💡Okay, so I have this equation to solve: the cube root of (5 minus x over 3) equals negative 2. Hmm, let me write that down to visualize it better. It's like this: ∛(5 - x/3) = -2. Alright, cube roots can sometimes be tricky, but I remember that to get rid of a cube root, you can cube both sides of the equation. That should help me solve for x.So, if I cube both sides, what happens? On the left side, cubing the cube root should just give me the expression inside, right? So, that would be 5 - x/3. And on the right side, cubing -2. Let me calculate that. (-2) cubed is -8 because (-2) times (-2) is 4, and then 4 times (-2) is -8. So, after cubing both sides, my equation simplifies to 5 - x/3 = -8.Now, I need to solve for x. Let me rewrite the equation: 5 - (x/3) = -8. I think the next step is to isolate the term with x. So, I should get rid of that 5 on the left side. To do that, I can subtract 5 from both sides. Let's see, 5 - (x/3) - 5 equals -8 - 5. Simplifying both sides, the 5 and -5 on the left cancel out, leaving me with -x/3. On the right, -8 minus 5 is -13. So, now I have -x/3 = -13.Hmm, okay, so I have -x divided by 3 equals -13. I need to solve for x, so I should get rid of the fraction. Maybe I can multiply both sides by 3 to eliminate the denominator. Let's try that. Multiplying both sides by 3 gives me (-x/3) * 3 = -13 * 3. On the left, the 3s cancel out, leaving me with -x. On the right, -13 times 3 is -39. So, now I have -x = -39.Wait, so -x equals -39. To solve for x, I need to get rid of that negative sign in front of x. I can do that by multiplying both sides by -1. Let's see, (-x) * (-1) equals (-39) * (-1). On the left, that gives me x, and on the right, positive 39. So, x equals 39.Let me double-check my work to make sure I didn't make any mistakes. Starting from the original equation: ∛(5 - x/3) = -2. If I plug x = 39 back into the equation, does it hold true? Let's see. Inside the cube root, it would be 5 - 39/3. 39 divided by 3 is 13, so 5 - 13 is -8. The cube root of -8 is indeed -2, which matches the right side of the equation. So, that checks out.I think that's the only solution because cube roots are one-to-one functions, meaning each input has exactly one output. So, there's only one value of x that satisfies the equation. Therefore, x = 39 is the solution.Just to recap the steps: I started by cubing both sides to eliminate the cube root, which gave me a linear equation. Then, I solved that linear equation by isolating x through inverse operations—subtracting 5 and then multiplying by 3. Finally, I verified my solution by plugging it back into the original equation to ensure it works.I don't think I made any calculation errors because each step led logically to the next, and the final check confirmed the solution. It's always good to verify, especially when dealing with roots and exponents, because sometimes extraneous solutions can pop up, but in this case, everything seems fine.So, yeah, I'm pretty confident that x = 39 is the correct and only solution to the equation.

