💡First, I need to calculate the product of 98 and 102.Next, I will divide the resulting product by 11 to find the remainder.Alternatively, I can use the divisibility rule for 11 by alternating the sum of the digits of the product and checking if the result is divisible by 11. This will help confirm the remainder.

💡Okay, so I have this problem here where I'm told that functions f(x) and g(x) = 2^x are inverse functions. I need to find the interval where f(4x - x²) is monotonically increasing. The options are A: (-∞, 2], B: (0, 2), C: [2, 4), and D: [2, +∞). Hmm, let me try to figure this out step by step.First, since f and g are inverses, that means f(g(x)) = x and g(f(x)) = x. Since g(x) is 2^x, which is an exponential function, its inverse should be a logarithmic function. So, f(x) must be log base 2 of x, right? So, f(x) = log₂x. That makes sense because the inverse of an exponential function is a logarithmic function with the same base.Now, the function we're looking at is f(4x - x²). Substituting f(x) into this, we get log₂(4x - x²). So, f(4x - x²) = log₂(4x - x²). Before I can talk about the monotonicity, I need to make sure I know the domain of this function. The argument of the logarithm must be positive because you can't take the log of a non-positive number. So, 4x - x² > 0. Let me solve that inequality.4x - x² > 0 Let's rewrite this as -x² + 4x > 0. Multiplying both sides by -1 (and remembering to flip the inequality sign), we get x² - 4x < 0. Factor this: x(x - 4) < 0. So, the product of x and (x - 4) is less than zero. That happens when one is positive and the other is negative. So, if x is between 0 and 4, then x is positive and (x - 4) is negative, so their product is negative. If x is less than 0, both are negative, so their product is positive. If x is greater than 4, both are positive, so their product is positive. Therefore, the inequality holds when x is in (0, 4). So, the domain of f(4x - x²) is (0, 4).Alright, now I need to figure out where this function is increasing. The function is log₂(4x - x²). Since logarithmic functions are monotonically increasing when their base is greater than 1, which it is here (base 2), the function log₂(u) is increasing when u is increasing, and decreasing when u is decreasing.So, let me consider the inner function u = 4x - x². I need to find where u is increasing or decreasing because that will determine the behavior of the entire function f(4x - x²).Let's find the derivative of u with respect to x to determine its monotonicity. The derivative of u is du/dx = 4 - 2x. Setting this equal to zero to find critical points: 4 - 2x = 0 ⇒ x = 2. So, at x = 2, the function u has a critical point. To determine if it's a maximum or minimum, I can look at the second derivative or analyze the sign of the first derivative around x = 2. The second derivative of u is d²u/dx² = -2, which is negative, so the function u has a maximum at x = 2.Therefore, u = 4x - x² is increasing on the interval (0, 2) and decreasing on the interval (2, 4). Since log₂(u) is an increasing function, the composition f(4x - x²) will be increasing when u is increasing and decreasing when u is decreasing. So, when x is in (0, 2), u is increasing, so log₂(u) is increasing. When x is in (2, 4), u is decreasing, so log₂(u) is decreasing. Therefore, the interval where f(4x - x²) is monotonically increasing is (0, 2). Looking back at the options, that's option B: (0, 2). Wait, let me double-check. The function f(4x - x²) is log₂(4x - x²). The domain is (0, 4). The inner function 4x - x² is a downward-opening parabola with vertex at x = 2. So, it increases up to x = 2 and then decreases after that. Since log is increasing, the composition will increase where the inner function increases and decrease where the inner function decreases. So, yes, increasing on (0, 2), decreasing on (2, 4). Therefore, the correct answer should be B: (0, 2).**Final Answer**boxed{B}

