💡Okay, so I have this problem here: The ratio of cats to dogs at the pet store is 3:4. There are 18 cats. How many dogs are there at the pet store? Alright, let's see. Ratios can sometimes be a bit tricky, but I think I can handle this.First, I need to understand what the ratio 3:4 means. I remember that a ratio compares two quantities. So, in this case, for every 3 cats, there are 4 dogs. That means the number of dogs is a bit more than the number of cats. Got it.Now, the problem tells me there are 18 cats. I need to find out how many dogs there are. Since the ratio is 3:4, I can set up a proportion to solve this. Let me recall how proportions work. If two ratios are equal, then their cross products are equal. So, I can write the ratio of cats to dogs as 3/4 and set that equal to 18/x, where x is the number of dogs I'm trying to find.Wait, let me make sure I'm setting this up correctly. The ratio of cats to dogs is 3:4, which can be written as 3/4. So, cats/dogs = 3/4. That means 18/x = 3/4. Yeah, that seems right.Now, to solve for x, I can cross-multiply. So, 3 times x should equal 4 times 18. Let me write that down: 3x = 4 * 18. Okay, 4 times 18 is... let me calculate that. 4 times 10 is 40, and 4 times 8 is 32, so 40 + 32 is 72. So, 3x = 72.Now, to find x, I need to divide both sides by 3. So, x = 72 / 3. Let me do that division. 72 divided by 3 is 24. So, x equals 24. That means there are 24 dogs at the pet store.Wait, let me double-check my work to make sure I didn't make any mistakes. I set up the proportion as 3/4 = 18/x, which seems correct because the ratio of cats to dogs is 3:4. Then I cross-multiplied to get 3x = 72, and dividing both sides by 3 gives x = 24. That makes sense.Just to be thorough, I can also think about it in terms of groups. The ratio 3:4 means that for every group of 3 cats, there are 4 dogs. If there are 18 cats, how many groups of 3 cats are there? Well, 18 divided by 3 is 6. So, there are 6 groups of 3 cats. Since each group corresponds to 4 dogs, I can multiply 6 by 4 to find the total number of dogs. 6 times 4 is 24. Yep, that matches what I got earlier.Another way to look at it is by using fractions. If 3 parts correspond to 18 cats, then one part is 18 divided by 3, which is 6. So, each part is 6. Then, since dogs are 4 parts, I multiply 6 by 4 to get 24 dogs. This also confirms my previous answer.I think I've approached this from a couple of different angles now, and each time I end up with 24 dogs. That gives me confidence that 24 is the correct answer. I don't see any mistakes in my calculations or reasoning, so I'm pretty sure that's the right number of dogs at the pet store.

💡Okay, so I have this problem here with two propositions, p and q, and I need to figure out the range of values for 'a' based on the truth values of these propositions. Let me try to break it down step by step.First, let's understand what each proposition is saying.**Proposition p**: The solution set of the inequality ( x^{2} - (a+1)x + 1 > 0 ) is all real numbers, ( mathbb{R} ). That means, no matter what x I plug in, the quadratic expression should always be positive. So, this inequality should hold true for every real number x.