🧠:[<think>Okay, let's see. The problem is about shifting the graph of y = sin(ωx) to the left by π/3 units and making it coincide with y = cos(ωx). We need to find possible values of ω from the given options. Hmm. First, I remember that shifting a function left by a certain amount involves replacing x with (x + shift) in the function. So, shifting sin(ωx) left by π/3 would mean we replace x with x + π/3. That should give us sin(ω(x + π/3)). The problem states that this shifted function is equal to cos(ωx). So, the equation we need to solve is:sin(ω(x + π/3)) = cos(ωx)Now, I need to recall some trigonometric identities to see if I can simplify this equation. I remember that sin(A + B) = sinA cosB + cosA sinB. Let me apply that here. Let's let A = ωx and B = ω*(π/3). Then,sin(ωx + ωπ/3) = sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3)So, the left side becomes sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3). And this is equal to cos(ωx). So:sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3) = cos(ωx)Let me rearrange terms:sin(ωx)cos(ωπ/3) = cos(ωx) - cos(ωx)sin(ωπ/3)Factor out cos(ωx) on the right side:sin(ωx)cos(ωπ/3) = cos(ωx)(1 - sin(ωπ/3))Hmm, not sure if that's helpful. Maybe bring all terms to one side:sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3) - cos(ωx) = 0Factor terms with cos(ωx):sin(ωx)cos(ωπ/3) + cos(ωx)(sin(ωπ/3) - 1) = 0Alternatively, maybe express this as a single sine or cosine function. Let me think. The left side is a combination of sin(ωx) and cos(ωx), which can be written as a single sinusoidal function. Maybe that's a good approach.Let me denote θ = ωx to simplify notation. Then the equation becomes:sinθ cos(ωπ/3) + cosθ sin(ωπ/3) - cosθ = 0Combine the first two terms using the sine addition formula:sin(θ + ωπ/3) - cosθ = 0So, sin(θ + ωπ/3) = cosθBut θ is ωx, so substituting back:sin(ωx + ωπ/3) = cos(ωx)Alternatively, I can use the identity that sin(A) = cos(B) implies that A = B + π/2 + 2πn or A = -B + π/2 + 2πn for some integer n. Maybe that's useful here.So, set ωx + ωπ/3 = ±(ωx) + π/2 + 2πnWait, but this has to hold for all x, right? Because the functions need to coincide for all x. So, the equation sin(ωx + ωπ/3) = cos(ωx) must hold for all x. Which means that the functions are identical, so their expressions must be equivalent. Therefore, the coefficients of x and the constants must satisfy certain conditions.Alternatively, perhaps we can use another identity. Since sin(α) = cos(β), and we know that sin(α) = cos(π/2 - α). So, cos(β) = cos(π/2 - α), which implies that β = π/2 - α + 2πn or β = - (π/2 - α) + 2πn for some integer n.But in this case, we have sin(ωx + ωπ/3) = cos(ωx). Let me write that as:cos(π/2 - ωx - ωπ/3) = cos(ωx)Therefore, the equation holds if:π/2 - ωx - ωπ/3 = ωx + 2πnorπ/2 - ωx - ωπ/3 = -ωx + 2πnfor some integer n, and for all x. However, since this must be true for all x, the coefficients of x in the equations must be equal, and the constants must also match.Let's take the first case:π/2 - ωx - ωπ/3 = ωx + 2πnSimplify:-ωx - ωπ/3 + π/2 = ωx + 2πnBring like terms together:-ωx - ωπ/3 - ωx = 2πn - π/2Combine the x terms:-2ωx - ωπ/3 = 2πn - π/2But this must hold for all x, which is only possible if the coefficient of x is zero. So:-2ω = 0 ⇒ ω = 0But ω can't be zero because then the original functions would be constant functions. The problem likely expects a non-zero ω. So this case is invalid.Now consider the second case:π/2 - ωx - ωπ/3 = -ωx + 2πnSimplify:π/2 - ωπ/3 = 2πnAgain, the x terms cancel out, so:π/2 - ωπ/3 = 2πnDivide both sides by π:1/2 - ω/3 = 2nSo,-ω/3 = 2n - 1/2Multiply both sides by -3:ω = -3(2n - 1/2) = -6n + 3/2Thus, ω = 3/2 - 6nNow, n is an integer, so let's find possible ω values from the options given.The options are A: 3/2, B:1/2, C:1/3, D:2/3.Let's plug n=0: ω = 3/2 - 0 = 3/2, which is option A.n=1: ω = 3/2 -6(1)= 3/2 -6= -9/2. Not among options.n=-1: ω=3/2 -6(-1)=3/2 +6=15/2. Not in options.So the only possible ω from this case is 3/2, which is option A.Wait, but are there other possibilities? Let me check.Alternatively, perhaps there's another approach. Let me consider shifting the sine function left by π/3, so the function becomes sin(ω(x + π/3)) = sin(ωx + ωπ/3). And this is equal to cos(ωx). So:sin(ωx + ωπ/3) = cos(ωx)We can use the identity that sin(α) = cos(β) implies α = β + π/2 + 2πn or α = -β + π/2 + 2πn, for integer n.But since this equation must hold for all x, the arguments of the sine and cosine must differ by a multiple of 2π plus π/2. So, setting:ωx + ωπ/3 = ωx + π/2 + 2πnorωx + ωπ/3 = -ωx + π/2 + 2πnFirst equation:ωx + ωπ/3 = ωx + π/2 + 2πnSubtract ωx from both sides:ωπ/3 = π/2 + 2πnDivide by π:ω/3 = 1/2 + 2nMultiply by 3:ω = 3/2 + 6nPossible ω values here would be 3/2 (n=0), 15/2 (n=1), etc. So 3/2 is an option (A), which is one of the choices.Second equation:ωx + ωπ/3 = -ωx + π/2 + 2πnBring terms with x to one side:ωx + ωx = π/2 + 2πn - ωπ/32ωx = π/2 + 2πn - ωπ/3But this must hold for all x, which implies that the coefficient of x must be zero, so 2ω = 0 ⇒ ω=0, which is invalid. The remaining equation is 0 = π/2 + 2πn - ωπ/3. If ω=0, this becomes 0 = π/2 + 2πn, which is impossible. Hence, no solution from this case.Therefore, the only possible solution is ω=3/2 +6n. Given the options, the only valid value is ω=3/2 (option A). Let me check with the options.Wait, but the answer options include 1/2, 1/3, 2/3. Maybe I made a mistake here.Alternatively, perhaps there's another way to approach this problem.Let me recall that shifting a sine function left by π/3 to get a cosine function. The standard identity is that sin(x + π/2) = cos(x). So shifting sin(x) left by π/2 gives cos(x). But in this case, it's sin(ωx) shifted left by π/3. So maybe the shift required to turn sine into cosine is π/(2ω). Because if we have sin(ω(x + c)) = cos(ωx), then ωc = π/2, so c = π/(2ω). But in the problem, the shift is given as π/3. So π/(2ω) = π/3 ⇒ 1/(2ω) = 1/3 ⇒ ω = 3/2. Which is option A. That seems straightforward.Therefore, ω must be 3/2. So answer is A.But let me check with another method.Suppose we set the shifted sine function equal to the cosine function:sin(ω(x + π/3)) = cos(ωx)Expanding the left side:sin(ωx + ωπ/3) = sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3)Set equal to cos(ωx):sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3) = cos(ωx)Bring all terms to one side:sin(ωx)cos(ωπ/3) + cos(ωx)(sin(ωπ/3) - 1) = 0For this equation to hold for all x, the coefficients of sin(ωx) and cos(ωx) must both be zero. Therefore:cos(ωπ/3) = 0andsin(ωπ/3) - 1 = 0But cos(ωπ/3) = 0 implies ωπ/3 = π/2 + kπ for some integer k, so ω = (π/2 + kπ)*(3/π) = 3/2 + 3kSimilarly, sin(ωπ/3) - 1 = 0 implies sin(ωπ/3) = 1 ⇒ ωπ/3 = π/2 + 2mπ ⇒ ω = (π/2 + 2mπ)*(3/π) = 3/2 + 6mSo both conditions must be satisfied. So from cos(ωπ/3)=0, ω=3/2 +3kFrom sin(ωπ/3)=1, ω=3/2 +6mSo the solutions are the intersection of these two sets. So ω must be 3/2 +6m, since 3/2 +6m is a subset of 3/2 +3k (when k is even). Therefore, the solutions are ω=3/2 +6m, m integer.Therefore, the smallest positive solution is 3/2. If m=0, ω=3/2. For m=1, ω=15/2, which is larger. So from the options given, 3/2 is an option (A). So the answer is A.But the problem says "possible value(s)", and the options include other fractions. Let me check if I missed something.Wait, but if we use the identity sin(θ + φ) = cosθ, then θ + φ = π/2 - θ + 2πn or θ + φ = π/2 + θ + 2πn?Wait, earlier approach had using sin(α) = cos(β) leading to α = π/2 - β + 2πn or α = π/2 + β + 2πn.Wait, let's re-examine that. If sin(α) = cos(β), then since sin(α) = cos(π/2 - α), we have cos(π/2 - α) = cos(β). Therefore, π/2 - α = ±β + 2πn. Therefore:Case 1: π/2 - α = β + 2πn ⇒ α + β = π/2 - 2πnCase 2: π/2 - α = -β + 2πn ⇒ α - β = π/2 - 2πnIn our equation, α = ωx + ωπ/3, β = ωx.So, substituting into the two cases:Case 1: (ωx + ωπ/3) + ωx = π/2 - 2πn ⇒ 2ωx + ωπ/3 = π/2 - 2πnBut this must hold for all x, so the coefficient of x must be zero: 2ω = 0 ⇒ ω=0, which is invalid. Then, even if ω=0, the constants would not match. So this case is impossible.Case 2: (ωx + ωπ/3) - ωx = π/2 - 2πn ⇒ ωπ/3 = π/2 - 2πn ⇒ ω/3 = 1/2 - 2n ⇒ ω = 3/2 -6nSo ω = 3/2 -6n. For integer n.Possible values:n=0: ω=3/2 (option A)n=1: ω=3/2 -6= -9/2 (negative, but maybe ω is allowed to be negative? The problem doesn't specify. However, if ω is negative, it's equivalent to a positive frequency with a phase shift. But since the question asks for possible values, and options are positive, maybe only 3/2 is valid.Wait, but the options include 1/2, 1/3, 2/3. Could there be another solution?Alternatively, perhaps I made a mistake in the previous steps.Wait, another approach: if we have sin(ωx + ωπ/3) = cos(ωx), this can be rewritten as:sin(ωx + ωπ/3) = sin(π/2 - ωx)Because cos(θ) = sin(π/2 - θ). So,sin(ωx + ωπ/3) = sin(π/2 - ωx)The general solution for sin(A) = sin(B) is A = B + 2πn or A = π - B + 2πn, where n is integer.Therefore, either:1. ωx + ωπ/3 = π/2 - ωx + 2πnor2. ωx + ωπ/3 = π - (π/2 - ωx) + 2πn = π/2 + ωx + 2πnLet's consider case 1:ωx + ωπ/3 = π/2 - ωx + 2πnBring terms with x to one side:ωx + ωx = π/2 - ωπ/3 + 2πn2ωx = π/2 - ωπ/3 + 2πnAgain, this must hold for all x, so the coefficient of x must be zero:2ω = 0 ⇒ ω=0, which is invalid. So discard this case.Case 2:ωx + ωπ/3 = π/2 + ωx + 2πnSubtract ωx from both sides:ωπ/3 = π/2 + 2πnDivide by π:ω/3 = 1/2 + 2nMultiply by 3:ω = 3/2 +6nWhich is the same result as before. So possible ω values are 3/2 +6n. Since the options include 3/2 (A), that's the valid one.But wait, maybe there's another identity or another approach where ω could be different? Let's test the given options.Suppose we check option A: ω=3/2Shift sin(3/2 x) left by π/3: sin(3/2(x + π/3)) = sin(3/2 x + 3/2 * π/3) = sin(3/2 x + π/2) = sin(3/2 x + π/2)But sin(θ + π/2) = cosθ, so this becomes cos(3/2 x). Which is exactly the function we're comparing to: cos(ωx) with ω=3/2. So it works. Therefore, option A is correct.Now check option B: ω=1/2Shift sin(1/2 x) left by π/3: sin(1/2(x + π/3)) = sin(1/2 x + π/6)Compare to cos(1/2 x). So is sin(1/2 x + π/6) = cos(1/2 x)?Let’s see. sin(θ + π/6) = sinθ cos(π/6) + cosθ sin(π/6) = (sqrt(3)/2) sinθ + (1/2) cosθ. For this to equal cosθ, we need:(sqrt(3)/2) sinθ + (1/2) cosθ = cosθ ⇒ (sqrt(3)/2) sinθ = (1/2) cosθ ⇒ sqrt(3) sinθ = cosθ ⇒ tanθ = 1/sqrt(3) ⇒ θ = π/6 + kπ. But this must hold for all x, which it does not. Therefore, option B is invalid.Similarly, check option D: ω=2/3Shift sin(2/3 x) left by π/3: sin(2/3(x + π/3)) = sin(2/3 x + 2π/9)Compare to cos(2/3 x). So sin(2/3 x + 2π/9) = cos(2/3 x)Again, use identity sin(α) = cos(β). Let α = 2/3 x + 2π/9, β = 2/3 x.Then, sin(α) = cos(β) ⇒ α = π/2 - β + 2πn or α = π/2 + β + 2πn.Case 1: 2/3 x + 2π/9 = π/2 - 2/3 x + 2πn ⇒ (2/3 + 2/3)x = π/2 - 2π/9 + 2πn ⇒ (4/3)x = (9π/18 - 4π/18) + 2πn ⇒ (4/3)x = 5π/18 + 2πn. But this can't hold for all x. So invalid.Case 2: 2/3 x + 2π/9 = π/2 + 2/3 x + 2πn ⇒ 2π/9 = π/2 + 2πn ⇒ 2π/9 - π/2 = 2πn ⇒ (4π/18 - 9π/18) = 2πn ⇒ (-5π/18) = 2πn ⇒ n = -5/36. Not an integer, so invalid. Hence, option D is out.Similarly, option C: ω=1/3Shift sin(1/3 x) left by π/3: sin(1/3(x + π/3)) = sin(1/3 x + π/9)Compare to cos(1/3 x). So sin(1/3 x + π/9) = cos(1/3 x)Again, let’s see:sin(θ + π/9) = cosθ ⇒ sinθ cos(π/9) + cosθ sin(π/9) = cosθRearrange:sinθ cos(π/9) = cosθ (1 - sin(π/9))Divide both sides by cosθ (assuming cosθ ≠0):tanθ cos(π/9) = 1 - sin(π/9)But this must hold for all θ =1/3 x, which is impossible since tanθ varies with θ. Thus, option C is invalid.Therefore, only option A:3/2 works.Final Answerboxed{A}</think>To solve the problem, we start by noting that shifting the graph of ( y = sin(omega x) ) left by (frac{pi}{3}) units results in the function ( y = sin(omega(x + frac{pi}{3})) ). This shifted function is given to coincide with ( y = cos(omega x) ). Therefore, we need to solve the equation:[sin(omega(x + frac{pi}{3})) = cos(omega x)]Using the trigonometric identity for sine addition, we expand the left-hand side:[sin(omega x + omega frac{pi}{3}) = sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) sin(omega frac{pi}{3})]This must equal (cos(omega x)). Rearranging terms, we get:[sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) sin(omega frac{pi}{3}) = cos(omega x)]Subtracting (cos(omega x)) from both sides:[sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) (sin(omega frac{pi}{3}) - 1) = 0]For this equation to hold for all (x), the coefficients of (sin(omega x)) and (cos(omega x)) must be zero. This gives us two conditions:1. (cos(omega frac{pi}{3}) = 0)2. (sin(omega frac{pi}{3}) - 1 = 0)Solving the first condition (cos(omega frac{pi}{3}) = 0):[omega frac{pi}{3} = frac{pi}{2} + kpi implies omega = frac{3}{2} + 3k quad text{for integer } k]Solving the second condition (sin(omega frac{pi}{3}) = 1):[omega frac{pi}{3} = frac{pi}{2} + 2mpi implies omega = frac{3}{2} + 6m quad text{for integer } m]The intersection of these solutions occurs when (k) is even, giving (omega = frac{3}{2} + 6m). The smallest positive solution is (omega = frac{3}{2}).Testing the given options, we find that (omega = frac{3}{2}) (option A) satisfies the conditions, while the other options do not. Thus, the correct answer is:[boxed{A}],<think>Okay, so I need to solve this problem where shifting the graph of y = sin(ωx) to the left by π/3 units makes it coincide with y = cos(ωx). And then find the possible value(s) of ω from the given options. Let me think about how to approach this.First, shifting a function left by π/3 means replacing x with x + π/3 in the function. So the shifted sine function would be y = sin[ω(x + π/3)]. The problem states that this shifted sine function is equal to the cosine function y = cos(ωx). So the equation we get is:sin[ω(x + π/3)] = cos(ωx)Hmm, okay. Now I need to find the value(s) of ω that satisfy this equation for all x. Because if the graphs coincide, the equation must hold for every x in their domain. So this identity must be true for all x, which means we can use trigonometric identities to manipulate it.I remember that sin(A) = cos(B) can be rewritten using co-function identities. For example, sin(θ) = cos(π/2 - θ). Alternatively, there's a phase shift relationship between sine and cosine. Maybe I can express one as a shifted version of the other.Alternatively, maybe I can use the sine addition formula on the left side. Let's try that. The left side is sin[ωx + ωπ/3]. So expanding that using the sine addition formula:sin(ωx + ωπ/3) = sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3)And this is supposed to equal cos(ωx). So we have:sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3) = cos(ωx)Now, let's rearrange terms:sin(ωx)cos(ωπ/3) + cos(ωx)sin(ωπ/3) - cos(ωx) = 0Factor out cos(ωx) from the last two terms:sin(ωx)cos(ωπ/3) + cos(ωx)[sin(ωπ/3) - 1] = 0For this equation to hold for all x, the coefficients of sin(ωx) and cos(ωx) must both be zero. Because sin and cos are linearly independent functions; you can't express one as a multiple of the other unless the coefficients are zero. So that gives us two equations:1. cos(ωπ/3) = 02. sin(ωπ/3) - 1 = 0Wait, but let me check that. If we have A sin(ωx) + B cos(ωx) = 0 for all x, then A must be zero and B must be zero. So yes, that's correct.Therefore, setting coefficients to zero:1. Coefficient of sin(ωx): cos(ωπ/3) = 02. Coefficient of cos(ωx): sin(ωπ/3) - 1 = 0But wait, these are two equations. Let me write them down:1. cos(ωπ/3) = 02. sin(ωπ/3) = 1So we need to find ω such that both cos(ωπ/3) = 0 and sin(ωπ/3) = 1.But cos(θ) = 0 and sin(θ) = 1 simultaneously only when θ = π/2 + 2πk, where k is an integer. Because that's where cosine is zero and sine is one.So, ωπ/3 = π/2 + 2πkSolving for ω:ω = (π/2 + 2πk) * (3/π) = (3/2) + 6kWhere k is any integer.Therefore, possible values of ω are 3/2 + 6k. Now, looking at the answer choices: A is 3/2, B is 1/2, C is 1/3, D is 2/3.So 3/2 is option A. But is there another possible value? If k = 0, we get 3/2. If k = -1, then ω = 3/2 - 6 = -9/2, which is negative, and since ω is a coefficient in sin(ωx), typically we consider ω positive for frequency. So maybe only k=0 is valid here. Hence, ω = 3/2 is the only solution. So answer A.But wait, let me double-check. Let's substitute ω = 3/2 into the original equation and see if it works.Original shifted sine function: sin[(3/2)(x + π/3)] = sin[(3/2)x + (3/2)(π/3)] = sin[(3/2)x + π/2]And the cosine function is cos[(3/2)x]But sin(θ + π/2) = cos(θ). So sin[(3/2)x + π/2] = cos[(3/2)x]. Therefore, yes, it works. So ω = 3/2 is indeed a solution.Are there any other possible solutions? For example, if k=1, ω = 3/2 + 6 = 15/2, which is not among the options. So the only possible value in the options is 3/2, which is option A.Wait, but let me check if there's another way to approach the problem that might give another answer. Maybe using phase shifts differently.Alternatively, we know that sin(θ + π/2) = cos(θ). So, if we can write the shifted sine function as sin(ωx + ωπ/3) = cos(ωx). Comparing to sin(ωx + π/2) = cos(ωx). Therefore, we need:ωπ/3 = π/2 + 2πk, where k is any integer.Then solving for ω gives:ω = (π/2 + 2πk) * 3/π = (3/2 + 6k)Which is the same as before. So only k=0 gives ω=3/2, which is option A. So that seems correct.But maybe there's another identity or another approach? Let me think.Alternatively, we can use the identity that sin(α + β) = cos(β) when α = π/2 - β + 2πk? Wait, not exactly. Wait, if sin(A) = cos(B), then A = π/2 - B + 2πk or A = π/2 + B + 2πk, for some integer k. So in this case, we have:sin(ωx + ωπ/3) = cos(ωx)Therefore, setting A = ωx + ωπ/3 and B = ωx, then:Either:1. ωx + ωπ/3 = π/2 - ωx + 2πkor2. ωx + ωπ/3 = π/2 + ωx + 2πkfor some integer k, and this must hold for all x.Let's consider both cases.Case 1:ωx + ωπ/3 = π/2 - ωx + 2πkBring terms with x to one side:ωx + ωx = π/2 - ωπ/3 + 2πk2ωx = π(1/2 - ω/3) + 2πkBut this has to hold for all x, which is only possible if the coefficient of x is zero and the constant term is also zero. So:2ω = 0 ⇒ ω = 0, but ω = 0 is not possible because the original functions would be constants, which can't coincide unless both are zero, which they aren't. So case 1 leads to no solution.Case 2:ωx + ωπ/3 = π/2 + ωx + 2πkSubtract ωx from both sides:ωπ/3 = π/2 + 2πkDivide both sides by π:ω/3 = 1/2 + 2kMultiply both sides by 3:ω = 3/2 + 6kWhich is the same result as before. So only possible ω is 3/2 + 6k. So again, the only valid answer in the options is 3/2, which is option A.Therefore, the answer is A.But wait, the problem says "possible value(s)", and the options include other fractions. Maybe I missed something. Let me check again.Wait, maybe the equation sin(ω(x + π/3)) = cos(ωx) can also be satisfied for other ω if we consider different identities. Let me try another approach.We can write the equation as:sin(ωx + ωπ/3) = cos(ωx)But using the identity sin(θ) = cos(π/2 - θ), so:sin(ωx + ωπ/3) = sin(π/2 - ωx)Therefore, this gives us:ωx + ωπ/3 = π/2 - ωx + 2πn or ωx + ωπ/3 = π - (π/2 - ωx) + 2πn, where n is integer.First equation:ωx + ωπ/3 = π/2 - ωx + 2πnBring terms with x to the left:ωx + ωx = π/2 - ωπ/3 + 2πn2ωx = π(1/2 - ω/3) + 2πnAgain, this must hold for all x, so coefficients of x must be zero and constants must be zero.So 2ω = 0 ⇒ ω = 0, which is invalid.Then constants: π(1/2 - ω/3) + 2πn = 0 ⇒ 1/2 - ω/3 + 2n = 0 ⇒ ω/3 = 1/2 + 2n ⇒ ω = 3/2 + 6nWhich is the same as before. So only solution ω = 3/2 + 6n, n integer.Second equation from the sine equality:ωx + ωπ/3 = π - π/2 + ωx + 2πn ⇒ ωx + ωπ/3 = π/2 + ωx + 2πnSubtract ωx:ωπ/3 = π/2 + 2πn ⇒ ω = (π/2 + 2πn) * 3/π = 3/2 + 6nSame result. So regardless of the approach, we get ω = 3/2 + 6n. So again, the only valid answer in the options is 3/2, which is A.Wait, but maybe there's another approach where the phase shift is considered as a horizontal shift. Let me think. The original function is sin(ωx). Shifting left by π/3 gives sin(ω(x + π/3)) = sin(ωx + ωπ/3). For this to be equal to cos(ωx), which is sin(ωx + π/2), since cos(θ) = sin(θ + π/2). So:sin(ωx + ωπ/3) = sin(ωx + π/2)For these two sine functions to be equal for all x, their arguments must differ by a multiple of 2π or be supplementary angles differing by a multiple of 2π. So:Either:1. ωx + ωπ/3 = ωx + π/2 + 2πnor2. ωx + ωπ/3 = π - (ωx + π/2) + 2πnfor some integer n.First case:Subtract ωx from both sides:ωπ/3 = π/2 + 2πn ⇒ ω = (π/2 + 2πn) * 3/π = 3/2 + 6nSame as before.Second case:ωx + ωπ/3 = π - ωx - π/2 + 2πnBring like terms together:ωx + ωx = π - π/2 - ωπ/3 + 2πn2ωx = π/2 - ωπ/3 + 2πnAgain, this must hold for all x. So coefficient of x must be zero:2ω = 0 ⇒ ω = 0, invalid.So only the first case gives valid solutions. Hence, ω = 3/2 + 6n. So again, same conclusion.Therefore, the answer is A: 3/2.Wait, but the options include B: 1/2, C: 1/3, D: 2/3. Is there a possibility that I made a mistake in considering all solutions? Let me plug in some of the other options and see.For example, let's check option B: ω = 1/2.Shifted function: sin[(1/2)(x + π/3)] = sin[(1/2)x + π/6]Compare to cos[(1/2)x].Is sin(θ + π/6) equal to cos(θ)? Let's check θ = 0: sin(π/6) = 1/2, cos(0) = 1. Not equal. So they are not the same function. Therefore, ω = 1/2 is not a solution.Option C: ω = 1/3Shifted function: sin[(1/3)(x + π/3)] = sin[(1/3)x + π/9]Compare to cos[(1/3)x]Again, at x = 0: sin(π/9) ≈ 0.3420, cos(0) = 1. Not equal. Not a solution.Option D: ω = 2/3Shifted function: sin[(2/3)(x + π/3)] = sin[(2/3)x + (2/3)(π/3)] = sin[(2/3)x + 2π/9]Compare to cos[(2/3)x]At x = 0: sin(2π/9) ≈ 0.6428, cos(0) = 1. Not equal. So not a solution.Therefore, only option A works.Alternatively, let's check another point for ω = 3/2.Take x = 0:Shifted function: sin[(3/2)(0 + π/3)] = sin[(3/2)(π/3)] = sin(π/2) = 1cos[(3/2)(0)] = cos(0) = 1. So equal here.Take x = π/(3ω). Wait, let's pick x = 2π/(3ω). Wait, let me pick x = something else.Take x = π/(3). Then:Shifted function: sin[(3/2)(π/3 + π/3)] = sin[(3/2)(2π/3)] = sin(π) = 0cos[(3/2)(π/3)] = cos(π/2) = 0. Equal here.Take x = π/6:Shifted function: sin[(3/2)(π/6 + π/3)] = sin[(3/2)(π/6 + 2π/6)] = sin[(3/2)(π/2)] = sin(3π/4) = √2/2cos[(3/2)(π/6)] = cos(π/4) = √2/2. Equal again. So seems to hold.Therefore, ω = 3/2 is correct. So answer is A.I think that's thorough. All approaches lead to ω = 3/2. The other options don't satisfy the equation when tested. So I'm confident the answer is A.Final Answerboxed{A}</think>To solve the problem, we start by noting that shifting the graph of ( y = sin(omega x) ) left by ( frac{pi}{3} ) units results in the function ( y = sin(omega (x + frac{pi}{3})) ). This shifted function must coincide with ( y = cos(omega x) ). Therefore, we need to find (omega) such that:[sin(omega (x + frac{pi}{3})) = cos(omega x)]Using the sine addition formula, we expand the left-hand side:[sin(omega x + omega frac{pi}{3}) = sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) sin(omega frac{pi}{3})]We want this to equal (cos(omega x)). Rearranging terms, we get:[sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) sin(omega frac{pi}{3}) - cos(omega x) = 0]Factoring out (cos(omega x)) from the last two terms:[sin(omega x) cos(omega frac{pi}{3}) + cos(omega x) (sin(omega frac{pi}{3}) - 1) = 0]For this equation to hold for all (x), the coefficients of (sin(omega x)) and (cos(omega x)) must be zero. This gives us two equations:1. (cos(omega frac{pi}{3}) = 0)2. (sin(omega frac{pi}{3}) - 1 = 0)Solving these, we find that (cos(omega frac{pi}{3}) = 0) and (sin(omega frac{pi}{3}) = 1). These conditions are satisfied when:[omega frac{pi}{3} = frac{pi}{2} + 2kpi quad text{for integer } k]Solving for (omega):[omega = frac{3}{2} + 6k]Considering the given options, the only valid solution is (omega = frac{3}{2}) (when (k = 0)). Testing this value confirms it satisfies the original equation for all (x).Thus, the possible value of (omega) is (boxed{A}).]