💡Okay, so I've got this sequence problem here. It says that there's a sequence {a_n} where each term is defined by a_{n+1} = 3a_n + 4, and the first term a_1 is 1. There are two parts to this problem: first, I need to prove that the sequence {a_n + 2} is a geometric progression, and second, I have to find the sum of the first n terms, S_n, of the sequence {a_n}.Alright, let's start with part (I). I need to show that {a_n + 2} is a geometric progression. A geometric progression is a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio. So, if I can show that (a_{n+1} + 2) divided by (a_n + 2) is a constant, then that would mean {a_n + 2} is a geometric progression.Given the recursive formula a_{n+1} = 3a_n + 4, I can try to manipulate this to see if adding 2 to each term makes it geometric. Let's compute (a_{n+1} + 2) / (a_n + 2):(a_{n+1} + 2) = 3a_n + 4 + 2 = 3a_n + 6So, (a_{n+1} + 2) / (a_n + 2) = (3a_n + 6) / (a_n + 2)Hmm, let's factor the numerator: 3a_n + 6 = 3(a_n + 2)So, substituting back in, we get:(3(a_n + 2)) / (a_n + 2) = 3Oh, that's nice! The ratio simplifies to 3, which is a constant. That means each term of {a_n + 2} is 3 times the previous term, so {a_n + 2} is indeed a geometric progression with a common ratio of 3.Cool, that wasn't too bad. Now, moving on to part (II). I need to find the sum of the first n terms of the sequence {a_n}, which is S_n.First, since I've already shown that {a_n + 2} is a geometric progression, maybe I can express a_n in terms of this geometric sequence. Let's denote b_n = a_n + 2. Then, from part (I), we know that b_n is a geometric progression with first term b_1 = a_1 + 2 = 1 + 2 = 3, and common ratio r = 3.So, the general term of a geometric progression is b_n = b_1 * r^{n-1}. Plugging in the values, we get:b_n = 3 * 3^{n-1} = 3^nBut since b_n = a_n + 2, that means a_n = b_n - 2 = 3^n - 2Alright, so now I have an explicit formula for a_n: a_n = 3^n - 2.Now, to find S_n, the sum of the first n terms, I can write:S_n = a_1 + a_2 + a_3 + ... + a_nSubstituting the formula for a_n:S_n = (3^1 - 2) + (3^2 - 2) + (3^3 - 2) + ... + (3^n - 2)I can split this sum into two separate sums:S_n = (3^1 + 3^2 + 3^3 + ... + 3^n) - (2 + 2 + 2 + ... + 2)The first sum is a geometric series with first term 3 and common ratio 3, and there are n terms. The second sum is just adding 2 n times, which is 2n.The formula for the sum of a geometric series is S = a*(r^n - 1)/(r - 1), where a is the first term, r is the common ratio, and n is the number of terms.Applying this to the first sum:Sum of 3^1 + 3^2 + ... + 3^n = 3*(3^n - 1)/(3 - 1) = (3^{n+1} - 3)/2So, putting it all together:S_n = (3^{n+1} - 3)/2 - 2nHmm, let me double-check that. The sum of the geometric series is indeed (3*(3^n - 1))/(3 - 1) = (3^{n+1} - 3)/2. Then subtracting 2n gives the total sum.Yeah, that seems right. So, S_n is equal to (3^{n+1} - 3)/2 minus 2n.Let me test this with a small n to make sure. Let's take n=1. Then S_1 should be a_1 = 1.Plugging into the formula: (3^{2} - 3)/2 - 2*1 = (9 - 3)/2 - 2 = 6/2 - 2 = 3 - 2 = 1. Correct.How about n=2. Then S_2 = a_1 + a_2 = 1 + (3*1 + 4) = 1 + 7 = 8.Using the formula: (3^{3} - 3)/2 - 2*2 = (27 - 3)/2 - 4 = 24/2 - 4 = 12 - 4 = 8. Correct.n=3: S_3 = 1 + 7 + (3*7 + 4) = 1 + 7 + 25 = 33.Formula: (3^{4} - 3)/2 - 2*3 = (81 - 3)/2 - 6 = 78/2 - 6 = 39 - 6 = 33. Correct.Alright, seems like the formula works for small n. So, I think I've got it right.**Final Answer**Ⅰ. The sequence {a_n + 2} is a geometric progression with common ratio 3.Ⅱ. The sum of the first n terms is boxed{frac{3^{n+1} - 3}{2} - 2n}.