**Proposition q**: The function ( f(x) = (a+1)^{x} ) is increasing in its domain. For an exponential function like this, it's increasing if the base is greater than 1. If the base is between 0 and 1, the function is decreasing. If the base is less than or equal to 0, the function isn't defined properly for all real numbers, right? So, for f(x) to be increasing, ( a+1 ) must be greater than 1.Now, the problem states that ( p land q ) is false, and ( p lor q ) is true. So, let me recall what these logical connectives mean.- ( p land q ) is true only if both p and q are true. Since it's false, at least one of p or q must be false.- ( p lor q ) is true if at least one of p or q is true. Since it's true, at least one of them must be true.So, combining these two, we can say that exactly one of p or q is true, and the other is false. They can't both be true because ( p land q ) is false, and they can't both be false because ( p lor q ) is true.Alright, so I need to find the values of 'a' where either p is true and q is false, or p is false and q is true.Let me tackle each proposition separately first.**Analyzing Proposition p:**The inequality is ( x^{2} - (a+1)x + 1 > 0 ) for all real x. For this quadratic to be always positive, it must never touch or cross the x-axis. That means the quadratic equation ( x^{2} - (a+1)x + 1 = 0 ) has no real roots.To find when a quadratic has no real roots, we look at the discriminant. The discriminant ( D ) of a quadratic ( ax^2 + bx + c ) is ( D = b^2 - 4ac ). If ( D < 0 ), there are no real roots.So, let's compute the discriminant for our quadratic:( D = [-(a+1)]^2 - 4 times 1 times 1 )( D = (a+1)^2 - 4 )For no real roots, we need:( (a+1)^2 - 4 < 0 )( (a+1)^2 < 4 )Taking square roots on both sides:( |a + 1| < 2 )Which means:( -2 < a + 1 < 2 )Subtracting 1 from all parts:( -3 < a < 1 )So, proposition p is true when ( -3 < a < 1 ).**Analyzing Proposition q:**The function ( f(x) = (a+1)^x ) is increasing. As I thought earlier, for an exponential function ( b^x ), it's increasing if ( b > 1 ) and decreasing if ( 0 < b < 1 ). Also, if ( b leq 0 ), the function isn't defined for all real x (since negative bases can cause issues with real exponents, especially for non-integer exponents).Therefore, for ( f(x) ) to be increasing, we need:( a + 1 > 1 )( a > 0 )So, proposition q is true when ( a > 0 ).**Now, let's consider the logical conditions:**We have ( p land q ) is false, and ( p lor q ) is true. So, as I concluded earlier, exactly one of p or q is true.So, let's consider two cases:1. **Case 1: p is true and q is false.** - p is true: ( -3 < a < 1 ) - q is false: ( a leq 0 ) - So, the intersection of these two is ( -3 < a leq 0 )2. **Case 2: p is false and q is true.** - p is false: ( a leq -3 ) or ( a geq 1 ) - q is true: ( a > 0 ) - So, the intersection here is ( a geq 1 )Therefore, combining both cases, the range of 'a' is ( -3 < a leq 0 ) or ( a geq 1 ).Wait, let me double-check if I missed anything.- For p being true, it's strictly between -3 and 1.- For q being true, it's strictly greater than 0.- So, when p is true and q is false, it's the overlap where a is between -3 and 0, including 0 because q is false at 0.- When p is false, a is either less than or equal to -3 or greater than or equal to 1. But since q is true, we only take the part where a is greater than 0, so that's a >=1.Yes, that seems correct. So, the final range is ( -3 < a leq 0 ) or ( a geq 1 ).**Final Answer**The range of values for ( a ) is boxed{-3 < a leq 0 text{ or } a geq 1}.

