is sqrt( 3/4 + 9/4 ) = sqrt(12/4) = sqrt(3). Correct.So part (1) answer: OM = (-√3/2, 3/2) and |OM| = sqrt(3).Moving on to part (2): The vector ON = (1, √3) has an adjoint function φ(x). We need to find the range of real number k such that the inequality φ(x) + k φ(x + π/2) > 0 holds for all x in [0, 11π/12].First, let's find φ(x). Since the adjoint function of vector (a, b) is a sin x + b cos x. Here, ON = (1, √3), so φ(x) = 1*sin x + √3*cos x.Let me write φ(x) = sin x + √3 cos x. Maybe we can express this in amplitude-phase form for easier manipulation. Let's recall that A sin x + B cos x = C sin(x + δ) or C cos(x - θ). Let's do that.Compute the amplitude C = sqrt(A² + B²) where A = 1, B = √3. So C = sqrt(1 + 3) = sqrt(4) = 2.Then, to write φ(x) = 2 sin(x + δ). Let's compute δ. Since A = C cos δ, B = C sin δ. Wait, if we use the identity:A sin x + B cos x = C sin(x + δ), then:C sin(x + δ) = C sin x cos δ + C cos x sin δ. Comparing coefficients:A = C cos δB = C sin δTherefore, tan δ = B/A. So tan δ = √3 / 1 = √3. So δ = π/3. Therefore, φ(x) = 2 sin(x + π/3).Alternatively, if written as cosine: A sin x + B cos x = C cos(x - θ). Let me check that as well.Expressing φ(x) as a cosine function:C cos(x - θ) = C cos x cos θ + C sin x sin θ.Comparing with φ(x) = sin x + √3 cos x:Coefficient of sin x: 1 = C sin θCoefficient of cos x: √3 = C cos θTherefore, dividing the two: tan θ = 1 / √3 => θ = π/6. Then C is sqrt( (1)^2 + (√3)^2 ) = 2, same as before. So φ(x) can also be written as 2 cos(x - π/6). Either way, amplitude is 2, phase shift is π/6 or π/3 depending on sine or cosine form. But maybe the sine form is better here. Let's stick with φ(x) = 2 sin(x + π/3). Let me verify:sin(x + π/3) = sin x cos(π/3) + cos x sin(π/3) = (1/2) sin x + (√3/2) cos x. Multiply by 2: 2 sin(x + π/3) = sin x + √3 cos x, which matches φ(x). Perfect.So φ(x) = 2 sin(x + π/3). Similarly, φ(x + π/2) would be 2 sin(x + π/2 + π/3) = 2 sin(x + 5π/6). So the inequality becomes:φ(x) + k φ(x + π/2) = 2 sin(x + π/3) + k * 2 sin(x + 5π/6) > 0 for all x in [0, 11π/12].Alternatively, factor out the 2: 2[ sin(x + π/3) + k sin(x + 5π/6) ] > 0. Since 2 is positive, the inequality reduces to sin(x + π/3) + k sin(x + 5π/6) > 0 for all x in [0, 11π/12].We need to find k such that sin(x + π/3) + k sin(x + 5π/6) > 0 for all x in [0, 11π/12].Let me try to simplify the expression sin(x + π/3) + k sin(x + 5π/6). Perhaps we can combine these terms into a single sine function or find a relation between them.First, note that 5π/6 is π - π/6, so sin(x + 5π/6) = sin(π - π/6 + x) = sin(π/6 - x). Wait, let's check:Wait, sin(π - θ) = sin θ. So sin(x + 5π/6) = sin(x + π - π/6) = sin(π - (π/6 - x)) = sin(π/6 - x). Hmm, that might not be directly helpful. Alternatively, maybe expand both terms.Let's expand sin(x + π/3) and sin(x + 5π/6):sin(x + π/3) = sin x cos(π/3) + cos x sin(π/3) = (1/2) sin x + (√3/2) cos xsin(x + 5π/6) = sin x cos(5π/6) + cos x sin(5π/6). Compute cos(5π/6) = cos(π - π/6) = -cos(π/6) = -√3/2sin(5π/6) = sin(π - π/6) = sin(π/6) = 1/2Therefore, sin(x + 5π/6) = sin x (-√3/2) + cos x (1/2) = (-√3/2) sin x + (1/2) cos xTherefore, the expression sin(x + π/3) + k sin(x + 5π/6) becomes:[ (1/2 sin x + √3/2 cos x ) ] + k [ (-√3/2 sin x + 1/2 cos x ) ]Let's combine the coefficients of sin x and cos x:Sin x: (1/2 - k√3/2 )Cos x: (√3/2 + k/2 )Therefore, the entire expression is:[ (1/2 - (k√3)/2 ) sin x + ( √3/2 + k/2 ) cos x ]Factor out 1/2:(1/2)[ (1 - k√3 ) sin x + ( √3 + k ) cos x ]So the inequality is (1/2)[ (1 - k√3 ) sin x + ( √3 + k ) cos x ] > 0. Multiply both sides by 2 (positive, so inequality direction doesn't change):(1 - k√3 ) sin x + ( √3 + k ) cos x > 0 for all x in [0, 11π/12].Let me denote A = 1 - k√3 and B = √3 + k. Then the inequality becomes A sin x + B cos x > 0.We need to ensure that A sin x + B cos x > 0 for all x in [0, 11π/12]. This expression can be written as C sin(x + φ) > 0 where C = sqrt(A² + B²) and φ is some phase shift. However, since we need this to hold for all x in the interval, maybe we can analyze the minimum value of A sin x + B cos x over [0, 11π/12] and set that minimum greater than 0.Alternatively, find the minimum of the expression A sin x + B cos x over the interval and set it greater than zero.The minimum of A sin x + B cos x is -sqrt(A² + B²). Wait, but that is the case when the expression can take any angle. However, since x is restricted to [0, 11π/12], the minimum might be higher. Hmm.Alternatively, perhaps we can parametrize the expression. Let me consider A sin x + B cos x. Let's write this as R sin(x + α), where R = sqrt(A² + B²) and α is the angle such that sin α = B/R and cos α = A/R.Wait, actually:A sin x + B cos x = R sin(x + α), where:R = sqrt(A² + B²)sin α = B / Rcos α = A / RWait, let me check:R sin(x + α) = R sin x cos α + R cos x sin α. Comparing to A sin x + B cos x:A = R cos αB = R sin αTherefore, tan α = B / A, and R = sqrt(A² + B²). So correct.Therefore, the expression can be written as R sin(x + α), and its minimum over an interval depends on the interval. The minimum value would be -R if the interval includes a point where sin(x + α) = -1. But since x is limited to [0, 11π/12], we need to check whether x + α can reach 3π/2 (where sin is -1) within this interval.Alternatively, we can find the critical points where the derivative is zero to find minima and maxima within the interval.But maybe it's better to proceed step by step.First, let's compute R and α. Let me compute R:R = sqrt(A² + B²) = sqrt( (1 - k√3 )² + ( √3 + k )² )Let's compute (1 - k√3 )² = 1 - 2k√3 + 3k²(√3 + k )² = 3 + 2k√3 + k²Therefore, A² + B² = (1 - 2k√3 + 3k²) + (3 + 2k√3 + k²) = 1 + 3 + (-2k√3 + 2k√3) + (3k² + k²) = 4 + 4k²Thus, R = sqrt(4 + 4k²) = 2 sqrt(1 + k²)Interesting, so R is 2 sqrt(1 + k²). So the amplitude is 2 sqrt(1 + k²). Now, sin α = B/R = (√3 + k)/(2 sqrt(1 + k²))cos α = A/R = (1 - k√3 )/(2 sqrt(1 + k²))Therefore, tan α = (√3 + k)/(1 - k√3 )But maybe instead of working with α, let's consider the expression R sin(x + α). The minimum value of sin(x + α) over x in [0, 11π/12] is sin(min_x (x + α)). Wait, no, the sine function can achieve its minimum of -1 if the interval covers the required angle. But since x is in [0, 11π/12], the angle x + α ranges from α to 11π/12 + α.Therefore, if the interval [α, 11π/12 + α] includes 3π/2 (where sin is -1), then the minimum is -1; otherwise, the minimum is the minimum of sin at the endpoints or at critical points inside the interval.But this seems complicated. Maybe another approach.Alternatively, the inequality is equivalent to:A sin x + B cos x > 0, which is the same as:(1 - k√3 ) sin x + ( √3 + k ) cos x > 0Let me write this as:sin x (1 - k√3 ) + cos x (√3 + k ) > 0Let me denote t = x. We need to ensure that for all t in [0, 11π/12], the inequality holds.Alternatively, we can write this as:[ (1 - k√3 ) sin t + ( √3 + k ) cos t ] > 0Let me think of this as a function h(t) = (1 - k√3 ) sin t + ( √3 + k ) cos t. We need h(t) > 0 for all t in [0, 11π/12].To ensure this, the minimum of h(t) over the interval should be greater than zero. Therefore, we need to find the minimum of h(t) on [0, 11π/12] and set it greater than zero.To find the minimum, we can take the derivative of h(t) and find critical points.Compute h'(t) = (1 - k√3 ) cos t - ( √3 + k ) sin t.Set h'(t) = 0:(1 - k√3 ) cos t - ( √3 + k ) sin t = 0Let's rearrange:(1 - k√3 ) cos t = ( √3 + k ) sin tDivide both sides by cos t (assuming cos t ≠ 0):(1 - k√3 ) = ( √3 + k ) tan tTherefore, tan t = (1 - k√3 ) / ( √3 + k )Let me denote tan t = [ (1 - k√3 ) / ( √3 + k ) ]So critical points occur at t = arctan( [ (1 - k√3 ) / ( √3 + k ) ] ) + nπ. But since t is in [0, 11π/12], we need to check if this arctangent is within the interval.But solving for t here might be complicated. Alternatively, perhaps we can express h(t) in terms of a single sine function.Earlier, we saw that h(t) = R sin(t + α), where R = 2 sqrt(1 + k²), and α is such that sin α = (√3 + k)/R and cos α = (1 - k√3)/R.But since R = 2 sqrt(1 + k²), we have:sin α = (√3 + k)/(2 sqrt(1 + k²))cos α = (1 - k√3 )/(2 sqrt(1 + k²))Therefore, h(t) = 2 sqrt(1 + k²) sin(t + α)We need this to be greater than 0 for all t in [0, 11π/12]. So sin(t + α) > 0 for all t in [0, 11π/12], multiplied by 2 sqrt(1 + k²), which is always positive. Therefore, the inequality reduces to sin(t + α) > 0 for all t in [0, 11π/12].Therefore, we need that for all t in [0, 11π/12], t + α ∈ (0, π). Because sin is positive in (0, π). So if the entire interval [α, 11π/12 + α] is within (0, π), then sin(t + α) > 0.So, the lower bound: α > 0 (which might not be necessary because t starts at 0, so t + α >= α. But if α < 0, then the lower part of the interval would be less than 0. Wait, but α is an angle defined by tan α = (√3 + k)/(1 - k√3 ). Hmm.Wait, maybe we need to be careful here. Let me think.Alternatively, since h(t) = 2 sqrt(1 + k²) sin(t + α), and we need sin(t + α) > 0 for all t ∈ [0, 11π/12]. Therefore, the argument t + α must lie in (0, π) for all t ∈ [0, 11π/12]. Therefore:For the minimum t = 0: 0 + α > 0 => α > 0For the maximum t = 11π/12: 11π/12 + α < π => α < π - 11π/12 = π/12Therefore, combining both:0 < α < π/12But α is determined by k. So we need to find k such that α ∈ (0, π/12). Then, since α is determined by:tan α = (√3 + k ) / (1 - k√3 )Wait, tan α = B / A = (√3 + k ) / (1 - k√3 )So, given that 0 < α < π/12, then tan α is between 0 and tan(π/12).Compute tan(π/12). π/12 is 15 degrees. tan(15°) = 2 - √3 ≈ 0.2679.Therefore, tan α ∈ (0, 2 - √3 )But tan α = (√3 + k ) / (1 - k√3 )So, 0 < (√3 + k ) / (1 - k√3 ) < 2 - √3We need to solve for k such that:0 < [ (√3 + k ) / (1 - k√3 ) ] < 2 - √3Note that (1 - k√3 ) is in the denominator. We need to check the sign of the denominator because the inequality direction depends on that.Let me denote D = 1 - k√3So tan α = (√3 + k)/DWe have to consider two cases: D > 0 and D < 0.Case 1: D > 0, i.e., 1 - k√3 > 0 => k < 1/√3 ≈ 0.577Case 2: D < 0, i.e., k > 1/√3First, let's check if tan α is positive. Since tan α = (√3 + k)/D, and we have 0 < tan α < 2 - √3.In Case 1: D > 0, so (√3 + k)/D > 0 => √3 + k > 0. Since k is a real number, but if k is very negative, √3 + k might be negative. So we need both D > 0 and √3 + k > 0. So:D > 0 => k < 1/√3√3 + k > 0 => k > -√3Therefore, in Case 1: -√3 < k < 1/√3Similarly, in Case 2: D < 0, so (√3 + k)/D > 0 requires that √3 + k < 0 (since D < 0). So:√3 + k < 0 => k < -√3But D < 0 => k > 1/√3 ≈ 0.577. But if k > 1/√3, then k >= 0.577 > -√3 ≈ -1.732. So √3 + k >= √3 - √3 = 0? Wait, if k < -√3, then √3 + k < 0. But D < 0 when k > 1/√3. So, for Case 2, we need both k > 1/√3 and k < -√3, which is impossible. Therefore, Case 2 is impossible. Therefore, only Case 1 is possible.Therefore, the valid range of k is -√3 < k < 1/√3, and in this case, D > 0 and √3 + k > 0. Now, within this range, we have:0 < (√3 + k ) / (1 - k√3 ) < 2 - √3Let me write the inequality:0 < (√3 + k ) / (1 - k√3 ) < 2 - √3Since D = 1 - k√3 > 0, we can multiply the inequalities without flipping the signs.First, left inequality:(√3 + k ) / (1 - k√3 ) > 0Which is already satisfied because in Case 1, both numerator and denominator are positive. So, the left inequality holds for all k in (-√3, 1/√3). Now, the right inequality:(√3 + k ) / (1 - k√3 ) < 2 - √3Multiply both sides by (1 - k√3 ) which is positive:√3 + k < (2 - √3)(1 - k√3 )Expand the right-hand side:(2 - √3)(1 - k√3 ) = 2(1) - 2k√3 - √3(1) + √3 * k√3= 2 - 2k√3 - √3 + k * 3= 2 - √3 + 3k - 2k√3Therefore, the inequality becomes:√3 + k < 2 - √3 + 3k - 2k√3Bring all terms to the left:√3 + k - 2 + √3 - 3k + 2k√3 < 0Combine like terms:(√3 + √3) + (k - 3k) + ( -2 ) + 2k√3 < 02√3 - 2k - 2 + 2k√3 < 0Factor terms:(-2k + 2k√3 ) + (2√3 - 2 ) < 0Factor out 2k from the first group and 2 from the second:2k(-1 + √3 ) + 2(√3 - 1 ) < 0Factor out 2(√3 - 1 ):Notice that -1 + √3 = √3 - 1, so:2k(√3 - 1 ) + 2(√3 - 1 ) < 0Factor out 2(√3 - 1 ):2(√3 - 1 )(k + 1 ) < 0Since √3 - 1 ≈ 1.732 - 1 = 0.732 > 0, so the inequality reduces to:(k + 1 ) < 0Therefore, k + 1 < 0 => k < -1But recall that in Case 1, k is in (-√3, 1/√3 ). So combining k < -1 and -√3 < k < 1/√3, we get:-√3 < k < -1Because -√3 ≈ -1.732, so -1.732 < k < -1.Therefore, the range of k is (-√3, -1). But wait, let's check if this is correct. Let me verify step by step.Starting from the inequality:(√3 + k ) / (1 - k√3 ) < 2 - √3After multiplying both sides by (1 - k√3 ), which is positive, we get:√3 + k < (2 - √3)(1 - k√3 )Expanding RHS:= 2*1 - 2k√3 - √3*1 + √3*k√3= 2 - 2k√3 - √3 + 3kThus, RHS = 2 - √3 + 3k - 2k√3Then moving all terms to the left:√3 + k - (2 - √3 + 3k - 2k√3 ) < 0= √3 + k - 2 + √3 - 3k + 2k√3 < 0= 2√3 - 2 - 2k + 2k√3 < 0Factor:2(√3 -1 ) + 2k(√3 -1 ) < 0= 2(√3 -1 )(1 + k ) < 0Since √3 -1 > 0, divide both sides by 2(√3 -1 ):1 + k < 0 => k < -1Thus, the solution is k < -1. However, in Case 1, k is in (-√3, 1/√3 ). Therefore, the intersection is (-√3, -1). Hence, the range is -√3 < k < -1.But we need to check if this is indeed correct. Let's test a value in this interval, say k = -1.5 (which is between -√3 ≈ -1.732 and -1). Let's compute h(t) = (1 - k√3 ) sin t + ( √3 + k ) cos t.For k = -1.5:A = 1 - (-1.5)√3 = 1 + 1.5√3 ≈ 1 + 2.598 ≈ 3.598B = √3 + (-1.5) ≈ 1.732 - 1.5 ≈ 0.232Therefore, h(t) ≈ 3.598 sin t + 0.232 cos tWe need to check if this is always positive on [0, 11π/12]. Let's compute h(t) at t = 11π/12, which is 165 degrees.11π/12 ≈ 165 degrees. So sin(11π/12) = sin(π - π/12) = sin(π/12) ≈ 0.2588cos(11π/12) = -cos(π/12) ≈ -0.9659Thus, h(11π/12) ≈ 3.598*0.2588 + 0.232*(-0.9659) ≈ 0.931 - 0.224 ≈ 0.707 > 0At t = 0:h(0) = 0.232*1 ≈ 0.232 > 0At t = π/2 (90 degrees):h(π/2) = 3.598*1 + 0.232*0 ≈ 3.598 > 0At t = π (180 degrees), but 11π/12 is less than π. The maximum t is 11π/12 ≈ 165 degrees. So, seems like h(t) is positive throughout. Alternatively, let's test k = -1. Let's see:k = -1:A = 1 - (-1)√3 = 1 + √3 ≈ 2.732B = √3 + (-1) ≈ 0.732h(t) = 2.732 sin t + 0.732 cos tAt t = 11π/12 ≈ 165 degrees:sin(11π/12) ≈ 0.2588, cos ≈ -0.9659h ≈ 2.732*0.2588 + 0.732*(-0.9659) ≈ 0.707 - 0.707 ≈ 0. But the inequality is strict (>0), so k = -1 is not allowed. Hence, the upper bound is k > -√3 and k < -1, not including -1.But wait, according to our earlier result, k must be less than -1, but greater than -√3. So k ∈ (-√3, -1). However, let's verify with k approaching -1 from below. Let's take k = -1. Let's check if h(t) is ever zero. At t = 11π/12, h(t) ≈ 0. But as per the inequality, it needs to be greater than zero. Therefore, k = -1 is not allowed. So the range is open interval (-√3, -1).But let's confirm with another test point. Take k = -2, which is less than -√3 ≈ -1.732. Wait, but our Case 1 was for -√3 < k < 1/√3. If k = -2, which is less than -√3, but previously we concluded that Case 2 is invalid. However, let's check what happens if k = -2.If k = -2:A = 1 - (-2)√3 = 1 + 2√3 ≈ 1 + 3.464 ≈ 4.464B = √3 + (-2) ≈ 1.732 - 2 ≈ -0.268h(t) = 4.464 sin t - 0.268 cos tAt t = 11π/12:h ≈ 4.464*0.2588 -0.268*(-0.9659) ≈ 1.155 + 0.259 ≈ 1.414 > 0At t = 0:h(0) = -0.268*1 ≈ -0.268 < 0. Which violates the inequality. Therefore, k = -2 is invalid.Therefore, k must be greater than -√3 to keep B = √3 + k > 0. Wait, when k approaches -√3 from above:B = √3 + k. If k approaches -√3, then B approaches 0. So when k = -√3, B = 0, and h(t) = (1 - (-√3)*√3 ) sin t + 0* cos t = (1 + 3 ) sin t = 4 sin t. Then h(t) = 4 sin t. But x ∈ [0, 11π/12], which includes t = 0, where sin t = 0. Therefore, h(0) = 0, which again violates the strict inequality. Therefore, k cannot be -√3. Thus, the lower bound is k > -√3. But our earlier solution was k ∈ (-√3, -1). However, when k approaches -1 from below, say k = -1.1:A = 1 - (-1.1)√3 ≈ 1 + 1.1*1.732 ≈ 1 + 1.905 ≈ 2.905B = √3 + (-1.1) ≈ 1.732 - 1.1 ≈ 0.632h(t) = 2.905 sin t + 0.632 cos tAt t = 11π/12:h ≈ 2.905*0.2588 + 0.632*(-0.9659) ≈ 0.752 - 0.611 ≈ 0.141 > 0At t = 0:h(0) ≈ 0.632 > 0At critical points? Let's check if there is a minima in between. Take derivative h'(t) = 2.905 cos t - 0.632 sin tSet to zero:2.905 cos t = 0.632 sin t => tan t = 2.905 / 0.632 ≈ 4.6So t ≈ arctan(4.6) ≈ 77.5 degrees ≈ 1.35 radians. Since 77.5 degrees is about 1.35 rad, which is less than 11π/12 ≈ 165 degrees (≈ 2.88 rad). So critical point at t ≈ 1.35 rad. Let's compute h(t) at this point:tan t ≈ 4.6, so sin t ≈ 4.6 / sqrt(1 + 4.6²) ≈ 4.6 / sqrt(22.16) ≈ 4.6 / 4.707 ≈ 0.977cos t ≈ 1 / sqrt(1 + 4.6²) ≈ 0.213Therefore, h(t) ≈ 2.905*0.977 + 0.632*0.213 ≈ 2.839 + 0.135 ≈ 2.974 > 0Therefore, even at the critical point, h(t) is positive. So for k = -1.1, h(t) remains positive. Therefore, the range seems to be (-√3, -1). But let's check when k approaches -1 from the left, say k = -1 + ε where ε approaches 0 from the negative side. Wait, k approaches -1 from the left would be k = -1 - ε. Wait, confusing. Let me take k approaching -1 from above (i.e., k = -0.9). Wait, but our interval is k < -1. Let me take k approaching -1 from below (k approaching -1 from the left, i.e., k = -1.1, -1.01, etc.)But in our solution, k is approaching -1 from the lower side (i.e., k approaches -1 from values less than -1, like -1.1, -1.05, etc.) As k approaches -1, the upper limit of t + α approaches 11π/12 + α. When k approaches -1, let's compute α:tan α = (√3 + k ) / (1 - k√3 )For k approaching -1 from below (k → -1⁻):tan α ≈ (√3 - 1 ) / (1 - (-1)√3 ) = (√3 - 1 ) / (1 + √3 )Multiply numerator and denominator by (√3 -1 ):[(√3 -1 )² ] / [ (1 + √3)(√3 -1 ) ] = (3 - 2√3 +1 ) / (3 -1 ) = (4 - 2√3 ) / 2 = 2 - √3 ≈ 0.2679Therefore, tan α approaches 2 - √3, which is tan(π/12). Therefore, α approaches π/12. Therefore, when k approaches -1 from below, α approaches π/12. Then, the upper limit of t + α is 11π/12 + π/12 = π. Therefore, sin(t + α) approaches sin(π) = 0. Hence, h(t) approaches zero at t = 11π/12 when k approaches -1. Therefore, k must be strictly less than -1 to keep h(t) > 0. However, as we saw when k = -1, h(t) = 0 at t = 11π/12. Therefore, the upper bound is k > -√3 and k < -1.Similarly, when k approaches -√3 from above (k → -√3⁺), tan α = (√3 + k ) / (1 - k√3 ). Let k = -√3 + ε where ε → 0⁺:tan α ≈ (√3 + (-√3 + ε )) / (1 - (-√3 + ε )√3 ) = (ε ) / (1 + √3*√3 - ε√3 ) = ε / (1 + 3 - ε√3 ) = ε / (4 - ε√3 ) ≈ ε/4 → 0. Therefore, α approaches 0. Hence, the lower bound of t + α approaches 0 + 0 = 0, and the upper bound approaches 11π/12 + 0 = 11π/12. However, sin(0) = 0 and sin(11π/12) ≈ sin(165°) ≈ 0.2588 > 0. But h(t) in this case is approaching h(t) = (1 - (-√3 + ε )√3 ) sin t + ( √3 + (-√3 + ε ) ) cos t = (1 + 3 - ε√3 ) sin t + ε cos t ≈ (4 - ε√3 ) sin t + ε cos t. As ε →0⁺, this approaches 4 sin t, which at t =0 is 0. Therefore, h(t) approaches zero at t =0. Therefore, for k approaching -√3 from above, h(t) approaches zero at t=0. Hence, k must be strictly greater than -√3.Therefore, combining both limits, the range of k is -√3 < k < -1.But let's check if there are any other critical points. Wait, when k is in (-√3, -1), then the expression h(t) is positive everywhere on [0, 11π/12]. But to ensure this, we might need another check. Alternatively, since we already transformed the inequality to k ∈ (-√3, -1), and verified with test points, I think this is the correct range. Therefore, the answer to part (2) is the range of k is (-√3, -1), written as -√3 < k < -1.Final Answer(1) The adjoint vector is boxed{left( -dfrac{sqrt{3}}{2}, dfrac{3}{2} right)} and its magnitude is boxed{sqrt{3}}; (2) The range of k is boxed{(-sqrt{3}, -1)}.</think>(1) To find the adjoint vector (overrightarrow{OM}) of the function (g(x) = cos(x + frac{pi}{3}) + cos x), we use the cosine addition formula:[cos(x + frac{pi}{3}) = cos x cos frac{pi}{3} - sin x sin frac{pi}{3} = frac{1}{2} cos x - frac{sqrt{3}}{2} sin x]Thus,[g(x) = left( frac{1}{2} cos x - frac{sqrt{3}}{2} sin x right) + cos x = frac{3}{2} cos x - frac{sqrt{3}}{2} sin x]The adjoint vector (overrightarrow{OM}) is (left( -frac{sqrt{3}}{2}, frac{3}{2} right)). The magnitude is:[|overrightarrow{OM}| = sqrt{left( -frac{sqrt{3}}{2} right)^2 + left( frac{3}{2} right)^2} = sqrt{frac{3}{4} + frac{9}{4}} = sqrt{3}](2) The adjoint function of (overrightarrow{ON} = (1, sqrt{3})) is (varphi(x) = sin x + sqrt{3} cos x). This can be written as:[varphi(x) = 2 sin left( x + frac{pi}{3} right)]We need to ensure:[varphi(x) + k varphi left( x + frac{pi}{2} right) > 0 quad text{for all} quad x in [0, frac{11pi}{12}]]Expressing (varphi(x + frac{pi}{2})):[varphi left( x + frac{pi}{2} right) = 2 sin left( x + frac{pi}{2} + frac{pi}{3} right) = 2 sin left( x + frac{5pi}{6} right)]The inequality becomes:[2 sin left( x + frac{pi}{3} right) + 2k sin left( x + frac{5pi}{6} right) > 0]Simplifying, we find:[(1 - ksqrt{3}) sin x + (sqrt{3} + k) cos x > 0]This can be written as (R sin(x + alpha)) where (R = 2sqrt{1 + k^2}) and (alpha) is determined by (tan alpha = frac{sqrt{3} + k}{1 - ksqrt{3}}). Ensuring this is positive for all (x in [0, frac{11pi}{12}]) leads to solving:[0 < frac{sqrt{3} + k}{1 - ksqrt{3}} < 2 - sqrt{3}]This results in the range (-sqrt{3} < k < -1). Final Answer(1) The adjoint vector is (boxed{left( -dfrac{sqrt{3}}{2}, dfrac{3}{2} right)}) and its magnitude is (boxed{sqrt{3}}); (2) The range of (k) is (boxed{(-sqrt{3}, -1)}).,<think>Okay, let's see. I have this problem about adjoint vectors and functions. It's divided into two parts. Let me start with part (1). So, the function given is g(x) = cos(x + π/3) + cos x. I need to find its adjoint vector OM, which is (a, b), where the adjoint function is defined as f(x) = a sin x + b cos x. Then, compute the magnitude of OM, |OM|.Wait, the adjoint function of vector OM is f(x) = a sin x + b cos x. But here, they want the adjoint vector of the function g(x). So, reverse engineering it. If a function is expressed as a sin x + b cos x, its adjoint vector is (a, b). So, if I can express g(x) in the form a sin x + b cos x, then a and b will be the components of the adjoint vector.But g(x) is given as cos(x + π/3) + cos x. Let me expand cos(x + π/3) using the cosine addition formula. Recall that cos(A + B) = cos A cos B - sin A sin B. So, cos(x + π/3) = cos x cos(π/3) - sin x sin(π/3). We know that cos(π/3) is 1/2 and sin(π/3) is √3/2. So substituting in, we get:cos(x + π/3) = cos x * (1/2) - sin x * (√3/2) = (1/2) cos x - (√3/2) sin x.Therefore, g(x) = (1/2 cos x - √3/2 sin x) + cos x. Let's combine like terms. The cos x terms: (1/2 cos x + cos x) = (1/2 + 1) cos x = (3/2) cos x.The sin x term: -√3/2 sin x.So putting it all together, g(x) = (-√3/2) sin x + (3/2) cos x.Therefore, comparing this to the adjoint function f(x) = a sin x + b cos x, we can see that a = -√3/2 and b = 3/2. Hence, the adjoint vector OM is (-√3/2, 3/2). Now, we need to find the magnitude |OM|. The magnitude of a vector (a, b) is sqrt(a² + b²). Let's compute that.First, square each component:a² = (-√3/2)² = (3/4)b² = (3/2)² = 9/4Adding them: 3/4 + 9/4 = 12/4 = 3Then the magnitude is sqrt(3). So that's part (1) done. Let me verify once again:Original function: cos(x + π/3) + cos x. Expanded using cosine addition: (1/2 cos x - √3/2 sin x) + cos x = (3/2 cos x - √3/2 sin x). Thus, coefficients are -√3/2 and 3/2. The vector is (-√3/2, 3/2), magnitude sqrt( (3/4) + (9/4) ) = sqrt(12/4) = sqrt(3). Yep, that seems right.Moving on to part (2). The vector ON is (1, √3), and its adjoint function is φ(x). So, φ(x) = 1 * sin x + √3 * cos x. That simplifies to sin x + √3 cos x. Then, the inequality given is φ(x) + k φ(x + π/2) > 0 for all x in [0, 11π/12]. Need to find the range of k.First, let's write φ(x) and φ(x + π/2). φ(x) = sin x + √3 cos xφ(x + π/2) = sin(x + π/2) + √3 cos(x + π/2)We can use trigonometric identities to simplify these.Recall that sin(x + π/2) = cos x and cos(x + π/2) = -sin x. So:φ(x + π/2) = cos x + √3 (-sin x) = cos x - √3 sin xTherefore, the inequality becomes:[sin x + √3 cos x] + k [cos x - √3 sin x] > 0Let's distribute k:sin x + √3 cos x + k cos x - k√3 sin x > 0Now, group like terms:sin x (1 - k√3) + cos x (√3 + k) > 0So, this is an expression of the form A sin x + B cos x > 0, where A = (1 - k√3) and B = (√3 + k)Alternatively, we can write this as C sin(x + θ) > 0 or C cos(x - φ) > 0, but maybe it's easier to analyze the expression A sin x + B cos x.Alternatively, we can combine the terms into a single sine or cosine function. Let me recall that any expression of the form A sin x + B cos x can be written as R sin(x + φ), where R = sqrt(A² + B²) and φ is the phase shift. So, let's compute R and φ here. Let me denote:A = 1 - k√3B = √3 + kThen, R = sqrt(A² + B²) = sqrt( (1 - k√3)^2 + (√3 + k)^2 )Let me expand these squares:(1 - k√3)^2 = 1 - 2k√3 + 3k²(√3 + k)^2 = 3 + 2k√3 + k²Adding them together:1 - 2k√3 + 3k² + 3 + 2k√3 + k² = 1 + 3 + (3k² + k²) + (-2k√3 + 2k√3)Simplify:4 + 4k² + 0 = 4(1 + k²)Thus, R = sqrt(4(1 + k²)) = 2 sqrt(1 + k²)So, the amplitude is 2 sqrt(1 + k²). Then, the phase shift φ can be found by tan φ = B/A. Wait, no, actually, if we write it as R sin(x + φ), then the formula is:A sin x + B cos x = R sin(x + φ), where R = sqrt(A² + B²) and φ = arctan(B/A) if A ≠ 0.Wait, let's verify. Let me recall that:R sin(x + φ) = R sin x cos φ + R cos x sin φ. Comparing to A sin x + B cos x, so:A = R cos φB = R sin φTherefore, tan φ = B / A. So, φ = arctan(B/A)So, in our case, φ = arctan( (√3 + k) / (1 - k√3) )But perhaps we don't need the exact phase shift. The important part is that the inequality A sin x + B cos x > 0 can be written as R sin(x + φ) > 0. The minimum value of R sin(x + φ) is -R, but since we need it to be greater than 0 for all x in [0, 11π/12], we need to ensure that the expression never dips below zero in that interval.But since R is positive (it's 2 sqrt(1 + k²)), the expression R sin(x + φ) will have a minimum value of -R. But if we can guarantee that even the minimum value in the interval [0, 11π/12] is greater than 0, then the inequality holds.Alternatively, maybe it's easier to analyze the original expression A sin x + B cos x and find for which k the expression is positive for all x in the given interval. Let's think about this.Given x ∈ [0, 11π/12], we need (1 - k√3) sin x + (√3 + k) cos x > 0 for all x in this interval.Let me denote the left-hand side as E(x) = (1 - k√3) sin x + (√3 + k) cos xWe need E(x) > 0 for all x ∈ [0, 11π/12]. To find the range of k, we can consider the minimum of E(x) over the interval and ensure that the minimum is positive.To find the minimum, we can take the derivative and find critical points, but given that this is a trigonometric function, perhaps it's easier to write E(x) in the form C sin(x + θ) or C cos(x + θ), find its amplitude and phase, then check where it's positive.Wait, we already found that E(x) = 2 sqrt(1 + k²) sin(x + φ). So, the maximum value is 2 sqrt(1 + k²), and the minimum is -2 sqrt(1 + k²). However, since the interval is [0, 11π/12], which is less than π (since 11π/12 is 165 degrees, and π is 180 degrees), we need to check if within this interval, the function ever dips below zero.Alternatively, the function E(x) is a sine wave with amplitude 2 sqrt(1 + k²) and phase shift φ. But perhaps it's better to approach this by considering the original coefficients.Alternatively, set E(x) > 0 for all x ∈ [0, 11π/12]. So, for each x in that interval, (1 - k√3) sin x + (√3 + k) cos x > 0.We can treat this as a linear inequality in k. Let's rearrange terms to solve for k.E(x) = (1 - k√3) sin x + (√3 + k) cos x > 0Let's group terms with k:= sin x - k√3 sin x + √3 cos x + k cos x > 0= [sin x + √3 cos x] + k [cos x - √3 sin x] > 0But wait, this is exactly the original inequality given. Hmm, maybe that's not helpful. Let's instead isolate k.E(x) = sin x + √3 cos x + k (cos x - √3 sin x) > 0Then, solving for k:k (cos x - √3 sin x) > - (sin x + √3 cos x)If (cos x - √3 sin x) > 0, then:k > [ - (sin x + √3 cos x) ] / (cos x - √3 sin x )If (cos x - √3 sin x) < 0, then:k < [ - (sin x + √3 cos x) ] / (cos x - √3 sin x )But since x is in [0, 11π/12], we need to check the sign of (cos x - √3 sin x) over this interval.Alternatively, let's compute (cos x - √3 sin x). Let's factor out a 2:= 2 [ (1/2) cos x - (√3/2) sin x ]Notice that (1/2) = cos(π/3) and (√3/2) = sin(π/3). So:= 2 [ cos(π/3) cos x - sin(π/3) sin x ] = 2 cos(x + π/3)Therefore, cos x - √3 sin x = 2 cos(x + π/3)Therefore, the inequality becomes:E(x) = sin x + √3 cos x + k * 2 cos(x + π/3) > 0But sin x + √3 cos x can also be written as 2 sin(x + π/3). Let's check:Using the identity A sin x + B cos x = C sin(x + φ), where C = sqrt(A² + B²). Here, A = 1, B = √3, so C = sqrt(1 + 3) = 2. Then, φ = arctan(B/A) = arctan(√3) = π/3. So yes, sin x + √3 cos x = 2 sin(x + π/3)Therefore, E(x) = 2 sin(x + π/3) + 2k cos(x + π/3) > 0Factor out the 2:2 [ sin(x + π/3) + k cos(x + π/3) ] > 0Divide both sides by 2 (since 2 is positive, inequality direction remains):sin(x + π/3) + k cos(x + π/3) > 0Let me set θ = x + π/3. Then, since x ∈ [0, 11π/12], θ ∈ [π/3, 11π/12 + π/3] = [π/3, 11π/12 + 4π/12] = [π/3, 15π/12] = [π/3, 5π/4]So, the inequality becomes sin θ + k cos θ > 0 for all θ ∈ [π/3, 5π/4]So, now the problem reduces to finding k such that sin θ + k cos θ > 0 for all θ in [π/3, 5π/4]Let me denote F(θ) = sin θ + k cos θ. We need F(θ) > 0 for all θ ∈ [π/3, 5π/4]To find the range of k such that this holds, we can analyze F(θ). Let's consider the expression sin θ + k cos θ. Let's express this as R sin(θ + φ), but perhaps it's easier to solve for k.Let's rearrange the inequality:k cos θ > - sin θIf cos θ ≠ 0, then:k > - sin θ / cos θ = - tan θBut cos θ is positive or negative in the interval [π/3, 5π/4]. Let's check:θ ∈ [π/3, 5π/4]. π/3 is 60 degrees, 5π/4 is 225 degrees. So, in this interval, θ crosses π/2 (90 degrees) where cos θ = 0, and goes into the third quadrant (θ > π). So cos θ is positive in [π/3, π/2), negative in (π/2, 5π/4].Therefore, when cos θ > 0 (θ ∈ [π/3, π/2)), the inequality k > - tan θ must hold. When cos θ < 0 (θ ∈ (π/2, 5π/4]), the inequality direction reverses, so k < - tan θ.But at θ = π/2, cos θ = 0, so the original inequality becomes sin(π/2) + k*0 > 0 => 1 > 0, which is true regardless of k. So, θ = π/2 is okay.Therefore, we need to ensure:For θ ∈ [π/3, π/2): k > - tan θFor θ ∈ (π/2, 5π/4]: k < - tan θSo, to satisfy both conditions, k must be greater than all the values -tan θ in [π/3, π/2) and less than all the values -tan θ in (π/2, 5π/4]. Therefore, the lower bound for k is the maximum of -tan θ over θ ∈ [π/3, π/2), and the upper bound for k is the minimum of -tan θ over θ ∈ (π/2, 5π/4].So, let's compute these:First interval: θ ∈ [π/3, π/2). In this interval, tan θ is increasing from tan(π/3) = √3 to tan(π/2) which approaches infinity. Therefore, -tan θ is decreasing from -√3 to -infinity. So, the maximum of -tan θ in this interval is -√3 (at θ = π/3). Therefore, k must be greater than -√3.Second interval: θ ∈ (π/2, 5π/4]. In this interval, tan θ is negative (since θ is in second and third quadrants). Let's compute tan θ here:For θ ∈ (π/2, π], tan θ is negative, from 0 to -infty. For θ ∈ [π, 5π/4], tan θ is positive again? Wait, no. Wait, tan θ is sin θ / cos θ. For θ in (π/2, π), cos θ is negative, sin θ is positive, so tan θ is negative. For θ in (π, 5π/4], both sin θ and cos θ are negative, so tan θ is positive. Wait, wait.Wait, θ ∈ (π/2, π): sin θ positive, cos θ negative → tan θ negative.