💡Okay, so I have this function f(x) = x² + ln x - 1, and there are three parts to the problem. Let me try to tackle each part one by one.Starting with part (1): Find the maximum and minimum values of f(x) in the interval [1, e]. Hmm, okay. I remember that to find maxima and minima on a closed interval, I need to check the critical points inside the interval and also evaluate the function at the endpoints.First, I should find the derivative of f(x) to locate the critical points. The derivative of x² is 2x, the derivative of ln x is 1/x, and the derivative of -1 is 0. So, f'(x) = 2x + 1/x.Now, to find critical points, I set f'(x) equal to zero: 2x + 1/x = 0. Let me solve this equation. Multiplying both sides by x to eliminate the denominator gives 2x² + 1 = 0. That simplifies to 2x² = -1, which leads to x² = -1/2. Wait, that can't be right because x² is always non-negative, and here it's equal to a negative number. So, there are no real solutions. That means there are no critical points where the derivative is zero.Hmm, so does that mean the function has no critical points in the interval [1, e]? It seems so. Therefore, the extrema must occur at the endpoints of the interval. So, I need to evaluate f(x) at x = 1 and x = e.Calculating f(1): f(1) = (1)² + ln(1) - 1 = 1 + 0 - 1 = 0.Calculating f(e): f(e) = (e)² + ln(e) - 1 = e² + 1 - 1 = e².So, in the interval [1, e], the function f(x) has a minimum value of 0 at x = 1 and a maximum value of e² at x = e. That seems straightforward.Moving on to part (2): Prove that on the interval (1, +∞), the graph of f(x) is below the graph of g(x) = x³. So, I need to show that f(x) < g(x) for all x > 1.Let me define a new function h(x) = f(x) - g(x) = x² + ln x - 1 - x³. If I can show that h(x) < 0 for all x > 1, then f(x) < g(x) on that interval.First, let's compute h(1): h(1) = 1 + 0 - 1 - 1 = -1. So, at x = 1, h(x) is negative.Next, let's look at the behavior of h(x) as x approaches infinity. The dominant term in h(x) is -x³, which goes to negative infinity as x increases. So, h(x) tends to negative infinity as x approaches infinity.But to ensure that h(x) is always negative for x > 1, I should check if h(x) is decreasing on (1, +∞). For that, I can compute the derivative h'(x).h'(x) = derivative of h(x) = derivative of f(x) - derivative of g(x) = (2x + 1/x) - 3x².So, h'(x) = 2x + 1/x - 3x².I need to determine the sign of h'(x) for x > 1. Let's analyze h'(x):For x > 1, 3x² is a rapidly increasing term, while 2x and 1/x are much smaller in comparison. So, intuitively, h'(x) should be negative for x > 1, meaning h(x) is decreasing.To confirm, let's evaluate h'(x) at x = 1: h'(1) = 2(1) + 1/1 - 3(1)² = 2 + 1 - 3 = 0.So, at x = 1, the derivative is zero. Let's check the derivative just above 1, say at x = 1.1:h'(1.1) = 2(1.1) + 1/1.1 - 3(1.1)² ≈ 2.2 + 0.9091 - 3(1.21) ≈ 2.2 + 0.9091 - 3.63 ≈ 3.1091 - 3.63 ≈ -0.5209.Negative. So, h'(x) is negative just above 1. Since h'(x) is continuous and we saw that as x increases, h'(x) becomes more negative, h(x) is decreasing on (1, +∞).Since h(1) = -1 and h(x) is decreasing beyond that, h(x) will stay below zero for all x > 1. Therefore, f(x) < g(x) on (1, +∞).Alright, that seems solid.Now, part (3): Prove that [f'(x)]ⁿ - f'(xⁿ) ≥ 2ⁿ - 2 for n ∈ ℕ⁺. Hmm, this seems a bit more abstract. Let's parse this.First, f'(x) is 2x + 1/x, as we found earlier. So, [f'(x)]ⁿ is (2x + 1/x)ⁿ. Then, f'(xⁿ) would be the derivative of f at xⁿ, which is 2(xⁿ) + 1/(xⁿ).So, the expression is [ (2x + 1/x)ⁿ ] - [ 2xⁿ + 1/xⁿ ] ≥ 2ⁿ - 2.We need to show this inequality holds for all positive integers n.Let me test for n = 1: [f'(x)]¹ - f'(x¹) = (2x + 1/x) - (2x + 1/x) = 0. And 2¹ - 2 = 0. So, equality holds for n = 1.For n = 2: [f'(x)]² - f'(x²) = (2x + 1/x)² - (2x² + 1/x²).Let's compute (2x + 1/x)²: 4x² + 4 + 1/x². Then subtract (2x² + 1/x²): 4x² + 4 + 1/x² - 2x² - 1/x² = 2x² + 4.We need to show that 2x² + 4 ≥ 2² - 2 = 4 - 2 = 2. Well, 2x² + 4 is always greater than or equal to 4, since x² ≥ 0, so 2x² + 4 ≥ 4 ≥ 2. So, the inequality holds for n = 2.Wait, but 2x² + 4 is actually greater than 4, which is more than 2. So, the inequality holds, but it's not tight. Maybe for higher n, the inequality becomes tighter.Let me try n = 3: [f'(x)]³ - f'(x³) = (2x + 1/x)³ - (2x³ + 1/x³).First, expand (2x + 1/x)³. Using the binomial theorem:(2x)³ + 3*(2x)²*(1/x) + 