💡First, I need to multiply the coefficients of the two numbers. The coefficients are -2 and 4. Multiplying these gives -8.Next, I should add the exponents of the powers of 10. The exponents are 4 and 5, which add up to 9. This means the power of 10 in the final result is 10^9.Combining these results, the product of (-2 × 10^4) and (4 × 10^5) is -8 × 10^9.Looking at the answer choices, option D matches this result.

💡Okay, so I need to factorize the expression ( a^2 + 3a ). Hmm, factorizing usually means breaking down an expression into simpler parts that can be multiplied together to get the original expression. I remember that when factorizing, I should look for common factors in all the terms.Let me look at the terms in ( a^2 + 3a ). The first term is ( a^2 ), which is ( a ) multiplied by itself, and the second term is ( 3a ), which is 3 multiplied by ( a ). I notice that both terms have an ( a ) in them. So, ( a ) is a common factor.If I factor out the ( a ) from both terms, what do I get? Let's see:- From ( a^2 ), factoring out ( a ) leaves me with ( a ).- From ( 3a ), factoring out ( a ) leaves me with 3.So, if I take ( a ) as the common factor, the expression becomes ( a times (a + 3) ). Let me check if that's correct by expanding it back:( a times (a + 3) = a times a + a times 3 = a^2 + 3a ).Yes, that matches the original expression. So, I think I've factorized it correctly. There doesn't seem to be any other common factors between ( a ) and 3, so this should be the simplest form.Wait, just to make sure, could there be another way to factorize this? Maybe by grouping or something else? Let me think. Since there are only two terms, grouping doesn't really apply here. The only common factor is ( a ), so I don't think there's another way to factorize this expression further.Therefore, I'm confident that the factorized form of ( a^2 + 3a ) is ( a(a + 3) ).

💡First, I need to find the derivative of the function ( y = x^3 - 1 ) to determine the slope of the tangent line at ( x = 1 ).Calculating the derivative:[frac{dy}{dx} = 3x^2]Next, I evaluate the derivative at ( x = 1 ) to find the slope at that specific point:[frac{dy}{dx}Big|_{x=1} = 3(1)^2 = 3]Now that I have the slope of the tangent line, I need to find a point on the curve where ( x = 1 ). Substituting ( x = 1 ) into the original function:[y = (1)^3 - 1 = 0]So, the point is ( (1, 0) ).Using the point-slope form of a line, ( y - y_1 = m(x - x_1) ), where ( m ) is the slope and ( (x_1, y_1) ) is the point on the line:[y - 0 = 3(x - 1)]Simplifying this equation:[y = 3x - 3]Therefore, the equation of the tangent line to the curve ( y = x^3 - 1 ) at ( x = 1 ) is ( y = 3x - 3 ).

💡Okay, so I have this problem here: In the Cartesian coordinate system xOy, the line y = 2x + b is the tangent line of the curve y = a ln x. When a > 0, I need to find the minimum value of the real number b. Hmm, okay, let's break this down step by step.First, I remember that for a line to be tangent to a curve, two conditions must be satisfied at the point of tangency. The first condition is that the line and the curve must intersect at that point. The second condition is that their slopes must be equal at that point. So, I need to find a point (m, n) where both the curve and the line pass through, and the derivatives of both the curve and the line at that point are equal.Let me write down the given equations:1. The curve: y = a ln x2. The tangent line: y = 2x + bSo, at the point of tangency (m, n), both equations must hold true. That means:n = a ln m ...