💡Okay, so I have this problem here about a function and its tangent line, and then another part about another function being increasing. Let me try to figure this out step by step.First, part (I). The function is given as f(x) = (ln x)/a + x. They say that the tangent line at x=1 is described by 2x - y + b = 0. I need to find a and b.Hmm, tangent line at x=1. So, I remember that the tangent line at a point has two things: it passes through the point (1, f(1)) and has a slope equal to f’(1). So, I can use these two pieces of information to find a and b.Let me write down what I know:1. The equation of the tangent line is 2x - y + b = 0. Let me rearrange this to slope-intercept form to make it easier. So, subtract 2x and add y: y = 2x + b. Wait, no, actually, if I rearrange 2x - y + b = 0, it becomes y = 2x + b. So, the slope of the tangent line is 2, and the y-intercept is b.2. The function f(x) is (ln x)/a + x. So, f(1) would be (ln 1)/a + 1. But ln 1 is 0, so f(1) is just 1. So, the point (1, 1) lies on the tangent line.3. The derivative f’(x) is the slope of the tangent line at any point x. So, let me compute f’(x). The derivative of (ln x)/a is (1/(a x)) and the derivative of x is 1. So, f’(x) = 1/(a x) + 1.4. At x=1, the slope of the tangent line is f’(1) = 1/a + 1. But we know the slope is 2, so 1/a + 1 = 2. Let me solve for a.1/a + 1 = 2 Subtract 1 from both sides: 1/a = 1 So, a = 1.Okay, got a=1.Now, since the tangent line passes through (1,1), let's plug that into the tangent line equation y = 2x + b. So, when x=1, y=1.1 = 2(1) + b 1 = 2 + b Subtract 2: b = -1.So, a=1 and b=-1. That seems straightforward.Now, moving on to part (II). The function g(x) is given as f(x) + (1/2)x² - kx. Since f(x) is (ln x)/a + x, and we found a=1, so f(x) is ln x + x. Therefore, g(x) = ln x + x + (1/2)x² - kx.Simplify that: g(x) = (1/2)x² + (1 - k)x + ln x.They say that g(x) is an increasing function on its domain. The domain of g(x) would be x > 0 because of the ln x term.For a function to be increasing on its domain, its derivative should be non-negative for all x in the domain. So, let's compute g’(x).g’(x) = derivative of (1/2)x² is x, derivative of (1 - k)x is (1 - k), derivative of ln x is 1/x. So, putting it all together:g’(x) = x + (1 - k) + 1/x.So, g’(x) = x + 1/x + (1 - k).We need this to be greater than or equal to zero for all x > 0.So, x + 1/x + (1 - k) ≥ 0 for all x > 0.Let me rearrange this:x + 1/x + 1 - k ≥ 0 So, k ≤ x + 1/x + 1 for all x > 0.Therefore, k must be less than or equal to the minimum value of the function h(x) = x + 1/x + 1 on x > 0.So, I need to find the minimum of h(x) = x + 1/x + 1.I remember that for functions like x + 1/x, the minimum occurs at x=1. Let me verify that.Compute the derivative of h(x):h’(x) = 1 - 1/x².Set h’(x) = 0:1 - 1/x² = 0 1 = 1/x² x² = 1 x = 1 (since x > 0).So, the minimum occurs at x=1.Compute h(1):h(1) = 1 + 1/1 + 1 = 1 + 1 + 1 = 3.Therefore, the minimum value of h(x) is 3. So, k must be less than or equal to 3.Hence, the range of k is k ≤ 3.Wait, let me double-check.We had g’(x) = x + 1/x + (1 - k) ≥ 0.So, rearranged, it's x + 1/x + 1 ≥ k.So, k must be less than or equal to the minimum of x + 1/x + 1, which is 3. So, yes, k ≤ 3.I think that's correct.So, summarizing:(I) a=1, b=-1.(II) k ≤ 3.**Final Answer**(I) The values are ( a = boxed{1} ) and ( b = boxed{-1} ).(II) The range of ( k ) is ( boxed{(-infty, 3]} ).

