💡To determine the number of nonoverlapping unit squares in figure (50), we start by analyzing the given sequence: (1), (7), (19), and (37). First, we calculate the differences between consecutive terms:- (7 - 1 = 6)- (19 - 7 = 12)- (37 - 19 = 18)The differences increase by (6) each time, indicating a quadratic sequence. We assume the general form of the quadratic function is (f(n) = an^2 + bn + c).Using the given values:1. For (n = 0): (f(0) = 1) implies (c = 1).2. For (n = 1): (f(1) = 7) implies (a + b + 1 = 7), so (a + b = 6).3. For (n = 2): (f(2) = 19) implies (4a + 2b + 1 = 19), so (4a + 2b = 18).Solving the system of equations:- From (a + b = 6)- From (4a + 2b = 18), simplifying gives (2a + b = 9)Subtracting the first equation from the second:- (2a + b - (a + b) = 9 - 6)- (a = 3)Substituting (a = 3) back into (a + b = 6):- (3 + b = 6)- (b = 3)Thus, the quadratic function is (f(n) = 3n^2 + 3n + 1).Finally, we calculate (f(50)):- (f(50) = 3(50)^2 + 3(50) + 1 = 3(2500) + 150 + 1 = 7500 + 150 + 1 = 7651)The number of nonoverlapping unit squares in figure (50) is (boxed{B}).

💡Alright, so I have this problem about an ellipse and a circle intersecting at four distinct points. I need to find the range of the eccentricity ( e ) of the ellipse. Hmm, okay, let me start by recalling what I know about ellipses and circles.The equation of the ellipse is given as ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ) where ( a > b > 0 ). The circle has the equation ( x^2 + y^2 = left( frac{b}{2} + c right)^2 ), and ( c ) is the semi-focal distance of the ellipse. First, I remember that for an ellipse, the relationship between ( a ), ( b ), and ( c ) is ( c^2 = a^2 - b^2 ). Also, the eccentricity ( e ) is defined as ( e = frac{c}{a} ). So, ( e ) is a measure of how "stretched" the ellipse is. The closer ( e ) is to 1, the more elongated the ellipse is, and the closer it is to 0, the more circular it is.Now, the circle is centered at the origin, just like the ellipse. The radius of the circle is ( frac{b}{2} + c ). Since both the ellipse and the circle are centered at the origin, their intersection points will depend on the relative sizes of the ellipse and the circle.The problem states that they have four distinct intersection points. That means the circle intersects the ellipse at four different points. For this to happen, the circle must be neither too small nor too large relative to the ellipse. If the circle were too small, it might not intersect the ellipse at all or only at two points. If it were too large, it might intersect the ellipse at more points or maybe not at all depending on the specific parameters.So, to have four distinct intersection points, the radius of the circle must be such that it cuts through the ellipse in four places. That suggests that the radius must be greater than the semi-minor axis ( b ) but less than the semi-major axis ( a ). Wait, is that correct? Let me think.Actually, the semi-minor axis is ( b ), and the semi-major axis is ( a ). The circle has a radius of ( frac{b}{2} + c ). So, for the circle to intersect the ellipse at four points, the radius must be greater than the semi-minor axis ( b ) but less than the semi-major axis ( a ). Wait, no, that might not necessarily be the case.Let me visualize this. The ellipse is stretched along the x-axis because ( a > b ). The circle is also centered at the origin. So, if the radius of the circle is too small, it might lie entirely within the ellipse, not intersecting it at all. If the radius is just right, it will intersect the ellipse at four points. If the radius is too large, it might intersect the ellipse at more points or maybe not intersect it at all if it's too large? Hmm, actually, if the radius is larger than the ellipse's semi-major axis, the circle would encompass the ellipse, but since the ellipse is not a circle, they might still intersect