(1)andn = 2m + b ...(2)Additionally, the derivatives at that point must be equal. Let's find the derivative of the curve y = a ln x. The derivative dy/dx is a/x. So, at x = m, the slope of the curve is a/m.The slope of the tangent line y = 2x + b is 2, since it's in the form y = mx + c where m is the slope.Therefore, setting the slopes equal:a/m = 2 ...(3)So, from equation (3), I can solve for m in terms of a:a/m = 2 => m = a/2Okay, so m is equal to a/2. Now, let's substitute m back into equation (1) and equation (2) to find n and b.From equation (1):n = a ln m = a ln (a/2)From equation (2):n = 2m + b = 2*(a/2) + b = a + bSo, setting these two expressions for n equal to each other:a ln (a/2) = a + bTherefore, solving for b:b = a ln (a/2) - aSo, b is expressed in terms of a as b = a ln (a/2) - a.Now, the problem asks for the minimum value of b when a > 0. So, I need to find the minimum of the function b(a) = a ln (a/2) - a.To find the minimum, I can use calculus. I'll find the derivative of b with respect to a, set it equal to zero, and solve for a. Then, I'll check if that critical point is indeed a minimum.Let's compute the derivative of b with respect to a:b(a) = a ln (a/2) - aFirst, let's simplify ln (a/2). Using logarithm properties, ln (a/2) = ln a - ln 2. So,b(a) = a (ln a - ln 2) - a = a ln a - a ln 2 - aSo, b(a) = a ln a - a ln 2 - aNow, let's take the derivative:db/da = d/da [a ln a - a ln 2 - a]Compute term by term:1. d/da [a ln a] = ln a + a*(1/a) = ln a + 12. d/da [-a ln 2] = -ln 23. d/da [-a] = -1So, putting it all together:db/da = (ln a + 1) - ln 2 - 1 = ln a + 1 - ln 2 - 1 = ln a - ln 2Simplify:db/da = ln(a/2)So, the derivative of b with respect to a is ln(a/2). To find critical points, set db/da = 0:ln(a/2) = 0Solving for a:ln(a/2) = 0 => a/2 = e^0 = 1 => a = 2So, a critical point occurs at a = 2. Now, we need to determine whether this critical point is a minimum or a maximum.To do that, we can analyze the sign of the derivative around a = 2.- For a < 2: Let's pick a = 1. Then, ln(1/2) = ln(0.5) ≈ -0.693 < 0. So, db/da < 0 when a < 2.- For a > 2: Let's pick a = 4. Then, ln(4/2) = ln(2) ≈ 0.693 > 0. So, db/da > 0 when a > 2.Therefore, the function b(a) is decreasing for a < 2 and increasing for a > 2. Hence, the critical point at a = 2 is a local minimum. Since the function tends to infinity as a approaches 0 and as a approaches infinity, this local minimum is also the absolute minimum.Now, let's compute the value of b at a = 2:b(2) = 2 ln(2/2) - 2 = 2 ln(1) - 2 = 2*0 - 2 = -2So, the minimum value of b is -2.Let me just recap to make sure I didn't make any mistakes:1. Found the point of tangency by equating the curves and their derivatives.2. Expressed b in terms of a: b = a ln(a/2) - a.3. Took the derivative of b with respect to a, found the critical point at a = 2.4. Checked the intervals around a = 2 and confirmed it's a minimum.5. Calculated b at a = 2 and got -2.Everything seems to check out. I don't see any errors in my reasoning or calculations.**Final Answer**The minimum value of ( b ) is boxed{-2}.