🔑:Analysis:This problem examines the translation transformations of trigonometric function graphs and the application of trigonometric function properties. It is of medium difficulty.Step 1: Shift the graph of y=sin omega x to the left by frac{pi }{3} units, resulting in the function y=sin left(omegaleft(x+ frac{pi }{3}right)right)=sin left(omega x+ frac{omega pi }{3}right).Step 2: Since the graphs of y=sin left(omega x+ frac{omega pi }{3}right) and y=cos omega x coincide, we have frac{omega pi }{3}=2kpi + frac{pi }{2}, where kin mathbb{Z}.Step 3: Solve for omega:omega =6k+frac{3}{2}, where kin mathbb{Z}.Step 4: Among the given choices, only omega =frac{3}{2} satisfies the equation.Hence, the correct answer is boxed{text{A}}.

🧠:[<think>Okay, so I need to figure out the range of m and n such that the graphs of f(x) = (1 - m)ln x and g(x) = - (m/2)x² - (m² - m -1)x - n intersect at three distinct points. The answer choices are given as A through D. Let me start by understanding what the problem is asking.First, the intersection points of two graphs occur where f(x) = g(x). So, setting the two functions equal:(1 - m)ln x = - (m/2)x² - (m² - m -1)x - nI need to find the values of m and n for which this equation has three distinct real solutions. Since ln x is only defined for x > 0, we can restrict our attention to x > 0.Let me rearrange the equation:(1 - m)ln x + (m/2)x² + (m² - m -1)x + n = 0Let me denote h(x) = (1 - m)ln x + (m/2)x² + (m² - m -1)x + nSo, the problem reduces to finding m and n such that h(x) = 0 has three distinct positive real roots.To analyze the number of roots, I can consider the behavior of h(x) as x approaches 0+, x approaches infinity, and check its critical points to see when it can cross the x-axis three times.First, let's analyze the limits:1. As x approaches 0+:ln x tends to -infinity. So, the term (1 - m)ln x will dominate if (1 - m) ≠ 0.If (1 - m) > 0, i.e., m < 1, then (1 - m)ln x approaches -infinity.If (1 - m) < 0, i.e., m > 1, then (1 - m)ln x approaches +infinity.But we also have the other terms. Let's see:h(x) = (1 - m)ln x + (m/2)x² + (m² - m -1)x + nSo, as x approaches 0+:If m < 1: h(x) approaches -infinity (since (1 - m)ln x dominates and goes to -infinity)If m > 1: h(x) approaches +infinity (since (1 - m)ln x dominates and goes to +infinity)If m = 1: Then the ln x term disappears. Then h(x) = (1/2)x² + (1 -1 -1)x + n = (1/2)x² - x + n. As x approaches 0+, h(x) approaches n. But m=1 is not in the answer options, so maybe we can ignore this case for now.2. As x approaches infinity:The term (m/2)x² will dominate if m ≠ 0.If m > 0, then (m/2)x² tends to +infinity.If m < 0, then (m/2)x² tends to -infinity.Also, the term (m² - m -1)x is linear, but compared to the quadratic term, it becomes negligible as x approaches infinity.So, as x approaches infinity:If m > 0: h(x) approaches +infinityIf m < 0: h(x) approaches -infinityIf m = 0: Then h(x) = (1 - 0)ln x + 0 + (-0 -0 -1)x + n = ln x - x + n. Here, as x approaches infinity, ln x grows slower than -x, so h(x) approaches -infinity. But m=0 is also a possible boundary value. Let's check answer options; the options given are ranges, so m is either <0, between 0 and 1, or >1. Since m=0 and m=1 are boundaries, but the options are open intervals.So, summarizing the behavior:Case 1: m < 0As x→0+, h(x) approaches -infinity (since m < 0 <1, so (1 - m) >1, so ln x dominates to -infinity)As x→∞, h(x) approaches -infinity (since m <0, quadratic term dominates to -infinity)Therefore, for m <0, h(x) starts at -infinity and goes to -infinity. So, to have three roots, h(x) must have some maxima above zero and minima below zero. But since both ends go to -infinity, it's possible only if the function has at least two critical points (a local maximum and a local minimum), and the local maximum is above zero, and the local minimum is below zero. Then, the function would cross the x-axis three times. But let's verify this.Case 2: 0 < m <1As x→0+, h(x) approaches -infinity (since (1 - m) >0, so ln x dominates to -infinity)As x→∞, h(x) approaches +infinity (since m >0, quadratic term dominates to +infinity)Therefore, the function goes from -infinity to +infinity. So, for three roots, it must have two critical points (a local minimum and a local maximum), such that between them, the function crosses the x-axis three times. Wait, but if it starts from -infinity and ends at +infinity, it must cross at least once. To have three crossings, it needs to have two critical points where it turns around: first increasing to a local maximum above zero, then decreasing to a local minimum below zero, then increasing again to +infinity. So, if the local maximum is above zero and the local minimum is below zero, then three crossings. So, in this case, possible.Case 3: m >1As x→0+, h(x) approaches +infinity (since (1 - m) <0, so ln x term dominates to +infinity)As x→∞, h(x) approaches +infinity (since m >0, quadratic term dominates to +infinity)So, the function starts at +infinity and ends at +infinity. Therefore, to have three roots, it must dip below the x-axis twice, but comes back up. So, it needs two minima and one maximum, but both minima have to be below zero, and the maximum in between. However, starting and ending at +infinity, so two minima below zero and a maximum in between. Then, there can be two or three roots. Hmm, but this is more complicated. However, given the answer options, the choices are A: m <0, n>0; B:0 <m<1, n<0; C: m>0,n>0; D:0 <m<1,n>0.So, perhaps among these, the correct answer is one of these. Let's check the answer options.First, note that n is a vertical shift parameter in h(x). Because n is the constant term. So, if we adjust n, we can shift the graph of h(x) up or down. The number of roots depends on the interaction between the shape of h(x) and the value of n.But in the problem statement, the functions f(x) and g(x) have three intersection points. So, the equation h(x) =0 must have three solutions. So, depending on m, the shape of h(x) changes, and n shifts it vertically. So, for a given m, there may be a range of n where h(x) crosses the x-axis three times.So, the key is to find for which m the function h(x) (for that m) can have three real roots when shifted appropriately by n. Then, the range of n would correspond to shifting h(x) such that it crosses three times.But the answer options give specific ranges for both m and n. So, we need to find the relationship between m and n such that three roots exist.Alternatively, maybe we can consider the equation h(x) =0, which is (1 - m)ln x + (m/2)x² + (m² - m -1)x + n =0If we solve for n, we get n = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]So, for each x >0, n is determined by this equation. The set of all possible n for which there exists an x such that n = - [ ... ] is the range of the function k(x) = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]Therefore, for three distinct x's, n must be such that it is the image of three different x's under k(x). So, the function k(x) must attain a particular value three times. Therefore, k(x) must have a local maximum and a local minimum, and n must lie between these two extrema. Therefore, for three solutions, n must be between the local maximum and local minimum values of k(x). Wait, actually, if k(x) has a local maximum and a local minimum, then the graph of k(x) would look like an "N" shape or an inverted "N" shape, and if n is between the local maximum and local minimum, then the horizontal line y = n would intersect the graph three times.But in our case, k(x) = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]But let's see the derivative of k(x) to find its extrema.First, compute k'(x):k'(x) = - [ (1 - m)/x + m x + (m² - m -1) ]Set k'(x) =0:- [ (1 - m)/x + m x + (m² - m -1) ] =0=> (1 - m)/x + m x + (m² - m -1) =0Multiply both sides by x (since x >0):(1 - m) + m x² + (m² - m -1)x =0So, m x² + (m² - m -1)x + (1 - m) =0This is a quadratic equation in x:m x² + (m² - m -1)x + (1 - m) =0Let's denote coefficients:A = mB = m² - m -1C = 1 - mThe discriminant D = B² -4ACCompute D:D = (m² - m -1)^2 -4 * m * (1 - m)Expand (m² - m -1)^2:= (m²)^2 + (-m)^2 + (-1)^2 + 2*(m²*(-m) + m²*(-1) + (-m)*(-1))Wait, actually, (a - b - c)^2 = a² + b² + c² - 2ab - 2ac + 2bc. Wait, no. Wait, (m² - m -1)^2 = (m²)^2 + (-m)^2 + (-1)^2 + 2*(m²*(-m) + m²*(-1) + (-m)*(-1))= m^4 + m² +1 + 2*(-m^3 -m² + m)= m^4 + m² +1 -2m^3 -2m² + 2m= m^4 -2m^3 -m² +2m +1Then, subtract 4AC =4*m*(1 - m) =4m -4m²Thus, D = m^4 -2m^3 -m² +2m +1 -4m +4m²Simplify:m^4 -2m^3 +3m² -2m +1Hmm, let's check if this factors:Let me see if m^4 -2m^3 +3m² -2m +1 factors.Let me try to factor it as (m² + a m + b)(m² + c m + d)Multiply out:= m^4 + (a + c)m³ + (b + d + a c)m² + (a d + b c)m + b dSet equal to m^4 -2m³ +3m² -2m +1So,a + c = -2b + d + a c =3a d + b c = -2b d =1Looking at b d =1, possible integer solutions are b=1, d=1 or b=-1, d=-1.Try b=1, d=1:Then,a + c = -21 +1 + a c =3 => a c =1a d + b c = a*1 +1*c = a + c = -2, which matches the third equation.So, from a + c = -2 and a c =1.Thus, solving:a and c are roots of t² +2t +1=0 => t = (-2 ± sqrt(4 -4))/2 = (-2)/2 = -1. So, a = c = -1.Therefore, the quartic factors as (m² -m +1)(m² -m +1) = (m² -m +1)^2Check:(m² -m +1)^2 = m^4 -2m³ +3m² -2m +1. Yes, that's correct.So, discriminant D = (m² -m +1)^2Therefore, the discriminant is a perfect square, so the quadratic equation in x has roots:x = [ -B ± sqrt(D) ]/(2A) = [ - (m² -m -1) ± (m² -m +1) ]/(2m)Compute both roots:First root with '+':Numerator: - (m² -m -1) + (m² -m +1) = -m² +m +1 +m² -m +1 = 2So, x = 2/(2m) = 1/mSecond root with '-':Numerator: - (m² -m -1) - (m² -m +1) = -m² +m +1 -m² +m -1 = -2m² +2mThus, x = (-2m² +2m)/(2m) = (-2m(m -1))/(2m) = -(m -1) = 1 - mTherefore, the critical points of k(x) are at x = 1/m and x = 1 - m.But x must be positive, so 1/m >0 and 1 - m >0.Thus, 1/m >0 implies m >01 - m >0 implies m <1Therefore, critical points x =1/m and x=1 - m are positive only when 0 <m <1.If m >1, then x=1 - m would be negative, which is invalid since x >0.If m <0, then x=1/m is negative, which is invalid.Therefore, for m <0 or m >1, there is only one valid critical point?Wait, wait. Wait, when m <0, in the quadratic equation for critical points:m x² + (m² - m -1)x + (1 - m) =0We found roots x=1/m and x=1 - m. But if m <0, then x=1/m is negative, and x=1 - m is positive (since m <0, 1 - m >1). So, only x=1 - m is positive. Similarly, if m >1, x=1 - m is negative, so only x=1/m is positive (since m >1, 1/m is positive but less than1). Wait, m >1, so 1/m is positive but less than1.But in the case of m >1, the critical points are x=1/m (positive) and x=1 - m (negative). So, only x=1/m is valid. Similarly, when m <0, x=1/m is negative, x=1 - m is positive.Therefore, depending on m:- If 0 <m <1: two critical points at x=1/m and x=1 - m, both positive.- If m >1: only one critical point at x=1/m (since x=1 - m is negative)- If m <0: only one critical point at x=1 - m (since x=1/m is negative)Therefore, in the intervals 0 <m <1, the function k(x) has two critical points, so possibly a local maximum and a local minimum, leading to the possibility of three intersection points if n is between these two extrema. For m outside this interval, only one critical point, so the function is monotonic except for one turn, hence possibly only one intersection point or two.Therefore, the key case is 0 <m <1, where k(x) has two critical points. Therefore, in this case, the function k(x) can have a local maximum and a local minimum. Then, the values of n must lie between these two extrema to have three intersection points.Therefore, the answer is likely related to 0 <m <1. So, options B and D have 0 <m <1. Now, we need to find the range of n.Given that n is the vertical shift, and k(x) = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]So, the extrema of k(x) correspond to the critical points x=1/m and x=1 - m.So, to find the local maximum and minimum values of k(x), we need to evaluate k(x) at x=1/m and x=1 - m.Let me compute k(1/m) and k(1 - m).First, compute k(1/m):k(1/m) = - [ (1 - m)ln(1/m) + (m/2)(1/m)^2 + (m² - m -1)(1/m) ]Simplify:ln(1/m) = -ln m(1/m)^2 = 1/m²(m² - m -1)(1/m) = (m²/m - m/m -1/m) = m -1 -1/mThus:k(1/m) = - [ (1 - m)(-ln m) + (m/2)(1/m²) + (m -1 -1/m) ]Simplify term by term:First term: - [ - (1 - m)ln m ] = (1 - m)ln mSecond term: (m/2)(1/m²) = (1)/(2m)Third term: (m -1 -1/m)Therefore:k(1/m) = (1 - m)ln m + 1/(2m) + m -1 -1/mCombine like terms:= (1 - m)ln m + m -1 + (1/(2m) -1/m)= (1 - m)ln m + m -1 -1/(2m)Similarly, compute k(1 - m):x=1 - m, so plug into k(x):k(1 - m) = - [ (1 - m)ln(1 - m) + (m/2)(1 - m)^2 + (m² - m -1)(1 - m) ]Simplify term by term:First term: - (1 - m)ln(1 - m)Second term: (m/2)(1 - 2m + m²)Third term: (m² - m -1)(1 - m) = m²(1 - m) - m(1 - m) -1*(1 - m) = m² - m³ -m + m² -1 + m = 2m² - m³ -1Wait, let me compute step by step:(m² - m -1)(1 - m) = m²*(1 - m) - m*(1 - m) -1*(1 - m)= m² - m³ - m + m² -1 + m= (m² + m²) + (-m³) + (-m + m) + (-1)= 2m² - m³ -1Therefore, third term is 2m² - m³ -1Thus, putting together:k(1 - m) = - [ (1 - m)ln(1 - m) + (m/2)(1 - 2m + m²) + 2m² - m³ -1 ]Simplify:First term: - (1 - m)ln(1 - m)Second term: - (m/2)(1 - 2m + m²) = - (m/2 - m² + (m³)/2 )Third term: - (2m² - m³ -1 ) = -2m² + m³ +1Therefore, combining all terms:k(1 - m) = - (1 - m)ln(1 - m) - m/2 + m² - (m³)/2 -2m² + m³ +1Simplify:Combine like terms:- m/2m² -2m² = -m²- (m³)/2 + m³ = (m³)/2+1Therefore:k(1 - m) = - (1 - m)ln(1 - m) - m/2 - m² + (m³)/2 +1= - (1 - m)ln(1 - m) + (m³)/2 - m² - m/2 +1This seems complicated, but maybe we can compare k(1/m) and k(1 - m).But perhaps instead of computing exact expressions, we can analyze the behavior.Given that for 0 <m <1, the function k(x) has a local maximum at x=1/m and a local minimum at x=1 - m (or vice versa). Wait, actually, to determine which critical point is a maximum or a minimum, we can check the second derivative or analyze the sign changes.Alternatively, since we have two critical points, we can note that between the two critical points, the function changes from increasing to decreasing or vice versa.But given the original function's behavior: when 0 <m <1, as x→0+, k(x) tends to infinity? Wait, wait, k(x) is defined as - [ ... ].Wait, k(x) = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]So, as x→0+, (1 - m)ln x tends to -infinity (since 1 - m >0), so inside the brackets, (1 - m)ln x approaches -infinity, and the other terms are negligible. Therefore, k(x) = - [ -infinity + ... ] = +infinity.As x→∞, (m/2)x² dominates (since m >0), so inside the brackets, it tends to +infinity, so k(x) = - [ +infinity ] = -infinity.Therefore, the function k(x) starts at +infinity when x approaches 0, goes to -infinity as x approaches infinity. With two critical points in between (0 <m <1). Therefore, the first critical point at x=1/m (since 1/m >1 because 0 <m <1) and the second at x=1 - m (since 0 <m <1, so 1 - m is between 0 and1). So, x=1/m is larger than1, and x=1 -m is less than1.Therefore, moving from x=0 to x=1 -m (which is <1), the function decreases from +infinity to k(1 -m). Then, from x=1 -m to x=1/m, the function increases to k(1/m), and then from x=1/m to infinity, it decreases to -infinity.Wait, but how do we know whether the critical points are maxima or minima?Let me compute the second derivative.First, k(x) = - [ (1 - m)ln x + (m/2)x² + (m² - m -1)x ]First derivative k’(x) = - [ (1 - m)/x + m x + (m² - m -1) ]Second derivative k''(x) = - [ - (1 - m)/x² + m ]Therefore, k''(x) = (1 - m)/x² - mAt critical points x=1/m and x=1 -m:Compute k''(1/m):= (1 - m)/( (1/m)^2 ) - m = (1 - m) m² - m = m² - m³ -mSimilarly, at x=1 -m:k''(1 -m) = (1 - m)/( (1 -m)^2 ) - m = 1/(1 - m) - m= [1 - m(1 - m)] / (1 - m)= [1 - m + m²] / (1 - m)But for 0 <m <1, denominator 1 - m >0. The numerator 1 - m + m². Let's check if it's positive:1 - m + m² = m² -m +1. The discriminant is (-1)^2 -4*1*1=1 -4= -3 <0, so it's always positive. Therefore, k''(1 -m) = positive / positive = positive. Therefore, at x=1 -m, the second derivative is positive, so it's a local minimum.At x=1/m, the second derivative is m² -m³ -m. Let's factor:m² -m³ -m = -m³ +m² -m = -m(m² -m +1). Again, m² -m +1 has discriminant 1 -4= -3 <0, so m² -m +1 >0. Therefore, since 0 <m <1, -m(m² -m +1) is negative. Therefore, k''(1/m) <0, so x=1/m is a local maximum.Therefore, for 0 <m <1, k(x) has a local maximum at x=1/m and a local minimum at x=1 -m.Therefore, the graph of k(x) starts at +infinity, decreases to a local minimum at x=1 -m, then increases to a local maximum at x=1/m, then decreases to -infinity.Therefore, the local minimum at x=1 -m is lower than the local maximum at x=1/m. Therefore, for the equation k(x) =0 (i.e., n = k(x)) to have three real roots, the horizontal line n must intersect the graph of k(x) three times. This occurs when n is between the local minimum value and the local maximum value. Therefore, for three solutions, n must satisfy k(1 -m) <n <k(1/m)Therefore, the range of n is k(1 -m) <n <k(1/m). So, we need to find the expressions for k(1/m) and k(1 -m) and determine their signs.But this might be complicated. Alternatively, maybe we can reason based on the answer choices. The options for n are n >0, n <0, etc.From the options given, B:0 <m <1, n <0; D:0 <m <1, n >0.So, if for three intersection points, n must lie between k(1 -m) and k(1/m), we need to determine whether these values are positive or negative.Given that k(x) = - [ ... ], let's see. Let me think about the behavior.At x=1/m (local maximum), k(1/m) is a local maximum. Similarly, at x=1 -m (local minimum), k(1 -m) is a local minimum.Given that as x approaches 0+, k(x) approaches +infinity, then decreases to the local minimum at x=1 -m, then increases to the local maximum at x=1/m, then decreases to -infinity.Therefore, the local minimum at x=1 -m is the lowest point on the graph, and the local maximum at x=1/m is the highest point.Therefore, the range of n for which there are three solutions is the interval between the local minimum and the local maximum.But we need to determine the signs of these extrema.Compute k(1/m) and k(1 -m) for 0 <m <1.First, compute k(1/m). Earlier, we had:k(1/m) = (1 - m)ln m + m -1 -1/(2m)Similarly, let's check the sign of this expression.Since 0 <m <1, ln m <0, so (1 - m)ln m is negative (because 1 -m >0, ln m <0). Therefore, the first term is negative.The next terms: m -1 -1/(2m). Since 0 <m <1, m -1 is negative, and -1/(2m) is negative. Therefore, the sum m -1 -1/(2m) is negative.Therefore, k(1/m) is negative + negative = negative.Wait, but let's compute for a specific m, say m=1/2.m=1/2:k(1/m)=k(2)= (1 -1/2)ln(1/2) +1/2 -1 -1/(2*(1/2))= (1/2)(-ln2) + (-1/2) -1= (-ln2)/2 -3/2 ≈ (-0.3466) -1.5 ≈ -1.8466 <0Similarly, m=1/3:k(3)= (2/3)ln(1/3) +1/3 -1 -3/2= (2/3)(-ln3) -2/3 -3/2≈ -2/3*1.0986 -0.6667 -1.5 ≈ -0.7324 -0.6667 -1.5 ≈ -2.899 <0So, k(1/m) is negative for 0 <m <1.Now, compute k(1 -m). Let's take m=1/2:k(1 -1/2)=k(1/2)= - [ (1 -1/2)ln(1/2) + (1/2)/2*(1/2)^2 + ( (1/2)^2 -1/2 -1 )(1/2) ]Wait, this might get too messy. Alternatively, use the earlier expression:k(1 -m) = - (1 - m)ln(1 - m) + (m³)/2 - m² - m/2 +1But let's take m=1/2:k(1 -1/2)=k(1/2)= - [ (1 -1/2)ln(1/2) + (1/2)/2*(1/2)^2 + ((1/2)^2 -1/2 -1)(1/2) ]Compute term by term:First term: - (1/2)ln(1/2) = - (1/2)(-ln2) = (ln2)/2 ≈0.3466Second term: (1/2)/2*(1/2)^2 = (1/4)*(1/4) =1/16 ≈0.0625Third term: ((1/4 -1/2 -1))(1/2) = (-5/4)(1/2) = -5/8 ≈-0.625Therefore, total:=0.3466 +0.0625 -0.625 ≈-0.2169 <0Wait, but k(1 -m) is supposed to be the local minimum. If k(1 -m) is also negative, then the range of n would be between two negative numbers and the maximum is negative, so n would have to be between k(1 -m) and k(1/m), but both are negative. Therefore, n would need to be between two negative numbers, which would mean n is negative. But in the answer options, B says n <0 and D says n >0.Wait, but when m=1/2, k(1/m)=k(2)≈-1.8466 and k(1 -m)=k(1/2)≈-0.2169. Therefore, the range of n for three solutions is -1.8466 <n < -0.2169. So, n must be negative. But in this case, the answer would be B:0 <m <1, n <0. But let's check another value.Take m approaching 0, say m=0.1:Compute k(1/m)=k(10)= (0.9)ln0.1 +0.1 -1 -1/(2*0.1)=0.9*(-2.3026) + (-0.9) -5≈-2.0723 -0.9 -5≈-7.9723k(1 -m)=k(0.9)= -0.9ln0.9 + (0.1^3)/2 -0.1^2 -0.1/2 +1First term: -0.9ln0.9≈-0.9*(-0.1054)=0.0949Second term:0.001/2=0.0005Third term:-0.01Fourth term:-0.05Fifth term:+1Total≈0.0949 +0.0005 -0.01 -0.05 +1≈0.0949 +0.0005=0.0954; 0.0954 -0.01=0.0854; 0.0854 -0.05=0.0354; 0.0354 +1=1.0354>0So, for m=0.1, k(1 -m)=k(0.9)≈1.0354>0But earlier, for m=1/2, k(1 -m)≈-0.2169<0This suggests that for m approaching0, k(1 -m) is positive, but for m=1/2, it's negative. Therefore, there must be a value of m between 0 and1 where k(1 -m) transitions from positive to negative. Therefore, the local minimum k(1 -m) changes sign from positive to negative as m increases from0 to1.Therefore, the range of n for which there are three solutions is when n is between k(1 -m) and k(1/m). But for m near0, k(1 -m) is positive and k(1/m) is very negative, so the interval would be from negative to positive, but since k(1/m) <k(1 -m) when m is near0. Wait, no, wait, for m approaching0, k(1/m) tends to -infinity (since m approaches0, 1/m approaches infinity, but k(1/m)= (1 -m)ln m + m -1 -1/(2m). As m approaches0, ln m approaches -infinity, so (1 -m)ln m ~ ln m approaches -infinity, and -1/(2m) approaches -infinity. Therefore, k(1/m) approaches -infinity. Meanwhile, k(1 -m)=k(1 -m)= - (1 -m)ln(1 -m) + ... For m approaching0, 1 -m approaches1, ln(1 -m)≈-m -m²/2 -..., so - (1 -m)ln(1 -m)≈(1 -m)(m +m²/2)≈m. Then, the other terms: (m³)/2 -m² -m/2 +1≈0 -0 -0 +1=1. Therefore, k(1 -m)≈m +1≈1. So, as m approaches0, k(1 -m)≈1, and k(1/m) approaches -infinity. Therefore, for m approaching0, the range of n is (-infty, 1). But since n has to be between k(1 -m) and k(1/m), but since k(1/m) is less than k(1 -m) in this case (k(1/m) is very negative and k(1 -m) is ~1), the valid interval is k(1/m) <n <k(1 -m). But since k(1/m) is approaching -infty and k(1 -m) approaches1, so n can be between -infty and1. But since the original problem requires three distinct intersection points, n must lie between the local minimum and local maximum. Wait, but in this case, when m is approaching0, the local maximum is at x=1/m (which is very large), and the local minimum is at x=1 -m≈1.But when m approaches0, the local minimum at x≈1 is k(1 -m)≈1, and the local maximum at x=1/m is k(1/m)≈-infty. Therefore, the function k(x) decreases from +infty to ~1 at x≈1, then increases to a local maximum of ~-infty (wait, this doesn't make sense). There must be an error in my earlier reasoning.Wait, no. Wait, when m approaches0, the critical points are at x=1/m (very large) and x=1 -m≈1. The function k(x) tends to +infty as x approaches0, decreases to a local minimum at x=1 -m≈1 with k≈1, then increases to a local maximum at x=1/m (which is very large), but as x approaches infinity, k(x) tends to -infty. Therefore, after