θ ∈ (π, 3π/2): sin θ negative, cos θ negative → tan θ positive.But our interval is (π/2, 5π/4], so θ ∈ (π/2, π) ∪ [π, 5π/4]. So in (π/2, π), tan θ is negative; in [π, 5π/4], tan θ is positive (since both sin and cos are negative).Therefore, in θ ∈ (π/2, π), tan θ is negative, so -tan θ is positive. In θ ∈ [π, 5π/4], tan θ is positive, so -tan θ is negative.But we need to find k < -tan θ for θ ∈ (π/2, 5π/4].So, splitting into two parts:1. θ ∈ (π/2, π): here, -tan θ is positive, so k < positive number. Since θ ranges from π/2 to π, tan θ ranges from -infty to 0 (approaching from the left), so -tan θ ranges from +infty to 0. Therefore, to satisfy k < -tan θ for all θ in (π/2, π), k must be less than the infimum of -tan θ over this interval, which is 0. So, k < 0.2. θ ∈ [π, 5π/4]: here, tan θ is positive (as both sin and cos are negative), so -tan θ is negative. Therefore, the inequality k < -tan θ becomes k < negative number. Since θ ∈ [π, 5π/4], tan θ is positive, increasing from 0 to tan(5π/4) = 1. Therefore, -tan θ is decreasing from 0 to -1. So, the minimum of -tan θ in this interval is -1 (at θ = 5π/4). Therefore, to satisfy k < -tan θ for all θ in [π, 5π/4], k must be less than the minimum value of -tan θ, which is -1.But combining these two parts:From θ ∈ (π/2, π): k < 0From θ ∈ [π, 5π/4]: k < -1So, the stricter condition is k < -1. Therefore, the upper bound for k is -1.Therefore, combining both intervals:k must satisfy k > -√3 (from first interval) and k < -1 (from second interval). Wait, but how can k be both greater than -√3 and less than -1? That would mean -√3 < k < -1. Since -√3 ≈ -1.732, and -1 is greater than that, so the range is -√3 < k < -1. But wait, hold on. Let me verify. So in the first interval [π/3, π/2), we have k > -tan θ. Since θ ∈ [π/3, π/2), tan θ is increasing from √3 to +infty, so -tan θ is decreasing from -√3 to -infty. Therefore, the maximum of -tan θ in this interval is -√3. Hence, to satisfy k > -tan θ for all θ in this interval, k must be greater than the maximum of -tan θ, which is -√3. So k > -√3.In the second interval (π/2, 5π/4], we split into two parts:- For (π/2, π), tan θ is negative, so -tan θ is positive. Here, k must be less than all these positive numbers. The minimum of -tan θ in this sub-interval is 0 (as θ approaches π/2 from the right, tan θ approaches -infty, so -tan θ approaches +infty; as θ approaches π from the left, tan θ approaches 0 from the negative side, so -tan θ approaches 0). Therefore, to satisfy k < -tan θ for all θ in (π/2, π), k must be less than the infimum of -tan θ, which is 0. So k < 0.- For [π, 5π/4], tan θ is positive (since both sin and cos are negative), so -tan θ is negative. Here, k must be less than -tan θ for all θ in this sub-interval. The minimum of -tan θ here is -tan(5π/4) = -1 (since tan(5π/4) = 1). Therefore, k must be less than -1 to satisfy k < -tan θ for all θ in [π, 5π/4].Therefore, combining the two sub-intervals in the second interval:From (π/2, π): k < 0From [π, 5π/4]: k < -1The stricter condition is k < -1. So overall, combining with the first interval:k > -√3 and k < -1.Hence, the range of k is (-√3, -1).But we need to verify this. Let's check for some k in this interval and see if the inequality holds.Take k = -1.5, which is between -√3 ≈ -1.732 and -1.Compute E(x) = (1 - k√3) sin x + (√3 + k) cos x.For k = -1.5:A = 1 - (-1.5)(√3) = 1 + 1.5√3 ≈ 1 + 2.598 ≈ 3.598B = √3 + (-1.5) ≈ 1.732 - 1.5 ≈ 0.232So E(x) ≈ 3.598 sin x + 0.232 cos x. This is a sine-dominated function, so likely positive for all x. Wait, but x is in [0, 11π/12]. Let's check x = 5π/4 - π/3 = ?Wait, perhaps instead of plugging in specific values, we can check if k = -1.5 (which is within the proposed range) satisfies the inequality.Alternatively, let's check the endpoints.First, check k = -1. Let's see:At k = -1, E(x) = (1 - (-1)√3) sin x + (√3 + (-1)) cos x = (1 + √3) sin x + (√3 - 1) cos x.We need to check if this is positive for all x ∈ [0, 11π/12].Alternatively, since when k = -1, the upper bound, we need to check if E(x) = (1 + √3) sin x + (√3 - 1) cos x is positive over [0, 11π/12].Similarly, at k = -√3, the lower bound, E(x) would be (1 - (-√3)*√3) sin x + (√3 + (-√3)) cos x = (1 + 3) sin x + 0 cos x = 4 sin x. Then, over x ∈ [0, 11π/12], sin x is non-negative (since 11π/12 is less than π), so 4 sin x ≥ 0. However, the inequality is strict: > 0. So at x = 0, 4 sin 0 = 0, which is not greater than 0. Therefore, k cannot be equal to -√3. Hence, the open interval.But wait, when k approaches -√3 from above, say k = -√3 + ε, then E(x) becomes:(1 - (-√3 + ε)√3) sin x + (√3 + (-√3 + ε)) cos x= [1 + 3 - ε√3] sin x + [ε] cos x= (4 - ε√3) sin x + ε cos xAt x = 0, this becomes ε cos 0 = ε. If ε approaches 0 from the positive side, then at x = 0, E(x) approaches 0, which is not > 0. Therefore, k must be strictly greater than -√3.Similarly, at k = -1, let's check the original inequality.E(x) = (1 - (-1)√3) sin x + (√3 + (-1)) cos x = (1 + √3) sin x + (√3 - 1) cos xLet me compute E(x) at x = 5π/4, which is within the interval [0, 11π/12]? Wait, 5π/4 is 225 degrees, but 11π/12 is 165 degrees. Wait, 11π/12 is approximately 2.879 radians, while 5π/4 is approximately 3.927 radians. Wait, hold on: x is in [0, 11π/12], so θ = x + π/3 is in [π/3, 11π/12 + π/3] = [π/3, 15π/12] = [π/3, 5π/4]. But x itself is up to 11π/12. So, when x approaches 11π/12, θ approaches 11π/12 + π/3 = 11π/12 + 4π/12 = 15π/12 = 5π/4.Therefore, the maximum θ is 5π/4. So, testing at θ = 5π/4 (x = 5π/4 - π/3 = 15π/12 - 4π/12 = 11π/12), which is the upper limit.At θ = 5π/4, sin θ = -√2/2, cos θ = -√2/2.So, F(θ) = sin θ + k cos θ = (-√2/2) + k (-√2/2) = (-√2/2)(1 + k)For k = -1, F(θ) = (-√2/2)(1 -1) = 0. So at k = -1, the inequality becomes 0 > 0, which is false. Therefore, k cannot be equal to -1. Hence, the upper bound must be k > -1? Wait, this contradicts our previous conclusion.Wait, but in our earlier analysis, when θ ∈ [π, 5π/4], we needed k < -tan θ. For θ = 5π/4, tan θ = 1, so k < -1. So if k is approaching -1 from below (i.e., k = -1 - ε), then F(θ) = sin θ + k cos θ. At θ = 5π/4, that's -√2/2 + (-1 - ε)(-√2/2) = (-√2/2) + (√2/2 + ε√2/2) = ε√2/2. For ε > 0, this is positive. Wait, but if k approaches -1 from above (i.e., k = -1 + ε), then at θ = 5π/4:F(θ) = sin θ + k cos θ = -√2/2 + (-1 + ε)(-√2/2) = -√2/2 + √2/2 - ε√2/2 = -ε√2/2 < 0. Therefore, k must be less than -1 to make F(θ) positive at θ = 5π/4.But when k = -1.5, which is less than -1, then F(θ) at θ = 5π/4 is -√2/2 + (-1.5)(-√2/2) = (-√2/2) + (1.5√2/2) = (0.5√2/2) = √2/4 > 0. So yes, when k is less than -1, F(θ) at θ = 5π/4 is positive.But when k approaches -1 from above (k = -1 + ε), F(θ) at θ = 5π/4 becomes negative, which violates the inequality. Therefore, k must be strictly less than -1.However, when θ approaches π/2 from above, for k approaching -infty, but we have a lower bound of k > -√3.But earlier, we had this conflict where at k = -1, F(θ) = 0 at θ = 5π/4, which is not allowed. So k must be less than -1. But how does this reconcile with the other interval?Wait, let's think again. Our conclusion was that k must be greater than -√3 and less than -1. But when k is less than -1, how does that interact with the first interval condition?Wait, perhaps there's a miscalculation here. Let's retrace.Original transformed inequality after substitution θ = x + π/3: sin θ + k cos θ > 0 for all θ ∈ [π/3, 5π/4]So, for θ ∈ [π/3, 5π/4], sin θ + k cos θ > 0.We need to find k such that this inequality holds for all θ in that interval.Let's analyze the function F(θ) = sin θ + k cos θ.We need F(θ) > 0 for all θ ∈ [π/3, 5π/4].The idea is to find the minimum of F(θ) over this interval and set it to be greater than 0.To find the minimum, let's compute the derivative:F’(θ) = cos θ - k sin θSet F’(θ) = 0:cos θ - k sin θ = 0 ⇒ cos θ = k sin θ ⇒ tan θ = 1/kSo critical points occur where tan θ = 1/k.We need to check if these critical points are in the interval [π/3, 5π/4].Depending on k, tan θ = 1/k may have solutions in this interval.So, the possible minima can be at critical points or at the endpoints θ = π/3 or θ = 5π/4.Therefore, to find the minimum, evaluate F(θ) at θ = π/3, θ = 5π/4, and at any critical points within the interval.Let's compute F(π/3) = sin(π/3) + k cos(π/3) = (√3/2) + k*(1/2)F(5π/4) = sin(5π/4) + k cos(5π/4) = (-√2/2) + k*(-√2/2) = -√2/2 (1 + k)Now, check critical points where tan θ = 1/k. Let θ0 be such that θ0 = arctan(1/k). If θ0 ∈ [π/3, 5π/4], then we need to evaluate F(θ0).But depending on k, θ0 can be in different quadrants.First, consider k > 0:If k > 0, then 1/k > 0, so θ0 is in the first or third quadrant. However, our interval [π/3, 5π/4] spans from 60 degrees to 225 degrees. So θ0 could be in the first or third quadrant.But if k > 0, tan θ = 1/k is positive, so θ0 is in the first or third quadrant. However, in the interval [π/3, 5π/4], possible θ0 would be in [π/3, π/2) or (π, 5π/4]. Let's check:If 1/k > tan(5π/4) = 1, then θ0 < 5π/4. Wait, if k > 0, then 1/k can be any positive number. So:- If 0 < k < 1, then 1/k > 1, so θ0 = arctan(1/k) ∈ (0, π/2). But our interval starts at π/3 ≈ 1.047, so if 1/k > tan(π/3) ≈ 1.732, i.e., k < 1/1.732 ≈ 0.577, then arctan(1/k) ∈ (0, π/3), which is outside our interval. If k ≥ 0.577, then arctan(1/k) ∈ [π/3, π/2).- If k ≥ 1, then 1/k ≤ 1, so θ0 = arctan(1/k) ∈ (0, π/4], which is outside our interval.Wait, this seems complicated. Maybe split into cases based on the value of k.But since we derived earlier that k must be between -√3 and -1, which is negative, perhaps k is negative. Let's focus on k < 0.If k < 0, then 1/k < 0, so tan θ = 1/k is negative. Therefore, θ0 is in the fourth or second quadrant. But our interval [π/3, 5π/4] is from 60 degrees to 225 degrees. In this interval, tan θ is negative in (π/2, π), so θ0 ∈ (π/2, π) if 1/k is negative.So, θ0 = π + arctan(1/k). Wait, no. The general solution for tan θ = 1/k is θ = arctan(1/k) + nπ. But since 1/k is negative (as k < 0), arctan(1/k) is in (-π/2, 0). So θ0 = arctan(1/k) + π, which places θ0 in (π/2, π). Therefore, θ0 ∈ (π/2, π).Therefore, for k < 0, the critical point θ0 = arctan(1/k) + π is in (π/2, π), which is within our interval [π/3, 5π/4]. So, we need to evaluate F(θ0) at this critical point.But calculating F(θ0) where θ0 = arctan(1/k) + π.Wait, alternatively, since tan θ = 1/k, then sin θ = 1 / sqrt(1 + k²) and cos θ = k / sqrt(1 + k²), but considering the quadrant.Wait, θ is in (π/2, π), so sin θ is positive, cos θ is negative.Given tan θ = 1/k (which is negative, since k < 0), so 1/k is negative. Let’s set 1/k = -m, where m > 0. Then tan θ = -m. So θ is in (π/2, π), so we can write θ = π - arctan(m).Then, sin θ = sin(π - arctan(m)) = sin(arctan(m)) = m / sqrt(1 + m²)cos θ = cos(π - arctan(m)) = -cos(arctan(m)) = -1 / sqrt(1 + m²)But m = -1/k, since m = -1/k (because 1/k = -m). Therefore:sin θ = (-1/k) / sqrt(1 + (1/k)^2) = (-1/k) / sqrt((k² + 1)/k²) = (-1/k) / (sqrt(k² + 1)/|k|) = (-1/k) * (|k| / sqrt(k² + 1))Since k < 0, |k| = -k, so:sin θ = (-1/k) * (-k / sqrt(k² + 1)) = 1 / sqrt(k² + 1)Similarly, cos θ = -1 / sqrt(k² + 1)Therefore, F(θ0) = sin θ0 + k cos θ0 = [1 / sqrt(k² + 1)] + k [ -1 / sqrt(k² + 1) ] = [1 - k] / sqrt(k² + 1)But since k is negative, 1 - k > 1, so F(θ0) is positive. Wait, this suggests that at the critical point θ0, F(θ0) is positive. Therefore, the minimum must be at one of the endpoints.Wait, but this contradicts our previous analysis. Let me check this again.If F(θ0) = [1 - k]/sqrt(k² + 1), and k < 0, then since k is negative, 1 - k > 1, so F(θ0) is positive. Therefore, the minimum of F(θ) in the interval [π/3, 5π/4] must occur at one of the endpoints θ = π/3 or θ = 5π/4.Therefore, to ensure F(θ) > 0 for all θ ∈ [π/3, 5π/4], we just need to ensure that F(π/3) > 0 and F(5π/4) > 0.Compute F(π/3) = sin(π/3) + k cos(π/3) = (√3/2) + k*(1/2)F(5π/4) = sin(5π/4) + k cos(5π/4) = (-√2/2) + k*(-√2/2) = (-√2/2)(1 + k)Thus, the conditions are:1. (√3/2) + (k/2) > 02. (-√2/2)(1 + k) > 0Let's solve these inequalities.First inequality:(√3/2) + (k/2) > 0 ⇒ k > -√3Second inequality:(-√2/2)(1 + k) > 0 ⇒ (1 + k) < 0 ⇒ k < -1Therefore, combining both conditions: -√3 < k < -1Therefore, the range of k is (-√3, -1)So, that's the answer. Let me verify once again.If k is greater than -√3, then F(π/3) = √3/2 + k/2 > 0. If k < -1, then F(5π/4) = (-√2/2)(1 + k) > 0 since (1 + k) < 0. So both conditions are satisfied. Therefore, the range is indeed (-√3, -1).Final Answer(1) The adjoint vector is boxed{left( -dfrac{sqrt{3}}{2}, dfrac{3}{2} right)} and its magnitude is boxed{sqrt{3}};(2) The range of k is boxed{(-sqrt{3}, -1)}.</think>(1) To find the adjoint vector (overrightarrow{OM}) of the function (g(x) = cosleft(x + frac{pi}{3}right) + cos x), we use the cosine addition formula:[cosleft(x + frac{pi}{3}right) = cos x cosleft(frac{pi}{3}right) - sin x sinleft(frac{pi}{3}right) = frac{1}{2} cos x - frac{sqrt{3}}{2} sin x]Adding (cos x) to this result, we get:[g(x) = left(frac{1}{2} cos x - frac{sqrt{3}}{2} sin xright) + cos x = frac{3}{2} cos x - frac{sqrt{3}}{2} sin x]Thus, the adjoint vector (overrightarrow{OM}) is (left(-frac{sqrt{3}}{2}, frac{3}{2}right)). The magnitude of this vector is:[left|overrightarrow{OM}right| = sqrt{left(-frac{sqrt{3}}{2}right)^2 + left(frac{3}{2}right)^2} = sqrt{frac{3}{4} + frac{9}{4}} = sqrt{3}](2) The adjoint function of the vector (overrightarrow{ON} = (1, sqrt{3})) is (varphi(x) = sin x + sqrt{3} cos x). We need to ensure the inequality (varphi(x) + k varphileft(x + frac{pi}{2}right) > 0) holds for all (x in [0, frac{11pi}{12}]). First, we express (varphileft(x + frac{pi}{2}right)):[varphileft(x + frac{pi}{2}right) = sinleft(x + frac{pi}{2}right) + sqrt{3} cosleft(x + frac{pi}{2}right) = cos x - sqrt{3} sin x]Substituting into the inequality, we get:[(sin x + sqrt{3} cos x) + k (cos x - sqrt{3} sin x) > 0]Grouping like terms:[sin x (1 - ksqrt{3}) + cos x (sqrt{3} + k) > 0]We analyze this expression by converting it to the form (R sin(x + theta)) and ensuring it is positive for all (x) in the given interval. After evaluating the critical points and endpoints, we find the conditions:1. (k > -sqrt{3})2. (k < -1)Combining these conditions, the range of (k) is:[(-sqrt{3}, -1)]Final answers:(1) The adjoint vector is (boxed{left( -dfrac{sqrt{3}}{2}, dfrac{3}{2} right)}) and its magnitude is (boxed{sqrt{3}});(2) The range of (k) is (boxed{(-sqrt{3}, -1)}).]

🔑:Let's break down the solution step by step, following the given problem parts: Part (1): Finding the Adjoint Vector overrightarrow{OM} and its MagnitudeGiven the function g(x)=cos(x+frac{pi}{3})+cos x, we can use trigonometric identities to rewrite it as:begin{align*}g(x) &= cos(x+frac{pi}{3})+cos x &= left(cos x cos frac{pi}{3} - sin x sin frac{pi}{3}right) + cos x &= cos x left(frac{1}{2}right) - sin x left(frac{sqrt{3}}{2}right) + cos x &= -frac{sqrt{3}}{2}sin x + frac{3}{2}cos x.end{align*}Thus, the adjoint vector overrightarrow{OM} is (-frac{sqrt{3}}{2},frac{3}{2}).To find the magnitude |overrightarrow{OM}|, we calculate:begin{align*}|overrightarrow{OM}| &= sqrt{left(-frac{sqrt{3}}{2}right)^2 + left(frac{3}{2}right)^2} &= sqrt{frac{3}{4} + frac{9}{4}} &= sqrt{3}.end{align*}Therefore, the magnitude of the adjoint vector overrightarrow{OM} is boxed{sqrt{3}}. Part (2): Finding the Range of Real Number kThe adjoint function of overrightarrow{ON}=(1,sqrt{3}) is varphi(x)=sin x + sqrt{3}cos x = 2sin(x+frac{pi}{3}).Given varphi(x)+kvarphi(x+frac{pi}{2}) > 0, we substitute varphi(x) and simplify:begin{align*}2sin(x+frac{pi}{3}) + 2kcos(x+frac{pi}{3}) &> 0 sin(x+frac{pi}{3}) + kcos(x+frac{pi}{3}) &> 0.end{align*}We analyze this inequality in three intervals:1. For 0 leqslant x 0. - Thus, k > -tan(x+frac{pi}{3}). - The maximum of -tan(x+frac{pi}{3}) is -sqrt{3}, hence k > -sqrt{3}.2. When x = frac{pi}{6}: - The inequality simplifies to 1 > 0, which always holds.3. For frac{pi}{6} < x leqslant frac{11pi}{12}: - frac{pi}{2} < x+frac{pi}{3} leqslant frac{5pi}{4}, so cos(x+frac{pi}{3}) < 0. - Thus, k < -tan(x+frac{pi}{3}). - The minimum of -tan(x+frac{pi}{3}) is -1, hence k < -1.Combining these conditions, we find the range of k is boxed{(-sqrt{3}, -1)}.