💡Okay, so I have this problem where I need to find the smallest positive integer ( n ) such that ( S_n ) is an integer. ( S_n ) is defined as the sum of the reciprocals of the non-zero digits of all integers from 1 to ( 10^n ) inclusive. Hmm, that sounds a bit complicated, but let me break it down step by step.First, I need to understand what exactly ( S_n ) represents. It's the sum of reciprocals of non-zero digits. So, for each number from 1 to ( 10^n ), I look at each digit, ignore the zeros, and take the reciprocal of each non-zero digit, then sum all those reciprocals up. Got it.Let me consider a small example to get a feel for it. Let's say ( n = 1 ). Then, we're looking at numbers from 1 to 10. The non-zero digits are 1 through 9, each appearing once, and then 10, which has a non-zero digit 1. So, the reciprocals would be ( frac{1}{1}, frac{1}{2}, ldots, frac{1}{9}, frac{1}{1} ). Adding these up, ( S_1 ) would be ( 2 times frac{1}{1} + frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} ). Wait, actually, no. Each number from 1 to 9 contributes one reciprocal, and 10 contributes another ( frac{1}{1} ). So, ( S_1 = left( frac{1}{1} + frac{1}{2} + ldots + frac{1}{9} right) + frac{1}{1} ). That simplifies to ( 2 times frac{1}{1} + frac{1}{2} + ldots + frac{1}{9} ). So, ( S_1 ) is ( 2 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} ). Calculating that, ( 2 + frac{1}{2} = 2.5 ), ( 2.5 + frac{1}{3} approx 2.8333 ), ( + frac{1}{4} = 0.25 ), so ( 3.0833 ), ( + frac{1}{5} = 0.2 ), so ( 3.2833 ), ( + frac{1}{6} approx 0.1667 ), so ( 3.45 ), ( + frac{1}{7} approx 0.1429 ), so ( 3.5929 ), ( + frac{1}{8} = 0.125 ), so ( 3.7179 ), ( + frac{1}{9} approx 0.1111 ), so ( 3.829 ). So, ( S_1 approx 3.829 ), which is not an integer. So, ( n = 1 ) is out.Now, moving on to ( n = 2 ). So, numbers from 1 to 100. Each number can be considered as a two-digit number with leading zeros allowed, so from 00 to 99, plus 100. But wait, 00 isn't included since we start from 1, so actually from 01 to 99, and then 100. Hmm, but 01 is just 1, so maybe it's better to think of all numbers from 1 to 99, and then 100.For numbers from 1 to 99, each number can be thought of as having two digits, possibly with a leading zero. So, for each digit place (tens and units), each digit from 0 to 9 appears 10 times. But since we're ignoring zeros, each non-zero digit from 1 to 9 appears 10 times in each place. So, in the tens place, each non-zero digit appears 10 times, and similarly in the units place. So, for each non-zero digit ( d ), it appears 20 times in total from 1 to 99.Then, for 100, the non-zero digits are just 1. So, adding that, the total count for each digit is 20 for digits 1-9, except for digit 1, which is 21. Wait, no. Wait, from 1 to 99, each non-zero digit appears 20 times. Then, 100 adds another 1, so digit 1 appears 21 times, and digits 2-9 still appear 20 times.Therefore, ( S_2 ) would be ( 21 times frac{1}{1} + 20 times left( frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} right) ).Calculating that, ( 21 times 1 = 21 ). Then, ( 20 times left( frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} right) ). Let's compute the sum inside the parentheses first. ( frac{1}{2} = 0.5 ), ( frac{1}{3} approx 0.3333 ), ( frac{1}{4} = 0.25 ), ( frac{1}{5} = 0.2 ), ( frac{1}{6} approx 0.1667 ), ( frac{1}{7} approx 0.1429 ), ( frac{1}{8} = 0.125 ), ( frac{1}{9} approx 0.1111 ). Adding these up: 0.5 + 0.3333 = 0.8333, +0.25 = 1.0833, +0.2 = 1.2833, +0.1667 = 1.45, +0.1429 ≈ 1.5929, +0.125 = 1.7179, +0.1111 ≈ 1.829. So, the sum is approximately 1.829.Therefore, ( 20 times 1.829 ≈ 36.58 ). Adding the 21 from the digit 1, ( S_2 ≈ 21 + 36.58 = 57.58 ), which is still not an integer. So, ( n = 2 ) is also out.Hmm, maybe I need a better approach than calculating each ( S_n ) individually. Let's think about the general case for ( S_n ).For ( S_n ), we're considering numbers from 1 to ( 10^n ). Each number can be represented as an ( n )-digit number with leading zeros, so from 00...01 to 99...99, plus ( 10^n ). But ( 10^n ) only contributes a single 1, so it's similar to the previous cases.In each digit place (units, tens, hundreds, ..., up to the ( 10^{n-1} ) place), each digit from 0 to 9 appears equally often. Since we're ignoring zeros, each non-zero digit from 1 to 9 appears ( 10^{n-1} ) times in each of the ( n ) places. Therefore, each non-zero digit ( d ) appears ( n times 10^{n-1} ) times in total.But wait, except for the number ( 10^n ), which adds an extra 1. So, digit 1 appears ( n times 10^{n-1} + 1 ) times, and digits 2-9 appear ( n times 10^{n-1} ) times each.Therefore, ( S_n ) can be expressed as:[S_n = left( n times 10^{n-1} + 1 right) times frac{1}{1} + sum_{d=2}^{9} left( n times 10^{n-1} right) times frac{1}{d}]Simplifying, that's:[S_n = left( n times 10^{n-1} + 1 right) + n times 10^{n-1} times left( frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} right)]Let me denote ( H_9 ) as the 9th harmonic number, which is ( 1 + frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} ). So, ( H_9 approx 2.828968254 ). Then, the sum ( frac{1}{2} + frac{1}{3} + ldots + frac{1}{9} = H_9 - 1 approx 1.828968254 ).So, substituting back into ( S_n ):[S_n = n times 10^{n-1} + 1 + n times 10^{n-1} times (H_9 - 1)]Simplify further:[S_n = n times 10^{n-1} times (1 + H_9 - 1) + 1 = n times 10^{n-1} times H_9 + 1]Wait, that doesn't seem right. Let me check my steps again.Wait, no. Let's go back. The expression was:[S_n = left( n times 10^{n-1} + 1 right) + n times 10^{n-1} times (H_9 - 1)]So, expanding that:[S_n = n times 10^{n-1} + 1 + n times 10^{n-1} times H_9 - n times 10^{n-1}]Ah, the ( n times 10^{n-1} ) and ( -n times 10^{n-1} ) cancel each other out, leaving:[S_n = 1 + n times 10^{n-1} times (H_9 - 1)]So, ( S_n = 1 + n times 10^{n-1} times (H_9 - 1) ).But ( H_9 - 1 ) is approximately 1.828968254, as calculated earlier. So, ( S_n ) is 1 plus ( n times 10^{n-1} ) times approximately 1.828968254.We need ( S_n ) to be an integer. So, ( n times 10^{n-1} times (H_9 - 1) ) must be an integer minus 1, which is still an integer. Therefore, ( n times 10^{n-1} times (H_9 - 1) ) must be an integer.But ( H_9 - 1 ) is not an integer. It's approximately 1.828968254, which is a fraction. So, for ( n times 10^{n-1} times (H_9 - 1) ) to be an integer, ( n times 10^{n-1} ) must be a multiple of the denominator of ( H_9 - 1 ) when expressed in its simplest fractional form.So, I need to find the fractional representation of ( H_9 - 1 ). Let me compute ( H_9 ) exactly.Calculating ( H_9 ):[H_9 = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + frac{1}{5} + frac{1}{6} + frac{1}{7} + frac{1}{8} + frac{1}{9}]Let's find a common denominator. The least common multiple (LCM) of denominators 1 through 9 is 2520. So, converting each term:[1 = frac{2520}{2520}][frac{1}{2} = frac{1260}{2520}][frac{1}{3} = frac{840}{2520}][frac{1}{4} = frac{630}{2520}][frac{1}{5} = frac{504}{2520}][frac{1}{6} = frac{420}{2520}][frac{1}{7} = frac{360}{2520}][frac{1}{8} = frac{315}{2520}][frac{1}{9} = frac{280}{2520}]Adding all these up:2520 + 1260 = 37803780 + 840 = 46204620 + 630 = 52505250 + 504 = 57545754 + 420 = 61746174 + 360 = 65346534 + 315 = 68496849 + 280 = 7129So, ( H_9 = frac{7129}{2520} ). Therefore, ( H_9 - 1 = frac{7129}{2520} - 1 = frac{7129 - 2520}{2520} = frac{4609}{2520} ).So, ( H_9 - 1 = frac{4609}{2520} ). Let me check if this fraction can be simplified. The numerator is 4609, and the denominator is 2520. Let's see if they have any common factors.2520 factors: 2^3 * 3^2 * 5 * 74609: Let's check divisibility. 4609 ÷ 7 = 658.428... Not integer. 4609 ÷ 3 = 1536.333... Not integer. 4609 ÷ 2 = 2304.5, not integer. 4609 ÷ 5 = 921.8, not integer. So, it seems 4609 and 2520 are coprime. Therefore, ( H_9 - 1 = frac{4609}{2520} ) is in its simplest form.Therefore, ( S_n = 1 + n times 10^{n-1} times frac{4609}{2520} ). For ( S_n ) to be an integer, ( n times 10^{n-1} times frac{4609}{2520} ) must be an integer. Since 4609 and 2520 are coprime, ( n times 10^{n-1} ) must be a multiple of 2520.So, ( n times 10^{n-1} ) must be divisible by 2520. Let's factorize 2520 to understand its prime factors.2520 = 2^3 * 3^2 * 5 * 7So, ( n times 10^{n-1} ) must have at least these prime factors. Let's factorize ( 10^{n-1} ):10^{n-1} = (2 * 5)^{n-1} = 2^{n-1} * 5^{n-1}Therefore, ( n times 10^{n-1} = n times 2^{n-1} times 5^{n-1} )We need this product to have at least 2^3, 3^2, 5^1, and 7^1.Looking at the factors:- For 2^3: Since ( 2^{n-1} ) is already present, as long as ( n-1 geq 3 ), which means ( n geq 4 ), we have enough 2s.- For 3^2: The factor of 3^2 must come from ( n ). So, ( n ) must be divisible by 9.- For 5^1: Since ( 5^{n-1} ) is present, as long as ( n-1 geq 1 ), which is ( n geq 2 ), we have enough 5s.- For 7^1: The factor of 7 must come from ( n ). So, ( n ) must be divisible by 7.Additionally, since ( n ) must be divisible by both 9 and 7, the least common multiple (LCM) of 9 and 7 is 63. Therefore, ( n ) must be a multiple of 63.But we also need to ensure that ( n times 10^{n-1} ) has enough factors. Let's check if 63 satisfies all the conditions.n = 63:- 63 is divisible by 9 and 7, so it covers the 3^2 and 7^1.- ( 10^{62} = 2^{62} * 5^{62} ), which certainly covers 2^3 and 5^1.Therefore, ( n = 63 ) is the smallest positive integer that satisfies all the conditions. Let me verify this.Calculating ( S_{63} ):[S_{63} = 1 + 63 times 10^{62} times frac{4609}{2520}]Simplify:[S_{63} = 1 + frac{63 times 10^{62} times 4609}{2520}]Simplify the fraction:63 / 2520 = 63 / (63 * 40) = 1 / 40So,[S_{63} = 1 + frac{10^{62} times 4609}{40}]Simplify ( 10^{62} / 40 = 10^{62} / (10 * 4) = 10^{61} / 4 ). Since 10^{61} is divisible by 4 (as 100 is divisible by 4), ( 10^{61} / 4 ) is an integer.Therefore, ( S_{63} = 1 + text{integer} ), which is an integer.Hence, the smallest positive integer ( n ) for which ( S_n ) is an integer is 63.

