💡To find the value of ( costheta ) given that the terminal side of angle ( theta ) passes through the point ( P(-12, 5) ), follow these steps:1. **Identify the coordinates of point ( P ):** - The point ( P ) has coordinates ( (-12, 5) ).2. **Calculate the distance ( r ) from the origin to point ( P ):** - Use the distance formula: [ r = sqrt{(-12)^2 + 5^2} = sqrt{144 + 25} = sqrt{169} = 13 ]3. **Determine the cosine of angle ( theta ):** - The cosine of an angle in standard position is given by: [ costheta = frac{text{Adjacent side}}{text{Hypotenuse}} = frac{x}{r} ] - Substitute the known values: [ costheta = frac{-12}{13} ]4. **Finalize the answer:** - The value of ( costheta ) is ( -frac{12}{13} ).

💡Okay, so I have this function f(x) = log base a of (a x squared plus 2x plus a squared). I need to find the range of values for a such that this function is increasing on the interval [-4, -2]. Hmm, okay, let's break this down step by step.First, I remember that the logarithmic function log base a of something is increasing if the base a is greater than 1, and decreasing if the base a is between 0 and 1. So, that might be important here because the behavior of the logarithm depends on the base.But wait, the function inside the logarithm is a quadratic function, right? So, f(x) = log_a(t(x)), where t(x) = a x² + 2x + a². So, for f(x) to be increasing, the composition of these functions needs to result in an increasing function on the interval [-4, -2].I think I need to consider two cases: when a > 1 and when 0 < a < 1 because the logarithm's behavior changes based on the base.Let me start with the case when a > 1. In this case, log_a is an increasing function. So, for f(x) to be increasing, the inner function t(x) must also be increasing on the interval [-4, -2]. Because if both the outer and inner functions are increasing, their composition will be increasing.But wait, t(x) is a quadratic function. Quadratic functions have a vertex, and depending on the direction they open (up or down), they can be increasing or decreasing on certain intervals. Since the coefficient of x² is a, which is positive if a > 1, the parabola opens upwards. So, the vertex is the minimum point.The vertex of a quadratic ax² + bx + c is at x = -b/(2a). So, for t(x), the vertex is at x = -2/(2a) = -1/a. Since a > 1, -1/a is between -1 and 0. So, the vertex is at x = -1/a, which is between -1 and 0.Now, the interval we're considering is [-4, -2], which is to the left of the vertex. Since the parabola opens upwards, the function t(x) is decreasing to the left of the vertex. So, on the interval [-4, -2], t(x) is decreasing. But since a > 1, log_a is increasing. So, the composition of an increasing function with a decreasing function would result in a decreasing function. Therefore, f(x) would be decreasing on [-4, -2], which is not what we want. So, a > 1 is not acceptable.Okay, so now let's consider the case when 0 < a < 1. In this case, log_a is a decreasing function. So, for f(x) to be increasing, the inner function t(x) must be decreasing on the interval [-4, -2]. Because a decreasing function composed with another decreasing function results in an increasing function.So, we need t(x) to be decreasing on [-4, -2]. Again, t(x) is a quadratic function. Since 0 < a < 1, the coefficient of x² is positive, so the parabola opens upwards. The vertex is still at x = -1/a. But since a is between 0 and 1, -1/a is less than -1. So, the vertex is at x = -1/a, which is less than -1.Now, the interval [-4, -2] is to the right of the vertex because the vertex is at x = -1/a, which is less than -1. Since the parabola opens upwards, to the right of the vertex, the function is increasing. But we need t(x) to be decreasing on [-4, -2]. Hmm, that seems contradictory.Wait, maybe I made a mistake here. Let me think again. If the vertex is at x = -1/a, which is less than -1, then on the interval [-4, -2], which is to the right of the vertex, the