💡First, I need to determine the selling price of the article after a 10% discount is applied to the list price. The selling price is given as Rs. 216.67.Next, I'll calculate the list price by reversing the 10% discount. Since the selling price is 90% of the list price, I can find the list price by dividing the selling price by 0.90.Once I have the list price, I'll calculate the profit made on the cost price by subtracting the cost price from the selling price.Finally, to find the profit percentage, I'll divide the profit by the cost price and multiply by 100 to get the percentage.

💡Alright, so I need to prove that the sequence ( {a_n} ) defined by ( a_n = lfloor n sqrt{2} rfloor ) contains infinitely many powers of 2. Hmm, okay. Let's break this down step by step.First, I know that ( lfloor x rfloor ) denotes the floor function, which gives the greatest integer less than or equal to ( x ). So, ( a_n ) is essentially taking the integer part of ( n sqrt{2} ). Let me start by computing the first few terms of the sequence to see if I can spot a pattern or get some intuition.- For ( n = 1 ): ( a_1 = lfloor 1 times sqrt{2} rfloor = lfloor 1.4142 rfloor = 1 ), which is ( 2^0 ).- For ( n = 2 ): ( a_2 = lfloor 2 times sqrt{2} rfloor = lfloor 2.8284 rfloor = 2 ), which is ( 2^1 ).- For ( n = 3 ): ( a_3 = lfloor 3 times sqrt{2} rfloor = lfloor 4.2426 rfloor = 4 ), which is ( 2^2 ).- For ( n = 4 ): ( a_4 = lfloor 4 times sqrt{2} rfloor = lfloor 5.6568 rfloor = 5 ), which isn't a power of 2.- For ( n = 5 ): ( a_5 = lfloor 5 times sqrt{2} rfloor = lfloor 7.0710 rfloor = 7 ), not a power of 2.- For ( n = 6 ): ( a_6 = lfloor 6 times sqrt{2} rfloor = lfloor 8.4852 rfloor = 8 ), which is ( 2^3 ).Okay, so up to ( n = 6 ), we have powers of 2 at ( n = 1, 2, 3, ) and ( 6 ). That's interesting. So, it seems like every so often, the floor function gives us a power of 2. But how do I prove that this happens infinitely often?I think I need to show that for infinitely many ( n ), ( lfloor n sqrt{2} rfloor = 2^k ) for some integer ( k ). That is, I need to find infinitely many pairs ( (n, k) ) such that ( 2^k leq n sqrt{2} < 2^k + 1 ).Rewriting this inequality, we get:[frac{2^k}{sqrt{2}} leq n < frac{2^k + 1}{sqrt{2}}]Simplifying ( frac{2^k}{sqrt{2}} ), since ( sqrt{2} = 2^{1/2} ), we have:[frac{2^k}{2^{1/2}} = 2^{k - 1/2}]So, the inequality becomes:[2^{k - 1/2} leq n < 2^{k - 1/2} + frac{1}{sqrt{2}}]Now, ( n ) has to be an integer, so the length of the interval ( [2^{k - 1/2}, 2^{k - 1/2} + frac{1}{sqrt{2}}) ) must be at least 1 for there to exist an integer ( n ) in this interval. Let's check the length:[left(2^{k - 1/2} + frac{1}{sqrt{2}}right) - 2^{k - 1/2} = frac{1}{sqrt{2}} approx 0.7071]Hmm, the length is less than 1, which means it's possible that for some ( k ), there is no integer ( n ) in that interval. So, this approach might not directly work. Maybe I need a different strategy.I remember something about Beatty sequences and how they can be used to show that certain sequences contain infinitely many terms of another sequence. A Beatty sequence is of the form ( lfloor n alpha rfloor ) where ( alpha ) is irrational. Since ( sqrt{2} ) is irrational, ( {a_n} ) is a Beatty sequence.One property of Beatty sequences is that if ( alpha ) and ( beta ) are positive irrationals such that ( frac{1}{alpha} + frac{1}{beta} = 1 ), then the sequences ( lfloor n alpha rfloor ) and ( lfloor n beta rfloor ) partition the natural numbers. But I'm not sure if that directly helps here.Alternatively, maybe I can use the concept of density. The sequence ( {a_n} ) has density ( frac{1}{sqrt{2}} ) in the natural numbers, meaning that the proportion of numbers less than ( x ) in the sequence is roughly ( frac{x}{sqrt{2}} ). Since the powers of 2 grow exponentially, their density decreases, but maybe the intersection still occurs infinitely often.Wait, that might not be precise. Let me think differently.Perhaps I can use the fact that ( sqrt{2} ) is irrational, so the fractional parts ( {n sqrt{2}} ) are dense in [0,1). This is due to equidistribution theory. So, for any interval in [0,1), there are infinitely many ( n ) such that ( {n sqrt{2}} ) lies in that interval.If I can find an interval such that when ( {n sqrt{2}} ) is in that interval, ( lfloor n sqrt{2} rfloor ) is a power of 2, then I can conclude there are infinitely many such ( n ).Let me formalize this. Suppose ( 2^k leq n sqrt{2} < 2^k + 1 ). Then, ( lfloor n sqrt{2} rfloor = 2^k ). So, ( n ) must satisfy:[frac{2^k}{sqrt{2}} leq n < frac{2^k + 1}{sqrt{2}}]Let me denote ( m = 2^k ). Then, the inequality becomes:[frac{m}{sqrt{2}} leq n < frac{m + 1}{sqrt{2}}]So, the length of the interval is:[frac{m + 1}{sqrt{2}} - frac{m}{sqrt{2}} = frac{1}{sqrt{2}} approx 0.7071]Again, the length is less than 1, so it's possible that there is no integer ( n ) in this interval for some ( m ). But maybe for infinitely many ( m ), there is an integer ( n ) in this interval.Alternatively, perhaps I can consider the fractional parts ( { frac{m}{sqrt{2}} } ). If ( { frac{m}{sqrt{2}} } ) is less than ( 1 - frac{1}{sqrt{2}} ), then ( lfloor frac{m}{sqrt{2}} rfloor + 1 ) would be the next integer, and maybe that relates to ( n ).Wait, let's think about it differently. Let me define ( n = lfloor frac{2^k}{sqrt{2}} rfloor + 1 ). Then, ( n ) is the smallest integer greater than ( frac{2^k}{sqrt{2}} ). So, ( frac{2^k}{sqrt{2}} < n leq frac{2^k}{sqrt{2}} + 1 ).Multiplying all parts by ( sqrt{2} ), we get:[2^k < n sqrt{2} leq 2^k + sqrt{2}]Since ( sqrt{2} approx 1.4142 ), the upper bound is ( 2^k + 1.4142 ). Therefore, ( lfloor n sqrt{2} rfloor ) could be ( 2^k ) or ( 2^k + 1 ).But we want ( lfloor n sqrt{2} rfloor = 2^k ). So, we need:[n sqrt{2} < 2^k + 1]Which implies:[n < frac{2^k + 1}{sqrt{2}}]But since ( n = lfloor frac{2^k}{sqrt{2}} rfloor + 1 ), we have:[frac{2^k}{sqrt{2}} < n leq frac{2^k}{sqrt{2}} + 1]So, combining the two inequalities:[frac{2^k}{sqrt{2}} < n < frac{2^k + 1}{sqrt{2}}]Which simplifies to:[2^{k - 1/2} < n < 2^{k - 1/2} + frac{1}{sqrt{2}}]Again, the length of this interval is ( frac{1}{sqrt{2}} approx 0.7071 ), which is less than 1. So, it's possible that there is no integer ( n ) in this interval for some ( k ). But maybe for infinitely many ( k ), there is an integer ( n ) in this interval.To ensure that, I