💡Okay, so I'm trying to solve this problem about the function f(x) = 9/sin²x + 4/cos²x for x in (0, π/2). The first part asks for the maximum value of t such that f(x) is always greater than or equal to t. Then, when t is at its maximum, I need to solve another inequality involving absolute values.Starting with part (1). I know that f(x) is given, and I need to find the minimum value of f(x) because t has to be less than or equal to f(x) for all x in that interval. So, the maximum t can be is the minimum value of f(x). Hmm, how do I find the minimum of this function? It's a combination of 9 over sin squared x and 4 over cos squared x. Maybe I can use some inequality here, like the Cauchy-Schwarz inequality or AM-GM inequality. Let me think about AM-GM because it's often useful for expressions with fractions.Wait, another approach is to express f(x) in terms of sin²x and cos²x. Maybe I can set sin²x = a and cos²x = b, so a + b = 1 since sin²x + cos²x = 1. Then f(x) becomes 9/a + 4/b. So, I need to minimize 9/a + 4/b with the constraint that a + b = 1 and a, b > 0.Yes, that sounds right. So, using the method of Lagrange multipliers or substitution. Since a + b = 1, I can express b as 1 - a. Then f(x) becomes 9/a + 4/(1 - a). Now, I can take the derivative with respect to a and set it to zero to find the minimum.Let me compute the derivative. Let f(a) = 9/a + 4/(1 - a). The derivative f’(a) is -9/a² + 4/(1 - a)². Setting this equal to zero:-9/a² + 4/(1 - a)² = 0 => 4/(1 - a)² = 9/a² => 4a² = 9(1 - a)² Taking square roots on both sides (since a and 1 - a are positive):2a = 3(1 - a) => 2a = 3 - 3a => 5a = 3 => a = 3/5So, a = 3/5, which means b = 1 - 3/5 = 2/5. Plugging back into f(x):f(x) = 9/(3/5) + 4/(2/5) = 9*(5/3) + 4*(5/2) = 15 + 10 = 25.So, the minimum value of f(x) is 25, which means the maximum t is 25.Wait, let me check if I did that correctly. I used substitution and calculus, but maybe there's a smarter way using inequalities. Let me try the Cauchy-Schwarz inequality.Cauchy-Schwarz says that (u1v1 + u2v2)^2 ≤ (u1² + u2²)(v1² + v2²). Maybe I can set u1 = 3/sinx, u2 = 2/cosx, and v1 = sinx, v2 = cosx. Then:(u1v1 + u2v2)^2 ≤ (u1² + u2²)(v1² + v2²) => (3 + 2)^2 ≤ (9/sin²x + 4/cos²x)(sin²x + cos²x) => 25 ≤ (9/sin²x + 4/cos²x)(1) => 25 ≤ f(x)So, f(x) is always greater than or equal to 25, which confirms the minimum is 25. So, t_max = 25.Alright, part (1) is done. Now, part (2). When t is 25, we have to solve |x + 25/5| + |2x - 1| ≤ 6. Wait, 25/5 is 5, so the inequality becomes |x + 5| + |2x - 1| ≤ 6.I need to solve |x + 5| + |2x - 1| ≤ 6. To handle absolute values, I should consider different cases based on the critical points where the expressions inside the absolute values change sign. These occur at x = -5 and x = 1/2.So, the critical points divide the real line into three intervals:1. x < -52. -5 ≤ x ≤ 1/23. x > 1/2I'll analyze each interval separately.**Case 1: x < -5**In this interval, x + 5 is negative, so |x + 5| = -(x + 5) = -x -5. Similarly, 2x - 1 is also negative because x < -5, so 2x -1 < -10 -1 = -11 < 0. Hence, |2x -1| = -(2x -1) = -2x +1.So, the inequality becomes:(-x -5) + (-2x +1) ≤ 6 => -3x -4 ≤ 6 => -3x ≤ 10 => x ≥ -10/3 ≈ -3.333...But in this case, x < -5. So, we have x ≥ -10/3 and x < -5. But -10/3 is approximately -3.333, which is greater than -5. So, there's no overlap here. Therefore, no solution in this interval.**Case 2: -5 ≤ x ≤ 1/2**In this interval, x + 5 is non-negative because x ≥ -5, so |x +5| = x +5. However, 2x -1 is still negative because x ≤ 1/2, so 2x ≤1, hence 2x -1 ≤0. Therefore, |2x -1| = -(2x -1) = -2x +1.So, the inequality becomes:(x +5) + (-2x +1) ≤ 6 => (-x +6) ≤6 => -x ≤0 => x ≥0But in this case, x is between -5 and 1/2. So, combining x ≥0 and -5 ≤x ≤1/2, we get 0 ≤x ≤1/2.**Case 3: x > 1/2**In this interval, both x +5 and 2x -1 are positive. So, |x +5| = x +5 and |2x -1| = 2x -1.So, the inequality becomes:(x +5) + (2x -1) ≤6 => 3x +4 ≤6 => 3x ≤2 => x ≤2/3 ≈0.666...But in this case, x >1/2. So, combining x ≤2/3 and x >1/2, we get 1/2 <x ≤2/3.**Combining all cases:**From Case 2: 0 ≤x ≤1/2 From Case 3: 1/2 <x ≤2/3 So, overall, 0 ≤x ≤2/3.Wait, but in Case 2, x starts at -5, but we only get solutions from 0 to 1/2. So, the solution set is [0, 2/3].Let me double-check if x=0 satisfies the inequality:|0 +5| + |0 -1| =5 +1=6 ≤6. Yes, equality holds.x=2/3:|2/3 +5| + |4/3 -1| = |17/3| + |1/3| =17/3 +1/3=18/3=6 ≤6. Also equality holds.What about x=1/2:|1/2 +5| + |1 -1|=5.5 +0=5.5 ≤6. Yes, it's within the limit.And x=0.5 is included in the solution set.So, the solution set is all real numbers x such that 0 ≤x ≤2/3.**Final Answer**(1) The maximum value of ( t ) is (boxed{25}).(2) The solution set of the inequality is (boxed{left[0, dfrac{2}{3}right]}).