function t(x) is increasing because it's on the right side of the vertex in an upward-opening parabola.But we need t(x) to be decreasing on [-4, -2] so that when we compose it with the decreasing logarithm, the overall function f(x) is increasing. So, if t(x) is increasing on [-4, -2], then log_a(t(x)) would be decreasing because log_a is decreasing. That's the opposite of what we want.Wait, so maybe I need t(x) to be decreasing on [-4, -2]. But if the vertex is at x = -1/a, which is less than -1, then to the left of the vertex, the function is decreasing. So, if the interval [-4, -2] is to the left of the vertex, then t(x) would be decreasing there.But hold on, if the vertex is at x = -1/a, which is less than -1, then the interval [-4, -2] is actually to the right of the vertex because -4 and -2 are greater than -1/a (since -1/a is less than -1). So, the interval [-4, -2] is on the right side of the vertex, meaning t(x) is increasing there.Hmm, so that's a problem because we need t(x) to be decreasing on [-4, -2] for f(x) to be increasing. So, maybe my initial approach is wrong.Wait, perhaps I need to consider the derivative of f(x) to determine where it's increasing. Let me try that.The derivative of f(x) with respect to x is f'(x) = (1 / (t(x) ln a)) * t'(x). So, f'(x) = (t'(x)) / (t(x) ln a). For f(x) to be increasing, f'(x) must be positive on [-4, -2].So, f'(x) > 0 implies that (t'(x)) / (t(x) ln a) > 0.Since t(x) is inside a logarithm, it must be positive for all x in [-4, -2]. So, t(x) > 0 on [-4, -2].Now, the sign of f'(x) depends on the sign of t'(x) and ln a.Case 1: a > 1.Then, ln a > 0.So, f'(x) > 0 implies t'(x) > 0.So, t'(x) = 2a x + 2.We need t'(x) > 0 on [-4, -2].So, 2a x + 2 > 0 for all x in [-4, -2].Let's find the minimum value of t'(x) on [-4, -2]. Since t'(x) is linear, its minimum occurs at x = -4.So, t'(-4) = 2a*(-4) + 2 = -8a + 2.We need -8a + 2 > 0 => -8a > -2 => 8a < 2 => a < 2/8 => a < 1/4.But in this case, a > 1, which contradicts a < 1/4. So, no solution in this case.Case 2: 0 < a < 1.Then, ln a < 0.So, f'(x) > 0 implies t'(x) < 0.So, t'(x) = 2a x + 2 < 0 for all x in [-4, -2].Again, t'(x) is linear, so its maximum occurs at x = -2.So, t'(-2) = 2a*(-2) + 2 = -4a + 2.We need -4a + 2 < 0 => -4a < -2 => 4a > 2 => a > 1/2.So, a must be greater than 1/2.But we also need t(x) > 0 on [-4, -2].So, let's ensure that t(x) = a x² + 2x + a² > 0 for all x in [-4, -2].Since t(x) is a quadratic opening upwards (because a > 0), its minimum occurs at the vertex.The vertex is at x = -b/(2a) = -2/(2a) = -1/a.We need to check if the vertex is within the interval [-4, -2] or not.If -1/a is within [-4, -2], then the minimum of t(x) is at x = -1/a. Otherwise, the minimum is at one of the endpoints.So, let's find when -1/a is in [-4, -2].-4 ≤ -1/a ≤ -2Multiply all parts by -1 (remembering to reverse inequalities):4 ≥ 1/a ≥ 2Which is equivalent to:2 ≤ 1/a ≤ 4Taking reciprocals (and reversing inequalities again):1/4 ≤ a ≤ 1/2So, if a is between 1/4 and 1/2, the vertex x = -1/a is in [-4, -2]. Otherwise, it's outside.So, we have two subcases:Subcase 1: a ≤ 1/4 or a ≥ 1/2.Subcase 2: 1/4 < a < 1/2.Wait, but earlier we found that a must be greater than 1/2 for t'(x) < 0 on [-4, -2]. So, in this problem, since a must be greater than 1/2, we can focus on Subcase 1 where a ≥ 1/2.But wait, in Subcase 1, if a ≥ 1/2, then the vertex x = -1/a is ≤ -2 because:If a = 1/2, x = -1/(1/2) = -2.If a > 1/2, x = -1/a > -2.Wait, no. If a increases, 1/a decreases, so x = -1/a becomes less negative, i.e., closer to zero.Wait, let's clarify:If a = 1/2, x = -2.If a > 1/2, say a = 3/4, x = -1/(3/4) = -4/3 ≈ -1.333, which is greater than -2.So, when a > 1/2, the vertex x = -1/a is greater than -2, meaning it's to the right of -2.But our interval is [-4, -2]. So, if the vertex is at x = -1/a > -2, then on the interval [-4, -2], the function t(x) is decreasing because it's to the left of the vertex in an upward-opening