3*(2x)*(1/x)² + (1/x)³ = 8x³ + 3*(4x²)*(1/x) + 3*(2x)*(1/x²) + 1/x³ = 8x³ + 12x + 6/x + 1/x³.Subtract (2x³ + 1/x³): 8x³ + 12x + 6/x + 1/x³ - 2x³ - 1/x³ = 6x³ + 12x + 6/x.We need to show that 6x³ + 12x + 6/x ≥ 2³ - 2 = 8 - 2 = 6.Well, 6x³ + 12x + 6/x is clearly greater than 6 for x > 0, since each term is positive and at least 6 when x = 1: 6 + 12 + 6 = 24 ≥ 6. So, again, the inequality holds.Hmm, but I need a general proof for any n ∈ ℕ⁺. Maybe induction can be used here.Let me consider using mathematical induction.Base case: n = 1, which we saw gives 0 ≥ 0, which holds.Assume that for some k ≥ 1, [f'(x)]ᵏ - f'(xᵏ) ≥ 2ᵏ - 2.Now, we need to show that [f'(x)]^{k+1} - f'(x^{k+1}) ≥ 2^{k+1} - 2.But I'm not sure how to proceed with the induction step here because [f'(x)]^{k+1} = [f'(x)]ᵏ * f'(x), and f'(x^{k+1}) = 2x^{k+1} + 1/x^{k+1}.It might not be straightforward to relate this to the induction hypothesis. Maybe there's another approach.Alternatively, perhaps we can analyze the expression [f'(x)]ⁿ - f'(xⁿ) and find its minimum value.Given that f'(x) = 2x + 1/x, let's denote y = x + 1/(2x). Wait, but f'(x) = 2x + 1/x = 2(x + 1/(2x)). Maybe that substitution can help.Wait, actually, f'(x) = 2x + 1/x. Let me set t = x + 1/(2x). Then f'(x) = 2t.But I'm not sure if this substitution helps directly. Alternatively, perhaps consider that 2x + 1/x ≥ 2√(2x * 1/x) = 2√2 by AM-GM inequality, but that might not directly help.Wait, let's think about the expression [f'(x)]ⁿ - f'(xⁿ). Maybe we can find a lower bound for [f'(x)]ⁿ and an upper bound for f'(xⁿ), then subtract.But I'm not sure. Alternatively, perhaps consider that for x = 1, f'(1) = 2*1 + 1/1 = 3. So, [f'(1)]ⁿ - f'(1ⁿ) = 3ⁿ - 3. We need to show that 3ⁿ - 3 ≥ 2ⁿ - 2.But 3ⁿ grows faster than 2ⁿ, so 3ⁿ - 3 is certainly greater than 2ⁿ - 2 for n ≥ 1. But this is just at x = 1. We need to show it for all x > 0.Wait, but the problem doesn't specify the domain of x. It just says n ∈ ℕ⁺. So, perhaps x is in the domain where f is defined, which is x > 0 since ln x is involved.Wait, but in part (3), it's a theoretical question, so maybe x is any positive real number. So, we need to show that for any x > 0 and n ∈ ℕ⁺, [f'(x)]ⁿ - f'(xⁿ) ≥ 2ⁿ - 2.Hmm, let's consider x = 1. As above, [f'(1)]ⁿ - f'(1ⁿ) = 3ⁿ - 3. Since 3ⁿ - 3 ≥ 2ⁿ - 2 for all n ≥ 1, because 3ⁿ grows faster than 2ⁿ, and 3 - 3 = 0 ≥ 2 - 2 = 0 for n=1.But we need to show this for all x > 0. Maybe we can find the minimum of [f'(x)]ⁿ - f'(xⁿ) over x > 0 and show that it's at least 2ⁿ - 2.Alternatively, perhaps consider that [f'(x)]ⁿ is minimized when x is such that f'(x) is minimized, and f'(xⁿ) is maximized or something like that.Wait, f'(x) = 2x + 1/x. Let's find its minimum. The derivative of f'(x) with respect to x is 2 - 1/x². Setting this equal to zero: 2 - 1/x² = 0 ⇒ x² = 1/2 ⇒ x = 1/√2.So, the minimum of f'(x) is at x = 1/√2: f'(1/√2) = 2*(1/√2) + 1/(1/√2) = √2 + √2 = 2√2 ≈ 2.828.So, [f'(x)]ⁿ is minimized at x = 1/√2 with value (2√2)ⁿ.On the other hand, f'(xⁿ) = 2xⁿ + 1/xⁿ. Let's see, for x > 0, xⁿ + 1/xⁿ ≥ 2 by AM-GM, but here it's 2xⁿ + 1/xⁿ, which is different.Wait, 2xⁿ + 1/xⁿ. Let me denote y = xⁿ, so f'(xⁿ) = 2y + 1/y. The minimum of 2y + 1/y occurs at y where derivative is zero: 2 - 1/y² = 0 ⇒ y² = 1/2 ⇒ y = 1/√2. So, minimum value is 2*(1/√2) + √2 = √2 + √2 = 2√2.Wait, so f'(xⁿ) has a minimum value of 2√2, same as f'(x). Interesting.But in our expression, we have [f'(x)]ⁿ - f'(xⁿ). So, if both f'(x) and f'(xⁿ) have a minimum of 2√2, but [f'(x)]ⁿ would be (2√2)ⁿ, and f'(xⁿ) is at least 2√2.So, [f'(x)]ⁿ - f'(xⁿ) ≥ (2√2)ⁿ - 2√2.We need to show that (2√2)ⁿ - 2√2 ≥ 2ⁿ - 2.Is (2√2)ⁿ - 2√2 ≥ 2ⁿ - 2 for all n ∈ ℕ⁺?Let me check for n=1: (2√2) - 2√2 = 0 ≥ 2 - 2 = 0. Equality holds.n=2: (2√2)² - 2√2 = 8 - 2√2 ≈ 8 - 2.828 ≈ 5.172 ≥ 4 - 2 = 2. True.n=3: (2√2)³ - 2√2 = 16√2 - 2√2 = 14√2 ≈ 19.798 ≥ 8 - 2 = 6. True.So, it seems that (2√2)ⁿ grows much faster than 2ⁿ, so the inequality holds.But wait, is (2√2)ⁿ - 2√2 ≥ 2ⁿ - 2 for all n?Let me see for n=4: (2√2)^4 = (8)^2 = 64. 64 - 2√2 ≈ 64 - 2.828 ≈ 61.172 ≥ 16 - 2 = 14. True.Yes, it seems that (2√2)ⁿ grows exponentially with base ~2.828, which is larger than 2, so the left side grows much faster than the right side. Therefore, the inequality holds for all n ∈ ℕ⁺.But wait, this is only considering the minimum of [f'(x)]ⁿ - f'(xⁿ). But actually, [f'(x)]ⁿ - f'(xⁿ) could be larger elsewhere. So, the minimum is (2√2)ⁿ - 2√2, which is greater than or equal to 2ⁿ - 2. Therefore, [f'(x)]ⁿ - f'(xⁿ) ≥ 2ⁿ - 2 for all x > 0 and n ∈ ℕ⁺.I think that's the way to go. So, by finding the minimum of the expression, we can establish the inequality.