💡Alright, so I have this problem about a hyperbola, and I need to find the minimum value of |PF| + |PA| where P is a moving point on the right branch of the hyperbola. The hyperbola is given by the equation x²/4 - y²/12 = 1, and F is the left focus. Point A is at (1,4). The options are 8, 9, 10, or 12.First, I need to recall some properties of hyperbolas. A hyperbola has two foci, and for any point on the hyperbola, the absolute difference of the distances to the two foci is constant, equal to 2a, where a is the distance from the center to a vertex.Looking at the equation x²/4 - y²/12 = 1, I can see that it's a standard hyperbola centered at the origin (0,0). The term with x² is positive, so it opens left and right. The value under x² is 4, so a² = 4, which means a = 2. The value under y² is 12, so b² = 12, which means b = 2√3.Now, to find the foci, I remember that for hyperbolas, c² = a² + b², where c is the distance from the center to each focus. So, c² = 4 + 12 = 16, which means c = 4. Therefore, the foci are at (±4, 0). Since F is the left focus, F is at (-4, 0).So, F is (-4, 0), and the other focus F' is (4, 0).Now, point P is on the right branch of the hyperbola. So, P is somewhere on the right side, meaning its x-coordinate is positive.We need to find the minimum value of |PF| + |PA|, where A is (1,4). So, we're looking for the point P on the right branch such that the sum of the distances from P to F and from P to A is minimized.Hmm, this seems similar to the problem of finding the shortest path that reflects off a curve, but I'm not sure if that applies here. Maybe I can think of it in terms of reflections or use some properties of hyperbolas.Wait, hyperbolas have a reflection property, but it's different from ellipses. For hyperbolas, the reflection property is that the tangent at any point P on the hyperbola makes equal angles with the lines from P to each focus. But I'm not sure if that helps directly here.Alternatively, maybe I can use the definition of a hyperbola. For any point P on the hyperbola, |PF| - |PF'| = 2a, where F and F' are the two foci. Since P is on the right branch, and F is the left focus, the difference |PF| - |PF'| should be equal to 2a, which is 4.So, |PF| - |PF'| = 4.Therefore, |PF| = |PF'| + 4.So, if I substitute this into the expression |PF| + |PA|, I get |PF'| + 4 + |PA|.So, |PF| + |PA| = |PF'| + |PA| + 4.Therefore, to minimize |PF| + |PA|, I need to minimize |PF'| + |PA|, and then add 4.So, the problem reduces to finding the minimum of |PF'| + |PA|, where P is on the right branch of the hyperbola.Hmm, that seems more manageable.So, we need to minimize |PF'| + |PA|, where F' is (4,0) and A is (1,4).Wait a second, this looks like a classic problem where you have to find the shortest path from A to F' via a point P on a curve. In such cases, the minimal path is achieved when P lies on the straight line connecting A and F', but since P has to lie on the hyperbola, we need to see if the line AF' intersects the hyperbola.If it does, then the minimal value would be |AF'|, otherwise, we might have to find the minimal distance in another way.Let me calculate |AF'| first.Point A is (1,4), and F' is (4,0). The distance between A and F' is sqrt[(4-1)^2 + (0-4)^2] = sqrt[3^2 + (-4)^2] = sqrt[9 + 16] = sqrt[25] = 5.So, |AF'| = 5.Now, if the line AF' intersects the hyperbola, then the minimal |PF'| + |PA| would be 5, achieved when P is the intersection point. Then, adding the 4 from earlier, the total minimal |PF| + |PA| would be 5 + 4 = 9.But I need to check if the line AF' intersects the