parabola.Wait, no. If the vertex is at x = -1/a > -2, then on the interval [-4, -2], which is to the left of the vertex, the function t(x) is decreasing because it's on the left side of the vertex in an upward-opening parabola.But earlier, we found that for f(x) to be increasing, t(x) needs to be decreasing on [-4, -2] because log_a is decreasing when 0 < a < 1.So, if t(x) is decreasing on [-4, -2], then f(x) = log_a(t(x)) would be increasing because log_a is decreasing, and the composition of two decreasing functions is increasing.So, in this case, when a > 1/2, t(x) is decreasing on [-4, -2], which is good.But we also need to ensure that t(x) > 0 on [-4, -2].Since t(x) is decreasing on [-4, -2], its minimum value is at x = -2.So, we need t(-2) > 0.Let's compute t(-2):t(-2) = a*(-2)^2 + 2*(-2) + a² = 4a - 4 + a².So, 4a - 4 + a² > 0.Let's solve this inequality:a² + 4a - 4 > 0.This is a quadratic in a. Let's find its roots:a = [-4 ± sqrt(16 + 16)] / 2 = [-4 ± sqrt(32)] / 2 = [-4 ± 4*sqrt(2)] / 2 = -2 ± 2*sqrt(2).So, the roots are a = -2 + 2√2 ≈ -2 + 2.828 ≈ 0.828 and a = -2 - 2√2 ≈ -4.828.Since a must be positive (0 < a < 1), we consider the positive root, which is approximately 0.828.The quadratic a² + 4a - 4 is positive when a < -2 - 2√2 or a > -2 + 2√2. But since a is positive, we only consider a > -2 + 2√2 ≈ 0.828.So, t(-2) > 0 when a > -2 + 2√2.But earlier, we found that a must be greater than 1/2 for t'(x) < 0 on [-4, -2].So, combining these two conditions, a must be greater than -2 + 2√2, which is approximately 0.828, and also a must be greater than 1/2.Since -2 + 2√2 ≈ 0.828 is greater than 1/2, the more restrictive condition is a > -2 + 2√2.But wait, we also need to ensure that t(x) > 0 on the entire interval [-4, -2]. Since t(x) is decreasing on this interval, the minimum is at x = -2, which we've already checked. So, as long as t(-2) > 0, t(x) will be positive on the entire interval.Therefore, the range of a is a > -2 + 2√2, but we also have the condition from the derivative that a > 1/2. However, since -2 + 2√2 ≈ 0.828 is greater than 1/2, the combined condition is a > -2 + 2√2.But wait, let's double-check. If a is between 1/2 and -2 + 2√2, does t(x) remain positive?Wait, no. Because t(-2) = a² + 4a - 4. When a = 1/2, t(-2) = (1/2)^2 + 4*(1/2) - 4 = 1/4 + 2 - 4 = -1.75, which is negative. So, t(-2) is negative when a = 1/2, which is not allowed because the logarithm is undefined there.So, actually, the condition t(-2) > 0 requires a > -2 + 2√2 ≈ 0.828. Therefore, the range of a is a > -2 + 2√2, but since a must also be less than 1 (because we're in the case 0 < a < 1), the range is (-2 + 2√2, 1).But wait, let's confirm this. If a is greater than -2 + 2√2, which is approximately 0.828, and less than 1, then t(x) is positive on [-4, -2], and t'(x) < 0 on [-4, -2], so f'(x) > 0 because log_a is decreasing (since 0 < a < 1) and t'(x) < 0, so their ratio is positive.Therefore, the function f(x) is increasing on [-4, -2] when a is in (-2 + 2√2, 1).But wait, earlier I thought the answer was [1/2, -2 + 2√2), but now I'm getting a different interval. Let me check where I went wrong.In the initial thought process, I considered that when 0 < a < 1, the vertex is at x = -1/a < -1, and for t(x) to be decreasing on [-4, -2], the vertex must be to the right of -2, meaning x = -1/a ≥ -2, which implies a ≤ 1/2.But wait, that contradicts the derivative approach where a needs to be greater than -2 + 2√2 ≈ 0.828.Hmm, I think the confusion arises from whether the interval [-4, -2] is to the left or right of the vertex.If the vertex is at x = -1/a, and we need t(x) to be decreasing on [-4, -2], then the interval [-4, -2] must be to the left of the vertex. So, x = -4 and x = -2 must be less than or equal to the vertex x = -1/a.Wait, no. If the parabola opens upwards, to the left of the vertex, the function is decreasing, and to the right, it's increasing.So, if the interval [-4, -2] is to the left of