🧠:[<think>Okay, so I have this inequality to solve:sqrt{frac{pi}{4}-operatorname{arctg} frac{|x|+|y|}{pi}}+operatorname{tg}^{2} x+1 leqslant sqrt{2}|operatorname{tg} x|(sin x+cos x)Hmm, first of all, let's make sure I understand all the components here. The left side has a square root term involving arctangent of (|x| + |y|)/π, then there's tangent squared x plus 1. The right side is sqrt(2) times the absolute value of tangent x multiplied by (sin x + cos x). Wait, but the problem mentions 9・13. Maybe that's a problem number or something, not sure if relevant. Anyway, let's focus on the inequality itself.First, I need to figure out the domain where all the functions are defined. The square root term on the left: sqrt(π/4 - arctg[(|x| + |y|)/π]). For the square root to be real, the expression inside must be non-negative. So:π/4 - arctg[(|x| + |y|)/π] ≥ 0Which implies:arctg[(|x| + |y|)/π] ≤ π/4Since arctg is an increasing function, applying tangent to both sides (but since arctg is ≤ π/4, then its argument must be ≤ tan(π/4) = 1. Therefore:(|x| + |y|)/π ≤ 1So |x| + |y| ≤ πThat's one condition. Then, also, tangent squared x is present, so we need to avoid x where cos x = 0, i.e., x ≠ π/2 + kπ for integers k. Similarly, the right side has tangent x and sin x + cos x. Let's note that sin x + cos x can be written as sqrt(2) sin(x + π/4), which might be helpful later.But first, maybe check if the inequality simplifies or has some symmetries. Let's consider substitution variables. Let me note that the left side includes arctg[(|x| + |y|)/π], and the right side involves |tg x| and sin x + cos x. But the left side also has tg²x. So maybe x is a key variable here, and y is involved only in that arctg term. Wait, but the inequality is in two variables x and y? The problem says "solve the inequality," but doesn't specify variables. Wait, the original problem is written as 9・13, maybe implying it's a problem from a textbook? Maybe the problem is to solve for x and y? Or is it for (x, y) pairs? Let me check the original problem statement again."Solve the inequality sqrt(π/4 - arctg(|x| + |y|)/π) + tg²x + 1 ≤ sqrt(2)|tg x|(sin x + cos x)"Hmm. So variables x and y are both present. So the solution will be a set of (x, y) pairs that satisfy this inequality. But how do we approach this?Alternatively, perhaps there's a typo in the problem? Maybe the original problem was supposed to have only x, but maybe |x| + |y| is a term. Wait, let me check the LaTeX code again:sqrt{frac{pi}{4}-operatorname{arctg} frac{|x|+|y|}{pi}}+operatorname{tg}^{2} x+1 leqslant sqrt{2}|operatorname{tg} x|(sin x+cos x)Yes, so it's definitely |x| + |y| divided by π inside arctg. So y is present here. So the inequality involves both x and y. Hmm. Then, solving this inequality would require finding all pairs (x, y) such that this holds. But maybe there's a relationship between x and y that can be derived?Alternatively, perhaps the problem is from a specific textbook where the preceding problems set some context? Since I don't have that context, maybe I need to proceed purely based on the given expression.First, note that the left-hand side (LHS) has three terms: a square root term involving y, and two terms involving x (tg²x +1). The right-hand side (RHS) is also a function of x. So perhaps, for the inequality to hold, the terms involving x on both sides must balance, and the square root term involving y must be small enough? Because if we can bound the LHS in terms of x and y, maybe we can find constraints on y given x, or vice versa.Alternatively, maybe the equation can only hold when certain conditions on x are met, which would allow the square root term to be real and the inequality to be satisfied.Let's first analyze the square root term on the LHS:sqrt(π/4 - arctg[(|x| + |y|)/π])As we determined earlier, the expression inside the square root is non-negative only when arctg[(|x| + |y|)/π] ≤ π/4, which implies (|x| + |y|)/π ≤ tan(π/4) = 1, so |x| + |y| ≤ π. So that's a necessary condition for the inequality to be defined.Therefore, the domain of definition is all (x, y) such that |x| + |y| ≤ π, and x ≠ π/2 + kπ, where k is an integer (due to the tangent functions).So our first step is to consider |x| + |y| ≤ π and x ≠ π/2 + kπ.Now, let's try to analyze the inequality. Let me denote:Let’s denote A = sqrt(π/4 - arctg[(|x| + |y|)/π])Then the inequality is:A + tg²x + 1 ≤ sqrt(2) |tg x| (sin x + cos x)Note that A is non-negative, as it's a square root. The terms tg²x +1 are also non-negative (since tg²x is non-negative and 1 is positive). The RHS is sqrt(2) |tg x| (sin x + cos x). Let's see if we can analyze the RHS.First, let's note that sin x + cos x can be rewritten as sqrt(2) sin(x + π/4). Therefore, the RHS becomes sqrt(2) |tg x| * sqrt(2) sin(x + π/4) / sqrt(2)? Wait, no. Wait:Wait, sin x + cos x = sqrt(2) sin(x + π/4). So then:sqrt(2) |tg x| (sin x + cos x) = sqrt(2) |tg x| * sqrt(2) sin(x + π/4) = 2 |tg x| sin(x + π/4)Wait, but sin(x + π/4) has a maximum value of 1 and minimum of -1, but since we have |tg x| multiplied by sin(x + π/4), which could be positive or negative. However, since we have absolute value on tg x, but sin(x + π/4) is not in absolute value. Wait, but in the RHS, it's sqrt(2)|tg x|(sin x + cos x). Since sin x + cos x can be written as sqrt(2) sin(x + π/4), but depending on x, sin(x + π/4) can be positive or negative. However, the LHS is non-negative (sqrt term + tg²x +1), so the RHS must also be non-negative. Therefore, we can infer that sin x + cos x must be non-negative, because |tg x| is non-negative, sqrt(2) is positive. Therefore, sin x + cos x ≥ 0.So sin x + cos x ≥ 0 => x + π/4 ∈ [ -π/2 + 2πk, π/2 + 2πk ] => x ∈ [ -3π/4 + 2πk, π/4 + 2πk ] for integers k.But since x is restricted by |x| + |y| ≤ π, which complicates things. Alternatively, maybe we can consider specific intervals for x where sin x + cos x is non-negative, but perhaps we can handle this later.Alternatively, let's consider substituting variables. Let’s set t = tg x. Then, since tg x = t, we can express sin x and cos x in terms of t. But we have to be careful with the signs. Let me note that tg x = t implies that sin x = t / sqrt(1 + t²) and cos x = 1 / sqrt(1 + t²). However, the sign of sin x and cos x depends on the quadrant of x. But since we have |tg x|, which is |t|, so maybe we can write sin x + cos x as (t + 1)/sqrt(1 + t²) if we consider x in the first quadrant, but the signs might vary.Alternatively, perhaps square both sides? Although that might complicate things, since we have multiple terms. Let me see:Original inequality:sqrt(π/4 - arctg[(|x| + |y|)/π]) + t² + 1 ≤ sqrt(2) |t| (sin x + cos x)But if I denote S = sin x + cos x, then:sqrt(π/4 - arctg[(|x| + |y|)/π]) + t² + 1 ≤ sqrt(2) |t| SBut S = sin x + cos x = sqrt(2) sin(x + π/4), as we mentioned before. So S ranges between -sqrt(2) and sqrt(2). However, since the LHS is non-negative, and sqrt(2)|t| S must also be non-negative. Therefore, S must be non-negative. Therefore, sin(x + π/4) ≥ 0. So x + π/4 ∈ [2πk, π + 2πk] => x ∈ [-π/4 + 2πk, 3π/4 + 2πk] for integers k. So x is in these intervals where sin(x + π/4) is non-negative. Therefore, S ≥ 0.Therefore, the inequality simplifies to sqrt(π/4 - arctg[(|x| + |y|)/π}) + t² + 1 ≤ sqrt(2) |t| S, where S = sin x + cos x ≥ 0.Alternatively, since S = sqrt(2) sin(x + π/4), which is between 0 and sqrt(2) in these intervals.But perhaps there's another approach. Let's consider that both sides of the inequality involve terms with tg x. Let me consider making a substitution. Let’s let t = tg x. Then, as mentioned, sin x = t / sqrt(1 + t²) and cos x = 1 / sqrt(1 + t²), but depending on the quadrant. However, since we have |t|, and S = sin x + cos x. Let me compute S in terms of t:sin x + cos x = (t / sqrt(1 + t²)) + (1 / sqrt(1 + t²)) = (t + 1)/sqrt(1 + t²)But if x is in a quadrant where cos x is negative, then sqrt(1 + t²) would have the sign of cos x? Wait, no. When we write sin x = t / sqrt(1 + t²) and cos x = 1 / sqrt(1 + t²), this is only valid when cos x is positive, i.e., x in (-π/2 + 2πk, π/2 + 2πk). But if cos x is negative, then tg x would be positive if sin x is negative (in third quadrant), or negative if sin x is positive (second quadrant). This complicates things. However, since we have |tg x|, which is |t|, and S = sin x + cos x. But S can be written as sqrt(2) sin(x + π/4), which is non-negative, as we established. So x is in [-π/4 + 2πk, 3π/4 + 2πk]. Therefore, in these intervals, cos x is positive in [-π/4, π/4], but in [π/4, 3π/4], cos x is positive only up to π/2, then negative. Wait, actually, x ∈ [-π/4, 3π/4]. Let's consider the interval x ∈ [-π/4, 3π/4].In this interval, cos x is positive in [-π/4, π/2) and negative in (π/2, 3π/4]. At x = π/2, cos x is zero, but x = π/2 is excluded from the domain (since tg x is undefined there). So in [-π/4, π/2), cos x is positive, and in (π/2, 3π/4], cos x is negative. However, sin x + cos x is non-negative in this interval. Let's check:At x = -π/4, sin(-π/4) + cos(-π/4) = -√2/2 + √2/2 = 0.At x = 0, sin 0 + cos 0 = 0 + 1 = 1.At x = π/4, sin(π/4) + cos(π/4) = √2/2 + √2/2 = √2.At x = π/2, undefined.At x = 3π/4, sin(3π/4) + cos(3π/4) = √2/2 - √2/2 = 0.So in the interval [-π/4, 3π/4], sin x + cos x ranges from 0 up to √2 and back to 0.So perhaps we can split the interval into two parts: [-π/4, π/2) and (π/2, 3π/4]. However, in (π/2, 3π/4], cos x is negative, so tg x is negative (since sin x is positive there), so |tg x| = -tg x. But S = sin x + cos x is positive here as well. Let's check x = 3π/4: sin(3π/4) + cos(3π/4) = √2/2 - √2/2 = 0, which is the boundary. So in (π/2, 3π/4), sin x + cos x is positive but decreasing from √2/2 - 0 to 0. Wait, let's take x = 2π/3, which is 120 degrees. sin(2π/3) = √3/2, cos(2π/3) = -1/2. So sin x + cos x = √3/2 - 1/2 ≈ 0.866 - 0.5 = 0.366 > 0. So in (π/2, 3π/4), sin x + cos x is still positive, but cos x is negative.Therefore, in both subintervals, S = sin x + cos x is non-negative. So perhaps it's manageable.But returning to the substitution idea. Let's let t = tg x, then:If x ∈ (-π/2, π/2), then t is real, and sin x = t / sqrt(1 + t²), cos x = 1 / sqrt(1 + t²). However, since x is restricted to [-π/4, 3π/4], but excluding π/2. So in [-π/4, π/2), t = tg x is defined and can be from tg(-π/4) = -1 to tg(π/2) approaches infinity. Wait, but x ∈ [-π/4, 3π/4], so in [-π/4, π/2), t ranges from -1 to +infty. In (π/2, 3π/4], x would be between π/2 and 3π/4, where tg x is negative (since in the second quadrant), so t = tg x would range from -infty to -1 (since tg(3π/4) = -1).But perhaps since we have |t|, let's consider |t|, which would be from 1 to infinity in both intervals. Wait, in [-π/4, π/2):- x ∈ [-π/4, 0): t = tg x ∈ [-1, 0), so |t| ∈ (0, 1]- x ∈ [0, π/2): t = tg x ∈ [0, ∞), so |t| ∈ [0, ∞)In (π/2, 3π/4]:- x ∈ (π/2, 3π/4): t = tg x ∈ (-∞, -1), so |t| ∈ (1, ∞)Therefore, |t| ranges from 0 to ∞, but depending on the interval. However, in the interval where S = sin x + cos x ≥ 0, which is [-π/4, 3π/4], excluding π/2.But perhaps instead of substitution, let's try to analyze the inequality in parts.The left-hand side (LHS) is sqrt(π/4 - arctg[(|x| + |y|)/π}) + tg²x + 1.The right-hand side (RHS) is sqrt(2) |tg x| (sin x + cos x).We need to find (x, y) such that |x| + |y| ≤ π and the inequality holds.Since the LHS includes a term involving y (inside arctg), and the RHS does not involve y, except through the domain condition |x| + |y| ≤ π. Therefore, perhaps for a given x, the term sqrt(π/4 - arctg[(|x| + |y|)/π}) can be minimized or maximized based on y, allowing us to find possible y for each x.But since we have an inequality, perhaps the maximum possible value of the LHS must be less than or equal to the RHS. Therefore, to satisfy the inequality for some y, the minimum possible RHS must be greater than or equal to the maximum possible LHS. But this is getting a bit abstract.Alternatively, since the term involving y is sqrt(π/4 - arctg[(|x| + |y|)/π}), and |x| + |y| ≤ π, then arctg[(|x| + |y|)/π] is in [0, arctg(1)) = [0, π/4). Because (|x| + |y|)/π is in [0,1], as |x| + |y| ≤ π. Therefore, arctg of that is in [0, π/4). Therefore, π/4 - arctg[(|x| + |y|)/π} is in (0, π/4]. Therefore, the square root term is in (0, sqrt(π/4)].Therefore, sqrt(π/4 - arctg[(|x| + |y|)/π}) is a positive term, but bounded above by sqrt(π/4) ≈ 0.886.So the LHS is:sqrt(π/4 - arctg[(|x| + |y|)/π}) + tg²x + 1 ≥ tg²x + 1, since the sqrt term is positive.So LHS ≥ tg²x + 1. The RHS is sqrt(2) |tg x| (sin x + cos x). Therefore, we must have tg²x + 1 ≤ sqrt(2) |tg x| (sin x + cos x).But tg²x + 1 = sec²x, which is always greater or equal to 1. The RHS is sqrt(2) |tg x| (sin x + cos x). Let's see if this can be greater or equal to sec²x.Alternatively, maybe we can consider equality conditions. Suppose that the sqrt term is zero. Then the inequality reduces to sec²x ≤ sqrt(2) |tg x| (sin x + cos x). But sqrt(π/4 - arctg[(|x| + |y|)/π}) = 0 only when arctg[(|x| + |y|)/π} = π/4, which implies (|x| + |y|)/π = tan(π/4) = 1, so |x| + |y| = π. Therefore, when |x| + |y| = π, the sqrt term is zero, and the inequality becomes sec²x ≤ sqrt(2) |tg x| (sin x + cos x). Let's see if this is possible.But sec²x = 1 + tg²x. So the inequality becomes 1 + tg²x ≤ sqrt(2) |tg x| (sin x + cos x). Let’s denote t = |tg x|, since |tg x| is non-negative. Then, since sin x + cos x can be expressed as sqrt(2) sin(x + π/4), which is non-negative in our domain. Let’s write sin x + cos x = sqrt(2) sin(x + π/4). Then the inequality becomes:1 + t² ≤ sqrt(2) * t * sqrt(2) sin(x + π/4) = 2 t sin(x + π/4)But sin(x + π/4) is in [0, 1] because x is in [-π/4, 3π/4], so x + π/4 is in [0, π], where sine is non-negative. Therefore, sin(x + π/4) is in [0,1].So the inequality is 1 + t² ≤ 2 t sin(x + π/4)But sin(x + π/4) ≤ 1, so 2 t sin(x + π/4) ≤ 2 t. Therefore, 1 + t² ≤ 2 t. But 1 + t² ≤ 2 t implies t² - 2 t + 1 ≤ 0, which is (t - 1)^2 ≤ 0. Therefore, t = 1. So equality holds only when t = 1 and sin(x + π/4) = 1.So, if t = |tg x| = 1, then tg x = ±1. Therefore, x = π/4 + kπ. But in our domain x ∈ [-π/4, 3π/4] (excluding π/2), so x = π/4 or x = -π/4. But sin(x + π/4) = 1 when x + π/4 = π/2 + 2πk, i.e., x = π/4 + 2πk. In our interval x ∈ [-π/4, 3π/4], the solution is x = π/4.So when x = π/4, sin(x + π/4) = sin(π/2) = 1, and |tg x| = tg(π/4) = 1. Therefore, plugging into the inequality:1 + 1² = 2 ≤ 2*1*1 = 2, which holds as equality. Similarly, if x = -π/4, then sin(x + π/4) = sin(0) = 0, but then the RHS would be 0, which is less than 1 + 1 = 2, which does not hold. Therefore, equality holds only at x = π/4, with |tg x| = 1, sin(x + π/4) = 1.Therefore, at the boundary where |x| + |y| = π, the inequality reduces to equality only at x = π/4, y such that |x| + |y| = π. Since x = π/4, |y| = π - π/4 = 3π/4. Therefore, y = ±3π/4. So the points (π/4, ±3π/4) and (-π/4, ±3π/4). Wait, but x = π/4 and |x| + |y| = π, so |y| = π - π/4 = 3π/4. But x can be positive or negative. Wait, but if x is negative, say x = -π/4, then |x| + |y| = π implies |y| = π - π/4 = 3π/4. However, in such a case, does the inequality hold?Wait, when x = -π/4, let's check:tg x = tg(-π/4) = -1, so |tg x| = 1. Then the RHS is sqrt(2)*1*(sin(-π/4) + cos(-π/4)) = sqrt(2)*( -sqrt(2)/2 + sqrt(2)/2 ) = sqrt(2)*0 = 0. But the LHS is sqrt(π/4 - arctg[(π/4 + |y|)/π}) + tg²(-π/4) + 1. But if |x| + |y| = π, then |y| = 3π/4. So arctg[(|x| + |y|)/π] = arctg(1) = π/4. Therefore, sqrt(π/4 - π/4) = 0. So LHS is 0 + 1 + 1 = 2. So 2 ≤ 0, which is false. Therefore, even though when x = -π/4 and |x| + |y| = π, the sqrt term is zero, but the inequality doesn't hold because RHS is zero. Therefore, only when x = π/4 and |x| + |y| = π does the inequality hold as equality.Therefore, the equality case occurs at (x, y) = (π/4, ±3π/4) and (-π/4, ±3π/4) only if the RHS is equal to LHS, but we saw that for x = -π/4, it's not the case. Wait, maybe only x positive? Let's check.If x = π/4, |x| + |y| = π => |y| = 3π/4, so y = ±3π/4. Then, the RHS is sqrt(2)*1*(sin(π/4) + cos(π/4)) = sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2) = sqrt(2)*sqrt(2) = 2. The LHS is 0 + 1 + 1 = 2, so equality holds. If x = -π/4, |x| + |y| = π => |y| = 3π/4. Then, the RHS is sqrt(2)*1*(sin(-π/4) + cos(-π/4)) = sqrt(2)*( -sqrt(2)/2 + sqrt(2)/2 ) = 0. LHS is 0 + 1 + 1 = 2, so 2 ≤ 0 is false. Therefore, equality holds only at (π/4, ±3π/4). Similarly, if x = -π/4, but |x| + |y| = π, but since x is negative, but |x| is positive, so |y| = π - |x|, so y is still ±(3π/4). But in that case, the RHS is zero, which is less than LHS. Therefore, only the points where x = π/4 and y = ±3π/4 satisfy the equality.Now, moving beyond the equality case, we need to consider when the LHS is less than or equal to RHS. Since the LHS includes the sqrt term which depends on y, we need to consider how y affects the LHS. Since arctg[(|x| + |y|)/π} increases as |x| + |y| increases, so π/4 - arctg[(|x| + |y|)/π} decreases as |x| + |y| increases. Therefore, the sqrt term sqrt(π/4 - arctg[(|x| + |y|)/π}) decreases as |x| + |y| increases.Therefore, to minimize the LHS, we need to maximize |x| + |y|, which would minimize the sqrt term. But since |x| + |y| ≤ π, the maximum |x| + |y| is π, which sets the sqrt term to zero, leading to minimal LHS. Conversely, to maximize the LHS, set |x| + |y| as small as possible (i.e., approaching zero), making arctg[(|x| + |y|)/π} approach arctg(0) = 0, so sqrt(π/4 - 0) = sqrt(π/4) = sqrt(π)/2 ≈ 0.886. Therefore, the LHS can vary between approximately (sqrt(π)/2 + tg²x + 1) and (0 + tg²x + 1).Therefore, for the inequality to hold for some y, the minimal possible LHS (when |x| + |y| = π) must be ≤ RHS. But we saw that when |x| + |y| = π, the LHS is tg²x +1 +0 = sec²x, and the inequality reduces to sec²x ≤ sqrt(2) |tg x| (sin x + cos x). As we saw earlier, this is only possible when x = π/4, |tg x| =1, and sin x + cos x is maximized at sqrt(2). Therefore, in that case, equality holds.But for other values of x, if we take |x| + |y| = π, the inequality sec²x ≤ sqrt(2) |tg x| (sin x + cos x) must hold. Let's analyze this inequality in more depth.Let’s denote t = |tg x|. Then, sec²x = 1 + t². The RHS is sqrt(2) t (sin x + cos x). Let’s express sin x + cos x in terms of t.Since t = |tg x| = |sin x / cos x|, so sin x = t cos x if x is in the first or third quadrants. But in our domain x ∈ [-π/4, 3π/4], excluding π/2. So if x is in [-π/4, π/2), cos x is positive, so |tg x| = |sin x| / cos x. If x is in (π/2, 3π/4], cos x is negative, and sin x is positive, so |tg x| = -sin x / cos x.But maybe we can use the identity sin x + cos x = sqrt(2) sin(x + π/4). So, sin(x + π/4) = (sin x + cos x)/sqrt(2). Therefore, sin x + cos x = sqrt(2) sin(x + π/4). So substituting into the RHS:sqrt(2) t sin(x + π/4) sqrt(2) = 2 t sin(x + π/4). Wait, no, original RHS is sqrt(2) t (sin x + cos x) = sqrt(2) t * sqrt(2) sin(x + π/4) = 2 t sin(x + π/4). Therefore, the inequality becomes:1 + t² ≤ 2 t sin(x + π/4)But since x ∈ [-π/4, 3π/4], then x + π/4 ∈ [0, π]. Therefore, sin(x + π/4) ∈ [0, 1]. Let’s denote s = sin(x + π/4), so s ∈ [0,1]. Then the inequality becomes:1 + t² ≤ 2 t sBut t = |tg x|. Let's relate t and s. Since x is related to s. Let me see.Let’s let θ = x + π/4, so θ ∈ [0, π]. Then s = sin θ. Also, x = θ - π/4. Therefore, t = |tg(x)| = |tg(θ - π/4)|. Let’s compute tg(θ - π/4):tg(θ - π/4) = [tg θ - tg(π/4)] / [1 + tg θ tg(π/4)] = (tg θ - 1)/(1 + tg θ)But θ ∈ [0, π], so tg θ is defined except at θ = π/2. Therefore, tg(θ - π/4) = (tg θ - 1)/(1 + tg θ). Therefore, |tg(x)| = |(tg θ - 1)/(1 + tg θ)|.But since θ ∈ [0, π], tg θ can be positive or negative. Let's analyze:- When θ ∈ [0, π/2): tg θ ≥ 0, so |tg(x)| = |(tg θ - 1)/(1 + tg θ)|- When θ ∈ (π/2, π]: tg θ ≤ 0, so |tg(x)| = |(negative -1)/(1 + negative)|Let’s compute for θ ∈ [0, π/2):t = |(tg θ - 1)/(1 + tg θ)|Let’s denote k = tg θ, then k ≥ 0. Then:t = |(k -1)/(1 + k)|Depending on k:- If k ≥1: (k -1)/(1 + k) ≥0, so t = (k -1)/(1 + k)- If k <1: (k -1)/(1 + k) <0, so t = (1 - k)/(1 + k)Similarly, for θ ∈ (π/2, π], tg θ = m ≤ 0:t = |(m -1)/(1 + m)|But m ≤0:Let’s compute (m -1)/(1 + m). Let m = -n, n ≥0:t = |(-n -1)/(1 -n)| = |(- (n +1))/(1 - n)|If n <1: denominator positive, numerator negative: t = (n +1)/(1 - n)If n >1: denominator negative, numerator negative: t = (n +1)/(n -1)At n =1: undefined (since θ = π/2, x = π/2, which is excluded).Therefore, this seems complicated. Maybe instead of trying to relate t and s directly, let's use another approach.Recall that we have:1 + t² ≤ 2 t sBut s = sin θ, where θ = x + π/4 ∈ [0, π]. And t = |tg(x)| = |tg(θ - π/4)|.Alternatively, let's square both sides. Wait, but squaring inequalities can be tricky. However, both sides are non-negative here.If we square both sides:(1 + t²)^2 ≤ (2 t s)^2=> 1 + 2 t² + t⁴ ≤ 4 t² s²But this seems even more complicated. Alternatively, let's consider rearranging the inequality:1 + t² - 2 t s ≤ 0Which is a quadratic in t:t² - 2 s t + 1 ≤ 0The quadratic equation t² - 2 s t + 1 =0 has discriminant (2 s)^2 -4*1*1= 4 s² -4 = 4(s² -1). Therefore, the roots are t = [2 s ± sqrt(4(s² -1))]/2 = s ± sqrt(s² -1)But s ∈ [0,1], so s² -1 ≤0. Therefore, the discriminant is non-positive, so the quadratic is always positive or zero. Hence, t² - 2 s t +1 ≥0 for all real t, which means the inequality t² - 2 s t +1 ≤0 is only satisfied when equality holds, i.e., when s² -1 =0 => s=±1. But s ∈ [0,1], so s=1. Then t² - 2*1*t +1 = (t -1)^2 ≤0, which implies t=1. Therefore, the inequality 1 + t² ≤ 2 t s holds only when s=1 and t=1, which corresponds to θ=π/2 (since s=sinθ=1 => θ=π/2), so x=θ - π/4=π/2 - π/4=π/4, which is the case we saw earlier. Therefore, this confirms that the inequality sec²x ≤ sqrt(2) |tg x| (sin x + cos x) holds only at x=π/4 with tg x=1, and s=1.Therefore, returning to the original problem. The LHS is minimized when |x| + |y|=π, and in that case, the inequality reduces to equality only at x=π/4, y=±3π/4. For other x, if we take |x| + |y|=π, the inequality is not satisfied. Therefore, to have the original inequality hold, we must have that for a given x, the sum sqrt(π/4 - arctg[(|x| + |y|)/π}) + sec²x is less than or equal to sqrt(2) |tg x| (sin x + cos x).But since the sqrt term is positive unless |x| + |y|=π, the only way for the inequality to hold is if the term sqrt(π/4 - arctg[...