💡Okay, so I have this triangle ABC, and I need to find the length of side BC. Let me see what information I have. First, angle A is π/4, which is 45 degrees. That's a nice angle because I remember the sine and cosine of 45 degrees are both √2/2. Next, there's this equation: sin A + sin(B - C) = 2√2 sin 2C. Hmm, that looks a bit complicated, but maybe I can break it down. Also, the area of the triangle is 1. I know that the area of a triangle can be calculated using the formula (1/2)*ab*sin C, where a and b are sides and C is the included angle. Maybe that will come in handy later.Let me start by writing down what I know:- Angle A = π/4- sin A + sin(B - C) = 2√2 sin 2C- Area = 1I need to find the length of side BC, which is side a opposite angle A.First, let's tackle the trigonometric equation: sin A + sin(B - C) = 2√2 sin 2C.Since I know angle A is π/4, sin A is √2/2. So, substituting that in:√2/2 + sin(B - C) = 2√2 sin 2C.Hmm, okay. Maybe I can express sin(B - C) using the sine subtraction formula. I remember that sin(B - C) = sin B cos C - cos B sin C. Let me write that down:√2/2 + sin B cos C - cos B sin C = 2√2 sin 2C.Now, sin 2C is 2 sin C cos C, so let's substitute that in:√2/2 + sin B cos C - cos B sin C = 2√2 * 2 sin C cos C.Wait, that would be 4√2 sin C cos C. So the equation becomes:√2/2 + sin B cos C - cos B sin C = 4√2 sin C cos C.Hmm, this seems a bit messy. Maybe I can rearrange terms:sin B cos C - cos B sin C = 4√2 sin C cos C - √2/2.I wonder if I can factor something out or use some identities here. Let me think.I recall that sin(B - C) is equal to sin B cos C - cos B sin C, which is exactly the left side. So, sin(B - C) = 4√2 sin C cos C - √2/2.But wait, that's just restating the original equation. Maybe I need another approach.Alternatively, maybe I can express sin(B - C) in terms of angles. Since in a triangle, the sum of angles is π, so B + C = π - A = π - π/4 = 3π/4.So, B = 3π/4 - C.Therefore, sin(B - C) = sin(3π/4 - C - C) = sin(3π/4 - 2C).Hmm, that might be useful. Let me write that:sin(B - C) = sin(3π/4 - 2C).So, substituting back into the equation:√2/2 + sin(3π/4 - 2C) = 2√2 sin 2C.Let me compute sin(3π/4 - 2C). Using the sine subtraction formula:sin(3π/4 - 2C) = sin(3π/4) cos 2C - cos(3π/4) sin 2C.I know that sin(3π/4) is √2/2 and cos(3π/4) is -√2/2. So:sin(3π/4 - 2C) = (√2/2) cos 2C - (-√2/2) sin 2C = (√2/2) cos 2C + (√2/2) sin 2C.So, substituting back into the equation:√2/2 + (√2/2) cos 2C + (√2/2) sin 2C = 2√2 sin 2C.Let me factor out √2/2 on the left side:√2/2 [1 + cos 2C + sin 2C] = 2√2 sin 2C.Divide both sides by √2:1/2 [1 + cos 2C + sin 2C] = 2 sin 2C.Multiply both sides by 2:1 + cos 2C + sin 2C = 4 sin 2C.Subtract sin 2C from both sides:1 + cos 2C = 3 sin 2C.Hmm, okay, so now I have 1 + cos 2C = 3 sin 2C.I can use the double-angle identities here. Remember that cos 2C = 1 - 2 sin²C, but that might not help directly. Alternatively, I can write 1 + cos 2C as 2 cos²C.So, 2 cos²C = 3 sin 2C.But sin 2C is 2 sin C cos C, so:2 cos²C = 3 * 2 sin C cos C.Simplify:2 cos²C = 6 sin C cos C.Divide both sides by 2 