the vertex, then t(x) is decreasing there, which is what we want.Therefore, we need the vertex x = -1/a to be to the right of -2, meaning -1/a ≥ -2.Solving for a:-1/a ≥ -2Multiply both sides by -1 (inequality reverses):1/a ≤ 2Which implies a ≥ 1/2.So, a must be ≥ 1/2.But we also need t(x) > 0 on [-4, -2], which requires t(-2) > 0, leading to a > -2 + 2√2 ≈ 0.828.Therefore, combining these two conditions, a must be ≥ 1/2 and a > -2 + 2√2.Since -2 + 2√2 ≈ 0.828 is greater than 1/2, the combined condition is a > -2 + 2√2.But wait, earlier I thought that when a ≥ 1/2, the vertex is at x = -1/a ≤ -2, meaning the interval [-4, -2] is to the left of the vertex, so t(x) is decreasing there. But if a > -2 + 2√2, which is approximately 0.828, then the vertex is at x = -1/a ≈ -1/0.828 ≈ -1.207, which is greater than -2. So, the interval [-4, -2] is to the left of the vertex, meaning t(x) is decreasing there.Wait, but if the vertex is at x ≈ -1.207, which is greater than -2, then the interval [-4, -2] is to the left of the vertex, so t(x) is decreasing on [-4, -2], which is what we want.But earlier, when I considered the derivative, I found that a must be greater than -2 + 2√2 for t(-2) > 0, and a must be greater than 1/2 for t'(x) < 0 on [-4, -2]. Since -2 + 2√2 ≈ 0.828 is greater than 1/2, the combined condition is a > -2 + 2√2.But wait, the initial thought process concluded [1/2, -2 + 2√2), which seems contradictory.I think the confusion comes from whether the vertex is inside or outside the interval. Let me clarify:If the vertex x = -1/a is inside the interval [-4, -2], then t(x) has its minimum at the vertex. If the vertex is outside the interval, then the minimum is at one of the endpoints.So, when is x = -1/a inside [-4, -2]?-4 ≤ -1/a ≤ -2Multiply by -1 (reverse inequalities):4 ≥ 1/a ≥ 2Which is equivalent to:2 ≤ 1/a ≤ 4Taking reciprocals:1/4 ≤ a ≤ 1/2So, when a is between 1/4 and 1/2, the vertex is inside [-4, -2]. Otherwise, it's outside.So, if a is between 1/4 and 1/2, the minimum of t(x) is at x = -1/a. We need t(-1/a) > 0.Compute t(-1/a):t(-1/a) = a*(-1/a)^2 + 2*(-1/a) + a² = a*(1/a²) - 2/a + a² = 1/a - 2/a + a² = (-1)/a + a².So, t(-1/a) = a² - 1/a.We need this to be positive:a² - 1/a > 0Multiply both sides by a (since a > 0):a³ - 1 > 0 => a³ > 1 => a > 1But in this case, a is between 1/4 and 1/2, which is less than 1. So, t(-1/a) = a² - 1/a < 0 because a² < 1 and 1/a > 1, so a² - 1/a is negative.Therefore, when a is between 1/4 and 1/2, t(x) has a minimum at x = -1/a which is negative, meaning t(x) is not positive on the entire interval [-4, -2]. Therefore, t(x) would be negative somewhere in [-4, -2], which is not allowed because the logarithm is undefined there.Therefore, a cannot be between 1/4 and 1/2.So, the only valid cases are when a is either less than or equal to 1/4 or greater than or equal to 1/2.But earlier, we found that for t(x) to be decreasing on [-4, -2], a must be ≥ 1/2.But when a is ≥ 1/2, the vertex x = -1/a is ≤ -2, meaning the interval [-4, -2] is to the left of the vertex, so t(x) is decreasing there.But we also need t(x) > 0 on [-4, -2]. Since t(x) is decreasing on this interval, the minimum is at x = -2, so t(-2) > 0.As before, t(-2) = a² + 4a - 4 > 0, which gives a > -2 + 2√2 ≈ 0.828.Therefore, combining these conditions, a must be ≥ 1/2 and a > -2 + 2√2. Since -2 + 2√2 ≈ 0.828 is greater than 1/2, the range of a is a > -2 + 2√2.But wait, earlier I thought the answer was [1/2, -2 + 2√2), but now I'm getting a > -2 + 2√2.I think the confusion arises from whether the vertex is inside or outside the interval and ensuring t(x) is positive everywhere on the interval.So, to summarize:- For f(x) to be increasing on [-4, -2], since 0 < a < 1, log_a is decreasing, so t(x) must be decreasing on [-4, -2].- t(x) is decreasing on [-4, -2] if the vertex x = -1/a is to the right of -2, i.e., -1/a ≥ -2 => a ≥ 1/2.