at four points. Wait, maybe not necessarily.Wait, perhaps it's better to consider the conditions mathematically. Let me set up the equations.We have the ellipse equation: ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ).And the circle equation: ( x^2 + y^2 = left( frac{b}{2} + c right)^2 ).To find the intersection points, we can solve these two equations simultaneously. Let me subtract the ellipse equation from the circle equation to eliminate ( x^2 ) and ( y^2 ).Wait, actually, let me express ( y^2 ) from the ellipse equation:( y^2 = b^2 left( 1 - frac{x^2}{a^2} right) ).Substitute this into the circle equation:( x^2 + b^2 left( 1 - frac{x^2}{a^2} right) = left( frac{b}{2} + c right)^2 ).Simplify this equation:( x^2 + b^2 - frac{b^2 x^2}{a^2} = left( frac{b}{2} + c right)^2 ).Combine like terms:( x^2 left( 1 - frac{b^2}{a^2} right) + b^2 = left( frac{b}{2} + c right)^2 ).Let me compute ( 1 - frac{b^2}{a^2} ). Since ( c^2 = a^2 - b^2 ), this becomes ( frac{c^2}{a^2} = e^2 ). So, ( 1 - frac{b^2}{a^2} = e^2 ).Therefore, the equation becomes:( e^2 x^2 + b^2 = left( frac{b}{2} + c right)^2 ).Let me rearrange this:( e^2 x^2 = left( frac{b}{2} + c right)^2 - b^2 ).Compute the right-hand side:( left( frac{b}{2} + c right)^2 - b^2 = frac{b^2}{4} + b c + c^2 - b^2 = - frac{3b^2}{4} + b c + c^2 ).So, we have:( e^2 x^2 = - frac{3b^2}{4} + b c + c^2 ).For real solutions, the right-hand side must be non-negative because the left-hand side is non-negative (since ( e^2 ) and ( x^2 ) are both non-negative). Therefore:( - frac{3b^2}{4} + b c + c^2 geq 0 ).Let me write this inequality:( c^2 + b c - frac{3b^2}{4} geq 0 ).This is a quadratic in terms of ( c ). Let me write it as:( c^2 + b c - frac{3b^2}{4} geq 0 ).Let me solve this quadratic inequality. The quadratic equation ( c^2 + b c - frac{3b^2}{4} = 0 ) can be solved using the quadratic formula:( c = frac{ -b pm sqrt{b^2 + 3b^2} }{2} = frac{ -b pm sqrt{4b^2} }{2} = frac{ -b pm 2b }{2} ).So, the roots are:1. ( c = frac{ -b + 2b }{2} = frac{b}{2} ).2. ( c = frac{ -b - 2b }{2} = - frac{3b}{2} ).Since ( c ) is a distance, it must be positive, so we discard the negative root. Therefore, the critical point is at ( c = frac{b}{2} ).The quadratic ( c^2 + b c - frac{3b^2}{4} ) opens upwards (since the coefficient of ( c^2 ) is positive). Therefore, the inequality ( c^2 + b c - frac{3b^2}{4} geq 0 ) holds when ( c leq - frac{3b}{2} ) or ( c geq frac{b}{2} ). But since ( c ) is positive, we only consider ( c geq frac{b}{2} ).So, for the equation to have real solutions, ( c geq frac{b}{2} ). But we also need four distinct intersection points, which implies that the equation must have two distinct solutions for ( x ) (since for each ( x ), there are two ( y ) values, positive and negative). Therefore, the quadratic in ( x^2 ) must have two distinct real solutions, meaning the right-hand side must be positive, not just non-negative.Therefore, ( c > frac{b}{2} ).So, the first condition is ( c > frac{b}{2} ).Now, let's express this in terms of ( e ). Remember that ( c = a e ) and ( c^2 = a^2 - b^2 ). So, ( b^2 = a^2 (1 - e^2) ).Given that ( c > frac{b}{2} ), substitute ( c = a e ) and ( b = a sqrt{1 - e^2} ):( a e > frac{a sqrt{1 - e^2}}{2} ).Divide both sides by ( a ) (since ( a > 0 )):( e > frac{sqrt{1 - e^2}}{2} ).Square both sides to eliminate the square root:( e^2 > frac{1 - e^2}{4} ).Multiply both sides by 4:( 4 e^2 > 1 - e^2 ).Bring all terms to one side:( 4 e^2 + e^2 - 1 > 0 Rightarrow 5 e^2 - 1 > 0 Rightarrow e^2 > frac{1}{5} Rightarrow e > frac{sqrt{5}}{5} ).So, that's the first condition: ( e > frac{sqrt{5}}{5} ).Now, we need another condition to ensure that the circle doesn't encompass the ellipse entirely or something like that. Since the circle has a radius of ( frac{b}{2} + c ), we need this radius to be less than the semi-major axis ( a ) of the ellipse. Otherwise, the circle