🧠:[<think>Okay, so I have this problem here about finding the number closest to P_{1991}, where P_n is a product of fractions from T_2/(T_2 -1) up to T_n/(T_n -1). And T_n is the nth triangular number, which is 1 + 2 + 3 + ... + n. The answer choices are 2.0, 2.3, 2.6, 2.9, and 3.2. Hmm, let me try to figure this out step by step.First, I remember that the triangular number T_n is given by the formula T_n = n(n + 1)/2. Let me confirm that. Yeah, if n=1, T_1=1(2)/2=1, which is correct. For n=2, T_2=2(3)/2=3. So T_2=3, T_3=6, T_4=10, and so on. That makes sense.So the problem is about the product P_n, which is a product starting from T_2/(T_2 -1) multiplied by T_3/(T_3 -1), up to T_n/(T_n -1). So for each term in the product, it's T_k over (T_k - 1), starting from k=2 up to k=n.Let me write that out more formally. P_n = product from k=2 to k=n of [T_k / (T_k -1)]. And we need to compute P_{1991} and see which of the given options it's closest to.First, let's try to express each term T_k / (T_k -1) in terms of k. Since T_k = k(k + 1)/2, so substituting into the term:T_k / (T_k -1) = [k(k + 1)/2] / [k(k + 1)/2 -1]Let me compute the denominator: k(k + 1)/2 - 1 = [k(k +1) - 2]/2. So the entire fraction becomes:[k(k +1)/2] / [ (k(k +1) - 2)/2 ] = [k(k +1)/2] * [2 / (k(k +1) - 2)] = k(k +1) / [k(k +1) - 2]So each term simplifies to k(k +1) divided by [k(k +1) - 2]. Let me compute that for some small k to see if there's a pattern.Take k=2: T_2=3, so term is 3/(3-1)=3/2. According to the formula above: 2*3/(2*3 - 2)=6/(6 -2)=6/4=3/2. Correct.k=3: T_3=6, so term is 6/(6-1)=6/5. Using formula: 3*4/(3*4 - 2)=12/10=6/5. Correct.k=4: T_4=10, so term is 10/(10 -1)=10/9. Formula: 4*5/(4*5 -2)=20/18=10/9. Correct.So the formula works. Therefore, each term is [k(k +1)] / [k(k +1) - 2]. Let's see if we can factor the denominator.The denominator is k(k +1) -2. Let's expand that: k^2 +k -2. Let's factor this quadratic. Looking for two numbers that multiply to -2 and add to +1. Those numbers are +2 and -1. Therefore, k^2 +k -2 = (k +2)(k -1). Let me check: (k +2)(k -1) = k^2 -k +2k -2 =k^2 +k -2. Perfect.So denominator factors as (k +2)(k -1). Therefore, the term [k(k +1)] / [ (k +2)(k -1) ].So each term in the product is [k(k +1)] / [(k -1)(k +2)]. Therefore, P_n is the product from k=2 to k=n of [k(k +1)] / [(k -1)(k +2)].Let me write that out for a few terms to see if there's a telescoping product.For k=2: [2*3] / [1*4] = 6/4 = 3/2For k=3: [3*4] / [2*5] = 12/10 = 6/5For k=4: [4*5] / [3*6] = 20/18 = 10/9For k=5: [5*6]/[4*7] = 30/28 = 15/14So each term is [k(k +1)] / [(k -1)(k +2)]. Let's see if there's a telescoping pattern here.Let me separate the terms into numerators and denominators.The product is:Product from k=2 to n of [k(k +1)] / [(k -1)(k +2)]Let me split the product into numerator product and denominator product:Numerator product: Product from k=2 to n of k(k +1) = [Product from k=2 to n of k] * [Product from k=2 to n of (k +1)]Denominator product: Product from k=2 to n of (k -1)(k +2) = [Product from k=2 to n of (k -1)] * [Product from k=2 to n of (k +2)]So compute each part:First, numerator:Product from k=2 to n of k is 2*3*4*...*n = n! / 1! = n!Product from k=2 to n of (k +1) = 3*4*5*...*(n +1) = (n +1)! / 2! = (n +1)! / 2Therefore, numerator product is n! * (n +1)! / 2Denominator:Product from k=2 to n of (k -1) = 1*2*3*...*(n -1) = (n -1)!Product from k=2 to n of (k +2) = 4*5*6*...*(n +2) = (n +2)! / (3! ) = (n +2)! / 6Therefore, denominator product is (n -1)! * (n +2)! / 6Therefore, putting numerator and denominator together:P_n = [n! * (n +1)! / 2] / [ (n -1)! * (n +2)! / 6 ] = [n! * (n +1)! / 2] * [6 / ( (n -1)! * (n +2)! ) ]Simplify this expression.First, note that n! = n * (n -1)! , so we can write n! = n * (n -1)!.Similarly, (n +1)! = (n +1) * n!But maybe let's substitute n! = n * (n -1)! into the numerator:Numerator: n! * (n +1)! = [n * (n -1)! ] * [ (n +1) * n! ] = n * (n -1)! * (n +1) * n! = n*(n +1) * (n -1)! * n!But perhaps this isn't the simplest way. Let me see:Alternatively, let's separate the terms:Numerator: n! * (n +1)! = n! * (n +1) * n! = (n!)^2 * (n +1)Denominator: (n -1)! * (n +2)! = (n -1)! * (n +2)(n +1)n! )Wait, perhaps step by step:(n +2)! = (n +2)(n +1)n!Therefore, denominator: (n -1)! * (n +2)! = (n -1)! * (n +2)(n +1)n! )So putting back into the expression:P_n = [ (n! * (n +1)! ) / 2 ] * [ 6 / ( (n -1)! * (n +2)! ) ]Substitute the expressions:= [ (n! * (n +1) * n! ) / 2 ] * [ 6 / ( (n -1)! * (n +2)(n +1) n! ) ) ]Simplify numerator and denominator:Numerator: n! * n! * (n +1) / 2Denominator: (n -1)! * (n +2)(n +1) n! )So P_n = [ (n! * n! * (n +1) ) / 2 ] * [ 6 / ( (n -1)! * (n +2)(n +1) n! ) ) ]Let me cancel out terms:First, the n! in the numerator of the first fraction cancels with the n! in the denominator of the second fraction.So now we have:= [ (n! * (n +1) ) / 2 ] * [ 6 / ( (n -1)! * (n +2)(n +1) ) ]Note that n! = n * (n -1)! so we can substitute that:= [ (n * (n -1)! ) * (n +1) / 2 ] * [ 6 / ( (n -1)! * (n +2)(n +1) ) ]Now, (n -1)! in numerator and denominator cancels.So:= [ n * (n +1) / 2 ] * [ 6 / ( (n +2)(n +1) ) ]Cancel out (n +1):= [ n / 2 ] * [ 6 / (n +2) ) ]Multiply the terms:= (n * 6) / (2(n +2)) ) = (6n) / (2(n +2)) ) = (3n) / (n +2)So after all that cancellation, P_n simplifies to 3n / (n +2).Wait, that's a huge simplification! Let me check this with the initial terms.Let me check n=2. Then P_2 = T_2/(T_2 -1) = 3/(3 -1)=3/2=1.5. According to the formula 3n/(n +2)=3*2/(2 +2)=6/4=1.5. Correct.For n=3, P_3 = [3/2] * [6/5] = (3/2)*(6/5)=18/10=9/5=1.8. According to the formula, 3*3/(3 +2)=9/5=1.8. Correct.For n=4, P_4 = [3/2] * [6/5] * [10/9] = (3/2)*(6/5)*(10/9). Let's compute:(3/2)*(6/5) = (18/10) = 9/5, then 9/5 *10/9 = 10/5=2. Formula gives 3*4/(4 +2)=12/6=2. Correct.So the formula works. Therefore, in general, P_n = 3n / (n +2). So then, P_{1991} = 3*1991 / (1991 +2) = 3*1991 /1993.So compute 3*1991 /1993. Let me compute that.First, 3*1991=5973. Then, 5973 divided by 1993.Let me compute 1993*3=5979. So 5973 is 5979 -6=1993*3 -6. Therefore, 5973/1993=3 -6/1993.Therefore, 3 - 6/1993≈3 -0.00301≈2.99699.Wait, but the answer choices are 2.0, 2.3, 2.6, 2.9, 3.2. So 2.99699 is approximately 3.0, but the closest option is 2.9 or 3.2. Since 3.0 is exactly between 2.9 and 3.2, but 2.99699 is 0.00301 away from 3.0, so it's 2.99699, which is approximately 3.0. But 3.0 is not one of the options. Wait, the options are 2.0, 2.3, 2.6, 2.9, 3.2. So 3.0 would be closest to 2.9 or 3.2?Wait, 2.99699 is 2.99699, which is 2.99699. So 2.99699 is 0.00301 away from 3.0. The difference between 2.99699 and 2.9 is 0.09699, and the difference to 3.2 is 0.20301. Therefore, 2.99699 is closer to 3.0, but since 3.0 is not an option, the closest given option is 2.9. But wait, 2.99699 is almost 3.0. Wait, the answer choices are (A) 2.0, (B) 2.3, (C) 2.6, (D) 2.9, (E) 3.2. So 2.99699 is 0.09699 above 2.9 and 0.203 below 3.2. Therefore, 0.09699 is smaller than 0.203, so it is closer to 2.9. Therefore, the answer is D) 2.9.Wait, but wait, hold on, maybe I made a mistake here. Let me check the arithmetic again.Wait, we have P_n = 3n / (n + 2). So for n=1991, that's 3*1991/(1991 + 2)=5973/1993.Let me compute 1993 divided into 5973. Let's see:1993*3=5979, which is 6 more than 5973, so 5973=1993*3 -6. Therefore, 5973/1993=3 -6/1993. So 3 minus a small fraction.Compute 6/1993. Let's compute that as a decimal.1993 divided by 6 is approximately 332.1666, but wait, 6/1993 is approximately 0.00301. So 3 -0.00301=2.99699. So approximately 2.997. So 2.997 is very close to 3.0, but the options are 2.9 and 3.2. Since 2.997 is 0.003 away from 3.0 and 0.097 away from 2.9. So 0.003 is much closer. But wait, 3.0 is not an option. Wait, but maybe the options are approximate, but according to the problem statement, the answer must be one of the given choices. Wait, but 2.997 is closest to 3.0, which isn't an option, but 2.9 is 0.097 less, and 3.2 is 0.203 higher. Therefore, 0.097 is smaller than 0.203, so 2.9 is the closest. So answer D.But wait, the problem is from the 42nd American High School Mathematics Examination, 1991. Maybe there's a trick here. Wait, maybe I made a mistake in simplifying the product. Let me double-check the simplification steps.We had P_n = product from k=2 to n of [k(k +1)] / [(k -1)(k +2)]. Then we split into numerator and denominator products.Numerator: product of k(k +1) from k=2 to n.Denominator: product of (k -1)(k +2) from k=2 to n.Expressed as:Numerator: [2*3*...*n] * [3*4*...*(n +1)] = (n! /1!) * ((n +1)!)/2! ) = n! * (n +1)! / 2.Denominator: [1*2*...*(n -1)] * [4*5*...*(n +2)] = ( (n -1)! ) * ( (n +2)! ) / (3! ) ) = (n -1)! * (n +2)! ) / 6.Therefore, P_n = [n! (n +1)! / 2 ] / [ (n -1)! (n +2)! / 6 ] = [n! (n +1)! * 6 ] / [ 2 (n -1)! (n +2)! ) ]Simplify numerator and denominator:Cancel 6 and 2 to get 3.So P_n = [3 n! (n +1)! ] / [ (n -1)! (n +2)! ) ]Now, note that n! = n (n -1)! and (n +2)! = (n +2)(n +1)!.Therefore, substitute:P_n = [3 * n (n -1)! * (n +1)! ] / [ (n -1)! * (n +2)(n +1)! ) ] = [3n ] / (n +2 )Yes, that's correct. So P_n = 3n / (n +2). Therefore, for n=1991, P_{1991} = 3*1991 /1993 ≈ 2.997, which is approximately 3.0, but since that's not an option, the closest is 2.9.But wait, the problem says "the number closest to P_{1991}" among the options given. The options are 2.0, 2.3, 2.6, 2.9, 3.2. So 2.997 is very close to 3.0, but since 3.0 isn't there, 2.9 is closer than 3.2. Because 2.997 -2.9=0.097, and 3.2 -2.997=0.203, so 0.097 <0.203, so 2.9 is closer.Therefore, the answer is D) 2.9. But wait, let me check if I made a mistake in the algebra.Wait, 3n/(n +2). For n=1991, that's 3*1991/(1991 +2)=5973/1993. Let's compute 5973 divided by 1993.Alternatively, note that 1993*3=5979, which is 6 more than 5973, so 5973=1993*3 -6. Therefore, 5973/1993=3 -6/1993.So, 6/1993 is approximately 0.00301, so 3 -0.00301≈2.99699. So approximately 2.997. So this is 2.997, which is 0.003 away from 3.0. But since the options are 2.9 and 3.2, the closest is 2.9. Because 2.997 is 0.097 above 2.9 and 0.203 below 3.2. So 0.097 <0.203, so 2.9 is closer. Therefore, the answer is D) 2.9.But wait, let me verify once again with another approach. Let me compute the exact decimal of 5973/1993.Compute 1993*2=3986. 1993*3=5979. So 5973 is 5979 -6. Therefore, 5973/1993=3 -6/1993≈3 -0.00301≈2.99699. So approximately 2.997, which is 2.997. So 2.997 is indeed closest to 3.0, but since 3.0 isn't an option, among the given choices, 2.9 is closer. Therefore, the answer is D.Alternatively, maybe the original problem has a typo, but given the options, D) 2.9 is the closest. Therefore, the correct answer should be D) 2.9.Final Answerboxed{D}</think>Given the problem, we need to find the number closest to ( P_{1991} ), where ( P_n ) is defined as the product:[P_n = frac{T_2}{T_2 - 1} cdot frac{T_3}{T_3 - 1} cdot frac{T_4}{T_4 - 1} cdots frac{T_n}{T_n - 1}]and ( T_n ) is the ( n )-th triangular number, given by ( T_n = frac{n(n+1)}{2} ).First, we express each term ( frac{T_k}{T_k - 1} ) in terms of ( k ):[frac{T_k}{T_k - 1} = frac{frac{k(k+1)}{2}}{frac{k(k+1)}{2} - 1} = frac{k(k+1)}{k(k+1) - 2} = frac{k(k+1)}{(k-1)(k+2)}]This simplifies each term in the product to:[frac{k(k+1)}{(k-1)(k+2)}]We then consider the product ( P_n ):[P_n = prod_{k=2}^{n} frac{k(k+1)}{(k-1)(k+2)}]This product can be simplified by recognizing it as a telescoping product. Breaking it down into separate products for the numerator and denominator, we get:[P_n = frac{prod_{k=2}^{n} k(k+1)}{prod_{k=2}^{n} (k-1)(k+2)}]Simplifying the numerator and denominator separately:- The numerator product is ( prod_{k=2}^{n} k cdot prod_{k=2}^{n} (k+1) = n! cdot frac{(n+1)!}{2} )- The denominator product is ( prod_{k=2}^{n} (k-1) cdot prod_{k=2}^{n} (k+2) = (n-1)! cdot frac{(n+2)!}{6} )Putting these together, we get:[P_n = frac{n! cdot frac{(n+1)!}{2}}{(n-1)! cdot frac{(n+2)!}{6}} = frac{3n}{n+2}]For ( n = 1991 ):[P_{1991} = frac{3 cdot 1991}{1991 + 2} = frac{5973}{1993}]Approximating this value:[frac{5973}{1993} approx 3 - frac{6}{1993} approx 3 - 0.00301 approx 2.997]The closest value to 2.997 among the given options (2.0, 2.3, 2.6, 2.9, 3.2) is 2.9. Therefore, the answer is:[boxed{D}],<think>Okay, so I have this problem here where I need to find the number closest to ( P_{1991} ). The options are (A) 2.0, (B) 2.3, (C) 2.6, (D) 2.9, and (E) 3.2. Let me try to break this down step by step.First, let's understand what ( T_n ) is. The problem says ( T_n = 1 + 2 + 3 + cdots + n ). Oh, that's the nth triangular number. I remember the formula for the nth triangular number is ( T_n = frac{n(n+1)}{2} ). Let me verify that real quick. For example, if n=2, then ( T_2 = 1 + 2 = 3 ), and using the formula, ( frac{2 times 3}{2} = 3 ). Yep, that works. Okay, so ( T_n = frac{n(n+1)}{2} ).Now, ( P_n ) is defined as the product from k=2 to n of ( frac{T_k}{T_k - 1} ). So, ( P_n = prod_{k=2}^n frac{T_k}{T_k - 1} ). I need to compute this product up to n=1991 and see which of those options it's closest to. Hmm, that seems like a huge product. There must be a way to simplify this product before trying to compute it for such a large n.Let me write out the first few terms to see if I can spot a pattern. Let's start with n=2:For n=2: ( P_2 = frac{T_2}{T_2 - 1} = frac{3}{3 - 1} = frac{3}{2} = 1.5 ).For n=3: ( P_3 = frac{T_2}{T_2 - 1} cdot frac{T_3}{T_3 - 1} ). Let's compute each term:( T_2 = 3 ), so first term is ( frac{3}{2} ).( T_3 = 6 ), so second term is ( frac{6}{5} ).Thus, ( P_3 = frac{3}{2} times frac{6}{5} = frac{18}{10} = 1.8 ).For n=4: ( P_4 = P_3 times frac{T_4}{T_4 - 1} ).( T_4 = 10 ), so the next term is ( frac{10}{9} ).Thus, ( P_4 = 1.8 times frac{10}{9} = 1.8 times 1.overline{1} approx 2.0 ).Wait, so P_4 is approximately 2.0. Hmm, but let me compute exactly:( frac{18}{10} times frac{10}{9} = frac{18}{9} = 2 ). Oh, that's exactly 2. So P_4 = 2.Interesting, that's exactly one of the answer choices. Let's do n=5.( T_5 = 15 ), so term is ( frac{15}{14} ).Thus, P_5 = 2 × 15/14 ≈ 2.142857.n=6: T_6 = 21, term is 21/20.P_6 ≈ 2.142857 × 1.05 ≈ 2.25.n=7: T_7=28, term is 28/27≈1.037. So P_7≈2.25 × 1.037≈2.333.n=8: T_8=36, term is 36/35≈1.0286. So P_8≈2.333×1.0286≈2.4.n=9: T_9=45, term is 45/44≈1.0227. So P_9≈2.4×1.0227≈2.4545.n=10: T_10=55, term is 55/54≈1.0185. So P_10≈2.4545×1.0185≈2.5.Hmm, so as n increases, P_n seems to be increasing, but how much does it increase each time? The terms are getting closer to 1, so the increments are getting smaller. The question is, does this product approach a finite limit as n approaches infinity, or does it diverge? If it converges, then perhaps for n=1991, it's very close to that limit. If it diverges, it might be going to infinity, but given the answer choices, the closest numbers are up to 3.2, so maybe it converges to a value around there.But first, let's try to find a general expression for P_n. Let's see:Given that ( T_k = frac{k(k+1)}{2} ), so ( T_k - 1 = frac{k(k+1)}{2} - 1 = frac{k(k+1) - 2}{2} = frac{k^2 + k - 2}{2} ). Let's factor that denominator:( k^2 + k - 2 = (k + 2)(k - 1) ). Let's check: (k + 2)(k - 1) = k^2 -k + 2k -2 = k^2 +k -2. Yes, correct. So ( T_k - 1 = frac{(k + 2)(k - 1)}{2} ).Therefore, each term in the product is ( frac{T_k}{T_k - 1} = frac{frac{k(k+1)}{2}}{frac{(k + 2)(k - 1)}{2}} = frac{k(k +1)}{(k + 2)(k - 1)} ).So, ( P_n = prod_{k=2}^n frac{k(k +1)}{(k + 2)(k - 1)} ).Let me write that out:( P_n = prod_{k=2}^n frac{k(k +1)}{(k - 1)(k + 2)} ).Hmm, this looks like a telescoping product. Telescoping products are products where terms cancel out when written out. Let me try to write each term as separate fractions to see if there's cancellation.Let's note that each term is:( frac{k}{k - 1} times frac{k +1}{k + 2} ).So, each term can be split into two fractions: one that is ( frac{k}{k - 1} ) and another that is ( frac{k +1}{k + 2} ).Therefore, the product becomes:( prod_{k=2}^n left( frac{k}{k - 1} times frac{k +1}{k + 2} right ) = left( prod_{k=2}^n frac{k}{k - 1} right ) times left( prod_{k=2}^n frac{k +1}{k + 2} right ) ).Let's compute each product separately.First product: ( prod_{k=2}^n frac{k}{k - 1} ).This is a telescoping product. Let's write out the terms:For k=2: 2/1For k=3: 3/2For k=4: 4/3...For k=n: n/(n -1)When we multiply all these terms together, each numerator cancels with the denominator of the next term:2/1 × 3/2 × 4/3 × ... × n/(n -1) = n/1 = n.So, the first product simplifies to n.Second product: ( prod_{k=2}^n frac{k +1}{k + 2} ).Similarly, let's write out the terms:For k=2: (2 +1)/(2 +2) = 3/4For k=3: 4/5For k=4: 5/6...For k=n: (n +1)/(n +2)Again, this is a telescoping product. Each numerator cancels with the denominator of the previous term:3/4 × 4/5 × 5/6 × ... × (n +1)/(n +2) = 3/(n +2).Wait, let's check:First term: 3/4Second term: 4/5Third term:5/6...Last term: (n +1)/(n +2)So, multiplying all these terms:3/4 × 4/5 × 5/6 × ... × (n +1)/(n +2) = 3/(n +2). Because the 4 in the denominator cancels with the 4 in the numerator of the next term, the 5 in the denominator cancels with the 5 in the numerator, etc., leaving only the first numerator 3 and the last denominator (n +2). So, the second product simplifies to 3/(n +2).Therefore, combining both products:( P_n = n times frac{3}{n + 2} = frac{3n}{n + 2} ).Wait, really? Let me verify this with the earlier examples.For n=2: ( P_2 = frac{3*2}{2 +2} = 6/4 = 3/2 = 1.5 ). Which matches our previous calculation.For n=3: ( P_3 = frac{3*3}{3 +2} = 9/5 = 1.8 ). Also matches.For n=4: ( 3*4/(4 +2) = 12/6 = 2 ). Correct.For n=5: 15/7 ≈ 2.1428. Which matches our earlier result.n=10: ( 3*10/(10 +2) = 30/12 = 2.5 ). Yep, that's what we had. So this formula seems correct.Therefore, general formula is ( P_n = frac{3n}{n + 2} ). That simplifies everything!So, the problem is asking for ( P_{1991} ). Let's compute that.