x=1/m, the function decreases to -infty. Therefore, the local maximum at x=1/m is actually a high point before decreasing to -infty.Wait, but if m approaches0, x=1/m is very large. Let's compute k(x) at x=1/m where m approaches0.k(1/m)= (1 -m)ln m + m -1 -1/(2m)As m approaches0+, ln m approaches -infty, so (1 -m)ln m ≈ ln m approaches -infty, and -1/(2m) approaches -infty. Therefore, k(1/m) approaches -infty -infty = -infty. So, even though x=1/m is a local maximum, its value is still -infty. Therefore, the function decreases from +infty at x=0 to a local minimum of ~1 at x≈1, then increases to a local maximum of -infty at x≈infty, which is impossible. Therefore, there must be an error in my analysis.Wait, this suggests that my previous conclusion that x=1/m is a local maximum is incorrect for m approaching0. Wait, but earlier, we found that for 0 <m <1, x=1/m is a local maximum (since k''(1/m) <0). But if m approaches0, x=1/m becomes very large, and at that point, k(x) is already approaching -infty as x approaches infinity, so the local maximum at x=1/m must be a finite value. But according to the expression for k(1/m), it tends to -infty as m approaches0.Therefore, perhaps my earlier analysis is correct, but there's a mistake in interpreting the results.Alternatively, maybe I need to visualize the function for m approaching0.For m approaching0+, the function k(x) = - [ (1 -m)ln x + (m/2)x² + (m² -m -1)x ].As m approaches0, this becomes k(x) ≈ - [ ln x + (-1)x ] = -ln x +x.So, k(x) ≈ x - ln x.This function's derivative is 1 -1/x. Setting to zero: x=1. So, the critical point at x=1. Then, k(1) =1 -ln1=1. Then, as x approaches0+, k(x) approaches +infty; as x approachesinfty, k(x) approaches infinity (since x dominates). Wait, but for m=0, the quadratic term disappears, and we have k(x)= -lnx +x -x +n ? Wait, no, let's see:Wait, original functions for m=0:f(x) = (1 -0)lnx = lnxg(x) = -0 - (0 -0 -1)x -n = x -nSo, the equation is lnx =x -n, which rearranges to n =x -lnx. The function n =x -lnx.Which has a minimum at x=1 (derivative 1 -1/x=0 =>x=1), n=1 -0=1. Therefore, for m approaching0, the function k(x)=n= x -lnx - (terms with m). Therefore, the minimum of n is1, but when m is slightly positive, perhaps the minimum n becomes less than1?Wait, this is getting too convoluted. Let's step back.The problem is giving us four options, and through analysis, we determined that the case 0 <m <1 is the only one where three intersection points are possible. Now, we need to determine whether n should be positive or negative.From the specific example with m=1/2, the range of n was between approximately -1.8466 and -0.2169, which are both negative. So, n must be negative.However, for m approaching0, the local minimum k(1 -m) approaches1, and the local maximum k(1/m) approaches -infty. Therefore, the valid n would have to be between -infty and1. But this contradicts the previous example. This suggests that my previous analysis is incomplete.Alternatively, maybe I need to re-express k(x).Given that k(x) = - [ (1 -m)lnx + (m/2)x² + (m² -m -1)x ]Let me factor out m/2:k(x) = - (m/2)x² - (m² -m -1)x - (1 -m)lnx -nWait, no, wait, originally, the equation is f(x)=g(x):(1 -m)lnx = - (m/2)x² - (m² -m -1)x -nTherefore, rearranged:n = - (m/2)x² - (m² -m -1)x - (1 -m)lnxSo, n is expressed in terms of x. For three distinct x's, this equation must hold for three different x's. Therefore, the curve of n as a function of x must have a local maximum and a local minimum, and the value of n must be between these two for three solutions.Therefore, when 0 <m <1, the function n(x) = - (m/2)x² - (m² -m -1)x - (1 -m)lnx has a local maximum and a local minimum. The values of n must lie between these two extrema.To determine the sign of these extrema, let's consider:If the local minimum is negative and the local maximum is also negative, then n must be negative. If the local minimum is negative and the local maximum is positive, then n can be either. But according to the answer choices, n is either positive or negative.From the m=1/2 example, both extrema were negative. For m approaching1, let's check:Take m approaching1 from below, say m=0.9:k(1/m)=k(10/9)≈compute:k(10/9)= (1 -0.9)ln(0.9) +0.9 -1 -1/(2*(10/9))=0.1*(-0.1054) -0.1 -9/20≈-0.01054 -0.1 -0.45≈-0.5605 <0k(1 -m)=k(0.1)= -0.9ln0.1 + (0.9^3)/2 -0.9^2 -0.9/2 +1= -0.9*(-2.3026) + 0.729/2 -0.81 -0.45 +1≈2.0723 +0.3645 -0.81 -0.45 +1≈2.0723 +0.3645=2.4368; 2.4368 -0.81=1.6268; 1.6268 -0.45=1.1768; 1.1768 +1=2.1768>0So, k(1 -m)=2.1768>0, and k(1/m)=≈-0.5605<0. Therefore, for m=0.9, the range of n is between ≈-0.5605 and≈2.1768. Therefore, n can be both negative and positive. But wait, but the three intersection points would require n to be between the local minimum and local maximum. For m=0.9, the local minimum is at x=1/m≈1.111, but no, wait:Wait, when m=0.9, local maximum is at x=1/m≈1.111, and local minimum is at x=1 -m=0.1. Then, the function k(x) starts at +infty, decreases to k(0.1)=≈2.1768 (local minimum), then increases to k(1.111)=≈-0.5605 (local maximum), then decreases to -infty. Therefore, the local maximum is lower than the local minimum. Wait, that can't be. If the function decreases to a local minimum, then increases to a local maximum, then decreases again, the local maximum would be higher than the local minimum. But in this case, the local maximum is lower than the local minimum. That doesn't make sense. There must be an error in computation.Wait, no, because k(x) is defined as n = - [ ... ].Wait, if at x=0.1 (local minimum), k(x)=≈2.1768, and at x=1.111 (local maximum), k(x)=≈-0.5605, this would mean that the function decreases from +infty to 2.1768, then increases to -0.5605, then decreases to -infty. But increasing from 2.1768 to -0.5605 is actually decreasing, because -0.5605 <2.1768. Therefore, there must be a mistake in the calculation.Wait, no. If the function has a local minimum at x=0.1 with k=2.1768 and a local maximum at x=1.111 with k≈-0.5605, then the function is decreasing from +infty to x=0.1 (k=2.1768), then decreasing further to x=1.111 (k=-0.5605), then decreasing to -infty. But this contradicts the earlier analysis that between x=0.1 and x=1.111, the function is increasing.Therefore, there must be a mistake in the computation of k(1/m) and k(1 -m). Let's redo the calculation for m=0.9.First, compute k(1/m)=k(10/9):k(10/9) = - [ (1 -0.9)ln(10/9) + (0.9/2)(10/9)^2 + (0.9² -0.9 -1)(10/9) ]Compute each term:First term: (0.1)ln(10/9) ≈0.1*0.1054≈0.01054Second term: (0.45)(100/81)≈0.45*1.2345≈0.5556Third term: (0.81 -0.9 -1)(10/9)= (-1.09)(10/9)≈-1.2111Thus, inside the brackets:≈0.01054 +0.5556 -1.2111≈-0.645Thus, k(10/9)= -(-0.645)=0.645>0Wait, this contradicts the previous result. Therefore, my earlier calculation was wrong.Let me recompute k(1/m) for m=0.9:k(1/m) = - [ (1 -m)ln(1/m) + (m/2)(1/m)^2 + (m² -m -1)(1/m) ]= - [ 0.1*ln(10/9) + (0.9/2)(1/ (0.9)^2 ) + (0.81 -0.9 -1)(1/0.9) ]Compute term by term:First term:0.1*ln(10/9)≈0.1*0.1054≈0.01054Second term:0.45*(1/0.81)≈0.45*1.2345≈0.5556Third term:(0.81 -0.9 -1)/0.9= (-1.09)/0.9≈-1.2111Total inside brackets:0.01054 +0.5556 -1.2111≈-0.645Therefore, k(1/m)= -(-0.645)=0.645>0Similarly, compute k(1 -m)=k(0.1):k(0.1)= - [ (1 -0.9)ln0.1 + (0.9/2)(0.1)^2 + (0.81 -0.9 -1)(0.1) ]= - [0.1*(-2.3026) +0.045*0.01 + (-1.09)*0.1 ]= - [ -0.23026 +0.00045 -0.109 ]= - [ -0.33881 ] =0.33881>0Wait, so both k(1/m) and k(1 -m) are positive for m=0.9? That can't be. Because if the function decreases from +infty to a local minimum, then increases to a local maximum, then decreases to -infty, then the local minimum must be lower than the local maximum.But according to this, k(1/m)=0.645 and k(0.1)=0.338, so the local minimum is at x=0.1 with k=0.338, and the local maximum at x=10/9 with k=0.645. Therefore, the function decreases from +infty to 0.338 at x=0.1, then increases to 0.645 at x=10/9, then decreases to -infty. Therefore, for three intersections, n must be between 0.338 and0.645. Therefore, n must be positive. But this contradicts the previous example with m=1/2 where n needed to be negative.Wait, there is inconsistency here. It seems that depending on m, the local extrema can be both positive or both negative.Therefore, perhaps there is a value of m in (0,1) where k(1 -m)=0. Let's find m such that k(1 -m)=0.Set k(1 -m)=0:- (1 -m)ln(1 -m) + (m³)/2 - m² -m/2 +1=0This equation likely does not have an analytical solution, but we can attempt to find approximate values.Let's try m=0.5:k(1 -0.5)=k(0.5):= -0.5*ln0.5 + (0.125)/2 -0.25 -0.25 +1= -0.5*(-0.6931) +0.0625 -0.25 -0.25 +1≈0.3466 +0.0625 -0.25 -0.25 +1≈0.3466+0.0625=0.4091; 0.4091-0.25=0.1591; 0.1591-0.25= -0.0909; -0.0909+1=0.9091>0Wait, but earlier calculation for m=0.5 gave k(1 -m)=k(0.5)≈-0.2169<0. Now, this is inconsistent. There must be an error in computation.Wait, let's recalculate k(1 -m) for m=0.5:k(1 -m)=k(0.5)= - [ (1 -0.5)ln0.5 + (0.5/2)(0.5)^2 + ((0.5)^2 -0.5 -1)(0.5) ]= - [0.5*(-0.6931) +0.25*0.25 + (0.25 -0.5 -1)*0.5 ]= - [ -0.3466 +0.0625 + (-1.25)*0.5 ]= - [ -0.3466 +0.0625 -0.625 ]= - [ -0.3466 +0.0625= -0.2841; -0.2841 -0.625= -0.9091 ]= - (-0.9091)=0.9091>0Wait, this contradicts my previous calculation where I thought k(1 -m)≈-0.2169 for m=0.5. So, where was the mistake?Ah, earlier I think I miscalculated the third term:((0.5)^2 -0.5 -1)*0.5= (0.25 -0.5 -1)*0.5= (-1.25)*0.5= -0.625Then, total inside the brackets:-0.3466 +0.0625 -0.625= -0.9091Then, k(0.5)= -(-0.9091)=0.9091>0Therefore, previous calculation was wrong. Therefore, for m=0.5, k(1 -m)=0.9091>0 and k(1/m)=k(2)= - [0.5*ln2 +0.5*(4) + (-1.25)*2 ]Wait, compute k(2):k(2)= - [ (1 -0.5)ln2 +0.5/2*(4) + (0.25 -0.5 -1)*2 ]= - [0.5*0.6931 +0.25*4 + (-1.25)*2 ]= - [0.3466 +1 + (-2.5) ]= - [0.3466 +1 -2.5 ]= - [ -1.1534 ]=1.1534>0Therefore, k(1/m)=k(2)=1.1534>0 and k(1 -m)=0.9091>0. Therefore, for m=0.5, both local extrema are positive, and the function decreases from +infty to 0.9091, then increases to1.1534, then decreases to -infty. Therefore, to have three intersection points, n must be between0.9091 and1.1534, which are both positive. Therefore, n>0.But earlier when I thought m=0.5 had negative n range, I must have made a mistake in calculation. Therefore, correcting that, for m=0.5, both extrema are positive, so n must be between them, hence positive. For m approaching1, say m=0.9, we saw that k(1/m)=0.645>0 and k(1 -m)=0.338>0. Wait, no, for m=0.9, earlier calculation showed k(1/m)=0.645>0 and k(1 -m)=0.338>0. Therefore, n must be between0.338 and0.645, so still positive.For m approaching0, say m=0.1, we saw k(1/m) is -infty and k(1 -m) approaches1. Therefore, the range of n is between -infty and1. But since the local maximum at x=1/m is -infty, and the local minimum at x=1 -m is1, the valid interval is from -infty to1, but since the function goes to -infty, the actual interval where three solutions exist is when n <1. However, since the local minimum is1, and the function approaches -infty, if n is less than1, the line n would intersect the graph twice: once after the local minimum and once before. Wait, this is confusing.Wait, for m approaching0, the function k(x) = - [ (1 -m)lnx + (m/2)x² + (m² -m -1)x ] ≈ - [ lnx + (-1)x ]So, k(x)≈x -lnx. The derivative is1 -1/x, critical point atx=1, k(1)=1 -0=1. Asx approaches0+, k(x) approaches+infty; asx approachesinfty, k(x)≈x -lnx approaches+infty. Therefore, for m=0, the graph of k(x) has a minimum atx=1, n=1. Therefore, the equation n=x -lnx has two solutions when n>1, and one solution whenn=1, and no solutions whenn<1. But this contradicts the previous statement. Wait, but actually, when m=0, the equation islnx= x -n, which rearranged isx -lnx= n. The functionx -lnx has a minimum atx=1 ofn=1. Therefore, for n>1, there are two solutions; forn=1, one solution; forn<1, no solutions.But when m is slightly positive, say m=0.1, the behavior changes. The function k(x)= - [0.9lnx +0.05x² + (-1.09)x]. This function has a local minimum atx=1 -m=0.9 and a local maximum atx=1/m=10. As computed earlier, k(0.9)≈0.338>0 and k(10)=≈-7.9723<0. Therefore, the function decreases from+infty to0.338 atx=0.9, then increases to-7.9723 atx=10, then decreases to-infty. Wait, increasing from0.338 to-7.9723 is actually decreasing. Therefore, this suggests that there's only a local minimum atx=0.9, and then decreases further. This contradicts the earlier analysis. Therefore, there must be a mistake in critical points.Wait, earlier, we determined that for0 <m <1, there are two critical points atx=1/m andx=1 -m. For m=0.1, x=1/m=10 andx=1 -m=0.9. But when m=0.1, the quadratic equation for critical points:m x² + (m² -m -1)x + (1 -m)=0=0.1x² + (0.01 -0.1 -1)x +0.9=0=0.1x² -1.09x +0.9=0Solving:x=(1.09±sqrt(1.09² -4*0.1*0.9))/(2*0.1)= (1.09±sqrt(1.1881 -0.36))/0.2= (1.09±sqrt(0.8281))/0.2= (1.09±0.91)/0.2Thus:x=(1.09+0.91)/0.2=2/0.2=10x=(1.09-0.91)/0.2=0.18/0.2=0.9Therefore, critical points atx=10 andx=0.9.The second derivative test atx=0.9:k''(0.9)= (1 -0.1)/(0.9)^2 -0.1=0.9/0.81 -0.1≈1.111 -0.1=1.011>0 → local minimum.Atx=10:k''(10)= (1 -0.1)/100 -0.1=0.9/100 -0.1=0.009 -0.1= -0.091<0 → local maximum.Therefore, for m=0.1, k(x) has a local minimum atx=0.9 withk≈0.338 and a local maximum atx=10 withk≈-7.9723. Therefore, the function decreases from+infty to0.338, then increases to-7.9723, then decreases to-infty. This means that the local maximum is below the local minimum. Therefore, the valid interval for n to have three solutions is between the local maximum and local minimum, i.e., -7.9723 <n <0.338. Therefore, n can be negative or positive, but less than0.338.But when n=0.3, which is less than0.338, the line n=0.3 would intersect the graph three times: once between0 and0.9, once between0.9 and10, and once after10. But asx approaches0+, k(x) approaches+infty, so there's definitely a solution near0. Between0.9 and10, since k(x) goes from0.338 down to-7.9723, then up to-7.9723 and down to-infty, wait, no, after the local minimum at0.9, the function increases to the local maximum at10 withk≈-7.9723 (which is lower than the local minimum), then decreases to-infty. Therefore, between x=0.9 andx=10, the function increases from0.338 to-7.9723, which is actually impossible. This suggests a mistake in the analysis.Wait, the function can't increase from0.338 to-7.9723, since increasing means going up in value. Therefore, there must be a misunderstanding. If the local minimum atx=0.9 is0.338, and the local maximum atx=10 is-7.9723, this would mean that the function decreases from+infty to0.338, then decreases further to-7.9723, then decreases to-infty. Therefore, there are no increasing intervals. Therefore, only one critical point?But earlier, we had two critical points. This inconsistency arises because the local maximum atx=10 is lower than the local minimum atx=0.9. Therefore, this is impossible. Therefore, there must be a mistake in the computation.Wait, perhaps I made a mistake in assigning the maximum and minimum. Earlier, we found that for0 <m <1, x=1/m is a local maximum andx=1 -m is a local minimum. But if atx=1/m, k(x) is less than atx=1 -m, then the local maximum is lower than the local minimum, which is impossible.Therefore, there must be a mistake in the second derivative test. Let me recompute the second derivative.k''(x) = (1 -m)/x² -mAt x=1/m:k''(1/m) = (1 -m)/(1/m²) -m = (1 -m)m² -m = m² -m³ -m= m² -m -m³For m=0.1:k''(10) =0.1² -0.1 -0.1³=0.01 -0.1 -0.001= -0.091<0 → Local maximum.Atx=0.9:k''(0.9)= (1 -0.1)/0.81 -0.1=0.9/0.81 -0.1≈1.111 -0.1=1.011>0 → Local minimum.Therefore, the local maximum atx=10 is indeed a maximum, but its value is less than the local minimum atx=0.9. This implies that the function is decreasing from+infty tox=0.9 (local minimum), then decreasing further tox=10 (local maximum), which is impossible because a local maximum cannot be lower than a local minimum.Therefore, this indicates that my earlier analysis is flawed. The correct interpretation is that betweenx=0.9 andx=10, the function must increase to the local maximum atx=10. However, since the local maximum atx=10 is lower than the local minimum atx=0.9, this is contradictory.Therefore, this suggests that for m=0.1, the function k(x) has a local minimum atx=0.9 and a local maximum atx=10, but since k(x) atx=10 is less than atx=0.9, this implies that the function decreases from the local minimum to the local maximum, which is impossible. Therefore, there must be a miscalculation ink(x) atx=10 for m=0.1.Let me recomputek(10) for m=0.1:k(10) = - [ (1 -0.1)ln10 + (0.1/2)(10)^2 + (0.1² -0.1 -1)(10) ]= - [0.9*2.3026 +0.05*100 + (0.01 -0.1 -1)*10 ]= - [2.0723 +5 + (-1.09)*10 ]= - [2.0723 +5 -10.9 ]= - [ (2.0723 +5) -10.9 ] = - [7.0723 -10.9 ] = - [ -3.8277 ] =3.8277>0Wait, this is different from earlier calculations. So, k(10)=3.8277>0, which is higher thank(0.9)=0.338>0. Therefore, the function decreases from+infty to0.338 atx=0.9, then increases to3.8277 atx=10, then decreases to-infty. Therefore, the local minimum is0.338, local maximum is3.8277. Therefore, the valid range of n for three solutions is0.338 <n <3.8277, which is positive.This makes sense. Therefore, for m=0.1, n must be positive. For m=0.5, n must be between0.9091 and1.1534, positive. For m=0.9, n must be between0.338 and0.645, positive. Therefore, for all0 <m <1, the range of n is between two positive numbers, so n must be positive. Therefore, the correct answer is D:0 <m <1, n >0.Earlier miscalculations were due to arithmetic errors. Correcting them shows that for0 <m <1, the range of n is positive. Therefore, the answer is D.Final Answerboxed{D}</think>To determine the range of real numbers (m) and (n) such that the graphs of (f(x) = (1 - m)ln x) and (g(x) = -frac{m}{2}x^2 - (m^2 - m - 1)x - n) have three distinct intersection points, we need to analyze the equation ((1 - m)ln x = -frac{m}{2}x^2 - (m^2 - m - 1)x - n).1. Rearranging the Equation: [ (1 - m)ln x + frac{m}{2}x^2 + (m^2 - m - 1)x + n = 0 ] Denote this function as (h(x)).2. Behavior Analysis: - As (x to 0^+), ((1 - m)ln x) dominates. If (0 < m < 1), (h(x) to -infty). If (m > 1), (h(x) to +infty). - As (x to infty), (frac{m}{2}x^2) dominates. If (m > 0), (h(x) to +infty). If (m < 0), (h(x) to -infty).3. Critical Points: - The critical points of (h(x)) are found by solving the quadratic equation derived from the first derivative: [ m x^2 + (m^2 - m - 1)x + (1 - m) = 0 ] - The solutions are (x = frac{1}{m}) and (x = 1 - m). These are positive if (0 < m < 1).4. Second Derivative Test: - The second derivative (h''(x)) determines the nature of the critical points. For (0 < m < 1), (x = frac{1}{m}) is a local maximum and (x = 1 - m) is a local minimum.5. Range of (n): - For three distinct intersection points, (n) must lie between the local maximum and local minimum values of (h(x)). - Calculations show that for (0 < m < 1), the local minimum and maximum values of (h(x)) are positive, hence (n) must be positive.Thus, the correct answer is (boxed{D}): (0 < m < 1) and (n > 0).,<think>Okay, so I need to find the range of real numbers m and n such that the graphs of f(x) = (1 - m) ln x and g(x) = - (m/2)x² - (m² - m - 1)x - n intersect at three distinct points. The answer choices are given as A, B, C, D. Let me try to work through this step by step.First, to find the intersection points of f(x) and g(x), I need to set them equal to each other and solve for x. So:(1 - m) ln x = - (m/2)x² - (m² - m - 1)x - nLet me rearrange this equation to bring all terms to one side:(1 - m) ln x + (m/2)x² + (m² - m - 1)x + n = 0Hmm, this equation needs to have three distinct real solutions for x. Since ln x is only defined for x > 0, we are only considering x in the positive real numbers. So the problem reduces to finding m and n such that this equation has three distinct positive roots.I need to analyze how the functions f(x) and g(x) behave and where they can intersect. Let me consider the left-hand side as a function h(x) = (1 - m) ln x + (m/2)x² + (m² - m - 1)x + n. Then the problem is equivalent to h(x) = 0 having three distinct positive roots.To have three distinct roots, the function h(x) must cross the x-axis three times. For that to happen, h(x) must have certain critical points (local maxima and minima) which allow it to cross the axis three times. So I should analyze the derivative of h(x) to find its critical points and then use the second derivative test or other methods to check the conditions for three roots.First, let's compute h'(x):h'(x) = d/dx [ (1 - m) ln x + (m/2)x² + (m² - m - 1)x + n ]Differentiating term by term:- The derivative of (1 - m) ln x is (1 - m)/x- The derivative of (m/2)x² is m x- The derivative of (m² - m - 1)x is (m² - m - 1)- The derivative of n is 0So, h'(x) = (1 - m)/x + m x + (m² - m - 1)To find critical points, set h'(x) = 0:(1 - m)/x + m x + (m² - m - 1) = 0Multiply through by x to eliminate the denominator (since x > 0, multiplying is allowed without changing inequality direction):(1 - m) + m x² + (m² - m - 1)x = 0So, m x² + (m² - m - 1)x + (1 - m) = 0This is a quadratic equation in x. Let's denote coefficients:A = mB = m² - m - 1C = 1 - mSo the quadratic equation is A x² + B x + C = 0.For this quadratic to have two distinct real roots (so that h(x) has two critical points, leading to possible three roots for h(x)=0), the discriminant must be positive.The discriminant D = B² - 4ACCompute D:D = (m² - m - 1)^2 - 4 * m * (1 - m)Let me expand (m² - m - 1)^2:= (m²)^2 + (-m)^2 + (-1)^2 + 2*(m²*(-m) + m²*(-1) + (-m)*(-1))Wait, actually, (a + b + c)^2 is a² + b² + c² + 2ab + 2ac + 2bc. Wait, but here it's (m² - m -1)^2, so treat as (a + b)^2 where a = m², b = -m -1. So:= (m²)^2 + 2*(m²)*(-m -1) + (-m -1)^2= m^4 - 2 m³ - 2 m² + (m² + 2m +1)= m^4 - 2 m³ - 2 m² + m² + 2m +1= m^4 - 2 m³ - m² + 2m +1Now subtract 4AC = 4*m*(1 - m):= 4m(1 - m) = 4m - 4m²Therefore, D = [m^4 - 2 m³ - m² + 2m +1] - [4m - 4m²]= m^4 - 2 m³ - m² + 2m +1 -4m +4m²Combine like terms:- m^4 term: m^4- m³ term: -2 m³- m² terms: (-1 +4)m² = 3 m²- m terms: (2m -4m) = -2m- constants: +1So D = m^4 - 2 m³ + 3 m² - 2 m +1Hmm, need to check if this discriminant D is positive. Let's factor this quartic polynomial if possible.Looking at D = m^4 - 2 m³ + 3 m² - 2 m +1. Let me see if this factors into quadratics.Suppose D = (m² + a m + b)(m² + c m + d). Let's try to find integers a, b, c, d.Multiply out:= m^4 + (a + c)m³ + (b + d + a c)m² + (a d + b c)m + b dSet equal to m^4 -2 m³ +3 m² -2 m +1. Therefore:1. a + c = -22. b + d + a c = 33. a d + b c = -24. b d = 1Since b d =1, possible integer solutions are b=1 and d=1, or b=-1 and d=-1.Try b=1, d=1:Then from equation 1: a + c = -2Equation 3: a*1 + 1*c = a + c = -2. But equation 3 is supposed to be -2. So that's consistent.Equation 2: 1 +1 + a c = 2 + a c =3 => a c =1So we have a + c = -2, a c =1. Solving quadratic equation: t² +2 t +1=0. Which factors as (t+1)^2=0, so a = c = -1.Therefore, D factors as (m² - m +1)(m² -m +1) = (m² -m +1)^2.Wait, let me check:(m² - m +1)^2 = m^4 -2 m³ +3 m² -2 m +1. Yes, that's exactly D.So D = (m² - m +1)^2. Since m² -m +1 is a quadratic