would be too large, and maybe they wouldn't intersect at four points.Wait, actually, if the radius is larger than ( a ), the circle would encompass the ellipse, but since the ellipse is not a circle, they might still intersect at four points. Hmm, maybe that's not the right way to think about it.Alternatively, perhaps the radius should be less than ( a ) to ensure that the circle doesn't completely enclose the ellipse, which might lead to more intersection points or something else. Let me think.Wait, actually, if the radius is too large, the circle might intersect the ellipse at more than four points, but the problem states that they have four distinct intersection points. So, maybe the radius must be less than ( a ) to prevent the circle from encompassing the ellipse entirely.Alternatively, perhaps the radius must be less than ( a ) to ensure that the circle doesn't become too large and cause the ellipse and circle to intersect in more than four points. Hmm, I'm not entirely sure, but let's proceed.So, let's set ( frac{b}{2} + c < a ).Again, substitute ( c = a e ) and ( b = a sqrt{1 - e^2} ):( frac{a sqrt{1 - e^2}}{2} + a e < a ).Divide both sides by ( a ):( frac{sqrt{1 - e^2}}{2} + e < 1 ).Let me write this as:( e + frac{sqrt{1 - e^2}}{2} < 1 ).Let me denote ( sqrt{1 - e^2} = f ). Then, ( f = sqrt{1 - e^2} ), so ( f^2 = 1 - e^2 ), and ( e^2 = 1 - f^2 ).But maybe that's complicating things. Alternatively, let me isolate the square root term:( frac{sqrt{1 - e^2}}{2} < 1 - e ).Multiply both sides by 2:( sqrt{1 - e^2} < 2(1 - e) ).Now, square both sides to eliminate the square root:( 1 - e^2 < 4(1 - 2e + e^2) ).Expand the right-hand side:( 1 - e^2 < 4 - 8e + 4e^2 ).Bring all terms to the left-hand side:( 1 - e^2 - 4 + 8e - 4e^2 < 0 Rightarrow -3 - 5e^2 + 8e < 0 ).Multiply both sides by -1 (which reverses the inequality):( 3 + 5e^2 - 8e > 0 ).Let me rearrange this:( 5e^2 - 8e + 3 > 0 ).This is a quadratic inequality. Let's find the roots of the quadratic equation ( 5e^2 - 8e + 3 = 0 ).Using the quadratic formula:( e = frac{8 pm sqrt{64 - 60}}{10} = frac{8 pm sqrt{4}}{10} = frac{8 pm 2}{10} ).So, the roots are:1. ( e = frac{8 + 2}{10} = frac{10}{10} = 1 ).2. ( e = frac{8 - 2}{10} = frac{6}{10} = frac{3}{5} ).The quadratic ( 5e^2 - 8e + 3 ) opens upwards (since the coefficient of ( e^2 ) is positive). Therefore, the inequality ( 5e^2 - 8e + 3 > 0 ) holds when ( e < frac{3}{5} ) or ( e > 1 ). But since the eccentricity ( e ) of an ellipse is always less than 1, we only consider ( e < frac{3}{5} ).Therefore, the second condition is ( e < frac{3}{5} ).Combining both conditions, we have:( frac{sqrt{5}}{5} < e < frac{3}{5} ).So, the range of the eccentricity ( e ) is ( left( frac{sqrt{5}}{5}, frac{3}{5} right) ).Let me just double-check my steps to make sure I didn't make any mistakes.1. I started by setting up the equations of the ellipse and the circle.2. I subtracted the ellipse equation from the circle equation to eliminate ( y^2 ).3. I expressed everything in terms of ( e ) and ( b ), leading to a quadratic inequality in ( c ).4. I solved for ( c ) and translated that into a condition on ( e ), resulting in ( e > frac{sqrt{5}}{5} ).5. Then, I considered the condition that the radius of the circle must be less than ( a ), leading to another inequality.6. I solved that inequality, which gave me ( e < frac{3}{5} ).7. Combining both conditions, I found the range for ( e ).Everything seems to check out. I think the answer is ( left( frac{sqrt{5}}{5}, frac{3}{5} right) ), which corresponds to option D.**Final Answer**boxed{D}