( P_{1991} = frac{3 times 1991}{1991 + 2} = frac{5973}{1993} ).Now, let's compute this division. Let's see, 1993 × 3 = 5979. So 5973 is 6 less than 5979, so:( frac{5973}{1993} = 3 - frac{6}{1993} ).Compute ( 6 / 1993 ). Let's approximate that.1993 is approximately 2000. So 6/2000 = 0.003. Therefore, 6/1993 ≈ 0.003 - a little bit more. Let's compute it more precisely.1993 × 0.003 = 5.979. So 6 - 5.979 = 0.021. Therefore, 0.021 / 1993 ≈ 0.0000105. So, 6/1993 ≈ 0.003 + 0.0000105 ≈ 0.0030105. Therefore, approximately 0.00301.Thus, ( 3 - 0.00301 ≈ 2.99699 ).So, ( P_{1991} ≈ 2.997 ), which is approximately 2.997. Comparing this to the answer choices:(A) 2.0(B) 2.3(C) 2.6(D) 2.9(E) 3.2So, 2.997 is very close to 3.0, which is not listed, but the closest given option is (D) 2.9 and (E) 3.2. Wait, 2.997 is 0.003 less than 3.0, so it's approximately 2.997, which is 2.997 - 2.9 = 0.097 above 2.9, and 3.2 - 2.997 = 0.203. So it's closer to 2.9 by 0.097, and to 3.2 by 0.203. Therefore, 2.9 is closer.But wait, let me check the exact value. Let's compute 5973 divided by 1993.Let me perform the division step by step.Compute 5973 ÷ 1993.First, note that 1993 × 3 = 5979, as above. Therefore, 1993 × 2.997 = 5979 - 1993 × 0.003.Wait, maybe another approach. Let's compute 5973 ÷ 1993.Since 1993 × 3 = 5979, so 5973 is 6 less than 5979. So,5973 = 1993 × 3 - 6.Therefore,5973/1993 = 3 - 6/1993 ≈ 3 - 0.00301 ≈ 2.99699.So, approximately 2.997, which is 2.997. So, if the options are 2.0, 2.3, 2.6, 2.9, 3.2, then the closest is 3.0, which is not listed, but between 2.9 and 3.2, which is it closer to?Compute the differences:Difference between 2.997 and 2.9: 0.097Difference between 3.2 and 2.997: 0.203Since 0.097 < 0.203, 2.997 is closer to 2.9.But wait, hold on. Wait, 2.997 is almost 3.0. Maybe the options are in increments, and 2.9 is option D, while 3.2 is E. But 2.997 is just 0.003 away from 3.0. However, since 3.0 isn't an option, we have to choose between D and E. So 2.997 is 0.097 away from D (2.9) and 0.203 away from E (3.2). Therefore, the closest is D, 2.9.Wait, but let me double-check my calculations. Maybe there's a mistake in deriving the general formula.Wait, according to the earlier steps, we split the product into two telescoping products:First product: product_{k=2}^n [k/(k-1)] = n/1 = n.Second product: product_{k=2}^n [(k+1)/(k+2)] = 3/(n+2).Therefore, multiplying these gives 3n/(n +2). So, P_n = 3n/(n +2). Therefore, for n=1991,P_{1991} = (3*1991)/(1991 +2) = 5973/1993 ≈ 2.997.Yes, that seems correct. So then, 2.997 is indeed approximately 3.0, but since that's not an option, we have to pick the closest. The options given are 2.0, 2.3, 2.6, 2.9, 3.2. So, 2.997 is between 2.9 and 3.2. Let's compute the exact differences:2.997 - 2.9 = 0.0973.2 - 2.997 = 0.203Since 0.097 is smaller than 0.203, the closest is 2.9. So the answer is D.But wait, hold on. Let me check with n=4, which we calculated as exactly 2.0. According to the formula, 3*4/(4 +2)=12/6=2. Correct. So formula holds.Similarly, n=10, 3*10/12=30/12=2.5. Correct. So formula is correct.Therefore, for n=1991, 3*1991/1993≈2.997. So answer is D, 2.9.But wait, 2.997 is almost 3.0. The options don't have 3.0, but 3.2 is the next one. Maybe there's a mistake here? Wait, no. Wait, 2.997 is 2.997, which is 2.9 when rounded to one decimal place. But the options are given as exact numbers: 2.0, 2.3, 2.6, 2.9, 3.2. So 2.997 is 0.097 away from 2.9, which is 9.7% of the distance to 3.2. Wait, but the options are not in percentages, but absolute differences. So, since 0.097 is less than 0.203, 2.9 is closer.Alternatively, maybe the question expects an approximate value? But given that 2.997 is approximately 3.0, but 3.0 is not an option, but 2.9 is. Wait, but 2.997 is 2.997, so to one decimal place, that would round to 3.0. But perhaps the answer choices are given in increments, and the closest one is 2.9. Hmm.Alternatively, maybe I made a miscalculation. Let me verify the formula once more.Original product: ( prod_{k=2}^n frac{T_k}{T_k -1} ). We transformed each term to ( frac{k(k+1)}{(k-1)(k+2)} ), then split into two products: ( prod frac{k}{k-1} ) and ( prod frac{k+1}{k+2} ). The first product telescopes to n, the second telescopes to 3/(n+2). Therefore, multiplying them gives 3n/(n+2). Yes, that seems correct.Therefore, the formula is correct. So, 3n/(n+2). For n=1991, that is 3*1991/(1993). Let me compute this division more accurately.Compute 5973 ÷ 1993.1993 × 3 = 5979So, 5973 = 5979 - 6Therefore, 5973/1993 = (5979 -6)/1993 = 3 - 6/1993Compute 6/1993:1993 × 3 = 5979, so 1993 is approximately 2000, so 6/1993 ≈ 6/2000 = 0.003, but since 1993 is 7 less than 2000, the actual value is slightly more than 0.003. Let's compute 6 ÷ 1993.1993 × 0.003 = 5.979So, 0.003 × 1993 = 5.979Then, 6 - 5.979 = 0.021Therefore, 0.021 ÷ 1993 ≈ 0.00001054Therefore, 6/1993 ≈ 0.003 + 0.00001054 ≈ 0.00301054Therefore, 3 - 0.00301054 ≈ 2.99698946So, approximately 2.996989, which is approximately 2.9970. So, 2.9970 is the value of P_{1991}.Looking at the options: (A) 2.0, (B) 2.3, (C) 2.6, (D) 2.9, (E) 3.2.Compute the absolute differences:|2.997 - 2.9| = 0.097|2.997 - 3.2| = 0.203Therefore, 0.097 < 0.203, so 2.9 is closer. Therefore, the answer is D) 2.9.But let me check once more if there was any miscalculation.Wait, 3n/(n +2). Let's take n approaching infinity. Then, 3n/(n +2) ≈ 3n/n = 3. So as n increases, the limit is 3. Therefore, for very large n, like 1991, the value is approaching 3 from below. So, 1991 is a large number, so it's very close to 3. But 3 is not an option. The closest is 2.9 and 3.2. Since it's approaching 3 from below, 2.997 is just shy of 3, so 0.003 less. So, 2.997 is 2.997, which is 2.9 when rounded to one decimal place, but 3.0 when rounded to the nearest tenth. However, since 2.997 is 2.997, which is 0.003 less than 3.0. But in the options given, there's no 3.0. Therefore, we have to pick between the given options. Since 2.997 is 2.997, which is 2.9 (if we truncate after one decimal), but actually, 2.997 is 3.0 when rounded to one decimal place because the second decimal is 9, which would round up. Wait, 2.997 to one decimal place:2.997 ≈ 3.0 (since the second decimal is 9, which is 5 or above, we round up the first decimal. But the first decimal is 9, so 2.9 becomes 3.0 when rounded to one decimal place. But the options are given as 2.0, 2.3, 2.6, 2.9, 3.2. So, if we follow standard rounding rules, 2.997 rounded to one decimal is 3.0, but that's not an option. Therefore, perhaps the question is asking for the closest number without rounding? That is, just the closest given value. In that case, 2.997 is 0.097 away from 2.9 and 0.203 away from 3.2, so 2.9 is closer. Therefore, the answer is D) 2.9.Alternatively, maybe the problem was presented differently? Wait, no, the problem says "the number closest to ( P_{1991} )" is one of the options. So, even though 2.997 is very close to 3, if 3 isn't an option, then we pick the closest given number, which is 2.9.Alternatively, maybe there is a mistake in the telescoping product?Wait, let me re-examine the splitting of the product.Original term: ( frac{k(k +1)}{(k -1)(k +2)} ).Split into two fractions: ( frac{k}{k -1} times frac{k +1}{k + 2} ).Then, product from k=2 to n of that is:( prod_{k=2}^n frac{k}{k -1} times prod_{k=2}^n frac{k +1}{k +2} ).First product: telescopes to n/1.Second product: telescopes to 3/(n +2).Multiply together: 3n/(n +2). That seems correct.Yes, so the formula is correct.Therefore, P_{1991} = 3*1991/(1991 +2) = 5973/1993 ≈ 2.997.Thus, 2.997 is closest to 2.9 (D). Therefore, the answer is D) 2.9.Final Answerboxed{D}</think>Given ( T_n = 1 + 2 + 3 + cdots + n ), which is the nth triangular number, we know ( T_n = frac{n(n+1)}{2} ). The problem requires finding the value of ( P_n ) defined as the product:[P_n = prod_{k=2}^n frac{T_k}{T_k - 1}]We start by expressing each term in the product in terms of ( k ):[frac{T_k}{T_k - 1} = frac{frac{k(k+1)}{2}}{frac{k(k+1)}{2} - 1} = frac{k(k+1)}{(k-1)(k+2)}]This can be split into two fractions:[frac{k}{k-1} times frac{k+1}{k+2}]We then consider the product from ( k=2 ) to ( n ):[P_n = left( prod_{k=2}^n frac{k}{k-1} right) times left( prod_{k=2}^n frac{k+1}{k+2} right)]The first product is a telescoping product:[prod_{k=2}^n frac{k}{k-1} = frac{2}{1} times frac{3}{2} times cdots times frac{n}{n-1} = n]The second product is also a telescoping product:[prod_{k=2}^n frac{k+1}{k+2} = frac{3}{4} times frac{4}{5} times cdots times frac{n+1}{n+2} = frac{3}{n+2}]Combining these results, we get:[P_n = n times frac{3}{n+2} = frac{3n}{n+2}]For ( n = 1991 ):[P_{1991} = frac{3 times 1991}{1991 + 2} = frac{5973}{1993} approx 2.997]The closest number to 2.997 among the given options is 2.9. Therefore, the answer is:[boxed{D}]]

🔑:［Solution］From the given, we knowfrac{T_{n}}{T_{n}-1}=frac{frac{n(n+1)}{2}}{frac{n(n+1)}{2}-1}=frac{n(n+1)}{n^{2}+n-2}=frac{n(n+1)}{(n+2)(n-1)},thenbegin{aligned}P_{1991} & =frac{2 cdot 3}{4 cdot 1} cdot frac{3 cdot 4}{5 cdot 2} cdot frac{4 cdot 5}{6 cdot 3} cdots frac{1990 cdot 1991}{1992 cdot 1989} cdot frac{1991 cdot 1992}{1993 cdot 1990} & =frac{3 cdot 1991}{1993} approx 2.997 .end{aligned}Therefore, the answer is (D).

🧠:[<think>Okay, so I have this problem with two parts. Let me start with part (1). I need to find the equations of an ellipse C₁ and a hyperbola C₂. The given information is that the foci of the ellipse C₁ are F₁(-1, 0) and F₂(1, 0), which are also the vertices of the hyperbola C₂. Additionally, one of their intersection points is M(2√3/3, √3/3). First, let's tackle the ellipse C₁. For an ellipse, the standard form is (x²/a²) + (y²/b²) = 1, where 2a is the major axis, 2b is the minor axis, and the distance between the center and each focus is c, with the relationship a² = b² + c². Given the foci are at (-1, 0) and (1, 0), that tells me the center of the ellipse is at the origin (0,0), since foci are symmetric about the center. The distance from the center to each focus is c = 1. Therefore, for the ellipse, c = 1. So, we need to find a and b. We know that a² = b² + c². But we also have a point M(2√3/3, √3/3) that lies on the ellipse. Let's plug that point into the ellipse equation to get an equation involving a and b. Plugging in x = 2√3/3 and y = √3/3:( ( (2√3/3)² ) / a² ) + ( ( (√3/3)² ) / b² ) = 1Calculating those squares:( (4*3)/9 ) / a² + ( (3)/9 ) / b² = 1Simplify:(12/9)/a² + (3/9)/b² = 1Which simplifies to:(4/3)/a² + (1/3)/b² = 1So, equation (1): (4)/(3a²) + 1/(3b²) = 1But we also know that for the ellipse, a² = b² + c², and c = 1, so a² = b² + 1.So, equation (2): a² = b² + 1Now, substitute equation (2) into equation (1):4/(3(b² + 1)) + 1/(3b²) = 1Multiply both sides by 3b²(b² + 1) to eliminate denominators:4b² + (b² + 1) = 3b²(b² + 1)Simplify left side: 4b² + b² + 1 = 5b² + 1Right side: 3b²(b² + 1) = 3b⁴ + 3b²Set equal:5b² + 1 = 3b⁴ + 3b²Bring all terms to one side:3b⁴ + 3b² -5b² -1 = 0 ⇒ 3b⁴ - 2b² -1 = 0Let me set u = b², so equation becomes 3u² -2u -1 = 0Solving quadratic equation:u = [2 ± √(4 + 12)] / 6 = [2 ± √16]/6 = [2 ± 4]/6So, u = (2 + 4)/6 = 6/6 = 1 or u = (2 -4)/6 = -2/6 = -1/3Since u = b² must be positive, u = 1. Hence, b² = 1 ⇒ b = 1Then from equation (2): a² = 1 + 1 = 2 ⇒ a = √2Therefore, the equation of the ellipse C₁ is x²/2 + y²/1 = 1, or x²/2 + y² = 1.Now, moving on to the hyperbola C₂. The vertices of the hyperbola are at (-1, 0) and (1, 0), so similar to the ellipse, the center is at (0,0). For a hyperbola, the standard form is x²/a_h² - y²/b_h² = 1, where 2a_h is the distance between the vertices, and the relationship between a_h, b_h, and c_h (distance from center to foci) is c_h² = a_h² + b_h².Here, the vertices are at (±1, 0), so a_h = 1. So, the equation of the hyperbola is x²/1² - y²/b_h² = 1, or x² - y²/b_h² = 1.We need to find b_h. Again, we have the intersection point M(2√3/3, √3/3) which lies on both the ellipse and the hyperbola. So, let's plug M into the hyperbola equation.Plugging in x = 2√3/3 and y = √3/3:( (2√3/3)² ) / 1 - ( (√3/3)² ) / b_h² = 1Calculating:( (4*3)/9 ) - ( (3)/9 ) / b_h² = 1 ⇒ (12/9) - (1/3)/b_h² = 1 ⇒ (4/3) - (1)/(3b_h²) = 1Subtract 4/3 from both sides:-1/(3b_h²) = 1 - 4/3 = -1/3Multiply both sides by -3:1/b_h² = 1 ⇒ b_h² = 1 ⇒ b_h = 1Therefore, the equation of the hyperbola C₂ is x² - y²/1 = 1, or x² - y² = 1.Wait, but let me check that again. If b_h² = 1, then the equation is x² - y²/1 = 1, so x² - y² = 1. That seems correct. But let's confirm with the point M.Plugging M into x² - y² = 1:( (2√3/3)^2 ) - ( (√3/3)^2 ) = (4*3)/9 - (3)/9 = 12/9 - 3/9 = 9/9 = 1. Correct.So that's the hyperbola.Therefore, part (1) answers:Ellipse C₁: x²/2 + y² = 1Hyperbola C₂: x² - y² = 1Now, moving on to part (2). If point P is a moving point on the right branch of the hyperbola (so x ≥ 1), and point Q is a moving point on the y-axis such that F₁P ⊥ F₁Q. We need to determine whether line PQ passes through a fixed point. If it does, find the coordinates; otherwise, explain.Let me parse this. F₁ is (-1, 0). So, given P on hyperbola C₂ (right branch, so x ≥ 1), and Q on y-axis (so Q has coordinates (0, q) for some q). The condition is that vector F₁P is perpendicular to vector F₁Q. Then, we need to check if the line PQ passes through a fixed point.First, let's recall that two vectors are perpendicular if their dot product is zero.Let me denote coordinates:F₁ = (-1, 0)Point P = (p_x, p_y), on hyperbola C₂, so p_x² - p_y² = 1, and p_x ≥ 1.Point Q = (0, q), on y-axis.Vectors F₁P and F₁Q:Vector F₁P = P - F₁ = (p_x - (-1), p_y - 0) = (p_x +1, p_y)Vector F₁Q = Q - F₁ = (0 - (-1), q - 0) = (1, q)Their dot product should be zero:(p_x +1)(1) + (p_y)(q) = 0So:(p_x +1) + p_y q = 0 ⇒ p_y q = - (p_x +1) ⇒ q = - (p_x +1)/p_yAssuming p_y ≠ 0. If p_y = 0, then from hyperbola equation x² - 0 = 1 ⇒ x = ±1, but since P is on right branch, x=1. So P=(1,0). Then vector F₁P would be (1 +1, 0 -0)=(2,0), and vector F₁Q=(1, q). Their dot product is 2*1 +0*q=2 ≠0, so perpendicularity is impossible. So points with p_y=0 cannot satisfy F₁P ⊥ F₁Q. Therefore, for points P where p_y ≠0, we can define q= - (p_x +1)/p_y.Therefore, point Q is (0, - (p_x +1)/p_y )Now, line PQ connects point P(p_x, p_y) and Q(0, - (p_x +1)/p_y ). Let's find the equation of line PQ.First, compute the slope m:m = [ - (p_x +1)/p_y - p_y ] / [0 - p_x ] = [ - (p_x +1)/p_y - p_y ] / (-p_x )Let me compute numerator:- (p_x +1)/p_y - p_y = [ - (p_x +1) - p_y² ] / p_ySo numerator is [ - (p_x +1) - p_y² ] / p_yTherefore, slope m is [ - (p_x +1 + p_y² ) / p_y ] / (-p_x ) = [ - (p_x +1 + p_y² ) / p_y ] * (-1/p_x ) = (p_x +1 + p_y² ) / (p_x p_y )So slope m = (p_x +1 + p_y² ) / (p_x p_y )Alternatively, maybe parametrize the line. Alternatively, use point-slope form.Using point Q(0, q) to write the equation:(y - q) = m (x - 0 )So y = m x + qBut m and q are in terms of p_x and p_y. Let's substitute m and q.But since P lies on the hyperbola C₂, which is x² - y² =1, so p_x² - p_y² =1 ⇒ p_y² = p_x² -1Therefore, in the slope m, we can substitute p_y² = p_x² -1:m = (p_x +1 + (p_x² -1 )) / (p_x p_y ) = (p_x +1 + p_x² -1 ) / (p_x p_y ) = (p_x² + p_x ) / (p_x p_y ) = [p_x (p_x +1 ) ] / (p_x p_y ) = (p_x +1 ) / p_yTherefore, slope m = (p_x +1)/p_yBut we have q = - (p_x +1)/p_y. So m = - qTherefore, the equation of PQ is y = m x + q = - q x + q = q ( -x +1 )Alternatively, since m = (p_x +1)/p_y and q = - (p_x +1)/p_y, then m = -q. So substituting back into the equation:y = (-q) x + q = q ( -x +1 )Alternatively, maybe parametrize differently. Let me check.Wait, maybe I made a miscalculation here. Let me verify the slope again.Earlier, I had:Slope m = (p_x +1 + p_y² ) / (p_x p_y )But since p_y² = p_x² -1,m = (p_x +1 + p_x² -1 ) / (p_x p_y ) = (p_x² + p_x ) / (p_x p_y ) = (p_x (p_x +1 )) / (p_x p_y ) = (p_x +1 ) / p_yYes, correct. Then q = - (p_x +1 ) / p_y, so m = - q. Therefore, equation is:y = m x + q = (-q) x + q = q (1 - x )Alternatively, express in terms of variables. Wait, perhaps we can write parametric equations for PQ. Let's think.Alternatively, consider parametric form. Let me take a parameter t such that when t=0, we are at Q(0, q) and t=1, we are at P(p_x, p_y). Then the parametric equations are:x = p_x * ty = q + (p_y - q ) * tBut perhaps not necessary. Alternatively, express line PQ as y = m x + q, where m = (p_y - q)/(p_x - 0 ) = (p_y - q)/p_x. But we already have another expression for m. Wait, maybe there is a confusion here.Wait, slope can also be calculated as (p_y - q)/ (p_x - 0 ) = (p_y - q)/p_x. But earlier, we had m = (p_x +1 ) / p_y. Therefore, equate the two expressions:(p_y - q)/p_x = (p_x +1 ) / p_yCross-multiplying:(p_y - q ) p_y = p_x (p_x +1 )But q = - (p_x +1 ) / p_y, so substitute:(p_y - ( - (p_x +1 ) / p_y )) p_y = p_x (p_x +1 )Left side:[ p_y + (p_x +1 ) / p_y ] p_y = p_y * p_y + (p_x +1 ) = p_y² + p_x +1Therefore, left side is p_y² + p_x +1, right side is p_x² + p_xSince p_y² = p_x² -1 (from hyperbola equation), substitute:(p_x² -1 ) + p_x +1 = p_x² + p_xLeft side: p_x² -1 + p_x +1 = p_x² + p_x, which equals right side. Therefore, the equations are consistent. So that's good.Therefore, the equation of PQ is y = m x + q, where m = (p_x +1)/p_y and q = - (p_x +1)/p_y. Therefore, substituting:y = ( (p_x +1)/p_y ) x - (p_x +1)/p_y = (p_x +1)/p_y (x -1 )Alternatively, factor out (p_x +1)/p_y:y = [ (p_x +1)/p_y ] (x -1 )So, for any point P(p_x, p_y) on hyperbola C₂ (right branch), the line PQ can be written as y = k(x -1 ), where k = (p_x +1)/p_y. The question is: does this line PQ pass through a fixed point, regardless of P? If so, find that point.To check if PQ passes through a fixed point, we need to see if there exists a point (a, b) such that for all P on the hyperbola (with p_x ≥1), the line PQ passes through (a, b). Alternatively, since the line PQ is y = k(x -1 ), where k varies depending on P, we need to see if all such lines pass through a common point. A family of lines y = k(x -1 ) passes through (1, 0) for any k. But wait, when x=1, y=0 regardless of k. But (1,0) is F₂. However, let's check if that's the case here.Wait, but in our case, k is not arbitrary; k is determined by the point P on the hyperbola. So, the lines PQ are a subset of all lines passing through (1,0). But actually, if we write y = k(x -1 ), then regardless of k, (1,0) is on the line. Therefore, if all such PQ lines are of the form y = k(x -1 ), then they all pass through (1,0). Therefore, the fixed point is (1,0). But let me verify this with specific points.Take point M(2√3/3, √3/3). Let's see if line PQ for this point passes through (1,0). First, compute Q. Q = (0, q) where q = - (p_x +1)/p_y. For P = M(2√3/3, √3/3):q = - (2√3/3 +1 ) / ( √3/3 ) = - [ (2√3 +3)/3 ] / ( √3 /3 ) = - (2√3 +3 ) / √3 = - (2 + 3/√3 ) = - (2 + √3 )So Q = (0, -2 -√3 )Then, the line PQ connects (2√3/3, √3/3 ) and (0, -2 -√3 )Let's compute the equation of this line.Slope m = ( √3/3 - (-2 -√3 ) ) / (2√3/3 - 0 ) = ( √3/3 +2 +√3 ) / (2√3/3 ) = (2 + (√3 + 3√3)/3 ) / (2√3/3 )Simplify numerator:2 + (4√3)/3 = (6 +4√3)/3So slope m = (6 +4√3)/3 divided by (2√3)/3 = (6 +4√3)/3 * 3/(2√3 ) = (6 +4√3 )/(2√3 ) = (3 +2√3 )/√3Rationalize denominator:(3 +2√3 )/√3 = 3/√3 + 2√3 /√3 = √3 + 2So slope m = √3 + 2Equation of line PQ: y - (-2 -√3 ) = (√3 +2)(x -0 )So y +2 +√3 = (√3 +2 )x ⇒ y = (√3 +2 )x -2 -√3Check if (1,0) is on this line:Plug x=1, y= (√3 +2 )*1 -2 -√3 = √3 +2 -2 -√3 = 0. Yes, (1,0) is on the line.Another example: take point P(1,0) on hyperbola. But wait, P(1,0) is on the hyperbola x² - y²=1, yes. Then Q would be (0, - (1 +1)/0 ), but division by zero. So as before, this point is problematic. But since P=(1,0) is on the hyperbola, but with p_y=0, we saw earlier that such points cannot satisfy F₁P ⊥ F₁Q. So perhaps P=(1,0) is excluded. Let's take another point on hyperbola C₂.Take P=(sqrt(2),1). Since (sqrt(2))² -1²=2 -1=1, so yes. Then Q=(0, - (sqrt(2)+1)/1 )= (0, -sqrt(2)-1). Then line PQ is from (sqrt(2),1) to (0, -sqrt(2)-1).Slope m=(1 - (-sqrt(2)-1 ))/(sqrt(2)-0 )=(1 +sqrt(2)+1)/sqrt(2)=(2 +sqrt(2))/sqrt(2)= (2/sqrt(2)) + (sqrt(2)/sqrt(2))= sqrt(2) +1Equation: y - (-sqrt(2)-1 ) = (sqrt(2)+1)x ⇒ y + sqrt(2)+1 = (sqrt(2)+1)x ⇒ y=(sqrt(2)+1)x -sqrt(2)-1Check if (1,0) is on this line:Left side: y=0Right side: (sqrt(2)+1)*1 -sqrt(2)-1 = sqrt(2)+1 -sqrt(2)-1=0. Correct.Therefore, line passes through (1,0). So even though we took another point, it still passes through (1,0). Therefore, it seems that all such lines PQ pass through the fixed point (1,0).Wait, but (1,0) is F₂, which is one of the foci of the ellipse and a vertex of the hyperbola. So seems like the fixed point is F₂.But let me confirm with another point. Let's take P=(sqrt(5), 2). Since (sqrt(5))² -2²=5-4=1. Then Q=(0, - (sqrt(5)+1)/2 ). Then line PQ connects (sqrt(5),2) and (0, -(sqrt(5)+1)/2 ). Let's compute the equation.Slope m=(2 - (-(sqrt(5)+1)/2 ))/(sqrt(5)-0 )=(2 + (sqrt(5)+1)/2 )/sqrt(5)= ( (4 + sqrt(5)+1 ) /2 ) /sqrt(5)= ( (5 + sqrt(5)) /2 ) /sqrt(5)= (5 + sqrt(5)) / (2 sqrt(5)) Rationalize:Multiply numerator and denominator by sqrt(5):(5 sqrt(5) +5 ) / (2*5 )= (5 sqrt(5) +5 )/10 = (sqrt(5)+1)/2So slope m=(sqrt(5)+1)/2Equation of line PQ: y - ( -(sqrt(5)+1)/2 ) = (sqrt(5)+1)/2 (x -0 )Thus, y + (sqrt(5)+1)/2 = (sqrt(5)+1)/2 x ⇒ y = [(sqrt(5)+1)/2] x - (sqrt(5)+1)/2Check if (1,0) is on the line:Left side: y=0Right side: [(sqrt(5)+1)/2]*1 - (sqrt(5)+1)/2 =0. Correct.Therefore, indeed, all lines PQ pass through the fixed point (1,0), which is F₂.Therefore, the answer to part (2) is that line PQ passes through the fixed point F₂(1,0).But just to make sure there isn't a mistake in the reasoning. Let's think again.We parametrized the line PQ as y = [ (p_x +1 ) / p_y ] (x -1 ). So for any point P on hyperbola, the line PQ is of this form. Therefore, regardless of the value of [ (p_x +1 ) / p_y ], the line will pass through (1,0), since when x=1, y=0. Therefore, regardless of the slope, the line passes through (1,0). Therefore, yes, all such lines pass through (1,0). Therefore, the fixed point is (1,0), which is F₂.Therefore, part (2) answer is the fixed point is (1,0).Final Answer(1) The equations are boxed{dfrac{x^2}{2} + y^2 = 1} for ellipse C_1 and boxed{x^2 - y^2 = 1} for hyperbola C_2.(2) The line PQ passes through the fixed point boxed{(1, 0)}.</think>Given the foci of ellipse (C_1) are (F_1(-1,0)) and (F_2(1,0)), which are the vertices of hyperbola (C_2), and one of the intersection points of (C_1) and (C_2) is (Mleft( frac{2 sqrt{3}}{3}, frac{ sqrt{3}}{3}right)):(1) Equations of Ellipse (C_1) and Hyperbola (C_2):- Ellipse (C_1): - The foci are at ((-1, 0)) and ((1, 0)), so the center is at the origin ((0, 0)) and (c = 1). - The standard form of the ellipse is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1) with (a^2 = b^2 + c^2). - Using the point (Mleft( frac{2 sqrt{3}}{3}, frac{ sqrt{3}}{3}right)): [ frac{left( frac{2 sqrt{3}}{3} right)^2}{a^2} + frac{left( frac{ sqrt{3}}{3} right)^2}{b^2} = 1 ] Simplifying, we find (a^2 = 2) and (b^2 = 1). Thus, the equation of (C_1) is: [ boxed{frac{x^2}{2} + y^2 = 1} ]- Hyperbola (C_2): - The vertices are at ((-1, 0)) and ((1, 0)), so the center is at the origin ((0, 0)) and (a_h = 1). - The standard form of the hyperbola is (frac{x^2}{a_h^2} - frac{y^2}{b_h^2} = 1). - Using the point (Mleft( frac{2 sqrt{3}}{3}, frac{ sqrt{3}}{3}right)): [ frac{left( frac{2 sqrt{3}}{3} right)^2}{1} - frac{left( frac{ sqrt{3}}{3} right)^2}{b_h^2} = 1 ] Simplifying, we find (b_h^2 = 1). Thus, the equation of (C_2) is: [ boxed{x^2 - y^2 = 1} ](2) Fixed Point for Line (PQ):- Point (P) is on the right branch of hyperbola (C_2) and point (Q) is on the (y)-axis such that (F_1P perp F_1Q).- The coordinates of (Q) are ((0, q)) where (q = -frac{p_x + 1}{p_y}).- The line (PQ) has the equation (y = frac{p_x + 1}{p_y}(x - 1)), which passes through the fixed point ((1, 0)) for all valid (P).Thus, the fixed point is:[boxed{(1, 0)}],<think>Okay, let's try to tackle this problem step by step. First, part (1) asks for the equations of the ellipse C₁ and the hyperbola C₂. Let me start with the ellipse.Given that the foci of ellipse C₁ are F₁(-1, 0) and F₂(1, 0). So, the major axis is along the x-axis because the foci are on the x-axis. For an ellipse, the standard form is (x²/a²) + (y²/b²) = 1, where 2a is the length of the major axis, 2b is the minor axis, and the distance between the center and each focus is c, with c² = a² - b².Since the foci are at (-1, 0) and (1, 0), the center of the ellipse is at the midpoint, which is (0, 0). The distance from the center to each focus is c = 1. So, c = 1. Therefore, c² = 1.We need to find a² and b². Also, we know that one of the intersection points is M(2√3/3, √3/3). Since this point lies on the ellipse, plugging it into the ellipse equation should satisfy it. Let's do that once we have the variables.But we need another equation to find a and b. Since c² = a² - b², if we can find a², then we can find b². Alternatively, if we can use the fact that the sum of distances from any point on the ellipse to the two foci is 2a, we can use point M to compute 2a.Let me compute the sum of distances from M to F₁ and F₂.First, distance from M to F₁(-1, 0):Distance formula: sqrt[( (2√3/3 + 1)^2 + (√3/3 - 0)^2 )]Compute (2√3/3 + 1). Let's convert 1 to 3/3: 2√3/3 + 3/3 = (3 + 2√3)/3. Wait, but actually, no. Wait, 2√3/3 + 1 = (2√3 + 3)/3. Then square that:[(2√3 + 3)/3]^2 = ( (2√3)^2 + 2*2√3*3 + 3² ) / 9 = (12 + 12√3 + 9)/9 = (21 + 12√3)/9.Then the y-component: (√3/3)^2 = (3)/9 = 1/3.So total distance squared to F₁ is (21 + 12√3)/9 + 1/3 = (21 + 12√3 + 3)/9 = (24 + 12√3)/9 = (8 + 4√3)/3.Therefore, distance from M to F₁ is sqrt[(8 + 4√3)/3]. Hmm, this is getting complicated. Maybe there's a better way. Alternatively, perhaps compute the actual distances numerically.Wait, let's compute coordinates numerically to check:Point M: (2√3/3, √3/3). Let's approximate √3 ≈ 1.732, so:x-coordinate: 2*1.732 / 3 ≈ 3.464 / 3 ≈ 1.155y-coordinate: 1.732 / 3 ≈ 0.577F₁ is (-1, 0), so distance from M to F₁:sqrt[(1.155 + 1)^2 + (0.577)^2] = sqrt[(2.155)^2 + (0.577)^2] ≈ sqrt[4.643 + 0.333] ≈ sqrt[4.976] ≈ 2.23.Similarly, distance from M to F₂(1, 0):sqrt[(1.155 - 1)^2 + (0.577)^2] = sqrt[(0.155)^2 + 0.333] ≈ sqrt[0.024 + 0.333] ≈ sqrt[0.357] ≈ 0.597.Sum of distances: 2.23 + 0.597 ≈ 2.827. So 2a ≈ 2.827, so a ≈ 1.413. Hmm, but these are approximate values. Let's compute symbolically.Wait, maybe the coordinates are exact. Let me try again.Distance from M to F₁: sqrt[( (2√3/3 + 1)^2 + (√3/3)^2 )]First, 2√3/3 + 1 = (2√3 + 3)/3. Squaring that:( (2√3 + 3)^2 ) / 9 = ( (4*3) + 12√3 + 9 ) / 9 = (12 + 9 + 12√3)/9 = (21 + 12√3)/9.Then adding the y-component (√3/3)^2 = 3/9 = 1/3. So total:sqrt[(21 + 12√3)/9 + 1/3] = sqrt[(21 + 12√3 + 3)/9] = sqrt[(24 + 12√3)/9] = sqrt[(24 + 12√3)/9] = sqrt[ (12*(2 + √3))/9 ] = sqrt[ (4*(2 + √3))/3 ] = (2/√3)*sqrt(2 + √3).Hmm, not sure if this helps. Maybe better to compute the sum of the two distances.Similarly, distance from M to F₂ is sqrt[( (2√3/3 - 1)^2 + (√3/3)^2 )]Compute 2√3/3 - 1 = (2√3 - 3)/3. Squaring that:( (2√3 - 3)^2 ) / 9 = (12 - 12√3 + 9)/9 = (21 - 12√3)/9.Adding the y-component: 1/3, so total:sqrt[(21 - 12√3)/9 + 1/3] = sqrt[(21 - 12√3 + 3)/9] = sqrt[(24 - 12√3)/9] = sqrt[ (12*(2 - √3))/9 ] = sqrt[ (4*(2 - √3))/3 ] = (2/√3)*sqrt(2 - √3).So the sum of the two distances is:(2/√3)*(sqrt(2 + √3) + sqrt(2 - √3)).Let me compute sqrt(2 + √3) + sqrt(2 - √3).Let me square this sum:[sqrt(2 + √3) + sqrt(2 - √3)]² = (2 + √3) + 2*sqrt{(2 + √3)(2 - √3)} + (2 - √3) = 4 + 2*sqrt{4 - 3} = 4 + 2*sqrt{1} = 4 + 2 = 6. Therefore, sqrt(2 + √3) + sqrt(2 - √3) = sqrt{6}.Therefore, the sum is (2/√3)*sqrt{6} = (2/√3)*(√6) = 2*(√6/√3) = 2*√2. So 2a = 2√2, so a = √2. Therefore, a² = 2.Now, since c = 1, c² = 1. Then b² = a² - c² = 2 - 1 = 1.Wait, so b² = 1. Therefore, the equation of the ellipse is (x²/2) + y² = 1.Let me check if point M(2√3/3, √3/3) lies on this ellipse.Compute x²/2 + y²:( (2√3/3)^2 ) / 2 + ( (√3/3)^2 )= ( (4*3)/9 ) / 2 + (3/9 )= (12/9)/2 + 1/3= (4/3)/2 + 1/3= 2/3 + 1/3 = 1.Yes, that works. So the ellipse equation is x²/2 + y² = 1.Great, that's C₁.Now, moving on to hyperbola C₂. The problem states that the foci of the ellipse are the vertices of the hyperbola. So the vertices of hyperbola C₂ are at (-1, 0) and (1, 0), same as the foci of the ellipse. Therefore, the hyperbola's vertices are at (±1, 0), so its center is also at (0, 0), and its transverse axis is along the x-axis.For a hyperbola, the standard form is (x²/a_h²) - (y²/b_h²) = 1, where 2a_h is the distance between the vertices. Here, a_h is 1, so a_h² = 1.We need to find b_h². Also, since point M is an intersection point of the ellipse and hyperbola, it must satisfy both equations.So, the hyperbola equation is (x²/1) - (y²/b_h²) = 1. Let's plug in M's coordinates.M is (2√3/3, √3/3). Plugging into hyperbola equation:( (2√3/3)^2 ) - ( (√3/3)^2 ) / b_h² = 1Compute each term:( (4*3)/9 ) - ( (3)/9 ) / b_h² = 1Simplify:(12/9) - (1/3)/b_h² = 1Which is:4/3 - (1)/(3b_h²) = 1Subtract 4/3 from both sides:- (1)/(3b_h²) = 1 - 4/3 = -1/3Multiply both sides by -3:1/(b_h²) = 1Therefore, b_h² = 1. Hence, the hyperbola equation is x² - y² = 1.Wait, but let's check if point M satisfies x² - y² = 1:( (2√3/3)^2 ) - ( (√3/3)^2 ) = (12/9) - (3/9) = 9/9 = 1. Yes, correct. So the hyperbola equation is x² - y² = 1.Therefore, part (1) is solved: ellipse C₁ is x²/2 + y² = 1, hyperbola C₂ is x² - y² = 1.Now, part (2): If point P is a moving point on the right branch of the hyperbola (so x ≥ 1), and point Q is a moving point on the y-axis such that F₁P ⊥ F₁Q, determine whether line PQ passes through a fixed point.First, let's parse the problem. F₁ is (-1, 0). So for any point P on the hyperbola, we need to find Q on the y-axis such that the line segment F₁P is perpendicular to the line segment F₁Q. Then, we need to see if PQ passes through a fixed point regardless of the choice of P and Q.Let me try to visualize this. F₁ is at (-1, 0). P is on the right branch of the hyperbola x² - y² =1, so P has coordinates (x_p, y_p) where x_p ≥ 1. Q is on the y-axis, so Q has coordinates (0, q) for some q.The condition is that F₁P is perpendicular to F₁Q. So vectors F₁P and F₁Q must be perpendicular.First, compute vectors F₁P and F₁Q.F₁ is (-1, 0). P is (x_p, y_p). So vector F₁P is (x_p - (-1), y_p - 0) = (x_p + 1, y_p).Similarly, Q is (0, q), so vector F₁Q is (0 - (-1), q - 0) = (1, q).The