hyperbola.Let me find the equation of the line AF'.Points A(1,4) and F'(4,0).The slope m = (0 - 4)/(4 - 1) = (-4)/3.So, the equation is y - 4 = (-4/3)(x - 1).Simplify:y = (-4/3)x + (4/3) + 4Convert 4 to thirds: 4 = 12/3So, y = (-4/3)x + (4/3 + 12/3) = (-4/3)x + 16/3.So, the equation is y = (-4/3)x + 16/3.Now, let's find the intersection points of this line with the hyperbola x²/4 - y²/12 = 1.Substitute y = (-4/3)x + 16/3 into the hyperbola equation.So,x²/4 - [(-4/3 x + 16/3)^2]/12 = 1.Let me compute [(-4/3 x + 16/3)^2]:First, factor out 4/3: [(-4/3)(x - 4)]^2 = (16/9)(x - 4)^2.So, [(-4/3 x + 16/3)^2] = (16/9)(x² - 8x + 16).Therefore, the equation becomes:x²/4 - [(16/9)(x² - 8x + 16)]/12 = 1.Simplify the second term:(16/9)(x² - 8x + 16)/12 = (16/108)(x² - 8x + 16) = (4/27)(x² - 8x + 16).So, now the equation is:x²/4 - (4/27)(x² - 8x + 16) = 1.Multiply through by 108 (the least common multiple of 4 and 27) to eliminate denominators:108*(x²/4) - 108*(4/27)(x² - 8x + 16) = 108*1.Simplify each term:108/4 = 27, so first term is 27x².108*(4/27) = 16, so second term is 16(x² - 8x + 16).Third term is 108.So, the equation becomes:27x² - 16(x² - 8x + 16) = 108.Expand the second term:27x² - 16x² + 128x - 256 = 108.Combine like terms:(27x² - 16x²) = 11x²So, 11x² + 128x - 256 = 108.Bring 108 to the left:11x² + 128x - 256 - 108 = 0Simplify:11x² + 128x - 364 = 0.Now, solve this quadratic equation for x.Use quadratic formula:x = [-b ± sqrt(b² - 4ac)] / (2a)Where a = 11, b = 128, c = -364.Compute discriminant:b² - 4ac = 128² - 4*11*(-364)128² = 163844*11 = 4444*364 = let's compute 44*364:First, 40*364 = 14,5604*364 = 1,456So, total is 14,560 + 1,456 = 16,016But since c is negative, it's -4ac = -4*11*(-364) = +16,016So, discriminant is 16384 + 16016 = 32400.sqrt(32400) = 180.So, x = [-128 ± 180]/(2*11) = [-128 ± 180]/22.Compute both solutions:First solution: (-128 + 180)/22 = (52)/22 = 26/11 ≈ 2.36Second solution: (-128 - 180)/22 = (-308)/22 = -14.So, x = 26/11 and x = -14.Now, since P is on the right branch of the hyperbola, x must be positive. So, x = 26/11 is the valid solution.So, x = 26/11 ≈ 2.36.Find y-coordinate:y = (-4/3)x + 16/3Plug in x = 26/11:y = (-4/3)*(26/11) + 16/3= (-104/33) + (176/33)= (72)/33= 24/11 ≈ 2.18.So, the point P is at (26/11, 24/11).Now, since this point lies on the line AF', the minimal |PF'| + |PA| is indeed |AF'| = 5.Therefore, the minimal value of |PF| + |PA| is 5 + 4 = 9.So, the answer should be 9, which is option B.But wait, let me double-check if there's any mistake in my calculations.First, the hyperbola equation: x²/4 - y²/12 = 1. Correct.Foci at (±4, 0). Correct.Definition of hyperbola: |PF| - |PF'| = 2a = 4. Correct.Then, |PF| + |PA| = |PF'| + |PA| + 4. Correct.Then, to minimize |PF'| + |PA|, we need to see if the line AF' intersects the hyperbola. Correct.Equation of line AF' is y = (-4/3)x + 16/3. Correct.Substituted into hyperbola equation, got quadratic equation, solved, found x = 26/11 and x = -14. Correct.Since P is on the right branch, x = 26/11 is valid. Correct.Computed y = 24/11. Correct.Therefore, minimal |PF'| + |PA| is |AF'| = 5. Correct.Hence, minimal |PF| + |PA| is 5 + 4 = 9. Correct.So, I think my reasoning is solid. The answer is 9, which is option B.**Final Answer**boxed{B}