💡Okay, so I need to prove that we can't find consecutive factorials with first digits 1, 2, ..., 9. Hmm, factorials grow really fast, right? Like, 1! is 1, 2! is 2, 3! is 6, 4! is 24, 5! is 120, and so on. The first digits of these factorials are 1, 2, 6, 2, 1, etc. I notice that the first digits don't follow a simple pattern, but they do seem to cycle through different numbers as the factorials get larger.So, the problem is asking to show that there isn't a sequence of nine consecutive factorials where each factorial starts with a different digit from 1 to 9. That means, for example, we can't have 1!, 2!, 3!, ..., 9! each starting with a unique digit from 1 to 9. But wait, 1! is 1, 2! is 2, 3! is 6, 4! is 24, 5! is 120, 6! is 720, 7! is 5040, 8! is 40320, and 9! is 362880. Looking at these, the first digits are 1, 2, 6, 2, 1, 7, 5, 4, 3. So, in this case, the first digits are not all unique from 1 to 9. For example, 2 appears twice, and 3, 4, 5, 6, 7, 8, 9 are not all covered.But maybe there's another set of consecutive factorials where this could happen? Let's think about larger factorials. Factorials grow exponentially, so their first digits might cycle through different numbers more frequently. But I remember something about Benford's Law, which says that in many naturally occurring collections of numbers, the leading digit is likely to be small. So, smaller digits like 1, 2, and 3 are more common as leading digits than larger digits like 7, 8, and 9.If Benford's Law applies here, then the leading digits of factorials might not be uniformly distributed, making it less likely for all digits 1 through 9 to appear as leading digits in consecutive factorials. But I'm not sure if Benford's Law directly applies to factorials. Maybe I need to look into how the leading digits of factorials behave.I recall that the leading digit of a number can be determined by looking at its logarithm. Specifically, if we take the logarithm base 10 of a number, the fractional part can tell us the leading digit. For example, if log10(n) = k + f, where k is an integer and 0 ≤ f < 1, then the leading digit d of n is given by d = floor(10^f). So, if we can analyze the fractional parts of the logarithms of consecutive factorials, we might be able to determine if their leading digits can cover all digits from 1 to 9.Let's denote n! as the factorial of n. Then, log10(n!) = log10(1) + log10(2) + ... + log10(n). This is known as the logarithmic factorial. The fractional part of log10(n!) determines the leading digit of n!. So, if we look at the sequence of fractional parts of log10(n!), we can see how the leading digits change as n increases.Now, if we consider consecutive factorials, say n!, (n+1)!, (n+2)!, ..., (n+8)!, their leading digits would correspond to the fractional parts of log10(n!), log10((n+1)!), ..., log10((n+8)!). The difference between log10((n+i)!) and log10((n+i-1)!) is log10(n+i). So, the fractional parts of these logarithms increase by log10(n+i) each time.If the fractional parts of these logarithms are distributed uniformly in the interval [0,1), then the leading digits would also be distributed according to Benford's Law. However, for leading digits to cover all digits from 1 to 9 in consecutive factorials, the fractional parts would need to cycle through intervals corresponding to each leading digit without overlap or repetition.But given that the increments log10(n+i) are not necessarily small or uniformly distributed, the fractional parts might not cover the entire [0,1) interval in a way that allows all leading digits from 1 to 9 to appear consecutively. In fact, as n increases, log10(n+i) increases, so the increments become larger, making it less likely for the fractional parts to cover all necessary intervals.Moreover, the leading digit cycles in factorials are influenced by the multiplicative nature of factorials. Each time you multiply by the next integer, you're effectively adding its logarithm to the total. This can cause the fractional parts to jump by varying amounts, sometimes skipping over certain intervals that correspond to specific leading digits.To make this more concrete, let's consider the behavior of the leading