💡Okay, so I have this problem about an absent-minded scientist who built a device with a sensor and a transmitter. The sensor has an average lifespan of 3 years, and the transmitter has an average lifespan of 5 years. The scientist calculated that the average lifespan of the entire device is 3 years and 8 months. I need to figure out if the scientist made a mistake in his calculations.Alright, first, let's break down what's given. The sensor has an expected lifespan of 3 years, and the transmitter has an expected lifespan of 5 years. The device consists of both components, so it will stop working when either the sensor or the transmitter fails. That means the lifespan of the device is determined by the shorter of the two lifespans.So, if I denote the lifespan of the sensor as a random variable ξ and the lifespan of the transmitter as η, then the lifespan of the device T is the minimum of ξ and η. In mathematical terms, T = min(ξ, η).Now, the expected value of T, which is E[T], should be less than or equal to the expected value of the shorter-lived component. Since the sensor has a shorter expected lifespan (3 years) compared to the transmitter (5 years), the expected lifespan of the device should be less than or equal to 3 years.But the scientist claims that the average lifespan of the entire device is 3 years and 8 months. Let me convert that into years to make it easier to compare. 8 months is 8/12 of a year, which is approximately 0.6667 years. So, 3 years and 8 months is roughly 3.6667 years.Wait a minute, 3.6667 years is greater than 3 years. That doesn't make sense because the device's lifespan should be determined by the shorter-lived component, which is the sensor. If the sensor only lasts an average of 3 years, the device shouldn't last longer than that on average.But maybe I'm oversimplifying. Perhaps the scientist considered some other factors, like the distributions of the lifespans. If the sensor and transmitter have different distributions, maybe the expected minimum could be higher? Hmm, let's think about that.If both components have exponential distributions, which are commonly used for lifespans, the expected minimum of two independent exponential variables is given by 1/(λ1 + λ2), where λ1 and λ2 are the rates (1 over the mean). So, for the sensor, λ1 = 1/3, and for the transmitter, λ2 = 1/5. Then, the expected minimum would be 1/(1/3 + 1/5) = 1/(8/15) = 15/8 = 1.875 years. That's way less than 3 years.But wait, the scientist said 3 years and 8 months, which is longer than both 1.875 years and 3 years. That seems contradictory. Maybe the components aren't independent? Or perhaps they have different distributions, like Weibull distributions or something else?If the sensor and transmitter have different distributions, the calculation might be more complicated. For example, if the sensor has a higher variance, maybe the minimum could be higher? I'm not sure. Let me try to recall some probability theory.The expected value of the minimum of two random variables can be calculated using the formula:E[min(ξ, η)] = ∫₀^∞ P(min(ξ, η) > t) dtWhich is the same as:E[min(ξ, η)] = ∫₀^∞ P(ξ > t and η > t) dtSo, it's the integral of the probability that both components survive beyond time t.If I assume that the lifespans are independent, then P(ξ > t and η > t) = P(ξ > t) * P(η > t).But without knowing the specific distributions, it's hard to compute this integral. However, I can reason that the expected minimum should still be less than or equal to the smaller of the two expected lifespans.In this case, since the sensor has an expected lifespan of 3 years, the device's expected lifespan should be less than or equal to 3 years. The scientist's calculation of 3 years and 8 months is greater than 3 years, which seems impossible.Wait, unless the scientist considered the device to continue working even if one component fails, but that contradicts the problem statement which says the device consists of a sensor and a transmitter, implying both are necessary.Alternatively, maybe the scientist made a mistake in assuming that