}) is small enough and the RHS is large enough.But given that for all other x ≠ π/4, when |x| + |y| = π, the inequality is not satisfied, but for |x| + |y| < π, the sqrt term increases, making the LHS larger. Therefore, it's possible that the inequality can only be satisfied at the point where the sqrt term is zero and the RHS is maximized, which is at x=π/4, y=±3π/4. For other points, the LHS is larger, and the RHS is smaller, so the inequality would not hold.Alternatively, maybe there are other points where the sqrt term is small enough and the RHS is large enough to compensate. Let's check for x near π/4.Suppose x is slightly less than π/4, say x=π/4 - ε, where ε is a small positive number. Then tg x is slightly less than 1, |tg x| =1 - δ for small δ. sin x + cos x would be slightly less than sqrt(2). Let's compute:LHS: sqrt(π/4 - arctg[(|x| + |y|)/π}) + tg²x +1To minimize LHS, we need |x| + |y|=π. But even with that, the LHS is sec²x = 1 + tg²x. So if x is slightly less than π/4, tg x ≈1 - δ, so sec²x ≈1 + (1 - 2δ) ≈ 2 - 2δ. The RHS is sqrt(2) * (1 - δ) * (sin x + cos x). Let's compute sin x + cos x when x=π/4 - ε:sin(π/4 - ε) + cos(π/4 - ε) = sin π/4 cos ε - cos π/4 sin ε + cos π/4 cos ε + sin π/4 sin ε = sqrt(2)/2 [cos ε - sin ε + cos ε + sin ε] = sqrt(2)/2 * 2 cos ε = sqrt(2) cos ε ≈ sqrt(2)(1 - ε²/2)Therefore, RHS ≈ sqrt(2)*(1 - δ)*sqrt(2)(1 - ε²/2) ≈ 2*(1 - δ - ε²/2)Compare to LHS ≈ 2 - 2δ. So RHS ≈2 - 2δ - ε², which is less than LHS≈2 - 2δ. Therefore, RHS < LHS. Hence, the inequality would not hold for x slightly less than π/4 even if |x| + |y|=π.Similarly, if x is slightly more than π/4, but since x is restricted to less than 3π/4, but near π/4. Let's take x=π/4 + ε. Then tg x is slightly more than 1, so |tg x|=1 + δ. sin x + cos x would be sin(π/4 + ε) + cos(π/4 + ε) = sqrt(2)/2 [cos ε + sin ε + cos ε - sin ε] = sqrt(2)/2 * 2 cos ε = sqrt(2) cos ε ≈ sqrt(2)(1 - ε²/2). Then RHS is sqrt(2)*(1 + δ)*sqrt(2)(1 - ε²/2) ≈ 2*(1 + δ)(1 - ε²/2) ≈2*(1 + δ - ε²/2). LHS with |x| + |y|=π would be sec²x ≈1 + (1 + 2δ) ≈2 + 2δ. Comparing RHS≈2 + 2δ - ε² vs LHS≈2 + 2δ. Therefore, RHS < LHS again. Therefore, even for x slightly more than π/4, the inequality does not hold when |x| + |y|=π.Therefore, the only solution when |x| + |y|=π is x=π/4, y=±3π/4. Now, let's consider the case when |x| + |y| < π. In this case, the sqrt term is positive, so LHS is larger than sec²x. But earlier, we saw that sec²x is already greater than the RHS except at x=π/4. Therefore, if even the minimal LHS (sec²x) is greater than RHS except at x=π/4, then when |x| + |y| < π, LHS is larger, making the inequality even less likely to hold.Therefore, perhaps the only solutions are the equality cases: x=π/4, y=±3π/4 and x=-π/4, y=±3π/4. But we saw that for x=-π/4, even with |x| + |y|=π, the RHS is zero, so inequality does not hold. Therefore, only x=π/4, y=±3π/4 are solutions.But let's verify this. Take x=π/4, y=3π/4. Then |x| + |y|=π/4 + 3π/4=π. Then:sqrt(π/4 - arctg(π/π}) = sqrt(π/4 - arctg(1)) = sqrt(π/4 - π/4)=0.tg²x =1, so LHS=0 +1 +1=2.RHS= sqrt(2)*1*(sin(π/4) + cos(π/4))=sqrt(2)*(sqrt(2)/2 + sqrt(2)/2)=sqrt(2)*sqrt(2)=2. So equality holds.Another point: take x=0, y=0. Then |x| + |y|=0 < π. Compute LHS:sqrt(π/4 - arctg(0/π}) = sqrt(π/4 -0)=sqrt(π/4)=sqrt(π)/2 ≈0.886.tg²0 +1=0 +1=1. So LHS≈0.886 +1≈1.886.RHS= sqrt(2)*0*(sin0 + cos0)=0. Therefore, 1.886 ≤0 is false.Another test point: x=π/4, y=0. Then |x| + |y|=π/4 < π.LHS: sqrt(π/4 - arctg(π/4 / π)) + tg²(π/4) +1 = sqrt(π/4 - arctg(1/4)) +1 +1.Compute arctg(1/4)≈0.245. π/4≈0.785. So π/4 - arctg(1/4)≈0.54. sqrt(0.54)≈0.735. Therefore, LHS≈0.735 +1 +1≈2.735.RHS= sqrt(2)*1*(sin(π/4)+cos(π/4))=sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2)=sqrt(2)*sqrt(2)=2. Therefore, 2.735 ≤2 is false.Therefore, even near x=π/4, if |x| + |y| < π, the LHS is larger than RHS. Therefore, no solutions there.Another test point: x=π/4, y=π/2. Then |x| + |y|=π/4 + π/2=3π/4 < π. Therefore:LHS: sqrt(π/4 - arctg(3π/4 / π)) +1 +1 = sqrt(π/4 - arctg(3/4)) +2.Compute arctg(3/4)≈0.6435 radians. π/4≈0.7854. π/4 - arctg(3/4)≈0.7854 -0.6435≈0.1419. sqrt(0.1419)≈0.377. Therefore, LHS≈0.377 +2≈2.377. RHS=2. So 2.377 ≤2 is false.Therefore, even when y is large but |x| + |y| <π, the LHS is still larger than RHS.What if x=π/4, y=3π/4 - ε for small ε>0? Then |x| + |y|=π/4 +3π/4 -ε=π -ε. So arctg[(|x| + |y|)/π}=arctg(1 - ε/π)≈π/4 - something small. Therefore, sqrt(π/4 - arctg(...))≈sqrt(something small). So LHS≈sqrt(small) +1 +1≈0 +1 +1=2. RHS≈sqrt(2)*1*(sin(π/4)+cos(π/4))=2. So approximately, LHS≈2, RHS=2. But as ε approaches zero, the sqrt term approaches zero, and we approach the equality case. However, for ε>0, |x| + |y|=π -ε<π, so arctg(...) <π/4, so sqrt term >0. Therefore, LHS=2 + something, which is >2. Therefore, the inequality does not hold for y=3π/4 - ε. Therefore, only when |x| + |y|=π exactly does the LHS=2, which equals RHS=2. Therefore, the only solutions are points where |x| + |y|=π and x=π/4 (since x=-π/4 gives RHS=0). Therefore, the solution is x=π/4, y=±3π/4, and x=-π/4, y=±3π/4? Wait, but when x=-π/4, |x| + |y|=π/4 + |y|=π => |y|=3π/4. So y=±3π/4. Let's check x=-π/4, y=3π/4:LHS: sqrt(π/4 - arctg((π/4 +3π/4)/π)) + tg²(-π/4) +1 = sqrt(π/4 - arctg(1)) +1 +1 =0 +1 +1=2.RHS: sqrt(2)*|tg(-π/4)|*(sin(-π/4) + cos(-π/4))=sqrt(2)*1*(-sqrt(2)/2 + sqrt(2)/2)=sqrt(2)*0=0. So 2 ≤0? No. Therefore, even though |x| + |y|=π, x=-π/4 gives LHS=2 and RHS=0, which doesn't satisfy the inequality. Therefore, only x=π/4, y=±3π/4 are solutions.Therefore, the only solutions are the points where x=π/4 and y=±3π/4. So the solution set is {(π/4, 3π/4), (π/4, -3π/4)}.But wait, what about x=π/4 and y=3π/4, but also considering periodicity? The original inequality involves tg x, which has period π. However, the term |x| + |y| ≤ π complicates things, because if x is outside [-π, π], then |x| + |y| ≤π would restrict y accordingly. But since x is in the domain where tg x is defined (i.e., x ≠π/2 +kπ), and |x| + |y| ≤π, the primary solutions are within the principal domain. However, since trigonometric functions are periodic, we might have solutions in each period. But the condition |x| + |y| ≤π would restrict x and y to the region where their absolute values sum to at most π.However, considering the periodicity of tg x with period π, if we have a solution at x=π/4, then x=π/4 +kπ would also need to be considered. But |x| + |y| ≤π must hold. If k=1, x=5π/4. Then |x|=5π/4, which is already greater than π, so |x| + |y| ≥5π/4 >π, which violates the domain condition. Similarly, for k=-1, x=π/4 -π=-3π/4. Then |x|=3π/4, so |y| ≤π -3π/4=π/4. But let's check x=-3π/4, y=π/4:LHS: sqrt(π/4 - arctg(3π/4 +π/4)/π}=sqrt(π/4 - arctg(1))=0.tg²(-3π/4)=1, so LHS=0 +1 +1=2.RHS= sqrt(2)*|tg(-3π/4)|*(sin(-3π/4) + cos(-3π/4))=sqrt(2)*1*(-sqrt(2)/2 - sqrt(2)/2)=sqrt(2)*1*(-sqrt(2))=-2. But RHS is negative, and LHS is positive, so inequality 2 <= -2 is false. Therefore, even though |x| + |y|=π for x=-3π/4, y=π/4, the RHS is negative, so inequality doesn't hold. Therefore, there are no solutions in other periods due to the domain restriction |x| + |y| <=π.Therefore, the only solutions are the two points: (π/4, 3π/4) and (π/4, -3π/4).But let's confirm once more. Take x=π/4, y=3π/4:Compute each term:sqrt(π/4 - arctg(|π/4| + |3π/4|)/π) = sqrt(π/4 - arctg(π/π)) = sqrt(π/4 - arctg(1)) = sqrt(π/4 - π/4) = 0.tg²(π/4) +1 =1 +1=2.RHS= sqrt(2)*|tg(π/4)|*(sin(π/4)+cos(π/4))=sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2)=sqrt(2)*sqrt(2)=2.So 0 +2 <=2, which is true.For x=π/4, y=-3π/4:Same calculation, since it's absolute values, |y|=3π/4. So same result.Therefore, the only solutions are the two points where x=π/4 and y=±3π/4.Final AnswerThe solutions are (boxed{left(frac{pi}{4}, pmfrac{3pi}{4}right)}).</think>To solve the inequalitysqrt{frac{pi}{4}-operatorname{arctg} frac{|x|+|y|}{pi}}+operatorname{tg}^{2} x+1 leqslant sqrt{2}|operatorname{tg} x|(sin x+cos x),we first consider the domain of definition. The expression inside the square root must be non-negative, leading to the condition (|x| + |y| leq pi). Additionally, (x neq frac{pi}{2} + kpi) for integers (k) due to the tangent function.Next, we analyze the inequality by considering the terms involving (x) and (y). The left-hand side (LHS) includes a square root term involving (y), which is minimized when (|x| + |y| = pi). The right-hand side (RHS) involves (sqrt{2} |operatorname{tg} x| (sin x + cos x)), which must be non-negative, implying (sin x + cos x geq 0).We find that the inequality simplifies significantly when (|x| + |y| = pi). In this case, the square root term becomes zero, and the inequality reduces to (sec^2 x leq sqrt{2} |operatorname{tg} x| (sin x + cos x)). This holds true only when (x = frac{pi}{4}) and (|operatorname{tg} x| = 1), leading to (sin x + cos x = sqrt{2}). For (x = frac{pi}{4}), we find (|y| = frac{3pi}{4}), giving the solutions ((frac{pi}{4}, pmfrac{3pi}{4})). Checking other values of (x) within the domain (|x| + |y| leq pi) shows that the LHS exceeds the RHS, confirming no additional solutions.Thus, the solutions to the inequality are:boxed{left(frac{pi}{4}, pmfrac{3pi}{4}right)}.,<think>Okay, so I need to solve this inequality:sqrt{frac{pi}{4}-operatorname{arctg} frac{|x|+|y|}{pi}}+operatorname{tg}^{2} x+1 leqslant sqrt{2}|operatorname{tg} x|(sin x+cos x)First, let me parse this step by step. The left side has a square root term, a tangent squared term, and a constant 1. The right side is a product of sqrt(2), the absolute value of tangent x, and (sin x + cos x). Hmm, this seems complicated, but maybe I can break it down.Starting with the square root term on the left: sqrt(π/4 - arctg[(|x| + |y|)/π]). Since arctg is involved, the argument of the square root must be non-negative. So, first, I need to ensure that:π/4 - arctg[(|x| + |y|)/π] ≥ 0Which implies that:arctg[(|x| + |y|)/π] ≤ π/4Since arctg is an increasing function, this would mean:(|x| + |y|)/π ≤ tan(π/4) = 1Therefore:|x| + |y| ≤ π * 1 = πSo, the domain of this inequality is all (x, y) such that |x| + |y| ≤ π. That's a diamond-shaped region centered at the origin with vertices at (±π, 0) and (0, ±π). So, that's the first condition. Any solution must lie within this diamond.Now, moving on to the rest of the inequality. Let's denote the square root term as A, so:A = sqrt(π/4 - arctg[(|x| + |y|)/π])Then the left side is A + tan²x + 1, and the right side is sqrt(2) |tan x| (sin x + cos x).So, the inequality is:A + tan²x + 1 ≤ sqrt(2) |tan x| (sin x + cos x)Hmm. Let me think. Maybe it's helpful to analyze this inequality by considering variables x and y. But there's a |y| term inside the arctg, so y is present here. However, the rest of the terms only involve x. So, perhaps y is a parameter here? Wait, no, the problem is to solve the inequality for (x, y), right? So, both variables are involved. But most of the terms relate to x, except for the arctg term which involves both |x| and |y|.This seems tricky. Maybe I can first look for possible x and y that satisfy the domain |x| + |y| ≤ π, and then analyze the rest of the inequality. Let's consider that.But since the square root term A is non-negative, and tan²x is non-negative, and 1 is positive, the left side is at least 1. The right side is sqrt(2) |tan x| (sin x + cos x). Let's see when this right side is real. Since |tan x| is involved, tan x must be defined, so x ≠ π/2 + kπ for integers k.But given that |x| ≤ π (since |x| + |y| ≤ π), x must be in [-π, π]. However, tan x is defined for x ≠ ±π/2. So, within the domain |x| + |y| ≤ π, x can be in (-π/2, π/2) to avoid the vertical asymptotes of tan x. Wait, but |x| + |y| ≤ π. If x is near π/2, then |y| must be ≤ π - π/2 = π/2. So, even if x is approaching π/2, y can adjust to make |x| + |y| ≤ π. So, maybe the domain includes x values beyond (-π/2, π/2), but tan x is problematic at x = ±π/2.But perhaps the problem implicitly assumes that tan x is defined, so x must be in (-π/2, π/2). Hmm. The problem statement didn't specify, but since tan x is in the inequality, x must be such that tan x is defined. So, x ∈ (-π/2, π/2) plus multiples of π, but given |x| + |y| ≤ π, x can't be outside of [-π, π]. But tan x has period π, so tan x is the same as tan(x + kπ). So, maybe x is considered modulo π, but given the domain |x| + |y| ≤ π, x is in [-π, π], but tan x is undefined at ±π/2. Therefore, perhaps the problem is considering x in (-π/2, π/2). Let me check.If x is in (-π/2, π/2), then |x| < π/2, so |y| ≤ π - |x| > π - π/2 = π/2. So, y can be up to π/2. But maybe the problem is intended for x in (-π/2, π/2). Let me assume that, as otherwise tan x is undefined.So, assuming x ∈ (-π/2, π/2). Then, we can proceed.Now, let's analyze the inequality:sqrt(π/4 - arctg[(|x| + |y|)/π]) + tan²x + 1 ≤ sqrt(2) |tan x| (sin x + cos x)First, perhaps consider the case when x and y are non-negative. Since we have |x| and |y|, the inequality is symmetric in x and y with respect to sign. So, maybe we can assume x ≥ 0 and y ≥ 0, solve the inequality in the first quadrant, and then extend the solution to other quadrants by symmetry. That might simplify things.So, let's suppose x ≥ 0 and y ≥ 0. Then |x| = x and |y| = y, so the inequality becomes:sqrt(π/4 - arctg[(x + y)/π]) + tan²x + 1 ≤ sqrt(2) tan x (sin x + cos x)Because tan x is positive when x is in (0, π/2), so |tan x| = tan x.Similarly, in other quadrants, due to absolute values, we can mirror the solutions.So, perhaps focusing on x ≥ 0 and y ≥ 0 is sufficient, and then the solution can be extended via symmetry.Now, let's consider variables x and y such that x + y ≤ π, x ≥ 0, y ≥ 0.But how are x and y related here? The left side has a term with arctg[(x + y)/π], which complicates things. However, perhaps we can consider substituting variables.Let me denote t = (x + y)/π, so t ≤ 1, since x + y ≤ π. Then arctg(t) is in [0, arctg(1)) = [0, π/4). Therefore, π/4 - arctg(t) is in (0, π/4]. So the square root term is sqrt(π/4 - arctg(t)) which is real and positive.But how does this help? Hmm.Alternatively, maybe fix x and see what y must satisfy. Let's see.But the left side includes both x and y in the sqrt term, but the rest is in x. The right side is only in x. So, for given x, the inequality relates to y through the sqrt term.Wait, but the sqrt term is decreasing in (x + y), since arctg is increasing, so π/4 - arctg[(x + y)/π] is decreasing as (x + y) increases, so the sqrt term is decreasing in (x + y). Therefore, for fixed x, as y increases, the sqrt term decreases, making the left side smaller, which might make the inequality easier to satisfy. Conversely, for fixed x, as y decreases, the sqrt term increases, making the left side larger, which might make the inequality harder to satisfy.But the problem is to find all (x, y) such that the inequality holds. So perhaps for each x, find the range of y such that the inequality holds. But this seems complicated. Alternatively, maybe there's a particular relation between x and y that must hold, or perhaps the inequality is only satisfied at specific points.Alternatively, maybe the inequality can only hold when some terms are balanced. Let me look at the left side and the right side.Left side: sqrt(π/4 - arctg[(x + y)/π}) + tan²x + 1Right side: sqrt(2) tan x (sin x + cos x)Since the left side has a sqrt term that's at most sqrt(π/4) (since arctg[(x + y)/π] ≥ 0, so π/4 - arctg[(x + y)/π] ≤ π/4). So sqrt(π/4) = sqrt(π)/2 ≈ 0.886. Then tan²x + 1 is sec²x, which is always ≥1.So the left side is at least 1 + 0 ≈ 1 (if the sqrt term is near 0) and up to sqrt(π)/2 + sec²x + 1. Wait, no, sqrt(π/4) ≈ 0.886, so the sqrt term is up to around 0.886, and then tan²x +1 = sec²x, which is 1 + tan²x. So the left side is sqrt(π/4 - arctg(...)) + sec²x.The right side is sqrt(2) |tan x| (sin x + cos x). Let's analyze this term. Let's denote t = tan x, then sin x + cos x can be written as sqrt(2) sin(x + π/4). Wait, since sin x + cos x = sqrt(2) sin(x + π/4). Therefore, the right side is sqrt(2) |t| * sqrt(2) sin(x + π/4) = 2 |t| sin(x + π/4). Wait, hold on:Wait, no. sin x + cos x = sqrt(2) sin(x + π/4). So, sqrt(2) |tan x| (sin x + cos x) = sqrt(2) |tan x| * sqrt(2) sin(x + π/4) = 2 |tan x| sin(x + π/4). But that's only if the original expression is written correctly.Wait, original right side is sqrt(2) |tan x| (sin x + cos x). So, it's sqrt(2) multiplied by |tan x| multiplied by (sin x + cos x). If we write sin x + cos x as sqrt(2) sin(x + π/4), then the right side becomes sqrt(2) |tan x| * sqrt(2) sin(x + π/4) = 2 |tan x| sin(x + π/4). So, that's a simplification.But maybe this helps. So, the right side is 2 |tan x| sin(x + π/4). Let's note that sin(x + π/4) is between -1 and 1, but since x is in (-π/2, π/2), x + π/4 ranges from (-π/2 + π/4) = -π/4 to (π/2 + π/4) = 3π/4. Therefore, sin(x + π/4) ranges from sin(-π/4) = -sqrt(2)/2 to sin(3π/4) = sqrt(2)/2. Wait, but if x is in (-π/2, π/2), x + π/4 is in (-π/4, 3π/4), so sin(x + π/4) is in (-sqrt(2)/2, 1]. Wait, sin(3π/4) is sqrt(2)/2, but at x = π/2, which is excluded, sin(3π/4) is sqrt(2)/2. Wait, but x can approach π/2. So, sin(x + π/4) can approach sin(3π/4) = sqrt(2)/2. So actually, sin(x + π/4) is in (-sqrt(2)/2, sqrt(2)/2]. Wait, no. Wait, sin(-π/4) = -sqrt(2)/2, sin(π/2) = 1. Wait, hold on. If x is approaching π/2, then x + π/4 approaches 3π/4, whose sine is sqrt(2)/2. If x is approaching -π/2, then x + π/4 approaches -π/4, sine is -sqrt(2)/2. But when x = 0, sin(π/4) = sqrt(2)/2. Wait, no. Let's compute:If x is in (-π/2, π/2), then x + π/4 is in (-π/4, 3π/4). The sine function on (-π/4, 3π/4) reaches its maximum at π/2, which is 1, and its minimum at -π/4, which is -sqrt(2)/2. So sin(x + π/4) ranges from -sqrt(2)/2 to 1. Therefore, sin(x + π/4) is in [-sqrt(2)/2, 1].But in the right side, we have 2 |tan x| sin(x + π/4). Since |tan x| is non-negative, and sin(x + π/4) can be negative or positive. However, since the left side is non-negative, the right side must also be non-negative for the inequality to hold. Therefore, we must have 2 |tan x| sin(x + π/4) ≥ 0. Since |tan x| ≥ 0, this implies that sin(x + π/4) ≥ 0. Therefore, sin(x + π/4) ≥ 0 implies that x + π/4 ∈ [0, π], which given x ∈ (-π/2, π/2), translates to x ∈ [-π/4, 3π/4). But since x ∈ (-π/2, π/2), the intersection is x ∈ [-π/4, π/2). Therefore, the inequality can only hold when x ∈ [-π/4, π/2). But since we are assuming x ≥ 0 for simplicity, then x ∈ [0, π/2). Wait, but x can also be in [-π/4, 0), but given our initial assumption of x ≥ 0, we can restrict to x ∈ [0, π/2).Therefore, in the right side, sin(x + π/4) is non-negative for x ∈ [-π/4, π/2). So, in the original problem, for the inequality to hold, x must be in [-π/4, π/2). Similarly for the negative side due to absolute values.But since we are considering x ≥ 0, y ≥ 0, we can focus on x ∈ [0, π/2), and y ∈ [0, π - x].But this is getting complicated. Maybe instead of splitting into quadrants, consider x in [-π/4, π/2) to ensure sin(x + π/4) ≥ 0.Alternatively, let's consider specific cases where the inequality simplifies. For example, maybe when x = 0. Let's check x = 0.If x = 0, then tan x = 0, so the right side becomes 0. The left side becomes sqrt(π/4 - arctg(|0| + |y|)/π) + 0 + 1. So:sqrt(π/4 - arctg(|y|/π)) + 1 ≤ 0But sqrt(...) + 1 is always positive, so this inequality can't hold. Therefore, x = 0 is not a solution.Similarly, if x is near 0, the right side is small because tan x is small, while the left side has at least 1 + sqrt(π/4 - arctg(...)), which is more than 1. So near x = 0, the inequality doesn't hold.Another case: suppose y = 0. Then the sqrt term becomes sqrt(π/4 - arctg(|x|/π}). Let's see if that simplifies things. Maybe not directly, but perhaps.Alternatively, let's consider when y is as large as possible, given x. Since |x| + |y| ≤ π, if we fix x, then |y| ≤ π - |x|. So, to minimize the sqrt term (since larger y would make arctg[(|x| + |y|)/π] larger, hence π/4 - arctg(...) smaller, so sqrt term smaller). Therefore, for a fixed x, the minimal left side occurs when |y| is maximal, i.e., |y| = π - |x|. Conversely, the minimal |y| is 0, which would make the sqrt term largest. Therefore, if for a given x, the inequality holds when |y| is maximal (i.e., sqrt term is minimal), then it would hold for all |y| ≤ π - |x|. Conversely, if the inequality doesn't hold when |y| is maximal, then there may be some y where it holds.But this seems like a possible approach. So, for each x, check if the inequality holds when y is chosen to minimize the left side (i.e., y maximal), and if so, then all y with |y| ≤ π - |x| would satisfy the inequality. Otherwise, there may be some subset.But this requires solving for y, which complicates things. Alternatively, maybe the inequality is only satisfied when the sqrt term is zero? Because sqrt(...) is non-negative, so if the sqrt term is zero, then the left side is tan²x + 1, which is sec²x. So, if sqrt(π/4 - arctg(...)) = 0, then π/4 - arctg(...) = 0, so arctg(...) = π/4, so (|x| + |y|)/π = tan(π/4) = 1, so |x| + |y| = π. So, on the boundary of the domain.Then, in that case, the left side is 0 + tan²x + 1 = sec²x. The right side is sqrt(2) |tan x| (sin x + cos x). So, the inequality becomes:sec²x ≤ sqrt(2) |tan x| (sin x + cos x)But sec²x = 1 + tan²x. Let me write this in terms of tan x. Let t = tan x, so sec²x = 1 + t². Also, sin x + cos x can be written as sqrt(1 + t²) (sin x + cos x) / sqrt(1 + t²). Wait, sin x + cos x = sqrt(2) sin(x + π/4), as before. But let's express sin x + cos x in terms of t.If t = tan x, then sin x = t / sqrt(1 + t²), cos x = 1 / sqrt(1 + t²). Therefore, sin x + cos x = (t + 1)/sqrt(1 + t²). Therefore, the right side is sqrt(2) |t| * (t + 1)/sqrt(1 + t²) = sqrt(2) |t| (t + 1) / sqrt(1 + t²)Therefore, the inequality becomes:1 + t² ≤ sqrt(2) |t| (t + 1) / sqrt(1 + t²)Multiply both sides by sqrt(1 + t²):(1 + t²)^(3/2) ≤ sqrt(2) |t| (t + 1)Assuming t ≠ 0 (since t=0 leads to left side 1, right side 0, which doesn't hold). Let's square both sides to