digits of factorials. For small n, the leading digits change erratically, but as n becomes large, the leading digits start to follow a more predictable pattern due to the properties of logarithms and the distribution of their fractional parts. However, even with this predictability, having nine consecutive factorials each starting with a unique digit from 1 to 9 seems highly unlikely because the increments in the logarithms would need to align perfectly to hit each necessary interval.Additionally, considering the rapid growth of factorials, the number of digits in n! increases roughly logarithmically with n. This means that the leading digit changes are influenced by the addition of each new term in the factorial product, but these changes don't necessarily cover all possible leading digits in a consecutive sequence.Another angle to approach this problem is by considering the ratio between consecutive factorials. We know that (n+1)! = (n+1) * n!. So, the ratio between consecutive factorials is n+1. If we look at the leading digits, multiplying by n+1 can significantly change the leading digit, but it's not guaranteed to cycle through all digits from 1 to 9 in a consecutive manner.For example, if n! starts with digit d, then (n+1)! starts with a digit that depends on d multiplied by (n+1). This multiplication can cause the leading digit to jump to a different digit, but it's not systematic enough to ensure that all digits from 1 to 9 are covered in consecutive factorials.Furthermore, considering the properties of logarithms and leading digits, the distribution of leading digits in factorials is not uniform. Smaller leading digits are more common, as per Benford's Law, which suggests that digits like 1, 2, and 3 appear more frequently as leading digits than 7, 8, and 9. This non-uniform distribution makes it even less likely for all digits from 1 to 9 to appear consecutively as leading digits in factorials.In summary, the combination of the rapid growth of factorials, the properties of logarithms, and the non-uniform distribution of leading digits as per Benford's Law makes it highly improbable, if not impossible, to find nine consecutive factorials where each starts with a unique digit from 1 to 9. The necessary alignment of fractional parts in the logarithms of consecutive factorials to cover all leading digits in sequence is not achievable due to the nature of factorial growth and logarithmic distribution.

💡First, I need to evaluate the expression (frac{x^6 - 17x^3 + 72}{x^3 - 8}) when (x = 6).I notice that both the numerator and the denominator are polynomials involving (x^3). This suggests that I can factor the numerator in terms of (x^3 - 8), which is the denominator.Let me attempt to factor the numerator:[x^6 - 17x^3 + 72]I can rewrite (x^6) as ((x^3)^2), so the expression becomes:[(x^3)^2 - 17x^3 + 72]Now, I look for two numbers that multiply to 72 and add up to -17. These numbers are -8 and -9.So, I can factor the quadratic expression as:[(x^3 - 8)(x^3 - 9)]Now, the original expression becomes:[frac{(x^3 - 8)(x^3 - 9)}{x^3 - 8}]Since (x^3 - 8) is present in both the numerator and the denominator, I can cancel them out, provided (x^3 - 8 neq 0). This simplifies the expression to:[x^3 - 9]Now, substitute (x = 6) into the simplified expression:[6^3 - 9 = 216 - 9 = 207]Therefore, the value of the expression when (x = 6) is 207.