💡Okay, so I've got this problem here about sets A and B, and I need to find the intersection of the complement of A in R and B. Hmm, let me try to break this down step by step.First, let me understand what set A is. The problem says A is the set of all real numbers x such that the absolute value of x is greater than 1. So, A = {x | |x| > 1, x ∈ R}. That means A includes all real numbers less than -1 and greater than 1. So, in interval notation, A would be (-∞, -1) ∪ (1, ∞). Got that.Now, the complement of A in R, which is denoted as C_R A, would be all the real numbers that are not in A. Since A is everything less than -1 and greater than 1, the complement would be everything between -1 and 1, including -1 and 1 themselves. So, C_R A = [-1, 1]. That makes sense because it's the closure of the interval between -1 and 1.Next, let's look at set B. B is defined as {y | y = x², x ∈ R}. So, B is the set of all real numbers y such that y is the square of some real number x. Since squaring any real number gives a non-negative result, B is the set of all non-negative real numbers. In interval notation, that's [0, ∞). So, B = [0, ∞).Now, the problem asks for the intersection of the complement of A and B. That is, (C_R A) ∩ B. Let's write that out. C_R A is [-1, 1], and B is [0, ∞). The intersection of these two sets would be the set of all real numbers that are in both [-1, 1] and [0, ∞). Visually, if I imagine the number line, [-1, 1] covers from -1 to 1, and [0, ∞) covers from 0 to the right indefinitely. The overlapping part of these two intervals is from 0 to 1. So, the intersection should be [0, 1].Let me double-check that. If I take any number between 0 and 1, it's definitely in both sets. Numbers less than 0 are in C_R A but not in B, and numbers greater than 1 are in B but not in C_R A. So, the overlap is exactly [0, 1].Looking back at the options given:A: {x | -1 ≤ x ≤ 1} – That's the complement of A, but we need the intersection with B, which narrows it down further.B: {x | x ≥ 0} – That's set B itself, but we need the intersection with the complement of A, which is narrower.C: {x | 0 ≤ x ≤ 1} – That's exactly the intersection we found.D: ∅ – That's the empty set, which isn't the case here.So, the correct answer should be option C.**Final Answer**boxed{C}

💡Alright, so I have this problem here with two parts, (I) and (II). Let me try to understand what it's asking for.First, we're given two vectors, **a** and **b**, defined as:- **a** = (cos(λθ), cos((10 - λ)θ))- **b** = (sin((10 - λ)θ), sin(λθ))And we're told that λ and θ are real numbers.Part (I) asks for the value of |**a**|² + |**b**|². Okay, so I need to find the magnitude squared of vector **a** and vector **b**, then add them together.Let me recall that the magnitude squared of a vector (x, y) is x² + y². So, for vector **a**, that would be [cos(λθ)]² + [cos((10 - λ)θ)]². Similarly, for vector **b**, it would be [sin((10 - λ)θ)]² + [sin(λθ)]².So, |**a**|² = cos²(λθ) + cos²((10 - λ)θ)And |**b**|² = sin²((10 - λ)θ) + sin²(λθ)Therefore, |**a**|² + |**b**|² = cos²(λθ) + cos²((10 - λ)θ) + sin²((10 - λ)θ) + sin²(λθ)Hmm, I remember that cos²(x) + sin²(x) = 1 for any x. So, maybe I can group terms accordingly.Looking at the