with discriminant (-1)^2 -4*1*1=1 -4 = -3 <0, so it's always positive. Therefore, D is a square of a positive number, hence D ≥0, but since m² -m +1 is never zero for real m, D is always positive. Therefore, the quadratic equation for h'(x)=0 has two distinct real roots for any real m. Wait, but wait, A = m. If m=0, the quadratic becomes 0 x² + (0 -0 -1)x +1 -0= -x +1=0, which is linear. But when m=0, the original equation for h'(x) would be different. Wait, but in our discriminant calculation, we assumed m ≠0 because A = m. So if m=0, the original equation for h'(x)=0 would be (1 - 0)/x +0 x + (0 -0 -1) = 1/x -1 =0 → x=1. So only one critical point. Therefore, when m=0, h'(x)=0 has only one critical point, so the analysis is different.But in the discriminant calculation, when m≠0, the quadratic equation has two distinct real roots. So in general, except when m=0, h'(x) has two critical points. But for h(x)=0 to have three distinct roots, we need h(x) to cross the x-axis three times, which requires that h(x) has a local maximum above the x-axis and a local minimum below the x-axis (or vice versa), so that the graph crosses the axis three times.Therefore, the function h(x) must have two critical points (since h'(x)=0 has two roots when m≠0) and the function values at these critical points must have opposite signs (so that the graph crosses the x-axis between them). Wait, actually, to have three roots, the function must go from one direction, reach a local maximum above the x-axis, then a local minimum below the x-axis, and then go to infinity, crossing the x-axis three times. Alternatively, the other way around. So the key is that the function has a local maximum and a local minimum, and that the local maximum is positive and the local minimum is negative (assuming the function tends to +infinity as x approaches infinity). Wait, let's analyze the behavior of h(x) as x approaches 0+ and as x approaches infinity.First, as x approaches 0+:The term (1 - m) ln x will dominate. Since ln x approaches -infinity as x approaches 0+, the sign of this term depends on (1 - m). If (1 - m) >0, i.e., m <1, then (1 - m) ln x approaches -infinity. If (1 - m) <0, i.e., m >1, then (1 - m) ln x approaches +infinity. If m=1, then the ln x term disappears.Next, as x approaches infinity:The dominant term in h(x) is (m/2)x². The coefficient (m/2) determines the behavior. If m >0, then (m/2)x² approaches +infinity. If m <0, then (m/2)x² approaches -infinity. If m=0, then the dominant term is the linear term (m² - m -1)x. But m=0 here would make that coefficient (0 -0 -1) = -1, so linear term is -x, which approaches -infinity as x approaches infinity.So, putting together the behavior as x→0+ and x→∞:Case 1: m <0Then (1 - m) >1, so as x→0+, h(x) approaches -infinity. As x→∞, h(x) approaches -infinity because m/2 is negative. So the function goes from -infty to -infty, but if it has two critical points (since m≠0), maybe a local maximum and a local minimum. For three roots, we need the function to cross the x-axis three times, but if both ends go to -infinity, then to have three crossings, the function would need to rise from -infty, cross the x-axis, reach a local maximum above the x-axis, then a local minimum below the x-axis, and then go back to -infty. But in this case, the local maximum has to be above the x-axis and the local minimum below. So between 0 and the first critical point (local maximum), the function goes from -infty to a maximum, then decreases to a minimum, then increases again? Wait, no, wait. Wait, h(x) has two critical points when m≠0. Depending on the concavity.Wait, maybe I need to sketch the graph based on the derivative. Let me think. Wait, if m <0, then the leading term (m/2)x² is negative, so as x→∞, h(x) tends to -infty. As x→0+, h(x) tends to -infty because (1 - m) >0, so ln x tends to -infty. So h(x) starts at -infty, comes up, maybe has a local maximum, then a local minimum, then goes back to -infty. So if the local maximum is above the x-axis and the local minimum is below, then there can be two crossings. But to have three crossings, we would need the function to cross three times, but if both ends are going to -infty, it's difficult. Wait, perhaps in this case, if the local maximum is above the x-axis and the local minimum is also above, but then h(x) would only cross once. If the local maximum is above and the local minimum is below, then the function crosses three times: once before the maximum, once between max and min, and once after the min. Wait, but if the function tends to -infty on both ends, then after the minimum, which is below the x-axis, the function goes to -infty, so it can't cross again. Hmm, maybe only two crossings. Wait, let me think again.Suppose h(x) approaches -infty as x approaches 0, then rises to a local maximum above the x-axis, then decreases to a local minimum below the x-axis, and then continues to decrease to -infty. So between 0 and the local maximum, it crosses the x-axis once, between the maximum and minimum, crosses again, and then after the minimum, it's going to -infty, so maybe doesn't cross again. Wait, actually, in this case, there might be two crossings: one between 0 and the maximum, and another between the maximum and minimum. But if the minimum is below the x-axis, then after the minimum, it's going to -infty, so it might not cross again. So total two crossings. To have three crossings, we need the function to go from -infty, cross up, then cross down, then cross up again, but since it's going to -infty at both ends, that might not happen. Alternatively, maybe if the minimum is above the x-axis, then it could have one crossing, but that seems inconsistent.Wait, perhaps when m <0, the function can't have three crossings because both ends go to -infty, so maximum two crossings. Therefore, maybe m <0 is not the right answer. Let's check the answer options: Option A is m <0, n >0. Maybe not. Let's see.Case 2: 0 < m <1Here, (1 - m) is positive, so as x→0+, h(x) approaches -infty. As x→∞, since m >0, h(x) approaches +infty (because (m/2)x² dominates and is positive). So the function goes from -infty to +infty. If it has two critical points (local maximum and local minimum), then depending on their positions relative to the x-axis, we can have three crossings. For example, if the function starts at -infty, rises to a local maximum above the x-axis, then decreases to a local minimum below the x-axis, then increases again to +infty. In this case, there would be three crossings: one before the maximum, one between max and min, and one after the min. So this could give three roots. But this requires that the local maximum is above the x-axis and the local minimum is below. So the values of h(x) at these critical points must be positive and negative, respectively. The value of n would affect the vertical shift of h(x), so n would determine whether the local max is above and local min is below the x-axis.Case 3: m >1Here, (1 - m) is negative, so as x→0+, h(x) approaches +infty (since (1 - m) ln x approaches +infty because ln x approaches -infty and multiplied by (1 - m) negative). As x→∞, since m >0, h(x) approaches +infty. So the function goes from +infty to +infty, but may dip down in between. To have three crossings, it would need to come down from +infty, cross the x-axis, reach a local minimum below, then a local maximum above, and then go back to +infty. But this would require two critical points, and the minimum to be below the x-axis and maximum above. However, since the function approaches +infty on both ends, if it dips below, it might cross twice. Wait, similar to case 2, but mirrored. Hmm. So starting at +infty, decreasing to a local minimum below the x-axis, then increasing to a local maximum above, then decreasing again to +infty? Wait, no, if m >1, as x approaches infinity, h(x) tends to +infty. So if it has a local minimum below the x-axis and a local maximum above, then it might cross the x-axis three times: once before the minimum, once between min and max, and once after the max. But since the function tends to +infty at both ends, after the maximum, it would start increasing again towards +infty, so maybe only two crossings. Hmm, this is getting complicated.Alternatively, perhaps focusing on 0 < m <1 is more promising since in that case, the function goes from -infty to +infty, which allows for an odd number of crossings. If there are two critical points, a maximum and a minimum, then depending on their positions, there can be three crossings. So maybe 0 < m <1 is a candidate. The answer options B and D have 0 < m <1, with n <0 and n >0 respectively. So need to figure out the role of n.Looking back at h(x) = (1 - m) ln x + (m/2)x² + (m² - m -1)x + n. The term n is a constant term, so it vertically shifts the graph of h(x). For h(x)=0 to have three solutions, the constant term n must be such that after the vertical shift, the local maximum is above the x-axis and the local minimum is below. Therefore, n affects the vertical position of h(x). If we can express the conditions on n such that h(x) at its local maximum is positive and at its local minimum is negative, then that would give the required range for n.Alternatively, if we think of h(x) = H(x) + n, where H(x) = (1 - m) ln x + (m/2)x² + (m² - m -1)x. Then the equation H(x) + n =0 is equivalent to H(x) = -n. So the number of solutions is the number of intersections between H(x) and the horizontal line y = -n. For three intersections, H(x) must have a shape that allows a horizontal line to cross it three times. For that, H(x) must have a local maximum above -n and a local minimum below -n, so that the line y = -n cuts through the graph three times.Therefore, the condition is that the maximum value of H(x) is greater than -n and the minimum value of H(x) is less than -n. Which translates to -n lying between the minimum and maximum values of H(x). Therefore, n must lie between - (maximum of H(x)) and - (minimum of H(x)). But since we need both the maximum and minimum to straddle the line y = -n, it's actually that -n is between the local minimum and local maximum of H(x). Therefore, the local maximum of H(x) must be greater than -n and the local minimum must be less than -n. But since H(x) is defined as h(x) - n, actually, maybe I need to compute the critical values.Alternatively, since h(x) = H(x) + n, the critical points of h(x) are the same as those of H(x), since the derivative h'(x) = H'(x). Therefore, at the critical points x1 and x2 (the solutions to h'(x)=0), the values h(x1) and h(x2) must satisfy h(x1) >0 and h(x2) <0 (or vice versa), so that the function crosses the x-axis three times.Therefore, the conditions are h(x1) >0 and h(x2) <0 (or the other way around). Since h(x) = H(x) + n, this would mean H(x1) + n >0 and H(x2) +n <0. Therefore, n > -H(x1) and n < -H(x2). Therefore, n must lie between -H(x1) and -H(x2), assuming H(x1) > H(x2). Wait, but x1 and x2 are critical points, which one is the maximum and which is the minimum?Since h'(x) =0 at x1 and x2. Let's assume x1 < x2. Then, depending on the second derivative or the behavior of h'(x), we can determine which is a maximum and which is a minimum.Alternatively, since as x approaches 0+, h(x) approaches -infty (for 0 < m <1), and as x approaches infinity, h(x) approaches +infty. The first critical point (smaller x) is likely a local maximum, and the second critical point (larger x) is a local minimum. Because the function is coming up from -infty, reaches a maximum, then decreases to a minimum, then increases to +infty. Therefore, x1 is the local maximum, x2 is the local minimum.Therefore, h(x1) >0 and h(x2) <0. Therefore:H(x1) + n >0 → n > -H(x1)H(x2) + n <0 → n < -H(x2)Therefore, the range of n is -H(x1) < n < -H(x2). Therefore, n must be positive if -H(x1) is positive and -H(x2) is positive? Wait, need to compute H(x1) and H(x2).But H(x1) = h(x1) -n. Wait, this is getting confusing. Let me think again.At the critical points x1 and x2 (local max and min), h(x1) and h(x2) are the function values. For three crossings, h(x1) must be positive and h(x2) must be negative. Therefore:h(x1) = (1 - m) ln x1 + (m/2)x1² + (m² - m -1)x1 + n >0h(x2) = (1 - m) ln x2 + (m/2)x2² + (m² - m -1)x2 + n <0Therefore, n > - [ (1 - m) ln x1 + (m/2)x1² + (m² - m -1)x1 ]andn < - [ (1 - m) ln x2 + (m/2)x2² + (m² - m -1)x2 ]So n must be greater than some value and less than another. The problem is that x1 and x2 are functions of m, so this condition must hold for some m.But this seems too abstract. Maybe there's another approach. Let's consider specific answer choices. The options are:A: m <0, n >0B:0 <m <1, n <0C:m >0, n >0D:0 <m <1, n >0From previous analysis, when 0 <m <1, the function h(x) goes from -infty to +infty, which can have three crossings if the local maximum is above the x-axis and the local minimum is below. So in this case, n would need to be such that the vertical shift allows this. If n is positive, then the entire function h(x) is shifted upwards. Wait, but h(x) = (1 - m) ln x + ... + n. If n is positive, that would raise the graph, potentially making the local minimum still below the x-axis and the local maximum higher above. Wait, but maybe the key is that with 0 <m <1, the coefficient of the x² term is positive (since m/2 >0), so as x increases, h(x) tends to +infty. If n is positive, then the constant term is adding a positive value, which might help in keeping the local minimum below the axis. Alternatively, if n is negative, it might pull the entire graph down, making the local maximum also below the axis.Wait, let's test with m=1/2 (which is in 0 <m <1). Then:h(x) = (1 - 1/2) ln x + (1/4)x² + ((1/4 - 1/2 -1))x + nSimplify:= (1/2) ln x + (1/4)x² + (-5/4)x + nThen h'(x) = (1 - 1/2)/x + (1/2)x + ( (1/4 - 1/2 -1) )Wait, h'(x) was previously found as:h'(x) = (1 - m)/x + m x + (m² - m -1)So substituting m=1/2:= (1 - 1/2)/x + (1/2)x + ((1/4) - 1/2 -1)= (1/2)/x + (1/2)x + (-5/4)Set h'(x) =0:(1/2)/x + (1/2)x -5/4 =0Multiply by 4x:2 + 2x² -5x =0 → 2x² -5x +2=0Solve: x = [5 ± sqrt(25 -16)]/4 = [5 ±3]/4 → x=2 or x=1/2So critical points at x=1/2 and x=2.Now compute h(x) at these points.First, x=1/2:h(1/2) = (1/2) ln(1/2) + (1/4)(1/2)^2 + (-5/4)(1/2) + n= (1/2)(-ln2) + (1/4)(1/4) + (-5/4)(1/2) + n= - (ln2)/2 + 1/16 -5/8 + nConvert to sixteenths:= -8 ln2 /16 +1/16 -10/16 +n= (-8 ln2 +1 -10)/16 +n= (-8 ln2 -9)/16 +nSimilarly, x=2:h(2) = (1/2) ln2 + (1/4)(4) + (-5/4)(2) +n= (1/2) ln2 +1 + (-10/4) +n= (1/2) ln2 +1 -2.5 +n= (1/2) ln2 -1.5 +nTo have three roots, we need h(1/2) >0 and h(2) <0.Compute h(1/2):= (-8 ln2 -9)/16 +n >0 → n > (8 ln2 +9)/16 ≈ (8*0.693 +9)/16 ≈ (5.544 +9)/16 ≈14.544/16≈0.909h(2):= (0.5*0.693) -1.5 +n ≈0.3465 -1.5 +n ≈-1.1535 +n <0 → n <1.1535Therefore, for m=1/2, n must be between approximately 0.909 and1.1535. However, the answer choices don't specify ranges for n, just n >0 or n <0, etc. Wait, but in this specific case with m=1/2, n needs to be positive (since 0.909 >0). So for m in (0,1), n needs to be positive. But in the answer choices, option D is 0 <m <1, n >0, and option B is 0 <m <1, n <0. So in this case, with m=1/2, n must be positive, which would correspond to option D.But wait, maybe other values of m in (0,1) require n >0? Let's check another m. Let me take m approaching 0, say m=0.1.Compute h'(x)=0:h'(x) = (1 -0.1)/x +0.1x + (0.01 -0.1 -1)=0.9/x +0.1x -1.09=0Multiply by x:0.9 +0.1x² -1.09x=0 →0.1x² -1.09x +0.9=0Multiply by10: x² -10.9x +9=0Discriminant D= (10.9)^2 -36=118.81 -36=82.81>0, roots:x=(10.9 ±sqrt(82.81))/2=(10.9 ±9.1)/2→ (20)/2=10 and (1.8)/2=0.9So critical points x=0.9 and x=10.Compute h(0.9):h(0.9) =0.9 ln0.9 +0.05*(0.81) + (0.01 -0.1 -1)*0.9 +nCompute term by term:0.9 ln0.9 ≈0.9*(-0.10536)= -0.0948240.05*0.81=0.0405(0.01 -0.1 -1)= -1.09, multiplied by0.9= -0.981Total so far: -0.094824 +0.0405 -0.981 ≈-1.0353So h(0.9)= -1.0353 +n >0 → n >1.0353h(10):h(10)=0.9 ln10 +0.05*100 + (-1.09)*10 +n=0.9*2.3026 +5 + (-10.9) +n≈2.0723 +5 -10.9 +n≈-3.8277 +n <0→n <3.8277So for m=0.1, n must be greater than ~1.035 and less than ~3.8277. Again, positive n.So for m in (0,1), n needs to be positive. Therefore, answer D:0 <m <1, n >0.But let's check why option B is 0 <m <1, n <0. Maybe if n is negative, for some m in (0,1), but from our examples, n needs to be positive. Let's see. If n is negative, then in the previous example with m=1/2, h(1/2) = (-8 ln2 -9)/16 +n. If n is negative, then h(1/2) would be even smaller, so even more negative. Similarly, h(2) = (1/2 ln2 -1.5) +n. If n is negative, h(2) would be less than (1/2 ln2 -1.5). Since 1/2 ln2≈0.346, so 0.346 -1.5≈-1.154. Then adding a negative n would make it even more negative. Therefore, both critical points would have h(x) negative, which would result in only one crossing. Hence, n needs to be positive to raise the graph so that the local maximum is above the x-axis.Therefore, the correct answer is D:0 <m <1, n >0.Final Answerboxed{D}</think>To determine the range of real numbers ( m ) and ( n ) such that the graphs of ( f(x) = (1 - m) ln x ) and ( g(x) = -frac{m}{2}x^2 - (m^2 - m - 1)x - n ) have three distinct intersection points, we analyze the function ( h(x) = (1 - m) ln x + frac{m}{2}x^2 + (m^2 - m - 1)x + n ). 1. Critical Points Analysis: - The derivative ( h'(x) ) is calculated and set to zero to find critical points: [ h'(x) = frac{1 - m}{x} + mx + (m^2 - m - 1) ] - Solving ( h'(x) = 0 ) leads to a quadratic equation in ( x ). The discriminant of this quadratic equation is always positive for ( m neq 0 ), indicating two distinct critical points.2. Behavior of ( h(x) ): - For ( 0 < m < 1 ), ( h(x) ) approaches (-infty) as ( x to 0^+ ) and (+infty) as ( x to infty ). This suggests the function can have three intersection points if it crosses the x-axis three times. - For ( m < 0 ) or ( m > 1 ), the behavior of ( h(x) ) at the extremes (0 and (infty)) does not allow three intersection points.3. Role of ( n ): - The constant term ( n ) vertically shifts the graph. For three intersections, ( h(x) ) must have a local maximum above the x-axis and a local minimum below the x-axis. This requires ( n ) to be positive, shifting the graph upwards sufficiently.4. Verification: - Testing specific values of ( m ) in the range ( 0 < m < 1 ) (e.g., ( m = 1/2 ) and ( m = 0.1 )) confirms that ( n ) must be positive to ensure the local maximum is above the x-axis and the local minimum is below.Thus, the correct range of ( m ) and ( n ) is ( 0 < m < 1 ) and ( n > 0 ).[boxed{D}]]

🔑:To solve the problem, we first establish the function h(x) by subtracting g(x) from f(x), which gives us:[h(x) = (1-m)ln x + frac{m}{2}x^{2} + (m^{2}-m-1)x + n]The derivative of h(x), denoted as h'(x), is calculated to find the critical points of h(x):[h'(x) = frac{d}{dx}[(1-m)ln x + frac{m}{2}x^{2} + (m^{2}-m-1)x + n] = frac{m{x^2} + (m^2-m-1)x + 1-m}{x}]For h(x) to have three distinct zeros, h'(x) must have two distinct real zeros in the interval (0, +infty). This leads us to analyze the conditions under which this is possible based on the value of m.Case 1: When m 0, for h'(x) = 0 to have two zeros in (0, +infty), we need 1-m > 0, which simplifies to 0 0, we analyze the value of h(x) at one of the critical points, say x_1, where 0 0 for the function to have three distinct zeros, we find that:[n > - (1-m)ln{x_1} + frac{m}{2}x_1^2 + 1 - m]Given 0 0.Therefore, combining the conditions on m and n, we conclude that 0 0. Thus, the correct answer is:[boxed{D}]

🧠:[<think>Okay, so I have this problem here where I need to find all pairs of polynomials f and g, both of degree 3, such that f(f(x)) is equal to (g(x))^3, and f(g(x)) is equal to (f(x))^3. Also, it's given that f(0) = 1. Hmm, that sounds pretty abstract, but let me try to break it down step by step.First, both f and g are cubic polynomials, right? So degree 3. Let me denote them as:f(x) = a x³ + b x² + c x + dg(x) = e x³ + f x² + g x + hBut since they're both cubic, their leading coefficients can't be zero. So a ≠ 0 and e ≠ 0.Now, the problem states that f(f(x)) = (g(x))³ and f(g(x)) = (f(x))³. Also, f(0) = d = 1. So d is 1. That's helpful.First, let's think about the degrees of these composite polynomials. Let's compute the degrees of both sides of the equations to get some constraints.For the first equation: f(f(x)) = (g(x))³.The left-hand side (LHS) is f composed with f. Since f is degree 3, f(f(x)) would have degree 3^2 = 9.The right-hand side (RHS) is (g(x))³. Since g is degree 3, raising it to the 3rd power gives degree 9. So degrees match up here, which is good.Similarly, for the second equation: f(g(x)) = (f(x))³.LHS is f(g(x)), which is degree 3*3 = 9.RHS is (f(x))³, which is degree 3*3 = 9. So degrees match here as well. So no immediate contradictions there.Now, let's think about the leading coefficients. Maybe that can give us some information.Starting with the first equation: f(f(x)) = (g(x))³.The leading term of f(x) is a x³. So f(f(x)) would have a leading term of a*(a x³)^3 = a*(a³ x^9) = a^4 x^9.On the RHS, (g(x))³ has leading term (e x³)^3 = e³ x^9.Therefore, equating the leading coefficients: a^4 = e³.Similarly, for the second equation: f(g(x)) = (f(x))³.LHS leading term: a*(e x³)^3 = a*e³ x^9.RHS leading term: (a x³)^3 = a³ x^9.Therefore, a*e³ = a³. Assuming a ≠ 0 (since f is cubic), we can divide both sides by a, getting e³ = a².So now we have two equations from leading coefficients:1. a^4 = e³2. e³ = a²So substituting equation 2 into equation 1: a^4 = a². Then, a^4 - a² = 0 => a²(a² - 1) = 0. Therefore, a² = 0 or a² = 1. But a ≠ 0, so a² = 1. Thus, a = 1 or a = -1.Therefore, possible a values are 1 or -1. Let's consider each case.Case 1: a = 1.From equation 2: e³ = a² = 1. So e = 1 (if we are working over real numbers; but the problem doesn't specify, so maybe complex? But perhaps they want real polynomials. Let me assume real for now. If e is real, then e = 1 is the only real cube root of 1. If complex, e could be 1, ω, ω² where ω is a primitive cube root of 1. But the problem might expect real coefficients. Let's proceed assuming real coefficients unless something forces us otherwise. Maybe the answer is real.)So if a = 1, then e = 1.Case 2: a = -1.Then e³ = a² = (-1)^2 = 1. So again, e = 1 (assuming real coefficients). Wait, but if a = -1, then e is still 1. Hmm.So regardless of a being 1 or -1, e = 1. Wait, but let's check:From equation 2, e³ = a². If a = 1, then e³ = 1 => e = 1 (real). If a = -1, e³ = (-1)^2 = 1, so e = 1 again. So in both cases, e = 1. So the leading coefficient of g is 1.So now, for both cases, e = 1. Therefore, the leading term of g(x) is x³.So now, let's note that:g(x) is a cubic polynomial with leading coefficient 1. So g(x) = x³ + p x² + q x + r.Similarly, f(x) has leading coefficient a = 1 or -1, and constant term d = 1. So f(x) is either:Case 1: f(x) = x³ + b x² + c x + 1Case 2: f(x) = -x³ + b x² + c x + 1Now, perhaps we can proceed to compare coefficients in the equations f(f(x)) = g(x)^3 and f(g(x)) = f(x)^3.But this seems complicated. Maybe there's a functional equation approach here.Given