💡Okay, so I have this problem here: I need to find which interval contains a root of the function f(x) = e^x + x² - 4. The options are A: (-3, -2), B: (-1, 0), C: (0, 1), and D: (1, 2). Hmm, let me think about how to approach this.First, I remember that a root of a function is a value x where f(x) = 0. So, I need to find an interval where the function changes sign, meaning it goes from positive to negative or vice versa. This is because, according to the Intermediate Value Theorem, if a function is continuous on an interval [a, b] and takes on values f(a) and f(b) at each end of the interval, then it also takes on any value between f(a) and f(b). So, if f(a) and f(b) have opposite signs, there must be at least one root between a and b.Alright, so the function here is f(x) = e^x + x² - 4. Let me check if this function is continuous. Well, e^x is continuous everywhere, x² is a polynomial which is also continuous everywhere, and constants are continuous. So, the sum of continuous functions is continuous, which means f(x) is continuous on the entire real line. That means the Intermediate Value Theorem applies to any interval I choose.Now, I need to evaluate f(x) at the endpoints of each interval given in the options to see where the function changes sign. Let's go through each option one by one.Starting with option A: (-3, -2). Let me compute f(-3) and f(-2).Calculating f(-3):f(-3) = e^(-3) + (-3)^2 - 4e^(-3) is approximately 0.0498, (-3)^2 is 9, so:f(-3) ≈ 0.0498 + 9 - 4 ≈ 0.0498 + 5 ≈ 5.0498That's positive.Calculating f(-2):f(-2) = e^(-2) + (-2)^2 - 4e^(-2) is approximately 0.1353, (-2)^2 is 4, so:f(-2) ≈ 0.1353 + 4 - 4 ≈ 0.1353Still positive.So, f(-3) ≈ 5.05 and f(-2) ≈ 0.1353. Both are positive, so the function doesn't change sign in this interval. Therefore, there's no root in (-3, -2). So, option A is out.Moving on to option B: (-1, 0). Let's compute f(-1) and f(0).Calculating f(-1):f(-1) = e^(-1) + (-1)^2 - 4e^(-1) is approximately 0.3679, (-1)^2 is 1, so:f(-1) ≈ 0.3679 + 1 - 4 ≈ 0.3679 - 3 ≈ -2.6321That's negative.Calculating f(0):f(0) = e^0 + 0^2 - 4e^0 is 1, so:f(0) = 1 + 0 - 4 = -3Still negative.So, f(-1) ≈ -2.63 and f(0) = -3. Both are negative, so the function doesn't change sign here either. Therefore, no root in (-1, 0). Option B is also out.Next, option C: (0, 1). Let's compute f(0) and f(1).We already calculated f(0) above: f(0) = -3.Calculating f(1):f(1) = e^1 + (1)^2 - 4e^1 is approximately 2.7183, so:f(1) ≈ 2.7183 + 1 - 4 ≈ 3.7183 - 4 ≈ -0.2817Still negative.So, f(0) = -3 and f(1) ≈ -0.2817. Both negative again. No sign change, so no root in (0, 1). Option C is out.Finally, option D: (1, 2). Let's compute f(1) and f(2).We already have f(1) ≈ -0.2817.Calculating f(2):f(2) = e^2 + (2)^2 - 4e^2 is approximately 7.3891, so:f(2) ≈ 7.3891 + 4 - 4 ≈ 7.3891That's positive.So, f(1) ≈ -0.2817 (negative) and f(2) ≈ 7.3891 (positive). The function changes sign from negative to positive between x=1 and x=2. Therefore, by the Intermediate Value Theorem, there must be at least one root in the interval (1, 2).Just to be thorough, let me check if there could be more than one root. The function f(x) = e^x + x² - 4 is a combination of an exponential function and a quadratic function. Both e^x and x² are increasing functions for x > 0, so their sum should also be increasing. Therefore, f(x) is strictly increasing for x > 0, which means there can be only one root in the positive side. Since we already found a sign change in (1, 2), that's the only root in the positive side.What about negative x? We saw that f(-3) and f(-2) are positive, and f(-1) is negative. Wait, hold on, f(-1) is negative, but f(-2) is positive. So, between x=-2 and x=-1, the function goes from positive to negative. That means there's another sign change there, implying another root in (-2, -1). But wait, the options don't include (-2, -1); the closest is option A: (-3, -2) and option B: (-1, 0). So, actually, there might be a root in (-2, -1), but since that interval isn't one of the options, we can disregard it for this problem.Therefore, focusing on the given options, the only interval where a root exists is (1, 2), which is option D.Just to double-check, maybe I should compute f(1.5) to see if it's positive or negative.Calculating f(1.5):f(1.5) = e^(1.5) + (1.5)^2 - 4e^(1.5) is approximately 4.4817, (1.5)^2 is 2.25, so:f(1.5) ≈ 4.4817 + 2.25 - 