cos C (assuming cos C ≠ 0, which it can't be in a triangle because angles are between 0 and π, so cos C can't be zero unless C is π/2, but let's check that later):cos C = 3 sin C.So, cos C = 3 sin C.Divide both sides by cos C:1 = 3 tan C.So, tan C = 1/3.Therefore, angle C is arctan(1/3). Let me compute that. Well, arctan(1/3) is approximately 18.43 degrees, but I'll keep it as arctan(1/3) for exactness.Now, knowing that tan C = 1/3, I can find sin C and cos C.In a right triangle, opposite side is 1, adjacent is 3, so hypotenuse is √(1² + 3²) = √10.Therefore, sin C = 1/√10 and cos C = 3/√10.So, sin C = √10/10 and cos C = 3√10/10.Wait, actually, sin C = 1/√10 = √10/10, and cos C = 3/√10 = 3√10/10. Yeah, that's correct.Now, since we have angle C, we can find angle B because B + C = 3π/4.So, B = 3π/4 - C.Therefore, sin B = sin(3π/4 - C).Using the sine subtraction formula:sin(3π/4 - C) = sin(3π/4) cos C - cos(3π/4) sin C.We know sin(3π/4) = √2/2, cos(3π/4) = -√2/2.So,sin B = (√2/2)(3√10/10) - (-√2/2)(√10/10)= (√2/2)(3√10/10) + (√2/2)(√10/10)= (√2 * 3√10)/(20) + (√2 * √10)/(20)= (3√20)/20 + (√20)/20= (4√20)/20= (4 * 2√5)/20= (8√5)/20= (2√5)/5.So, sin B = 2√5/5.Now, using the Law of Sines, which states that a/sin A = b/sin B = c/sin C.We need to find side a (which is BC), but let's see what we can do.First, let's denote the sides:- a is opposite angle A (π/4)- b is opposite angle B- c is opposite angle CFrom the Law of Sines:a / sin(π/4) = b / sin B = c / sin C.We can write ratios:a / (√2/2) = b / (2√5/5) = c / (√10/10).Let me compute each ratio:First, a / (√2/2) = 2a / √2 = √2 a.Second, b / (2√5/5) = (5b)/(2√5) = (5b√5)/(2*5) = (b√5)/2.Third, c / (√10/10) = 10c / √10 = (10c√10)/10 = c√10.So, all these ratios are equal:√2 a = (b√5)/2 = c√10.Let me denote this common ratio as k.So,√2 a = k,(b√5)/2 = k,c√10 = k.Therefore,a = k / √2,b = 2k / √5,c = k / √10.Now, we also know the area of the triangle is 1. The area can be calculated using (1/2)*b*c*sin A.Let me write that:Area = (1/2)*b*c*sin A = 1.Substituting the known values:(1/2)*(2k / √5)*(k / √10)*sin(π/4) = 1.Simplify step by step.First, multiply the constants:(1/2)*(2k / √5)*(k / √10) = (1/2)*(2k * k) / (√5 * √10) = (k²) / (√5 * √10).Simplify √5 * √10 = √50 = 5√2.So, (k²) / (5√2).Now, multiply by sin(π/4) which is √2/2:(k²) / (5√2) * (√2/2) = (k²) / (5√2) * (√2/2) = (k²) / (10).So, (k²)/10 = 1.Therefore, k² = 10, so k = √10.Since k is a length, it must be positive, so k = √10.Now, recall that a = k / √2 = √10 / √2 = √(10/2) = √5.So, side a, which is BC, is √5.Let me double-check my steps to make sure I didn't make a mistake.1. Started with the given equation, used angle A = π/4, expressed sin(B - C) in terms of angle C, ended up with tan C = 1/3.2. Found sin C and cos C, then found sin B using angle subtraction.3. Applied Law of Sines, set up ratios, introduced k as the common ratio.4. Used the area formula, substituted b and c in terms of k, solved for k, then found a.Everything seems to check out. So, the length of side BC is √5.**Final Answer**The length of side ( BC ) is (boxed{sqrt{5}}).

