💡Okay, so I have this problem where I'm given two vectors, vector a which is (3, -4) and vector b which is (cos α, sin α). I need to find the range of values for the magnitude of vector a plus twice vector b, which is written as |a + 2b|. The options given are A: [1,4], B: [2,6], C: [3,7], and D: [2√2, 4√2].Alright, let's start by understanding what each part means. Vector a is fixed at (3, -4), so that's straightforward. Vector b is a unit vector because it's given in terms of cosine and sine of some angle α, which means its magnitude is 1. So when we multiply vector b by 2, we're scaling it up to have a magnitude of 2.So, the vector a + 2b is essentially vector a plus a vector of magnitude 2 in some direction determined by α. The question is asking for the range of possible magnitudes of this resultant vector as α varies. That is, when we add vector a and 2 times vector b, how big or small can the resulting vector be?I think the way to approach this is to first express the vector a + 2b in terms of its components. So, vector a is (3, -4), and vector b is (cos α, sin α). So, 2b would be (2cos α, 2sin α). Therefore, adding a and 2b, we get:a + 2b = (3 + 2cos α, -4 + 2sin α)Now, to find the magnitude of this vector, we can use the formula for the magnitude of a vector (x, y), which is sqrt(x² + y²). So, plugging in the components we have:|a + 2b| = sqrt[(3 + 2cos α)² + (-4 + 2sin α)²]Let me expand this expression step by step. First, let's compute (3 + 2cos α)²:(3 + 2cos α)² = 9 + 12cos α + 4cos² αSimilarly, (-4 + 2sin α)²:(-4 + 2sin α)² = 16 - 16sin α + 4sin² αNow, adding these two results together:9 + 12cos α + 4cos² α + 16 - 16sin α + 4sin² αCombine like terms:9 + 16 = 2512cos α - 16sin α4cos² α + 4sin² αSo, the expression becomes:25 + 12cos α - 16sin α + 4(cos² α + sin² α)But wait, cos² α + sin² α is equal to 1, right? So that simplifies to:25 + 12cos α - 16sin α + 4(1) = 25 + 12cos α - 16sin α + 4 = 29 + 12cos α - 16sin αSo, the magnitude squared is 29 + 12cos α - 16sin α. Therefore, the magnitude itself is sqrt(29 + 12cos α - 16sin α).Now, we need to find the range of this expression as α varies. That is, we need to find the maximum and minimum values of sqrt(29 + 12cos α - 16sin α).Since sqrt is a monotonically increasing function, the maximum and minimum of the expression inside the sqrt will correspond to the maximum and minimum of the entire expression. So, we can instead focus on finding the range of 29 + 12cos α - 16sin α.Let me denote the expression inside the sqrt as E = 29 + 12cos α - 16sin α.So, E = 29 + 12cos α - 16sin α.To find the range of E, we can think of 12cos α - 16sin α as a single sinusoidal function. Remember that any expression of the form A cos α + B sin α can be written as C cos(α + φ), where C = sqrt(A² + B²) and φ is some phase shift.So, let's compute C for our expression:C = sqrt(12² + (-16)²) = sqrt(144 + 256) = sqrt(400) = 20.So, 12cos α - 16sin α can be written as 20 cos(α + φ), where φ is an angle such that cos φ = 12/20 and sin φ = -16/20. Simplifying, cos φ = 3/5 and sin φ = -4/5.Therefore, E = 29 + 20 cos(α + φ).Now, the maximum and minimum values of cos(α + φ) are 1 and -1, respectively. Therefore, the maximum value of E is 29 + 20(1) = 49, and the minimum value is 29 + 20(-1) = 9.Thus, E ranges