need to show that the fractional part ( { frac{2^k}{sqrt{2}} } ) is less than ( 1 - frac{1}{sqrt{2}} ) for infinitely many ( k ). Because if ( { frac{2^k}{sqrt{2}} } < 1 - frac{1}{sqrt{2}} ), then:[frac{2^k}{sqrt{2}} = lfloor frac{2^k}{sqrt{2}} rfloor + { frac{2^k}{sqrt{2}} } < lfloor frac{2^k}{sqrt{2}} rfloor + 1 - frac{1}{sqrt{2}}]So, adding 1 to both sides:[frac{2^k}{sqrt{2}} + 1 < lfloor frac{2^k}{sqrt{2}} rfloor + 2 - frac{1}{sqrt{2}}]But I'm not sure if that helps directly. Maybe I need to consider the properties of the fractional parts of ( frac{2^k}{sqrt{2}} ).Since ( sqrt{2} ) is irrational, the sequence ( { frac{2^k}{sqrt{2}} } ) is dense in [0,1). Therefore, for any subinterval of [0,1), there are infinitely many ( k ) such that ( { frac{2^k}{sqrt{2}} } ) lies in that subinterval. In particular, the subinterval ( [0, 1 - frac{1}{sqrt{2}}) ) has length ( 1 - frac{1}{sqrt{2}} approx 0.2929 ), which is positive. Hence, there are infinitely many ( k ) such that ( { frac{2^k}{sqrt{2}} } < 1 - frac{1}{sqrt{2}} ).For such ( k ), we have:[frac{2^k}{sqrt{2}} = lfloor frac{2^k}{sqrt{2}} rfloor + { frac{2^k}{sqrt{2}} } < lfloor frac{2^k}{sqrt{2}} rfloor + 1 - frac{1}{sqrt{2}}]Adding 1 to both sides:[frac{2^k}{sqrt{2}} + 1 < lfloor frac{2^k}{sqrt{2}} rfloor + 2 - frac{1}{sqrt{2}}]But I'm not sure if this is leading me anywhere. Maybe I should consider the next term in the sequence.Let me define ( n_k = lfloor frac{2^k}{sqrt{2}} rfloor + 1 ). Then, ( n_k ) is the smallest integer greater than ( frac{2^k}{sqrt{2}} ). So, ( n_k ) satisfies:[frac{2^k}{sqrt{2}} < n_k leq frac{2^k}{sqrt{2}} + 1]Multiplying by ( sqrt{2} ):[2^k < n_k sqrt{2} leq 2^k + sqrt{2}]Since ( sqrt{2} approx 1.4142 ), the upper bound is ( 2^k + 1.4142 ). Therefore, ( lfloor n_k sqrt{2} rfloor ) could be ( 2^k ) or ( 2^k + 1 ).But we want ( lfloor n_k sqrt{2} rfloor = 2^k ). So, we need:[n_k sqrt{2} < 2^k + 1]Which implies:[n_k < frac{2^k + 1}{sqrt{2}}]But since ( n_k = lfloor frac{2^k}{sqrt{2}} rfloor + 1 ), we have:[frac{2^k}{sqrt{2}} < n_k leq frac{2^k}{sqrt{2}} + 1]So, combining the two inequalities:[frac{2^k}{sqrt{2}} < n_k < frac{2^k + 1}{sqrt{2}}]Which simplifies to:[2^{k - 1/2} < n_k < 2^{k - 1/2} + frac{1}{sqrt{2}}]Again, the length of this interval is ( frac{1}{sqrt{2}} approx 0.7071 ), which is less than 1. So, it's possible that there is no integer ( n_k ) in this interval for some ( k ). But since the fractional parts ( { frac{2^k}{sqrt{2}} } ) are dense in [0,1), there are infinitely many ( k ) such that ( { frac{2^k}{sqrt{2}} } < 1 - frac{1}{sqrt{2}} ). For these ( k ), the interval ( [2^{k - 1/2}, 2^{k - 1/2} + frac{1}{sqrt{2}}) ) will contain an integer ( n_k ), ensuring that ( lfloor n_k sqrt{2} rfloor = 2^k ).Therefore, there are infinitely many ( k ) such that ( a_{n_k} = 2^k ), meaning the sequence ( {a_n} ) contains infinitely many powers of 2.I think this makes sense. The key idea is using the density of fractional parts to ensure that for infinitely many ( k ), the interval containing ( n ) has an integer, thus making ( a_n ) a power of 2. I should probably write this up more formally to make sure all the steps are clear.