💡Okay, so I need to find the equation of the normal line to the curve given by ( y = frac{4x - x^2}{4} ) at the point where the abscissa (which is the x-coordinate) is ( x_0 = 2 ). Hmm, let me think about how to approach this step by step.First, I remember that to find the normal line to a curve at a particular point, I need two things: the slope of the tangent line at that point and the coordinates of the point itself. Once I have the slope of the tangent, the slope of the normal line will be the negative reciprocal of that. Then, using the point-slope form of a line, I can write the equation of the normal line.Alright, so let's start by finding the coordinates of the point on the curve where ( x = 2 ). I'll plug ( x = 2 ) into the equation of the curve to find the corresponding y-coordinate.So, ( y = frac{4(2) - (2)^2}{4} ). Let's compute that:( 4(2) = 8 )( (2)^2 = 4 )So, ( y = frac{8 - 4}{4} = frac{4}{4} = 1 ).Okay, so the point is ( (2, 1) ).Next, I need to find the slope of the tangent line to the curve at this point. To do that, I'll find the derivative of the function ( y = frac{4x - x^2}{4} ).Let me rewrite the function to make differentiation easier:( y = frac{4x - x^2}{4} = frac{4x}{4} - frac{x^2}{4} = x - frac{x^2}{4} ).Now, I'll differentiate term by term.The derivative of ( x ) with respect to ( x ) is 1.The derivative of ( frac{x^2}{4} ) with respect to ( x ) is ( frac{2x}{4} = frac{x}{2} ).So, putting it all together, the derivative ( y' ) is:( y' = 1 - frac{x}{2} ).Now, I need to evaluate this derivative at ( x = 2 ) to find the slope of the tangent line at that point.So, ( y'(2) = 1 - frac{2}{2} = 1 - 1 = 0 ).Hmm, the slope of the tangent line is 0. That means the tangent line is horizontal at this point.But I need the slope of the normal line, which is the negative reciprocal of the tangent slope. However, if the tangent slope is 0, the normal line would be vertical because the reciprocal of 0 is undefined, and the negative reciprocal would essentially be an infinite slope, which corresponds to a vertical line.So, a vertical line passing through the point ( (2, 1) ) would have the equation ( x = 2 ).Wait, let me double-check that. If the tangent is horizontal, then the normal should indeed be vertical. And since it passes through ( x = 2 ), the equation is simply ( x = 2 ).Yes, that makes sense. So, the equation of the normal line is ( x = 2 ).But just to make sure I didn't make any mistakes, let me recap:1. Found the point on the curve at ( x = 2 ): ( (2, 1) ).2. Differentiated the function to find the slope of the tangent: ( y' = 1 - frac{x}{2} ).3. Evaluated the derivative at ( x = 2 ): slope is 0.4. Determined the normal slope: since tangent slope is 0, normal is vertical.5. Wrote the equation of the vertical line through ( (2, 1) ): ( x = 2 ).Everything seems to check out. I don't think I made any calculation errors. The logic flows correctly, and the result makes sense geometrically.So, I'm confident that the equation of the normal line is ( x = 2 ).**Final Answer**The equation of the normal line is boxed{x = 2}.

💡Okay, so I have this math problem here: I need to find the value of x that makes the expression (x - 3)/(4x) equal to zero. Hmm, let's see. I remember that for a fraction to be zero, the numerator has to be zero, right? Because if the top part is zero, no matter what the bottom is (as long as it's not zero too, because that would be undefined), the whole fraction becomes zero.So, first, I should set the numerator equal to zero. That means x - 3 = 0. Solving for x, I just add 3 to both sides, and I get x = 3. Okay, that seems straightforward.But wait, I also need to make sure that the denominator isn't zero when x is 3, because if the denominator is zero, the expression would be undefined, not zero. So, let's check the denominator: 4x. If x is 3, then 4 times 3 is 12, which is definitely not zero. So, that's good. The denominator is 12 when x is 3, which means the expression is defined and equals zero.Let me double-check my work. If I plug x = 3 back into the original expression: (3 - 3)/(4*3) = 0/12 = 0. Yep, that works out. So, x = 3 is the correct solution.I think that's all there is to it. I set the numerator equal to zero, solved for x, checked that the denominator isn't zero at that x value, and confirmed by plugging it back in. Everything seems to add up. I don't think I missed anything here. It was a bit of a brain teaser at first, but breaking it down step by step made it manageable.Just to be thorough, let me consider if there are any other possible solutions. Could there be another value of x that makes the numerator zero? Well, the numerator is x - 3, which only equals zero when x is 3. There's no other solution to that equation. So, x = 3 is the only value that satisfies the condition.Also, I should remember that if the denominator were zero, the expression would be undefined, not zero. So, it's important to check that. In this case, since 4x is 12 when x is 3, we're safe. If, for some reason, x were 0, the denominator would be zero, but that's not the case here.I guess another way to think about it is graphing the function. If I graph y = (x - 3)/(4x), I would see that it crosses the x-axis at x = 3 because that's where the numerator is zero. The graph would have a vertical asymptote at x = 0 because the denominator is zero there, making the function undefined. So, visually, it makes sense that x = 3 is where the function equals zero.Overall, I feel confident that x = 3 is the correct answer. It makes sense mathematically, and checking it by substitution confirms that it works. I don't see any errors in my reasoning, so I think I'm all set.