eliminate the square roots:(1 + t²)^3 ≤ 2 t² (t + 1)^2But this seems messy. Let's consider t > 0 (since x ∈ [0, π/2), so t = tan x ≥ 0). Then |t| = t, so the inequality is:(1 + t²)^3 ≤ 2 t² (t + 1)^2Expand both sides:Left side: (1 + t²)^3 = 1 + 3t² + 3t^4 + t^6Right side: 2 t² (t + 1)^2 = 2 t² (t² + 2t + 1) = 2 t^4 + 4 t^3 + 2 t²So, bringing all terms to the left:1 + 3t² + 3t^4 + t^6 - 2 t^4 - 4 t^3 - 2 t² ≥ 0Simplify:1 + (3t² - 2t²) + (3t^4 - 2t^4) + t^6 - 4t^3= 1 + t² + t^4 + t^6 - 4t^3So the inequality reduces to:t^6 + t^4 - 4 t^3 + t² + 1 ≥ 0This is a sixth-degree polynomial. Hmm, seems difficult to factor. Let me check for t = 1:1 + 1 - 4 + 1 + 1 = 0. So t=1 is a root.Similarly, check t=1: the polynomial equals 0. Therefore, (t - 1) is a factor. Let's perform polynomial division or factor it.Let me try to factor t^6 + t^4 - 4 t^3 + t² + 1. Let's group terms:= t^6 + t^4 + t² + 1 - 4 t^3Hmm, maybe group as (t^6 + t^4) + (t² + 1) - 4 t^3= t^4(t² + 1) + 1(t² + 1) - 4 t^3= (t^4 + 1)(t² + 1) - 4 t^3Not obvious. Alternatively, since t=1 is a root, divide by (t - 1):Using polynomial long division or synthetic division. Let's use synthetic division for t=1.The coefficients are 1 (t^6), 0 (t^5), 1 (t^4), -4 (t^3), 1 (t^2), 0 (t), 1 (constant).So, coefficients: 1, 0, 1, -4, 1, 0, 1Divide by (t - 1):Start with 1. Bring down 1.Multiply by 1: 1. Add to next: 0 +1=1.Multiply by 1:1. Add to next:1 +1=2.Multiply by1:2. Add to next: -4 +2= -2.Multiply by1:-2. Add to next:1 + (-2)= -1.Multiply by1:-1. Add to next:0 + (-1)= -1.Multiply by1:-1. Add to next:1 + (-1)=0. So the remainder is 0, as expected.So the quotient polynomial is t^5 + t^4 + 2 t^3 - 2 t^2 - t -1.So, t^6 + t^4 -4 t^3 + t^2 +1 = (t -1)(t^5 + t^4 + 2 t^3 -2 t^2 -t -1)Now, let's check if t=1 is a root of the quintic:Plug t=1: 1 +1 +2 -2 -1 -1=0. Yes, t=1 is another root.So, divide the quintic by (t -1):Coefficients:1,1,2,-2,-1,-1Synthetic division:1 | 1 1 2 -2 -1 -1Bring down 1.Multiply by1:1. Add to next:1 +1=2.Multiply by1:2. Add to next:2 +2=4.Multiply by1:4. Add to next:-2 +4=2.Multiply by1:2. Add to next:-1 +2=1.Multiply by1:1. Add to next:-1 +1=0. Remainder 0.So the quintic factors into (t -1)(t^4 + 2 t^3 +4 t^2 +2 t +1). Therefore, original polynomial is:(t -1)^2 (t^4 +2 t^3 +4 t^2 +2 t +1)Now, check if the quartic factors further. Let's try to factor t^4 +2 t^3 +4 t^2 +2 t +1.Check for rational roots using Rational Root Theorem. Possible roots are ±1. Let's test t=-1:1 -2 +4 -2 +1=2 ≠0. t=1:1 +2 +4 +2 +1=10≠0. So no rational roots. Maybe it's a biquadratic or can be factored into quadratics.Assume t^4 +2 t^3 +4 t^2 +2 t +1 = (t^2 + a t + b)(t^2 +c t +d)Multiply out: t^4 + (a + c) t^3 + (b + d + a c) t^2 + (a d + b c) t + b dSet coefficients equal:a + c =2b + d + a c =4a d + b c =2b d =1Since b d=1, possible b=d=1 or b=d=-1. Let's try b=d=1.Then:a + c =21 +1 +a c =4 ⇒ a c=2a*1 +1*c =2 ⇒ a +c=2, which is the first equation. Since a + c=2 and a c=2, solving quadratic equation x² -2x +2=0, which has discriminant 4 -8= -4. So no real solutions. Therefore, not possible.Try b=d=-1. Then b d=1, but (-1)(-1)=1. Then:a + c=2(-1)+(-1)+a c=4 ⇒ a c=6a*(-1) + (-1)*c=2 ⇒ -a -c=2 ⇒ a +c= -2. But earlier a +c=2. Contradiction. Therefore, no solution.Therefore, the quartic is irreducible over rationals. So the factorization is (t -1)^2 (t^4 +2 t^3 +4 t^2 +2 t +1). Now, we need to check if the quartic is always positive. Let's check t^4 +2 t^3 +4 t^2 +2 t +1.For real t, t^4 is always non-negative, 2 t^3 can be negative or positive, but let's check the entire expression. Maybe write it as t^4 + 2 t^3 +4 t^2 +2 t +1. Let's group terms:= t^4 +2 t^3 + t^2 +3 t^2 +2 t +1= t^2(t^2 +2 t +1) + (3 t^2 +2 t +1)= t^2(t +1)^2 + (3 t^2 +2 t +1)Both terms are positive for all real t, since 3 t^2 +2 t +1 has discriminant 4 -12= -8 <0, so always positive. Therefore, the quartic is always positive.Therefore, the polynomial t^6 + t^4 -4 t^3 + t^2 +1 = (t -1)^2 * positive. Therefore, the polynomial is ≥0 for all real t, and equals zero only at t=1. Therefore, the inequality t^6 + t^4 -4 t^3 + t^2 +1 ≥0 holds for all real t, with equality only at t=1. Therefore, going back, the original inequality (after squaring) is always true, but equality holds when t=1.But we squared both sides, which can introduce extraneous solutions. Therefore, we need to check when equality occurs. So, equality in the squared inequality occurs when t=1, i.e., tan x =1. Therefore, x= π/4. So, when x=π/4, and |x| + |y|=π (since we considered the case when the sqrt term is zero), so |y|=π - |x|=π - π/4=3π/4. Therefore, y= ±3π/4. But given our initial assumption of x ≥0, y ≥0, we have y=3π/4.Therefore, the point (x, y)=(π/4, 3π/4) is a solution. Also, due to symmetry, other points like (-π/4, 3π/4), (π/4, -3π/4), etc., but considering the original problem's absolute values, maybe these points are included.But wait, in the case when the sqrt term is zero, we have |x| + |y|=π. So, for the sqrt term to be zero, we need |x| + |y|=π. Then, in such cases, when tan x=1, i.e., x= π/4 +kπ. But since |x| ≤π, the possible x values are x= π/4, x= -π/4, x= 3π/4, x= -3π/4. However, tan x is 1 at π/4 and -3π/4, and -1 at -π/4 and 3π/4. Wait, tan(π/4)=1, tan(-π/4)=-1, tan(3π/4)=-1, tan(-3π/4)=1. But we had earlier restricted x to [-π/4, π/2) to have sin(x + π/4) ≥0.But if x=3π/4, that's outside of (-π/2, π/2), so tan x is undefined. Therefore, the only solutions in the domain where tan x is defined are x= π/4 and x= -π/4. But x= -π/4 is in [-π/4, π/2), so valid.But if x= π/4, then |x| + |y|=π implies |y|= π - π/4=3π/4. Similarly, if x= -π/4, |y|=π - π/4=3π/4. So, the solutions when sqrt term is zero are (x, y)= (±π/4, ±3π/4) and (±3π/4, ±π/4), but considering the domain |x| + |y| ≤π and tan x is defined.Wait, but x=3π/4 is not in (-π/2, π/2), so tan x is undefined there. Therefore, the only valid solutions are when x= π/4 and y=3π/4, and x= -π/4 and y=3π/4, but considering |x| + |y|=π and the domain. Wait, but if x= -π/4, then |x| + |y|=π/4 + |y|=π, so |y|=3π/4. Therefore, the points (x, y)= (π/4, 3π/4), (-π/4, 3π/4), (π/4, -3π/4), (-π/4, -3π/4). However, y=3π/4 is allowed since |x| + |y|=π/4 +3π/4=π, which is within the domain. Similarly for the others.But in these points, we need to check if the original inequality holds with equality. Let's check (π/4, 3π/4):Left side: sqrt(π/4 - arctg[(π/4 +3π/4)/π}) + tan²(π/4) +1 = sqrt(π/4 - arctg(1)) +1 +1= sqrt(π/4 - π/4) +2= sqrt(0)+2=0+2=2.Right side: sqrt(2) |tan(π/4)| (sin(π/4) + cos(π/4))= sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2)= sqrt(2)*1*(sqrt(2))=2. So equality holds.Similarly, for (-π/4, 3π/4):Left side: sqrt(π/4 - arctg(|-π/4| + |3π/4|)/π}) + tan²(-π/4) +1= sqrt(π/4 - arctg(π/4 +3π/4)/π) + tan²(π/4) +1= same as before, sqrt(0) +1 +1=2.Right side: sqrt(2)|tan(-π/4)|(sin(-π/4)+cos(-π/4))= sqrt(2)*1*( -sqrt(2)/2 + sqrt(2)/2 )= sqrt(2)*1*(0)=0. Wait, this contradicts. Wait, this is a problem.Wait, no. Wait, if x=-π/4, then tan x = tan(-π/4)=-1, so |tan x|=1. Then sin x + cos x= sin(-π/4)+cos(-π/4)= -sqrt(2)/2 + sqrt(2)/2=0. Therefore, right side is sqrt(2)*1*0=0. Left side is 2. So 2 ≤0, which is false. Therefore, (-π/4, 3π/4) is not a solution.Ah! So even though we thought it might be a solution, in reality, when x=-π/4, the right side becomes zero, but left side is 2, which doesn't satisfy the inequality. Therefore, only the points where x=π/4, y=3π/4 and x=π/4, y=-3π/4 (but considering the domain) might be solutions, but when x=π/4, |y|=3π/4, and if y is positive or negative. Wait, let's check x=π/4, y=-3π/4.Then, the left side is sqrt(π/4 - arctg(|π/4| + | -3π/4 |)/π) + tan²(π/4) +1= same as before, sqrt(0) +1 +1=2. Right side: sqrt(2)|tan(π/4)|(sin(π/4)+cos(π/4))=2, same as before. So, equality holds. So the points (π/4, ±3π/4) are solutions.But for x=π/4, y=-3π/4, |x| + |y|=π/4 +3π/4=π, so it's on the boundary.But in this case, the right side is sqrt(2)*1*(sin(π/4) + cos(π/4))= sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2)= sqrt(2)*sqrt(2)=2, which matches the left side.Similarly, if x=-π/4, but then y would have to be 3π/4, but as we saw, that gives a right side of 0, which doesn't equal 2. Therefore, only the points where x=π/4 and |y|=3π/4 are solutions. Similarly, by symmetry, if y=π/4 and |x|=3π/4, but then we have to check if the roles of x and y can be swapped. However, in the original inequality, y only appears in the arctg term, while x appears in tan x and sin/cos terms. So, swapping x and y would not necessarily satisfy the inequality unless the terms align.Wait, let's check the point (3π/4, π/4). However, x=3π/4 is outside of (-π/2, π/2), so tan x is undefined. Therefore, such points are not in the domain. Therefore, the only solutions are (π/4, ±3π/4) and (-π/4, ±3π/4). But as we saw, (-π/4, 3π/4) doesn't work because the right side becomes zero. Similarly, (-π/4, -3π/4): x=-π/4, y=-3π/4.Left side: sqrt(π/4 - arctg(π/4 +3π/4 /π)) + tan²(-π/4) +1= sqrt(0)+1+1=2. Right side: sqrt(2)*|tan(-π/4)|*(sin(-π/4)+cos(-π/4))=sqrt(2)*1*(-sqrt(2)/2 + sqrt(2)/2)=0. So again, 2 ≤0 is false. Therefore, only the points where x=π/4 and y=±3π/4 satisfy the equality.Similarly, if x=π/4 and y=-3π/4, then sin x + cos x= sin(π/4)+cos(π/4)=sqrt(2)/2 + sqrt(2)/2=sqrt(2). Therefore, the right side is sqrt(2)*1*sqrt(2)=2, which equals the left side.Therefore, the points (π/4, 3π/4) and (π/4, -3π/4) are solutions. Similarly, if we consider x=-pi/4 but with y adjusted, but as we saw, those do not work. Therefore, these two points are the only solutions where the sqrt term is zero.But earlier, we saw that the inequality reduces to 2 ≤ 2, which holds. So, these points are solutions.But are there other solutions where the sqrt term is positive? That is, where |x| + |y| < pi, so the sqrt term is positive, and the inequality holds.To check this, we need to see if there are other (x, y) pairs where the left side is less than or equal to the right side, apart from the two points we found.Let me consider x=pi/4, y=0. Then |x| + |y|=pi/4 < pi. Then the left side becomes sqrt(pi/4 - arctg(pi/4 / pi)) + tan²(pi/4) +1= sqrt(pi/4 - arctg(1/4)) +1 +1≈ sqrt(0.785 - 0.245) +2≈ sqrt(0.54)+2≈0.735+2≈2.735. The right side is sqrt(2)*1*(sin(pi/4)+cos(pi/4))=sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2)=sqrt(2)*sqrt(2)=2. So 2.735 ≤2? No, this doesn't hold. Therefore, (pi/4, 0) is not a solution.Alternatively, take x=pi/4, y=pi/4. Then |x| + |y|=pi/2 <pi. Left side: sqrt(pi/4 - arctg(pi/2 / pi)) + tan²(pi/4)+1= sqrt(pi/4 - arctg(1/2)) +1+1≈ sqrt(0.785 - 0.464) +2≈sqrt(0.321)+2≈0.566+2≈2.566. Right side: same as before, 2. So 2.566 ≤2? No.Therefore, when x=pi/4 but y is less than 3pi/4, the left side exceeds the right side. Therefore, only when y=3pi/4 (making the sqrt term zero) does the inequality hold.Similarly, take x=pi/3, which is approximately 1.047, within (-pi/2, pi/2). Let's set y=pi -pi/3=2pi/3≈2.094. So, |x| + |y|=pi/3 +2pi/3=pi. Then sqrt term is zero. Left side: 0 + tan²(pi/3)+1= (sqrt(3))² +1=3 +1=4. Right side: sqrt(2)*|tan(pi/3)|*(sin(pi/3) + cos(pi/3))= sqrt(2)*sqrt(3)*(sqrt(3)/2 +1/2)= sqrt(6)*( (sqrt(3)+1)/2 ). Compute this value: sqrt(6)≈2.449, sqrt(3)≈1.732, so (1.732 +1)/2≈1.366. Then, 2.449*1.366≈3.344. So 4 ≤3.344? No. Therefore, even though we are on the boundary |x| + |y|=pi, if x is not pi/4, the inequality does not hold. Therefore, only when x=pi/4 (and y=3pi/4) does the equality hold.Therefore, these points are the only solutions.But wait, let's check another x. For example, x=0. Let's say x=0, but earlier we saw that when x=0, the inequality cannot hold. But let's try x approaching pi/4. Take x=pi/4 - epsilon, and y=3pi/4 + delta, such that |x| + |y|=pi. But as epsilon approaches 0, delta approaches 0. Let's compute the left and right sides.Left side: sqrt(pi/4 - arctg( (pi - epsilon - delta)/pi )) + tan²(pi/4 - epsilon) +1. As epsilon and delta approach 0, arctg(1 - (epsilon + delta)/pi )≈ arctg(1) - (epsilon + delta)/pi * derivative of arctg at 1. The derivative of arctg(t) at t=1 is 1/(1 +1^2)=1/2. Therefore, arctg(1 - h) ≈ pi/4 - h/2. So pi/4 - arctg(1 - h)≈h/2. Therefore, sqrt(pi/4 - arctg(...))≈sqrt(h/2)=sqrt( (epsilon + delta)/ (2 pi) ). Therefore, sqrt term approaches zero as epsilon and delta approach zero.tan(pi/4 - epsilon)= (1 - tan epsilon)/(1 + tan epsilon)≈1 - 2 epsilon (using tan epsilon ≈ epsilon for small epsilon). Therefore, tan²x≈1 -4 epsilon. Therefore, tan²x +1≈2 -4 epsilon.Right side: sqrt(2) |tan x| (sin x + cos x). tan x≈1 -2 epsilon, |tan x|=1 -2 epsilon. sin x + cos x= sin(pi/4 - epsilon) + cos(pi/4 - epsilon)= [sin(pi/4)cos epsilon - cos(pi/4)sin epsilon] + [cos(pi/4)cos epsilon + sin(pi/4)sin epsilon]= sqrt(2)/2 cos epsilon - sqrt(2)/2 sin epsilon + sqrt(2)/2 cos epsilon + sqrt(2)/2 sin epsilon= sqrt(2) cos epsilon≈sqrt(2)(1 - epsilon²/2). Therefore, the right side≈sqrt(2)(1 -2 epsilon)(sqrt(2)(1 - epsilon²/2))=2(1 -2 epsilon)(1 - epsilon²/2)≈2(1 -2 epsilon - epsilon²/2)≈2 -4 epsilon - epsilon².Left side: sqrt(h/2) +2 -4 epsilon≈0 +2 -4 epsilon=2 -4 epsilon.Comparing to right side≈2 -4 epsilon - epsilon². So left side≈2 -4 epsilon vs right≈2 -4 epsilon -epsilon². Therefore, for small epsilon, left side is greater than right side. Therefore, near the point (pi/4, 3pi/4), for x slightly less than pi/4 and y slightly more than 3pi/4, the left side is greater than the right side, so the inequality does not hold. Hence, the only solution is exactly at (pi/4, 3pi/4) and (pi/4, -3pi/4).Similarly, check x slightly more than pi/4. But x cannot exceed pi/2. But suppose x=pi/4 + epsilon, then |x| + |y|=pi implies y=3pi/4 - epsilon. Then, sqrt term is zero (since |x| + |y|=pi). Then left side=0 + tan²(pi/4 + epsilon) +1. tan(pi/4 + epsilon)= (1 + tan epsilon)/(1 - tan epsilon)≈1 +2 epsilon (for small epsilon). Therefore, tan²x≈1 +4 epsilon. Therefore, left side≈0 +1 +4 epsilon +1=2 +4 epsilon. Right side= sqrt(2)|tan x|(sin x + cos x). tan x≈1 +2 epsilon. sin x + cos x= sin(pi/4 + epsilon)+cos(pi/4 + epsilon)= [sin(pi/4)cos epsilon + cos(pi/4)sin epsilon] + [cos(pi/4)cos epsilon - sin(pi/4)sin epsilon]= sqrt(2)/2 cos epsilon + sqrt(2)/2 sin epsilon + sqrt(2)/2 cos epsilon - sqrt(2)/2 sin epsilon= sqrt(2) cos epsilon≈sqrt(2)(1 - epsilon²/2). Therefore, right side≈sqrt(2)*(1 +2 epsilon)*sqrt(2)(1 - epsilon²/2)=2*(1 +2 epsilon)(1 - epsilon²/2)≈2*(1 +2 epsilon - epsilon²/2)≈2 +4 epsilon - epsilon². Therefore, left side≈2 +4 epsilon vs right≈2 +4 epsilon - epsilon². So left side > right side, so inequality does not hold. Therefore, even if x is slightly more than pi/4, the inequality fails.Therefore, the only solutions are the exact points where x=pi/4 and |y|=3pi/4. Similarly, due to the absolute values, the points where x=-pi/4 and |y|=3pi/4 would be considered, but as we saw earlier, those don't work because the right side becomes zero. Therefore, the only valid solutions are (pi/4, 3pi/4) and (pi/4, -3pi/4).But let's check for y=3pi/4, x=pi/4. But also, y=-3pi/4, x=pi/4. Similarly, if we take y=pi/4, x=3pi/4, but x=3pi/4 is outside of (-pi/2, pi/2), so invalid.Therefore, the only solutions are the two points where x=pi/4, y=±3pi/4.But wait, the problem is to solve the inequality for all (x, y) such that |x| + |y| ≤pi. So, are these the only solutions?Earlier, we saw that when the sqrt term is zero (i.e., |x| + |y|=pi), the inequality reduces to sec²x ≤ sqrt(2) |tan x| (sin x + cos x). And we found that equality holds only when tan x=1, i.e., x=pi/4. For other x, even on the boundary |x| + |y|=pi, the inequality does not hold. Therefore, only those two points are solutions.But perhaps there are solutions inside the domain where |x| + |y| < pi. Let's check x=pi/4, y=0. As before, left side≈2.735, right side=2. Not okay. What if we take x=pi/3, y=0. Then |x| + |y|=pi/3 <pi.Left side: sqrt(pi/4 - arctg(pi/3 /pi)) + tan²(pi/3) +1= sqrt(pi/4 - arctg(1/3)) + (sqrt(3))^2 +1≈ sqrt(0.785 -0.321)+3 +1≈sqrt(0.464)+4≈0.681+4≈4.681. Right side: sqrt(2)*sqrt(3)*(sin(pi/3)+cos(pi/3))=sqrt(6)*(sqrt(3)/2 +1/2)≈2.449*(1.366)≈3.344. So 4.681 ≤3.344? No.Alternatively, take x=pi/6, y=0. Left side: sqrt(pi/4 - arctg(pi/6 / pi)) + tan²(pi/6) +1= sqrt(pi/4 - arctg(1/6)) + (1/sqrt(3))^2 +1≈sqrt(0.785 -0.165)+1/3 +1≈sqrt(0.62)+1.333≈0.787+1.333≈2.12. Right side: sqrt(2)*(1/sqrt(3))*(sin(pi/6)+cos(pi/6))≈1.414*0.577*(0.5 +0.866)≈0.816*(1.366)≈1.115. So 2.12 ≤1.115? No.Another try: maybe x=pi/4, y=3pi/4 - delta, with small delta. Then, sqrt term is sqrt(pi/4 - arctg( (pi - delta)/pi ))≈sqrt(pi/4 - arctg(1 - delta/pi ))≈sqrt(pi/4 - (pi/4 - delta/(2 pi)) ))≈sqrt( delta/(2 pi) ). So sqrt term≈sqrt(delta/(2 pi)). Left side≈sqrt(delta/(2 pi)) +2. Right side= sqrt(2)*1*(sqrt(2)/2 + sqrt(2)/2 - something). Wait, x=pi/4, y=3pi/4 - delta. Then sin x + cos x= sqrt(2)/2 + sqrt(2)/2= sqrt(2). Therefore, right side= sqrt(2)*1*sqrt(2)=2. Left side≈sqrt(delta/(2 pi)) +2. So, for small delta, left side≈2 + something, which is greater than 2. Therefore, inequality does not hold.Therefore, it seems that the only solutions are the points where |x| + |y|=pi, tan x=1, and y=±3pi/4. Which gives (pi/4, 3pi/4) and (pi/4, -3pi/4). But need to check if there are other points where x is negative. For example, x=-pi/4, y=3pi/4. But as before, sin x + cos x= sin(-pi/4) + cos(-pi/4)= -sqrt(2)/2 + sqrt(2)/2=0. So right side=0, left side=2. Doesn't work.Therefore, the only solutions are the two points where x=pi/4, y=±3pi/4.But the problem says "Solve the inequality". The solutions are these two points. But wait, let me confirm once more.Wait, let's check another point. Suppose x=pi/4, y=3pi/4. Then, the original inequality is 2 <=2, which holds. Similarly, x=pi/4, y=-3pi/4. Also 2 <=2. Any other points? Let's suppose x=pi/4, y=3pi/4 + delta. But then |x| + |y|=pi/4 +3pi/4 +delta=pi + delta, which exceeds pi, so outside the domain. Therefore, no.Alternatively, x=pi/4 + epsilon, y=3pi/4 - epsilon. But then |x| + |y|=pi/4 + epsilon +3pi/4 -epsilon=pi. So on the boundary. But as we saw earlier, for x=pi/4 +epsilon, the left side becomes larger than the right side. Hence, no solutions there.Therefore, the only solutions are the two points (pi/4, 3pi/4) and (pi/4, -3pi/4). But wait, in terms of coordinates, since |x| + |y| <= pi, and y can be positive or negative, but x is pi/4 in both cases, and y is ±3pi/4.But wait, considering the problem's original inequality with |x| and |y|, the equation is symmetric in x and y in the sqrt term, but the rest of the terms only involve x. Therefore, perhaps there are solutions where y=pi/4 and |x|=3pi/4. But as before, if y=pi/4, |x|=3pi/4, then x=±3pi/4. But tan(3pi/4)=tan(pi - pi/4)= -1. So, let's check x=3pi/4, y=pi/4. But x=3pi/4 is outside of (-pi/2, pi/2), so tan x is undefined. Similarly, x=-3pi/4 is outside the domain. Therefore, no solutions there.Therefore, the only valid solutions are (pi/4, ±3pi/4).But to ensure completeness, let's check another quadrant. Suppose x is negative, say x=-pi/4, y=3pi/4. But then tan x=-1, sin x + cos x= sin(-pi/4)+cos(-pi/4)= -sqrt(2)/2 + sqrt(2)/2=0, so right side=0, left side=2. Doesn't work.Similarly, x=pi/4, y=-3pi/4. |x| + |y|=pi/4 +3pi/4=pi. Left side=2, right side= sqrt(2)*1*(sin(pi/4)+cos(pi/4))=2. So equality holds. Therefore, this point is a solution.Similarly, x=pi/4, y=3pi/4. Also works. But x=-pi/4, y=-3pi/4. Then left side=2, right side= sqrt(2)*|tan(-pi/4)|*(sin(-pi/4)+cos(-pi/4))= sqrt(2)*1*(-sqrt(2)/2 + sqrt(2)/2)=0. Doesn't work.Therefore, the valid solutions are (pi/4, 3pi/4), (pi/4, -3pi/4), (-pi/4, -3pi/4), (-pi/4, 3pi/4). Wait, no. Wait, x=-pi/4, y=-3pi/4: left side=2, right side= sqrt(2)*|tan(-pi/4)|*(sin(-pi/4)+cos(-pi/4))= sqrt(2)*1*(-sqrt(2)/2 + (-sqrt(2)/2))= sqrt(2)*1*(-sqrt(2))= -2. But the right side is negative, and left side is positive, so 2 <=-2? No. So even though |x| + |y|=pi, the right side can be negative, which doesn't satisfy the inequality. Therefore, only the points where x=pi/4 and y=±3pi/4 are valid.Therefore, the solutions are (pi/4, 3pi/4) and (pi/4, -3pi/4). But wait, if we take x=pi/4, y=-3pi/4, then sin x + cos x= sin(pi/4)+cos(pi/4)=sqrt(2)/2 + sqrt(2)/2=sqrt(2). So the right side is sqrt(2)*1*sqrt(2)=2. Left side is 2. So equality holds.Similarly, if x=-pi/4, y=-3pi/4, then sin x + cos x= sin(-pi/4)+cos(-pi/4)=-sqrt(2)/2 + sqrt(2)/2=0. Right side= sqrt(2)*1*0=0. Left side=2. Doesn't work.Therefore, the only valid solutions are the two points where x=pi/4, y=±3pi/4.Thus, after thorough analysis, the only solutions are the points (pi/4, 3pi/4) and (pi/4, -3pi/4).Final AnswerThe solutions are the points (boxed{left(frac{pi}{4}, frac{3pi}{4}right)}) and (boxed{left(frac{pi}{4}, -frac{3pi}{4}right)}).</think>To solve the inequalitysqrt{frac{pi}{4}-operatorname{arctg} frac{|x|+|y|}{pi}}+operatorname{tg}^{2} x+1 leqslant sqrt{2}|operatorname{tg} x|(sin x+cos x),we start by analyzing the domain. The argument of the square root must be non-negative, leading to the condition:|x| + |y| leq pi.Next, we consider the inequality itself. The left side includes a square root term, (tan^2 x), and 1, while the right side involves (sqrt{2} |tan x| (sin x + cos x)). We analyze specific cases and symmetry.1. Boundary Case: When (|x| + |y| = pi), the square root term becomes zero. This simplifies the inequality to: tan^2 x + 1 leq sqrt{2} |tan x| (sin x + cos x). Substituting (tan x = 1) (i.e., (x = frac{pi}{4})), we find (|y| = pi - frac{pi}{4} = frac{3pi}{4}). This gives the points ((frac{pi}{4}, frac{3pi}{4})) and ((frac{pi}{4}, -frac{3pi}{4})).2. Verification: For these points, we check: - Left side: (0 + tan^2 frac{pi}{4} + 1 = 2). - Right side: (sqrt{2} cdot 1 cdot (sin frac{pi}{4} + cos frac{pi}{4}) = 2).3. Other Cases: For other values of (x) and (y) within the domain, the inequality does not hold due to the left side being larger than the right side. This is confirmed through analysis and specific examples.Thus, the only solutions are the points where the inequality holds with equality, leading to the final answer: Final AnswerThe solutions are the points (boxed{left(frac{pi}{4}, frac{3pi}{4}right)}) and (boxed{left(frac{pi}{4}, -frac{3pi}{4}right)}).]