from 9 to 49. Therefore, sqrt(E) ranges from sqrt(9) = 3 to sqrt(49) = 7.So, the magnitude |a + 2b| ranges from 3 to 7. Therefore, the correct answer is option C: [3,7].Wait, let me double-check my steps to make sure I didn't make any mistakes.1. Expressed a + 2b correctly as (3 + 2cos α, -4 + 2sin α).2. Calculated the magnitude squared correctly: (3 + 2cos α)² + (-4 + 2sin α)².3. Expanded both squares correctly: 9 + 12cos α + 4cos² α and 16 - 16sin α + 4sin² α.4. Combined terms correctly: 25 + 12cos α - 16sin α + 4(cos² α + sin² α).5. Simplified cos² α + sin² α to 1, leading to 29 + 12cos α - 16sin α.6. Recognized that 12cos α - 16sin α can be written as 20 cos(α + φ) with C = 20.7. Therefore, E ranges from 9 to 49, so sqrt(E) ranges from 3 to 7.Yes, that all seems correct. So, the range is [3,7], which is option C.**Final Answer**boxed{C}

💡Alright, so I have this problem here where I need to find the interval in which the zero of the function ( f(x) = ln x + x - 4 ) is located. The options are:A: ( (0,1) )B: ( (1,2) )C: ( (2,3) )D: ( (3,4) )Okay, let's start by understanding what the problem is asking. A "zero" of a function is a value of ( x ) where ( f(x) = 0 ). So, I need to find between which two numbers the function crosses the x-axis.First, I know that ( ln x ) is only defined for ( x > 0 ), so the domain of ( f(x) ) is ( (0, infty) ). That means I don't have to worry about negative values of ( x ).Next, I recall that ( ln x ) is a slowly increasing function, and ( x ) itself is a linear function increasing at a constant rate. So, ( f(x) = ln x + x - 4 ) is a combination of these two functions.I think it might help to analyze the behavior of ( f(x) ) at the endpoints of the given intervals. If I can find two points where ( f(x) ) changes sign from negative to positive or vice versa, then by the Intermediate Value Theorem, there must be a zero in that interval.Let me start by evaluating ( f(x) ) at the endpoints of each interval.1. **Interval A: ( (0,1) )** - At ( x = 0 ): Wait, ( ln 0 ) is undefined, so we can't evaluate at 0. Let's try a value close to 0, say ( x = 0.1 ): [ f(0.1) = ln(0.1) + 0.1 - 4 approx (-2.3026) + 0.1 - 4 approx -6.2026 ] So, ( f(0.1) ) is negative. - At ( x = 1 ): [ f(1) = ln(1) + 1 - 4 = 0 + 1 - 4 = -3 ] So, ( f(1) ) is also negative. Since both ends of the interval ( (0,1) ) give negative values, there's no sign change here. Therefore, there might not be a zero in this interval.2. **Interval B: ( (1,2) )** - At ( x = 1 ): We already calculated ( f(1) = -3 ). - At ( x = 2 ): [ f(2) = ln(2) + 2 - 4 approx 0.6931 + 2 - 4 approx -1.3069 ] So, ( f(2) ) is still negative. Again, both ends of the interval ( (1,2) ) give negative values. No sign change means no zero in this interval either.3. **Interval C: ( (2,3) )** - At ( x = 2 ): We have ( f(2) approx -1.3069 ) (negative). - At ( x = 3 ): [ f(3) = ln(3) + 3 - 4 approx 1.0986 + 3 - 4 approx 0.0986 ] So, ( f(3) ) is positive. Ah, here we have a sign change from negative at ( x = 2 ) to positive at ( x = 3 ). This indicates that the function crosses the x-axis somewhere between 2 and 3. Therefore, by the Intermediate Value Theorem, there must be a zero in this interval.4. **Interval D: ( (3,4) )** - At ( x = 3 ): We have ( f(3) approx 0.0986 ) (positive). - At ( x = 4 ): [ f(4) = ln(4) + 4 - 4 approx 1.3863 + 4 - 4 approx 1.3863 ] So, ( f(4) ) is positive. Both ends of the interval ( (3,4) ) give positive values. No sign change means no zero in this interval.Just to be thorough, I should also consider the behavior of the function beyond these intervals. For ( x > 4 ), since ( ln x ) grows slower than ( x ), the function ( f(x) = ln x + x - 4 ) will continue to increase and stay positive. Similarly, for ( x < 0 ), the function isn't defined, so we don't need to consider that.Additionally, I can check the derivative of ( f(x) ) to ensure that it's monotonic, meaning it doesn't have any turning points that could create multiple zeros.The derivative ( f'(x) ) is:[f'(x) = frac{1}{x} + 1]Since ( frac{1}{x} > 0 ) for all ( x > 0 ), and 1 is always positive, ( f'(x) ) is always positive. This means ( f(x) ) is strictly increasing on its entire domain. Therefore, there can only be one zero crossing, and we've already identified it in interval ( (2,3) ).Just to double-check, let me pick a value in the middle of interval ( (2,3) ), say ( x = 2.5 ), and see what ( f(2.5) ) is:[f(2.5) = ln(2.5) + 2.5 - 4 approx 0.9163 + 2.5 - 4 approx -0.5837]So, ( f(2.5) ) is still negative. That means the zero is somewhere between 2.5 and 3.Let's try ( x = 2.8 ):[f(2.8) = ln(2.8) + 2.8 - 4 approx 1.0296 + 2.8 - 4 approx -0.1704]Still negative, but closer to zero.At ( x = 2.9 ):[f(2.9) = ln(2.9) + 2.9 - 4 approx 1.0647 + 2.9 - 4 approx -0.0353]Almost there.At ( x = 2.95 ):[f(2.95) = ln(2.95) + 2.95 - 4 approx 1.0818 + 2.95 - 4 approx 0.0318]Now, ( f(2.95) ) is positive. So, the zero is between 2.9 and 2.95.This further confirms that the zero is indeed in the interval ( (2,3) ).I think I've thoroughly checked all the intervals and even looked into the behavior of the function beyond the given options. Everything points to the zero being in interval C: ( (2,3) ).

💡Alright, so I'm trying to solve this problem where I have two polynomials, p₁ and p₂. The conditions given are that the derivative of their ratio, (p₂/p₁)', equals 2x. From this, it's concluded that p₂(x) is equal to (x² + c) times p₁(x), where c is some constant. Additionally, the sum of p₁ and p₂ is given as (x - 1)(x + 2)(x² - 3).Okay, let me break this down. First, the derivative condition: (p₂/p₁)' = 2x. I remember that the derivative of a quotient can be found using the quotient rule. So, if I let f = p₂ and g = p₁, then the derivative f'/g - f g'/g² equals 2x. That is,(p₂' p₁ - p₂ p₁') / p₁² = 2x.But the problem says that from this condition, it follows that p₂ = (x² + c) p₁. Hmm, maybe I can use this to find a relationship between p₂ and p₁.Let me try substituting p₂ = (x² + c) p₁ into the derivative condition. Then p₂' would be the derivative of (x² + c) p₁, which is 2x p₁ + (x² + c) p₁'. Plugging this back into the derivative condition:[2x p₁ + (x² + c) p₁' - (x² + c) p₁'] / p₁² = 2x.Wait, the p₁' terms cancel out? So I get:2x p₁ / p₁² = 2x.Simplifying, that's 2x / p₁ = 2x. So, 2x / p₁ = 2x implies that 1/p₁ = 1, so p₁ = 1. But that can't be right because p₁ is a polynomial, and if p₁ is 1, then p₂ would be x² + c, and their sum would be x² + c + 1, which is supposed to equal (x - 1)(x + 2)(x² - 3). But (x - 1)(x + 2)(x² - 3) is a quartic