💡Okay, so I have this problem where I need to prove that if a certain function f(x) equals zero at two points x₁ and x₂, then the difference between x₂ and x₁ must be an integer multiple of π. The function f(x) is given as a sum of cosines with different phase shifts and coefficients that decrease by a factor of 1/2 each time.First, I should write down the function to get a clear picture:f(x) = cos(a₁ + x) + (1/2)cos(a₂ + x) + (1/4)cos(a₃ + x) + ... + (1/2^{n-1})cos(aₙ + x).So, it's a finite sum of cosines, each with a different constant phase shift a_k and a coefficient that halves each time. The variable is x, and the a_k's are constants.I need to show that if f(x₁) = 0 and f(x₂) = 0, then x₂ - x₁ = mπ for some integer m.Hmm, okay. Maybe I can express f(x) in a more manageable form. Since each term is a cosine function with the same argument x plus a constant phase, perhaps I can factor out the x part.Using the cosine addition formula: cos(a + x) = cos(a)cos(x) - sin(a)sin(x). So, each term in f(x) can be expanded like that.Let me try that:f(x) = [cos(a₁)cos(x) - sin(a₁)sin(x)] + (1/2)[cos(a₂)cos(x) - sin(a₂)sin(x)] + (1/4)[cos(a₃)cos(x) - sin(a₃)sin(x)] + ... + (1/2^{n-1})[cos(aₙ)cos(x) - sin(aₙ)sin(x)].Now, I can factor out cos(x) and sin(x):f(x) = [cos(a₁) + (1/2)cos(a₂) + (1/4)cos(a₃) + ... + (1/2^{n-1})cos(aₙ)]cos(x) - [sin(a₁) + (1/2)sin(a₂) + (1/4)sin(a₃) + ... + (1/2^{n-1})sin(aₙ)]sin(x).Let me denote the coefficients of cos(x) and sin(x) as A and B respectively:A = cos(a₁) + (1/2)cos(a₂) + (1/4)cos(a₃) + ... + (1/2^{n-1})cos(aₙ),B = sin(a₁) + (1/2)sin(a₂) + (1/4)sin(a₃) + ... + (1/2^{n-1})sin(aₙ).So, f(x) can be written as:f(x) = A cos(x) - B sin(x).Now, if f(x₁) = 0 and f(x₂) = 0, then:A cos(x₁) - B sin(x₁) = 0,A cos(x₂) - B sin(x₂) = 0.I need to find the relationship between x₁ and x₂.Let me consider two cases:Case 1: A ≠ 0.If A is not zero, I can divide both equations by A:cos(x₁) - (B/A) sin(x₁) = 0,cos(x₂) - (B/A) sin(x₂) = 0.Let me denote (B/A) as some constant, say C.So, cos(x₁) = C sin(x₁),cos(x₂) = C sin(x₂).This implies that tan(x₁) = 1/C,and tan(x₂) = 1/C.Therefore, tan(x₁) = tan(x₂).The solutions to tan(x₁) = tan(x₂) are x₂ = x₁ + mπ, where m is an integer.Therefore, x₂ - x₁ = mπ.Case 2: A = 0.If A is zero, then from the equations:- B sin(x₁) = 0,- B sin(x₂) = 0.Since A = 0, but from the expression of A, it's a sum of cosines with coefficients. Similarly, B is a sum of sines with coefficients.But wait, can both A and B be zero?Let me check. Suppose A = 0 and B = 0.Then, f(x) = 0 for all x, which is not the case because f(x) is a sum of cosines with different phases and coefficients. Unless all the a_k's are chosen such that A and B both vanish, but that's a special case.But in the problem statement, it's given that f(x₁) = f(x₂) = 0, but not necessarily that f(x) is identically zero.So, if A = 0, then from f(x) = -B sin(x), so f(x) = -B sin(x).Therefore, if f(x₁) = 0, then -B sin(x₁) = 0, so sin(x₁) = 0, which implies x₁ = kπ.Similarly, x₂ = lπ, so x₂ - x₁ = (l - k)π = mπ, where m is an integer.Therefore, in this case as well, x₂ - x₁ is an integer multiple of π.Hence, regardless of whether A is zero or not, the difference x₂ - x₁ must be an integer multiple of π.Wait, but in the case where A = 0, we need to ensure that B ≠ 0. Because if both A and B are zero, then f(x) is identically zero, which is a trivial case.But in the problem statement, it's given that f(x₁) = f(x₂) = 0, but it's not stated that f(x) is not identically zero. So, perhaps I need to consider whether A and B can both be zero.Let me see. Suppose A = 0 and B = 0.Then, f(x) = 0 for all x. So, in that case, any x₁ and x₂ would satisfy f(x₁) = f(x₂) = 0, but the difference x₂ - x₁ can be arbitrary, not necessarily an integer multiple of π.But the problem statement says "if f(x₁) = f(x₂) = 0, then x₂ - x₁ = mπ". So, perhaps in the problem, it's