the device's lifespan is the average of the two components? That would be (3 + 5)/2 = 4 years, which is still less than 3 years and 8 months. Hmm, no, that doesn't add up.Another thought: perhaps the scientist used the harmonic mean instead of the minimum? The harmonic mean of 3 and 5 is 2/(1/3 + 1/5) = 2/(8/15) = 30/8 = 3.75 years, which is approximately 3 years and 9 months. That's close to the 3 years and 8 months the scientist calculated. Maybe the scientist used the harmonic mean by mistake, thinking it represents the expected minimum?But actually, the harmonic mean is not the same as the expected minimum. The expected minimum of two independent exponential variables is 1/(λ1 + λ2), which we calculated earlier as 1.875 years. So, the harmonic mean is not the correct approach here.Alternatively, maybe the scientist considered the device to have a lifespan equal to the sum of the sensor and transmitter lifespans? That would be 3 + 5 = 8 years, which is way higher than 3 years and 8 months. So, that doesn't make sense either.Wait, perhaps the scientist used the concept of parallel systems in reliability theory. In a parallel system, the system fails only when all components fail. But in this case, the device fails when either component fails, which is a series system. For a series system, the reliability is the product of the reliabilities of each component.But the expected lifespan of a series system is not simply the minimum of the expected lifespans. It depends on the specific distributions. However, it's generally true that the expected lifespan of a series system is less than or equal to the minimum expected lifespan of its components.So, in this case, since the sensor has an expected lifespan of 3 years, the device's expected lifespan should be less than or equal to 3 years. The scientist's calculation of 3 years and 8 months is greater than 3 years, which contradicts this principle.Therefore, it seems like the scientist made a mistake in his calculations. He probably used an incorrect method or assumption, such as averaging the lifespans or using the harmonic mean, which led him to an incorrect result.Another angle to consider: maybe the scientist didn't account for the fact that the device fails when either component fails, and instead thought of it as a system that can continue operating as long as at least one component is working. But that would be a parallel system, and the expected lifespan would be higher than both components, which is not the case here.Wait, no, in a parallel system, the system fails only when all components fail. So, the expected lifespan would be higher than both components. But the scientist's result is 3 years and 8 months, which is higher than the sensor's 3 years but lower than the transmitter's 5 years. That doesn't fit a parallel system either.I'm getting a bit confused here. Let me try to summarize:1. The device fails when either the sensor or the transmitter fails. So, it's a series system.2. The expected lifespan of a series system is less than or equal to the minimum expected lifespan of its components.3. The sensor has an expected lifespan of 3 years, so the device's expected lifespan should be ≤ 3 years.4. The scientist claims it's 3 years and 8 months, which is > 3 years.5. Therefore, the scientist must have made a mistake.Alternatively, if the device required both components to fail for it to stop working, it would be a parallel system, and the expected lifespan would be higher than both components. But that's not the case here.Wait, maybe the scientist confused series and parallel systems? If he thought it was a parallel system, he might have calculated the expected lifespan incorrectly as higher than both components, but in reality, it's a series system.Alternatively, maybe the scientist used the wrong formula for the expected minimum. For example, if he used the formula for independent variables with different distributions, but made a calculation error.But regardless of the specific distributions, the key point is that the expected minimum cannot exceed the minimum expected value of the individual components. Since the sensor has an expected lifespan of 3 years, the device's expected lifespan cannot be more than 3 years.Therefore, the scientist's calculation of 3 years and 8 months is incorrect. He must have made a mistake in his calculations.