💡Alright, so I have this problem here about an even function f(x). It says that f(x) is defined on all real numbers, and when x is greater than or equal to zero, f(x) equals x squared minus 4x. Then, I need to solve the inequality f(2x + 3) is less than or equal to 5. The options are given as sets of numbers, so I need to figure out which set is the solution.First, I remember that an even function satisfies f(x) = f(-x) for all x. That means the graph of the function is symmetric with respect to the y-axis. So, if I know f(x) for x >= 0, I automatically know it for x < 0 because it's just the mirror image.Given that f(x) = x² - 4x when x >= 0, then for x < 0, f(x) should be equal to f(-x). Let me compute that. If x is negative, then -x is positive, so f(-x) = (-x)² - 4*(-x) = x² + 4x. So, f(x) = x² + 4x when x < 0.Okay, so now I have f(x) defined for all real numbers. Now, I need to solve f(2x + 3) <= 5. That means I need to substitute 2x + 3 into the function f and set it less than or equal to 5.But since f is defined differently depending on whether its argument is positive or negative, I need to consider two cases:Case 1: 2x + 3 >= 0. In this case, f(2x + 3) = (2x + 3)² - 4*(2x + 3).Case 2: 2x + 3 < 0. In this case, f(2x + 3) = (2x + 3)² + 4*(2x + 3).So, I need to solve both cases separately and then combine the solutions.Let me start with Case 1: 2x + 3 >= 0.First, solve 2x + 3 >= 0. Subtract 3: 2x >= -3. Divide by 2: x >= -3/2. So, in this case, x is greater than or equal to -1.5.Now, compute f(2x + 3):f(2x + 3) = (2x + 3)² - 4*(2x + 3).Let me expand this:(2x + 3)² = 4x² + 12x + 9.Then, subtract 4*(2x + 3): 4x² + 12x + 9 - 8x - 12.Combine like terms:4x² + (12x - 8x) + (9 - 12) = 4x² + 4x - 3.So, f(2x + 3) = 4x² + 4x - 3.Set this less than or equal to 5:4x² + 4x - 3 <= 5.Subtract 5 from both sides:4x² + 4x - 8 <= 0.Factor out a 4:4(x² + x - 2) <= 0.Divide both sides by 4 (since 4 is positive, the inequality sign doesn't change):x² + x - 2 <= 0.Now, factor the quadratic:Looking for two numbers that multiply to -2 and add to 1. That would be 2 and -1.So, (x + 2)(x - 1) <= 0.To solve this inequality, find the critical points: x = -2 and x = 1.These divide the number line into three intervals:1. x < -22. -2 < x < 13. x > 1Test each interval:1. For x < -2, say x = -3: (-3 + 2)(-3 - 1) = (-1)(-4) = 4 > 0.2. For -2 < x < 1, say x = 0: (0 + 2)(0 - 1) = 2*(-1) = -2 < 0.3. For x > 1, say x = 2: (2 + 2)(2 - 1) = 4*1 = 4 > 0.Since the inequality is <= 0, the solution is the interval where the expression is negative or zero, which is [-2, 1].But remember, this is under Case 1 where x >= -1.5. So, the solution from this case is the overlap between [-2, 1] and x >= -1.5.So, the overlap is [-1.5, 1].Wait, but [-2,1] includes x from -2 to 1, but in Case 1, x must be >= -1.5. Therefore, the solution for Case 1 is x in [-1.5, 1].Now, let's move to Case 2: 2x + 3 < 0.Solve 2x + 3 < 0: 2x < -3 => x < -1.5.In this case, f(2x + 3) = (2x + 3)² + 4*(2x + 3).Compute this:(2x + 3)² = 4x² + 12x + 9.Add 4*(2x + 3): 4x² + 12x + 9 + 8x + 12.Combine like terms:4x² + (12x + 8x) + (9 + 12) = 4x² + 20x + 21.So, f(2x + 3) = 4x² + 20x + 21.Set this less than or equal to 5:4x² + 20x + 21 <= 5.Subtract 5 from both sides:4x² + 20x + 16 <= 0.Factor out a 4:4(x² + 5x + 4) <= 0.Divide both sides by 4:x² + 5x + 4 <= 0.Factor the quadratic:Looking for two numbers that multiply to 4 and add to 5. That would be 1 and 4.So, (x + 1)(x + 4) <= 0.Find the critical points: x = -1 and x = -4.These divide the number line into three intervals:1. x < -42. -4 < x < -13. x > -1Test each interval:1. For x < -4, say x = -5: (-5 + 1)(-5 + 4) = (-4)(-1) = 4 > 0.2. For -4 < x < -1, say x = -2: (-2 + 1)(-2 + 4) = (-1)(2) = -2 < 0.3. For x > -1, say x = 0: (0 + 1)(0 + 4) = 1*4 = 4 > 0.Since the