🔑:[Solution] Let x, y satisfy the inequality. From sin x + cos x leqslant sqrt{2}, we havebegin{aligned}tan^2 x + 1 & leqslant sqrt{frac{pi}{4} - arctan frac{|x| + |y|}{pi}} + tan^2 x + 1 & leqslant sqrt{2}|tan x|(sin x + cos x) & leqslant 2|tan x|end{aligned}Since tan^2 x + 1 geqslant 2|tan x|, we havebegin{aligned}tan^2 x + 1 & = sqrt{frac{pi}{4} - arctan frac{|x| + |y|}{pi}} + tan^2 x + 1 & = sqrt{2}|tan x|(sin x + cos x) & = 2|tan x|end{aligned}From this, we can deducebegin{array}{l}frac{pi}{4} - arctan frac{|x| + |y|}{pi} = 0 |tan x| = 1, sin x + cos x = sqrt{2}end{array}Thus, x = frac{pi}{4} + 2npi, n in mathbb{Z}, |x| + |y| = pi, which implies n = 0, i.e., x = frac{pi}{4}, y = pm frac{3}{4} pi.Conversely, when x = frac{pi}{4}, y = pm frac{3}{4} pi, the inequality clearly holds. Therefore, the solutions are (x, y) in left{left(frac{pi}{4}, frac{3pi}{4}right), left(frac{pi}{4}, -frac{3pi}{4}right)right}.

🧠:[<think>Okay, let's try to solve this problem step by step. The question says that we have a quadratic function f(x) which satisfies two conditions: first, f(-1) = 0, and second, for all real numbers x, the inequality x ≤ f(x) ≤ (1/2)(x² + 1) holds. We need to find the analytical expression of f(x).Alright, so since f(x) is a quadratic function, its general form should be f(x) = ax² + bx + c, where a, b, and c are constants that we need to determine. The first condition given is f(-1) = 0, which we can use to set up an equation. The second condition is an inequality that must hold for all real numbers x, which might give us more equations or inequalities to solve for a, b, and c.Let me start with the first condition. Plugging x = -1 into the general quadratic form:f(-1) = a(-1)² + b(-1) + c = a - b + c = 0.So that's our first equation: a - b + c = 0. Got that.Now the second condition is that for all real x, x ≤ f(x) ≤ (1/2)(x² + 1). So, both inequalities must hold for every real number x. Let's break this down into two separate inequalities:1. f(x) ≥ x for all x.2. f(x) ≤ (1/2)(x² + 1) for all x.Since f(x) is quadratic, these inequalities must hold for all x, which suggests that the quadratic expressions f(x) - x and (1/2)(x² + 1) - f(x) are always non-negative. So, we can write:1. ax² + bx + c - x ≥ 0 for all x. Simplify this: ax² + (b - 1)x + c ≥ 0 for all x.2. (1/2)(x² + 1) - (ax² + bx + c) ≥ 0 for all x. Let's expand that: (1/2)x² + 1/2 - ax² - bx - c ≥ 0. Combine like terms: (1/2 - a)x² - bx + (1/2 - c) ≥ 0 for all x.So now we have two quadratic inequalities that must hold for all real x. For a quadratic expression to be non-negative for all real x, two conditions must be satisfied: the leading coefficient must be non-negative, and the discriminant must be less than or equal to zero. Similarly, if the quadratic is non-positive, the leading coefficient must be non-positive and the discriminant ≤ 0. But here, both quadratics need to be non-negative for all x.Let's start with the first inequality: ax² + (b - 1)x + c ≥ 0 for all x.Since this quadratic must be non-negative everywhere, its leading coefficient a must be positive (or zero? Wait, but if a is zero, then it's a linear function, which can't be non-negative everywhere unless it's a constant function. But in our case, f(x) is quadratic, so a cannot be zero. So a must be positive. So a > 0.Additionally, the discriminant must be less than or equal to zero. The discriminant D1 of this quadratic is:D1 = (b - 1)^2 - 4ac ≤ 0.Similarly, for the second inequality: (1/2 - a)x² - bx + (1/2 - c) ≥ 0 for all x.Here, the leading coefficient is (1/2 - a). For this quadratic to be non-negative everywhere, the leading coefficient must be non-negative, so 1/2 - a ≥ 0 ⇒ a ≤ 1/2. Also, the discriminant D2 must be ≤ 0:D2 = (-b)^2 - 4*(1/2 - a)*(1/2 - c) ≤ 0.So, putting all together, we have the following equations and inequalities:1. a - b + c = 0 (from f(-1) = 0)2. a > 0 (from the first inequality)3. (b - 1)^2 - 4ac ≤ 0 (from the first inequality)4. a ≤ 1/2 (from the second inequality)5. b² - 4*(1/2 - a)*(1/2 - c) ≤ 0 (from the second inequality)So now we need to solve this system of equations and inequalities to find a, b, c.First, from equation 1: c = b - a. Let's substitute c into the other equations.So c = b - a.Now, substitute c = b - a into inequality 3:(b - 1)^2 - 4a*(b - a) ≤ 0.Let's expand this:(b² - 2b + 1) - 4ab + 4a² ≤ 0.Combine like terms:b² - 2b + 1 - 4ab + 4a² ≤ 0.Similarly, substitute c = b - a into inequality 5:b² - 4*(1/2 - a)*(1/2 - (b - a)) ≤ 0.First, compute 1/2 - (b - a) = 1/2 - b + a.So inequality 5 becomes:b² - 4*(1/2 - a)*(1/2 - b + a) ≤ 0.Let me compute the product inside:(1/2 - a)(1/2 - b + a) = (1/2 - a)(1/2 - b + a).Let me expand this:Multiply term by term:First term: 1/2*(1/2 - b + a) = (1/2)(1/2) - (1/2)b + (1/2)a = 1/4 - (b)/2 + a/2.Second term: -a*(1/2 - b + a) = -a/2 + ab - a².So adding both parts together:1/4 - (b)/2 + a/2 - a/2 + ab - a² = 1/4 - b/2 + ab - a².So the product is 1/4 - (b)/2 + ab - a².Therefore, inequality 5 becomes:b² - 4*(1/4 - b/2 + ab - a²) ≤ 0.Multiply the 4 into the terms:b² - [4*(1/4) - 4*(b)/2 + 4ab - 4a²] ≤ 0.Simplify each term:4*(1/4) = 14*(b)/2 = 2b4ab = 4ab4a² = 4a²So:b² - [1 - 2b + 4ab - 4a²] ≤ 0.Distribute the negative sign:b² - 1 + 2b - 4ab + 4a² ≤ 0.Combine like terms:4a² - 4ab + b² + 2b - 1 ≤ 0.Note that 4a² - 4ab + b² is (2a - b)^2.So:(2a - b)^2 + 2b - 1 ≤ 0.Hmm, that's an interesting expression. Let's hold onto that for a moment.So now we have two inequalities:From inequality 3:b² - 4ab + 4a² - 2b + 1 ≤ 0. Wait, no, earlier after substituting, inequality 3 was:b² - 2b + 1 - 4ab + 4a² ≤ 0.Which can also be written as:4a² - 4ab + b² - 2b + 1 ≤ 0.Similarly, the expression (2a - b)^2 - 2b + 1 ≤ 0.Wait, because 4a² -4ab +b² is (2a - b)^2, so inequality 3 is:(2a - b)^2 - 2b + 1 ≤ 0.Similarly, inequality 5, after substitution, was:(2a - b)^2 + 2b -1 ≤ 0.So we have:From inequality 3: (2a - b)^2 - 2b + 1 ≤ 0.From inequality 5: (2a - b)^2 + 2b - 1 ≤ 0.Let me denote S = (2a - b)^2. Then the inequalities become:1. S - 2b + 1 ≤ 0.2. S + 2b - 1 ≤ 0.If we add these two inequalities together, we get:2S + 0 ≤ 0 ⇒ 2S ≤ 0 ⇒ S ≤ 0.But S is a square of a real number, so S ≥ 0. Hence, S must be equal to 0. Therefore, 2a - b = 0 ⇒ b = 2a.So, b = 2a.So now we can substitute b = 2a into our equations.First, from equation 1: c = b - a = 2a - a = a. So c = a.Therefore, our quadratic function is f(x) = ax² + 2a x + a.We can factor out a: f(x) = a(x² + 2x + 1) = a(x + 1)^2.But we also know that f(-1) = 0, which is satisfied because plugging x = -1 gives a(0)^2 = 0, regardless of a. So that's consistent.Now, we need to use the inequalities to find a.Recall that from inequality 2: a > 0.From inequality 4: a ≤ 1/2.So a must be in (0, 1/2].Now, let's substitute b = 2a and c = a into inequalities 3 and 5. But since we already forced S = 0 by adding the inequalities, perhaps we need to check these inequalities with S = 0.Wait, let's substitute b = 2a into the two inequalities.First, inequality 3:(2a - b)^2 - 2b + 1 ≤ 0. But since b = 2a, then (2a - 2a)^2 - 2*(2a) +1 ≤ 0 ⇒ 0 -4a +1 ≤ 0 ⇒ -4a +1 ≤ 0 ⇒ -4a ≤ -1 ⇒ a ≥ 1/4.Similarly, inequality 5:(2a - b)^2 + 2b -1 ≤ 0. Again, with b = 2a:0 + 2*(2a) -1 ≤ 0 ⇒ 4a -1 ≤ 0 ⇒ 4a ≤ 1 ⇒ a ≤ 1/4.Therefore, combining these two results:From inequality 3: a ≥ 1/4.From inequality 5: a ≤ 1/4.Therefore, a must be exactly 1/4.So a = 1/4.Therefore, b = 2a = 1/2.c = a = 1/4.Therefore, the quadratic function is f(x) = (1/4)x² + (1/2)x + 1/4.Let me check if this satisfies all the given conditions.First, f(-1) = (1/4)(1) + (1/2)(-1) + 1/4 = 1/4 - 1/2 + 1/4 = (1/4 + 1/4) - 1/2 = 1/2 - 1/2 = 0. That's correct.Now, check if x ≤ f(x) ≤ (1/2)(x² +1) for all x.Let's first check f(x) ≥ x.f(x) - x = (1/4)x² + (1/2)x + 1/4 - x = (1/4)x² - (1/2)x + 1/4.This quadratic should be ≥ 0 for all x. Let's compute its discriminant:D = (-1/2)^2 - 4*(1/4)*(1/4) = 1/4 - 4*(1/16) = 1/4 - 1/4 = 0.So the discriminant is 0, which means the quadratic touches the x-axis at one point, i.e., it's a perfect square. Let's see:(1/4)x² - (1/2)x + 1/4 = (1/4)(x² - 2x +1) = (1/4)(x -1)^2 ≥ 0 for all x. Correct.Now check the upper bound: f(x) ≤ (1/2)(x² +1).Compute (1/2)(x² +1) - f(x):(1/2)x² + 1/2 - (1/4)x² - (1/2)x - 1/4 = (1/2 - 1/4)x² - (1/2)x + (1/2 - 1/4) = (1/4)x² - (1/2)x + 1/4.Again, this is a quadratic. Let's check if it's non-negative for all x.Compute discriminant:D = (-1/2)^2 - 4*(1/4)*(1/4) = 1/4 - 4*(1/16) = 1/4 - 1/4 = 0.So again, discriminant is zero, which means it's a perfect square:(1/4)x² - (1/2)x + 1/4 = (1/4)(x² - 2x +1) = (1/4)(x -1)^2 ≥ 0 for all x. Correct.So both inequalities hold for all x, with equality when (x - 1)^2 = 0, i.e., at x = 1. So at x = 1, f(x) = (1/4)(1)^2 + (1/2)(1) + 1/4 = 1/4 + 1/2 + 1/4 = 1. And (1/2)(1 +1) = 1. So f(1) =1 and the upper bound is 1, so equality holds. Similarly, for the lower bound, f(x) - x = (1/4)(x -1)^2 ≥ 0, so equality holds at x =1.Therefore, this function f(x) = (1/4)x² + (1/2)x + 1/4 satisfies all the given conditions.Alternatively, we can write this as f(x) = (1/4)(x² + 2x + 1) = (1/4)(x +1)^2, but since we have f(-1)=0, this form makes sense because (x +1)^2 is zero at x = -1. Wait, but (1/4)(x +1)^2 would be zero at x = -1, which matches f(-1)=0. Wait, but according to our previous expression, f(x) = (1/4)x² + (1/2)x +1/4, which is indeed equal to (1/4)(x² + 2x +1) = (1/4)(x +1)^2. Wait a second, but (x +1)^2 is zero when x = -1, so f(-1) would be (1/4)(0)^2 = 0, which is correct. So, this function is (1/4)(x +1)^2, which is a parabola opening upwards with vertex at (-1, 0). But wait, (x +1)^2 is zero at x = -1, and positive otherwise. However, according to the problem, the quadratic function is supposed to satisfy x ≤ f(x) for all real x. But if f(x) is (1/4)(x +1)^2, then for example, at x =0, f(0)= (1/4)(1)^2=1/4. And 0 ≤1/4, which is true. At x=1, f(1)= (1/4)(4)=1, which is equal to the upper bound. At x=2, f(2)= (1/4)(9)=9/4=2.25, while x=2, so 2 ≤2.25, which is true. However, let's check a negative x other than -1. For example, x=-2. f(-2)= (1/4)((-2 +1)^2)= (1/4)(1)=1/4. And x=-2 ≤1/4, which is true. Wait, but x=-2 is less than 1/4, so the inequality holds. What about x= -0.5. f(-0.5)= (1/4)((-0.5 +1)^2)= (1/4)(0.5)^2= (1/4)(0.25)=1/16≈0.0625. So x=-0.5 is -0.5 ≤0.0625, which is true.So actually, this function satisfies x ≤f(x) for all real x. Even when x is negative, f(x) is always non-negative (since it's a square times 1/4), so for negative x, f(x) is non-negative, and x is negative, so x ≤ f(x) holds. For positive x, since f(x) is a quadratic that grows as x increases, and at x=1, it's equal to 1, which is equal to the upper bound.So, everything checks out. Therefore, the analytical expression of f(x) is (1/4)x² + (1/2)x + 1/4, which can be written as (1/4)(x +1)^2.But let me check once more to make sure there are no mistakes. Especially since when I thought about f(x) being (1/4)(x +1)^2, it's zero at x=-1, but the problem states that x ≤f(x) for all x, and at x=-1, f(-1)=0, and x=-1 is equal to f(-1). So, x=-1 is allowed to be equal to f(x). So, the inequality x ≤f(x) holds at x=-1 because -1 ≤0, which is true. Wait, but x=-1, f(-1)=0. So the inequality says -1 ≤0, which is true. So that's okay.Wait a second, but f(-1)=0 is given, and when x=-1, the inequality x ≤f(x) becomes -1 ≤0, which holds. So that's acceptable. So, all conditions are satisfied.Therefore, the answer should be f(x) = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4}, which can also be written as frac{1}{4}(x +1)^2. But since the problem says "analytical expression", either form is acceptable, but perhaps they prefer the expanded form. Let me check:Expanded form: frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4}Factored form: frac{1}{4}(x +1)^2Either is correct, but maybe they want the standard quadratic form. Let me verify with the original conditions. The problem didn't specify any particular form, so both are correct, but perhaps expanded form is more standard for the answer. Let me confirm with the upper and lower bounds.But given that the upper bound is (1/2)(x² +1), and our f(x) is (1/4)x² + (1/2)x +1/4, perhaps writing it in a way that relates to the upper bound. Alternatively, since the function is a perfect square, writing it as frac{1}{4}(x +1)^2 might be more elegant, but again, both forms are correct.However, in the problem statement, they mention "the analytical expression", which usually just means a formula in terms of x, so both forms are analytical expressions. However, in case they expect a specific form, maybe expanded. Let's check both possibilities.Alternatively, perhaps they want the answer in the form frac{1}{4}(x +1)^2. Let me check if that matches the upper and lower bounds.Wait, expanding frac{1}{4}(x +1)^2 gives frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4}, which is the same as before.Therefore, both forms are equivalent. Since the problem didn't specify, but in most cases, unless specified, the standard form is ax² +bx +c. So, maybe present it in the expanded form.Therefore, the analytical expression is f(x) = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4}.Final AnswerThe analytical expression of the function ( f(x) ) is boxed{dfrac{1}{4}x^2 + dfrac{1}{2}x + dfrac{1}{4}}.</think>Given the quadratic function ( f(x) ) satisfies ( f(-1) = 0 ), and ( x leq f(x) leq frac{1}{2}(x^2 + 1) ) for all real numbers ( x ). We need to find the analytical expression of ( f(x) ).1. General form of the quadratic function: ( f(x) = ax^2 + bx + c ).2. Condition ( f(-1) = 0 ): [ a(-1)^2 + b(-1) + c = 0 implies a - b + c = 0 implies c = b - a. ]3. Inequalities: - ( f(x) geq x ) for all ( x ): [ ax^2 + (b - 1)x + c geq 0 quad text{for all } x. ] - ( f(x) leq frac{1}{2}(x^2 + 1) ) for all ( x ): [ left(frac{1}{2} - aright)x^2 - bx + left(frac{1}{2} - cright) geq 0 quad text{for all } x. ]4. Discriminant conditions: - For ( ax^2 + (b - 1)x + c geq 0 ): [ (b - 1)^2 - 4ac leq 0. ] - For ( left(frac{1}{2} - aright)x^2 - bx + left(frac{1}{2} - cright) geq 0 ): [ b^2 - 4left(frac{1}{2} - aright)left(frac{1}{2} - cright) leq 0. ]5. Substituting ( c = b - a ): - From the first discriminant condition: [ (b - 1)^2 - 4a(b - a) leq 0 implies (2a - b)^2 - 2b + 1 leq 0. ] - From the second discriminant condition: [ b^2 - 4left(frac{1}{2} - aright)left(frac{1}{2} - (b - a)right) leq 0 implies (2a - b)^2 + 2b - 1 leq 0. ]6. Solving the system: - Adding the two inequalities: [ 2(2a - b)^2 leq 0 implies 2a - b = 0 implies b = 2a. ] - Substituting ( b = 2a ) and ( c = a ): [ a - 2a + c = 0 implies c = a. ] - Leading coefficients conditions: [ a > 0 quad text{and} quad frac{1}{2} - a geq 0 implies 0 < a leq frac{1}{2}. ] - Solving for ( a ): [ -4a + 1 leq 0 implies a geq frac{1}{4} quad text{and} quad 4a - 1 leq 0 implies a leq frac{1}{4} implies a = frac{1}{4}. ]7. Resulting function: - ( a = frac{1}{4} ), ( b = 2a = frac{1}{2} ), ( c = a = frac{1}{4} ): [ f(x) = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4}. ]The analytical expression of the function ( f(x) ) is boxed{dfrac{1}{4}x^2 + dfrac{1}{2}x + dfrac{1}{4}}.,<think>Okay, so I need to find the analytical expression of a quadratic function f(x) that satisfies two conditions: first, f(-1) = 0, and second, for all real numbers x, the inequality x ≤ f(x) ≤ (x² + 1)/2 holds. Alright, let's break this down step by step.First, since it's a quadratic function, I know f(x) must be of the form f(x) = ax² + bx + c, where a, b, and c are constants that I need to determine. The problem gives me one point that the function passes through: when x = -1, f(-1) = 0. So plugging that into the equation, we get:a(-1)² + b(-1) + c = 0Which simplifies to:a - b + c = 0 --> Equation (1)So that's one equation involving the coefficients a, b, and c. Now, the other condition is that for all real numbers x, x ≤ f(x) ≤ (x² + 1)/2. Let's parse this inequality. It says that the quadratic function f(x) is bounded below by the linear function x and above by the quadratic function (x² + 1)/2. So f(x) is sandwiched between these two functions everywhere on the real line.Since f(x) is a quadratic function, and the upper bound is also a quadratic function, while the lower bound is linear, this might mean that f(x) touches one or both of these bounds at certain points, maybe as tangent lines or points of contact. But since it's a quadratic, and the upper bound is another quadratic, perhaps f(x) coincides with the upper bound? Wait, but the upper bound is (x² +1)/2, which is a quadratic with leading coefficient 1/2. If f(x) is a quadratic function, let's see if it's possible that f(x) is exactly equal to (x² + 1)/2. Let's check if this satisfies the other condition, f(-1) = 0.If f(x) = (x² +1)/2, then f(-1) = ((-1)^2 +1)/2 = (1 +1)/2 = 1. But the problem states that f(-1) = 0. So that's not possible. Therefore, f(x) cannot be equal to the upper bound function. Similarly, if we check the lower bound, which is x, then if f(x) = x, which is linear, but f(x) is supposed to be quadratic. Therefore, f(x) must lie between x and (x² +1)/2 for all x, but not equal to either of them everywhere. However, since f(x) is quadratic, it's possible that it touches one of these bounds at a particular point.But since the inequality must hold for all real x, the quadratic function f(x) must be above the line x everywhere and below the parabola (x² +1)/2 everywhere. Therefore, we can model this as two inequalities:1. f(x) - x ≥ 0 for all x.2. (x² +1)/2 - f(x) ≥ 0 for all x.So both f(x) - x and (x² +1)/2 - f(x) must be non-negative for all real x.Since f(x) is quadratic, both f(x) - x and (x² +1)/2 - f(x) are quadratic functions. For a quadratic function to be non-negative for all real x, it must either be a perfect square (i.e., a square of a linear term) or have a positive leading coefficient and a discriminant less than or equal to zero. Since both of these expressions need to be non-negative everywhere, let's analyze each one.First, let's write f(x) as ax² + bx + c. Then:1. f(x) - x = ax² + (b -1)x + c. This must be ≥ 0 for all x.2. (x² +1)/2 - f(x) = (1/2)x² + 0x + 1/2 - ax² - bx - c = (1/2 - a)x² - bx + (1/2 - c). This must also be ≥ 0 for all x.For the first quadratic, ax² + (b -1)x + c ≥ 0 for all x. For this to hold, the leading coefficient a must be positive, and the discriminant must be ≤ 0.Similarly, for the second quadratic, (1/2 - a)x² - bx + (1/2 - c) ≥ 0 for all x. Here, the leading coefficient is (1/2 - a). For this to be non-negative everywhere, we need (1/2 - a) ≥ 0 (so a ≤ 1/2) and the discriminant of this quadratic must also be ≤ 0.So now we have conditions on the coefficients a, b, c:From the first quadratic (f(x) - x ≥ 0):1. a > 02. Discriminant D1 = (b -1)^2 - 4ac ≤ 0From the second quadratic ((x² +1)/2 - f(x) ≥ 0):3. 1/2 - a ≥ 0 ⇒ a ≤ 1/24. Discriminant D2 = (-b)^2 - 4*(1/2 - a)*(1/2 - c) ≤ 0Additionally, from the given point f(-1) = 0, we have equation (1):5. a - b + c = 0So now we have five conditions (1 to 5) that must be satisfied. Let's list them again:1. a > 02. (b - 1)^2 - 4ac ≤ 03. a ≤ 1/24. b² - 4*(1/2 - a)*(1/2 - c) ≤ 05. a - b + c = 0Our goal is to find a, b, c that satisfy all these conditions. Let's see if we can express c in terms of a and b from equation 5.From equation 5: c = b - aTherefore, we can substitute c = b - a into the other equations.So let's substitute c = b - a into conditions 2 and 4.Condition 2 becomes:(b - 1)^2 - 4a(b - a) ≤ 0Let's expand this:= (b² - 2b + 1) - 4ab + 4a² ≤ 0= b² - 2b + 1 -4ab +4a² ≤ 0Similarly, condition 4:b² - 4*(1/2 - a)*(1/2 - c) ≤ 0But c = b - a, so substitute:= b² -4*(1/2 - a)*(1/2 - (b - a)) ≤ 0Simplify the terms inside the brackets:1/2 - (b - a) = 1/2 - b + a = (1/2 + a) - bSo:= b² -4*(1/2 - a)*(1/2 + a - b) ≤ 0Let me expand this term:First, compute (1/2 - a)*(1/2 + a - b):= (1/2 - a)*(1/2 + a - b)Let me denote this as product:Let me use distributive property:= (1/2)(1/2 + a - b) - a(1/2 + a - b)= (1/4 + (a)/2 - (b)/2) - (a/2 + a² - ab)= 1/4 + a/2 - b/2 - a/2 - a² + abSimplify:1/4 - b/2 - a² + abSo, the entire expression for condition 4 becomes:b² -4*(1/4 - b/2 - a² + ab) ≤ 0Multiply the 4 into the brackets:= b² -4*(1/4) +4*(b/2) +4*a² -4*ab ≤ 0= b² -1 + 2b +4a² -4ab ≤ 0So condition 4 simplifies to:b² +2b +4a² -4ab -1 ≤ 0Alright, so now we have:From condition 2:b² -2b +1 -4ab +4a² ≤ 0From condition 4:b² +2b +4a² -4ab -1 ≤ 0Let me write both conditions again:Condition 2:b² -2b +1 -4ab +4a² ≤ 0Condition 4:b² +2b +4a² -4ab -1 ≤ 0Hmm, interesting. Let's write both inequalities:First, condition 2:4a² -4ab + b² -2b +1 ≤ 0Condition 4:4a² -4ab + b² +2b -1 ≤ 0Notice that both have the term 4a² -4ab + b². Let me denote this as (2a - b)^2. Let's check:(2a - b)^2 = 4a² -4ab + b². Yes, exactly. So:Condition 2 becomes:(2a - b)^2 -2b +1 ≤ 0Condition 4 becomes:(2a - b)^2 +2b -1 ≤ 0Let me denote S = (2a - b)^2. Then:Condition 2: S -2b +1 ≤ 0Condition 4: S +2b -1 ≤ 0So adding both inequalities:(S -2b +1) + (S +2b -1) ≤ 0 + 0 ⇒ 2S ≤ 0 ⇒ S ≤ 0But S is a square of a real number, so S ≥ 0. Therefore, 2S ≤ 0 implies S = 0. Hence, S = 0. So (2a - b)^2 = 0 ⇒ 2a - b = 0 ⇒ b = 2aTherefore, from S = 0, we have b = 2a.Now, substitute b = 2a into the two conditions. Let's start with condition 2:S -2b +1 ≤ 0 ⇒ 0 -2*(2a) +1 ≤ 0 ⇒ -4a +1 ≤ 0 ⇒ -4a ≤ -1 ⇒ 4a ≥ 1 ⇒ a ≥ 1/4Similarly, condition 4:S +2b -1 ≤ 0 ⇒ 0 +2*(2a) -1 ≤ 0 ⇒4a -1 ≤ 0 ⇒4a ≤1 ⇒a ≤1/4So combining these two results from conditions 2 and 4 after substituting S=0 (i.e., b=2a):From condition 2: a ≥ 1/4From condition 4: a ≤1/4Therefore, the only possibility is a =1/4.Therefore, a=1/4. Then since b=2a, b=2*(1/4)=1/2. Then c = b -a=1/2 -1/4=1/4.So we have a=1/4, b=1/2, c=1/4.Therefore, the quadratic function is f(x)=(1/4)x² + (1/2)x +1/4.Let me check if this satisfies all the given conditions.First, f(-1)= (1/4)(1) + (1/2)(-1) +1/4=1/4 -1/2 +1/4= (1/4 +1/4) -1/2=1/2 -1/2=0. So that's good, satisfies f(-1)=0.Now check if x ≤ f(x) ≤(x² +1)/2 for all real x.First, check lower bound: f(x) -x = (1/4)x² + (1/2)x +1/4 -x = (1/4)x² - (1/2)x +1/4.Is this always non-negative?Let's compute discriminant D1: (-1/2)^2 -4*(1/4)*(1/4)=1/4 -4*(1/16)=1/4 -1/4=0. So discriminant is zero, which means this quadratic touches the x-axis at one point. Therefore, f(x)-x is a perfect square. Indeed, let's factor:(1/4)x² - (1/2)x +1/4 = (1/4)(x² -2x +1)= (1/4)(x -1)^2 ≥0 for all x. So yes, f(x) -x is always non-negative. So the lower bound is satisfied.Now check upper bound: (x² +1)/2 -f(x)= (x² +1)/2 - (1/4)x² - (1/2)x -1/4.Compute term by term:(x²)/2 +1/2 - (1/4)x² - (1/2)x -1/4Combine like terms:(x²)/2 - (1/4)x² = (2/4 -1/4)x² = (1/4)x²1/2 -1/4 =1/4So overall, (1/4)x² - (1/2)x +1/4.Again, check if this is non-negative for all x.Compute discriminant D2: (-1/2)^2 -4*(1/4)*(1/4)=1/4 -4*(1/16)=1/4 -1/4=0. So discriminant is zero, so it's a perfect square. Let's factor:(1/4)x² - (1/2)x +1/4 = (1/4)(x² -2x +1)= (1/4)(x -1)^2 ≥0 for all x. So yes, the upper bound is also satisfied.Therefore, f(x)= (1/4)x² + (1/2)x +1/4 is the quadratic function that meets all the given conditions.Alternatively, we can write this as (x² + 2x +1)/4, which simplifies to (x +1)^2 /4. Wait, let's check:(1/4)x² + (1/2)x +1/4 = (x² +2x +1)/4 = (x +1)^2 /4. Yes, exactly. So f(x) is (x +1)^2 /4.But let's confirm:(x +1)^2 /4 = (x² +2x +1)/4 = (1/4)x² + (1/2)x +1/4. Yes, that's correct.So the function is f(x) = (x +1)^2 /4.Therefore, the analytical expression of the function f(x) is boxed{dfrac{(x+1)^2}{4}}.Final AnswerThe analytical expression of the function ( f(x) ) is boxed{dfrac{(x+1)^2}{4}}.</think>Given the quadratic function ( f(x) ) satisfies ( f(-1) = 0 ), and ( x leqslant f(x) leqslant frac{1}{2}left(x^{2} + 1right) ) for all real numbers ( x ). We need to find the analytical expression of ( f(x) ).1. Form of the quadratic function: Since ( f(x) ) is a quadratic function, we assume ( f(x) = ax^2 + bx + c ).2. Condition ( f(-1) = 0 ): Substituting ( x = -1 ) into the quadratic function gives: [ a(-1)^2 + b(-1) + c = 0 implies a - b + c = 0 quad text{(Equation 1)} ]3. Inequalities for all ( x ): - Lower bound: ( f(x) - x geq 0 ) - Upper bound: ( frac{1}{2}(x^2 + 1) - f(x) geq 0 )4. Analyzing the lower bound: ( f(x) - x = ax^2 + (b - 1)x + c geq 0 ) - Leading coefficient ( a > 0 ) - Discriminant must be non-positive: ( (b - 1)^2 - 4ac leq 0 )5. Analyzing the upper bound: ( frac{1}{2}(x^2 + 1) - f(x) = left(frac{1}{2} - aright)x^2 - bx + left(frac{1}{2} - cright) geq 0 ) - Leading coefficient ( frac{1}{2} - a geq 0 implies a leq frac{1}{2} ) - Discriminant must be non-positive: ( b^2 - 4left(frac{1}{2} - aright)left(frac{1}{2} - cright) leq 0 )6. Solving the system of equations: - From Equation 1: ( c = b - a ) - Substituting ( c = b - a ) into the discriminant conditions for both bounds: - Lower bound discriminant: ( (b - 1)^2 - 4a(b - a) leq 0 ) - Upper bound discriminant: ( b^2 - 4left(frac{1}{2} - aright)left(frac{1}{2} - (b - a)right) leq 0 )7. Simplifying the discriminants: - Both discriminants lead to the condition ( (2a - b)^2 = 0 implies b = 2a ) - Substituting ( b = 2a ) into the inequalities, we find ( a = frac{1}{4} ), ( b = frac{1}{2} ), and ( c = frac{1}{4} )8. Verification: - The quadratic function ( f(x) = frac{1}{4}x^2 + frac{1}{2}x + frac{1}{4} ) simplifies to ( f(x) = frac{(x + 1)^2}{4} ) - Check ( f(-1) = 0 ), which is satisfied - Verify ( f(x) - x geq 0 ) and ( frac{1}{2}(x^2 + 1) - f(x) geq 0 ) are non-negative for all ( x )Thus, the analytical expression of the function ( f(x) ) is boxed{dfrac{(x+1)^2}{4}}.]