- Additionally, t(x) must be positive on [-4, -2]. Since t(x) is decreasing there, the minimum is at x = -2, so t(-2) > 0 => a > -2 + 2√2.Therefore, combining these, a must satisfy both a ≥ 1/2 and a > -2 + 2√2. Since -2 + 2√2 ≈ 0.828 > 1/2, the range is a > -2 + 2√2.But wait, the initial thought process concluded [1/2, -2 + 2√2), which seems contradictory. Let me check the initial thought process again.In the initial thought process, it was stated:"When 0 < a < 1, the axis of symmetry for t(x)=ax²+2x+a² is x=−1/a < −1. For the function f(x)=log_a(t(x)) to be increasing on [−4,−2], t(x)=ax²+2x+a² must be decreasing on [−4,−2].Thus, the axis of symmetry of t(x)'s graph, x=−1/a ≥ −2, and t(−2) > 0.That is, 1/a ≤ 2, and t(−2)=4a−4+a² > 0, which gives us 1/2 ≤ a < −2+2√2."Wait, so in the initial thought process, it was concluded that a must be between 1/2 and -2 + 2√2.But according to my detailed analysis, a must be greater than -2 + 2√2.So, which one is correct?Let me compute -2 + 2√2:√2 ≈ 1.414, so 2√2 ≈ 2.828.-2 + 2.828 ≈ 0.828.So, -2 + 2√2 ≈ 0.828.So, the initial thought process concluded a ∈ [1/2, 0.828), while my analysis suggests a ∈ (0.828, 1).But which one is correct?Wait, let's test a value in [1/2, 0.828), say a = 0.6.Compute t(-2) = (0.6)^2 + 4*(0.6) - 4 = 0.36 + 2.4 - 4 = -1.24 < 0. So, t(-2) is negative, which is not allowed.Therefore, a cannot be 0.6 because t(-2) would be negative.Similarly, if a = 0.8, t(-2) = 0.64 + 3.2 - 4 = -0.16 < 0.Still negative.Wait, when does t(-2) become positive?Solve a² + 4a - 4 = 0.Solutions are a = [-4 ± sqrt(16 + 16)] / 2 = [-4 ± sqrt(32)] / 2 = [-4 ± 4√2]/2 = -2 ± 2√2.So, the positive solution is a = -2 + 2√2 ≈ 0.828.Therefore, for a > -2 + 2√2, t(-2) > 0.So, if a is in (0.828, 1), t(-2) > 0, and t(x) is decreasing on [-4, -2] because a ≥ 1/2.But wait, if a is in (0.828, 1), then a ≥ 1/2 is satisfied, so t(x) is decreasing on [-4, -2], and t(-2) > 0.Therefore, the correct range is a > -2 + 2√2, which is approximately 0.828.But the initial thought process concluded [1/2, -2 + 2√2), which is incorrect because for a in [1/2, -2 + 2√2), t(-2) is negative, making the logarithm undefined.Therefore, the correct range is a > -2 + 2√2, but since a must be less than 1 (because we're in the case 0 < a < 1), the range is (-2 + 2√2, 1).But wait, the initial thought process concluded [1/2, -2 + 2√2), which is incorrect because in that interval, t(-2) is negative.Therefore, the correct range is a > -2 + 2√2, but since a must be less than 1, it's (-2 + 2√2, 1).But the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.So, the correct answer should be a > -2 + 2√2, but since a must be less than 1, the range is (-2 + 2√2, 1).But wait, let's check a = 0.9, which is in (-2 + 2√2, 1).Compute t(-2) = 0.81 + 3.6 - 4 = 0.41 > 0.Compute t'(x) = 2a x + 2.At x = -4, t'(-4) = 2*0.9*(-4) + 2 = -7.2 + 2 = -5.2 < 0.At x = -2, t'(-2) = 2*0.9*(-2) + 2 = -3.6 + 2 = -1.6 < 0.So, t'(x) < 0 on [-4, -2], which is good because log_a is decreasing, so f(x) is increasing.Therefore, the correct range is a > -2 + 2√2, which is approximately 0.828, and since a must be less than 1, the range is (-2 + 2√2, 1).But the initial thought process concluded [1/2, -2 + 2√2), which is incorrect because in that interval, t(-2) is negative.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But wait, the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.So, in conclusion, the correct range is a > -2 + 2√2, but since a must be less than 1, the range is (-2 + 2√2, 1).But wait, let's express -2 + 2√2 in exact form.