polynomial, whereas x² + c + 1 is quadratic. So that doesn't make sense. Maybe I made a mistake in my substitution.Wait, let me double-check. If p₂ = (x² + c) p₁, then p₂' = 2x p₁ + (x² + c) p₁'. So when I plug into the derivative condition:(p₂' p₁ - p₂ p₁') / p₁² = [2x p₁ + (x² + c) p₁' - (x² + c) p₁'] / p₁² = 2x p₁ / p₁² = 2x / p₁.So, 2x / p₁ = 2x implies that 1/p₁ = 1, so p₁ = 1. But as I saw, that leads to a contradiction because p₁ + p₂ would be 1 + (x² + c), which is quadratic, but it's supposed to be a quartic.Hmm, maybe I need to approach this differently. Perhaps instead of assuming p₂ = (x² + c) p₁, I should derive it from the given condition.Starting again, (p₂/p₁)' = 2x. Let me write this as:d/dx [p₂/p₁] = 2x.Using the quotient rule:(p₂' p₁ - p₂ p₁') / p₁² = 2x.Let me rearrange this:p₂' p₁ - p₂ p₁' = 2x p₁².But I also know that p₂ = (x² + c) p₁, so p₂' = 2x p₁ + (x² + c) p₁'. Plugging this into the equation:[2x p₁ + (x² + c) p₁'] p₁ - (x² + c) p₁ p₁' = 2x p₁².Simplify the left side:2x p₁² + (x² + c) p₁ p₁' - (x² + c) p₁ p₁' = 2x p₁².So, the terms with p₁ p₁' cancel out, leaving:2x p₁² = 2x p₁².Which is an identity, meaning that the assumption p₂ = (x² + c) p₁ is valid for any c. So, c can be any constant, but we need to determine c such that p₁ + p₂ equals (x - 1)(x + 2)(x² - 3).Given that p₂ = (x² + c) p₁, then p₁ + p₂ = p₁ + (x² + c) p₁ = p₁ (1 + x² + c) = (x - 1)(x + 2)(x² - 3).So, p₁ (x² + c + 1) = (x - 1)(x + 2)(x² - 3).Therefore, p₁ must be a factor of the right-hand side. Let's factor the right-hand side:(x - 1)(x + 2)(x² - 3) = (x² + x - 2)(x² - 3).So, it's a product of two quadratics. Now, p₁ is a polynomial, and p₁ (x² + c + 1) equals this product. So, p₁ must be one of the factors, and x² + c + 1 must be the other.So, there are two cases:Case 1: p₁ = (x² + x - 2) and x² + c + 1 = (x² - 3).Case 2: p₁ = (x² - 3) and x² + c + 1 = (x² + x - 2).Wait, but in the original problem, it says:"Consider the cases where p₁(x) is a polynomial:Case 1(x² + c + 1) = (x - 1)(x + 2) = x² + x - 2 - this is not possible.Case 2(x² + c + 1) = x² - 3 → c = -4. Then p₁ = (x - 1)(x + 2) and p₂ = (x² - 4)(x - 1)(x + 2)."So, in Case 1, they set x² + c + 1 equal to x² + x - 2, which would imply c + 1 = x - 2, but c is a constant, so this is impossible because the right side is linear in x. Therefore, Case 1 is invalid.In Case 2, they set x² + c + 1 equal to x² - 3, which gives c + 1 = -3, so c = -4. Then p₁ must be the other factor, which is (x - 1)(x + 2). Then p₂ is (x² + c) p₁, which with c = -4 becomes (x² - 4)(x - 1)(x + 2).Let me verify this:p₁ = (x - 1)(x + 2) = x² + x - 2.p₂ = (x² - 4) p₁ = (x² - 4)(x² + x - 2).Then p₁ + p₂ = (x² + x - 2) + (x² - 4)(x² + x - 2).Factor out (x² + x - 2):p₁ + p₂ = (x² + x - 2)(1 + x² - 4) = (x² + x - 2)(x² - 3).Which is indeed equal to (x - 1)(x + 2)(x² - 3), as given.So, this works out. Therefore, the correct case is Case 2, with c = -4, p₁ = (x - 1)(x + 2), and p₂ = (x² - 4)(x - 1)(x + 2).I think that makes sense. The key was recognizing that p₁ must be one of the quadratic factors, and x² + c + 1 must be the other. Since setting x² + c + 1 equal to a linear term is impossible, the only valid case is when x² + c + 1 equals the other quadratic factor, leading to c = -4 and p₁ being the remaining factor.