that f(f(x)) = g(x)^3 and f(g(x)) = f(x)^3. Let's see if we can substitute one into the other.From the first equation, g(x)^3 = f(f(x)). Then, plugging into the second equation: f(g(x)) = f(x)^3.But f(g(x)) can be written as f(g(x)) = (f(x))^3. Also, from the first equation, g(x)^3 = f(f(x)), so maybe substituting x with f(x), we get g(f(x))^3 = f(f(f(x))).But not sure if that helps. Let me see.Alternatively, suppose we think of these as functional equations. If f and g are inverses in some way? But given that f(f(x)) = g(x)^3 and f(g(x)) = f(x)^3, maybe there is some iterative structure here.Alternatively, perhaps f is some kind of homomorphism or something with respect to composition or multiplication. But polynomials under composition are tricky.Alternatively, maybe f is a linear function, but f is cubic, so that's not possible. Wait, unless f is a scaled monomial. Let me check if monomial solutions exist.Suppose f(x) is a monomial: f(x) = a x³ + 0 x² + 0 x + d. But given that f(0) = d =1, so d=1. But if f is a monomial, then the constant term is 1, which conflicts unless the monomial is of the form a x³ +1, but then unless a=0, which it can't be. Wait, so maybe f is not a monomial. Hmm.Alternatively, perhaps f and g are related in a specific way, such that composing f with itself gives the cube of g, and composing f with g gives the cube of f.This seems quite restrictive. Let me consider possible simple functions. For example, suppose f(x) = x + 1, but that's degree 1, not 3. Not helpful.Alternatively, suppose f(x) is a cube root function, but cubed. Wait, if f(f(x)) = g(x)^3, then perhaps g(x) is some kind of f inverse? But since f is cubic, its inverse would not be a polynomial. So that might not help.Alternatively, maybe f(x) = x³, but then f(f(x)) = (x³)^3 = x^9, which would mean g(x)^3 = x^9, so g(x) = x³. Then check the second equation: f(g(x)) = f(x³) = (x³)^3 = x^9. On the other hand, (f(x))^3 = (x³)^3 = x^9. So that works. Also, f(0) = 0, but the problem states f(0) =1. So this doesn't satisfy the condition. So discard this.But if we adjust f(x) to have a constant term 1. Suppose f(x) = x³ + 1. Then f(f(x)) = (x³ +1)^3 +1. That's degree 9, but (g(x))^3 would need to equal that. However, (x³ +1)^3 +1 is not a perfect cube of a cubic polynomial. Let me check: Suppose g(x)^3 = (x³ +1)^3 +1. But expanding (x³ +1)^3 +1: (x^9 + 3x^6 + 3x³ +1) +1 = x^9 +3x^6 +3x³ +2. This is not a cube of a cubic polynomial, since the coefficients would have to match the expansion of (ax³ + bx² + cx + d)^3. The constant term here is 2, which would have to be d^3. So d = cube root of 2, but then other coefficients would need to match, which is unlikely. So this approach probably doesn't work.Alternatively, maybe f is a linear function composed with a monomial. For example, f(x) = (x + c)^3. But f needs to be cubic, which it is. Let's see.Suppose f(x) = (x + c)^3. Then f(f(x)) = ((x + c)^3 + c)^3. Then g(x)^3 = ((x + c)^3 + c)^3, so g(x) = (x + c)^3 + c. But g(x) must be a cubic polynomial, which it is. Then check the second equation: f(g(x)) = (g(x) + c)^3 = ((x + c)^3 + c + c)^3 = ((x + c)^3 + 2c)^3. On the other hand, (f(x))^3 = ((x + c)^3)^3 = (x + c)^9. These two would have to be equal. But ((x + c)^3 + 2c)^3 versus (x + c)^9. These are only equal if 2c =0, so c=0. Then f(x) = x³, but then f(0)=0 ≠1. So that doesn't work. So this approach also fails.Hmm, maybe f and g are related through a scaling factor. Suppose f(x) = k x³ +1 (since f(0)=1). Then compute f(f(x)) = k (k x³ +1)^3 +1. Let's see:f(f(x)) = k*(k³ x^9 + 3k² x^6 + 3k x³ +1) +1 = k^4 x^9 + 3k³ x^6 + 3k² x³ + k +1.This is supposed to be equal to (g(x))^3. Since (g(x))^3 is a cubic cubed, so a degree 9 polynomial. Let's suppose g(x) is of the form x³ + a x² + b x + c. Then (g(x))^3 would expand to x^9 + 3a x^8 + (3a² + 3b) x^7 + ... which has non-zero coefficients for x^9, x^8, x^7, etc. But f(f(x)) as computed above has non-zero coefficients only for x^9, x^6, x^3, and the constant term. Therefore, unless a, b, c are chosen such that the coefficients of x^8, x^7, x^5, etc., in (g(x))^3 are zero. That would require a = 0, then 3a² + 3b = 0 => b =0, and so on. Let's check:If g(x) = x³ (i.e., a=0, b=0, c=0), then (g(x))³ = x^9. Comparing to f(f(x)) as above, which is k^4 x^9 + 3k³ x^6 + 3k² x³ + k +1. For this to equal x^9, we need:k^4 =1,3k³ =0,3k²=0,k +1=0.From 3k³=0 and 3k²=0, we get k=0, but then k^4=0≠1. Contradiction. So this doesn't work. Therefore, even if we set g(x)=x³, f(f(x)) can't be x^9 unless f is x³, which again conflicts with f(0)=1.Hmm. Maybe the polynomials have only terms of degree multiples of 3? For example, f(x) = a x³ + c x + d. Similarly, g(x) = e x³ + g x + h. Then, composing them might preserve the structure of having only certain degrees. Let's see.Suppose f(x) = a x³ + c x +1 (since f(0)=1). Then f(f(x)) = a (a x³ + c x +1)^3 + c (a x³ + c x +1) +1. Let's expand that:First, (a x³ + c x +1)^3. That would be a³ x^9 + 3a² c x^7 + 3a² x^6 + 3a c² x^5 + 6a c x^4 + (3a + 3a c²) x³ + 3c² x² + 3c x +1. Wait, this seems messy. Then multiply by a:a^4 x^9 + 3a³ c x^7 + 3a³ x^6 + 3a² c² x^5 + 6a² c x^4 + (3a² + 3a² c²) x³ + 3a c² x² + 3a c x + a.Then add c(a x³ + c x +1):c a x³ + c² x + c.Then add 1 more.So combining all terms:a^4 x^9 + 3a³ c x^7 + 3a³ x^6 + 3a² c² x^5 + 6a² c x^4 + (3a² + 3a² c² + c a) x³ + 3a c² x² + (3a c + c²) x + (a + c +1).This is supposed to equal (g(x))³. But (g(x))³ would be (e x³ + g x + h)^3. Let's expand that:e³ x^9 + 3e² g x^7 + 3e² h x^6 + 3e g² x^5 + 6e g h x^4 + (3e h² + g³) x³ + 3g² h x² + 3g h² x + h³.Therefore, equating coefficients:For x^9: a^4 = e³x^7: 3a³ c = 3e² g ⇒ a³ c = e² gx^6: 3a³ = 3e² h ⇒ a³ = e² hx^5: 3a² c² = 3e g² ⇒ a² c² = e g²x^4: 6a² c = 6e g h ⇒ a² c = e g hx^3: 3a² + 3a² c² + a c = 3e h² + g³x^2: 3a c² = 3g² h ⇒ a c² = g² hx: 3a c + c² = 3g h²constant term: a + c +1 = h³So that's a system of equations. Let's see if we can solve this.First, from x^9: a^4 = e³From x^6: a³ = e² h ⇒ h = a³ / e²From x^7: a³ c = e² g ⇒ g = (a³ c)/e²From x^5: a² c² = e g². Substitute g from above:a² c² = e*( (a³ c)/e² )² = e*( a^6 c² / e^4 ) = a^6 c² / e^3Thus, a² c² = a^6 c² / e^3 ⇒ 1 = a^4 / e^3 ⇒ But from x^9, a^4 = e³ ⇒ e³ = e³. So this is redundant. So no new information here.From x^4: a² c = e g h. Substitute g and h:a² c = e * (a³ c / e²) * (a³ / e² )= e * (a³ c / e²) * (a³ / e² )= e * a^6 c / e^4= a^6 c / e^3Thus, a² c = a^6 c / e^3Assuming c ≠0 (we can check later if c=0 is possible), divide both sides by c:a² = a^6 / e^3 ⇒ e^3 = a^4, which is again the x^9 equation. So redundant.From x^2: a c² = g² h. Substitute g and h:a c² = (a³ c / e² )² * (a³ / e² )= (a^6 c² / e^4 ) * (a³ / e² )= a^9 c² / e^6Thus, a c² = a^9 c² / e^6 ⇒ 1 = a^8 / e^6 ⇒ But since a^4 = e³, substitute e³ = a^4 ⇒ e^6 = a^8. Therefore, 1 = a^8 / a^8 =1. So again, redundant.From x term: 3a c + c² = 3g h². Substitute g and h:3a c + c² = 3*(a³ c / e² )*(a³ / e² )²= 3*(a³ c / e² )*(a^6 / e^4 )= 3*(a^9 c / e^6 )But from e³ = a^4 ⇒ e^6 = (a^4)^2 = a^8. Thus:= 3*(a^9 c / a^8 ) = 3*a cThus, 3a c + c² = 3a c ⇒ c² =0 ⇒ c=0.So c must be 0.Now, if c=0, let's see what other equations say.From x term: 3a*0 +0² =0= 3g h² ⇒ 0=3g h². So either g=0 or h=0.But from previous equations:From x^7: g = (a³ c)/e² =0 since c=0.From x^6: h =a³ / e²From x^9: a^4 = e³From constant term: a +0 +1 = h³ ⇒ h³ = a +1But h =a³ / e². And from a^4 = e³ ⇒ e= a^(4/3). Since e is a real coefficient, if a is negative, a^(4/3) is not real. Wait, but a is either 1 or -1 from earlier? Wait, hold on.Earlier, we considered leading coefficients and found that a is either 1 or -1, and e=1 in both cases. Wait, but that was before assuming f and g might have only terms of degrees 3,1,0. Wait, but in this approach, we allowed f to have x³, x, and constant term. But if c=0, then f(x) =a x³ +1.But earlier, from leading coefficients, we had a is either 1 or -1. Wait, maybe that analysis was separate. Let me check.Earlier, leading term analysis for f(f(x)) =g(x)^3 gave a^4 = e³. Then for f(g(x))=f(x)^3, leading term gives e³ a =a³, so e³ =a². Thus, combining both, a^4 =a², so a²=1, so a=1 or -1, and e³=1, so e=1 (real). So regardless of a=1 or -1, e=1.But here, in this current approach where f(x)=a x³ +1 (since c=0), and g(x)=e x³ +h (since g(x) is cubic with leading coefficient e=1, and c=0 leading to g=0, so g(x)=x³ + h. Wait, need to check.Wait, in this case, f(x)=a x³ +1, with a=1 or -1. Then g(x)=x³ + h, since leading coefficient e=1.From the constant term equation: h³ =a +1.From x^6 coefficient: 3a³ =3e² h ⇒ since e=1, 3a³=3*1²*h ⇒ a³ = h.Thus, h= a³. But h³ =a +1. Therefore, (a³)^3 =a +1 ⇒ a^9 =a +1.But a is either 1 or -1.Check a=1: 1^9=1=1+1? 1≠2. Not valid.Check a=-1: (-1)^9= -1= -1 +1=0. -1≠0. Not valid.Therefore, contradiction. Hence, this approach where f has only x³ and constant term doesn't work. So c=0 leads to inconsistency, meaning our assumption that f can be written as a x³ +1 might be invalid, or perhaps we need a different approach.Hmm, maybe f and g have more terms. This is getting complicated. Let's try to think differently.Given the equations f(f(x)) = g(x)^3 and f(g(x)) = f(x)^3. Let's consider substituting the first into the second.From the first equation, g(x) = [f(f(x))]^{1/3}. But since g(x) is a polynomial, [f(f(x))]^{1/3} must also be a polynomial. Therefore, f(f(x)) must be a perfect cube of a polynomial, which is g(x)^3. Therefore, g(x) is the cube root of f(f(x)) in the polynomial ring. Similarly, f(g(x)) is equal to (f(x))^3.So if we define g(x) = [f(f(x))]^{1/3}, then substitute into the second equation: f(g(x)) = f([f(f(x))]^{1/3}) = [f(x)]^3.Hmm, this seems recursive. Maybe f is a kind of "cube root" function in composition? Not sure.Alternatively, suppose f(x) = x³ + ... but with f(0)=1. Wait, if f(0)=1, then the constant term is 1. If f is x³ + lower terms, then composing f(f(x)) would give (x³ + ...)^3 + ..., which could be messy.Alternatively, maybe f is an involution such that f(f(x)) = something, but since f is cubic, composing it with itself gives degree 9, which is different.Alternatively, maybe f and g are related through a substitution. For example, suppose that f(x) = ω x where ω³=1, but f is cubic, not linear. Not helpful.Wait, perhaps f(x) = x³ +1. Let me check this case.If f(x)=x³ +1, then f(f(x)) = (x³ +1)^3 +1 = x^9 + 3x^6 + 3x³ +1 +1 = x^9 + 3x^6 + 3x³ +2. Then g(x)^3 must equal this. But g(x) is cubic, so g(x)^3 is degree 9. However, x^9 +3x^6 +3x³ +2 is not a perfect cube. For example, the coefficients 1, 0, 0, 3, 0, 0, 3, 0, 0, 2 don't align with the expansion of a cubic cubed. So this is not a cube.Alternatively, maybe f is a scaled and shifted cube. Suppose f(x)= (x + k)^3 + c, such that f(0)=1. Then f(0)=k³ + c =1. Let's see what f(f(x)) would be.f(f(x)) = ( (x +k)^3 + c +k )^3 +cThis seems complicated, but perhaps choosing k and c such that this expression becomes a perfect cube. However, it's unclear how to choose them.Alternatively, maybe both f and g are permutations of each other or something. Wait, since f(g(x)) = (f(x))^3, and f(f(x)) = (g(x))^3, perhaps there's a symmetry here.Let me try to compose functions. Suppose I take the first equation f(f(x)) = g(x)^3. Then substitute x with g(x):f(f(g(x))) = g(g(x))^3.But from the second equation, f(g(x)) = (f(x))^3. Therefore, f(f(g(x))) = f((f(x))^3).So combining these:f((f(x))^3) = g(g(x))^3.But not sure if that helps. Alternatively, from the first equation, g(x) = [f(f(x))]^{1/3}. Then substitute into the second equation:f([f(f(x))]^{1/3}) = (f(x))^3.Let me denote y = f(x). Then the equation becomes:f([f(y)]^{1/3}) = y^3.So this is a kind of functional equation for f: f([f(y)]^{1/3}) = y³.This seems like f is invertible in some sense, but since f is a cubic polynomial, it's not injective over the reals, so its inverse isn't a function. But maybe in the complex plane or formally.Alternatively, suppose [f(y)]^{1/3} is another polynomial, say h(y), such that h(y)^3 = f(y). Then the equation becomes f(h(y)) = y³. So f composed with h is y³. So h would be a right inverse of f with respect to y³. But h(y)^3 = f(y), so f is a cube of h(y). Therefore, f(y) = [h(y)]³. Then substituting back, f(h(y)) = y³. But f(h(y)) = [h(h(y))]^3. Therefore, [h(h(y))]^3 = y³ ⇒ h(h(y)) = y. So h is an involution, h(h(y))=y, and f(y)= [h(y)]³.Therefore, if we can find such an involution h, then f(y)=[h(y)]³ and g(x) = [f(f(x))]^{1/3} = [ [h(f(x))]^3 ]^{1/3} = h(f(x)). Therefore, g(x) = h(f(x)).But since f(x)= [h(x)]³, then g(x) = h([h(x)]³).But also, from the second equation, f(g(x))= [f(x)]^3. Let's see:f(g(x)) = [h(g(x))]^3 = [h(h([h(x)]³ ))]^3 = [ [h(h([h(x)]³ )) ]^3. But since h is an involution, h(h(z))=z, so this simplifies to [ [h(x)]³ ]^3 = [h(x)]^9.But on the other hand, [f(x)]^3 = [ [h(x)]³ ]^3 = [h(x)]^9. Therefore, equality holds. So this seems consistent.Therefore, if we can find a polynomial involution h (i.e., h(h(x))=x) such that f(x)=[h(x)]³ and g(x)=h(f(x))=h([h(x)]³ ). Then, such f and g would satisfy the given conditions.Moreover, since f is a cubic polynomial, h(x) must be a linear polynomial because [h(x)]³ is cubic. Wait, because if h is degree n, then [h(x)]³ is degree 3n. But f is cubic, so 3n=3 ⇒ n=1. Therefore, h(x) must be a linear involution, i.e., h(h(x))=x, so h is an involutionary linear function.A linear involution is a function h(x) = m x + b such that h(h(x))=x.Let's find such h(x):h(h(x)) = m*(m x + b) + b = m² x + m b + b = x.Therefore, we need:m² =1,m b + b =0 ⇒ b(m +1)=0.So either m=1 and b=0, or m=-1 and b can be any value (since m+1=0).Therefore, possible linear involutions are:1. h(x) = x (when m=1, b=0),2. h(x) = -x + b, where b is any constant (since m=-1, and b(m +1)=b(0)=0 for any b).Wait, if m=-1, then b(m +1)=b*0=0, so no restriction on b. Therefore, h(x) = -x + b is an involution for any b.But in our case, h(x) must be such that f(x)=[h(x)]³ is a cubic polynomial with f(0)=1.Let's check both cases.Case 1: h(x)=x.Then f(x)=[h(x)]³ =x³. Then f(0)=0≠1. Doesn't satisfy the condition.Case 2: h(x)= -x + b.Then f(x)=(-x + b)^3. Let's compute f(0)= (-0 + b)^3 =b^3=1. Therefore, b=1 (assuming real coefficients). Therefore, h(x)= -x +1.Therefore, f(x)=(-x +1)^3. Let's expand this:(-x +1)^3 = -x³ + 3x² -3x +1.Check f(0)=1: yes, the constant term is 1. Good.Then g(x)=h(f(x))= -f(x) +1= -[ -x³ + 3x² -3x +1 ] +1= x³ -3x² +3x -1 +1= x³ -3x² +3x.Simplify: g(x)=x³ -3x² +3x= x(x² -3x +3). But let's check if this satisfies the required conditions.First, check f(f(x))=g(x)^3.Compute f(f(x)):f(f(x))=f(-x³ +3x² -3x +1)= (-(-x³ +3x² -3x +1) +1)^3= (x³ -3x² +3x -1 +1)^3= (x³ -3x² +3x)^3= [g(x)]^3. So that works.Then check f(g(x))= [f(x)]^3.Compute f(g(x))=f(x³ -3x² +3x)= [ - (x³ -3x² +3x) +1 ]^3= (-x³ +3x² -3x +1)^3= [f(x)]^3. Which matches. So this works.Therefore, one solution is:f(x)= -x³ +3x² -3x +1,g(x)=x³ -3x² +3x.But wait, f(x) is (-x +1)^3, which expands to -x³ +3x² -3x +1. Yes.But the problem states that f and g are both degree 3 polynomials. These are both cubic, so that's good.Is this the only solution? Let's see.We assumed that h(x) is a linear involution, leading to f(x)=(-x +1)^3 and g(x)=x³ -3x² +3x. But in the complex case, there might be other cube roots of unity for a, but earlier leading coefficient analysis gave a=1 or -1. However, with a=-1, we obtained a valid solution.Wait, but we considered h(x)=-x +1, which led to f(x)=(-x +1)^3. Let's check the leading coefficient.The leading term of f(x) is (-x)^3 = -x³, so leading coefficient a=-1. Then from earlier analysis, e=1. Which matches with g(x) having leading coefficient 1. So this is consistent.But could there be another solution with a=1?If a=1, then e=1. Then we need to find f and g such that f(f(x))=g(x)^3 and f(g(x))=f(x)^3, with f of degree 3 leading coefficient 1, and f(0)=1.But earlier, when trying the monomial case, a=1 didn't work. But perhaps another non-monomial solution exists.Wait, let's try the same approach. Suppose h(x) is a linear involution, but with a=1. Wait, but if h(x) is linear involution, then f(x)=[h(x)]^3. But if a=1, the leading coefficient of f(x) would be [leading coefficient of h(x)]^3. Since h(x) is linear, h(x)=m x +b. Then leading coefficient of f(x) is m³. If a=1, then m³=1. Therefore, m=1 (if real). Therefore, h(x)=x +b. But h must be an involution, so:h(h(x))=h(x +b)= (x +b) +b =x +2b. This must equal x, so 2b=0 ⇒b=0. Therefore, h(x)=x. But then f(x)=x³, which leads to f(0)=0≠1. So no solution in this case.Therefore, the only linear involution over reals that gives f(0)=1 is h(x)=-x +1, leading to f(x)=(-x +1)^3 and g(x)=x³ -3x² +3x.But wait, are there higher-degree involutions? For example, could h(x) be a cubic involution?An involution must satisfy h(h(x))=x. If h is a cubic polynomial, then h(h(x)) would be degree 9. Setting it equal to x would require all coefficients except linear to cancel out, which is very restrictive. Let's check if such a cubic involution exists.Suppose h(x) is a cubic polynomial such that h(h(x))=x. Let's write h(x)=p x³ +q x² +r x +s. Then h(h(x)) is a degree 9 polynomial, which must equal x. Therefore, all coefficients except for the x term must be zero, and the x term coefficient must be 1. This seems highly unlikely unless p=q=s=0 and r=1, but then h(x)=x, which is degree 1, not 3. Therefore, there are no non-trivial cubic involutions. So h must be linear.Hence, the only solution is the one found earlier with h(x)=-x +1. Therefore, the pair is:f(x) = -x³ + 3x² - 3x +1,g(x) =x³ -3x² +3x.But let me check if there are other solutions where f and g have cross terms. Suppose that f and g have cross terms (i.e., x² or x terms), but still satisfy the functional equations. It's possible, but given the leading coefficient analysis and the requirement that h must be linear, it's unlikely. Let's verify with the obtained solution.Compute f(x)= -x³ +3x² -3x +1. Then f(0)=1, correct.Compute g(x)=x³ -3x² +3x. Then g(0)=0.Check f(f(x)):f(f(x))=f(-x³ +3x² -3x +1) = -(-x³ +3x² -3x +1)^3 + 3(-x³ +3x² -3x +1)^2 -3(-x³ +3x² -3x +1) +1.Wait, instead of expanding this, note that earlier we used the relation f(f(x))=g(x)^3, and since g(x)=x³ -3x² +3x, g(x)^3=(x³ -3x² +3x)^3. If we compute f(f(x)) as per the functional equation, it should equal this.But given that we arrived at this solution through the involution method, and verified symbolically that it works, this must hold.Alternatively, note that g(x)=x³ -3x² +3x =x(x² -3x +3). Not sure if that helps, but if we compute g(x)^3:g(x)^3=(x³ -3x² +3x)^3. This would indeed be a degree 9 polynomial, which matches f(f(x)).Similarly, f(g(x))= [ -g(x) +1 ]^3= [ -x³ +3x² -3x +1 ]^3= (f(x))^3. Hence, it's correct.Therefore, the only solution is f(x)=(-x +1)^3 and g(x)=x³ -3x² +3x.But let me check for uniqueness. Suppose there is another solution. For example, if we take h(x)=-x +b, then f(x)=(-x +b)^3. Then f(0)=b³=1 ⇒b=1 (real). Therefore, this is the only solution over real coefficients. If we allow complex coefficients, then b could be other cube roots of 1, but the problem doesn't specify, and usually in such problems, real coefficients are expected unless stated otherwise.Therefore, the only pair is f(x)= -x³ +3x² -3x +1 and g(x)=x³ -3x² +3x.But wait, let me write them in a more standard form.f(x) = -x³ +3x² -3x +1. This can be written as -(x³ -3x² +3x -1) = -(x-1)^3. Wait, (x-1)^3= x³ -3x² +3x -1. So f(x)= - (x -1)^3. Therefore, f(x)= - (x -1)^3.Similarly, g(x)=x³ -3x² +3x= x³ -3x² +3x -1 +1= (x -1)^3 +1. Wait, no. (x -1)^3= x³ -3x² +3x -1, so g(x)= (x -1)^3 +1 -1 +0= just x³ -3x² +3x.Alternatively, factor g(x): x(x² -3x +3). Not particularly useful.But recognizing that f(x)= - (x -1)^3 might be a more compact way to write it.Therefore, the pair is:f(x) = - (x - 1)^3,g(x) = x³ - 3x² + 3x.These are both cubic polynomials, satisfy f(0)= - (-1)^3= -(-1)=1, which matches the given condition.Thus, this should be the unique solution.Final AnswerThe pairs of polynomials are (boxed{f(x) = -(x - 1)^3}) and (boxed{g(x) = x^3 - 3x^2 + 3x}).</think>Given polynomials ( f(x) ) and ( g(x) ) of degree 3 such that ( f(f(x)) = (g(x))^3 ), ( f(g(x)) = (f(x))^3 ), and ( f(0) = 1 ), we need to find all such pairs of polynomials ( f ) and ( g ).1. Leading Coefficients Analysis: - The leading term of ( f(f(x)) ) is ( a^4 x^9 ) and the leading term of ( (g(x))^3 ) is ( e^3 x^9 ). This gives ( a^4 = e^3 ). - The leading term of ( f(g(x)) ) is ( a e^3 x^9 ) and the leading term of ( (f(x))^3 ) is ( a^3 x^9 ). This gives ( a e^3 = a^3 ), leading to ( e^3 = a^2 ). - Combining ( a^4 = e^3 ) and ( e^3 = a^2 ), we get ( a^4 = a^2 ), which implies ( a^2 = 1 ). Thus, ( a = 1 ) or ( a = -1 ).2. Functional Equations and Involution: - Considering ( f(f(x)) = (g(x))^3 ) and ( f(g(x)) = (f(x))^3 ), we recognize that ( g(x) ) must be the cube root of ( f(f(x)) ). - We find that ( f ) must be related to an involution ( h ) such that ( f(x) = (h(x))^3 ) and ( h(h(x)) = x ).3. Linear Involution: - A linear involution ( h(x) = -x + b ) is considered. Given ( f(0) = 1 ), we find ( b = 1 ). - This leads to ( h(x) = -x + 1 ), and thus ( f(x) = (-x + 1)^3 ).4. Verification: - Expanding ( f(x) = - (x - 1)^3 ) gives ( f(x) = -x^3 + 3x^2 - 3x + 1 ), which satisfies ( f(0) = 1 ). - Substituting ( f(x) ) into the functional equations, we find ( g(x) = x^3 - 3x^2 + 3x ).5. Uniqueness: - The analysis shows that the only solution over real coefficients is derived from the linear involution ( h(x) = -x + 1 ).Thus, the pairs of polynomials are:[boxed{f(x) = -(x - 1)^3} quad text{and} quad boxed{g(x) = x^3 - 3x^2 + 3x}]]

🔑:Answer: f(x)=-x^{2}+3 x^{2}-3 x+1=(1-x)^{3}, g(x)=(x-1)^{3}+1. (In the 2nd variant: f(x)=-2-(x+2)^{3}, left.g(x)=(x+2)^{3}.right)Let f(x)=a x^{3}+b x^{2}+c x+1. By the condition, a f^{3}(x)+b f^{2}(x)+c f(x)+1=g^{3}(x) u a g^{3}(x)+b g^{2}(x)+c g(x)+1= f^{3}(x). Subtract the second equality from the first: (a+1)left(f^{3}(x)-g^{3}(x)right)+bleft(f^{2}(x)-g^{2}(x)right)+c(f(x)-g(x))=0. This expression can easily be factored into two factors, and since one of them f(x)-g(x) is a non-zero polynomial, the second factor must be zero: (a+1)left(f^{2}(x)+f(x) g(x)+g^{2}(x)right)+b(f(x)+g(x))+c=0.Note that f^{2}(x)+f(x) g(x)+g^{2}(x) is a polynomial of degree 6 (its coefficient at x^{6} is a^{2}+a a_{1}^{2}+a_{1}^{2} neq 0, where a_{1} is the leading coefficient of g(x)). Therefore, a+1 must be equal to 0. Thus, a=-1, and in this case b(f(x)+g(x))=-c. This is possible in two cases:1) b=c=0. Then -f^{3}(x)+1=g^{3}(x), i.e., f^{3}(x)+g^{3}(x)=1. But the polynomial on the left side is the product (f+g)left(f^{2}-f g+g^{2}right), and the second bracket is a polynomial of degree 6 (as discussed above). Therefore, this cannot happen.2) f(x)+g(x)=-c / b is a constant polynomial. Denote -c / b=T, g(x)=T-f(x). Substituting this into the first condition, we get -f^{3}(x)+b f^{2}(x)+c f(x)+1=(T-f(x))^{3}=-f^{3}(x)+3 T f^{2}(x)-3 T^{2} f(x)+T^{3}. From this, it follows that b=3 T, c=-3 T^{2}, 1=T^{3}, i.e., f(x)=-x^{2}+3 x^{2}-3 x+1=(1-x)^{3}, g(x)=1-(1-x)^{3}= (x-1)^{3}+1.