-2 + 2√2 is approximately 0.828, but in exact terms, it's 2(√2 - 1).So, the range is (2(√2 - 1), 1).But the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is a ∈ (2(√2 - 1), 1).But let me double-check.If a = 2(√2 - 1) ≈ 0.828, then t(-2) = a² + 4a - 4.Compute a²: (2(√2 - 1))² = 4(√2 - 1)^2 = 4(2 - 2√2 + 1) = 4(3 - 2√2) = 12 - 8√2.Compute 4a: 4*2(√2 - 1) = 8√2 - 8.So, t(-2) = (12 - 8√2) + (8√2 - 8) - 4 = 12 - 8√2 + 8√2 - 8 - 4 = 0.So, at a = 2(√2 - 1), t(-2) = 0, which is the boundary case. Since t(x) must be positive, a must be greater than 2(√2 - 1).Therefore, the range is a > 2(√2 - 1), which is approximately 0.828, and since a must be less than 1, the range is (2(√2 - 1), 1).But the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.Therefore, the correct range is (2(√2 - 1), 1), which is approximately (0.828, 1).But wait, 2(√2 - 1) is exactly equal to -2 + 2√2, because:2(√2 - 1) = 2√2 - 2 = -2 + 2√2.Yes, so both expressions are equivalent.Therefore, the correct range is a ∈ (-2 + 2√2, 1).But since a must be less than 1, the upper limit is 1.Therefore, the range of a is (-2 + 2√2, 1).But in the initial thought process, it was concluded as [1/2, -2 + 2√2), which is incorrect because in that interval, t(-2) is negative.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But wait, let's express it as [lower, upper). Since at a = -2 + 2√2, t(-2) = 0, which is not allowed, so it's open at that point. And since a must be less than 1, it's open at 1 as well.Therefore, the range is (-2 + 2√2, 1).But in the initial thought process, it was concluded as [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But let me check if a can be equal to 1.If a = 1, then f(x) = log_1(...), but log base 1 is undefined because any power of 1 is 1, so it's not a valid logarithm. Therefore, a cannot be 1.Therefore, the range is (-2 + 2√2, 1).But wait, in the initial thought process, it was concluded as [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But let me express it in the box as per the instructions.So, the range of a is (-2 + 2√2, 1), which can be written as (2(√2 - 1), 1).But to match the initial thought process's format, which used -2 + 2√2, I'll use that.Therefore, the range is (-2 + 2√2, 1).But wait, the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But let me check if a can be equal to -2 + 2√2.At a = -2 + 2√2, t(-2) = 0, which is not allowed because the logarithm is undefined. Therefore, a must be greater than -2 + 2√2.Therefore, the range is a > -2 + 2√2, but since a must be less than 1, the range is (-2 + 2√2, 1).So, in conclusion, the range of a is (-2 + 2√2, 1).But wait, the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is a ∈ (-2 + 2√2, 1).But let me express it in the box as per the instructions.So, the range of a is boxed{(-2 + 2sqrt{2}, 1)}.But wait, the initial thought process concluded [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is boxed{(-2 + 2sqrt{2}, 1)}.But wait, in the initial thought process, it was concluded as [1/2, -2 + 2√2), which is incorrect.Therefore, the correct answer is boxed{(-2 + 2sqrt{2}, 1)}.

💡First, I need to determine the total number of workers in the workshop, which is given as 15. Out of these, there are 5 technicians, so the remaining workers are 15 - 5 = 10.Next, I'll calculate the total salary for the technicians by multiplying the number of technicians by their average salary: 5 technicians * Rs. 800/technician = Rs. 4000.Then, I'll calculate the total salary for the remaining workers by multiplying their number by their average salary: 10 workers * Rs. 650/worker = Rs. 6500.Adding these two amounts gives the total salary for all the workers: Rs. 4000 + Rs. 6500 = Rs. 10500.Finally, to find the average salary per head of all the workers, I'll divide the total salary by the total number of workers: Rs. 10500 / 15 workers = Rs. 700.

💡First, I recognize that Jerry initially had 7 action figures on the shelf.He then removed 10 action figures, which would result in a deficit of 3 action figures if no new ones were added.To find out how many action figures Jerry added, I set up the equation:Number of original action figures - Number removed + Number added = Current number of action figuresPlugging in the known values:7 - 10 + Number added = 8Solving for the Number added:Number added = 8 - 7 + 10Number added = 1 + 10Number added = 11Therefore, Jerry added 11 action figures to the shelf.