dot product of F₁P and F₁Q must be zero.So:(x_p + 1)(1) + (y_p)(q) = 0Therefore:x_p + 1 + y_p q = 0So we can solve for q:q = - (x_p + 1)/y_pTherefore, for each point P on the hyperbola, Q is determined as (0, - (x_p + 1)/y_p )Now, we need to find the equation of line PQ and check if it passes through a fixed point.First, let's parametrize point P on the hyperbola. Since hyperbola is x² - y² =1, we can write parametric equations. For hyperbola x²/a² - y²/b² =1, the parametrization can be (a secθ, b tanθ). Here, a=1, b=1, so parametrization is (secθ, tanθ). But since it's the right branch, θ is in (-π/2, π/2).Alternatively, we can use coordinates (x_p, y_p) with x_p = sqrt(1 + y_p²). But maybe parametrization is easier.Let me use θ parametrization. Let P be (secθ, tanθ). Then, x_p = secθ, y_p = tanθ.Then, Q is (0, - (secθ + 1)/tanθ )Simplify that:Compute -(secθ +1)/tanθ = -( (1/cosθ) +1 ) / (sinθ/cosθ ) = -( (1 + cosθ)/cosθ ) / (sinθ/cosθ ) = -(1 + cosθ)/sinθThus, Q is (0, - (1 + cosθ)/sinθ )So Q is (0, - (1 + cosθ)/sinθ )Now, we need to find the equation of line PQ where P is (secθ, tanθ) and Q is (0, - (1 + cosθ)/sinθ )Let me write the coordinates again:P: (secθ, tanθ) = (1/cosθ, sinθ/cosθ)Q: (0, - (1 + cosθ)/sinθ )First, find the equation of the line passing through P and Q.The slope of PQ is [ y_q - y_p ] / [ x_q - x_p ] = [ - (1 + cosθ)/sinθ - sinθ/cosθ ] / [ 0 - 1/cosθ ]Compute numerator:- (1 + cosθ)/sinθ - sinθ/cosθ = - [ (1 + cosθ)/sinθ + sinθ/cosθ ]Let me combine these terms:Common denominator is sinθ cosθ:= - [ ( (1 + cosθ) cosθ + sin²θ ) / (sinθ cosθ ) ]Compute numerator inside:(1 + cosθ)cosθ + sin²θ = cosθ + cos²θ + sin²θ = cosθ + (cos²θ + sin²θ ) = cosθ + 1Therefore, numerator becomes:- [ (cosθ + 1 ) / (sinθ cosθ ) ]Therefore, slope is:[ - (cosθ + 1 ) / (sinθ cosθ ) ] / [ -1/cosθ ] = [ - (cosθ +1 ) / (sinθ cosθ ) ] * [ -cosθ /1 ] = (cosθ +1 ) / sinθSo the slope m = (cosθ + 1)/sinθAlternatively, we can write this as cot(θ/2), since (1 + cosθ)/sinθ = cot(θ/2). Because 1 + cosθ = 2cos²(θ/2) and sinθ = 2sin(θ/2)cos(θ/2), so (1 + cosθ)/sinθ = (2cos²(θ/2))/(2sin(θ/2)cos(θ/2)) ) = cot(θ/2). So slope m = cot(θ/2)But perhaps not necessary. Anyway, the slope is (1 + cosθ)/sinθ.So the equation of PQ can be written using point Q (0, - (1 + cosθ)/sinθ ) and slope m:y - y_q = m (x - x_q )So:y + (1 + cosθ)/sinθ = ( (1 + cosθ)/sinθ ) (x - 0 )Therefore:y = ( (1 + cosθ)/sinθ ) x - (1 + cosθ)/sinθFactor out (1 + cosθ)/sinθ:y = ( (1 + cosθ)/sinθ )(x - 1 )Alternatively, let's write this as:y = m x - m, where m = (1 + cosθ)/sinθBut maybe we can express this in terms of θ or try to eliminate the parameter.Alternatively, let's see if this line passes through a fixed point (a, b) regardless of θ.Suppose that (a, b) is a fixed point such that for all θ, the line PQ passes through (a, b). Then:b = ( (1 + cosθ)/sinθ ) a - (1 + cosθ)/sinθMultiply both sides by sinθ:b sinθ = (1 + cosθ)(a -1 )Rearrange:b sinθ - (a -1)(1 + cosθ ) = 0This equation must hold for all θ. Let's expand the left side:b sinθ - (a -1) - (a -1) cosθ = 0Group terms:[ - (a -1) ] + [ b sinθ - (a -1) cosθ ] = 0For this to hold for all θ, the coefficients of sinθ, cosθ, and the constant term must all be zero. Therefore:Coefficient of sinθ: b = 0Coefficient of cosθ: - (a -1 ) = 0 ⇒ a -1 = 0 ⇒ a =1Constant term: - (a -1 ) = 0 ⇒ same as above, a =1But if a=1, then coefficient of sinθ is b =0. Therefore, the fixed point is (a, b ) = (1, 0 )Wait, but F₁ is (-1, 0). The fixed point is (1, 0 ) which is F₂.Wait, but let me verify. If the line PQ passes through (1, 0 ) for any θ, then substituting x=1, y=0 into the line equation should hold.Let me check:From the line equation:y = ( (1 + cosθ)/sinθ ) (x -1 )So if x=1, then y=0, regardless of θ. So yes, (1, 0 ) is on the line PQ for any θ.Therefore, line PQ always passes through the fixed point (1, 0 ), which is F₂.Therefore, the answer is that PQ always passes through the fixed point (1, 0 ).But let me cross-verify this. Let's take a specific example with point M given in part (1). Since M is an intersection point, let's check if PQ passes through (1,0) when P is M.Wait, point M is (2√3/3, √3/3). So P = M. Let's compute Q.From earlier, Q is (0, - (x_p +1)/y_p )x_p = 2√3/3, y_p = √3/3.So Q's y-coordinate is - ( (2√3/3 +1 ) / (√3/3 ) ) = - ( (2√3 + 3)/3 ) / (√3/3 ) = - (2√3 + 3)/√3 = - [ 2 + 3/√3 ] = - [ 2 + √3 ].Therefore, Q is (0, - (2 + √3 ) )Now, line PQ connects M(2√3/3, √3/3) and Q(0, - (2 + √3 ) )Let me find the equation of PQ.Slope m = [ - (2 + √3 ) - (√3/3 ) ] / [ 0 - 2√3/3 ] = [ -2 - √3 - √3/3 ] / [ -2√3/3 ]Simplify numerator:-2 - (4√3)/3Denominator: -2√3/3So m = [ -2 - (4√3)/3 ] / [ -2√3/3 ] = [ (-6 - 4√3)/3 ] / [ -2√3/3 ] = ( -6 -4√3 ) / (-2√3 ) = (6 + 4√3 ) / (2√3 ) = (3 + 2√3 ) / √3Multiply numerator and denominator by √3:(3√3 + 2*3 ) / 3 = (3√3 +6)/3 = √3 + 2Therefore, slope m = √3 + 2.Equation of line PQ:Using point Q(0, - (2 + √3 )):y = (√3 + 2 )x - (2 + √3 )Check if (1, 0 ) is on this line:Left side: 0Right side: (√3 + 2)(1) - (2 + √3 ) = √3 + 2 -2 -√3 = 0. Yes, correct.Therefore, in this case, line PQ passes through (1,0 ). Since we proved earlier that for any θ, PQ passes through (1,0 ), this fixed point is indeed valid.Therefore, the answer to part (2) is that line PQ always passes through the fixed point (1,0 ).Final Answer(1) The equation of ellipse ( C_1 ) is (boxed{dfrac{x^2}{2} + y^2 = 1}) and the equation of hyperbola ( C_2 ) is (boxed{x^2 - y^2 = 1}).(2) The line ( PQ ) passes through the fixed point with coordinates (boxed{(1, 0)}).</think>Given the problem, we need to find the equations of the ellipse ( C_1 ) and hyperbola ( C_2 ), and determine whether line ( PQ ) passes through a fixed point. Part (1)Ellipse ( C_1 ):- The foci of the ellipse are ( F_1(-1, 0) ) and ( F_2(1, 0) ), so the center is at the origin ((0, 0)) and the distance between the foci is ( 2c = 2 ), hence ( c = 1 ).- The sum of distances from any point on the ellipse to the foci is ( 2a ). Using the intersection point ( Mleft( frac{2sqrt{3}}{3}, frac{sqrt{3}}{3} right) ), we calculate the sum of distances to the foci: - Distance from ( M ) to ( F_1 ): ( sqrt{left( frac{2sqrt{3}}{3} + 1 right)^2 + left( frac{sqrt{3}}{3} right)^2} ) - Distance from ( M ) to ( F_2 ): ( sqrt{left( frac{2sqrt{3}}{3} - 1 right)^2 + left( frac{sqrt{3}}{3} right)^2} ) - Summing these distances, we find ( 2a = 2sqrt{2} ), so ( a = sqrt{2} ) and ( a^2 = 2 ).- Using ( c^2 = a^2 - b^2 ), we get ( b^2 = 1 ).- The equation of the ellipse is ( frac{x^2}{2} + y^2 = 1 ).Hyperbola ( C_2 ):- The vertices of the hyperbola are the foci of the ellipse, ( (-1, 0) ) and ( (1, 0) ), so ( a_h = 1 ) and ( a_h^2 = 1 ).- Using the intersection point ( M left( frac{2sqrt{3}}{3}, frac{sqrt{3}}{3} right) ) on the hyperbola, we substitute into the hyperbola equation ( frac{x^2}{1} - frac{y^2}{b_h^2} = 1 ): - Solving, we find ( b_h^2 = 1 ).- The equation of the hyperbola is ( x^2 - y^2 = 1 ). Part (2)Fixed Point Analysis:- For a point ( P ) on the hyperbola ( x^2 - y^2 = 1 ) with coordinates ( (sectheta, tantheta) ), point ( Q ) on the y-axis is determined by the condition ( F_1P perp F_1Q ).- The coordinates of ( Q ) are found to be ( left( 0, -frac{1 + costheta}{sintheta} right) ).- The equation of line ( PQ ) is derived using the slope ( frac{1 + costheta}{sintheta} ) and point ( Q ): - The line equation simplifies to ( y = left( frac{1 + costheta}{sintheta} right)(x - 1) ), which passes through the fixed point ( (1, 0) ) for all ( theta ). Final Answers(1) The equation of ellipse ( C_1 ) is (boxed{dfrac{x^2}{2} + y^2 = 1}) and the equation of hyperbola ( C_2 ) is (boxed{x^2 - y^2 = 1}).(2) The line ( PQ ) passes through the fixed point with coordinates (boxed{(1, 0)}).]

🔑:(1) Let ellipse C_1 be represented by the equation frac{x^{2}}{a^{2}}+ frac{y^{2}}{b^{2}}=1 (a > b > 0), and the equation of hyperbola C_2 be x^{2}- frac{y^{2}}{b_1^{2}}=1.Then, 2a=|MF_1|+|MF_2|= sqrt{ frac{8+4 sqrt{3}}{3}}+ sqrt{ frac{8-4 sqrt{3}}{3}}=2 sqrt{2}, and frac{12}{9}- frac{frac{1}{3}}{b_1^{2}}=1.Thus, a=sqrt{2}, b=1, b_1=1.Hence, the equation of ellipse C_1 is frac{x^{2}}{2}+y^{2}=1, and the equation of hyperbola C_2 is x^{2}-y^{2}=1.(2) Let the coordinates of point P be (x_0, y_0). Then, x_0^{2}-y_0^{2}=1.The slope of line F_1P is frac{y_0}{x_0+1}, so the slope of line F_1Q is - frac{x_0+1}{y_0}.Thus, the equation of line F_1Q is y=- frac{x_0+1}{y_0}(x+1).Setting x=0, we get y=- frac{x_0+1}{y_0}, so the coordinates of Q are left(0, - frac{x_0+1}{y_0}right).Hence, the equation of line PQ is frac{y-y_0}{- frac{x_0+1}{y_0}-y_0}= frac{x-x_0}{-x_0}.Substituting y_0^{2}=1-x_0^{2} into the equation, we get x_0 y_0 y=(x_0+x_0^{2})(x-1).Thus, the line passes through the fixed point boxed{(1, 0)}.

🧠:[<think>Okay, so I need to show that this big expression simplifies to 1. Let me write it down again to make sure I have it right:frac{7x - 5y}{3x + 2y} + frac{5x - 8y}{3x - 2y} - frac{x - 9y}{3x + 2y} - frac{8x - 10y}{3x - 2y} = 1Hmm, the denominators are 3x + 2y and 3x - 2y. Maybe I can combine the fractions with the same denominator first. Let me group them:First, the terms with denominator 3x + 2y: (7x - 5y)/(3x + 2y) minus (x - 9y)/(3x + 2y).Then the terms with denominator 3x - 2y: (5x - 8y)/(3x - 2y) minus (8x - 10y)/(3x - 2y).So combining those should simplify things. Let's do that step by step.Starting with the 3x + 2y denominator group:frac{7x - 5y - (x - 9y)}{3x + 2y}Wait, that's (7x -5y) minus (x -9y) all over 3x + 2y. Let me compute the numerator:7x -5y -x +9y = (7x - x) + (-5y +9y) = 6x +4y.So that first combined fraction is (6x +4y)/(3x +2y).Now for the 3x -2y denominator group:frac{5x -8y - (8x -10y)}{3x -2y}Again, simplifying the numerator:5x -8y -8x +10y = (5x -8x) + (-8y +10y) = -3x +2y.So the second combined fraction is (-3x +2y)/(3x -2y).Now the original expression simplifies to:frac{6x +4y}{3x +2y} + frac{-3x +2y}{3x -2y}Hmm, maybe these can be simplified further. Let me look at the first fraction: (6x +4y)/(3x +2y). Factor numerator and denominator.6x +4y = 2*(3x +2y). Oh, wait! 3x +2y is the denominator. So:frac{2*(3x +2y)}{3x +2y} = 2As long as 3x +2y ≠ 0, which I guess we can assume here. So that first fraction simplifies to 2.Cool! So now we have 2 + [(-3x +2y)/(3x -2y)].Now let's look at the second fraction: (-3x +2y)/(3x -2y). Let me factor the numerator:-3x +2y = - (3x -2y). Because if you factor out a -1, you get - (3x -2y). So:frac{ - (3x -2y) }{3x -2y} = -1Again, assuming 3x -2y ≠ 0. So this fraction simplifies to -1.Therefore, the entire expression becomes 2 + (-1) = 1. Which is exactly what we needed to show. Yay!Wait, let me just check my steps again to make sure I didn't make any mistakes. So combining the first two fractions and the last two fractions. Then simplifying each combined fraction by factoring. First fraction had 6x +4y over 3x +2y, which factors to 2*(3x +2y) over 3x +2y, cancels to 2. Then the second combined fraction had -3x +2y over 3x -2y, which is -1*(3x -2y) over 3x -2y, cancels to -1. So 2 -1 =1. Looks right. I think that's correct. Maybe I should plug in some numbers for x and y to verify.Let me pick x=1 and y=1. Then check both the original expression and the simplified result.Original expression:First term: (7*1 -5*1)/(3*1 +2*1) = (7-5)/5 = 2/5Second term: (5*1 -8*1)/(3*1 -2*1) = (5-8)/1 = -3/1 = -3Third term: (1 -9*1)/(3*1 +2*1) = (-8)/5 = -8/5Fourth term: (8*1 -10*1)/(3*1 -2*1) = (-2)/1 = -2So total expression: 2/5 + (-3) - (-8/5) - (-2)Compute step by step:2/5 -3 +8/5 +2Combine fractions: (2/5 +8/5) = 10/5 =2Combine integers: -3 +2 = -1Total: 2 -1 =1. Which matches the right side. So when x=1, y=1, it works.Another test with x=2, y=1.First term: (14 -5)/(6 +2)=9/8Second term: (10 -8)/(6 -2)=2/4=1/2Third term: (2 -9)/(6 +2)=(-7)/8Fourth term: (16 -10)/(6 -2)=6/4=3/2So compute: 9/8 +1/2 - (-7/8) -3/2Convert to eighths:9/8 +4/8 +7/8 -12/8Adding up numerators: 9+4+7-12=88/8=1. Again, it works. So seems correct.Therefore, the original equation simplifies to 1.Final Answerboxed{1}</think>To show that:frac{7x - 5y}{3x + 2y} + frac{5x - 8y}{3x - 2y} - frac{x - 9y}{3x + 2y} - frac{8x - 10y}{3x - 2y} = 1we start by combining the fractions with the same denominators.For the terms with denominator (3x + 2y):[frac{7x - 5y}{3x + 2y} - frac{x - 9y}{3x + 2y} = frac{(7x - 5y) - (x - 9y)}{3x + 2y} = frac{6x + 4y}{3x + 2y}]Simplifying the numerator:[6x + 4y = 2(3x + 2y)]Thus, the fraction simplifies to:[frac{2(3x + 2y)}{3x + 2y} = 2]For the terms with denominator (3x - 2y):[frac{5x - 8y}{3x - 2y} - frac{8x - 10y}{3x - 2y} = frac{(5x - 8y) - (8x - 10y)}{3x - 2y} = frac{-3x + 2y}{3x - 2y}]Simplifying the numerator:[-3x + 2y = - (3x - 2y)]Thus, the fraction simplifies to:[frac{-(3x - 2y)}{3x - 2y} = -1]Combining these results, we have:[2 + (-1) = 1]Therefore, the given expression simplifies to:[boxed{1}],<think>Okay, so I need to show that this big expression simplifies to 1. Let me look at it again:frac{7 x-5 y}{3 x+2 y} + frac{5 x-8 y}{3 x-2 y} - frac{x-9 y}{3 x+2 y} - frac{8 x-10 y}{3 x-2 y} = 1Hmm, all these fractions have denominators of either (3x + 2y) or (3x - 2y). Maybe I can combine the ones with the same denominator first. That should make things simpler, right?Let me group them:For the denominators (3x + 2y), we have two fractions: (frac{7x -5y}{3x +2y}) and (-frac{x -9y}{3x +2y}).Similarly, for denominators (3x -2y), we have (frac{5x -8y}{3x -2y}) and (-frac{8x -10y}{3x -2y}).So combining each pair:First, ( frac{7x -5y}{3x +2y} - frac{x -9y}{3x +2y} ). Since the denominators are the same, I can subtract the numerators:( frac{(7x -5y) - (x -9y)}{3x +2y} )Let me compute the numerator:7x -5y -x +9y = (7x -x) + (-5y +9y) = 6x +4ySo that simplifies to ( frac{6x +4y}{3x +2y} )Hmm, maybe factor numerator? Let's see:6x +4y = 2*(3x +2y). Oh! That's exactly the denominator. So 2*(3x +2y)/(3x +2y) = 2. Nice, that part simplifies to 2.Now, the other pair: ( frac{5x -8y}{3x -2y} - frac{8x -10y}{3x -2y} )Again, same denominator, so combine numerators:(5x -8y) - (8x -10y) = 5x -8y -8x +10y = (5x -8x) + (-8y +10y) = -3x +2ySo that's ( frac{-3x +2y}{3x -2y} )Hmm, let's factor numerator or see if it's related to denominator:-3x +2y = - (3x -2y). Oh, that's the negative of the denominator. So:( frac{ - (3x -2y) }{3x -2y} = -1 )So that simplifies to -1.So now, combining the two simplified parts:First part was 2, second part was -1. So total expression is 2 + (-1) = 1.Therefore, the entire left-hand side simplifies to 1, which is equal to the right-hand side. Hence, proved.Wait, let me check my steps again to be sure.First combination:7x -5y minus (x -9y):7x -5y -x +9y = 6x +4y. Correct. Then 6x +4y = 2*(3x +2y). Denominator cancels, giving 2. Perfect.Second combination:5x -8y minus (8x -10y):5x -8y -8x +10y = -3x +2y. Then numerator is - (3x -2y), denominator is 3x -2y. So that's -1. Correct.Thus, 2 -1 =1. Yep, looks solid. So the equation holds true. Therefore, the original expression does indeed equal 1.Final Answerboxed{1}</think>To show that:frac{7 x-5 y}{3 x+2 y} + frac{5 x-8 y}{3 x-2 y} - frac{x-9 y}{3 x+2 y} - frac{8 x-10 y}{3 x-2 y} = 1we start by combining the fractions with the same denominators.For the denominators (3x + 2y):[frac{7x - 5y}{3x + 2y} - frac{x - 9y}{3x + 2y} = frac{(7x - 5y) - (x - 9y)}{3x + 2y}]Simplifying the numerator:[(7x - 5y) - (x - 9y) = 7x - 5y - x + 9y = 6x + 4y]This simplifies to:[frac{6x + 4y}{3x + 2y} = frac{2(3x + 2y)}{3x + 2y} = 2]For the denominators (3x - 2y):[frac{5x - 8y}{3x - 2y} - frac{8x - 10y}{3x - 2y} = frac{(5x - 8y) - (8x - 10y)}{3x - 2y}]Simplifying the numerator:[(5x - 8y) - (8x - 10y) = 5x - 8y - 8x + 10y = -3x + 2y]This simplifies to:[frac{-3x + 2y}{3x - 2y} = frac{-(3x - 2y)}{3x - 2y} = -1]Combining the results from both parts:[2 + (-1) = 1]Thus, the given expression simplifies to:[boxed{1}]]

🔑:The given expression can also be written as:begin{gathered}frac{7 x-5 y-x+9 y}{3 x+2 y}+frac{5 x-8 y-8 x+10 y}{3 x-2 y}= =frac{6 x+4 y}{3 x+2 y}+frac{-3 x+2 y}{3 x-2 y}=2-1=1end{gathered}Number of solutions: 25.(Zoltán Harsányi, Eger.)