💡Alright, so I have this problem here: Given ( a, b, c ) are positive real numbers, and the equation is ( a x + b[x] - c = 0 ). I need to figure out how many solutions this equation can have. The options are: (A) Must have no solution, (B) Can have at most one solution, (C) Can have finitely many solutions only, (D) Can have infinitely many solutions.First, let me recall what the notation [x] means. It's the integer part of x, also known as the floor function. So, for any real number x, [x] is the greatest integer less than or equal to x. For example, [3.7] is 3, and [5] is 5.Okay, so the equation is ( a x + b[x] - c = 0 ). Let me rewrite this equation to make it clearer: ( a x + b [x] = c ). Since a, b, and c are positive, I can think about how x and [x] relate here.Let me denote [x] as k, where k is an integer. So, k ≤ x < k + 1. Then, x can be written as k + y, where y is the fractional part, so 0 ≤ y < 1.Substituting x = k + y into the equation, I get:( a(k + y) + b k = c )Expanding this, it becomes:( a k + a y + b k = c )Combining like terms:( (a + b)k + a y = c )Now, I can solve for y:( a y = c - (a + b)k )( y = frac{c - (a + b)k}{a} )Since y must satisfy 0 ≤ y < 1, I can write:( 0 ≤ frac{c - (a + b)k}{a} < 1 )Multiplying all parts by a (since a is positive, the inequality signs don't change):( 0 ≤ c - (a + b)k < a )Let me rearrange these inequalities:First inequality: ( c - (a + b)k ≥ 0 )Which implies:( (a + b)k ≤ c )Second inequality: ( c - (a + b)k < a )Which implies:( c - a < (a + b)k )So combining both inequalities:( c - a < (a + b)k ≤ c )Dividing all parts by (a + b):( frac{c - a}{a + b} < k ≤ frac{c}{a + b} )Now, k is an integer, so I need to find integers k that satisfy this inequality.Let me denote:Lower bound: ( L = frac{c - a}{a + b} )Upper bound: ( U = frac{c}{a + b} )So, k must satisfy L < k ≤ U.Since L and U are real numbers, the number of integers k that satisfy this inequality depends on the interval (L, U].Given that a, b, c are positive, let's see what L and U represent.First, let's compute U - L:( U - L = frac{c}{a + b} - frac{c - a}{a + b} = frac{c - (c - a)}{a + b} = frac{a}{a + b} )Since a and b are positive, ( frac{a}{a + b} ) is less than 1. So, the length of the interval (L, U] is less than 1.This means that there can be at most one integer k in this interval because the interval is less than 1 unit long. So, either there is exactly one integer k or no integer k in this interval.If there is an integer k in this interval, then y is determined uniquely as ( y = frac{c - (a + b)k}{a} ), which will be between 0 and 1, so x = k + y will be a valid solution.If there is no integer k in this interval, then there is no solution.Therefore, the equation can have at most one solution.Wait, let me double-check. Suppose the interval (L, U] contains exactly one integer k. Then, x = k + y is a solution. If the interval doesn't contain any integer, then no solution exists. So, the equation can have either one solution or no solution, but not more than one.So, the correct answer should be (B) Can have at most one solution.But just to make sure, let me think of some examples.Suppose a = 1, b = 1, c = 2.Then, the equation is ( x + [x] = 2 ).Let me solve this. Let x = k + y, 0 ≤ y < 1.Then, equation becomes ( k + y + k = 2 ) => ( 2k + y = 2 ).So, y = 2 - 2k.Since y must be between 0 and 1, 0 ≤ 2 - 2k < 1.So, 2 - 2k ≥ 0 => k ≤ 1.And 2 - 2k < 1 => 2k > 1 => k > 0.5.Since k is integer, k must be 1.So, y = 2 - 2*1 = 0.Thus, x = 1 + 0 = 1.So, x = 1 is the solution.Another example: a = 1, b = 2, c = 3.Equation: ( x + 2[x] = 3 ).Let x = k + y.Equation becomes: ( k + y + 2k = 3 ) => ( 3k + y = 3 ).Thus, y = 3 - 3k.Since 0 ≤ y < 1, 0 ≤ 3 - 3k < 1.So, 3 - 3k ≥ 0 => k ≤ 1.And 3 - 3k < 1 => 3k > 2 => k > 2/3.Thus, k must be 1.Then, y = 3 - 3*1 = 0.So, x = 1 + 0 = 1.Another example: a = 2, b = 3, c = 5.Equation: ( 2x + 3[x] = 5 ).Let x = k + y.Then, 2(k + y) + 3k = 5 => 2k + 2y + 3k = 5 => 5k + 2y = 5.Thus, 2y = 5 - 5k => y = (5 - 5k)/2.Since 0 ≤ y < 1, 0 ≤ (5 - 5k)/2 < 1.Multiply all parts by 2:0 ≤ 5 - 5k < 2.First inequality: 5 - 5k ≥ 0 => k ≤ 1.Second inequality: 5 - 5k < 2 => -5k < -3 => k > 3/5.So, k must be 1.Then, y = (5 - 5*1)/2 = 0.Thus, x = 1 + 0 = 1.Hmm, so in these examples, when there is a solution, it's unique.What if I choose c such that there's no integer k in the interval?Let me pick a = 1, b = 1, c = 1.5.Equation: ( x + [x] = 1.5 ).Let x = k + y.Equation: k + y + k = 1.5 => 2k + y = 1.5.Thus, y = 1.5 - 2k.Since 0 ≤ y < 1, 0 ≤ 1.5 - 2k < 1.First inequality: 1.5 - 2k ≥ 0 => k ≤ 0.75.Second inequality: 1.5 - 2k < 1 => -2k < -0.5 => k > 0.25.So, k must satisfy 0.25 < k ≤ 0.75.But k is integer, so possible k is 0 or 1.But 0.25 < k ≤ 0.75, so k must be 0. But 0 is not greater than 0.25, so no solution.Thus, no solution exists.Another example: a = 1, b = 1, c = 0.5.Equation: ( x + [x] = 0.5 ).x = k + y.Equation: 2k + y = 0.5.Thus, y = 0.5 - 2k.0 ≤ y < 1 => 0 ≤ 0.5 - 2k < 1.First inequality: 0.5 - 2k ≥ 0 => k ≤ 0.25.Second inequality: 0.5 - 2k < 1 => -2k < 0.5 => k > -0.25.So, k must satisfy -0.25 < k ≤ 0.25.But k is integer, so k = 0.Then, y = 0.5 - 2*0 = 0.5.Thus, x = 0 + 0.5 = 0.5.So, x = 0.5 is a solution.Wait, but [0.5] is 0, so plugging back into the equation: 0.5 + 1*0 = 0.5, which is correct.So, in this case, there is a solution.Wait, but in the previous example with c = 1.5, there was no solution because k had to be between 0.25 and 0.75, but k had to be integer, so no solution.So, depending on c, sometimes there is a solution, sometimes not. But when there is a solution, it's unique.Hence, the equation can have at most one solution.Therefore, the correct answer is (B).**Final Answer**boxed{B}

💡Okay, so I have this equation here: |x - log y| = x + 2 log y. Hmm, I need to find the values of x and y that satisfy this equation. The options are given as A) (0,1), B) (0,10), C) (1,0), and D) (0,2). First, I remember that absolute value equations can often be split into two cases: one where the expression inside the absolute value is non-negative, and another where it's negative. So, let me try that approach here.Case 1: x - log y is greater than or equal to zero. In this case, the absolute value |x - log y| would just be x - log y. So, substituting that into the equation, I get:x - log y = x + 2 log yHmm, let me simplify this. If I subtract x from both sides, I get:- log y = 2 log yAdding log y to both sides gives:0 = 3 log yWhich simplifies to log y = 0. Since log y is base 10, unless specified otherwise, log y = 0 means y = 10^0 = 1. So, y is 1. Now, going back to the original condition for this case, x - log y >= 0. Since log y is 0, this simplifies to x >= 0. So, x can be any non-negative number, but looking at the options, x is 0 in option A. So, that seems possible.Case 2: x - log y is less than zero. In this case, the absolute value |x - log y| would be -(x - log y) = -x + log y. Substituting that into the equation, I get:-x + log y = x + 2 log yLet me simplify this. If I bring all terms to one side, I get:-x + log y - x - 2 log y = 0Combining like terms:-2x - log y = 0Which simplifies to:-2x = log yOr, x = - (log y)/2Now, I also need to remember the condition for this case, which is x - log y < 0. Substituting x from above:- (log y)/2 - log y < 0Combining the terms:- (log y)/2 - 2 (log y)/2 = - (3 log y)/2 < 0Multiplying both sides by -2 (which reverses the inequality):3 log y > 0So, log y > 0, which means y > 10^0 = 1. So, y must be greater than 1. Now, let's see if this aligns with the options given.Looking at the options, B is (0,10), D is (0,2). Both