🧠:[<think>Okay, let's see. I need to find the maximum value of the function f(x) = x³ + b x² -12x on the interval [-4, 4]. They told me that when x=2, the function reaches an extremum. The options are A: 8, B:12, C:16, D:32.First, since there's an extremum at x=2, that means the derivative at x=2 must be zero. Let me start by finding the derivative of f(x). The derivative of x³ is 3x², the derivative of b x² is 2b x, and the derivative of -12x is -12. So, f'(x) = 3x² + 2b x -12.Since there's an extremum at x=2, plugging that into the derivative should equal zero. Let's set up the equation:f'(2) = 3*(2)² + 2b*(2) -12 = 0Calculating that step by step:3*(4) + 4b -12 = 012 + 4b -12 = 0Simplify the 12 and -12: 4b = 0So, 4b = 0 => b = 0.Wait, so b is zero? That simplifies the original function to f(x) = x³ -12x. Let me check that again. If b is zero, then the function is indeed x³ -12x. Then, the derivative is 3x² -12. Setting that equal to zero gives 3x² -12 =0, so x²=4, so x=2 or x=-2. But the problem only mentions that there's an extremum at x=2. Hmm, but actually, if b=0, then x=2 and x=-2 are both critical points. But maybe in the problem statement, they just mentioned one of them. Anyway, the key here is that we found b=0.So now, the function is f(x) = x³ -12x. Now, we need to find its maximum on the interval [-4,4]. To find the maximum on a closed interval, we need to check the critical points within the interval and the endpoints.First, critical points are where the derivative is zero. We already found them: x=2 and x=-2. So, these are both in the interval [-4,4], so we need to evaluate the function at x=-4, x=-2, x=2, and x=4.Let me compute f(-4):f(-4) = (-4)³ -12*(-4) = -64 + 48 = -16f(-2) = (-2)³ -12*(-2) = -8 +24 = 16f(2) = (2)³ -12*(2) = 8 -24 = -16f(4) = (4)³ -12*(4) = 64 -48 = 16So, the function values at the critical points and endpoints are:-4: -16-2: 162: -164: 16So, the maximum value is 16, which occurs at both x=-2 and x=4. Since 16 is one of the options, option C. But wait, let me double-check my calculations.Wait, f(-2) is (-2)^3 which is -8, minus 12*(-2) which is +24. So, -8 +24 is indeed 16. Similarly, f(4) is 64 -48=16. So, both x=-2 and x=4 give 16, which is the highest value. The other points are -16 and -16 and 16. So, the maximum is 16, so answer C. But wait, let me confirm once more.Wait, maybe I made a mistake with the critical points. Let me check the derivative again. If f(x) =x³ -12x, then f'(x)=3x² -12. Setting to zero: 3x²=12, x²=4, so x=2 and x=-2. Correct. So critical points at x=2 and x=-2. Both in [-4,4]. So, the function at those points:x=-2: f(-2)= (-8) -12*(-2)= -8 +24=16x=2: f(2)=8 -24= -16So, endpoints:x=-4: (-64) -12*(-4)= -64 +48= -16x=4: 64 -48=16Therefore, the maximum values are 16 at x=-2 and x=4. So the maximum is 16. So, answer C:16.But wait, the options are A:8, B:12, C:16, D:32. So, 16 is there. Hmm, but wait, let me check the problem again. The function was given as f(x)=x³ +b x² -12x. And we found that b=0. So the function is indeed x³ -12x. So, all steps correct. Then, yes, the maximum is 16.But just to make sure, maybe I made a mistake in calculating the value at x=-4. Let's check again:x=-4: (-4)^3 = -64, -12*(-4)=48. So, -64 +48= -16. Correct.x=4: 64 -48=16. Correct.So, the maximum is 16. So, answer C.Wait, but the options are A:8, B:12, C:16, D:32. So 16 is there as option C. Therefore, the answer is C.But wait, the problem mentions that when x=2, the function reaches an extremum. So, x=2 is a critical point. Since we found that at x=2, the function is -16, which is a local minimum, because the second derivative would be 6x. At x=2, second derivative is 12, which is positive, so it's a local minimum. Therefore, the extremum at x=2 is a minimum, which makes sense. The other critical point at x=-2, second derivative is 6*(-2)= -12, which is negative, so it's a local maximum. So, the local maximum at x=-2 is 16, and at the endpoint x=4, the value is also 16. Therefore, the maximum on the interval is 16.Therefore, the correct answer is C:16.Final Answerboxed{C}</think>Given the function ( f(x) = x^3 + bx^2 - 12x ) reaches an extremum at ( x = 2 ), we need to find the maximum value of ( f(x) ) on the interval ([-4, 4]).First, we find the derivative of ( f(x) ):[ f'(x) = 3x^2 + 2bx - 12 ]Since there is an extremum at ( x = 2 ), we set ( f'(2) = 0 ):[3(2)^2 + 2b(2) - 12 = 0 12 + 4b - 12 = 0 4b = 0 b = 0]With ( b = 0 ), the function simplifies to ( f(x) = x^3 - 12x ). We then find the critical points by solving ( f'(x) = 0 ):[3x^2 - 12 = 0 x^2 = 4 x = 2 text{ or } x = -2]Next, we evaluate ( f(x) ) at the critical points and the endpoints of the interval ([-4, 4]):- At ( x = -4 ): [ f(-4) = (-4)^3 - 12(-4) = -64 + 48 = -16 ]- At ( x = -2 ): [ f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16 ]- At ( x = 2 ): [ f(2) = 2^3 - 12(2) = 8 - 24 = -16 ]- At ( x = 4 ): [ f(4) = 4^3 - 12(4) = 64 - 48 = 16 ]The maximum value of ( f(x) ) on the interval ([-4, 4]) is 16, which occurs at both ( x = -2 ) and ( x = 4 ).Thus, the maximum value is (boxed{C}).,<think>Okay, so I need to solve this calculus problem. Let me read it again and make sure I understand what's being asked. The function is f(x) = x³ + b x² - 12x. When x = 2, this function reaches an extremum. Then, we need to find the maximum value of f(x) on the interval [-4, 4]. The options are A: 8, B: 12, C: 16, D: 32. Alright, step by step.First, since there's an extremum at x = 2, that means the derivative at x = 2 should be zero. So, I need to find the derivative f'(x), set it equal to zero at x = 2, and solve for b. Once I have the value of b, I can write the complete function f(x) and then find its maximum on the interval [-4, 4]. That makes sense.Let me start by finding the derivative of f(x). The function is a cubic polynomial: f(x) = x³ + b x² - 12x. The derivative, f'(x), is straightforward. The derivative of x³ is 3x², the derivative of b x² is 2b x, and the derivative of -12x is -12. So, f'(x) = 3x² + 2b x - 12.Now, since there's an extremum at x = 2, substituting x = 2 into f'(x) should give zero. Let's set that up:f'(2) = 3*(2)² + 2b*(2) - 12 = 0Calculating each term step by step:3*(2)² = 3*4 = 122b*(2) = 4bThen, the equation becomes:12 + 4b - 12 = 0Simplifying that:12 - 12 + 4b = 0 => 0 + 4b = 0 => 4b = 0 => b = 0Wait, so b is zero? Let me check that again. If I plug x = 2 into f'(x):3*(2)^2 + 2b*2 - 12 = 03*4 + 4b -12 = 012 + 4b -12 = 0Yeah, the 12 and -12 cancel out, leaving 4b = 0, so b = 0. Hmm, that seems straightforward. So, the function is f(x) = x³ + 0x² -12x = x³ -12x. Let me confirm that. If b is zero, then the function simplifies to f(x) = x³ - 12x. Then, the extremum at x = 2 is a critical point. But since the question says "reaches an extremum", that's correct—it could be a maximum or a minimum. But maybe we need to check if it's a maximum or a minimum? Although, since the problem is asking for the maximum on the interval, perhaps we need to evaluate the function at all critical points and endpoints.Wait, but first, maybe confirm that x = 2 is indeed a maximum or a minimum. Let's check the second derivative to determine the concavity at that point. If the second derivative is positive, it's a local minimum; if it's negative, it's a local maximum.Compute the second derivative f''(x). The first derivative is f'(x) = 3x² -12 (since b = 0). So, f''(x) = 6x. Then, at x = 2, f''(2) = 6*2 = 12, which is positive. Therefore, x = 2 is a local minimum. Wait, that's interesting. So the extremum at x = 2 is a local minimum. Hmm. So the function has a local minimum at x = 2 and possibly another critical point where the derivative is zero. Let me check.Wait, if b = 0, then f'(x) = 3x² -12. Setting that equal to zero gives 3x² -12 = 0 => x² = 4 => x = ±2. So the critical points are at x = 2 and x = -2. Therefore, the function has critical points at x = -2 and x = 2, with x = 2 being a local minimum (since f''(2) = 12 > 0) and x = -2 being a local maximum (since f''(-2) = 6*(-2) = -12 < 0). So, the function has a local maximum at x = -2 and a local minimum at x = 2.But the problem statement mentions that when x = 2, the function reaches an extremum. Which is correct—it is an extremum, but specifically a minimum. So, maybe the problem is setting up that at x = 2, there is an extremum (which is a minimum), and given that, we need to find the maximum on the interval [-4, 4]. So, even though x = 2 is a minimum, the maximum on the interval could be at one of the endpoints or at the other critical point x = -2.So, to find the maximum value on the interval [-4, 4], we need to evaluate f(x) at the critical points x = -2 and x = 2, and at the endpoints x = -4 and x = 4.Given that, let's compute f(x) at these four points: x = -4, x = -2, x = 2, x = 4.First, let's confirm the function with b = 0: f(x) = x³ -12x.Compute f(-4):f(-4) = (-4)^3 -12*(-4) = -64 + 48 = -16.Compute f(-2):f(-2) = (-2)^3 -12*(-2) = -8 + 24 = 16.Compute f(2):f(2) = (2)^3 -12*(2) = 8 -24 = -16.Compute f(4):f(4) = (4)^3 -12*(4) = 64 -48 = 16.So, the function evaluated at the critical points and endpoints gives:At x = -4: -16At x = -2: 16At x = 2: -16At x = 4: 16Therefore, the maximum value of f(x) on the interval [-4, 4] is 16, achieved at both x = -2 and x = 4. So, the answer should be C: 16.Wait, but let me double-check my calculations to make sure I didn't make any arithmetic errors.First, f(-4):(-4)^3 = -64-12*(-4) = +48-64 +48 = -16. Correct.f(-2):(-2)^3 = -8-12*(-2) = +24-8 +24 = 16. Correct.f(2):2^3 =8-12*2 = -248 -24 = -16. Correct.f(4):4^3 =64-12*4 = -4864 -48 =16. Correct.So all calculations check out. So the maximum is 16. Therefore, the answer is C: 16.But wait, the options given are A:8, B:12, C:16, D:32. So C is 16, which is one of the options. So, that should be the correct answer.But let me make sure there's no trick here. The function is a cubic, so on the interval [-4,4], it can have a maximum either at critical points or endpoints. We found the critical points at x=-2 and x=2. The endpoints are x=-4 and x=4.The maximum value among these is 16, which occurs at x=-2 and x=4. So, 16 is the maximum. The options include 16 as option C, so that must be the answer.Alternatively, maybe there's a mistake in my process. Let me check again.Wait, the original problem says "the function f(x) =x³ +b x² -12x reaches an extremum when x=2". We found that for this to be true, b=0. Then, with b=0, the function becomes f(x)=x³ -12x. Then, the extremum at x=2 is a local minimum. Then, to find the maximum on the interval [-4,4], we check critical points and endpoints, which we did. So, the answer is 16. Hmm, that seems correct.Alternatively, maybe I misapplied the derivative. Let's double-check.Original function: f(x) =x³ +b x² -12x.Derivative: f'(x)=3x² +2b x -12. At x=2, f'(2)=0. So:3*(4) + 2b*(2) -12 = 012 +4b -12=0 => 4b=0 => b=0. That's correct.Therefore, b=0, which simplifies f(x) to x³ -12x.Critical points at x=2 and x=-2. Then, endpoints at x=-4 and x=4. Evaluations give -16,16,-16,16. So maximum 16. So, answer is C. That seems right.Alternatively, maybe the problem is expecting something else? Let me check the question again."When x=2, the function f(x)=x³ +b x² -12x reaches an extremum. The maximum value of f(x) on the interval [-4,4] is: A:8, B:12, C:16, D:32"No, the process seems correct. The key steps are:1. Use the extremum at x=2 to find b=0.2. Find critical points (x=-2 and x=2).3. Evaluate f(x) at critical points and endpoints.4. Compare the values to find the maximum.Everything checks out. Therefore, the maximum value is 16, which is option C. So, the answer is C.Final Answerboxed{C}</think>Given the function ( f(x) = x^3 + bx^2 - 12x ) reaches an extremum at ( x = 2 ), we need to find the maximum value of ( f(x) ) on the interval ([-4, 4]).1. Find the value of ( b ): - The first derivative of ( f(x) ) is ( f'(x) = 3x^2 + 2bx - 12 ). - Since there is an extremum at ( x = 2 ), we set ( f'(2) = 0 ): [ 3(2)^2 + 2b(2) - 12 = 0 implies 12 + 4b - 12 = 0 implies 4b = 0 implies b = 0 ]2. Determine the function ( f(x) ): - With ( b = 0 ), the function simplifies to ( f(x) = x^3 - 12x ).3. Find critical points: - The first derivative is ( f'(x) = 3x^2 - 12 ). - Setting ( f'(x) = 0 ): [ 3x^2 - 12 = 0 implies x^2 = 4 implies x = pm 2 ] - The critical points are ( x = -2 ) (local maximum) and ( x = 2 ) (local minimum).4. Evaluate ( f(x) ) at critical points and endpoints: - ( f(-4) = (-4)^3 - 12(-4) = -64 + 48 = -16 ) - ( f(-2) = (-2)^3 - 12(-2) = -8 + 24 = 16 ) - ( f(2) = 2^3 - 12(2) = 8 - 24 = -16 ) - ( f(4) = 4^3 - 12(4) = 64 - 48 = 16 )The maximum value of ( f(x) ) on the interval ([-4, 4]) is 16, which occurs at both ( x = -2 ) and ( x = 4 ).Thus, the answer is (boxed{C}).]

🔑:Given the function fleft(xright)=x^{3}+bx^{2}-12x, we are told that at x=2, the function reaches an extremum. To find the value of b and the maximum value of f(x) on the interval left[-4,4right], we follow these steps:1. Find the derivative of f(x):[f'(x) = 3x^{2} + 2bx - 12]2. Set f'(x) to 0 at x=2 to find b:[f'(2) = 3(2)^{2} + 2b(2) - 12 = 0][12 + 4b - 12 = 0][4b = 0][b = 0]3. Substitute b=0 back into f(x):[f(x) = x^{3} - 12x]4. Find the critical points of f(x) by setting its derivative to 0:[f'(x) = 3x^{2} - 12][3x^{2} - 12 = 0][x^{2} = 4][x = pm 2]5. Analyze the intervals determined by the critical points to find where f(x) is increasing or decreasing:- On left(-infty, -2right), f'(x) > 0, so f(x) is increasing.- On left(-2, 2right), f'(x) 0, so f(x) is increasing.6. Evaluate f(x) at the critical points and the endpoints of the interval left[-4,4right]:- f(-2) = (-2)^{3} - 12(-2) = -8 + 24 = 16- f(4) = 4^{3} - 12(4) = 64 - 48 = 167. Determine the maximum value of f(x) on the interval left[-4,4right]:Since f(x) reaches its maximum at both x = -2 and x = 4 within the interval, and both give f(x) = 16, the maximum value of f(x) on left[-4,4right] is 16.Therefore, the correct answer is boxed{C}.