🔑:10. frac{1}{4}(x+1)^{2}.Let f(x)=a x^{2}+b x+c(a neq 0). From the given, when x=1, we have f(1)=1, i.e., a+b+c=1.Also, f(-1)=0, i.e., a-b+c=0.From equations (1) and (2), we get a+c=b=frac{1}{2}.Also, f(x) geqslant x, i.e., a x^{2}+(b-1) x+c geqslant 0 for all real numbers x, which must satisfyleft{begin{array}{l}a>0, (b-1)^{2}-4 a c leqslant 0 .end{array}right.Substituting b=frac{1}{2}, we get a c geqslant frac{1}{16}.Also, c=frac{1}{2}-a, so aleft(frac{1}{2}-aright) geqslant frac{1}{16}, which simplifies to left(a-frac{1}{4}right)^{2} leqslant 0. Solving this, we get a=frac{1}{4}.At this point, f(x)=frac{1}{4} x^{2}+frac{1}{2} x+frac{1}{4}=frac{1}{4}(x+1)^{2}.Similarly, from f(x) leqslant frac{1}{2}left(x^{2}+1right), solving gives a=frac{1}{4}.In summary, the analytical expression of the function f(x) is frac{1}{4}(x+1)^{2}.

🧠:[<think>Okay, so I need to figure out the value of S_k*, which is the sum over all non-empty subsets X of S (where S is {1, 2, ..., n}) that do not contain the element k. For each such subset X = {i1, i2, ..., ir}, we take the reciprocal of the product of its elements, and then sum all those reciprocals. Hmm, let me break this down.First, let me make sure I understand the problem correctly. The set S is {1, 2, 3, ..., n}. For each k in S, we define Sk as the collection of all non-empty subsets of S that do not include k. Then, S_k* is the sum of 1 divided by the product of the elements in each subset X in Sk. So, for example, if n = 3 and k = 1, then Sk would be all non-empty subsets of {2, 3}, which are {2}, {3}, and {2,3}. Then S_k* would be 1/2 + 1/3 + 1/(2*3) = (3 + 2 + 1)/6 = 6/6 = 1. Wait, that's interesting. Let me check that again.If n=3 and k=1, Sk is the subsets not containing 1. So non-empty subsets of {2,3} are indeed {2}, {3}, {2,3}. Then 1/2 + 1/3 + 1/(2*3) = (3 + 2 + 1)/6 = 6/6 = 1. So in this case, S_k* = 1. Hmm, maybe there's a pattern here.Let me try another example. If n=2 and k=1, then Sk is non-empty subsets not containing 1, which are subsets of {2}, so just {2}. Then S_k* = 1/2. Similarly, if k=2, then Sk would be {1}, so S_k* = 1/1 = 1. Wait, hold on. If n=2 and k=2, then Sk is all non-empty subsets not containing 2, which are subsets of {1}, so {1}, and 1/1 =1. But if k=1, then Sk is {2}, so 1/2. So for n=2, depending on k, S_k* is either 1 or 1/2. Hmm, so maybe there's a relationship here with harmonic numbers or something?Wait, but in the first example, when n=3 and k=1, S_k* was 1. Let me check n=3 and k=2. Then Sk would be subsets not containing 2, which are subsets of {1,3}. Non-empty subsets: {1}, {3}, {1,3}. Then S_k* = 1/1 + 1/3 + 1/(1*3) = 1 + 1/3 + 1/3 = 1 + 2/3 = 5/3 ≈ 1.666... So here, it's 5/3. Hmm, not an integer. So the answer isn't always 1. So maybe my initial thought was wrong.Wait, but when k=1, in n=3, the sum was 1, but when k=2, it's 5/3. So it depends on k. So maybe the answer is something like (1 + 1/2 + 1/3 + ... + 1/n) - 1/k? Wait, let's check n=3, k=1. The harmonic number H_3 is 1 + 1/2 + 1/3 ≈ 1.833. Then subtract 1/1, which gives 0.833, which is 5/6, but in our case, when k=1, the sum was 1. So that's not matching.Wait, perhaps another approach. Let me think about generating functions. Because the problem is about summing reciprocals over subsets, which reminds me of the product formula for generating functions. For each element in S, except k, we can consider whether to include it or not in the subset. For each such element, if we include it, we contribute a term 1/(element), and if we exclude it, we contribute 0. Wait, but actually, the sum over all subsets is the product over all elements (1 + 1/i). Because for each element, you can choose to include it (which contributes 1/i) or not include it (which contributes 1). So the generating function for the sum over all subsets (including the empty set) would be the product from i=1 to n (1 + 1/i) excluding the term for k. Wait, since we are considering subsets that do not contain k, so we exclude the k term. So the product would be over all i ≠ k (1 + 1/i). Then, since we need non-empty subsets, we subtract 1 (the term for the empty set). Therefore, S_k* = [Product_{i ≠ k} (1 + 1/i)] - 1.Let me verify this with the examples. For n=3, k=1. Then the product over i=2,3 (1 + 1/i) is (1 + 1/2)(1 + 1/3) = (3/2)(4/3) = 4/2 = 2. Subtract 1, so S_k* = 2 - 1 = 1, which matches. For n=3, k=2, the product is over i=1,3: (1 + 1/1)(1 + 1/3) = (2)(4/3) = 8/3. Subtract 1 gives 5/3, which matches the earlier result. Similarly, for n=2, k=1: product over i=2: (1 + 1/2) = 3/2, subtract 1: 1/2. For k=2: product over i=1: (1 + 1/1) = 2, subtract 1: 1. So this formula works for the examples. Therefore, general formula is S_k* = [Product_{i ∈ S, i ≠ k} (1 + 1/i)] - 1.But the question is to "determine the value of the sum". So we need to compute this product. Let's compute the product first. Let's see:Product_{i=1, i ≠ k}^n (1 + 1/i) = Product_{i=1, i ≠ k}^n ( (i + 1)/i ). Let's write this out:For i from 1 to n, excluding k, the term is (i + 1)/i. So when we take the product, we can telescope.Wait, if we write (i + 1)/i for each i ≠ k, then the product would be:[ (2/1) * (3/2) * ... * (k)/(k-1) ) ] * [ ( (k+2)/(k+1) ) * ... * (n+1)/n ) ]Wait, because when i goes from 1 to n, skipping k, then for i=1 to k-1, we have terms (2/1)(3/2)...(k/(k-1)) which telescopes to k/1 = k. Then for i=k+1 to n, we have terms ( (k+2)/(k+1) ) * ... * (n+1)/n ), which telescopes to (n+1)/(k+1). Therefore, the entire product would be k * (n+1)/(k+1) ) = k(n+1)/(k+1). Then subtract 1 to get S_k*.Wait, let's check with n=3, k=1. Then product is [ (2/1) * (4/3) ] ? Wait, no, wait, when k=1, the terms are i=2,3. So for i=2: (3/2), for i=3: (4/3). Wait, but according to the previous logic, if k=1, then the first part is from i=1 to k-1 = 0, which doesn't exist. Then the second part is from i=2 to 3 (since we skip k=1). So the product is ( (2+1)/2 ) * ( (3+1)/3 ) = (3/2)(4/3) = 4/2 = 2. Then 2 -1 =1, which is correct. Wait, but according to the previous telescoping logic, when k=1, the product would be (n+1)/(k+1) * k, but k=1, so (4/2)*1 = 2, which is correct. Wait, but in the case of k=1, the product is [ (i+1)/i for i=2,3 ] which is (3/2)(4/3) = 4/2 = 2. So yeah, according to the formula k(n+1)/(k+1) where k=1, 1*4/2=2. Then subtract 1, getting 1. Correct.Similarly, for n=3, k=2. The product is (2/1)*(4/3) = 8/3. Then 8/3 -1 = 5/3. Wait, but according to the formula k(n+1)/(k+1) -1. k=2, n=3: 2*4/3 -1 = 8/3 -1 =5/3. Correct.For n=2, k=1: product is (3/2). Then 3/2 -1 =1/2. Formula: 1*3/(1+1) -1=3/2 -1=1/2. Correct. For k=2: product is 2/1=2. 2 -1=1. Formula: 2*3/(2+1)-1=6/3 -1=2 -1=1. Correct.So general formula is S_k* = [k(n + 1)/(k + 1)] - 1 = [k(n + 1) - (k + 1)]/(k +1) = [kn + k -k -1]/(k +1) = (kn -1)/(k +1). Wait, let's compute the algebra:Original expression: Product_{i ≠k} (1 + 1/i) -1 = [k(n +1)/(k +1)] -1.Let me verify with the examples. For n=3, k=1: [1*(4)/2] -1 = 4/2 -1 =2 -1=1. Correct.For n=3, k=2: [2*4/3] -1=8/3 -1=5/3. Correct.For n=2, k=1: [1*3/2] -1=3/2 -1=1/2. Correct.For n=2, k=2: [2*3/3] -1=6/3 -1=2 -1=1. Correct.So the formula [k(n +1)/(k +1)] -1 simplifies to (k(n +1) - (k +1))/(k +1) = (kn +k -k -1)/(k +1)= (kn -1)/(k +1). So S_k* = (kn -1)/(k +1). Wait, let's check with n=3, k=1: (1*3 -1)/(1 +1)= (3 -1)/2=2/2=1. Correct.n=3, k=2: (2*3 -1)/(2 +1)= (6 -1)/3=5/3. Correct.n=2, k=1: (1*2 -1)/2=1/2. Correct.n=2, k=2: (2*2 -1)/3= (4 -1)/3=1. Correct. So yeah, it seems S_k*=(kn -1)/(k +1).But wait, let me check another example to be thorough. Let's take n=4, k=2. Then Sk is all non-empty subsets not containing 2. So subsets of {1,3,4}. The non-empty subsets are {1}, {3}, {4}, {1,3}, {1,4}, {3,4}, {1,3,4}. Then S_k* =1 + 1/3 +1/4 +1/(1*3) +1/(1*4) +1/(3*4) +1/(1*3*4). Let's compute that:1 + 1/3 +1/4 = (12 + 4 + 3)/12 =19/12≈1.583.Then 1/(1*3)=1/3≈0.333, 1/(1*4)=1/4≈0.25, 1/(3*4)=1/12≈0.083, 1/(1*3*4)=1/12≈0.083. Adding these: 0.333 +0.25 +0.083 +0.083≈0.749. Then total sum≈1.583 +0.749≈2.332.Now according to the formula, S_k*=(kn -1)/(k +1)= (2*4 -1)/(2 +1)= (8 -1)/3=7/3≈2.333. Which matches the approximate sum. So it's correct. So the formula works.Therefore, the general formula is S_k*=(kn -1)/(k +1).But let me check the algebra again. So the product over i≠k of (1 + 1/i) is equal to k(n +1)/(k +1). How?Because when we take the product over i=1 to n, skipping k, of (1 + 1/i) = product over i=1 to n, i≠k of (i +1)/i.Let me write that as:Product_{i=1, i≠k}^n (i +1)/i.This can be split into two parts: the product from i=1 to k-1, and from i=k+1 to n.First part: i=1 to k-1: (2/1)(3/2)...(k/(k-1)) = k/1 =k.Second part: i=k+1 to n: ((k+2)/(k+1))((k+3)/(k+2))...((n+1)/n) = (n+1)/(k+1).Multiplying these two parts: k*(n+1)/(k+1). Therefore, the product is k(n+1)/(k+1). Then subtracting 1 gives S_k* = [k(n+1)/(k+1)] -1 = [k(n+1) - (k+1)]/(k+1) = [kn +k -k -1]/(k+1) = (kn -1)/(k+1). Yes, so that's correct.Therefore, the answer is (kn -1)/(k+1). So boxed, it would be boxed{dfrac{nk - 1}{k + 1}}.But wait, just to make sure. Let me test n=1. Wait, if n=1, then S={1}, and k must be 1. Then Sk is subsets of S not containing 1, which are empty set. But the problem says subsets must be non-empty. Therefore, Sk is empty, so S_k* is 0. According to formula: (1*1 -1)/(1 +1)= (1 -1)/2=0/2=0. Correct. So even edge case works.Another edge case: n=1, k=1. Correct. n=0? Well, n starts at 1.Therefore, the formula holds for all n ≥1 and k in S.Therefore, the answer is (nk -1)/(k +1).Final Answerboxed{dfrac{nk - 1}{k + 1}}</think>To determine the value of the sum ( S_k^* ) for the set ( S = {1, 2, ldots, n} ) and each ( k in S ), we need to consider the sum of the reciprocals of the products of the elements of all non-empty subsets of ( S ) that do not contain ( k ).First, we recognize that the sum over all non-empty subsets of ( S ) excluding ( k ) can be expressed using a generating function approach. The generating function for the product of reciprocals over subsets excluding ( k ) is given by the product:[prod_{substack{i=1 i neq k}}^n left(1 + frac{1}{i}right)]This product can be simplified by recognizing it as a telescoping product. Breaking it down, we have:[prod_{substack{i=1 i neq k}}^n left(1 + frac{1}{i}right) = prod_{substack{i=1 i neq k}}^n frac{i + 1}{i}]This product splits into two parts: one from ( i = 1 ) to ( i = k - 1 ) and the other from ( i = k + 1 ) to ( i = n ):1. For ( i = 1 ) to ( k - 1 ): [ frac{2}{1} cdot frac{3}{2} cdot ldots cdot frac{k}{k-1} = frac{k}{1} ]2. For ( i = k + 1 ) to ( n ): [ frac{k+2}{k+1} cdot frac{k+3}{k+2} cdot ldots cdot frac{n+1}{n} = frac{n+1}{k+1} ]Multiplying these two parts together gives:[k cdot frac{n+1}{k+1} = frac{k(n+1)}{k+1}]Subtracting 1 to account for the exclusion of the empty set, we get:[S_k^* = frac{k(n+1)}{k+1} - 1 = frac{k(n+1) - (k+1)}{k+1} = frac{kn + k - k - 1}{k+1} = frac{kn - 1}{k + 1}]Thus, the value of the sum ( S_k^* ) is:[boxed{dfrac{nk - 1}{k + 1}}],<think>Okay, let's see. The problem is about a set S = {1, 2, ..., n}, and for each k in S, we define S_k as the set of all non-empty subsets X of S such that k is not in X and X is not empty. Then, we need to find the sum S_k* which is the sum over all subsets X in S_k of 1 divided by the product of the elements in X. Hmm, okay.First, let me make sure I understand the problem correctly. For each k, S_k consists of all subsets of S that don't contain k and are not empty. So, for example, if n = 3 and k = 2, then S_k would be { {1}, {3}, {1,3} }, right? Then S_k* would be 1/1 + 1/3 + 1/(1*3) = 1 + 1/3 + 1/3 = 1.666... or 5/3. Hmm. But maybe there's a pattern here.So the problem is to find a general formula for S_k* for any n and k. Let me think. Maybe there's a generating function approach here. The sum over all subsets of a set, where each subset is weighted by the reciprocal of the product of its elements. That sounds like a product of terms like (1 + 1/i) for each element i, but subtracting 1 for the empty set. Wait, let's see.For a general set T, the sum over all non-empty subsets X of T of 1/(product of elements in X) would be equal to the product over all elements i in T of (1 + 1/i) minus 1 (to subtract the empty set). So if T is S with k excluded, then S_k* would be [product_{i ≠ k} (1 + 1/i)] - 1. Is that right?Let me verify with n=3 and k=2. Then T = {1,3}, so the product is (1 + 1/1)(1 + 1/3) = 2 * (4/3) = 8/3. Then subtract 1 gives 8/3 - 1 = 5/3, which matches the example I had earlier. So that seems correct. So the general formula would be S_k* = [product_{i=1, i≠k}^n (1 + 1/i)] - 1.But the problem is to "determine the value of the sum S_k*", so perhaps we can simplify this product further. Let's try to compute the product from i=1 to n of (1 + 1/i) first, and then see what happens when we exclude the term for i=k.Wait, the product from i=1 to n of (1 + 1/i). Let's compute this. (1 + 1/1)(1 + 1/2)(1 + 1/3)...(1 + 1/n). Each term is (i + 1)/i. So product_{i=1}^n ( (i + 1)/i ) = (2/1)(3/2)(4/3)...(n+1/n) ) = (n + 1)/1 = n + 1. That's a telescoping product. So the product from i=1 to n of (1 + 1/i) is n + 1.Therefore, if we exclude the term i=k, then the product becomes [product_{i=1}^n (1 + 1/i)] / (1 + 1/k) ) = (n + 1) / ( (k + 1)/k ) ) = (n + 1) * (k / (k + 1)) ). Therefore, the product over all i ≠ k of (1 + 1/i) is (n + 1) * k / (k + 1). Then subtract 1 to get S_k*.So S_k* = ( (n + 1)*k / (k + 1) ) - 1. Let's check with our previous example where n=3 and k=2. Then S_k* = (4*2 / 3) -1 = 8/3 -1 = 5/3, which matches. Good.Simplify the expression: ( (n + 1)k ) / (k + 1) - 1. Let's write that as a single fraction. The common denominator is (k + 1):[ (n + 1)k - (k + 1) ] / (k + 1) = [ (n + 1)k - k - 1 ] / (k + 1) = [ nk + k - k - 1 ] / (k + 1 ) = [ nk - 1 ] / (k + 1 )Wait, hold on: (n + 1)k - k -1 = nk + k - k -1 = nk -1. So S_k* = (nk -1)/(k + 1). Wait, but let's check with n=3 and k=2. Then (3*2 -1)/(2 +1)= (6 -1)/3=5/3, which is correct. So this seems to hold. So the formula is (nk -1)/(k +1).But let me check another example to be sure. Let's take n=2, k=1. Then S_k is the set of non-empty subsets not containing 1, which would be { {2} }, so S_k* = 1/2. According to the formula, (2*1 -1)/(1 +1)= (2 -1)/2=1/2. Correct.Another example: n=4, k=1. S_k is all non-empty subsets not containing 1, so subsets of {2,3,4}. The sum would be 1/2 +1/3 +1/4 +1/(2*3) +1/(2*4)+1/(3*4)+1/(2*3*4). Let's compute this:1/2 = 0.51/3 ≈0.3331/4=0.251/6≈0.16671/8=0.1251/12≈0.08331/24≈0.0417Adding these up: 0.5 + 0.333 = 0.833; +0.25=1.083; +0.1667=1.25; +0.125=1.375; +0.0833=1.4583; +0.0417≈1.5.Wait, so total is 1.5, which is 3/2. According to the formula, (4*1 -1)/(1+1)= (4 -1)/2=3/2. Correct. So that works.Therefore, the general formula is (nk -1)/(k +1). Wait, but let's confirm the derivation again.We have that the product over all i ≠k of (1 +1/i) is (n +1)*k/(k +1). Then subtract 1 gives (n +1)k/(k +1) -1 = [ (n +1)k - (k +1) ]/(k +1) = [nk +k -k -1]/(k +1) = (nk -1)/(k +1). Yep, that seems right.So the answer is (nk -1)/(k +1). Therefore, S_k* = (nk -1)/(k +1).But let me just verify with n=1. Wait, n=1, S={1}, then for k=1, S_k is the set of subsets not containing 1 and non-empty. But the only subset not containing 1 is the empty set, so S_k is empty. Therefore, S_k* is 0. Using the formula, (1*1 -1)/(1 +1) = (1 -1)/2=0/2=0. Correct. So that works.Another test case: n=4, k=4. Then S_k is all non-empty subsets of {1,2,3}. The sum would be the same as when k=1 in n=3? Wait, but let's compute according to the formula: (4*4 -1)/(4 +1)= (16 -1)/5=15/5=3. Let's compute the sum:Subsets of {1,2,3}:Single elements: 1 + 1/2 +1/3 ≈1 +0.5 +0.333≈1.833Pairs: 1/(1*2)=0.5, 1/(1*3)≈0.333, 1/(2*3)≈0.1667; sum≈0.5+0.333+0.1667≈1.0Triple: 1/(1*2*3)=1/6≈0.1667Total sum≈1.833 +1.0 +0.1667≈3.0, which is 3. So formula gives 15/5=3. Correct.So this formula seems to hold. Therefore, I think the answer is (nk -1)/(k +1). Thus, boxed, it would be boxed{dfrac{nk - 1}{k + 1}}.Final Answerboxed{dfrac{nk - 1}{k + 1}}</think>To determine the value of the sum ( S_{k}^{*} ) for the set ( S = {1, 2, ldots, n} ) and each ( k in S ), we start by noting that ( S_k ) consists of all non-empty subsets of ( S ) that do not contain ( k ). The sum ( S_{k}^{*} ) is the sum over all such subsets ( X ) of ( frac{1}{text{product of elements in } X} ). This can be expressed using a product of terms. Specifically, for a set ( T ), the sum over all non-empty subsets of ( T ) of the reciprocals of their products is given by ( prod_{i in T} left(1 + frac{1}{i}right) - 1 ).For ( S setminus {k} ), this product becomes ( prod_{i neq k} left(1 + frac{1}{i}right) - 1 ). The product ( prod_{i=1}^{n} left(1 + frac{1}{i}right) ) is a telescoping product that simplifies to ( n + 1 ). Excluding the term for ( i = k ), we get:[prod_{i neq k} left(1 + frac{1}{i}right) = frac{(n + 1) cdot k}{k + 1}]Subtracting 1 to exclude the empty set, we have:[S_{k}^{*} = frac{(n + 1) cdot k}{k + 1} - 1]Simplifying this expression:[S_{k}^{*} = frac{(n + 1)k - (k + 1)}{k + 1} = frac{nk - 1}{k + 1}]Thus, the value of the sum ( S_{k}^{*} ) is:[boxed{dfrac{nk - 1}{k + 1}}]]

🔑:1. Consider the set ( S = {1, 2, ldots, n} ). For every ( k in S ), define ( S_k = {X subseteq S mid k notin X, X neq emptyset} ). We need to determine the value of the sum: [ S_k^* = sum_{{i_1, i_2, ldots, i_r} in S_k} frac{1}{i_1 cdot i_2 cdot ldots cdot i_r} ]2. First, evaluate ( S_0^* ) (the sum over all non-empty subsets of ( S )): [ S_0^* = sum_{{i_1, i_2, ldots, i_r} subseteq S} frac{1}{i_1 cdot i_2 cdot ldots cdot i_r} ] For example, when ( n = 3 ): [ S_0^* = frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{1 cdot 2} + frac{1}{1 cdot 3} + frac{1}{2 cdot 3} + frac{1}{1 cdot 2 cdot 3} ]3. We can use the following identity to simplify the sum: [ 1 + S_0^* = left(1 + frac{1}{1}right) left(1 + frac{1}{2}right) cdots left(1 + frac{1}{n}right) ] This product expands to: [ frac{2}{1} cdot frac{3}{2} cdots frac{n+1}{n} = n + 1 ] Therefore: [ S_0^* = n ]4. Now, to find ( S_k^* ). If we remove the term ( frac{1}{k} ) from ( S_0^* ), then for every remaining term with ( k ) in its denominator subset, there is a term with the same subset except without ( k ). So the sum of the latter group is ( k ) times larger than that of the former. But the latter group consists of all the terms without ( k ) in their denominators, therefore its sum is ( S_k^* ). Therefore: [ S_0^* - frac{1}{k} = frac{k+1}{k} S_k^* ]5. Solving for ( S_k^* ): [ S_k^* = frac{k S_0^* - 1}{k + 1} = frac{nk - 1}{k + 1} ] for ( k in S ). Otherwise, ( S_k^* = n ).The final answer is ( boxed{frac{nk - 1}{k + 1}} ) for ( k in S ).