y values are greater than 1, so let's check if x is 0 in these cases.From x = - (log y)/2, if x is 0, then:0 = - (log y)/2Which implies log y = 0, so y = 1. But in this case, y must be greater than 1, so y = 1 is not allowed here. Therefore, there's a contradiction. So, in this case, x can't be 0 unless y is 1, which isn't allowed here. Therefore, this case doesn't give us a valid solution from the options provided.So, going back to Case 1, we have y = 1 and x >= 0. Among the options, only A) (0,1) fits because x is 0 and y is 1. The other options either have y not equal to 1 or x not equal to 0, which doesn't satisfy the condition from Case 1.Wait, let me double-check. If I plug in option A into the original equation:|x - log y| = |0 - log 1| = |0 - 0| = 0And the right side is x + 2 log y = 0 + 2*0 = 0So, 0 = 0, which works.What about option B: (0,10)Left side: |0 - log 10| = |0 - 1| = 1Right side: 0 + 2*1 = 2So, 1 ≠ 2, which doesn't work.Option C: (1,0)But log y would be log 0, which is undefined, so that's not valid.Option D: (0,2)Left side: |0 - log 2| = | - log 2| = log 2 ≈ 0.3010Right side: 0 + 2 log 2 ≈ 0.6020So, 0.3010 ≈ 0.6020, which isn't equal.Therefore, only option A works.I think that's thorough enough. I considered both cases, checked the conditions, and verified each option. So, the answer should be A) (0,1).

💡Alright, so I'm trying to solve part (a) of this problem. Let me read it again to make sure I understand.Inside the square (ABCD) with side length 1, there is a convex polygon (M) whose area exceeds (1/2). I need to prove that there exists a line (l), parallel to any chosen side of the square, that intersects the polygon (M) in a segment of length greater than (1/2).Hmm, okay. So, the square has side length 1, so its area is 1. The polygon (M) is convex and has an area greater than (1/2). I need to show that there's a line parallel to one of the sides (so either horizontal or vertical) that cuts through (M) and the segment of intersection is longer than (1/2).Let me think about convex polygons. Since (M) is convex, any line intersecting it will do so in at most two points, right? So, the intersection will be a segment. The length of this segment is what we're concerned with.I remember something about the average width of a convex set. Maybe that could be useful here. If the area is more than (1/2), maybe the average width is more than (1/2) as well. But I'm not sure about that. Let me think.Alternatively, maybe I can use the pigeonhole principle. If the area is more than (1/2), then in some direction, the width must be more than (1/2). But how do I formalize that?Wait, another idea: maybe I can use the fact that for a convex polygon, the area can be expressed as the integral of its width over all directions. But that might be too advanced for this problem.Let me try a simpler approach. Let's consider horizontal lines first. If I can show that there's a horizontal line cutting (M) with a segment longer than (1/2), that would suffice. Similarly, the same argument would apply for vertical lines.So, suppose I consider all horizontal lines (y = k) for (k) between 0 and 1. Each such line intersects (M) in some segment. The length of this segment varies with (k). I need to show that for some (k), the length is greater than (1/2).Since (M) is convex, the function that maps (k) to the length of the intersection segment is continuous and piecewise linear. So, it's a continuous function.Now, the area of (M) can be expressed as the integral of the lengths of these intersection segments from (k = 0) to (k = 1). So,[text{Area of } M = int_{0}^{1} text{length}(y = k) , dk]Given that the area is greater than (1/2), we have:[int_{0}^{1} text{length}(y = k) , dk > frac{1}{2}]If the maximum length of these segments is less than