🧠:[<think>Okay, let me try to figure out this problem. So, we have an odd function y = f(x) defined on all real numbers. When x is greater than or equal to 0, the function is given as f(x) = x² - 3x + b. We need to find f(-2) and choose the correct answer from the options A to D.First, let me recall what an odd function is. An odd function satisfies the property that f(-x) = -f(x) for all x in its domain. That means if we know the value of the function at some positive x, we can find the value at the corresponding negative x by just taking the negative of the positive value. So, for example, f(-2) should be equal to -f(2). That seems straightforward.But wait, we also have this constant term b in the expression for f(x) when x is non-negative. Since the function is odd, the behavior for x < 0 must be related to the behavior for x > 0. However, the problem only gives us the expression for x ≥ 0. So, maybe we need to use the property of odd functions to determine the value of b first?Hold on, maybe the function is defined piecewise. For x ≥ 0, it's given as x² - 3x + b, and for x < 0, it's defined such that f(-x) = -f(x). But for the function to be odd, certain conditions must hold, especially at x = 0. Let's check that.An odd function must satisfy f(0) = 0. Let me verify that. If x = 0, then according to the given expression, f(0) = (0)² - 3*(0) + b = 0 - 0 + b = b. But since the function is odd, f(0) must be 0. Therefore, b must equal 0. Wait, that seems important. So, the constant term b is zero. Let me confirm that again.Yes, because for an odd function, f(-0) = -f(0), but -0 is 0, so f(0) = -f(0), which implies that 2f(0) = 0, hence f(0) = 0. So, substituting x = 0 into the given expression for x ≥ 0, we get f(0) = 0² -3*0 + b = b. Therefore, b must be 0. So, the function for x ≥ 0 is f(x) = x² - 3x. That makes sense.Therefore, now that we know b = 0, we can proceed to find f(-2). Since the function is odd, f(-2) = -f(2). So, we need to compute f(2) first. Let's do that.Substituting x = 2 into the expression for x ≥ 0, we have f(2) = (2)² - 3*(2) + b. Wait, but we already determined that b = 0, right? So, f(2) = 4 - 6 + 0 = -2. Then, f(-2) = -f(2) = -(-2) = 2. Therefore, the answer should be 2, which is option B. Hmm, but let me check again to make sure I didn't make a mistake.Wait, let me verify the value of b again. When x = 0, f(0) must be 0 because the function is odd. So plugging x = 0 into x² - 3x + b gives 0 - 0 + b = b. Therefore, b must be 0. Correct. Then f(2) is indeed 4 - 6 = -2. Therefore, f(-2) is -(-2) = 2. So, option B. That seems right.But wait, let me think again. The problem says "defined on ℝ" and "when x ≥ 0, f(x) = x² - 3x + b". So, since the function is odd, the part for x < 0 is determined by the part for x > 0. But we had to use the condition at x = 0 to find that b = 0. Then, once we have that, f(-2) is just -f(2), and f(2) is computed as -2, so f(-2) is 2. So, the answer should be B.Alternatively, let's think if there's another way someone might approach this problem. For instance, maybe they could write the expression for f(-x) when x is positive and set it equal to -f(x). Let's see.Since f is odd, for x > 0, we have f(-x) = -f(x). Let's take x = 2, then f(-2) = -f(2). But f(2) is calculated using the expression for x ≥ 0, which is 2² - 3*2 + b = 4 - 6 + b = -2 + b. But since we already found that b must be 0, that gives f(2) = -2, so f(-2) = 2. So that's the same result.But maybe if someone didn't realize that b has to be 0, they might get confused. For example, if someone thought that b could be arbitrary, they might get stuck. But since f(0) must be 0, and when x = 0, f(0) = 0² -3*0 + b = b, so b has to be 0. Therefore, that's a crucial step here.Therefore, the answer should be B: 2. Let me check the options again. Yes, option B is 2. So, that's the correct answer.Wait a second, but looking back at the original problem statement: "when x ≥ 0, f(x) = x² - 3x + b". So, when x is non-negative, the function is a quadratic with a constant term b. But because the function is odd, f(0) must be 0, so substituting x=0 gives b=0. So, without that, the function wouldn't be odd. Therefore, that's a necessary condition.Alternatively, maybe there's a different approach. Suppose we didn't consider f(0) first. Let's see. The function is odd, so for any x, f(-x) = -f(x). Let's consider x > 0. Then, f(x) = x² - 3x + b, and f(-x) = ?But for x > 0, -x is negative. Since the function is odd, f(-x) = -f(x). But how is the function defined for x < 0? The problem doesn't give an explicit expression, but since the function is odd, we can write for x < 0, f(x) = -f(-x), where -x is positive. Therefore, for x < 0, f(x) = -[(-x)² - 3*(-x) + b] = -[x² + 3x + b]. So, in general, f(x) is piecewise defined:f(x) = x² - 3x + b, for x ≥ 0,f(x) = -x² - 3x - b, for x < 0.Wait, hold on, let me verify that. If x < 0, then let me set x = -t where t > 0. Then, f(x) = f(-t) = -f(t) = -[t² - 3t + b]. So, substituting back t = -x, we have f(x) = -[(-x)² - 3*(-x) + b] = -[x² + 3x + b]. So, yes, for x < 0, f(x) = -x² - 3x - b.But then, if we plug x = 0 into both expressions, the right-hand limit as x approaches 0 from above is f(0+) = 0² - 3*0 + b = b, and the left-hand limit as x approaches 0 from below is f(0-) = -0² - 3*0 - b = -b. But since the function is defined at x=0 and must be continuous there (as it's defined on all real numbers and presumably continuous, though the problem doesn't explicitly state continuity), then the left-hand limit, right-hand limit, and the function value at 0 must all be equal. However, for the function to be odd, f(0) must be 0. So, from the right-hand side, f(0) = b, from the left-hand side, f(0) = -b. Therefore, for these to be equal and also equal to 0, we must have b = 0. Therefore, that's another way to see that b must be 0. Therefore, even if we didn't consider x=0 initially, by ensuring the function is continuous at 0 (which is necessary for it to be defined on all real numbers without any points of discontinuity), we must have b = 0.Therefore, regardless of the approach, we end up with b=0. Then, computing f(-2) as -f(2) gives us -[2² -3*2 +0] = -[4 -6] = -[-2] = 2. Therefore, the answer is 2, which is option B.But let me check once more to make sure I didn't miss anything. The key steps are:1. Odd function implies f(0) = 0, which gives b = 0.2. For x = 2, compute f(2) = 2² -3*2 = 4 -6 = -2.3. Then, f(-2) = -f(2) = -(-2) = 2.All steps seem solid. The only possible confusion could be if someone forgets that odd functions require f(0) = 0, but once you remember that, it's straightforward. Alternatively, if someone thought that the function might not be continuous at 0, but since it's defined on all real numbers and there's no mention of discontinuities, we can assume it's continuous, especially since the problem gives a single expression for x ≥ 0. So, continuity at 0 is necessary here.Therefore, the answer is B: 2.Final Answerboxed{B}</think>Given that the function ( y = f(x) ) is an odd function defined on (mathbb{R}), and when ( x geq 0 ), ( f(x) = x^2 - 3x + b ), we need to find ( f(-2) ).1. Odd Function Property: An odd function satisfies ( f(-x) = -f(x) ) for all ( x ). This implies ( f(0) = 0 ).2. Determine ( b ): - For ( x = 0 ), the function value is ( f(0) = 0^2 - 3 cdot 0 + b = b ). - Since ( f(0) = 0 ), we have ( b = 0 ).3. Compute ( f(2) ): - Substitute ( x = 2 ) into the given expression for ( x geq 0 ): ( f(2) = 2^2 - 3 cdot 2 + 0 = 4 - 6 = -2 ).4. Find ( f(-2) ): - Using the odd function property: ( f(-2) = -f(2) = -(-2) = 2 ).Thus, the value of ( f(-2) ) is (boxed{B}).,<think>Okay, so I need to find f(-2) for an odd function f(x) that's defined on all real numbers. When x is greater than or equal to 0, the function is given as f(x) = x² - 3x + b. The options are A: -2, B: 2, C: 10, D: -10. Hmm, let's think about this step by step.First, I remember that an odd function has the property that f(-x) = -f(x) for all x in its domain. That means if I can find f(2), then f(-2) should just be -f(2), right? So maybe I can use that property here. But wait, the function for x ≥ 0 is given as x² - 3x + b. So I need to figure out what b is before I can compute f(2) and then f(-2). But how?Hmm, odd functions also have to satisfy f(0) = 0 because if you plug in x = 0, then f(-0) = -f(0) which simplifies to f(0) = -f(0), and the only number that is equal to its own negative is zero. So f(0) must be 0. Let me check if that helps. Given that when x ≥ 0, f(x) = x² - 3x + b. So plugging in x = 0, we get f(0) = 0² - 3*0 + b = b. But since f(0) must be 0, that means b = 0. Wait, so b is zero? Let me confirm. If the function is odd, then f(0) = 0, so substituting x=0 into the expression x² - 3x + b gives b, which must equal 0. Therefore, b = 0. So then for x ≥ 0, the function is f(x) = x² - 3x. Now, to find f(2), we substitute x = 2 into that expression: f(2) = (2)² - 3*(2) = 4 - 6 = -2. Then, since the function is odd, f(-2) = -f(2) = -(-2) = 2. So the answer should be B: 2. But wait, let me check again. Maybe I made a mistake.Wait, hold on. The problem says the function is defined on all real numbers, and it's odd. For x ≥ 0, it's given as x² - 3x + b. We concluded that f(0) = 0, so b must be zero. Therefore, for x ≥ 0, f(x) = x² - 3x. Then f(2) = 4 - 6 = -2. Therefore, f(-2) = -f(2) = 2. That's answer B. But let me check if there's another way this could be approached, maybe I missed something.Alternatively, maybe there's a different way to interpret the problem. For example, perhaps the function is defined piecewise, with x² - 3x + b for x ≥ 0 and some other expression for x < 0. But since it's an odd function, the expression for x < 0 is determined by the expression for x > 0. Specifically, for x < 0, f(x) = -f(-x) = -[(-x)² - 3*(-x) + b] because -x would be positive, so we use the x ≥ 0 expression. Wait, but hold on, if we use that, then f(-x) = (-x)^2 - 3*(-x) + b = x² + 3x + b. Then f(x) = -f(-x) = -x² - 3x - b. So for x < 0, f(x) = -x² - 3x - b. But wait, but in the problem statement, when x ≥ 0, f(x) = x² - 3x + b. So if that's the case, then for x < 0, the function is -x² - 3x - b. But then, does this function pass through the origin? Let's check x=0 from the right and left.From the right (x ≥ 0), f(0) = 0² - 3*0 + b = b. From the left (x < 0), approaching 0 from the left, f(0) would be -0² - 3*0 - b = -b. But since the function is continuous (assuming it's defined on all real numbers without any breaks), then the left-hand limit and the right-hand limit at x=0 must be equal. Therefore, b = -b, which implies that 2b = 0, so b = 0. Therefore, this confirms that b must be zero. So then, f(x) for x ≥ 0 is x² - 3x, and for x < 0, f(x) = -x² - 3x.Therefore, f(-2) would be -(-2)^2 - 3*(-2) = -4 + 6 = 2. So that's the same result as before. Therefore, the answer is B: 2. Let me check the options again: A is -2, B is 2, C is 10, D is -10. So B is correct.Wait a minute, but just to ensure I didn't skip a step, let me verify once more. The key points here are:1. The function is odd, so f(-x) = -f(x).2. At x=0, f(0) must be 0, which gives b=0.3. For x ≥ 0, f(x) = x² - 3x.4. Therefore, f(2) = 4 - 6 = -2.5. Thus, f(-2) = -f(2) = 2.Alternatively, using the expression for x < 0: f(-2) = -(-2)^2 -3*(-2) = -4 +6 = 2. Either way, same answer. So yeah, B is correct.But let me check the problem statement again to make sure I didn't misread anything. It says "the function y = f(x) is an odd function defined on ℝ, and when x ≥ 0, f(x) = x² - 3x + b". So the key part is that the function is odd, so we use that to find b. Then we use the oddness again to find f(-2) from f(2). Correct.Is there any chance that b is not zero? Suppose for some reason, the problem is not continuous? Wait, the problem states that the function is defined on all ℝ. If it's defined everywhere, then it must be continuous at x=0? Well, actually, not necessarily. Being defined on all real numbers doesn't automatically imply continuity unless specified. But in the case of odd functions, even if there's a discontinuity, f(0) must still be 0. Wait, but if the function isn't continuous at 0, how would that affect things?But the problem doesn't mention continuity, just that it's an odd function defined on ℝ. However, in standard practice, unless specified otherwise, functions defined piecewise like this are assumed to be continuous unless there's a discontinuity explicitly introduced. But here, since the problem gives f(x) for x ≥0 and says it's an odd function, then for x <0, it's determined by oddness. Therefore, the function is defined everywhere. However, unless there's a specific discontinuity, we can assume continuity. But even if we don't assume continuity, the value at x=0 must still be 0, because f(-0) = -f(0) implies f(0) = 0. Therefore, regardless of continuity, f(0) = 0. Therefore, substituting x=0 into the expression for x≥0 gives f(0)=b=0. Therefore, even if the function has a discontinuity elsewhere, at x=0 it must be 0. Therefore, b=0.Therefore, even if there is a jump discontinuity somewhere else, the value at x=0 is forced to be 0, so b=0. Therefore, we can safely say that b=0. Therefore, f(2) = -2, and f(-2) = 2. Therefore, the answer is B: 2. So I think that's solid.Wait, but let me check again if maybe the question is tricking me. For example, maybe when they say "defined on ℝ", they could have different expressions, but since it's odd, the expression for x <0 is determined by x≥0. Therefore, the function is fully defined once we know the expression for x≥0. So in that case, we don't have any other parameters except b. Since f(0) = 0, we have to set b=0. Therefore, f(-2) = -f(2) = -[4 -6] = -(-2) =2. Therefore, answer B.Alternatively, if someone thought that the function is f(x) = x² - 3x + b for all x, but that's not what the problem says. It says "when x ≥0, f(x)=x² -3x +b". So it's a piecewise function where for x≥0, it's x² -3x +b, and for x<0, it's determined by the oddness. Therefore, the only way to find b is through the condition at x=0. So yes, b must be 0, leading to f(-2)=2. Therefore, answer B.I don't see any other way this problem can be approached. So I think the answer is B.Final Answerboxed{B}</think>Given that the function ( y = f(x) ) is an odd function defined on (mathbb{R}), and when ( x geq 0 ), ( f(x) = x^2 - 3x + b ), we need to find ( f(-2) ).1. Odd Function Property: An odd function satisfies ( f(-x) = -f(x) ) for all ( x in mathbb{R} ). This implies ( f(0) = 0 ).2. Determine ( b ): - For ( x = 0 ), ( f(0) = 0^2 - 3 cdot 0 + b = b ). - Since ( f(0) = 0 ), we have ( b = 0 ).3. Function for ( x geq 0 ): - With ( b = 0 ), the function becomes ( f(x) = x^2 - 3x ) for ( x geq 0 ).4. Calculate ( f(2) ): - ( f(2) = 2^2 - 3 cdot 2 = 4 - 6 = -2 ).5. Apply Odd Function Property for ( f(-2) ): - ( f(-2) = -f(2) = -(-2) = 2 ).Thus, the value of ( f(-2) ) is (boxed{B}).]

🔑:Since the function y=f(x) is an odd function defined on mathbb{R}, and when x geqslant 0, f(x)=x^{2}-3x+b,we have f(0)=0, which implies b=0.Thus, when x geqslant 0, f(x)=x^{2}-3x. Therefore, f(2)=2^{2}-3 times 2=-2.Using the property of odd functions, we know that f(-x)=-f(x). Thus, f(-2)=-f(2)=2.Hence, the answer is boxed{B}.

🧠:[<think>Okay, so I need to find the equation of the set of points that are all at a distance of √3 from the point (-1, 0). Hmm, let's see. I remember that the set of points at a fixed distance from a given point forms a circle. Right, so this should be a circle with center at (-1, 0) and radius √3. Let me verify that.First, the general equation of a circle with center (h, k) and radius r is (x - h)^2 + (y - k)^2 = r^2. So if our center is (-1, 0), then h is -1 and k is 0. Plugging those into the equation, we get (x - (-1))^2 + (y - 0)^2 = (√3)^2. Simplifying that, it becomes (x + 1)^2 + y^2 = 3. That seems straightforward.Wait, but maybe I should double-check using the distance formula to make sure. The distance between any point (x, y) and (-1, 0) should be √3. The distance formula is √[(x2 - x1)^2 + (y2 - y1)^2]. Applying that here, it's √[(x - (-1))^2 + (y - 0)^2] = √3. Squaring both sides to eliminate the square root gives (x + 1)^2 + y^2 = 3. Yep, that matches the circle equation I wrote earlier. Is there any chance I could have mixed up the signs? Let me think. The center is (-1, 0), so when substituting into the circle equation, it's (x - h) where h is -1. So x - (-1) is x + 1. That's correct. And the y term is (y - 0)^2, which is just y squared. The radius squared is (√3)^2, which is 3. All right, that all checks out.Maybe I can test a point to see if it works. For example, take the point (-1 + √3, 0). The distance from (-1, 0) should be √3. Let's compute it: √[((-1 + √3) - (-1))^2 + (0 - 0)^2] = √[(√3)^2 + 0] = √3. That works. How about another point, like (-1, √3)? Distance is √[0^2 + (√3)^2] = √3. Also works. What about a point not on the axes? Let's say (0, 0). Distance from (-1, 0) is √[(0 - (-1))^2 + (0 - 0)^2] = √1 = 1, which is not √3, so (0,0) shouldn't be on the circle. Plugging into the equation: (0 + 1)^2 + 0^2 = 1 + 0 = 1 ≠ 3. Correct, so (0,0) is not on the circle. Another test: the point (-1 - √3, 0). Distance is √[(-1 - √3 - (-1))^2 + 0] = √[(-√3)^2] = √3. Plugging into the equation: (-1 - √3 +1)^2 +0 = (-√3)^2 = 3. Yep, that works too. So all the points that are √3 away from (-1,0) lie on this circle. Therefore, the equation is (x + 1)^2 + y^2 = 3. I think that's it. I don't see any mistakes in the reasoning. The key steps were recognizing it's a circle, using the center coordinates in the standard equation, and verifying with the distance formula and test points. Wait, just to be thorough, is there a possibility that it's not a circle? Well, in Euclidean geometry, the set of all points equidistant from a single point is indeed a circle. So unless there's some other constraint, it should be a circle. Since the problem doesn't mention any other conditions, I think this is safe. Therefore, the final answer should be (x + 1)^2 + y^2 = 3. Let me write that in a box as requested.Final AnswerThe equation is boxed{(x + 1)^2 + y^2 = 3}.</think>To find the equation of the set of points that are all at a distance of (sqrt{3}) from the point ((-1, 0)), we recognize that this set forms a circle. The general equation of a circle with center ((h, k)) and radius (r) is ((x - h)^2 + (y - k)^2 = r^2).Given the center ((-1, 0)) and radius (sqrt{3}):- The center coordinates (h = -1) and (k = 0).- The radius squared is ((sqrt{3})^2 = 3).Substituting these values into the general circle equation, we get:[(x - (-1))^2 + (y - 0)^2 = 3]Simplifying this, the equation becomes:[(x + 1)^2 + y^2 = 3]Verification using the distance formula confirms that any point ((x, y)) on this circle is indeed (sqrt{3}) units away from ((-1, 0)). Testing specific points such as ((-1 + sqrt{3}, 0)), ((-1, sqrt{3})), and ((-1 - sqrt{3}, 0)) also confirms the correctness of the equation.Thus, the equation of the set of points is (boxed{(x + 1)^2 + y^2 = 3}).,<think>Okay, let me try to figure this out. The problem is asking for the equation of the set of points that are all at a distance of √3 from the point (-1, 0). Hmm, so I need to find all the points (x, y) in the plane such that their distance from (-1, 0) is exactly √3. First, I remember that the distance formula between two points (x1, y1) and (x2, y2) is √[(x2 - x1)² + (y2 - y1)²]. So, if we want the distance between (x, y) and (-1, 0) to be √3, we can set up the equation using this formula. Let me write that out. The distance between (x, y) and (-1, 0) is √[(x - (-1))² + (y - 0)²] = √3. Simplifying inside the square root, that becomes √[(x + 1)² + y²] = √3. To get rid of the square root, I can square both sides of the equation. That would give [(x + 1)² + y²] = (√3)². Since squaring both sides removes the square roots. Calculating the right side, (√3)² is just 3. So now the equation is (x + 1)² + y² = 3. Let me check if this makes sense. This looks like the standard equation of a circle. The general form is (x - h)² + (y - k)² = r², where (h, k) is the center and r is the radius. In this case, comparing the two equations, the center should be at (-1, 0) and the radius squared is 3, so the radius is √3. Which matches the problem statement: all points at a distance of √3 from (-1, 0). So that checks out. Wait, let me make sure I didn't mix up any signs. The standard equation is (x - h)² + (y - k)² = r². Here, in the given problem, the center is (-1, 0), so h = -1 and k = 0. Therefore, substituting h and k into the equation, we get (x - (-1))² + (y - 0)² = (√3)², which simplifies to (x + 1)² + y² = 3. Yes, that seems correct. So the equation of the set of points is a circle with center (-1, 0) and radius √3. Let me just think if there's any other way this could be approached. Maybe if I tried expanding the equation? Let's see. If we expand (x + 1)² + y² = 3, we get x² + 2x + 1 + y² = 3. Then combining like terms, x² + y² + 2x + 1 - 3 = 0, which simplifies to x² + y² + 2x - 2 = 0. But that's the general form of the circle equation. However, the problem just asks for the equation, and both forms are correct. But usually, the standard form is more informative because it directly shows the center and radius. So unless specified otherwise, the standard form is preferable. Another check: pick a point that should be on this circle and verify. For example, the point (-1, √3). Let's compute its distance from (-1, 0): √[(-1 - (-1))² + (√3 - 0)²] = √[0 + 3] = √3, which matches. Similarly, the point (-1 + √3, 0) should be on the circle. Distance is √[(√3)² + 0] = √3, which is correct. What about a point not on the circle? Let's say the origin (0, 0). Distance to (-1, 0) is √[(0 +1)^2 + 0] = √1 = 1, which is less than √3, so (0, 0) shouldn't satisfy the equation. Plugging into the equation: (0 +1)^2 + 0^2 = 1 ≠ 3. Correct. Another test: point (-1 - √3, 0). Distance is √[(-√3)^2 + 0] = √3. Plugging into equation: (-√3)^2 + 0^2 = 3, which equals 3. So that works. Hmm, seems like all tests are passing. Therefore, I think the equation is (x + 1)^2 + y^2 = 3. I don't see any mistakes in my reasoning. The steps are straightforward: use the distance formula, square both sides to eliminate the square root, simplify, and confirm by testing points. So I think that's the answer.Final AnswerThe equation is boxed{(x + 1)^2 + y^2 = 3}.</think>To find the equation of the set of points in the plane that are all at a distance of (sqrt{3}) from the point ((-1, 0)), we start by using the distance formula. The distance between any point ((x, y)) and the point ((-1, 0)) is given by:[sqrt{(x - (-1))^2 + (y - 0)^2} = sqrt{3}]Simplifying inside the square root, we get:[sqrt{(x + 1)^2 + y^2} = sqrt{3}]Squaring both sides to eliminate the square root, we obtain:[(x + 1)^2 + y^2 = 3]This is the standard form of the equation of a circle with center ((-1, 0)) and radius (sqrt{3}). Verifying with test points and expanding the equation confirms the correctness of this form.Thus, the equation of the set of points is (boxed{(x + 1)^2 + y^2 = 3}).]

🔑:The set of points in a plane that are all a distance of sqrt{3} from the point (-1, 0) forms a circle centered at (-1, 0) with a radius of sqrt{3}.To find the equation of this circle, we use the standard equation of a circle centered at (h, k) with radius r:(x - h)^2 + (y - k)^2 = r^2Using the center (-1, 0) and radius sqrt{3}, we substitute:(x - (-1))^2 + (y - 0)^2 = (sqrt{3})^2Simplify and expand:(x + 1)^2 + y^2 = 3The equation of the circle in expanded form is:(x + 1)^2 + y^2 = 3Therefore, the equation of the set of points in the plane that are all at a distance of sqrt{3} from the point (-1, 0) is:[boxed{(x + 1)^2 + y^2 = 3}]This solution demonstrates the basic understanding of the definition of a circle. Knowing that the set of all points equidistant from a single point forms a circle is key.