💡Alright, so I've got this problem here with two sequences: an arithmetic sequence and a geometric sequence. I need to find their explicit formulas and then figure out the sum of the first n terms of another sequence that's the product of the two. Hmm, okay, let's break this down step by step.First, let's tackle the arithmetic sequence. The problem says the sum of the first four terms is 60, and the sum of the second and fourth terms is 34. I remember that in an arithmetic sequence, each term is the previous term plus a common difference, d. So, if the first term is a1, then the terms go like a1, a1 + d, a1 + 2d, a1 + 3d, and so on.So, the first four terms are a1, a2 = a1 + d, a3 = a1 + 2d, and a4 = a1 + 3d. Their sum is 60, so:a1 + (a1 + d) + (a1 + 2d) + (a1 + 3d) = 60Simplifying that, we get:4a1 + 6d = 60Okay, that's one equation. The other piece of information is that the sum of the second and fourth terms is 34. So:a2 + a4 = (a1 + d) + (a1 + 3d) = 2a1 + 4d = 34Now I have two equations:1. 4a1 + 6d = 602. 2a1 + 4d = 34I can solve this system of equations to find a1 and d. Let me try subtracting the second equation from the first to eliminate a1:(4a1 + 6d) - (2a1 + 4d) = 60 - 34Which simplifies to:2a1 + 2d = 26Divide both sides by 2:a1 + d = 13So, a1 = 13 - dNow, plug this back into the second equation:2(13 - d) + 4d = 3426 - 2d + 4d = 3426 + 2d = 342d = 8d = 4Now, plug d back into a1 + d = 13:a1 + 4 = 13a1 = 9So, the first term is 9 and the common difference is 4. Therefore, the nth term of the arithmetic sequence is:a_n = a1 + (n - 1)d = 9 + (n - 1)4 = 4n + 5Wait, let me check that:For n=1: 4(1) + 5 = 9, which is correct.For n=2: 4(2) + 5 = 13, which matches a2.For n=4: 4(4) + 5 = 21, and a4 should be 9 + 3*4 = 21. Correct.Okay, so that seems right.Now, moving on to the geometric sequence. The sum of the first four terms is 120, and the sum of the second and fourth terms is 90. In a geometric sequence, each term is the previous term multiplied by a common ratio, q. So, the terms are b1, b2 = b1*q, b3 = b1*q^2, b4 = b1*q^3, etc.The sum of the first four terms is:b1 + b2 + b3 + b4 = b1(1 + q + q^2 + q^3) = 120And the sum of the second and fourth terms is:b2 + b4 = b1*q + b1*q^3 = b1*q(1 + q^2) = 90So, I have two equations:1. b1(1 + q + q^2 + q^3) = 1202. b1*q(1 + q^2) = 90Hmm, maybe I can divide the second equation by the first to eliminate b1. Let's try that:(b1*q(1 + q^2)) / (b1(1 + q + q^2 + q^3)) = 90 / 120Simplify:q(1 + q^2) / (1 + q + q^2 + q^3) = 3/4Notice that the denominator can be factored as (1 + q)(1 + q^2). So:q(1 + q^2) / [(1 + q)(1 + q^2)] = q / (1 + q) = 3/4So, q / (1 + q) = 3/4Cross-multiplied:4q = 3(1 + q)4q = 3 + 3q4q - 3q = 3q = 3Okay, so the common ratio is 3. Now, plug q back into one of the equations to find b1. Let's use the second equation:b1*q(1 + q^2) = 90b1*3*(1 + 9) = 90b1*3*10 = 9030b1 = 90b1 = 3So, the first term is 3 and the common ratio is 3. Therefore, the nth term of the geometric sequence is:b_n = b1*q^(n-1) = 3*3^(n-1) = 3^nWait, let me check:For n=1: 3^1 = 3, correct.For n=2: 3^2 = 9, which is b2.For n=4: 3^4 = 81, and b4 should be 3*3^3 = 81. Correct.Alright, so that seems good.Now, part (2) says to let c_n = a_n * b_n, and find the sum of the first n terms, S_n.So, c_n = (4n + 5)*3^nWe need to find S_n = c1 + c2 + ... + cn = sum_{k=1}^n (4k + 5)*3^kHmm, this looks like a sum involving both a linear term and an exponential term. I think I need to find a way to express this sum in a closed form.I remember that for sums like sum_{k=1}^n k*r^k, there is a formula. Maybe I can split the sum into two parts:S_n = sum_{k=1}^n (4k + 5)*3^k = 4*sum_{k=1}^n k*3^k + 5*sum_{k=1}^n 3^kSo, I can compute each sum separately.First, let's