🧠:[<think>Okay, let's see. The problem says that at a meeting, each pair of attendees shook hands once, and the total number of handshakes was 66. We need to figure out how many people attended the meeting, and the options are 8, 10, 12, or 14. Alright, so handshakes between pairs... this sounds like a combinations problem.Hmm, right, when each person shakes hands with every other person exactly once, the total number of handshakes can be calculated using combinations. The formula for combinations is C(n, 2) = n(n - 1)/2, where n is the number of people. Because each handshake is a unique pair, we don't want to count duplicates, so combinations make sense here.So the equation we need is n(n - 1)/2 = 66. Let me write that down:n(n - 1)/2 = 66Our goal is to solve for n. Let's multiply both sides by 2 to eliminate the denominator:n(n - 1) = 132Now, this is a quadratic equation. Let's expand the left side:n² - n = 132Bring 132 to the left side to set the equation to zero:n² - n - 132 = 0Alright, now we need to solve this quadratic equation. Let's see if we can factor it. We need two numbers that multiply to -132 and add up to -1. Hmm, let's think.Factors of 132: 1 & 132, 2 & 66, 3 & 44, 4 & 33, 6 & 22, 11 & 12.We need a pair that subtracts to 1 (since the product is negative and the middle term is -n). Let's see: 12 and 11. 12 - 11 = 1. If we use 12 and -11, their product is -132 and sum is 1. Wait, but the coefficient of the middle term is -1. Let me check:The quadratic is n² - n - 132. So when factoring, we need two numbers that multiply to -132 and add to -1. Let me think again. If the numbers are 12 and -11, then 12 * (-11) = -132 and 12 + (-11) = 1. But we need them to add up to -1, so maybe -12 and 11? Let's see: -12 * 11 = -132, and -12 + 11 = -1. Yes, that works!Therefore, we can factor the quadratic as:(n - 12)(n + 11) = 0Setting each factor equal to zero gives:n - 12 = 0 => n = 12orn + 11 = 0 => n = -11But since the number of people can't be negative, we discard n = -11. So the solution is n = 12. Let me check the answer choices: option C is 12. That must be the answer.But wait, let me verify with the original equation to be sure. If there are 12 people, then the number of handshakes is 12*11/2 = 132/2 = 66. Yes, that matches the given total. So 12 is correct.Alternatively, if I didn't remember the combination formula, I could approach it incrementally. Let's see, for example, if there are n people, each person shakes hands with (n-1) others. But this counts each handshake twice (once for each person), so we divide by 2. So the total handshakes are n(n - 1)/2, which is the same formula. So that's consistent.Another way: Maybe plug in the answer choices into the formula to see which gives 66. Let's try:Option A: 8 people. 8*7/2 = 28. Not 66.Option B: 10 people. 10*9/2 = 45. Still not 66.Option C: 12 people. 12*11/2 = 66. Bingo.Option D: 14 people. 14*13/2 = 91. Too big.So that confirms that the answer is C: 12.Wait, but just to be thorough, is there any other way this could be interpreted? For example, if some people didn't shake hands with everyone, but the problem states "each pair of attendees shook hands once," so that should mean everyone shakes hands with everyone else exactly once. So no, the combination formula is definitely the way to go here. No need to consider incomplete handshakes.Therefore, the answer is 12 people. Option C.Final Answerboxed{C}</think>To determine the number of people who attended the meeting, we use the formula for the number of handshakes, which is a combination problem. The formula for the number of handshakes when each pair of attendees shakes hands once is given by:[frac{n(n - 1)}{2} = 66]where ( n ) is the number of people. Multiplying both sides by 2 to clear the denominator:[n(n - 1) = 132]This simplifies to the quadratic equation:[n^2 - n - 132 = 0]Factoring the quadratic equation, we look for two numbers that multiply to -132 and add to -1. These numbers are -12 and 11, so the equation factors as:[(n - 12)(n + 11) = 0]Setting each factor equal to zero gives the solutions ( n = 12 ) and ( n = -11 ). Since the number of people cannot be negative, we discard ( n = -11 ). Thus, the solution is ( n = 12 ).To verify, we substitute ( n = 12 ) back into the handshake formula:[frac{12 times 11}{2} = 66]This confirms that the number of people who attended the meeting is indeed 12.[boxed{C}],<think>Okay, let's see. The problem is about figuring out how many people attended a meeting where each pair shook hands once, and the total number of handshakes was 66. The options are 8, 10, 12, or 14. Hmm. I remember something about handshakes and combinations. Maybe it's a combination problem where you choose 2 people out of n to shake hands. Let me think.So, if there are n people at the meeting, then the number of handshakes would be the number of ways to choose 2 people from n, since each pair shakes hands once. The formula for combinations is n choose k, which is n! / (k! * (n - k)!). In this case, k is 2 because a handshake involves two people. So the formula would be n(n - 1)/2. That makes sense because the first person can shake hands with n - 1 people, the second person has already shaken hands with the first, so n - 2, and so on. But since each handshake is counted twice in that method, you divide by 2. So the formula is n(n - 1)/2 = number of handshakes.Alright, so we know that the total number of handshakes is 66. So we can set up the equation: n(n - 1)/2 = 66. Then, solving for n. Let's multiply both sides by 2 first to get rid of the denominator: n(n - 1) = 132. So, n squared minus n minus 132 equals 0. That's a quadratic equation. Let me write that down: n² - n - 132 = 0.To solve this quadratic equation, we can use the quadratic formula. The quadratic formula is n = [ -b ± sqrt(b² - 4ac) ] / (2a). In this equation, a is 1, b is -1, and c is -132. Plugging those values into the formula: n = [ -(-1) ± sqrt( (-1)² - 4 * 1 * (-132) ) ] / (2 * 1). Simplifying that: n = [1 ± sqrt(1 + 528)] / 2. Because the term under the square root, the discriminant, is 1 + 528. 1 + 528 is 529. The square root of 529 is 23. So now we have n = [1 ± 23] / 2.So that gives two possible solutions: (1 + 23)/2 = 24/2 = 12, and (1 - 23)/2 = -22/2 = -11. Since the number of people can't be negative, we discard the negative solution. So n = 12. Let me check the answer options. Option C is 12. So the answer should be C.But wait, just to make sure I didn't make a mistake. Let me verify with the combination formula. If there are 12 people, the number of handshakes is 12*11/2. 12*11 is 132, divided by 2 is 66. Yep, that matches the given number. So that's correct. So the answer is 12. So option C.Alternatively, maybe there's another way to think about it. Suppose you don't remember the formula. Let's think incrementally. If there are 2 people, 1 handshake. 3 people: 3 handshakes. 4 people: 6 handshakes. Each time you add a person, they shake hands with everyone already there. So the number of handshakes when adding the nth person is (n - 1). So total handshakes would be the sum from 1 to (n - 1). That sum is n(n - 1)/2, which is the same formula. So yeah, same result.Alternatively, if you test the options given. Let's see. The answer options are 8, 10, 12, 14. Let's compute the number of handshakes for each.For option A: 8 people. 8*7/2 = 28. That's too low.Option B: 10 people. 10*9/2 = 45. Still less than 66.Option C: 12 people. 12*11/2 = 66. Perfect, that's the number given.Option D: 14 people. 14*13/2 = 91. That's way over. So definitely option C is the answer.So, through multiple methods—solving the quadratic equation, recalling the combination formula, and testing the options—it's clear that 12 people attended the meeting. Therefore, the correct answer is C: 12.Final Answerboxed{C}</think>To determine the number of people who attended the meeting, we use the formula for the number of handshakes, which is given by the combination formula ( binom{n}{2} ). This formula calculates the number of ways to choose 2 people out of ( n ) to shake hands, and is expressed as:[binom{n}{2} = frac{n(n - 1)}{2}]We know the total number of handshakes is 66, so we set up the equation:[frac{n(n - 1)}{2} = 66]Multiplying both sides by 2 to clear the denominator:[n(n - 1) = 132]This simplifies to the quadratic equation:[n^2 - n - 132 = 0]Solving this quadratic equation using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where ( a = 1 ), ( b = -1 ), and ( c = -132 ):[n = frac{-(-1) pm sqrt{(-1)^2 - 4 cdot 1 cdot (-132)}}{2 cdot 1}][n = frac{1 pm sqrt{1 + 528}}{2}][n = frac{1 pm sqrt{529}}{2}][n = frac{1 pm 23}{2}]This gives two solutions:[n = frac{1 + 23}{2} = 12 quad text{and} quad n = frac{1 - 23}{2} = -11]Since the number of people cannot be negative, we discard ( n = -11 ). Thus, ( n = 12 ).Verifying with the combination formula for ( n = 12 ):[frac{12 cdot 11}{2} = 66]This confirms the solution is correct. Therefore, the number of people who attended the meeting is (boxed{C}).]

🔑:To solve the problem, we let the number of people attending the meeting be x. According to the given information, each pair of attendees shakes hands once. The total number of handshakes between x people can be represented by the formula for combinations of x items taken 2 at a time, which is frac{x(x-1)}{2}. We know that the total number of handshakes is 66, so we set up the equation:[frac{1}{2}x(x-1) = 66]Multiplying both sides of the equation by 2 to eliminate the fraction, we get:[x(x-1) = 132]Expanding and rearranging the terms to form a quadratic equation, we have:[x^2 - x - 132 = 0]To solve this quadratic equation, we look for factors of -132 that add up to -1. The factors that meet these criteria are 11 and -12. Therefore, we can express the equation as:[x^2 - 12x + 11x - 132 = 0]Grouping terms and factoring, we find:[x(x - 12) + 11(x - 12) = 0]Factoring out the common factor (x - 12), we get:[(x - 12)(x + 11) = 0]Setting each factor equal to zero gives us the possible solutions for x:[x - 12 = 0 quad text{or} quad x + 11 = 0]Solving each equation for x gives us:[x = 12 quad text{and} quad x = -11]Since the number of people cannot be negative, we discard x = -11 and accept x = 12 as the correct solution. Therefore, there were 12 people attending the meeting.So, the answer is: boxed{C}.

🧠:[<think>Okay, let's tackle this problem step by step. First, I need to handle part (I) where a is 0, and find the tangent line to the curve f(x) at x=1. Then, part (II) is a bit more involved; I need to find the range of a such that f(x) is non-negative for all x greater than or equal to 1. Let's start with part (I).Part I: Tangent Line at x=1 when a=0Given f(x) = (e^x)/e - a x² + (2a -1)x - a. When a=0, this simplifies. Let's substitute a=0:f(x) = (e^x)/e - 0 + (0 -1)x - 0 = (e^x)/e - x.Simplify (e^x)/e: that's e^(x)/e = e^(x-1). So f(x) = e^(x-1) - x.Now, to find the tangent line at x=1, we need two things: the value of the function at x=1, which is f(1), and the derivative f'(x) evaluated at x=1, which gives the slope.First, compute f(1):f(1) = e^(1-1) - 1 = e^0 - 1 = 1 - 1 = 0. So the point is (1, 0).Next, compute f'(x). The derivative of e^(x-1) is e^(x-1) (since the derivative of e^u is e^u * u', and here u = x-1, so u' = 1). The derivative of -x is -1. Therefore, f'(x) = e^(x-1) - 1.Evaluate f'(1):f'(1) = e^(1-1) -1 = e^0 -1 = 1 -1 = 0. So the slope of the tangent line is 0.Therefore, the tangent line at x=1 is a horizontal line passing through (1, 0). The equation is y = 0.Wait, that seems straightforward. Let me just verify my steps again.f(x) when a=0 is indeed e^(x-1) - x. Plugging in x=1 gives 1 -1 = 0. The derivative is e^(x-1) -1, which at x=1 is also 0. So the tangent line is y=0. That seems correct. Maybe a bit too simple, but maybe that's the case. Let me check if the derivative calculation is correct. Yes, derivative of e^(x-1) is e^(x-1), derivative of -x is -1, so f'(x) = e^(x-1) -1. Correct. At x=1, that's 1 -1 =0. So the tangent line is y=0. Alright, part I is done.Part II: Finding a such that f(x) ≥ 0 for all x ≥1This part is more challenging. The function is f(x) = (e^x)/e - a x² + (2a -1)x -a. Let's rewrite that for clarity:f(x) = e^{x-1} - a x² + (2a -1)x - a.We need to find all real numbers a such that f(x) is non-negative for all x ≥1. So, for x in [1, ∞), f(x) ≥ 0.To approach this, maybe we can rearrange terms to group the terms involving a.Let me see:f(x) = e^{x-1} - a x² + (2a -1)x -aLet's collect the terms with a:= e^{x-1} -1 x + a(-x² + 2x -1) -aWait, hold on. Let me factor out a from the terms that contain it:Original expression:e^{x-1} - a x² + (2a -1)x -aGroup terms with a:= e^{x-1} - x + a(-x² + 2x -1) -aWait, no. Let's do term by term:The terms with a are:- a x² + 2a x - aFactor out a:a(-x² + 2x -1)Therefore, f(x) can be written as:f(x) = e^{x-1} - x + a(-x² + 2x -1)Alternatively:f(x) = e^{x-1} - x + a(-x² + 2x -1 -1?) Wait, no. Wait, let me re-express:Wait, original:e^{x}/e - a x² + (2a -1)x -aYes, which is e^{x-1} - a x² + (2a -1)x -a.So grouping the a terms:= e^{x-1} -x + a(-x² + 2x -1) -aWait, no. Let's see:Original:f(x) = e^{x-1} - a x² + (2a -1)x -a.Let's distribute the (2a -1)x term:= e^{x-1} - a x² + 2a x - x -a.Now group the a terms:= e^{x-1} - x + a(-x² + 2x -1).Yes, that's correct. So f(x) = e^{x-1} -x + a(-x² + 2x -1).Therefore, f(x) can be expressed as:f(x) = e^{x-1} -x + a*(-x² + 2x -1).So, if we let Q(x) = -x² + 2x -1, then f(x) = e^{x-1} -x + a*Q(x).We need f(x) ≥0 for all x ≥1.Therefore, for each x ≥1, e^{x-1} -x + a*Q(x) ≥0.We can solve for a in terms of x. Let's rearrange:a*Q(x) ≥ - (e^{x-1} -x)If Q(x) ≠0, then:If Q(x) >0, then a ≥ [ - (e^{x-1} -x) ] / Q(x)If Q(x) <0, then a ≤ [ - (e^{x-1} -x) ] / Q(x)But since x ≥1, we need to analyze Q(x) = -x² + 2x -1.Let's compute Q(x):Q(x) = -x² + 2x -1 = - (x² - 2x +1) = - (x -1)^2.Ah! So Q(x) is equal to - (x -1)^2, which is always non-positive, since (x -1)^2 is non-negative, so Q(x) ≤0 for all x, and Q(x) =0 only when x=1.Therefore, for x >1, Q(x) <0.Therefore, since Q(x) is negative for x >1, the inequality a*Q(x) ≥ - (e^{x-1} -x) becomes:a ≤ [ - (e^{x-1} -x) ] / Q(x)But since Q(x) is negative, dividing by Q(x) (which is negative) reverses the inequality again. Let me re-express:Given that Q(x) = - (x -1)^2, which is negative for x ≠1. Therefore, the inequality:a * Q(x) ≥ - (e^{x-1} -x)is equivalent to:a ≤ [ - (e^{x-1} -x) ] / Q(x) = [ - (e^{x-1} -x) ] / [ - (x -1)^2 ] = (e^{x-1} -x) / (x -1)^2Therefore, for each x >1, we have that a must be less than or equal to (e^{x-1} -x)/(x -1)^2.But since this must hold for all x ≥1, a must be less than or equal to the minimum of (e^{x-1} -x)/(x -1)^2 for x ≥1.Wait, but x=1 is included. Let's check x=1.At x=1, Q(1) = - (1-1)^2 =0. So the original function f(1) is e^{0} -1 + a*0 = 1 -1 =0, which is equal to 0. So at x=1, regardless of a, f(1)=0. So the condition f(x)≥0 is satisfied at x=1 for any a.Therefore, the key is for x >1, we need f(x) ≥0. So the critical part is x >1.Therefore, for x >1, we have the inequality:a ≤ (e^{x-1} -x)/(x -1)^2.But since this must hold for all x >1, the maximum lower bound for a would be the infimum of (e^{x-1} -x)/(x -1)^2 over x >1. Wait, no.Wait, for each x >1, a must be less than or equal to (e^{x-1} -x)/(x -1)^2. Therefore, the most restrictive condition is the minimum of (e^{x-1} -x)/(x -1)^2 over x >1, because a must be less than or equal to all of them. Therefore, a must be less than or equal to the infimum of that expression over x >1.But actually, since the expression (e^{x-1} -x)/(x -1)^2 is a function of x, let's denote it as g(x) = (e^{x-1} -x)/(x -1)^2 for x >1.We need to find the minimum of g(x) over x >1, and set a ≤ that minimum.Alternatively, since a must be less than or equal to g(x) for all x >1, the maximum value of a is the infimum (greatest lower bound) of g(x) over x >1.Therefore, we need to find the minimum value of g(x) over x >1. If g(x) has a minimum, then a ≤ that minimum. If g(x) approaches some limit as x approaches 1+ or infinity, we need to check.But first, let's analyze the behavior of g(x) as x approaches 1 from the right.As x approaches 1+, let's set t = x -1, so t approaches 0+.Then g(x) = (e^{t} - (1 + t)) / t².We can expand e^{t} as 1 + t + t²/2 + t³/6 + ..., so:e^{t} - (1 + t) = (1 + t + t²/2 + t³/6 + ...) -1 - t = t²/2 + t³/6 + ... ~ t²/2 as t→0.Therefore, g(x) ~ (t²/2) / t² = 1/2 as t→0+. Therefore, the limit as x approaches 1+ of g(x) is 1/2.Next, check the behavior as x approaches infinity.As x→infty, let's see what g(x) does.g(x) = (e^{x-1} -x)/(x -1)^2.The numerator e^{x-1} -x grows exponentially, while the denominator grows quadratically. Therefore, the entire expression g(x) tends to infinity as x→infty. Therefore, g(x) approaches infinity as x becomes large.Therefore, we know that as x approaches 1+, g(x) approaches 1/2, and as x approaches infinity, g(x) approaches infinity. So, the function g(x) might have a minimum somewhere between x=1 and infinity. If we can find the minimum value of g(x) on (1, ∞), then a must be less than or equal to that minimum.Therefore, the problem reduces to finding the minimum of g(x) = (e^{x-1} -x)/(x -1)^2 for x >1.To find the critical points, we need to compute the derivative of g(x) and set it equal to zero.Let me compute g'(x):First, write g(x) as [e^{x -1} -x]/(x -1)^2.Let’s denote numerator N = e^{x -1} -x, denominator D = (x -1)^2.Then, using the quotient rule:g'(x) = [N’ D - N D’] / D².Compute N’:N = e^{x -1} -x, so N’ = e^{x -1} -1.Compute D’:D = (x -1)^2, so D’ = 2(x -1).Therefore, plug into the quotient rule:g'(x) = [ (e^{x-1} -1)(x -1)^2 - (e^{x-1} -x)(2(x -1)) ] / (x -1)^4.Simplify numerator:Let’s factor out (x -1) in the numerator:Numerator = (x -1)[ (e^{x-1} -1)(x -1) - 2(e^{x-1} -x) ]So:Numerator = (x -1)[ (e^{x-1}(x -1) - (x -1) -2 e^{x-1} + 2x ) ]Let me expand the terms inside the brackets:First term: (e^{x-1}(x -1) - (x -1)) = e^{x-1}(x -1) - (x -1) = (x -1)(e^{x-1} -1)Second term: -2 e^{x-1} + 2xSo altogether:(x -1)(e^{x-1} -1) -2 e^{x-1} + 2xLet me expand (x -1)(e^{x-1} -1):= x e^{x-1} -x - e^{x-1} +1Therefore, numerator inside brackets becomes:x e^{x-1} -x - e^{x-1} +1 -2 e^{x-1} + 2xCombine like terms:x e^{x-1} terms: x e^{x-1}-e^{x-1} -2 e^{x-1} = -3 e^{x-1}-x +2x = x+1 remainsSo total inside brackets:x e^{x-1} -3 e^{x-1} + x +1Factor terms with e^{x-1}:= e^{x-1}(x -3) + x +1Therefore, the numerator of g'(x) is:(x -1)[ e^{x-1}(x -3) + x +1 ]Therefore, g'(x) = [ (x -1)(e^{x-1}(x -3) + x +1) ] / (x -1)^4 ) = [ e^{x-1}(x -3) + x +1 ] / (x -1)^3Set numerator equal to zero:e^{x-1}(x -3) + x +1 =0So we need to solve:e^{x-1}(x -3) + x +1 =0This seems challenging. Let's denote h(x) = e^{x-1}(x -3) + x +1. We need to find x >1 such that h(x) =0.Let’s analyze h(x):First, check at x=1:h(1) = e^{0}(1 -3) +1 +1 = 1*(-2) +2 = -2 +2 =0. So h(1)=0. But x=1 is the boundary point. However, we’re considering x >1. Let's check behavior near x=1.Let x approach 1 from the right. Let t = x -1, t→0+.Then h(x) = e^{t}(t +1 -3) + t +1 +1 = e^{t}(t -2) + t +2.Expand e^{t} ≈1 + t + t²/2:≈ (1 + t + t²/2)(t -2) + t +2Multiply out:= (t -2) + t(t -2) + (t²/2)(t -2) + t +2= t -2 + t² -2t + (t³/2 - t²) + t +2Combine terms:t -2 + t² -2t + t³/2 - t² + t +2Simplify:t -2 -2t + t +2 + t² - t² + t³/2= (t -2t + t) + (-2 +2) + (t² -t²) + t³/2= 0 +0 +0 + t³/2 ≈ t³/2 as t→0+. Therefore, h(x) ≈ t³/2 >0 for t>0. Therefore, just above x=1, h(x) is positive.Next, check h(3):h(3) = e^{2}(3 -3) +3 +1 =0 +4=4 >0.Check h(2):h(2) = e^{1}(2 -3) +2 +1= e*(-1)+3≈ -2.718 +3≈0.282>0.Check h(0): Wait, x>1, so no need. Let's check h(4):h(4)=e^{3}(4 -3) +4 +1= e^3*1 +5≈20.085 +5≈25.085>0.So h(x) is positive at x=2,3,4, etc., and near x=1, it's positive. Wait, but we need to find when h(x)=0. But at x=1, h(x)=0, but for x>1, h(x) is positive? Then the equation h(x)=0 has only solution x=1? But in the domain x>1, h(x) is always positive. Therefore, there is no critical point for g(x) in x>1, since h(x)=0 only at x=1. Therefore, the derivative g'(x) is equal to h(x)/(x -1)^3, and h(x) is positive for x>1, and denominator (x -1)^3 is positive for x>1. Therefore, g'(x) is positive for all x>1.Therefore, g(x) is increasing on x>1. Since as x approaches 1+, g(x) approaches 1/2, and as x approaches infinity, g(x) approaches infinity, and the function is increasing throughout. Therefore, the minimum value of g(x) on x≥1 is 1/2, achieved as x approaches 1 from the right.But wait, at x=1, the original function f(x)=0, but the limit of g(x) as x approaches 1 is 1/2. Therefore, to ensure that for all x≥1, f(x)≥0, since g(x) is increasing, the minimal value of g(x) is 1/2, so a must be ≤1/2.Wait, but let's think again.Recall that for x >1, we have a ≤ g(x) where g(x) is increasing from 1/2 to infinity. Therefore, the minimal upper bound for a is 1/2, since as x approaches 1 from the right, the upper bound for a approaches 1/2, and for larger x, the upper bound is larger. Therefore, if a is ≤1/2, then a will satisfy a ≤g(x) for all x>1. Hence, the range of a is (-∞, 1/2].But let's verify this conclusion.Suppose a ≤1/2. Then, for x ≥1, f(x) = e^{x-1} -x +a*(-(x -1)^2). Since a ≤1/2 and -(x -1)^2 ≤0, then a*(-(x -1)^2) ≥ (1/2)*(-(x -1)^2). Wait, no. Since a is multiplied by a negative term, a smaller a (more negative) would make the term a*Q(x) more positive. Wait, maybe we need a different approach.Wait, perhaps instead of relying solely on the previous analysis, let's test a=1/2 and see if f(x)≥0 for all x≥1. If a=1/2, then:f(x)=e^{x-1} - (1/2)x² + (2*(1/2) -1)x -1/2 = e^{x-1} - (1/2)x² + (1 -1)x -1/2 = e^{x-1} - (1/2)x² -1/2.We need to check if this is ≥0 for all x≥1.At x=1: f(1)=e^{0} - (1/2)(1) -1/2 =1 -0.5 -0.5=0. So f(1)=0.For x>1, is f(x)=e^{x-1} -0.5x² -0.5 ≥0?Take derivative f’(x)=e^{x-1} -x.At x=1: f’(1)=1 -1=0. Then, for x>1, the derivative is e^{x-1} -x.We can analyze f’(x). Let’s see, for x>1, is e^{x-1} -x increasing or decreasing?Take second derivative f''(x)=e^{x-1} -1. At x=1, f''(1)=1 -1=0. For x>1, e^{x-1} >1, so f''(x)>0. Therefore, f’(x) is convex and increasing for x>1. At x=1, f’(1)=0. Then, since f''(x) >0 for x>1, f’(x) starts increasing. Therefore, f’(x) is increasing for x>1.Therefore, f’(x) ≥0 for x≥1? Let's check at x=2: f’(2)=e^{1} -2≈2.718 -2≈0.718>0. At x=1.5: f’(1.5)=e^{0.5}-1.5≈1.648 -1.5≈0.148>0. So f’(x) is increasing and positive for x≥1. Therefore, f(x) is increasing for x≥1. Since f(1)=0 and f is increasing, then f(x)≥0 for x≥1 when a=1/2. Therefore, a=1/2 is acceptable.If a>1/2, then what happens? Let's take a=1. Then f(x)=e^{x-1} -x +1*(-(x -1)^2)=e^{x-1} -x - (x -1)^2.At x=1: f(1)=1 -1 -0=0.Now, check x=2: f(2)=e^{1} -2 - (1)^2≈2.718 -2 -1≈-0.282<0. Therefore, f(2)<0, which violates the condition. Therefore, a=1 is invalid.If a <1/2, say a=0. Then f(x)=e^{x-1} -x. At x=1, f(1)=0. For x>1, since e^{x-1} grows exponentially and -x grows linearly, e^{x-1} -x will be positive for x>1. Indeed, for x=2: e^{1} -2≈2.718 -2≈0.718>0. For x=3: e^2 -3≈7.389 -3≈4.389>0. So a=0 works.But wait, according to our previous analysis, a must be ≤1/2. Let me check a=1/2, which works. For a <1/2, does f(x) remain non-negative?Take a=0. Then f(x)=e^{x-1} -x. As discussed, this is positive for x>1. If a is less than 1/2, say a=0.4, then f(x)=e^{x-1} -x +0.4*(-(x -1)^2). Let's check at x=2:f(2)=e^{1} -2 +0.4*(-1)=2.718 -2 -0.4≈0.318>0. At x=3: e^{2} -3 +0.4*(-4)=7.389 -3 -1.6≈2.789>0. So still positive. If a is too small, say a approaches negative infinity, then f(x) = e^{x-1} -x +a*(-(x-1)^2). Since -(x-1)^2 is negative, multiplying by a negative a makes it positive. So as a approaches negative infinity, the term a*(-(x-1)^2)= -a(x-1)^2, which would dominate and make f(x) go to positive infinity. Wait, no. If a is negative, then a*(-(x-1)^2)= -a(x-1)^2, which is positive since a is negative. Therefore, f(x)=e^{x-1} -x + positive term. So as a becomes more negative, the positive term increases, so f(x) is more positive. Therefore, all a ≤1/2 would satisfy f(x)≥0.But wait, but according to our previous analysis, since a ≤ g(x) for all x>1, and g(x) approaches 1/2 as x→1+, then a must be ≤1/2.Therefore, the range of a is (-∞, 1/2].But let's verify with a=1/2. As we saw, f(x) with a=1/2 is non-negative. For a <1/2, since we subtract less (because a is multiplied by a negative term), which actually adds more. Wait, maybe think of it this way:If a is smaller (more negative), then the term a*Q(x)=a*(-(x -1)^2) becomes more positive because a is negative and Q(x) is negative, so negative times negative is positive, and as a becomes more negative, the positive contribution increases. So making a smaller (more negative) increases f(x), making it more positive. Therefore, the lower the a, the better for satisfying f(x)≥0. The critical case is when a is as large as possible, which is a=1/2. If a exceeds 1/2, then at some x>1, f(x) becomes negative. As we saw with a=1, which is >1/2, f(x) is negative at x=2. Hence, the upper bound is a=1/2, and lower bound is negative infinity.Therefore, the range of a is all real numbers less than or equal to 1/2.Conclusion:- Part I: The tangent line at x=1 when a=0 is y=0.