expression, I see cos²(λθ) + sin²(λθ) which is 1, and cos²((10 - λ)θ) + sin²((10 - λ)θ) which is also 1.So, adding them together, 1 + 1 = 2.Wait, so |**a**|² + |**b**|² = 2? That seems straightforward. Maybe that's the answer for part (I).Now, moving on to part (II). It says, if θ = π/20, prove that **a** is parallel to **b**.Hmm, parallel vectors. I remember that two vectors are parallel if one is a scalar multiple of the other. So, **a** = k**b** for some scalar k.Alternatively, another way to check if two vectors are parallel is to see if their cross product is zero. In two dimensions, the cross product is essentially the determinant of a 2x2 matrix formed by the vectors, and if it's zero, they're parallel.So, maybe I can compute the determinant of **a** and **b** and see if it's zero when θ = π/20.Let me write down the vectors again:- **a** = (cos(λθ), cos((10 - λ)θ))- **b** = (sin((10 - λ)θ), sin(λθ))The determinant would be:cos(λθ) * sin(λθ) - cos((10 - λ)θ) * sin((10 - λ)θ)Wait, that looks like the sine of a difference identity. Let me recall that sin(A - B) = sinA cosB - cosA sinB.But here, I have cos(λθ) sin(λθ) - cos((10 - λ)θ) sin((10 - λ)θ). Hmm, maybe I can factor this differently.Alternatively, I can use the identity for sin(2x) = 2 sinx cosx. So, cosx sinx = (1/2) sin(2x).So, let's apply that:cos(λθ) sin(λθ) = (1/2) sin(2λθ)cos((10 - λ)θ) sin((10 - λ)θ) = (1/2) sin(2(10 - λ)θ)So, the determinant becomes:(1/2) sin(2λθ) - (1/2) sin(2(10 - λ)θ)Factor out 1/2:(1/2)[sin(2λθ) - sin(2(10 - λ)θ)]Now, I can use the sine subtraction formula:sin A - sin B = 2 cos((A + B)/2) sin((A - B)/2)Let me set A = 2λθ and B = 2(10 - λ)θSo, sin(2λθ) - sin(2(10 - λ)θ) = 2 cos[(2λθ + 2(10 - λ)θ)/2] sin[(2λθ - 2(10 - λ)θ)/2]Simplify the arguments:(2λθ + 20θ - 2λθ)/2 = (20θ)/2 = 10θ(2λθ - 20θ + 2λθ)/2 = (4λθ - 20θ)/2 = 2λθ - 10θSo, sin(2λθ) - sin(2(10 - λ)θ) = 2 cos(10θ) sin(2λθ - 10θ)Therefore, the determinant becomes:(1/2) * 2 cos(10θ) sin(2λθ - 10θ) = cos(10θ) sin(2λθ - 10θ)Now, we're given that θ = π/20. Let's substitute that in.First, compute 10θ:10θ = 10*(π/20) = π/2So, cos(10θ) = cos(π/2) = 0Therefore, the determinant is 0 * sin(2λθ - 10θ) = 0Since the determinant is zero, the vectors **a** and **b** are parallel when θ = π/20.Wait, that seems to work out. So, the cross product is zero, meaning they are parallel.Alternatively, I could have checked if one vector is a scalar multiple of the other. Let's see:If **a** = k**b**, then:cos(λθ) = k sin((10 - λ)θ)cos((10 - λ)θ) = k sin(λθ)So, from the first equation, k = cos(λθ) / sin((10 - λ)θ)From the second equation, k = cos((10 - λ)θ) / sin(λθ)Therefore, cos(λθ) / sin((10 - λ)θ) = cos((10 - λ)θ) / sin(λθ)Cross-multiplying:cos(λθ) sin(λθ) = cos((10 - λ)θ) sin((10 - λ)θ)Which is the same as the determinant being zero, so it's consistent.Therefore, when θ = π/20, the vectors are indeed parallel.So, summarizing my thoughts:For part (I), by expanding the magnitudes and using the Pythagorean identity, I found that |**a**|² + |**b**|² = 2.For part (II), by computing the determinant of the vectors and substituting θ = π/20, I found that the determinant is zero, implying that the vectors are parallel.I think that covers both parts of the problem.