💡Alright, so I've got this trapezoid problem here. Let me try to figure it out step by step. Okay, the trapezoid is named ABCD, and sides AB and CD are parallel. That's good to know because in trapezoids, the consecutive angles between the non-parallel sides are supplementary. I remember that from geometry class. So, that means angle A and angle D should add up to 180 degrees because they're on the same side and AB is parallel to CD.The problem also says that angle A is three times angle D, and angle C is two times angle D. Hmm, okay, so let me write that down:- Angle A = 3 * Angle D- Angle C = 2 * Angle DSince AB is parallel to CD, angle A and angle D are consecutive interior angles, so they should add up to 180 degrees. That makes sense because if two lines are parallel, the consecutive interior angles are supplementary. So, angle A + angle D = 180 degrees.But angle A is 3 times angle D, so I can substitute that into the equation. Let me do that:3 * angle D + angle D = 180 degreesThat simplifies to:4 * angle D = 180 degreesSo, if I divide both sides by 4, I get:angle D = 180 / 4angle D = 45 degreesOkay, so angle D is 45 degrees. Now, since angle A is three times angle D, angle A should be:angle A = 3 * 45 degreesangle A = 135 degreesWait, let me make sure I didn't make a mistake. So, angle D is 45 degrees, angle A is 135 degrees. Does that add up to 180 degrees? Yes, 45 + 135 is 180, so that checks out.But hold on, the problem also mentions angle C is two times angle D. So, angle C should be:angle C = 2 * 45 degreesangle C = 90 degreesHmm, okay, so angle C is 90 degrees. Since AB is parallel to CD, angle C and angle B should also be supplementary, right? Because they are consecutive interior angles as well.So, angle C + angle B = 180 degreesWe know angle C is 90 degrees, so:90 + angle B = 180angle B = 180 - 90angle B = 90 degreesWait, so angle B is also 90 degrees? That means both angle B and angle C are right angles. Interesting. So, in this trapezoid, two of the angles are right angles. That makes it a right trapezoid, I think. Yeah, a trapezoid with at least two right angles is called a right trapezoid.Let me just recap to make sure everything adds up. We have:- angle A = 135 degrees- angle D = 45 degrees- angle C = 90 degrees- angle B = 90 degreesLet me check if the sum of all angles in the trapezoid is 360 degrees because that's a property of quadrilaterals.135 + 45 + 90 + 90 = 360 degreesYes, that adds up correctly. So, all the angles are accounted for, and they sum up to 360 degrees as they should.I think that makes sense. So, angle A is 135 degrees. I don't see any mistakes in my calculations, so I think that's the correct answer.

💡Okay, so I'm trying to solve this problem where I need to find the range of the real number ( b ) such that the function ( f(x) = (x - 1)ln x - ax + a + b ) has two distinct zeros for any real number ( a ). The options given are A: ( (-infty, -1] ), B: ( (0, 1) ), C: ( (-infty, 0) ), and D: ( (0, +infty) ). First, I need to understand what it means for the function ( f(x) ) to have two distinct zeros. That means the equation ( f(x) = 0 ) has two different solutions for ( x ). So, I can rewrite the equation as:[ (x - 1)ln x - ax + a + b = 0 ]I think it might help to rearrange this equation to see if I can interpret it in terms of two functions intersecting. Let me try moving the linear terms to the other side:[ (x - 1)ln x = ax - a - b ]Hmm, so this equation is saying that the function ( g(x) = (x - 1)ln x ) is equal to the linear function ( h(x) = ax - a - b ). So, if I can analyze the behavior of ( g(x) ) and ( h(x) ), maybe I can figure out when they intersect twice.Let me first analyze ( g(x) = (x - 1)ln x ). I should find its derivative to understand its increasing or decreasing behavior. The derivative ( g'(x) ) is:[ g'(x) = frac{d}{dx}[(x - 1)ln x] ]Using the product rule, which states that ( frac{d}{dx}[u cdot v] = u'v + uv' ), where ( u = x - 1 ) and ( v = ln x ). So,[ u' = 1 ][ v' = frac{1}{x} ]Therefore,[ g'(x) = 1 cdot ln x + (x - 1) cdot