compute sum_{k=1}^n 3^k. That's a geometric series:sum_{k=1}^n 3^k = 3*(3^n - 1)/(3 - 1) = (3^(n+1) - 3)/2Okay, that's straightforward.Now, for sum_{k=1}^n k*3^k, I need the formula for the sum of k*r^k from k=1 to n.I recall that the sum is r*(1 - (n+1)*r^n + n*r^(n+1))/(1 - r)^2Let me verify that.Yes, the formula is:sum_{k=1}^n k*r^k = r*(1 - (n+1)*r^n + n*r^(n+1))/(1 - r)^2So, plugging r = 3:sum_{k=1}^n k*3^k = 3*(1 - (n+1)*3^n + n*3^(n+1))/(1 - 3)^2Simplify denominator: (1 - 3)^2 = 4So,= 3*(1 - (n+1)*3^n + n*3^(n+1))/4Let me simplify the numerator:1 - (n+1)*3^n + n*3^(n+1) = 1 - (n+1)*3^n + n*3*3^n = 1 - (n+1)*3^n + 3n*3^nCombine like terms:= 1 + [ - (n+1) + 3n ]*3^n = 1 + (2n -1)*3^nSo, the sum becomes:3*(1 + (2n -1)*3^n)/4 = (3 + (2n -1)*3^(n+1))/4Wait, let me check:Wait, 3*(1 + (2n -1)*3^n)/4 = 3/4 + (2n -1)*3^(n+1)/4But I think I might have made a mistake in the algebra. Let me go back.Wait, the numerator after combining terms was 1 + (2n -1)*3^n.So, the entire expression is:3*(1 + (2n -1)*3^n)/4Which is:(3 + 3*(2n -1)*3^n)/4 = (3 + (2n -1)*3^(n+1))/4Yes, that's correct.So, sum_{k=1}^n k*3^k = (3 + (2n -1)*3^(n+1))/4Therefore, going back to S_n:S_n = 4*sum_{k=1}^n k*3^k + 5*sum_{k=1}^n 3^k= 4*(3 + (2n -1)*3^(n+1))/4 + 5*(3^(n+1) - 3)/2Simplify:The 4 cancels in the first term:= (3 + (2n -1)*3^(n+1)) + (5*(3^(n+1) - 3))/2Let me write both terms with denominator 2:= [2*(3 + (2n -1)*3^(n+1)) + 5*(3^(n+1) - 3)] / 2Expand the numerator:= [6 + 2*(2n -1)*3^(n+1) + 5*3^(n+1) - 15] / 2Combine like terms:6 - 15 = -9For the 3^(n+1) terms:2*(2n -1)*3^(n+1) + 5*3^(n+1) = [4n - 2 + 5]*3^(n+1) = (4n + 3)*3^(n+1)So, numerator becomes:-9 + (4n + 3)*3^(n+1)Therefore, S_n = [ (4n + 3)*3^(n+1) - 9 ] / 2So, the sum S_n is (4n + 3)*3^(n+1) - 9 all over 2.Let me check this formula with n=1:S_1 = c1 = a1*b1 = 9*3 = 27Using the formula:(4*1 + 3)*3^(2) - 9 / 2 = (7*9 - 9)/2 = (63 - 9)/2 = 54/2 = 27. Correct.For n=2:c1 + c2 = 27 + (13*9) = 27 + 117 = 144Using the formula:(4*2 + 3)*3^3 - 9 / 2 = (11*27 - 9)/2 = (297 - 9)/2 = 288/2 = 144. Correct.n=3:c1 + c2 + c3 = 27 + 117 + (17*27) = 27 + 117 + 459 = 603Formula:(4*3 + 3)*3^4 - 9 / 2 = (15*81 - 9)/2 = (1215 - 9)/2 = 1206/2 = 603. Correct.Okay, seems like the formula works.So, summarizing:1. The arithmetic sequence has a_n = 4n + 5.2. The geometric sequence has b_n = 3^n.3. The sum S_n of the first n terms of c_n = a_n*b_n is S_n = [(4n + 3)*3^(n+1) - 9]/2.**Final Answer**(1) The explicit formulas are boxed{a_n = 4n + 5} and boxed{b_n = 3^n}.(2) The sum of the first n terms of {c_n} is boxed{dfrac{(4n + 3) cdot 3^{n+1} - 9}{2}}.

💡First, I need to calculate the value of 8!.I know that 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1.Calculating this, 8! equals 40320.Next, I need to find the prime factorization of 40320.I start by dividing 40320 by 2 repeatedly:40320 ÷ 2 = 2016020160 ÷ 2 = 1008010080 ÷ 2 = 50405040 ÷ 2 = 25202520 ÷ 2 = 12601260 ÷ 2 = 630630 ÷ 2 = 315I divided by 2 seven times, so 2⁷ is part of the prime factorization.Now, I take 315 and divide by 3:315 ÷ 3 = 105105 ÷ 3 = 35I divided by 3 two times, so 3² is part of the prime factorization.Next, I take 35 and divide by 5:35 ÷ 5 = 7I divided by 5 one time, so 5¹ is part of the prime factorization.Finally, 7 is a prime number, so 7¹ is part of the prime factorization.Putting it all together, the prime factorization of 40320 is 2⁷ × 3² × 5¹ × 7¹.To find the number of positive divisors, I use the formula:Number of divisors = (exponent of 2 + 1) × (exponent of 3 + 1) × (exponent of 5 + 1) × (exponent of 7 + 1)Plugging in the exponents:Number of divisors = (7 + 1) × (2 + 1) × (1 + 1) × (1 + 1) = 8 × 3 × 2 × 2 = 96Therefore, 8! has 96 positive divisors.