💡Alright, so I'm trying to solve this problem about Calen, Caleb, and Candy and their pencils. Let me try to understand what's going on here.First, the problem says that Calen originally had 5 more pencils than Caleb. Okay, so if I let the number of pencils Caleb has be some number, then Calen has that number plus 5. That seems straightforward.Next, it says that Caleb has 3 less than twice as many pencils as Candy. Hmm, okay, so if I let the number of pencils Candy has be another number, then Caleb has twice that number minus 3. So, if Candy has, say, x pencils, then Caleb has 2x - 3 pencils.Now, the problem also mentions that Calen lost 10 pencils and was left with 10 pencils. So, if Calen lost 10 pencils and ended up with 10, that means he originally had 20 pencils, right? Because 10 pencils lost plus 10 pencils remaining equals 20 pencils originally.Since Calen originally had 5 more pencils than Caleb, and we now know Calen had 20 pencils, that means Caleb must have had 15 pencils. Because 20 minus 5 equals 15.Okay, so Caleb has 15 pencils. But the problem wants to know how many pencils Candy has. Earlier, we established that Caleb has 3 less than twice as many pencils as Candy. So, if I let the number of pencils Candy has be x, then Caleb has 2x - 3 pencils.We know Caleb has 15 pencils, so I can set up the equation: 2x - 3 = 15. To solve for x, I'll add 3 to both sides: 2x = 18. Then, divide both sides by 2: x = 9.So, Candy has 9 pencils. Let me double-check that to make sure I didn't make a mistake. If Candy has 9 pencils, then twice that is 18, and 3 less than that is 15, which matches the number of pencils Caleb has. And since Calen has 5 more than Caleb, that's 20 pencils, which aligns with the information that Calen lost 10 pencils to be left with 10.Yeah, that all adds up. So, Candy has 9 pencils.

💡Alright, so I have this problem here about sets M and N, and I need to find their intersection. Let me try to understand what each set represents and then figure out where they overlap.First, set M is defined as {s | |s| < 4}. Okay, so this means M includes all real numbers s such that the absolute value of s is less than 4. In other words, s can be any number between -4 and 4, but not including -4 and 4 themselves because it's strictly less than 4. So, M is the interval (-4, 4).Next, set N is defined as {x | 3x ≥ -1}. Hmm, this is an inequality. To find out what values of x satisfy this, I can solve the inequality. If I divide both sides by 3, I get x ≥ -1/3. So, N includes all real numbers x that are greater than or equal to -1/3. That means N is the interval [-1/3, ∞).Now, I need to find the intersection of M and N, which is M ∩ N. The intersection of two sets consists of all elements that are common to both sets. So, I need to find all x that are in both M and N.Set M is (-4, 4), and set N is [-1/3, ∞). To find their intersection, I'll look for the overlap between these two intervals. Starting from the lower bound, M starts at -4, but N starts at -1/3. Since -1/3 is greater than -4, the lower bound of the intersection will be -1/3. For the upper bound, M ends at 4, and N goes to infinity. Since M only goes up to 4, the upper bound of the intersection will be 4.So, putting it all together, the intersection M ∩ N is the set of all x such that -1/3 ≤ x < 4. Let me double-check to make sure I didn't make any mistakes. - For set M: |s| < 4 translates to -4 < s < 4. That seems right.- For set N: 3x ≥ -1 simplifies to x ≥ -1/3. That also looks correct.- Intersection: The overlapping part between (-4, 4) and [-1/3, ∞) is indeed [-1/3, 4). I think that makes sense. So, the correct answer should be the set of x where x is greater than or equal to -1/3 and less than 4.