frac{1}{x} ][ g'(x) = ln x + frac{x - 1}{x} ][ g'(x) = ln x + 1 - frac{1}{x} ]Okay, so ( g'(x) = ln x + 1 - frac{1}{x} ). I need to analyze where this derivative is positive or negative to understand the monotonicity of ( g(x) ).Let me consider the domain of ( g(x) ). Since ( ln x ) is defined for ( x > 0 ), the domain is ( x > 0 ).Let me evaluate ( g'(x) ) at some points:1. At ( x = 1 ): [ g'(1) = ln 1 + 1 - frac{1}{1} = 0 + 1 - 1 = 0 ] 2. For ( x ) approaching 0 from the right: - ( ln x ) approaches ( -infty ) - ( 1 - frac{1}{x} ) approaches ( -infty ) because ( frac{1}{x} ) becomes very large. So overall, ( g'(x) ) approaches ( -infty ).3. For ( x ) approaching ( +infty ): - ( ln x ) approaches ( +infty ) but grows slower than linear terms. - ( 1 - frac{1}{x} ) approaches 1. So, ( g'(x) ) approaches ( +infty ).Now, let me analyze the sign of ( g'(x) ) in different intervals.- For ( x ) in ( (0, 1) ): Let's pick ( x = 0.5 ): [ g'(0.5) = ln 0.5 + 1 - frac{1}{0.5} ] [ ln 0.5 approx -0.693 ] [ 1 - 2 = -1 ] So, ( g'(0.5) approx -0.693 - 1 = -1.693 ), which is negative.- For ( x ) in ( (1, +infty) ): Let's pick ( x = 2 ): [ g'(2) = ln 2 + 1 - frac{1}{2} ] [ ln 2 approx 0.693 ] [ 1 - 0.5 = 0.5 ] So, ( g'(2) approx 0.693 + 0.5 = 1.193 ), which is positive.So, from this, I can conclude that ( g'(x) ) is negative for ( x in (0, 1) ) and positive for ( x in (1, +infty) ). Therefore, ( g(x) ) is decreasing on ( (0, 1) ) and increasing on ( (1, +infty) ). At ( x = 1 ), ( g(x) ) has a critical point. Since the derivative changes from negative to positive, this critical point is a minimum. Let's compute ( g(1) ):[ g(1) = (1 - 1)ln 1 = 0 times 0 = 0 ]So, the function ( g(x) ) has a minimum at ( (1, 0) ).Now, let's analyze the linear function ( h(x) = ax - a - b ). This is a straight line with slope ( a ) and y-intercept ( -a - b ). Notice that ( h(x) ) can be rewritten as:[ h(x) = a(x - 1) - b ]This shows that the line passes through the point ( (1, -b) ) regardless of the value of ( a ). Because when ( x = 1 ), ( h(1) = a(0) - b = -b ).So, the line ( h(x) ) always passes through ( (1, -b) ). The slope ( a ) can be any real number, meaning the line can be made steeper or flatter as needed.Now, the problem states that for any real number ( a ), the equation ( g(x) = h(x) ) has two distinct solutions. So, regardless of how we choose ( a ), the line ( h(x) ) must intersect the curve ( g(x) ) at two distinct points.Given that ( g(x) ) has a minimum at ( (1, 0) ), and it's decreasing before ( x = 1 ) and increasing after ( x = 1 ), the graph of ( g(x) ) looks like a "U" shape with the bottom at ( (1, 0) ).The line ( h(x) ) passes through ( (1, -b) ). So, the y-coordinate at ( x = 1 ) is ( -b ). To have two intersection points between ( g(x) ) and ( h(x) ), the line ( h(x) ) must cross the curve ( g(x) ) twice. Since ( g(x) ) is decreasing before ( x = 1 ) and increasing after ( x = 1 ), the line must intersect both the left and right sides of the curve.But wait, if ( h(x) ) passes through ( (1, -b) ), and ( g(1) = 0 ), then at ( x = 1 ), ( h(1) = -b ). So, the line is at ( y = -b ) when ( x = 1 ), while ( g(1) = 0 ). Therefore, the vertical distance between ( h(x) ) and ( g(x) ) at ( x = 1 ) is ( | -b - 0 | = |b| ). For the line ( h(x) ) to intersect ( g(x) ) at two distinct points, it must cross the curve ( g(x) ) on both sides of ( x = 1 ). But since ( g(x) ) has a minimum at ( x = 1 ), the line must be above this minimum to intersect both sides. Wait, actually, if the line is above the minimum, it might intersect twice, but if it's exactly at the minimum, it would be tangent, and if it's below, it might not intersect at all.Wait, no. Let me think again. Since ( g(x) ) is decreasing before ( x = 1 ) and increasing after ( x = 1 ), the line ( h(x) ) must cross ( g(x) ) once on ( (0, 1) ) and once on ( (1, +infty) ).But for this to happen, the line must be above the minimum value of ( g(x) ). Since the minimum value of ( g(x) ) is 0 at ( x = 1 ), the line must be above 0 at ( x = 1 ). But wait, ( h(1) = -b ). So, if ( h(1) > 0 ), then ( -b > 0 ), which implies ( b < 0 ). If ( h(1) = 0 ), then ( b = 0 ), and if ( h(1) < 0 ), then ( b > 0 ).But if ( h(1) = -b ) is above 0, meaning ( -b > 0 ) or ( b < 0 ), then the line is above the minimum point of ( g(x) ). Therefore, the line will intersect ( g(x) ) at two points: one on ( (0, 1) ) and one on ( (1, +infty) ).If ( h(1) = 0 ), meaning ( b = 0 ), the line passes through the minimum point ( (1, 0) ). In this case, depending on the slope ( a ), the line might be tangent to ( g(x) ) at ( x = 1 ) or intersect it at another point. But since ( a ) can be any real number, if ( b = 0 ), for some ( a ), the line might be tangent, resulting in only one intersection point. Therefore, ( b = 0 ) is not acceptable because the problem states that for any ( a ), there are two distinct zeros. So, ( b ) cannot be zero.If ( h(1) < 0 ), meaning ( -b < 0 ) or ( b > 0 ), then the line is below the minimum of ( g(x) ). In this case, depending on the slope ( a ), the line might not intersect ( g(x) ) at all or might intersect only once. For example, if ( a ) is very large, the line might intersect ( g(x) ) on the right side, but if ( a ) is very small, it might not intersect at all. Therefore, ( b > 0 ) is not acceptable because we need two intersections for any ( a ).Therefore, the only possibility is ( b < 0 ), which ensures that ( h(1) = -b > 0 ), placing the line above the minimum of ( g(x) ). This way, regardless of the slope ( a ), the line will intersect ( g(x) ) at two distinct points: one on ( (0, 1) ) and one on ( (1, +infty) ).So, the range of ( b ) is all real numbers less than 0, which is ( (-infty, 0) ). Therefore, the correct answer is option C.**Final Answer**boxed{C}

💡Okay, so I have this problem here: Given that |m + 3| = m + 3, I need to find the range of m for which |3m + 9| is greater than or equal to 4m - 3. Hmm, let me try to figure this out step by step.First, let's look at the condition given: |m + 3| = m + 3. I remember that the absolute value of a number is equal to the number itself only when the number is non-negative. So, if |m + 3| equals m + 3, that must mean that m + 3 is greater than or equal to zero. Let me write that down:m + 3 ≥ 0Subtracting 3 from both sides gives:m ≥ -3Okay, so that tells me that m has to be at least -3. That's our first condition.Now, moving on to the inequality we need to solve: |3m + 9| ≥ 4m - 3. Hmm, this looks a bit more complicated, but let's break it down.First, I notice that 3m + 9 can be factored. Let me factor out a 3:3m + 9 = 3(m + 3)So, |3(m + 3)| is the same as 3|m + 3|. Since we already know from the first condition that m + 3 is non-negative (because m ≥ -3), the absolute value of m + 3 is just m + 3. Therefore, |3m + 9| simplifies to:3(m + 3) = 3m + 9So, the inequality becomes:3m + 9 ≥ 4m - 3Alright, now let's solve this inequality for m. Let me subtract 3m from both sides to get:9 ≥ m - 3Then, I'll add 3 to both sides:12 ≥ mWhich is the same as:m ≤ 12So, putting it all together, we have two conditions:1. m ≥ -3 (from the absolute value condition)2. m ≤ 12 (from solving the inequality)Therefore, combining these two, m must be between -3 and 12, inclusive. So, the range of m is:-3 ≤ m ≤ 12Let me just double-check my steps to make sure I didn't make any mistakes. Starting from |m + 3| = m + 3, which implies m + 3 ≥ 0, so m ≥ -3. Then, substituting into |3m + 9|, since m + 3 is non-negative, 3m + 9 is also non-negative, so the absolute value can be removed, leading to 3m + 9. Then, setting up the inequality 3m + 9 ≥ 4m - 3, solving for m gives m ≤ 12. Combining both conditions, m is between -3 and 12. Yeah, that seems right.I don't think I made any errors here. The key was recognizing that because of the first condition, the expression inside the absolute value in the second inequality is non-negative, so we can drop the absolute value and solve the linear inequality straightforwardly. That simplifies things a lot.So, yeah, I'm confident that the range of m is from -3 to 12, including both endpoints.