🧠:[<think>Okay, let's tackle this problem step by step. First, there are two parts: part (1) is to find the range of the function f(x) on the interval [0, π/2], and part (2) is a triangle problem where we need to find the value of f(B). Let's start with part (1).Part 1: Finding the range of f(x)The function given is f(x) = 2√3 sinx cosx - 3 sin²x - cos²x + 3. Hmm, this looks like a trigonometric function. Maybe I can simplify it using some trigonometric identities.First, I notice that 2 sinx cosx is equal to sin(2x). So, 2√3 sinx cosx would be √3 sin(2x). Then, the term -3 sin²x - cos²x can maybe be rewritten using the identity sin²x = (1 - cos(2x))/2 and cos²x = (1 + cos(2x))/2. Let me try that.So, let's rewrite each term:- 2√3 sinx cosx = √3 sin(2x)- -3 sin²x = -3*(1 - cos(2x))/2 = -3/2 + (3/2)cos(2x)- -cos²x = - (1 + cos(2x))/2 = -1/2 - (1/2)cos(2x)- +3 remains as is.Now, combining all these terms:f(x) = √3 sin(2x) + (-3/2 + 3/2 cos2x) + (-1/2 - 1/2 cos2x) + 3Let's combine like terms:First, the constants: -3/2 -1/2 + 3 = (-3/2 -1/2) + 3 = (-2) + 3 = 1.Next, the cos(2x) terms: 3/2 cos2x - 1/2 cos2x = (3/2 - 1/2) cos2x = cos2x.So now, f(x) = √3 sin2x + cos2x + 1.Hmm, that's simpler: f(x) = √3 sin2x + cos2x + 1. This looks like a harmonic function of the form A sin2x + B cos2x + C. The range of A sinθ + B cosθ is [-√(A² + B²), √(A² + B²)], so then adding C would shift that range. So, perhaps the amplitude here is √( (√3)^2 + 1^2 ) = √(3 + 1) = 2. Therefore, the term √3 sin2x + cos2x can range between -2 and +2, so adding 1 would make the range from -1 to 3. But wait, we have to check if this is valid for x in [0, π/2]. Also, when x is in [0, π/2], 2x is in [0, π]. So the angle 2x ranges from 0 to π. Let me confirm if the maximum and minimum of the harmonic function occur within this interval.Alternatively, maybe I can write √3 sin2x + cos2x as a single sine or cosine function. Let's try that. Let's consider the expression √3 sin2x + cos2x.We can write this as 2 sin(2x + φ), where φ is a phase shift. To find φ:The amplitude is √( (√3)^2 + 1^2 ) = √(3 + 1) = 2. So, √3 sin2x + cos2x = 2 sin(2x + φ).To find φ, we can use:sinφ = B/amplitude = 1/2, cosφ = A/amplitude = √3/2. Therefore, φ is π/6. Because sinφ = 1/2 and cosφ = √3/2, so φ = π/6.Therefore, √3 sin2x + cos2x = 2 sin(2x + π/6). Therefore, f(x) = 2 sin(2x + π/6) + 1.Now, since x ∈ [0, π/2], 2x ∈ [0, π], so 2x + π/6 ∈ [π/6, 7π/6].The sine function sinθ reaches its maximum at θ = π/2 (which is within [π/6, 7π/6]) and minimum at θ = 7π/6 (which is 3π/2, but 7π/6 is in the third quadrant where sine is negative). Wait, let's check:sinθ is maximum at π/2 (which is 1), and minimum at 3π/2 (which is -1). But our interval is from π/6 to 7π/6. So, θ = π/2 is included (so maximum is 1), but the minimum in this interval would be at θ = 7π/6, where sin(7π/6) = -1/2. Wait, but is that the minimum?Wait, let's consider θ from π/6 to 7π/6. The sine function in this interval starts at sin(π/6) = 1/2, increases to sin(π/2) = 1 at θ = π/2, then decreases to sin(π) = 0 at θ = π, then goes to sin(7π/6) = -1/2. So in this interval, the maximum is 1, and the minimum is -1/2. Therefore, sinθ ∈ [-1/2, 1] for θ ∈ [π/6, 7π/6].Therefore, 2 sinθ ∈ [-1, 2], so 2 sinθ + 1 ∈ [0, 3]. Therefore, the range of f(x) is [0, 3].Wait, but let's verify this. Let me check critical points. Since f(x) = 2 sin(2x + π/6) + 1, the derivative is f’(x) = 2 * cos(2x + π/6) * 2 = 4 cos(2x + π/6). Setting derivative to zero gives cos(2x + π/6) = 0. Therefore, 2x + π/6 = π/2 + kπ. Solving for x:2x = π/2 - π/6 + kπ = (3π/6 - π/6) + kπ = (2π/6) + kπ = π/3 + kπ.Therefore, x = π/6 + kπ/2. Within x ∈ [0, π/2], possible solutions are:k=0: x = π/6.k=1: x = π/6 + π/2 = 2π/3, which is greater than π/2, so not in the interval.So only critical point is at x = π/6. Let's compute f(π/6):f(π/6) = 2 sin(2*(π/6) + π/6) + 1 = 2 sin(π/3 + π/6) + 1 = 2 sin(π/2) + 1 = 2*1 + 1 = 3.Then, check the endpoints:At x=0:f(0) = 2 sin(π/6) + 1 = 2*(1/2) + 1 = 1 + 1 = 2.Wait, but when x=0, the original function is:2√3*0*1 -3*0 -1 +3 = 0 -0 -1 +3 = 2. Which matches.At x=π/2:2x + π/6 = π + π/6 = 7π/6. So sin(7π/6) = -1/2. Therefore, f(π/2) = 2*(-1/2) +1 = -1 +1 = 0.But let's check with original function:f(π/2) = 2√3 sin(π/2)cos(π/2) -3 sin²(π/2) -cos²(π/2) +3. sin(π/2) =1, cos(π/2)=0. So:2√3*1*0 -3*1 -0 +3 = 0 -3 +0 +3 = 0. Correct.So the function reaches maximum at x=π/6 (value 3), minimum at x=π/2 (value 0). At x=0, it's 2. Therefore, the range is [0, 3]. So part (1)'s answer is [0,3].Part 2: Triangle ABC, find f(B)Given in triangle ABC: sides opposite angles A, B, C are a, b, c respectively. Given that b/a = √3, and [sin(2A + C)] / sinA = 2 + 2 cos(A + C). Need to find f(B).First, let's recall in a triangle, A + B + C = π. So maybe we can express angles in terms of each other.Given that (sin(2A + C)) / sinA = 2 + 2 cos(A + C)Let me denote some substitutions. Since in a triangle, A + B + C = π. So, A + C = π - B. Therefore, cos(A + C) = cos(π - B) = -cosB.Similarly, 2A + C can be written as A + (A + C) = A + (π - B). Let's see:2A + C = A + (A + C) = A + (π - B)But not sure if that helps. Let me try to compute sin(2A + C).2A + C = 2A + (π - A - B) [since A + B + C = π => C = π - A - B]So, 2A + C = 2A + π - A - B = A + π - B.Therefore, sin(2A + C) = sin(A + π - B) = sin(π + A - B). But sin(π + θ) = -sinθ, so this is -sin(A - B).Therefore, the equation becomes:[ -sin(A - B) ] / sinA = 2 + 2 cos(A + C) = 2 + 2*(-cosB) = 2 - 2 cosB.So,-sin(A - B)/sinA = 2 - 2 cosB.Multiply both sides by -1:sin(A - B)/sinA = -2 + 2 cosB.Hmm, let's see. Let's write sin(A - B) as sinA cosB - cosA sinB. So:[ sinA cosB - cosA sinB ] / sinA = cosB - cotA sinB = -2 + 2 cosB.So,cosB - cotA sinB = -2 + 2 cosBLet's rearrange:cosB - cotA sinB - 2 cosB + 2 = 0=> -cosB - cotA sinB + 2 = 0Multiply through by sinA to eliminate cotA (cotA = cosA/sinA):- cosB sinA - cosA sinB + 2 sinA = 0Note that cosA sinB + sinA cosB = sin(A + B). Wait, but here we have - (cosB sinA + cosA sinB) + 2 sinA = 0.So,- sin(A + B) + 2 sinA = 0But in a triangle, A + B = π - C. So sin(A + B) = sin(π - C) = sinC.Therefore, the equation becomes:- sinC + 2 sinA = 0 => 2 sinA = sinC.By the law of sines, sinA / a = sinB / b = sinC / c = 1/(2R), where R is the circumradius.Given that 2 sinA = sinC, so sinC = 2 sinA. Therefore, from the law of sines:sinC / c = sinA / a => (2 sinA) / c = sinA / a => 2/c = 1/a => c = 2a.So, c = 2a.Also, given that b/a = √3, so b = a√3.So, sides are a, b = a√3, c = 2a. So, sides are proportional to 1 : √3 : 2.Wait, that looks like a 30-60-90 triangle. Let me check.In a triangle with sides 1, √3, 2, the angles opposite would be 30°, 60°, 90°. Let's verify with the Law of Sines:sinA / a = sinB / b = sinC / c.Suppose a = 1, b = √3, c = 2.sinA / 1 = sinB / √3 = sinC / 2.If angle opposite 2 is 90°, sinC = 1, so 1/2 = sinA /1 => sinA = 1/2 => A = 30°, then sinB / √3 = 1/2 => sinB = (√3)/2 => B = 60°. So yes, that's a 30-60-90 triangle.So in this case, angles A = 30°, B = 60°, C = 90°, but wait, the sides are a=1, b=√3, c=2. So opposite angles A, B, C are 30°, 60°, 90°. So in this problem, the triangle seems to be a 30-60-90 triangle. Therefore, angle B is 60°, so f(B) = f(60°). But wait, the original function f(x) is in terms of radians or degrees? Since in the problem statement, the first part is for x ∈ [0, π/2], so radians. So angle B is in radians. If angle B is 60°, that's π/3 radians.So f(B) = f(π/3). Let's compute f(π/3).First, let's compute f(x) as per the original function:f(x) = 2√3 sinx cosx -3 sin²x - cos²x +3.But earlier, we simplified f(x) to 2 sin(2x + π/6) +1. So perhaps using that is easier.But let's verify.Compute f(π/3):2√3 sin(π/3) cos(π/3) -3 sin²(π/3) -cos²(π/3) +3.Compute each term:sin(π/3) = √3/2, cos(π/3) = 1/2.So:2√3*(√3/2)*(1/2) = 2√3*(√3/4) = 2*(3)/4 = 6/4 = 3/2.-3 sin²(π/3) = -3*(3/4) = -9/4.-cos²(π/3) = - (1/4).+3.Adding all together: 3/2 -9/4 -1/4 +3.Convert to quarters:3/2 = 6/4, 3 = 12/4.So 6/4 -9/4 -1/4 +12/4 = (6 -9 -1 +12)/4 = (8)/4 = 2.Alternatively, using the simplified f(x) = 2 sin(2x + π/6) +1.Compute 2x + π/6 where x = π/3: 2*(π/3) + π/6 = 2π/3 + π/6 = 5π/6. So sin(5π/6) = 1/2. Therefore, 2*(1/2) +1 = 1 +1 =2. Correct.Therefore, f(B) = 2. But let's confirm that the triangle is indeed 30-60-90.Wait, we concluded that sides are a, b = a√3, c=2a, so triangle with sides proportional to 1, √3, 2. So angles would be 30°, 60°, 90°, which would give B = 60°, hence f(B)=2.But let me check if there could be another configuration.Wait, the steps we followed: from the given equation, we derived that 2 sinA = sinC, and from the law of sines, sinA/a = sinC/c, so sinC = (c/a) sinA. Therefore, 2 sinA = (c/a) sinA => c = 2a, provided sinA ≠0, which it is in a triangle.Therefore, c=2a. Also, given that b/a = √3, so sides a, b=√3 a, c=2a. So the sides are fixed in proportion. Therefore, the angles are fixed as 30°,60°,90°. Therefore, angle B is 60°, so f(B)=2.But let me confirm with another approach. Suppose angle B is π/3. Then f(B) is 2. So the answer is 2. Therefore, part (2) answer is 2.But let's make sure there are no other possibilities. Suppose the triangle isn't right-angled. Wait, but with sides a, √3 a, 2a, it must be right-angled because (a)^2 + (√3 a)^2 = a² + 3a² = 4a² = (2a)^2. Therefore, it's a right-angled triangle with hypotenuse 2a, so angle opposite to 2a is 90°, which is angle C. Therefore, angles A=30°, B=60°, C=90°, as in a 30-60-90 triangle. Therefore, B=60°=π/3, and f(B)=2. So this seems correct.Final Answer(1) boxed{[0, 3]}(2) boxed{2}</think>Given the function ( f(x) = 2sqrt{3} sin x cos x - 3 sin^2 x - cos^2 x + 3 ):(1) To find the range of ( f(x) ) when ( x in [0, frac{pi}{2}] ):First, simplify the function using trigonometric identities:[f(x) = sqrt{3} sin 2x + cos 2x + 1]This can be rewritten as:[f(x) = 2 sin left(2x + frac{pi}{6}right) + 1]The range of ( sin theta ) for ( theta in [frac{pi}{6}, frac{7pi}{6}] ) is ([- frac{1}{2}, 1]). Therefore, the range of ( 2 sin left(2x + frac{pi}{6}right) ) is ([-1, 2]). Adding 1 shifts this range to ([0, 3]).Thus, the range of ( f(x) ) when ( x in [0, frac{pi}{2}] ) is (boxed{[0, 3]}).(2) In triangle ( ABC ), given ( frac{b}{a} = sqrt{3} ) and ( frac{sin(2A + C)}{sin A} = 2 + 2 cos(A + C) ), we need to find the value of ( f(B) ):Using the given equation and trigonometric identities, we derived that the triangle must be a 30-60-90 triangle with sides ( a, sqrt{3}a, 2a ). This results in angles ( A = 30^circ ), ( B = 60^circ ), and ( C = 90^circ ). Converting ( B = 60^circ ) to radians gives ( B = frac{pi}{3} ).Evaluating ( f(B) ) at ( x = frac{pi}{3} ):[fleft(frac{pi}{3}right) = 2 sin left(2 cdot frac{pi}{3} + frac{pi}{6}right) + 1 = 2 sin left(frac{5pi}{6}right) + 1 = 2 cdot frac{1}{2} + 1 = 2]Thus, the value of ( f(B) ) is (boxed{2}).,<think>Okay, let's tackle this problem step by step. First, part (1) asks for the range of the function ( f(x) = 2sqrt{3}sin x cos x - 3sin^2 x - cos^2 x + 3 ) when ( x in [0, frac{pi}{2}] ). Hmm, trigonometric function with multiple terms. I need to simplify this expression first, probably using some trigonometric identities.Starting with the first term: ( 2sqrt{3}sin x cos x ). I remember that ( sin 2x = 2sin x cos x ), so this term can be rewritten as ( sqrt{3}sin 2x ). That's a good start.Next, the second term is ( -3sin^2 x ). The third term is ( -cos^2 x ). Maybe I can combine the ( sin^2 x ) and ( cos^2 x ) terms. Let's see. If I factor out a negative sign, it becomes ( - (3sin^2 x + cos^2 x) ). Hmm, not sure yet. Let's also look at the constant term, which is +3.So, putting it all together so far:( f(x) = sqrt{3}sin 2x - 3sin^2 x - cos^2 x + 3 )Let me write all the terms:1. ( sqrt{3}sin 2x )2. ( -3sin^2 x )3. ( -cos^2 x )4. ( +3 )I need to combine terms where possible. The ( sin^2 x ) and ( cos^2 x ) terms can perhaps be simplified. Let's consider combining the ( -3sin^2 x - cos^2 x ). Let's factor:( -3sin^2 x - cos^2 x = - (3sin^2 x + cos^2 x) )Hmm, this can be written as ( - [2sin^2 x + (sin^2 x + cos^2 x)] = - [2sin^2 x + 1] ), since ( sin^2 x + cos^2 x = 1 ). So, substituting back:( f(x) = sqrt{3}sin 2x - [2sin^2 x + 1] + 3 )Simplify the constants:( -1 + 3 = 2 ), so:( f(x) = sqrt{3}sin 2x - 2sin^2 x + 2 )Now, the term ( -2sin^2 x ). Maybe express ( sin^2 x ) in terms of ( cos 2x ). Recall that ( sin^2 x = frac{1 - cos 2x}{2} ). So substituting that:( -2sin^2 x = -2 cdot frac{1 - cos 2x}{2} = - (1 - cos 2x) = -1 + cos 2x )Therefore, substituting back into the function:( f(x) = sqrt{3}sin 2x + (-1 + cos 2x) + 2 )Simplify constants again:( -1 + 2 = 1 ), so:( f(x) = sqrt{3}sin 2x + cos 2x + 1 )Ah, now this looks more manageable. So the function simplifies to:( f(x) = sqrt{3}sin 2x + cos 2x + 1 )Now, this is a linear combination of ( sin 2x ) and ( cos 2x ), which can be written as a single sine or cosine function with a phase shift. Let's try that.The general form is ( Asin theta + Bcos theta = Csin(theta + phi) ) or ( Ccos(theta + phi) ). Let's compute the amplitude ( C = sqrt{A^2 + B^2} ).Here, ( A = sqrt{3} ) and ( B = 1 ), so ( C = sqrt{ (sqrt{3})^2 + 1^2 } = sqrt{3 + 1} = 2 ).Therefore, ( sqrt{3}sin 2x + cos 2x = 2 sin(2x + phi) ), where ( phi ) is the phase shift. To find ( phi ):( sin phi = frac{B}{C} = frac{1}{2} ), ( cos phi = frac{A}{C} = frac{sqrt{3}}{2} ). Therefore, ( phi = frac{pi}{6} ).Alternatively, since ( sin phi = frac{1}{2} ) and ( cos phi = frac{sqrt{3}}{2} ), ( phi = 30^circ = frac{pi}{6} ).Therefore, ( sqrt{3}sin 2x + cos 2x = 2sinleft(2x + frac{pi}{6}right) ).Wait, let me check that. Let me expand ( 2sin(2x + frac{pi}{6}) ):Using the sine addition formula:( 2sin(2x + frac{pi}{6}) = 2left[ sin 2x cos frac{pi}{6} + cos 2x sin frac{pi}{6} right] )( = 2left[ sin 2x cdot frac{sqrt{3}}{2} + cos 2x cdot frac{1}{2} right] )( = 2 cdot frac{sqrt{3}}{2} sin 2x + 2 cdot frac{1}{2} cos 2x )( = sqrt{3}sin 2x + cos 2x )Yes, that's correct. So indeed, ( sqrt{3}sin 2x + cos 2x = 2sinleft(2x + frac{pi}{6}right) ).Therefore, the function ( f(x) ) becomes:( f(x) = 2sinleft(2x + frac{pi}{6}right) + 1 )Now, this is much simpler. To find the range of ( f(x) ) when ( x in [0, frac{pi}{2}] ), we can analyze the sine function within that interval.First, let's determine the range of the argument ( 2x + frac{pi}{6} ) when ( x in [0, frac{pi}{2}] ).When ( x = 0 ):( 2(0) + frac{pi}{6} = frac{pi}{6} )When ( x = frac{pi}{2} ):( 2(frac{pi}{2}) + frac{pi}{6} = pi + frac{pi}{6} = frac{7pi}{6} )So the argument ( 2x + frac{pi}{6} ) ranges from ( frac{pi}{6} ) to ( frac{7pi}{6} ).The sine function ( sintheta ) has a maximum of 1 and a minimum of -1 in general, but over the interval ( [frac{pi}{6}, frac{7pi}{6}] ), let's see where the maximum and minimum occur.The critical points of ( sintheta ) occur at ( theta = frac{pi}{2} ) (maximum) and ( theta = frac{3pi}{2} ) (minimum). However, our interval is from ( frac{pi}{6} ) to ( frac{7pi}{6} ), which includes ( frac{pi}{2} ) but not ( frac{3pi}{2} ).Wait, ( frac{7pi}{6} ) is ( pi + frac{pi}{6} ), which is in the third quadrant. The sine function in this interval ( [frac{pi}{6}, frac{7pi}{6}] ) starts at ( frac{pi}{6} ), goes up to ( frac{pi}{2} ), then down to ( pi ) (where sine is 0), and then continues down to ( frac{7pi}{6} ), where it's negative.So the maximum value of ( sintheta ) in this interval is 1 (at ( theta = frac{pi}{2} )), and the minimum is at ( theta = frac{7pi}{6} ), which is ( sin frac{7pi}{6} = -frac{1}{2} ).Therefore, ( sintheta ) ranges from -1/2 to 1 in this interval.Hence, ( 2sintheta + 1 ) will range from ( 2(-1/2) + 1 = -1 + 1 = 0 ) to ( 2(1) + 1 = 3 ).But wait, is that correct? Let me check. If the minimum of ( sintheta ) is -1/2, then multiplying by 2 gives -1, then adding 1 gives 0. The maximum is 1, multiplied by 2 is 2, plus 1 is 3. So yes, the range of ( f(x) ) should be [0, 3]. However, I need to confirm if the sine function actually reaches those extremes within the interval.Wait, when ( theta = frac{pi}{2} ), which is within ( [frac{pi}{6}, frac{7pi}{6}] ), so yes, the maximum of 1 is achieved. For the minimum, at ( theta = frac{7pi}{6} ), which is included in the interval, so ( sin frac{7pi}{6} = -1/2 ). Therefore, the minimum value of ( 2sintheta + 1 ) is indeed 0. So the range of ( f(x) ) is [0, 3].Wait, but hold on. Let's check the endpoints and critical points to be thorough.First, when ( x = 0 ):( f(0) = 2sqrt{3}sin 0 cos 0 - 3sin^2 0 - cos^2 0 + 3 )But plugging into the simplified version:( f(0) = 2sin(frac{pi}{6}) + 1 = 2 cdot frac{1}{2} + 1 = 1 + 1 = 2 )Alternatively, using the simplified ( f(x) = 2sin(2x + frac{pi}{6}) + 1 ):At ( x = 0 ):( 2sin(frac{pi}{6}) + 1 = 2 cdot frac{1}{2} + 1 = 2 )Similarly, at ( x = frac{pi}{2} ):( 2sin(2 cdot frac{pi}{2} + frac{pi}{6}) + 1 = 2sin(pi + frac{pi}{6}) + 1 = 2 cdot (-frac{1}{2}) + 1 = -1 + 1 = 0 )So at ( x = 0 ), f(x) = 2, and at ( x = pi/2 ), f(x) = 0. But earlier analysis suggested that the maximum is 3. So where does the maximum occur?The maximum of ( 2sintheta + 1 ) occurs when ( sintheta = 1 ), which is at ( theta = frac{pi}{2} ). Let's find the x corresponding to ( theta = frac{pi}{2} ):( 2x + frac{pi}{6} = frac{pi}{2} )Solving for x:( 2x = frac{pi}{2} - frac{pi}{6} = frac{3pi}{6} - frac{pi}{6} = frac{2pi}{6} = frac{pi}{3} )Therefore, ( x = frac{pi}{6} ).So check f(x) at ( x = pi/6 ):( f(pi/6) = 2sin(2 cdot frac{pi}{6} + frac{pi}{6}) + 1 = 2sin(frac{pi}{3} + frac{pi}{6}) + 1 = 2sin(frac{pi}{2}) + 1 = 2 cdot 1 + 1 = 3 )Therefore, the maximum is indeed 3 at ( x = pi/6 ), and the minimum is 0 at ( x = pi/2 ), but wait, at x = 0, the value is 2, and the function goes up to 3 at ( x = pi/6 ), then down to 0 at ( x = pi/2 ).So the range is from 0 to 3. But let's confirm if there are any other critical points.To be thorough, let's compute the derivative of f(x) and check for critical points in the interval [0, π/2].Original simplified function: ( f(x) = 2sin(2x + frac{pi}{6}) + 1 )Derivative: ( f'(x) = 2 cdot cos(2x + frac{pi}{6}) cdot 2 = 4cos(2x + frac{pi}{6}) )Setting derivative to zero:( 4cos(2x + frac{pi}{6}) = 0 )Which implies:( cos(2x + frac{pi}{6}) = 0 )Solutions for ( 2x + frac{pi}{6} = frac{pi}{2} + kpi ), where k is integer.So:( 2x = frac{pi}{2} - frac{pi}{6} + kpi )( 2x = frac{pi}{3} + kpi )( x = frac{pi}{6} + frac{kpi}{2} )Within [0, π/2], possible k values:k=0: x= π/6 ≈ 0.523, which is within [0, π/2].k=1: x= π/6 + π/2 = 2π/3 ≈ 2.094, which is outside [0, π/2].k=-1: x= π/6 - π/2 = -π/3, which is negative. So only critical point is at x= π/6, which we already found gives the maximum.Therefore, the function has only one critical point at x= π/6, which is a maximum. The endpoints at x=0 and x= π/2 give f(x)=2 and 0, respectively.Therefore, the range of f(x) is [0, 3].Wait, but hold on. The function goes from 2 at x=0, up to 3 at x=π/6, then down to 0 at x=π/2. So the minimum is 0 and maximum is 3. Therefore, the range is [0, 3]. Therefore, the answer to part (1) is [0, 3].Now, moving on to part (2). In triangle ABC, sides opposite angles A, B, C are a, b, c respectively. Given that ( frac{b}{a} = sqrt{3} ), and ( frac{sin(2A + C)}{sin A} = 2 + 2cos(A + C) ). Need to find the value of f(B).First, let's recall that in a triangle, the sum of angles is π. So A + B + C = π. Therefore, A + C = π - B.Given the equation:( frac{sin(2A + C)}{sin A} = 2 + 2cos(A + C) )Replace A + C with π - B:( frac{sin(2A + C)}{sin A} = 2 + 2cos(pi - B) )But ( cos(pi - B) = -cos B ), so:( frac{sin(2A + C)}{sin A} = 2 - 2cos B )Now, let's express 2A + C in terms of other angles. Since A + B + C = π, then C = π - A - B. So substituting into 2A + C:2A + C = 2A + (π - A - B) = A + π - BTherefore, ( sin(2A + C) = sin(A + π - B) )But ( sin(A + π - B) = sin(pi + A - B) = -sin(A - B) ), since ( sin(pi + x) = -sin x ).Therefore:( frac{ - sin(A - B) }{ sin A } = 2 - 2cos B )Multiply both sides by -1:( frac{ sin(A - B) }{ sin A } = -2 + 2cos B )Hmm, let's write that as:( frac{ sin(A - B) }{ sin A } = 2(cos B - 1) )Let me compute the left-hand side (LHS):( sin(A - B) = sin A cos B - cos A sin B )Therefore, LHS:( frac{ sin A cos B - cos A sin B }{ sin A } = cos B - cot A sin B )Therefore:( cos B - cot A sin B = 2(cos B - 1) )Let's rearrange terms:Left side: ( cos B - cot A sin B )Right side: ( 2cos B - 2 )Bring all terms to left side:( cos B - cot A sin B - 2cos B + 2 = 0 )Simplify:( - cos B - cot A sin B + 2 = 0 )Multiply through by -1:( cos B + cot A sin B - 2 = 0 )So:( cos B + cot A sin B = 2 )But ( cot A = frac{cos A}{sin A} ), so:( cos B + frac{cos A}{sin A} sin B = 2 )Combine terms:( cos B + frac{ cos A sin B }{ sin A } = 2 )Hmm, this expression can be related to the sine of some angle or maybe using the Law of Sines.Given that ( frac{b}{a} = sqrt{3} ), and by the Law of Sines, ( frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R ). Therefore, ( frac{b}{a} = frac{sin B}{sin A} = sqrt{3} ), so ( sin B = sqrt{3} sin A ).Let me note that down:From ( frac{b}{a} = sqrt{3} ), we have ( frac{sin B}{sin A} = sqrt{3} ), so ( sin B = sqrt{3} sin A ).So, ( sin B = sqrt{3} sin A ). Let's keep this in mind.Now, going back to the equation:( cos B + frac{ cos A sin B }{ sin A } = 2 )Since ( sin B = sqrt{3} sin A ), substitute that into the equation:( cos B + frac{ cos A cdot sqrt{3} sin A }{ sin A } = 2 )Simplify:The ( sin A ) cancels:( cos B + sqrt{3} cos A = 2 )So:( cos B + sqrt{3} cos A = 2 )Now, since in a triangle, A + B + C = π, and we need to relate angles A and B. Also, ( sin B = sqrt{3} sin A ).Let me use the identity ( sin B = sqrt{3} sin A ). Let's express cosines in terms of sines.We can use the Pythagorean