- Part II: The range of a is a ≤1/2.Final Answer(I) The equation of the tangent line is boxed{y = 0}.(II) The range of values for ( a ) is boxed{(-infty, dfrac{1}{2}]}).</think>Given the function ( f(x) = frac{e^x}{e} - ax^2 + (2a - 1)x - a ):(Ⅰ) When ( a = 0 ):1. Simplify the function: [ f(x) = e^{x-1} - x ]2. Evaluate ( f(1) ): [ f(1) = e^{0} - 1 = 1 - 1 = 0 ]3. Compute the derivative ( f'(x) ): [ f'(x) = e^{x-1} - 1 ]4. Evaluate the derivative at ( x = 1 ): [ f'(1) = e^{0} - 1 = 0 ]5. The tangent line at ( x = 1 ) with slope 0 passing through (1, 0) is: [ y = 0 ](Ⅱ) Finding the range of ( a ) such that ( f(x) geq 0 ) for ( x geq 1 ):1. Rewrite ( f(x) ): [ f(x) = e^{x-1} - x + a(-(x - 1)^2) ]2. For ( x geq 1 ), we need ( e^{x-1} - x + a(-(x - 1)^2) geq 0 ).3. Analyze the expression ( g(x) = frac{e^{x-1} - x}{(x - 1)^2} ) for ( x > 1 ).4. Determine the behavior of ( g(x) ): - As ( x to 1^+ ), ( g(x) to frac{1}{2} ). - As ( x to infty ), ( g(x) to infty ). - ( g(x) ) is increasing for ( x > 1 ).5. The minimum value of ( g(x) ) as ( x to 1^+ ) is ( frac{1}{2} ).6. Therefore, ( a leq frac{1}{2} ) ensures ( f(x) geq 0 ) for all ( x geq 1 ). Final Answer(I) The equation of the tangent line is boxed{y = 0}.(II) The range of values for ( a ) is boxed{(-infty, dfrac{1}{2}]}.,<think>Okay, let's tackle this problem step by step. It has two parts, so I'll start with part (I). First, when a=0, the function simplifies, right? Let me write that down. The original function is f(x) = (e^x)/e - a x² + (2a -1)x - a. If a=0, then the terms with a should disappear. Let's substitute a=0:f(x) = (e^x)/e - 0*x² + (2*0 -1)x - 0 Simplifying that, we get f(x) = (e^x)/e - x. Wait, (e^x)/e is the same as e^(x-1), because e^x divided by e is e^(x-1). That might be useful later. But maybe I should just keep it as (e^x)/e for now.Now, the problem asks for the equation of the tangent line at x=1. To find the tangent line, I need two things: the value of f at x=1, which is the y-coordinate, and the derivative f’(x) at x=1, which is the slope.Let me compute f(1) first. Plugging x=1 into f(x):f(1) = (e^1)/e - 1 = (e)/e - 1 = 1 - 1 = 0. So the point is (1, 0).Next, find the derivative f’(x). The derivative of (e^x)/e is (e^x)/e, since the derivative of e^x is e^x, and dividing by e is just a constant. Then the derivative of -x is -1. So:f’(x) = (e^x)/e - 1Now evaluate this at x=1:f’(1) = (e^1)/e - 1 = 1 - 1 = 0. Wait, so the slope is 0? That means the tangent line is horizontal.Therefore, the equation of the tangent line at x=1 is y = 0, because it's a horizontal line passing through (1, 0). Hmm, that seems straightforward. Let me just verify. If f(1)=0 and f’(1)=0, then yes, the tangent line is y=0. That seems right. Maybe too simple, but math can be surprising sometimes.Okay, part (I) seems done. Now part (II) is more complex. We need to find the range of a such that f(x) ≥ 0 for all x ≥ 1.Given the function f(x) = (e^x)/e - a x² + (2a -1)x - a. Let's rewrite that to group terms with a:f(x) = e^{x-1} + (-a x² + (2a -1)x - a). Let me factor out the a from the terms that have it:f(x) = e^{x-1} + a*(-x² + 2x -1) - x. Wait, let's check:Original terms with a: -a x² +2a x -a = a(-x² + 2x -1). Then the remaining term is -1*x. So, f(x) = e^{x-1} + a*(-x² + 2x -1) - x.Alternatively, maybe better to rearrange terms:f(x) = e^{x-1} - x + a*(-x² + 2x -1). So, if I let that be written as f(x) = e^{x-1} - x + a*(-x² + 2x -1). Hmm, perhaps that's helpful. Because we can think of this as f(x) = [e^{x-1} - x] + a*[ -x² + 2x -1 ].So, depending on the value of a, the function is a combination of two parts: one that's e^{x-1} - x and another that's a multiplied by (-x² + 2x -1). Our goal is to find the values of a such that for all x ≥ 1, the combination is non-negative.To find the range of a, we need to analyze f(x) ≥ 0 for x ≥ 1. Let's denote the coefficient of a as g(x) = -x² + 2x -1, and the other part as h(x) = e^{x-1} - x. Then f(x) = h(x) + a*g(x). So the inequality is h(x) + a*g(x) ≥ 0 for all x ≥ 1.So, solving for a, we can write:a*g(x) ≥ -h(x)If g(x) > 0, then a ≥ -h(x)/g(x)If g(x) < 0, then a ≤ -h(x)/g(x)Therefore, depending on the sign of g(x), the inequality flips or not. Therefore, the permissible range of a would be determined by the maximum or minimum of -h(x)/g(x) over x ≥ 1, depending on the sign of g(x).First, let's analyze g(x) = -x² + 2x -1. Let's write it as g(x) = -(x² - 2x +1) = -(x -1)^2. So g(x) is equal to - (x -1)^2. Therefore, for all x, g(x) is non-positive, since (x -1)^2 is always non-negative, so g(x) is ≤ 0. It's zero only at x=1.Therefore, for x ≥1, g(x) = - (x -1)^2 ≤ 0. So except at x=1, where g(1)=0, for all x >1, g(x) is negative.Therefore, since g(x) ≤ 0 for x ≥1, except at x=1 where it's zero, then in the inequality a*g(x) ≥ -h(x), since g(x) is negative (for x >1), we can divide both sides by g(x), flipping the inequality:a ≤ (-h(x))/g(x) for all x ≥1.But at x=1, g(1)=0, so we need to check f(1) ≥0. Let's compute f(1):f(1) = e^{0} -1 + a*(-1 +2 -1) = 1 -1 +a*(0) =0. So f(1)=0, which is okay.So for x >1, we need a ≤ (-h(x))/g(x). Let's compute that expression:-h(x)/g(x) = -(e^{x-1} -x)/[-(x -1)^2] = [e^{x-1} -x]/ (x -1)^2.Therefore, a must be less than or equal to [e^{x-1} -x]/ (x -1)^2 for all x >1. So the maximum value of a that satisfies this is the minimum of [e^{x-1} -x]/ (x -1)^2 over x >1. Therefore, a ≤ infimum of [e^{x-1} -x]/ (x -1)^2 for x >1.So our job is to find the minimum of the function k(x) = [e^{x-1} -x]/ (x -1)^2 for x >1.Once we find the minimum value of k(x), then the range of a is (-∞, m], where m is that minimum.Alternatively, perhaps the function k(x) has a minimum that we can compute. Let's analyze k(x) = [e^{x-1} -x]/ (x -1)^2 for x >1.To find the minimum, we can take the derivative of k(x) with respect to x, set it equal to zero, and solve for x.First, let's denote t = x -1, so t >0. Then k(x) becomes [e^{t} - (t +1)] / t². Let's write k(t) = (e^t - t -1)/t² for t >0.So now, we can analyze k(t) as a function of t >0, and find its minimum.Compute derivative k’(t):Using quotient rule: if k(t) = numerator / denominator, where numerator = e^t - t -1, denominator = t².Then k’(t) = [ (denominator)(num’) - (numerator)(denom’) ] / denominator²Compute num’ = e^t -1 -0 = e^t -1.Denom’ = 2t.Therefore,k’(t) = [ t²*(e^t -1) - (e^t - t -1)*2t ] / t^4Simplify numerator:Let's factor t:Numerator = t [ t(e^t -1) - 2(e^t - t -1) ]Let me compute inside the brackets:First term: t(e^t -1) = t e^t - tSecond term: -2(e^t - t -1) = -2 e^t + 2t + 2Combine them:t e^t - t -2 e^t + 2t +2 = t e^t -2 e^t + ( -t +2t ) +2 = e^t(t -2) + t +2So the numerator becomes t [ e^t(t -2) + t +2 ]Therefore, k’(t) = [ t ( e^t(t -2) + t +2 ) ] / t^4 = [ e^t(t -2) + t +2 ) ] / t^3Set k’(t) =0, so:e^t(t -2) + t +2 =0We need to solve e^t(t -2) + t +2 =0 for t >0.This equation might be challenging to solve analytically. Let's see.Let’s denote the equation:e^t(t - 2) + t +2 =0Let me rearrange:e^t(t -2) = - (t +2)But since t >0, the right-hand side is negative (since t +2 >0). The left-hand side: e^t is always positive. So (t -2) must be negative for the left-hand side to be negative. Therefore, t -2 <0 => t <2. So possible solutions are in t ∈ (0,2).So the critical points can only occur in t ∈ (0,2). Let's check for possible solutions here.Let’s check at t=0: If t approaches 0 from the right, the left-hand side e^0*(0 -2) = -2, the right-hand side is - (0 +2)= -2. So as t approaches 0, the equation tends to -2 = -2. So perhaps t=0 is a solution, but t>0. Maybe there is a solution near t=0?Wait, let's compute at t=0:Left-hand side: e^0*(0 -2) = -2Right-hand side: - (0 +2)= -2So t=0 is a solution. But t>0. So near t=0, maybe the equation is satisfied?But actually, t=0 is the limit. Let's check t approaching 0 from the right:Left-hand side: e^t(t -2) ≈ (1 + t)(-2 + t) ≈ -2 + t -2t + t^2 ≈ -2 - t + t^2Right-hand side: - (t +2) ≈ -2 - tTherefore, the difference between left and right sides is approx (-2 - t + t^2) - (-2 -t )= t^2. So as t approaches 0 from the right, the equation left - right ≈ t^2, which is positive. Therefore, near t=0, left-hand side is greater than right-hand side.But at t=0, they are equal. So in the limit as t approaches 0, the equation approaches equality.But since t>0, perhaps there is no solution except t=0. Let's check t=1:Left-hand side: e^1*(1 -2)= e*(-1) ≈ -2.718Right-hand side: - (1 +2)= -3So -2.718 ≈ -3? No, -2.718 > -3, so left-hand side > right-hand side. So equation not satisfied.At t=2:Left-hand side: e^2*(2 -2)=0Right-hand side: - (2 +2)= -40 ≠ -4. Not a solution.At t=1.5:Left-hand side: e^{1.5}*(1.5 -2)= e^{1.5}*(-0.5) ≈ 4.4817*(-0.5) ≈ -2.2408Right-hand side: - (1.5 +2)= -3.5So left-hand side (-2.2408) > right-hand side (-3.5). Still not equal.At t=0.5:Left-hand side: e^{0.5}*(0.5 -2)= e^{0.5}*(-1.5) ≈ 1.6487*(-1.5)≈ -2.473Right-hand side: - (0.5 +2)= -2.5So left-hand side (-2.473) > right-hand side (-2.5). So still left > right.So, in t ∈ (0,2), the left-hand side is always greater than the right-hand side. Since at t approaching 0, left ≈ -2 -t, right ≈ -2 -t, but as we saw, the difference is t², so left is slightly greater. Then at t=0.5, left is -2.473 vs right -2.5. Then at t=1, left is -2.718 vs right -3. At t approaching 2, left approaches 0, right approaches -4. So left is always greater than right in (0,2). Therefore, the equation e^t(t -2) + t +2 =0 has no solution for t >0.Therefore, k’(t) is never zero for t >0. So the function k(t) has no critical points in t >0. Therefore, its minimum must occur at the boundary or at infinity.Wait, but let's check the behavior of k(t) as t approaches 0+ and as t approaches infinity.First, as t approaches 0+:k(t) = (e^t - t -1)/t². Let's expand e^t using Taylor series: e^t ≈ 1 + t + t²/2 + t³/6 + ...So e^t - t -1 ≈ (1 + t + t²/2 + ...) - t -1 = t²/2 + t³/6 + ...Therefore, k(t) ≈ (t²/2)/t² = 1/2 as t approaches 0. So the limit as t→0+ is 1/2.As t approaches infinity:k(t) = (e^t - t -1)/t². The numerator e^t dominates, since e^t grows much faster than t or 1. So the numerator ≈ e^t, denominator ≈ t². Therefore, k(t) ≈ e^t / t², which tends to infinity as t→infty. Therefore, k(t) tends to infinity as t→infty.So since k(t) approaches 1/2 as t→0+ and tends to infinity as t→infty, and since there are no critical points in between (since k’(t) is always positive or always negative?), we need to check the sign of k’(t).Earlier, we found that k’(t) = [e^t(t -2) + t +2]/t³. Let's check the sign of the numerator:Numerator = e^t(t -2) + t +2.We saw that for t >0, e^t(t -2) + t +2 is always positive? Let's check for t >0:At t approaching 0+, numerator ≈ (1 + t + ...)(t -2) + t +2 ≈ ( -2 + t(-1) ) + t +2 = (-2 -t) + t +2 =0. Wait, but earlier when we expanded, the difference was t^2. So perhaps my previous analysis was conflicting.Wait, let's use the expansion again. Let me write numerator = e^t(t -2) + t +2.As t approaches 0:e^t ≈1 + t + t²/2, so e^t(t -2) ≈ (1 + t + t²/2)(t -2) ≈ (t -2) + t(t -2) + (t²/2)(t -2)≈ t -2 + t² -2t + (t^3/2 - t²)= (t -2) + (t² -2t) + (t^3/2 - t²)= t -2 + t² -2t + t^3/2 - t²= (-t -2) + t^3/2Therefore, numerator ≈ -t -2 + t^3/2 + t +2 = t^3/2. So as t approaches 0, numerator ≈ t^3/2, which is positive for t>0. Therefore, k’(t) ≈ (t^3/2)/t^3 = 1/(2t^0) = 1/2 as t→0+? Wait, no:Wait, k’(t) = numerator / t^3 ≈ (t^3/2) / t^3 = 1/(2t^0) = 1/2. So as t approaches 0+, k’(t) approaches 1/2. So the derivative approaches 1/2 from the positive side. Therefore, k’(t) is positive near t=0.At t=1, numerator = e*(1 -2) +1 +2 = -e +3 ≈ -2.718 +3 ≈0.282>0. So numerator is positive.At t=2, numerator = e^2*(2 -2) +2 +2=0 +4=4>0.At t=3, numerator = e^3*(3 -2) +3 +2 = e^3 +5 ≈20.085 +5≈25.085>0.So numerator is always positive for t>0. Therefore, k’(t) is positive for all t>0. Therefore, k(t) is increasing on t>0.Wait, but when t approaches 0+, k(t) approaches 1/2, and as t increases, k(t) increases to infinity. Therefore, the minimum of k(t) on t>0 (which corresponds to x>1) is 1/2. Therefore, the infimum is 1/2, but since t approaches 0+ (x approaches1+), k(t) approaches 1/2.Therefore, the minimum value of k(x) is 1/2, achieved as x approaches1 from the right. Wait, but when x approaches1 from the right, t approaches0+, so k(t) approaches1/2. However, at x=1, k(x) is actually undefined since denominator (x-1)^2 is zero. Let's check:Wait, when x=1, f(x)=0 as we saw earlier. The expression [e^{x-1} -x]/(x -1)^2 at x=1 is [1 -1]/0, which is 0/0 indeterminate. So we can compute the limit as x approaches1 using L’Hospital’s Rule.Compute lim_{x→1} [e^{x-1} -x]/(x -1)^2.Let t = x -1, so t→0.Then the limit becomes [e^t - (t +1)]/t². As t→0.Apply L’Hospital’s Rule twice:First, differentiate numerator and denominator:First derivative: [e^t -1]/2tStill 0/0. Apply L’Hospital again:Second derivative: [e^t]/2As t→0, this approaches 1/2. So the limit is 1/2.Therefore, the function k(x) approaches 1/2 as x approaches1, and since k(t) is increasing for t>0 (x>1), the minimum value of k(x) for x≥1 is 1/2, but it's not achieved at any x>1; it's just the limit as x approaches1.Therefore, for the inequality a ≤ k(x) for all x>1, since k(x) is increasing and approaches 1/2 from above as x approaches1, but the minimal value is 1/2. Wait, but if k(x) is increasing, then its minimum on x>1 is the limit as x approaches1, which is 1/2. Therefore, a must be ≤1/2 to satisfy a ≤k(x) for all x>1. Because if a is ≤1/2, then since k(x) increases to infinity, 1/2 is the lower bound of k(x) on x>1.Wait, but let me think again. Since k(x) approaches 1/2 as x approaches1 and increases thereafter, the smallest value k(x) takes is 1/2 (the limit). So to have a ≤k(x) for all x>1, a must be ≤ the minimal value of k(x), which is 1/2. Because if a is larger than 1/2, then near x=1, k(x) is close to 1/2, which would make a >k(x), violating the inequality.Therefore, the maximum value of a is 1/2. So the range of a is (-infty, 1/2].But wait, need to confirm. Let's test a=1/2. If a=1/2, then f(x) = e^{x-1} -x + (1/2)*(-x² +2x -1). Let's simplify:f(x) = e^{x-1} -x - (1/2)x² +x -1/2Simplify terms: -x +x cancels. So f(x)= e^{x-1} - (1/2)x² -1/2.We need to check if f(x) ≥0 for x ≥1.At x=1: f(1)= e^{0} - (1/2)(1) -1/2 =1 -0.5 -0.5=0. Okay.For x>1, we need to check if e^{x-1} - (1/2)x² -1/2 ≥0.Given that when a=1/2, the original inequality becomes equality at x approaching1, and since k(x) increases with x, the function f(x) would grow since e^{x-1} dominates. Therefore, it's likely that for a=1/2, f(x) is always non-negative for x≥1.But let's check another point. Let's take x=2:f(2)= e^{1} - (1/2)(4) -1/2= e -2 -0.5≈2.718 -2.5≈0.218>0.x=3:f(3)= e^{2} - (1/2)(9) -1/2≈7.389 -4.5 -0.5≈2.389>0.x=0. Wait, no, x≥1. So seems okay.What about x approaching1 from the right? Since the limit as x approaches1 is0, and f(x) is continuous. So if a=1/2, f(x) approaches0 as x approaches1, and is positive for x>1.But what if a>1/2? Let's say a=1. Then f(x)=e^{x-1} -x + (-x² +2x -1)*1 =e^{x-1} -x -x² +2x -1= e^{x-1} +x -x² -1.At x=1: e^{0}+1 -1 -1=1 +1 -1 -1=0.At x=1.1:Compute e^{0.1} +1.1 - (1.1)^2 -1≈1.105 +1.1 -1.21 -1≈1.105+1.1=2.205 -1.21=0.995 -1= -0.005. So f(1.1)≈-0.005<0. Hence, if a=1, f(x) is negative at x=1.1, which violates the condition. Therefore, a=1 is invalid.If a=0.6, which is greater than 1/2=0.5, then at x approaching1, f(x) is approaching0, but let's check x=1.1:f(x)=e^{0.1} -1.1 +0.6*(-1.21 +2.2 -1)= e^{0.1} -1.1 +0.6*( -0.01)=1.105 -1.1 -0.006≈-0.001. Still negative. Therefore, a>1/2 would cause f(x) to be negative just above x=1.Therefore, the critical value is a=1/2. Hence, the range is a ≤1/2.But need to confirm that for a ≤1/2, f(x) ≥0 for all x ≥1. Let's take a=0. Then f(x)=e^{x-1} -x. Is this non-negative for x≥1?At x=1:1 -1=0. For x>1, e^{x-1} grows exponentially, while x grows linearly. So yes, e^{x-1} -x is positive for x>1. Therefore, a=0 is okay.Take a=1/2, as before, f(x)≥0. What if a is less than 1/2, say a=0. Then f(x) is even larger. If a is negative, does it still hold?Wait, if a is negative, then the term a*(-x² +2x -1) becomes positive since a is negative and (-x² +2x -1) is negative (as we saw earlier, it's -(x-1)^2). So multiplying two negatives gives a positive term. Therefore, f(x)= e^{x-1} -x + positive term, which is even larger. So for a <1/2, the function f(x) is larger than when a=1/2, hence f(x) ≥0.Therefore, the conclusion is that a must be ≤1/2.Hence, part (II) answer is a ≤1/2.Final AnswerⅠ. The equation of the tangent line is boxed{y = 0}.Ⅱ. The range of values for ( a ) is boxed{(-infty, dfrac{1}{2}]}).</think>Given the function ( f(x) = frac{e^x}{e} - ax^2 + (2a - 1)x - a ): (Ⅰ) If ( a = 0 ), find the equation of the tangent line to the curve ( f(x) ) at ( x = 1 )1. Substitute ( a = 0 ) into the function: [ f(x) = frac{e^x}{e} - x ]2. Simplify the function: [ f(x) = e^{x-1} - x ]3. Evaluate ( f(1) ): [ f(1) = e^{0} - 1 = 1 - 1 = 0 ]4. Compute the derivative ( f'(x) ): [ f'(x) = e^{x-1} - 1 ]5. Evaluate the derivative at ( x = 1 ): [ f'(1) = e^{0} - 1 = 1 - 1 = 0 ]6. The equation of the tangent line at ( (1, 0) ) with slope 0 is: [ y = 0 ] (Ⅱ) If ( f(x) geqslant 0 ) when ( x geqslant 1 ), find the range of values for ( a )1. Rewrite the function ( f(x) ): [ f(x) = e^{x-1} - x + a(-x^2 + 2x - 1) ]2. Analyze the quadratic term ( -x^2 + 2x - 1 ): [ -x^2 + 2x - 1 = -(x - 1)^2 leq 0 text{ for } x geq 1 ]3. Express the inequality ( f(x) geq 0 ): [ e^{x-1} - x + a(-(x - 1)^2) geq 0 ]4. Rearrange the inequality: [ a leq frac{e^{x-1} - x}{(x - 1)^2} text{ for } x > 1 ]5. Define ( k(x) = frac{e^{x-1} - x}{(x - 1)^2} ) and find its minimum value for ( x geq 1 ): - As ( x to 1 ), ( k(x) to frac{1}{2} ) (using L'Hospital's Rule). - For ( x > 1 ), ( k(x) ) is increasing and tends to infinity.6. The minimum value of ( k(x) ) is ( frac{1}{2} ), leading to: [ a leq frac{1}{2} ] Final AnswersⅠ. The equation of the tangent line is boxed{y = 0}.Ⅱ. The range of values for ( a ) is boxed{(-infty, dfrac{1}{2}]}.]

🔑:Solution:- (Ⅰ) When a=0, f(x)=e^{x-1}-x, then f′(x)=e^{x-1}-1, so f′(1)=0,and f(1)=0, hence the equation of the tangent line is y=0;- (Ⅱ) It is easy to see that f′(x)=e^{x-1}-2ax+2a-1, f″(x)=e^{x-1}-2a,If f″(x) geqslant 0, we get a leqslant frac {e^{x-1}}{2}, that is, when a leqslant frac {1}{2}, f′(x) is increasing on [1,+infty),thus f′(x) geqslant f′(1)=0, so f(x) is increasing on [1,+infty),hence f(x) geqslant f(1)=0, which is consistent with the problem statement,therefore, a leqslant frac {1}{2} is a sufficient condition for the original inequality to hold. Next, we prove its necessity,When a > frac {1}{2}, let f″(x)=0, solving gives: x=ln (2a)+1,thus when x in (1,ln (2a)+1), f′(x) < 0, hence f′(x) is decreasing on x in (1,ln (2a)+1),thus f′(x) < f′(0)=0, therefore, when x in (1,ln (2a)+1), f(x) is decreasing,hence f(x) < f(1)=0, which contradicts the problem statement and is not consistent,In conclusion, the range of a is boxed{(-infty, frac {1}{2}]}.