💡Okay, so I have this problem where there's a function f(x) that satisfies the equation f(x + y) = f(x) f(y) for all real numbers x and y. They also tell me that f(2) = 4, and I need to find f(8). Hmm, interesting. I remember that functional equations often have specific types of functions as solutions, like exponential functions or something else. Let me think about this step by step.First, the equation f(x + y) = f(x) f(y) is a well-known functional equation. I think it's called the exponential Cauchy equation. The solutions to this equation are exponential functions, right? So, maybe f(x) is an exponential function. That makes sense because when you add the exponents, you multiply the bases, which kind of resembles the property here.But let me not jump to conclusions. Let me try to work it out step by step. Since I know f(2) = 4, maybe I can find f(4) first and then build up to f(8). Let's see.If I set x = 2 and y = 2 in the equation f(x + y) = f(x) f(y), then I get:f(2 + 2) = f(2) f(2)So, f(4) = f(2) * f(2) = 4 * 4 = 16.Okay, so f(4) is 16. That seems straightforward. Now, can I find f(8) using the same approach? Let's try setting x = 4 and y = 4 in the original equation:f(4 + 4) = f(4) f(4)So, f(8) = f(4) * f(4) = 16 * 16 = 256.Wait, so f(8) is 256? That seems pretty big, but considering it's an exponential function, it makes sense. Let me check if this aligns with the properties of exponential functions.If f(x) is an exponential function, it should be of the form f(x) = a^x for some base a. Let's test this. If f(x) = a^x, then f(x + y) = a^(x + y) = a^x * a^y = f(x) f(y), which fits the given functional equation. So, yes, f(x) is an exponential function.Given that f(2) = 4, we can find the base a. So, f(2) = a^2 = 4. Solving for a, we get a = sqrt(4) = 2 or a = -2. But since f(x) is defined for all real numbers, and exponential functions with negative bases can be tricky because they might not be real for all x (for example, if x is a fraction, a negative base could result in a complex number). So, it's safer to assume that a is positive. Therefore, a = 2.So, f(x) = 2^x. Let's verify this with the given information. f(2) = 2^2 = 4, which matches. Then f(4) = 2^4 = 16, which also matches. And f(8) = 2^8 = 256, which is consistent with what I found earlier.Wait, but what if a was -2? Let me check. If a = -2, then f(2) = (-2)^2 = 4, which is still 4. But then f(1) would be (-2)^1 = -2, and f(0.5) would be sqrt(-2), which is not a real number. Since the problem states that the function is defined for all real numbers x and y, we can't have a negative base because it would result in complex numbers for non-integer exponents. Therefore, a must be positive, so a = 2 is the correct base.Another way to think about this is to use the property of exponents. Since f(x + y) = f(x) f(y), it suggests that f(x) is an exponential function. So, if I let f(x) = e^{kx}, where k is some constant, then f(x + y) = e^{k(x + y)} = e^{kx} e^{ky} = f(x) f(y), which satisfies the equation. But in this case, f(2) = e^{2k} = 4. So, solving for k, we get 2k = ln(4), so k = (ln(4))/2 = ln(2). Therefore, f(x) = e^{ln(2) x} = 2^x, which is the same result as before.So, whether I approach it by iteratively finding f(4) and f(8) or by recognizing the form of the exponential function, I end up with f(x) = 2^x, and thus f(8) = 256.Just to make sure I haven't missed anything, let me consider if there are other possible functions that satisfy f(x + y) = f(x) f(y). I recall that the only continuous solutions to this functional equation are exponential functions. However, without the continuity condition, there might be other pathological solutions. But since the problem doesn't specify anything about continuity, I might need to consider that.But in most standard problems like this, especially in competitions or exams, they usually expect the exponential solution because it's the most straightforward and continuous. Plus, given that f(2) is a positive real number, it's likely that f(x) is positive for all x, which is consistent with an exponential function.Also, if I think about the function f(x) = 0 for all x, that would satisfy f(x + y) = 0 = 0 * 0 = f(x) f(y). But in this case, f(2) = 4, which isn't zero, so f(x) can't be the zero function.Another thought: what if f(x) is a constant function? If f(x) = c for all x, then f(x + y) = c = c * c = c^2. So, c = c^2, which implies c = 0 or c = 1. But again, f(2) = 4, so it can't be a constant function. Therefore, the only viable solution is the exponential function.So, putting it all together, f(x) must be 2^x, and thus f(8) is 256. I think that's solid. I don't see any mistakes in my reasoning, and I've checked it from multiple angles.**Final Answer**The value of ( f(8) ) is boxed{256}.