💡Alright, so I have this algebra problem here, and I need to find the value of a certain expression given that ( a - b = 2sqrt{3} ). The expression is:[left( frac{a^2 + b^2}{2a} - b right) cdot frac{a}{a - b}]Hmm, okay. Let me try to break this down step by step. First, I notice that the expression has two main parts: the first part is inside the parentheses, and the second part is multiplied by ( frac{a}{a - b} ). Maybe I can simplify the part inside the parentheses first.So, looking at the first part:[frac{a^2 + b^2}{2a} - b]I think I can combine these two terms into a single fraction to make it easier to handle. To do that, I'll need a common denominator. The first term already has a denominator of ( 2a ), and the second term, ( b ), can be written as ( frac{2ab}{2a} ) so that both terms have the same denominator.Let me write that out:[frac{a^2 + b^2}{2a} - frac{2ab}{2a}]Now that both terms have the same denominator, I can combine them:[frac{a^2 + b^2 - 2ab}{2a}]Okay, so that simplifies the first part to:[frac{a^2 + b^2 - 2ab}{2a}]Wait a minute, the numerator looks familiar. ( a^2 + b^2 - 2ab ) is actually the expansion of ( (a - b)^2 ). Let me check that:[(a - b)^2 = a^2 - 2ab + b^2]Yes, that's exactly the same as the numerator. So, I can rewrite the expression as:[frac{(a - b)^2}{2a}]Great, that seems simpler. Now, the entire original expression is this fraction multiplied by ( frac{a}{a - b} ). So, putting it all together:[frac{(a - b)^2}{2a} cdot frac{a}{a - b}]Hmm, I can see that ( a ) in the numerator of the second fraction will cancel out with the ( a ) in the denominator of the first fraction. Let's write that out:[frac{(a - b)^2}{2a} cdot frac{a}{a - b} = frac{(a - b)^2 cdot a}{2a cdot (a - b)}]Simplifying the ( a ) terms:[frac{(a - b)^2}{2 cdot (a - b)}]Now, I can cancel out one ( (a - b) ) term from the numerator and the denominator:[frac{(a - b)}{2}]So, the entire expression simplifies down to ( frac{a - b}{2} ). But wait, the problem gives me that ( a - b = 2sqrt{3} ). So, substituting that in:[frac{2sqrt{3}}{2}]The 2 in the numerator and the 2 in the denominator cancel out, leaving me with:[sqrt{3}]So, the value of the original expression is ( sqrt{3} ). Let me double-check my steps to make sure I didn't make any mistakes.1. I combined the terms inside the parentheses correctly by finding a common denominator.2. I recognized that ( a^2 + b^2 - 2ab ) is ( (a - b)^2 ), which seems right.3. I simplified the multiplication by canceling out the ( a ) terms and one ( (a - b) ) term.4. Substituted the given value of ( a - b ) and simplified.Everything seems to check out. I don't think I made any errors in the algebraic manipulations. So, I'm confident that the answer is ( sqrt{3} ).**Final Answer**boxed{A}

💡First, I need to determine the original cost of the car that Ramu bought. He spent Rs. 8000 on repairs and sold the car for Rs. 64900, making a profit of 29.8%.I'll denote the original cost of the car as ( C ).The total cost for Ramu is the sum of the original cost and the repair expenses:[text{Total Cost} = C + 8000]Ramu sold the car for Rs. 64900, which includes a 29.8% profit. This means the selling price is 129.8% of the total cost:[64900 = (C + 8000) times 1.298]To find ( C ), I'll divide the selling price by 1.298:[C + 8000 = frac{64900}{1.298} = 50000]Finally, I'll subtract the repair costs to determine the original cost:[C = 50000 - 8000 = 42000]