assumed that f(x) is not identically zero, which would require that A and B are not both zero.Alternatively, maybe the function f(x) cannot be identically zero unless all the a_k's are chosen in a specific way, which might not be the case here.Wait, let's think about whether A and B can both be zero.A = sum_{k=1}^n (1/2^{k-1}) cos(a_k),B = sum_{k=1}^n (1/2^{k-1}) sin(a_k).For A and B to both be zero, we would need:sum_{k=1}^n (1/2^{k-1}) cos(a_k) = 0,sum_{k=1}^n (1/2^{k-1}) sin(a_k) = 0.Is this possible? Yes, if the vectors (cos(a_k), sin(a_k)) are arranged in such a way that their weighted sum cancels out. But it's not guaranteed; it depends on the specific a_k's.However, in the problem statement, the a_k's are given as real constants, so they could be chosen such that A and B are both zero, making f(x) identically zero. But in that case, the conclusion that x₂ - x₁ = mπ would still hold because any x₁ and x₂ would satisfy f(x₁) = f(x₂) = 0, but the difference could be arbitrary. Wait, no, because if f(x) is identically zero, then any x₁ and x₂ would satisfy f(x₁) = f(x₂) = 0, but the difference x₂ - x₁ could be any real number, not necessarily an integer multiple of π.But the problem statement says "if f(x₁) = f(x₂) = 0, then x₂ - x₁ = mπ". So, perhaps the problem assumes that f(x) is not identically zero, meaning that A and B are not both zero.Alternatively, maybe even if f(x) is identically zero, the conclusion still holds vacuously because any x₁ and x₂ would satisfy f(x₁) = f(x₂) = 0, but the difference x₂ - x₁ could be arbitrary, which contradicts the conclusion.Hmm, perhaps I need to ensure that A and B are not both zero. Let me check if that's necessarily the case.Suppose I evaluate f(x) at x = -a₁.Then, f(-a₁) = cos(a₁ - a₁) + (1/2)cos(a₂ - a₁) + (1/4)cos(a₃ - a₁) + ... + (1/2^{n-1})cos(aₙ - a₁).Which simplifies to:f(-a₁) = 1 + (1/2)cos(a₂ - a₁) + (1/4)cos(a₃ - a₁) + ... + (1/2^{n-1})cos(aₙ - a₁).Since each cosine term is bounded between -1 and 1, the sum:(1/2)cos(a₂ - a₁) + (1/4)cos(a₃ - a₁) + ... + (1/2^{n-1})cos(aₙ - a₁) ≤ (1/2) + (1/4) + ... + (1/2^{n-1}).The sum of the geometric series 1/2 + 1/4 + ... + 1/2^{n-1} is 1 - 1/2^{n-1}.Therefore, f(-a₁) ≥ 1 - (1 - 1/2^{n-1}) = 1/2^{n-1} > 0.So, f(-a₁) is positive, which means that f(x) is not identically zero. Therefore, A and B cannot both be zero because if they were, f(x) would be identically zero, but we just found that f(-a₁) > 0.Therefore, A and B cannot both be zero. So, in the case where A = 0, B must be non-zero, and vice versa.Thus, going back to the two cases:Case 1: A ≠ 0. Then, as before, tan(x₁) = tan(x₂), leading to x₂ - x₁ = mπ.Case 2: A = 0, but then B ≠ 0, so f(x) = -B sin(x). Therefore, f(x) = 0 implies sin(x) = 0, so x = kπ, leading again to x₂ - x₁ = mπ.Therefore, in both cases, the difference x₂ - x₁ must be an integer multiple of π.I think that covers all possibilities. So, the conclusion is that if f(x₁) = f(x₂) = 0, then x₂ - x₁ = mπ for some integer m.

💡First, I need to determine the amount P invested in the business. I know that Q invested Rs 15,000 and that the profit is divided between P and Q in a 5:1 ratio.Since the profit is divided based on their investments, the ratio of their investments should be the same as the ratio of their profits. This means that P's investment to Q's investment is 5:1.Given that Q's investment is Rs 15,000, I can set up the proportion to find P's investment:P's Investment / 15,000 = 5 / 1.Solving for P's Investment:P's Investment = 15,000 * 5 = 75,000.Therefore, P invested Rs 75,000 in the business.