💡Okay, so I have this function ( f(x) = 2^{x} + 2^{ax + b} ) and I'm given two points: ( f(1) = frac{5}{2} ) and ( f(2) = frac{17}{4} ). I need to find the values of ( a ) and ( b ). Hmm, let's see. First, I can plug in ( x = 1 ) into the function. That gives me ( f(1) = 2^{1} + 2^{a(1) + b} = 2 + 2^{a + b} ). And this is equal to ( frac{5}{2} ). So, I can write the equation:[ 2 + 2^{a + b} = frac{5}{2} ]Subtracting 2 from both sides, I get:[ 2^{a + b} = frac{5}{2} - 2 = frac{1}{2} ]So, ( 2^{a + b} = frac{1}{2} ). Since ( frac{1}{2} ) is ( 2^{-1} ), this means:[ a + b = -1 ]Okay, that's one equation. Now, let's use the second point ( f(2) = frac{17}{4} ). Plugging ( x = 2 ) into the function:[ f(2) = 2^{2} + 2^{a(2) + b} = 4 + 2^{2a + b} ]This equals ( frac{17}{4} ), so:[ 4 + 2^{2a + b} = frac{17}{4} ]Subtracting 4 from both sides:[ 2^{2a + b} = frac{17}{4} - 4 = frac{1}{4} ]Again, ( frac{1}{4} ) is ( 2^{-2} ), so:[ 2a + b = -2 ]Now I have a system of two equations:1. ( a + b = -1 )2. ( 2a + b = -2 )I can solve this system. Let's subtract the first equation from the second:[ (2a + b) - (a + b) = -2 - (-1) ][ 2a + b - a - b = -2 + 1 ][ a = -1 ]Now, plug ( a = -1 ) back into the first equation:[ -1 + b = -1 ][ b = 0 ]So, ( a = -1 ) and ( b = 0 ). That seems straightforward. Let me double-check by plugging these back into the original function:For ( x = 1 ):[ f(1) = 2^{1} + 2^{-1(1) + 0} = 2 + 2^{-1} = 2 + frac{1}{2} = frac{5}{2} ]Which matches.For ( x = 2 ):[ f(2) = 2^{2} + 2^{-1(2) + 0} = 4 + 2^{-2} = 4 + frac{1}{4} = frac{17}{4} ]That also matches. Okay, so part (I) is solved.Moving on to part (II): Determine and prove the parity of ( f(x) ). Parity refers to whether the function is even, odd, or neither. An even function satisfies ( f(-x) = f(x) ) for all ( x ), and an odd function satisfies ( f(-x) = -f(x) ).Given that ( a = -1 ) and ( b = 0 ), the function simplifies to:[ f(x) = 2^{x} + 2^{-x} ]Let's compute ( f(-x) ):[ f(-x) = 2^{-x} + 2^{-(-x)} = 2^{-x} + 2^{x} ]Which is the same as ( f(x) ). Therefore, ( f(-x) = f(x) ), so the function is even. That seems clear.Now, part (III): Determine and prove the monotonic behavior of ( f(x) ) on ( (-infty, 0) ). Monotonic behavior means whether the function is increasing or decreasing on that interval.First, let's recall that ( f(x) = 2^{x} + 2^{-x} ). To determine if it's increasing or decreasing, we can look at its derivative.Compute the derivative ( f'(x) ):[ f'(x) = ln(2) cdot 2^{x} - ln(2) cdot 2^{-x} ]Simplify:[ f'(x) = ln(2) (2^{x} - 2^{-x}) ]Now, let's analyze the sign of ( f'(x) ) on ( (-infty, 0) ). For ( x < 0 ), ( 2^{x} ) is less than 1, and ( 2^{-x} ) is greater than 1. Therefore, ( 2^{x} - 2^{-x} ) is negative because ( 2^{x} < 2^{-x} ) when ( x < 0 ).Since ( ln(2) ) is positive, the entire derivative ( f'(x) ) is negative on ( (-infty, 0) ). A negative derivative implies that the function is decreasing on that interval.Alternatively, without calculus, we can consider the behavior of each term. ( 2^{x} ) is an increasing function, but for ( x < 0 ), it's increasing towards 1. ( 2^{-x} ) is a decreasing function for ( x < 0 ) because as ( x ) decreases, ( -x ) increases, making ( 2^{-x} ) increase. Wait, that seems conflicting. Let me think again.Actually, ( 2^{-x} ) can be rewritten as ( (1/2)^{x} ), which is an increasing function when ( x ) increases, but since ( x ) is negative, as ( x ) becomes more negative (i.e., decreases), ( (1/2)^{x} ) increases. So, both ( 2^{x} ) and ( 2^{-x} ) are increasing as ( x ) increases, but in the interval ( (-infty, 0) ), as ( x ) increases towards 0, both terms increase. However, the overall function ( f(x) = 2^{x} + 2^{-x} ) might not be straightforward.Wait, maybe it's better to consider specific values. Let's pick two points in ( (-infty, 0) ), say ( x = -2 ) and ( x = -1 ).Compute ( f(-2) = 2^{-2} + 2^{2} = frac{1}{4} + 4 = frac{17}{4} ).Compute ( f(-1) = 2^{-1} + 2^{1} = frac{1}{2} + 2 = frac{5}{2} ).So, as ( x ) increases from -2 to -1, ( f(x) ) decreases from ( frac{17}{4} ) to ( frac{5}{2} ). That suggests the function is decreasing on ( (-infty, 0) ).Alternatively, considering the derivative approach, since ( f'(x) < 0 ) for ( x < 0 ), the function is indeed decreasing on that interval.So, putting it all together, the function is even and decreasing on ( (-infty, 0) ).**Final Answer**Ⅰ. ( a = boxed{-1} ) and ( b = boxed{0} ).Ⅱ. ( f(x) ) is an even function, so the parity is boxed{text{even}}.Ⅲ. ( f(x) ) is boxed{text{decreasing}} on ((-∞,0)).