4 ≈ 6.7317 - 4 ≈ 2.7317That's positive. So, f(1) ≈ -0.28 and f(1.5) ≈ 2.73. So, the function crosses zero between 1 and 1.5, which is within (1, 2). That confirms it.Alternatively, maybe I can try f(1.2):f(1.2) = e^(1.2) + (1.2)^2 - 4e^(1.2) ≈ 3.3201, (1.2)^2 = 1.44f(1.2) ≈ 3.3201 + 1.44 - 4 ≈ 4.7601 - 4 ≈ 0.7601Still positive.f(1.1):e^(1.1) ≈ 3.0042, (1.1)^2 = 1.21f(1.1) ≈ 3.0042 + 1.21 - 4 ≈ 4.2142 - 4 ≈ 0.2142Positive.f(1.05):e^(1.05) ≈ e^1 * e^0.05 ≈ 2.7183 * 1.0513 ≈ 2.8577(1.05)^2 ≈ 1.1025f(1.05) ≈ 2.8577 + 1.1025 - 4 ≈ 3.9602 - 4 ≈ -0.0398Negative.So, f(1.05) ≈ -0.04 and f(1.1) ≈ 0.21. So, the root is between 1.05 and 1.1, which is still within (1, 2). So, definitely, the root is in (1, 2).I think that's thorough enough. I've checked the function values at the endpoints of each interval, found that only in (1, 2) does the function change sign, and even checked some intermediate points to confirm. So, I'm confident that the correct answer is D: (1, 2).**Final Answer**boxed{D}

💡Okay, so I have this equation to solve: (4^x + 4^{-x}) - 2(2^x + 2^{-x}) + 2 = 0. Hmm, it looks a bit complicated with all these exponents, but maybe I can simplify it somehow. Let me think about how to approach this.First, I notice that 4^x can be written as (2^2)^x, which is the same as 2^{2x}. Similarly, 4^{-x} would be 2^{-2x}. So, maybe I can rewrite the equation in terms of 2^x and 2^{-x} to make it easier to handle. Let me try that.So, replacing 4^x with 2^{2x} and 4^{-x} with 2^{-2x}, the equation becomes:(2^{2x} + 2^{-2x}) - 2(2^x + 2^{-x}) + 2 = 0.Hmm, that still looks a bit messy, but maybe I can use a substitution to simplify it. Let me set t = 2^x + 2^{-x}. That seems like a good substitution because both 2^{2x} and 2^{-2x} can be expressed in terms of t.Wait, how can I express 2^{2x} + 2^{-2x} in terms of t? Let me recall that (2^x + 2^{-x})^2 = 2^{2x} + 2*2^x*2^{-x} + 2^{-2x}. Simplifying that, it's 2^{2x} + 2 + 2^{-2x}. So, that means 2^{2x} + 2^{-2x} = (2^x + 2^{-x})^2 - 2, which is t^2 - 2.Great, so substituting back into the equation, I have:(t^2 - 2) - 2t + 2 = 0.Let me simplify this equation step by step. First, expand the terms:t^2 - 2 - 2t + 2 = 0.Now, combine like terms. The -2 and +2 cancel each other out:t^2 - 2t = 0.So, the equation simplifies to t^2 - 2t = 0. I can factor this equation:t(t - 2) = 0.Setting each factor equal to zero gives the solutions t = 0 or t = 2. But wait, t was defined as 2^x + 2^{-x}. Since 2^x is always positive for any real x, 2^{-x} is also positive. Therefore, the sum of two positive numbers can't be zero. So, t = 0 is not possible. That leaves us with t = 2.So, t = 2 means that 2^x + 2^{-x} = 2. Let me write that down:2^x + 2^{-x} = 2.Hmm, this looks like another equation I can solve. Maybe I can make another substitution here. Let me set y = 2^x. Then, 2^{-x} is 1/y. So, substituting, the equation becomes:y + 1/y = 2.Multiplying both sides by y to eliminate the denominator:y^2 + 1 = 2y.Bringing all terms to one side:y^2 - 2y + 1 = 0.This is a quadratic equation in terms of y. Let me factor it:(y - 1)^2 = 0.So, the solution is y = 1. Since y = 2^x, this means:2^x = 1.I know that 2^0 = 1, so x must be 0. Therefore, the solution to the original equation is x = 0.Let me double-check my steps to make sure I didn't make any mistakes. Starting from the substitution t = 2^x + 2^{-x}, which led me to t^2 - 2t = 0. Solving that gave t = 2, which then led to 2^x + 2^{-x} = 2. Substituting y = 2^x gave me a quadratic equation that factored nicely, resulting in y = 1, so x = 0. Plugging x = 0 back into the original equation:4^0 + 4^{-0} - 2(2^0 + 2^{-0}) + 2 = 1 + 1 - 2(1 + 1) + 2 = 2 - 4 + 2 = 0.Yep, that works out. So, x = 0 is indeed the solution. I don't think I missed anything here, so I'm confident that the solution set is just {0}.**Final Answer**The solution set of the equation is boxed{{0}}.