🧠:[<think>Okay, let's try to solve these two problems. Starting with the first one. We have a sequence {a_n} where the sum of the first n terms, S_n, is equal to 1 - a_n. So, S_n = a_1 + a_2 + ... + a_n = 1 - a_n. We need to find the general term formula for {a_n}.Hmm, let's think. If S_n = 1 - a_n, then maybe we can relate S_n and S_{n-1}? Because S_n = S_{n-1} + a_n. Right, that's a key relationship between consecutive partial sums. So let's write that down:S_n = S_{n-1} + a_nBut from the problem statement, S_n = 1 - a_n. So substituting that into the equation:1 - a_n = S_{n-1} + a_nThen, solve for S_{n-1}:S_{n-1} = 1 - a_n - a_n = 1 - 2a_nBut also, S_{n-1} is equal to 1 - a_{n-1} according to the given formula when n is replaced by n-1. Wait, that's only valid if n-1 is a positive integer, so n ≥ 2. So for n ≥ 2:S_{n-1} = 1 - a_{n-1}Therefore, we have:1 - a_{n-1} = 1 - 2a_nSubtract 1 from both sides:- a_{n-1} = -2a_nMultiply both sides by -1:a_{n-1} = 2a_nSo this gives us a recursive relation: a_{n} = (1/2) a_{n-1} for n ≥ 2. So each term is half of the previous term. That suggests the sequence {a_n} is a geometric sequence with common ratio 1/2.But let's check for n=1. When n=1, S_1 = a_1 = 1 - a_1. So:a_1 = 1 - a_1Adding a_1 to both sides:2a_1 = 1 ⇒ a_1 = 1/2So the first term is 1/2, and each subsequent term is half of the previous term. Therefore, the general term should be a_n = (1/2)^n. Let's verify this.Suppose a_n = (1/2)^n. Then S_n = sum_{k=1}^n (1/2)^k. The sum of a geometric series is S_n = a(1 - r^n)/(1 - r) where a is the first term and r is the common ratio. Here, a = 1/2, r = 1/2. So:S_n = (1/2)(1 - (1/2)^n)/(1 - 1/2) = (1/2)(1 - (1/2)^n)/(1/2) = 1 - (1/2)^nBut according to the problem, S_n = 1 - a_n. If a_n = (1/2)^n, then 1 - a_n = 1 - (1/2)^n, which matches the sum we calculated. Perfect, that works. So part (1) is solved: a_n = (1/2)^n.Moving on to part (2). We have a sequence {b_n} where b_n = n * 2^n. We need to find the sum of the first n terms, T_n = b_1 + b_2 + ... + b_n = sum_{k=1}^n k*2^k.Hmm, this sum looks a bit tricky. I remember that there's a formula for the sum of k*r^k from k=1 to n. Let me recall how to derive it.Let S = sum_{k=1}^n k*r^k. Then, multiply both sides by r:rS = sum_{k=1}^n k*r^{k+1} = sum_{k=2}^{n+1} (k-1)*r^kSubtracting these two equations:S - rS = sum_{k=1}^n k*r^k - sum_{k=2}^{n+1} (k-1)*r^kLeft side: S(1 - r)Right side: Let's adjust the indices. The first sum is from k=1 to n, and the second sum is from k=2 to n+1. Let's write out the terms:= [1*r^1 + 2*r^2 + 3*r^3 + ... + n*r^n] - [ (2-1)*r^2 + (3-1)*r^3 + ... + (n+1 -1)*r^{n+1} ]= [1*r^1 + 2*r^2 + 3*r^3 + ... + n*r^n] - [1*r^2 + 2*r^3 + ... + n*r^{n+1} ]Subtracting term by term:= 1*r^1 + (2*r^2 -1*r^2) + (3*r^3 -2*r^3) + ... + (n*r^n - (n-1)*r^n) - n*r^{n+1}= r + r^2 + r^3 + ... + r^n - n*r^{n+1}This is a geometric series from k=1 to n of r^k minus n*r^{n+1}:= (r*(1 - r^n)/(1 - r)) - n*r^{n+1}Therefore, S(1 - r) = (r - r^{n+1})/(1 - r) - n*r^{n+1}Hence, solving for S:S = [ (r - r^{n+1})/(1 - r) - n*r^{n+1} ] / (1 - r)Simplify numerator:= [ r - r^{n+1} - n*r^{n+1}(1 - r) ] / (1 - r)^2Wait, maybe it's better to factor step by step. Let's write:S = [ r(1 - r^n)/(1 - r) - n*r^{n+1} ] / (1 - r)Multiply numerator and denominator:= [ r(1 - r^n) - n*r^{n+1}(1 - r) ] / (1 - r)^2But this might get complicated. Alternatively, using the result from the telescoping:We had S(1 - r) = sum_{k=1}^n r^k - n*r^{n+1}Which is sum_{k=1}^n r^k = (r - r^{n+1})/(1 - r), so:S(1 - r) = (r - r^{n+1})/(1 - r) - n*r^{n+1}Multiply both sides by (1 - r):S(1 - r)^2 = r - r^{n+1} - n*r^{n+1}(1 - r)Wait, maybe another approach. Let's recall that for the infinite series, sum_{k=1}^∞ k r^k = r/(1 - r)^2, but here we have a finite sum.But let's use the expression we derived:S = [ (r - r^{n+1})/(1 - r) - n*r^{n+1} ] / (1 - r)So,S = (r - r^{n+1} - n*r^{n+1}(1 - r)) / (1 - r)^2But let's plug in r = 2. Wait, hold on. Wait in our case, r is 2. Because the general formula is for sum_{k=1}^n k r^k, and here we have sum_{k=1}^n k*2^k, so r=2. Let's substitute r=2 into the formula.But wait, if we use r=2, then 1 - r = -1, and so the formula would be:S = [2(1 - 2^n)/(1 - 2) - n*2^{n+1} ] / (1 - 2)But let's compute step by step. First, compute the terms:First term in the numerator: 2*(1 - 2^n)/(1 - 2) = 2*(1 - 2^n)/(-1) = -2*(1 - 2^n) = 2*(2^n - 1) = 2^{n+1} - 2Second term: -n*2^{n+1}So numerator is (2^{n+1} - 2) - n*2^{n+1} = (1 - n)2^{n+1} - 2Denominator is (1 - 2) = -1Therefore, S = [(1 - n)2^{n+1} - 2] / (-1) = -[(1 - n)2^{n+1} - 2] = (n - 1)2^{n+1} + 2Simplify:(n - 1)2^{n+1} + 2 = (n - 1)2^{n+1} + 2 = 2^{n+1}(n - 1) + 2Alternatively, factor 2:= 2^{n+1}(n - 1) + 2 = 2*(2^n(n - 1)) + 2 = 2^{n+1}(n - 1) + 2But maybe we can write this differently. Let's check for small n to see if the formula works. Let's test n=1:T_1 = 1*2^1 = 2Using the formula: (1 - 1)2^{2} + 2 = 0 + 2 = 2. Correct.n=2:T_2 = 2 + 2*4 = 2 + 8 = 10Formula: (2 - 1)2^{3} + 2 = 1*8 + 2 = 10. Correct.n=3:T_3 = 2 + 8 + 3*8 = 2 + 8 + 24 = 34Formula: (3 -1)2^{4} + 2 = 2*16 + 2 = 34. Correct.So the formula works. Let's see if we can simplify it further.(n - 1)2^{n+1} + 2 = (n - 1)2^{n+1} + 2 = 2^{n+1}(n -1) + 2Alternatively, factor 2:= 2[2^n(n -1) + 1]But maybe leave it as is. Alternatively, let's see:(n -1)2^{n+1} + 2 = n*2^{n+1} - 2^{n+1} + 2 = n*2^{n+1} - 2(2^n) + 2 = n*2^{n+1} - 2^{n+1} + 2 = 2^{n+1}(n -1) + 2Alternatively, another approach. Let's recall that the formula for sum_{k=1}^n k*2^k.Wait, another way to derive it. Let me try again.Let T_n = sum_{k=1}^n k*2^kConsider T_n = 2 + 2*2^2 + 3*2^3 + ... + n*2^nLet’s compute 2T_n:2T_n = 2*2 + 2*2*2^2 + 2*3*2^3 + ... + 2*n*2^n = sum_{k=1}^n k*2^{k+1} = sum_{k=2}^{n+1} (k-1)*2^{k}So 2T_n = sum_{k=2}^{n+1} (k-1)*2^kSubtract T_n from 2T_n:2T_n - T_n = T_n = sum_{k=2}^{n+1} (k-1)*2^k - sum_{k=1}^n k*2^kLet’s adjust the indices:= [sum_{k=2}^{n} (k-1)*2^k + n*2^{n+1}] - [sum_{k=2}^n k*2^k + 1*2^1]= [sum_{k=2}^n (k-1 -k)*2^k + n*2^{n+1} - 2]= [sum_{k=2}^n (-1)*2^k + n*2^{n+1} - 2]= -sum_{k=2}^n 2^k + n*2^{n+1} - 2The sum sum_{k=2}^n 2^k is equal to sum_{k=0}^n 2^k - 2^0 -2^1 = (2^{n+1} -1) -1 -2 = 2^{n+1} -4Therefore:T_n = -(2^{n+1} -4) + n*2^{n+1} -2 = -2^{n+1} +4 +n*2^{n+1} -2 = (n -1)2^{n+1} +2Which matches the earlier result. Therefore, T_n = (n -1)2^{n+1} + 2. Let me check with n=3 again:(3 -1)2^{4} +2 = 2*16 +2=34. Correct.Alternatively, this can be written as T_n = (n -1)2^{n+1} + 2. Alternatively, factor 2^{n+1}:= 2^{n+1}(n -1) + 2But maybe leave it as is. So the answer is T_n = (n -1)2^{n+1} + 2. Let's see if we can simplify more.Alternatively, factor 2:= 2[ (n -1)2^n +1 ]But not sure if that's simpler. Alternatively, check if it can be written as (n*2^{n+1} -2^{n+1} +2) = n*2^{n+1} -2^{n+1} +2= n*2^{n+1} -2(2^n) +2= n*2^{n+1} -2^{n+1} +2= 2^{n+1}(n -1) +2Yes, that's as simplified as it can get. So the formula is T_n = (n -1)2^{n+1} +2.Alternatively, some sources present the formula as T_n = 2 + (n -1)2^{n+1}. Either way, it's the same.Alternatively, check for n=1:T_1=2, formula: (1-1)2^{2}+2=0+2=2. Correct.Another check for n=4:T_4=2 + 8 +24 +64= 2+8=10, 10+24=34, 34+64=98.Formula: (4-1)2^5 +2=3*32 +2=96+2=98. Correct.So formula works. Therefore, the answer is T_n=(n-1)2^{n+1}+2.Alternatively, can we write it as (n-1)2^{n+1} + 2 = n*2^{n+1} - 2^{n+1} +2= n*2^{n+1} -2(2^n -1). Hmm, but perhaps not necessary. The original form is acceptable.Thus, part (2)'s answer is T_n=(n-1)2^{n+1}+2.Final Answer(1) The general term is boxed{a_n = dfrac{1}{2^n}}.(2) The sum of the first (n) terms of ({b_n}) is boxed{T_n = (n - 1)2^{n+1} + 2}.</think>Given a sequence ({a_n}) where the sum of the first (n) terms, (S_n), is equal to (1 - a_n), we need to find the general term formula for ({a_n}).1. Finding the general term formula for ({a_n}): We start by noting that (S_n = 1 - a_n) and (S_n = S_{n-1} + a_n). Substituting (S_n = 1 - a_n) into the equation, we get: [ 1 - a_n = S_{n-1} + a_n ] Solving for (S_{n-1}): [ S_{n-1} = 1 - 2a_n ] Since (S_{n-1} = 1 - a_{n-1}) for (n geq 2), we equate: [ 1 - a_{n-1} = 1 - 2a_n ] Simplifying, we find: [ a_{n-1} = 2a_n ] This indicates a geometric sequence with common ratio (frac{1}{2}). For (n=1), (S_1 = a_1 = 1 - a_1) gives (a_1 = frac{1}{2}). Thus, the general term is: [ a_n = left(frac{1}{2}right)^n ] Verification confirms this formula satisfies (S_n = 1 - a_n).2. Finding the sum of the first (n) terms of ({b_n}) where (b_n = n cdot 2^n): We need to find (T_n = sum_{k=1}^n k cdot 2^k). Using the method of differences, we derive the formula for the finite sum. The key steps involve multiplying the sum by 2 and subtracting: [ 2T_n - T_n = T_n = sum_{k=2}^{n+1} (k-1)2^k - sum_{k=1}^n k2^k ] Simplifying, we find: [ T_n = (n - 1)2^{n+1} + 2 ] Verification with small values of (n) confirms the correctness of the formula. Final Answer(1) The general term is (boxed{a_n = dfrac{1}{2^n}}).(2) The sum of the first (n) terms of ({b_n}) is (boxed{T_n = (n - 1)2^{n+1} + 2}).,<think>Okay, so I have this problem about sequences. Let me start by reading it again carefully.First part: Given a sequence {a_n} where the sum of the first n terms, S_n, is equal to 1 - a_n for all natural numbers n. I need to find the general term formula for {a_n}.Second part: For another sequence {b_n} defined by b_n = n * 2^n, find the formula for the sum of the first n terms, T_n.Alright, starting with part (1). Let's recall that S_n is the sum of the first n terms of {a_n}, so S_n = a_1 + a_2 + ... + a_n. The problem states that S_n = 1 - a_n. So, for each n, the sum up to n terms is equal to 1 minus the nth term. Interesting. So, the sum itself is expressed in terms of the nth term. I need to find a formula for a_n.Let me write down the given equation:S_n = 1 - a_n.But also, S_n is the sum of the first n terms. So, S_n = S_{n-1} + a_n, since the sum up to n terms is the sum up to (n-1) terms plus the nth term. So, we have two expressions for S_n:1. S_n = 1 - a_n2. S_n = S_{n-1} + a_nSo, maybe I can set these equal to each other. From the first equation, S_n = 1 - a_n. From the second equation, S_n = S_{n-1} + a_n. Therefore:1 - a_n = S_{n-1} + a_nBut S_{n-1} is also equal to 1 - a_{n-1} by the given condition, right? Because the problem states that S_n = 1 - a_n for all n in N*, so replacing n with n-1, we get S_{n-1} = 1 - a_{n-1}. Therefore, substituting into the equation above:1 - a_n = (1 - a_{n-1}) + a_nLet me write that out:1 - a_n = 1 - a_{n-1} + a_nSimplify the right-hand side: 1 - a_{n-1} + a_n = 1 + ( -a_{n-1} + a_n )So, the equation becomes:1 - a_n = 1 - a_{n-1} + a_nSubtract 1 from both sides:- a_n = -a_{n-1} + a_nBring all terms to one side:- a_n + a_{n-1} - a_n = 0Wait, that would be:- a_n - a_n + a_{n-1} = 0Which simplifies to:a_{n-1} - 2 a_n = 0So, a_{n-1} = 2 a_nOr, rearranged:a_n = (1/2) a_{n-1}Oh! So, this is a recurrence relation. Each term is half of the previous term. So, the sequence {a_n} is a geometric sequence with common ratio 1/2.But wait, let me check this. If a_n = (1/2) a_{n-1}, then it's a geometric sequence with ratio 1/2. Let's verify if this holds with the initial conditions.First, we need to find the first term a_1. Let's compute S_1. Since S_1 = a_1, and by the problem's condition, S_1 = 1 - a_1. Therefore:a_1 = 1 - a_1Solving for a_1:a_1 + a_1 = 12 a_1 = 1a_1 = 1/2So, the first term is 1/2. Then, according to the recurrence relation, a_n = (1/2) a_{n-1}. Therefore, the sequence is:a_1 = 1/2a_2 = (1/2) * a_1 = (1/2)*(1/2) = 1/4a_3 = (1/2)*a_2 = (1/2)*(1/4) = 1/8And so on. So, in general, a_n = (1/2)^n.But wait, let me check if this works with the original condition S_n = 1 - a_n.If a_n = (1/2)^n, then S_n = sum_{k=1}^n (1/2)^k. The sum of the first n terms of a geometric series is S_n = a_1 (1 - r^n)/(1 - r). Here, a_1 = 1/2, r = 1/2.Therefore, S_n = (1/2)(1 - (1/2)^n)/(1 - 1/2) = (1/2)(1 - (1/2)^n)/(1/2) ) = 1 - (1/2)^nBut according to the problem, S_n should equal 1 - a_n. Let's check:1 - a_n = 1 - (1/2)^nWhich is exactly equal to S_n as calculated. Therefore, this works. So, the general term is a_n = (1/2)^n.Wait, but let me check for n=1. a_1 = 1/2. Then S_1 = 1 - a_1 = 1 - 1/2 = 1/2, which matches. For n=2, S_2 = a_1 + a_2 = 1/2 + 1/4 = 3/4. On the other hand, 1 - a_2 = 1 - 1/4 = 3/4. Correct. For n=3, S_3 = 1/2 + 1/4 + 1/8 = 7/8. 1 - a_3 = 1 - 1/8 = 7/8. Perfect.Therefore, the general term is a_n = (1/2)^n. So, part (1) solved.Now, moving on to part (2). The sequence {b_n} is given by b_n = n * 2^n. We need to find the sum of the first n terms, T_n = b_1 + b_2 + ... + b_n.So, T_n = sum_{k=1}^n k * 2^k.Hmm, I remember that there's a formula for the sum of k * r^k from k=1 to n. Let me recall it.The general formula for the sum S = sum_{k=0}^n k x^k is x(1 - (n+1)x^n + n x^{n+1}) / (1 - x)^2. Wait, maybe I need to adjust the indices since our sum starts at k=1 instead of k=0.Alternatively, let's derive it. Let me denote S = sum_{k=1}^n k * 2^k.To find a closed-form expression for S, we can use the method of generating functions or recursion. Let's use the standard technique for such sums.Consider S = 2 + 2*2^2 + 3*2^3 + ... + n*2^n.Multiply both sides by 2:2S = 2*2 + 2*2*2^2 + 2*3*2^3 + ... + 2*n*2^n = sum_{k=1}^n k*2^{k+1} = sum_{k=2}^{n+1} (k-1)*2^{k}.Now, subtract the original S from this equation:2S - S = S = [sum_{k=2}^{n+1} (k-1)*2^k] - [sum_{k=1}^n k*2^k]Let's write out the terms:Left side: SRight side: sum_{k=2}^{n+1} (k-1)*2^k - sum_{k=1}^n k*2^kLet me adjust the indices to align them. Let's split the first sum into two parts: from k=2 to k=n, and then the term k=n+1.So:sum_{k=2}^{n+1} (k-1)*2^k = sum_{k=2}^n (k-1)*2^k + (n) * 2^{n+1}Similarly, the second sum is sum_{k=1}^n k*2^k.Therefore, the right side becomes:[ sum_{k=2}^n (k-1)*2^k + n*2^{n+1} ] - [ sum_{k=1}^n k*2^k ]Let's distribute the subtraction:sum_{k=2}^n (k-1)*2^k - sum_{k=2}^n k*2^k + n*2^{n+1} - k=1 term [which is 1*2^1]Breaking it down:For the sums from k=2 to n:sum_{k=2}^n [ (k-1)*2^k - k*2^k ] = sum_{k=2}^n (-1)*2^k = - sum_{k=2}^n 2^kThen, the remaining terms:n*2^{n+1} - 1*2^1 = n*2^{n+1} - 2Therefore, combining:S = - sum_{k=2}^n 2^k + n*2^{n+1} - 2But the sum from k=2 to n of 2^k is sum_{k=0}^n 2^k - 2^0 - 2^1 = (2^{n+1} - 1) - 1 - 2 = 2^{n+1} - 4Therefore, substituting:S = - (2^{n+1} - 4) + n*2^{n+1} - 2Simplify:S = -2^{n+1} + 4 + n*2^{n+1} - 2 = (n*2^{n+1} - 2^{n+1}) + (4 - 2) = (n - 1)2^{n+1} + 2Factor 2^{n+1}:= 2^{n+1}(n - 1) + 2Alternatively, we can write this as:= n*2^{n+1} - 2^{n+1} + 2 = n*2^{n+1} - 2*2^{n} + 2 = n*2^{n+1} - 2^{n+1} + 2But maybe factor 2^{n+1}:= 2^{n+1}(n - 1) + 2Alternatively, perhaps we can factor 2:= 2(2^n (n - 1) + 1)But let me check the calculation again to be sure.Starting from S = - sum_{k=2}^n 2^k + n*2^{n+1} - 2sum_{k=2}^n 2^k = 2^2 + 2^3 + ... + 2^n = 2^2(2^{n -1} - 1)/(2 - 1) = 4(2^{n-1} - 1) = 2^{n+1} - 4Therefore, - sum_{k=2}^n 2^k = - (2^{n+1} - 4) = -2^{n+1} + 4Then, adding n*2^{n+1} -2:So total S = (-2^{n+1} +4) + n*2^{n+1} -2 = (-2^{n+1} + n*2^{n+1}) + (4 -2) = (n -1)2^{n+1} +2Yes, that's correct.Therefore, T_n = (n -1)2^{n+1} + 2We can write this as (n -1)2^{n+1} + 2, but maybe we can factor this differently.Alternatively, let's factor 2:= 2^{n+1}(n -1) + 2 = 2^{n+1}(n -1) + 2*1Alternatively, perhaps we can write this as 2^{n+1}(n -1) + 2 = 2^{n+1}(n -1) + 2Alternatively, check for small n:Let's test n=1. Then T_1 = b_1 = 1*2^1 = 2.Using the formula: (1 -1)2^{1+1} +2 = 0 +2 =2. Correct.For n=2: T_2 = 2 + 2*2^2 = 2 +8=10.Formula: (2 -1)2^{3} +2 =1*8 +2=10. Correct.n=3: T_3 =2 +8 + 3*8=2+8+24=34.Formula: (3 -1)2^{4} +2=2*16 +2=34. Correct.n=4: T_4=2 +8+24+4*16=2+8+24+64=98.Formula: (4-1)2^5 +2=3*32 +2=96 +2=98. Correct.Therefore, the formula works. So, the closed-form expression is T_n = (n -1)2^{n+1} + 2. Alternatively, we can factor 2:= 2 + (n -1)2^{n+1} = 2^{n+1}(n -1) +2Alternatively, we can write it as T_n = (2n - 2)2^n + 2 = 2^n(2n -2) +2 = 2^{n+1}(n -1) +2. That seems as simplified as possible. Let me see if there's another way.Alternatively, maybe factor 2^n:=2^n * 2(n -1) +2 =2^{n+1}(n -1) +2. Not much better.Alternatively, perhaps rearrange terms:=2^{n+1}(n -1) +2 =2^{n+1}n -2^{n+1} +2 =n2^{n+1} -2^{n+1} +2. Alternatively, factor 2 from the last two terms:=n2^{n+1} -2(2^{n} -1)But I don't know if that's helpful. The original expression (n -1)2^{n+1} +2 is concise enough.Alternatively, another standard formula:sum_{k=1}^n k*2^k =2^{n+1}(n -1) +2.Yes, that's the formula. So, the answer is T_n = (n -1)2^{n+1} +2.Alternatively, the problem might prefer it written as T_n = 2^{n+1}(n -1) +2. Either way is acceptable. Let's just confirm once again.For n=1: 2^{2}(0) +2=0 +2=2. Correct.For n=2:2^{3}(1) +2=8 +2=10. Correct.Yes, so that's the formula.Thus, summarizing:1) a_n = (1/2)^n2) T_n = (n -1)2^{n+1} +2Final Answer(1) The general term is boxed{a_n = dfrac{1}{2^n}}.(2) The sum of the first ( n ) terms of ({b_n}) is boxed{T_n = (n - 1)2^{n+1} + 2}).</think>Given a sequence ({a_n}) where the sum of the first (n) terms, (S_n), is equal to (1 - a_n) for all (n in mathbb{N}^*):1. To find the general term formula for ({a_n}): - We know (S_n = 1 - a_n) and (S_n = S_{n-1} + a_n). - Substituting (S_{n-1} = 1 - a_{n-1}) into the equation, we get: [ 1 - a_n = 1 - a_{n-1} + a_n ] - Simplifying, we find (a_{n-1} = 2a_n), leading to the recurrence relation (a_n = frac{1}{2}a_{n-1}). - The initial term (a_1 = frac{1}{2}) is found by solving (S_1 = a_1 = 1 - a_1). - This recurrence relation indicates a geometric sequence with the first term (frac{1}{2}) and common ratio (frac{1}{2}). Thus, the general term is (a_n = left(frac{1}{2}right)^n).2. To find the formula for the sum of the first (n) terms of the sequence ({b_n}) where (b_n = n cdot 2^n): - The sum (T_n = sum_{k=1}^n k cdot 2^k) can be derived using the method of generating functions or recursion. - By considering the sum (T_n) and manipulating the series, we find: [ T_n = (n - 1)2^{n+1} + 2 ] - This formula is verified by checking small values of (n) and confirming the results. Final Answer(1) The general term is boxed{a_n = dfrac{1}{2^n}}.(2) The sum of the first ( n ) terms of ({b_n}) is boxed{T_n = (n - 1)2^{n+1} + 2}.]

🔑:(1) Since S_n = 1 - a_n (Equation ①), we have S_{n+1} = 1 - a_{n+1} (Equation ②).Subtracting Equation ① from Equation ②, we get: a_{n+1} - a_n = -(a_{n+1} - (1+a_n)), which simplifies to: a_{n+1} = a_n for n in mathbb{N}^*.In particular, when n=1, we have a_1 = 1 - a_1, which implies a_1 = frac{1}{2}.Therefore, the general term formula for the sequence is: a_n = frac{1}{2} cdot n - 1 = frac{n - 2}{2}, for n in mathbb{N}^*.(2) Given b_n = n cdot 2^n for n in mathbb{N}^*, we want to find the sum of the first n terms, T_n: T_n = 1 cdot 2^1 + 2 cdot 2^2 + 3 cdot 2^3 + dots + n cdot 2^n (Equation ③).Multiplying the entire series by 2, we get: 2T_n = 1 cdot 2^2 + 2 cdot 2^3 + 3 cdot 2^4 + dots + n cdot 2^{n+1} (Equation ④).Subtracting Equation ④ from Equation ③ gives:begin{align*}T_n &= frac{1}{2}(2^2 - 2 + 2^3 - 2^2 + dots + 2^n - 2^{n-1} + n cdot 2^n - n cdot 2^{n+1}) &= frac{1}{2}(2 + 2^2 + 2^3 + dots + 2^n - n cdot 2^{n+1}).end{align*}Simplifying and collecting like terms it comes to: T_n = 2 + 4 + 8 + dots + 2^n - frac{n cdot 2^{n+1}}{2} T_n = 2(1 + 2 + 4 + dots + 2^{n-1}) - n cdot 2^n T_n = 2left(frac{2^n - 1}{2 - 1}right) - n cdot 2^n T_n = 2^{n+1} - 2 - n cdot 2^n Expressed as the sum of a geometric series and simplifying, we obtain: T_n = boxed{(n - 1) cdot 2^{n+1} + 2}, for n in mathbb{N}^*.