identity:( cos A = sqrt{1 - sin^2 A} )( cos B = sqrt{1 - sin^2 B} = sqrt{1 - 3 sin^2 A} )But let's plug these into the equation:( sqrt{1 - 3 sin^2 A} + sqrt{3} sqrt{1 - sin^2 A} = 2 )This looks complicated. Let me set ( x = sin A ), so ( 0 < x < 1 ). Then:( sqrt{1 - 3x^2} + sqrt{3} sqrt{1 - x^2} = 2 )We need to solve for x.Let me denote this equation as:( sqrt{1 - 3x^2} + sqrt{3(1 - x^2)} = 2 )Let me square both sides to eliminate the square roots. Let me denote LHS as S, so S = 2. Then:( S^2 = (sqrt{1 - 3x^2} + sqrt{3(1 - x^2)})^2 )( = (1 - 3x^2) + 3(1 - x^2) + 2 sqrt{ (1 - 3x^2) cdot 3(1 - x^2) } )Simplify:First term: 1 - 3x^2Second term: 3 - 3x^2Sum of first two terms: 1 - 3x^2 + 3 - 3x^2 = 4 - 6x^2Third term: 2 sqrt{ 3(1 - x^2)(1 - 3x^2) }So:( (4 - 6x^2) + 2 sqrt{ 3(1 - x^2)(1 - 3x^2) } = 4 ) (since S = 2, S^2 = 4)Therefore:( 4 - 6x^2 + 2 sqrt{ 3(1 - x^2)(1 - 3x^2) } = 4 )Subtract 4 from both sides:( -6x^2 + 2 sqrt{ 3(1 - x^2)(1 - 3x^2) } = 0 )Divide both sides by 2:( -3x^2 + sqrt{ 3(1 - x^2)(1 - 3x^2) } = 0 )Bring the square root term to one side:( sqrt{ 3(1 - x^2)(1 - 3x^2) } = 3x^2 )Square both sides again:Left side squared: 3(1 - x^2)(1 - 3x^2)Right side squared: 9x^4Thus:3(1 - x^2)(1 - 3x^2) = 9x^4Expand the left side:First, compute (1 - x^2)(1 - 3x^2):= 1*(1) + 1*(-3x^2) - x^2*(1) + x^2*(3x^2)= 1 - 3x^2 - x^2 + 3x^4= 1 - 4x^2 + 3x^4Multiply by 3:= 3 - 12x^2 + 9x^4Set equal to 9x^4:3 - 12x^2 + 9x^4 = 9x^4Subtract 9x^4 from both sides:3 - 12x^2 = 0Thus:3 = 12x^2=> x^2 = 3/12 = 1/4=> x = 1/2 or x = -1/2But since x = sin A and A is an angle in a triangle, 0 < A < π, so sin A is positive. Thus, x = 1/2.Therefore, sin A = 1/2. Therefore, angle A is either π/6 or 5π/6. But since it's a triangle angle, and sum of angles is π, A must be π/6 because if A were 5π/6, the sum of angles would exceed π.Therefore, A = π/6.Then, sin B = √3 sin A = √3*(1/2) = √3/2. Therefore, B is either π/3 or 2π/3.But we need to check if this is possible. Let's see.Given that A = π/6, then angle B could be π/3 or 2π/3. Let's check if the sum A + B <= π.If B = π/3, then A + B = π/6 + π/3 = π/2, so angle C = π - π/2 = π/2. That's possible.If B = 2π/3, then A + B = π/6 + 2π/3 = 5π/6, so angle C = π - 5π/6 = π/6. That's also possible. So we need to determine which one is the case.Wait, but we can use the earlier equation: cos B + √3 cos A = 2.We have A = π/6, so cos A = √3/2. Therefore:cos B + √3*(√3/2) = 2cos B + (3/2) = 2cos B = 2 - 3/2 = 1/2Therefore, cos B = 1/2, which implies B = π/3 or 5π/3. But 5π/3 is more than 2π, which isn't possible in a triangle. Therefore, B = π/3.So angle B is π/3. Therefore, f(B) = f(π/3). Now, recall from part (1) that f(x) = 2 sin(2x + π/6) + 1.So compute f(π/3):f(π/3) = 2 sin(2*(π/3) + π/6) + 1 = 2 sin(2π/3 + π/6) + 1Convert 2π/3 to 4π/6, so 4π/6 + π/6 = 5π/6Therefore, f(π/3) = 2 sin(5π/6) + 1 = 2*(1/2) + 1 = 1 + 1 = 2Alternatively, check using the original function:f(x)=2√3 sinx cosx -3 sin²x -cos²x +3At x = π/3:sin(π/3) = √3/2, cos(π/3)=1/2So:2√3*(√3/2)*(1/2) = 2√3*(√3/4) = 2*(3/4) = 3/2-3*( (√3/2)^2 ) = -3*(3/4) = -9/4-cos²x = -(1/2)^2 = -1/4+3Total: 3/2 - 9/4 -1/4 +3Convert to quarters:3/2 = 6/4, so 6/4 -9/4 -1/4 +12/4 = (6 -9 -1 +12)/4 = (8)/4 = 2. Correct.Therefore, f(B) = 2.So the answer to part (2) is 2.Final Answer(1) The range of ( f(x) ) is boxed{[0, 3]}.(2) The value of ( f(B) ) is boxed{2}.</think>Given the function ( f(x) = 2sqrt{3}sin x cos x - 3sin^2 x - cos^2 x + 3 ):1. To find the range of ( f(x) ) when ( x in [0, frac{pi}{2}] ): - Simplify the function using trigonometric identities: [ f(x) = sqrt{3}sin 2x - 2sin^2 x + 2 ] Further simplification using the identity ( sin^2 x = frac{1 - cos 2x}{2} ): [ f(x) = sqrt{3}sin 2x + cos 2x + 1 ] This can be written as: [ f(x) = 2sinleft(2x + frac{pi}{6}right) + 1 ] - Determine the range of the argument ( 2x + frac{pi}{6} ) over ( x in [0, frac{pi}{2}] ): - The argument ranges from ( frac{pi}{6} ) to ( frac{7pi}{6} ). - The sine function over this interval ranges from -1/2 to 1. - Therefore, the range of ( 2sinleft(2x + frac{pi}{6}right) + 1 ) is [0, 3]. The range of ( f(x) ) is (boxed{[0, 3]}).2. In triangle ( ABC ), given ( frac{b}{a} = sqrt{3} ) and ( frac{sin(2A + C)}{sin A} = 2 + 2cos(A + C) ): - Use the given conditions and trigonometric identities: - From ( frac{b}{a} = sqrt{3} ), we get ( sin B = sqrt{3} sin A ). - Simplify the given equation using angle sum identities and solve for angles ( A ) and ( B ): - Find ( sin A = frac{1}{2} ), so ( A = frac{pi}{6} ). - Then ( sin B = sqrt{3} sin frac{pi}{6} = frac{sqrt{3}}{2} ), so ( B = frac{pi}{3} ). - Calculate ( f(B) ) using ( B = frac{pi}{3} ): [ fleft(frac{pi}{3}right) = 2sinleft(2 cdot frac{pi}{3} + frac{pi}{6}right) + 1 = 2sinleft(frac{5pi}{6}right) + 1 = 2 cdot frac{1}{2} + 1 = 2 ] The value of ( f(B) ) is (boxed{2}).]

🔑:(1) Since f(x)=2 sqrt {3}sin xcos x-3sin ^{2}x-cos ^{2}x+3= sqrt {3}sin 2x-cos 2x+1 (using double angle identities)= 2sin (2x+ frac {pi}{6})+1 (using sum of angles identity)Given xin[0, frac{pi}{2}], therefore 2x+ frac {pi}{6} in [frac {pi}{6}, frac {7pi}{6}].Hence, sin (2x+ frac {pi}{6}) in [- frac {1}{2},1].So, f(x)=2sin (2x+ frac {pi}{6})+1 in [0,3].(2) Given frac{sin (2A+C)}{sin A}=2+2cos (A+C),This implies sin (2A+C)=2sin A+2sin Acos (A+C).Using the sum of angles identity, this becomes sin Acos (A+C)+cos Asin (A+C)=2sin A+2sin Acos (A+C).Rearranging the terms, we get -sin Acos (A+C)+cos Asin (A+C)=2sin A, which simplifies to sin C=2sin A.By the sine rule, we have c=2a. Given that frac{b}{a}= sqrt{3}, we get b= sqrt{3}a.Using the cosine rule, we find cos A= frac {b^{2}+c^{2}-a^{2}}{2bc}= frac {3a^{2}+4a^{2}-a^{2}}{2cdot sqrt {3}acdot 2a}= frac { sqrt {3}}{2}.Hence, A=30^{circ}. Using the sine rule again, we find sin C=2sin A=1, which gives C=90^{circ}.Since the sum of the angles in a triangle is 180^{circ}, we have B=60^{circ}.Therefore, f(B)=f(60^{circ})= boxed{2}.

🧠:[<think>Okay, so I need to figure out whether the condition "cosA + sinA = cosB + sinB" in triangle ABC is a sufficient, necessary, both, or neither condition for angle C being 90 degrees. The options are A, B, C, D as given. Let me start by recalling some trigonometric identities and triangle properties.First, in any triangle, the sum of angles is 180 degrees. So if angle C is 90 degrees, then angles A and B must add up to 90 degrees. That means A + B = 90°, so B = 90° - A. Let's see if this relationship affects the equation cosA + sinA = cosB + sinB.If angle C is 90°, then B = 90° - A. Let's substitute B into the equation:cosA + sinA = cos(90° - A) + sin(90° - A).We know that cos(90° - A) is sinA and sin(90° - A) is cosA. So substituting those in:cosA + sinA = sinA + cosA.Which simplifies to the same thing on both sides. So, if angle C is 90°, then cosA + sinA must equal cosB + sinB. That means that C = 90° implies cosA + sinA = cosB + sinB. Therefore, the condition cosA + sinA = cosB + sinB is a necessary condition for C = 90°. But is it sufficient?To check if it's sufficient, we need to see whether if cosA + sinA = cosB + sinB, then does that necessarily mean angle C is 90°? Let's suppose that in some triangle ABC, cosA + sinA = cosB + sinB. Does this force angle C to be right?Let's try to work through the math. Let's denote angle A and angle B as variables. We can write the equation:cosA + sinA = cosB + sinB.Let me square both sides to see if that helps. Squaring both sides gives:(cosA + sinA)^2 = (cosB + sinB)^2.Expanding both sides:cos²A + 2sinA cosA + sin²A = cos²B + 2sinB cosB + sin²B.Simplify using the identity cos²x + sin²x = 1:1 + 2sinA cosA = 1 + 2sinB cosB.Subtract 1 from both sides:2sinA cosA = 2sinB cosB.Divide both sides by 2:sinA cosA = sinB cosB.Recall that 2sinx cosx = sin2x, so this equation becomes:sin2A = sin2B.So sin2A = sin2B. Which means that either:1. 2A = 2B + 360°k, or2. 2A = 180° - 2B + 360°k, for some integer k.But since angles in a triangle are between 0° and 180°, we can ignore the 360°k terms because 2A and 2B will be between 0° and 360°, so k=0.Therefore, either:1. 2A = 2B ⇒ A = B, or2. 2A = 180° - 2B ⇒ A + B = 90°.So, from the equation cosA + sinA = cosB + sinB, we can deduce that either angle A equals angle B, or angle A + angle B equals 90°. Now, in a triangle, angles A + B + C = 180°, so if A + B = 90°, then angle C = 90°, which is the case we want. However, the other possibility is that angle A = angle B. In this case, angle C would be 180° - 2A, which could be any angle depending on A. Therefore, if A = B, then angle C is not necessarily 90°, unless A = B = 45°, which would make angle C = 90°, but if A = B is something else, like 30°, then angle C would be 120°, which is not a right angle.Therefore, the condition cosA + sinA = cosB + sinB leads to two possibilities: either A = B or A + B = 90°. So, angle C being 90° is one of the possible outcomes, but not the only one. Therefore, the condition is not sufficient because even if cosA + sinA = cosB + sinB, angle C might not be 90° (if A = B instead). However, if angle C is 90°, then cosA + sinA must equal cosB + sinB, so it is a necessary condition.Therefore, the answer is B: Necessary but not sufficient condition.But wait, let me check again. Suppose angle A = angle B, then cosA + sinA = cosB + sinB is true, but angle C would be 180° - 2A. For angle C to be 90°, 180° - 2A = 90° ⇒ 2A = 90° ⇒ A = 45°, so in that case, angle A = angle B = 45°, which is the case where angle C is 90°. So if A = B, then angle C is 90° only if A and B are 45°, otherwise, angle C is not 90°. Wait, but in the equation sin2A = sin2B, if A = B, then angle C is 180° - 2A. So angle C can be 90° only if A = B = 45°, but in that case, A + B = 90°, which would also satisfy the other condition. Wait, but if A = B = 45°, then both possibilities are true: A = B and A + B = 90°, so that case is overlapping. So perhaps there's a case where angle A = angle B, but not 45°, leading to angle C not being 90°, but still satisfying cosA + sinA = cosB + sinB. For example, take angle A = angle B = 30°, then angle C = 120°, but cos30 + sin30 ≈ (√3/2 + 1/2) ≈ (0.866 + 0.5) ≈ 1.366, and cos30 + sin30 is same for angle B, so 1.366 = 1.366. So this satisfies the equation, but angle C is 120°, not 90°. Therefore, the condition can hold without angle C being 90°, so the condition is not sufficient. However, if angle C is 90°, then the equation must hold. Therefore, the condition is necessary but not sufficient. Hence, answer B.But wait, let me check another example. Suppose angle A = 60°, angle B = 30°, angle C = 90°, but this triangle would have angles adding up to 180°, right? Wait, 60 + 30 + 90 = 180. But in this case, angle C is 90°, so according to the problem's original statement, cosA + sinA should equal cosB + sinB. Let's check:cos60 + sin60 = 0.5 + (√3/2) ≈ 0.5 + 0.866 ≈ 1.366.cos30 + sin30 = (√3/2) + 0.5 ≈ 0.866 + 0.5 ≈ 1.366. So they are equal. So in this case, angle C is 90°, and the equation holds. If angle C is 90°, then A + B = 90°, so the equation holds. But if angle C is not 90°, but A = B, then the equation still holds. So the equation can hold in two cases: when C is 90°, or when A = B. Therefore, the equation is necessary for C = 90°, because if C is 90°, then the equation must hold. But it is not sufficient, because the equation can hold when C is not 90°, specifically when A = B. Therefore, the answer is B: Necessary but not sufficient condition.Final Answerboxed{B}</think>To determine whether the condition "cosA + sinA = cosB + sinB" in triangle ABC is a sufficient, necessary, both, or neither condition for angle C being 90 degrees, we start by analyzing the implications of this condition.1. Necessity Check: - If angle C is 90°, then A + B = 90°. Substituting B = 90° - A into the equation: [ cos A + sin A = cos(90° - A) + sin(90° - A) ] - Using trigonometric identities, this simplifies to: [ cos A + sin A = sin A + cos A ] - Both sides are equal, confirming that if C = 90°, the condition holds. Thus, the condition is necessary.2. Sufficiency Check: - Assume the condition holds: (cos A + sin A = cos B + sin B). - Squaring both sides and simplifying using trigonometric identities: [ sin 2A = sin 2B ] - This implies either (2A = 2B) (i.e., (A = B)) or (2A = 180° - 2B) (i.e., (A + B = 90°)). - If (A = B), angle C can be any value (e.g., if (A = B = 30°), then (C = 120°)), not necessarily 90°. - If (A + B = 90°), then (C = 90°). However, since (A = B) does not guarantee (C = 90°), the condition is not sufficient.Thus, the condition is necessary but not sufficient for angle C to be 90°. The answer is (boxed{B}).,<think>Okay, so I need to figure out whether the condition "cosA + sinA = cosB + sinB" in triangle ABC is sufficient, necessary, both, or neither for the angle C being 90 degrees. Let me start by recalling some trigonometric identities and properties of triangles.First, in any triangle, the sum of the angles is 180 degrees. So, A + B + C = 180°. If C is 90°, then A + B = 90°, meaning angles A and B are complementary. That could be useful.The given condition is cosA + sinA = cosB + sinB. I need to analyze what this equation implies about angles A and B. Let me try squaring both sides to see if that helps. Sometimes squaring helps simplify expressions involving sine and cosine.So, if I square both sides:(cosA + sinA)^2 = (cosB + sinB)^2Expanding both sides:cos²A + 2sinA cosA + sin²A = cos²B + 2sinB cosB + sin²BSimplify using the identity cos²θ + sin²θ = 1:1 + 2sinA cosA = 1 + 2sinB cosBSubtract 1 from both sides:2sinA cosA = 2sinB cosBDivide both sides by 2:sinA cosA = sinB cosBThis simplifies to:(1/2) sin2A = (1/2) sin2BMultiply both sides by 2:sin2A = sin2BSo, sin2A = sin2B. Now, when is sinα = sinβ? That happens when α = β + 2πn or α = π - β + 2πn for some integer n. But since we're dealing with angles in a triangle, which are between 0 and π radians (0° and 180°), we can ignore the 2πn part. So, either:1. 2A = 2B + 2πn (but since angles are between 0 and π, 2A and 2B are between 0 and 2π. The only possibilities here are n=0 or n=1, but n=1 would make angles exceed 2π, so likely only n=0. So 2A = 2B ⇒ A = B.Or,2. 2A = π - 2B + 2πn. Again, considering the principal values, n=0:2A = π - 2B ⇒ 2A + 2B = π ⇒ A + B = π/2 (90°). But in a triangle, A + B + C = π, so if A + B = π/2, then C = π/2. That would make triangle ABC right-angled at C. So, sin2A = sin2B implies either A = B or A + B = π/2 (i.e., C = 90°). Therefore, the original condition cosA + sinA = cosB + sinB is satisfied in two cases: either A = B or C = 90°. Wait, but the problem is stating whether this condition is a condition for "C = 90°". So, does the condition cosA + sinA = cosB + sinB imply that C is 90°, or can it also hold when A = B?If A = B, then in that case, angles A and B are equal, but angle C could be something else. For example, if A = B = 45°, then C would be 90°, but if A = B = 30°, then C would be 120°, which is not 90°. So, in the case where A = B, C could be 90°, but it could also be other angles. Therefore, the condition cosA + sinA = cosB + sinB can hold even when C is not 90°, such as in a triangle where A = B = 30°, and C = 120°. Let me check that example.If A = B = 30°, then cos30° + sin30° = (√3/2) + (1/2) ≈ 0.866 + 0.5 = 1.366. Similarly, cosB + sinB is the same, so cosA + sinA = cosB + sinB holds. But angle C here is 120°, which is not 90°. Therefore, the condition can hold even when C is not 90°, which means that the condition is not necessary for C = 90°, because even if C is 90°, the condition might hold, but the condition can also hold in other cases. Wait, actually, if C is 90°, then A + B = 90°, so in that case, the condition could hold if either A = B (so each is 45°) or if A + B = 90°, but sin2A = sin2B. Let me think again.Wait, when C is 90°, A + B = 90°, so B = 90° - A. Then, let's compute cosA + sinA and compare it with cosB + sinB, which is cos(90° - A) + sin(90° - A). cos(90° - A) = sinA and sin(90° - A) = cosA. Therefore, cosB + sinB = sinA + cosA. Therefore, cosA + sinA = cosB + sinB automatically holds when C = 90°, regardless of the values of A and B (as long as they add up to 90°). But in this case, even if A and B are different (for example, A = 30°, B = 60°), cosA + sinA ≈ 0.866 + 0.5 = 1.366, and cosB + sinB ≈ 0.5 + 0.866 = 1.366, so they are equal. Therefore, if C is 90°, then the condition holds. So C = 90° implies the condition. However, the condition can also hold when A = B even if C is not 90°, such as in the case where A = B = 30°, C = 120°. Therefore, the condition is both implied by C = 90°, but can also hold in other cases.Therefore, the condition is necessary for C = 90°? Wait, if C = 90°, then the condition must hold. So, necessity would mean that if C = 90°, then the condition is true, which is the case. However, sufficiency would mean that if the condition holds, then C must be 90°, which is not the case because we can have the condition hold when A = B. Therefore, the condition is a necessary condition for C = 90°, but it is not sufficient, because there are other cases where the condition holds but C is not 90°. So that would make the answer B: Necessary but not sufficient condition.Wait, but wait. Let me double-check. If the condition is necessary, that means that C = 90° can only happen if the condition holds. Which is true, because when C = 90°, we must have cosA + sinA = cosB + sinB. So, necessity holds. However, sufficiency would mean that whenever the condition holds, then C must be 90°, which is not the case because if A = B, then the condition holds regardless of C. So, the condition is necessary but not sufficient. Therefore, the answer is B.But wait, let me think again. Let's see. Suppose that in a triangle, cosA + sinA = cosB + sinB. Then, as we saw, either A = B or A + B = 90°. So, if the condition holds, then either A = B or C = 90°. Therefore, the condition holds if C = 90° (since that gives A + B = 90°, leading to the equality), or if A = B. So, the condition is necessary for C = 90° (since if C = 90°, the condition must hold), but it's not sufficient because the condition can hold in another case (A = B). Therefore, the answer is B: Necessary but not sufficient condition.Alternatively, maybe the answer is C: Necessary and sufficient condition. Wait, no, because there are cases where the condition holds but C is not 90°, as in the A = B case. Therefore, sufficiency fails.Therefore, the correct answer is B: Necessary but not sufficient condition.Wait, but wait, let's confirm. Suppose we have a triangle where C = 90°, then the condition holds. So, the condition is necessary. But if the condition holds, C may or may not be 90°. So, it's necessary but not sufficient. Hence, answer B.But maybe I made a mistake. Let me check again.Suppose that in a triangle, cosA + sinA = cosB + sinB. Then, sin2A = sin2B. Therefore, either 2A = 2B or 2A + 2B = π. Therefore, A = B or A + B = π/2 (i.e., C = π/2). So, the condition holds in two cases: when A = B or when C is right angle. Therefore, if the condition holds, C might be 90° or not. So, the condition is not sufficient to conclude that C is 90°, because there's another possibility (A = B). However, if C is 90°, then the condition must hold. Therefore, the condition is necessary for C = 90°, but not sufficient. Therefore, the answer is B.Alternatively, if someone thought that when A + B = π/2, then C = π/2, which is correct, but when A = B, C could be other angles. Therefore, the condition is not sufficient. Hence, answer B.Alternatively, maybe in the case where A = B, C is 90°, but no, as I saw earlier, if A = B = 30°, then C is 120°, which is not 90°, but cosA + sinA = cosB + sinB still holds. Therefore, the condition can hold even if C is not 90°, hence the condition is not sufficient. Therefore, the answer is B: Necessary but not sufficient condition.Final Answerboxed{B}</think>Given the condition "cosA + sinA = cosB + sinB" in triangle ABC, we need to determine if this is a sufficient, necessary, both, or neither condition for angle C being 90°.First, we square both sides of the given condition:[(cos A + sin A)^2 = (cos B + sin B)^2]Expanding both sides and using the Pythagorean identity:[cos^2 A + 2sin A cos A + sin^2 A = cos^2 B + 2sin B cos B + sin^2 B]Simplifying, we get:[1 + 2sin A cos A = 1 + 2sin B cos B]Subtracting 1 from both sides and dividing by 2:[sin A cos A = sin B cos B]Using the double-angle identity, this becomes:[frac{1}{2}sin 2A = frac{1}{2}sin 2B]Thus:[sin 2A = sin 2B]This implies two possibilities:1. (2A = 2B) (i.e., (A = B))2. (2A = pi - 2B) (i.e., (A + B = frac{pi}{2}))If (A + B = frac{pi}{2}), then angle (C = frac{pi}{2}) (90°), making the triangle right-angled at C. However, if (A = B), angle C can be any value such that (A + B + C = pi), which does not necessarily make C 90°.Therefore, the condition (cos A + sin A = cos B + sin B) is necessary for (C = 90^circ) (since if (C = 90^circ), the condition must hold), but it is not sufficient (since the condition can hold when (A = B) even if (C neq 90^circ)).Thus, the correct answer is (boxed{B}).]

🔑:When "C = 90°" holds, we have A + B = 90°, so "cosA + sinA = cosB + sinB" must hold. However, when A = B, "cosA + sinA = cosB + sinB" also holds, which means "cosA + sinA = cosB + sinB" does not imply "C = 90°". Therefore, "cosA + sinA = cosB + sinB" is a necessary but not sufficient condition for "C = 90°". Thus, the correct choice is B.Analysis: To determine the relationship between the two conditions, "cosA + sinA = cosB + sinB" and "C = 90°", we need to consider the definitions of trigonometric functions and evaluate the options accordingly.boxed{text{B}}