or equal to (1/2), then the integral would be at most:[int_{0}^{1} frac{1}{2} , dk = frac{1}{2}]But this contradicts the fact that the area is greater than (1/2). Therefore, there must be some (k) where the length is greater than (1/2).Wait, that seems too straightforward. Did I miss something? Let me double-check.I assumed that if all lengths were at most (1/2), the integral would be at most (1/2). But since the integral is greater than (1/2), there must be some (k) where the length is greater than (1/2). That makes sense.So, this argument works for horizontal lines. Similarly, it would work for vertical lines as well. Therefore, there exists a line parallel to any chosen side (either horizontal or vertical) that intersects (M) in a segment longer than (1/2).I think that's the solution. It uses the integral of the lengths of the intersection segments to relate to the area, and then applies the pigeonhole principle to show that at least one of those segments must be longer than (1/2).**Final Answer**a) boxed{text{Such a line } l text{ exists.}}b) boxed{text{Such a line } l text{ exists.}}

💡Alright, so I've got this problem here about a function f defined on all real numbers. It says that f is even, which I remember means that f(-x) = f(x) for all x. Okay, so the function is symmetric around the y-axis. Then it mentions that f(x + 2) = f(x), which tells me that the function is periodic with period 2. So every 2 units, the function repeats itself. Got it.Next, it says that f is monotonically increasing on the interval [0, 1]. Monotonically increasing means that as x increases, f(x) also increases. So from x = 0 to x = 1, the function goes up without any dips or peaks. That should be useful for comparing values.Now, the problem defines three values: a = f(3), b = f(√2), and c = f(2). I need to figure out the relationship between a, b, and c. The options are given, so I can check which one fits after I work through it.Let me start by recalling the properties of the function:1. It's even: f(-x) = f(x).2. It's periodic with period 2: f(x + 2) = f(x).3. It's increasing on [0, 1].Since f is periodic with period 2, I can use this to simplify the arguments of the function. For example, f(3) can be simplified because 3 is more than 2. Let me subtract 2 from 3 to get back into the principal period, which is from 0 to 2. So f(3) = f(3 - 2) = f(1). That gives me a = f(1).Similarly, for c = f(2), I can use the periodicity again. Since the period is 2, f(2) = f(0). So c = f(0).Now, for b = f(√2). √2 is approximately 1.414, which is between 1 and 2. Since the function is even, f(√2) = f(-√2). But since f is periodic, f(-√2) = f(-√2 + 2) because adding the period 2 doesn't change the value. So f(-√2 + 2) = f(2 - √2). 2 - √2 is approximately 0.586, which is between 0 and 1. So b = f(2 - √2).Now, I have:- a = f(1)- b = f(2 - √2)- c = f(0)Since the function is monotonically increasing on [0, 1], the values of f at points in [0, 1] increase as the input increases. So, f(0) < f(2 - √2) < f(1). Plugging back in terms of a, b, c:c < b < aWhich translates to a > b > c.Let me double-check to make sure I didn't make any mistakes. Starting with a = f(3), since the period is 2, subtracting 2 gives f(1). That seems right. For c = f(2), since the period is 2, f(2) = f(0). That also makes sense because f(0 + 2) = f(0). For b = f(√2), since √2 is about 1.414, which is more than 1 but less than 2. Using the even property, f(√2) = f(-√2), but since the function is periodic, f(-√2) = f(-√2 + 2) = f(2 - √2). 2 - √2 is approximately 0.586, which is between 0 and 1, so we can use the increasing property there. Since 0 < 2 - √2 < 1, f(0) < f(2 - √2) < f(1), so c < b < a. Therefore, a > b > c.Yes, that seems consistent. So the correct relationship is a > b > c, which is option C.