inequality is <= 0, the solution is the interval where the expression is negative or zero, which is [-4, -1].But remember, this is under Case 2 where x < -1.5. So, the solution from this case is the overlap between [-4, -1] and x < -1.5.So, the overlap is [-4, -1.5).Wait, but x < -1.5 is the condition, and the solution from the inequality is x between -4 and -1. So, the overlap is x between -4 and -1.5.Therefore, the solution for Case 2 is x in [-4, -1.5).Now, combining the solutions from both cases:From Case 1: x in [-1.5, 1]From Case 2: x in [-4, -1.5)So, the total solution set is x in [-4, 1].But wait, let me check if this makes sense.Wait, in Case 1, x >= -1.5, and the solution there was x in [-1.5, 1].In Case 2, x < -1.5, and the solution was x in [-4, -1.5).So, combining these, the total solution is x in [-4, 1].But looking at the answer choices, none of them are intervals. They are sets of specific numbers: A: {-5,5}, B: {-8,-2}, C: {-4,1}, D: {1,4}.Hmm, that's interesting. So, the solution set is actually the endpoints of the interval? Or maybe I made a mistake in interpreting the problem.Wait, the problem says "the solution set of the inequality f(2x+3) <= 5". So, if the solution is x in [-4,1], but the options are sets of numbers, not intervals. So, perhaps the question is asking for the boundary points where f(2x+3) equals 5, rather than the entire interval.Wait, let me think again. Maybe I need to find the values of x where f(2x+3) = 5, which would give me specific points, and then the inequality would include all points between them.But in my solution, I found that f(2x+3) <=5 for x in [-4,1]. So, the solution set is all real numbers between -4 and 1, inclusive.But the answer choices are sets of specific numbers, not intervals. So, perhaps I misunderstood the problem.Wait, looking back at the problem: "the solution set of the inequality f(2x+3) <= 5 is ( )." And the options are sets like {-5,5}, {-8,-2}, {-4,1}, {1,4}.So, maybe the solution set is the set of x values where f(2x+3) equals 5, which are the boundary points, so x = -4 and x =1, which is option C: {-4,1}.But wait, in my solution, the inequality holds for all x between -4 and 1, not just the endpoints. So, why are the answer choices only sets with two numbers?Maybe the problem is asking for the solution set in terms of the x-values where f(2x+3) equals 5, which are the endpoints of the interval where the inequality holds. So, the solution set is the set of x such that f(2x+3) =5, which are x = -4 and x =1.Alternatively, perhaps the problem is asking for the solution set in terms of the values of 2x+3, but that seems less likely.Wait, let me double-check my calculations.In Case 1: x >= -1.5, f(2x+3) =4x² +4x -3 <=5, leading to x in [-2,1]. But since x >= -1.5, the solution is x in [-1.5,1].In Case 2: x < -1.5, f(2x+3)=4x² +20x +21 <=5, leading to x in [-4,-1]. But since x < -1.5, the solution is x in [-4,-1.5).So, combining both, x in [-4,1].But the answer choices are sets of numbers, not intervals. So, perhaps the question is asking for the values of 2x+3 where f(2x+3)=5, which would be the boundary points.Wait, let's think about that. If f(2x+3) <=5, then the maximum value is 5, achieved at certain points. So, the solution set would include all x such that 2x+3 is in the domain where f(y) <=5, which is y in [-4,1], but y =2x+3.Wait, no, f(y) <=5, and f(y) is defined as y² -4y for y >=0, and y² +4y for y <0.So, f(y) <=5.Let me solve f(y) <=5.Case 1: y >=0, f(y)= y² -4y <=5.So, y² -4y -5 <=0.Factor: (y -5)(y +1) <=0.So, y in [-1,5]. But since y >=0, y in [0,5].Case 2: y <0, f(y)= y² +4y <=5.So, y² +4y -5 <=0.Factor: (y +5)(y -1) <=0.So, y in [-5,1]. But since y <0, y in [-5,0).Therefore, overall, y in [-5,5].So, f(y) <=5 when y in [-5,5].Therefore, for f(2x+3) <=5, we need 2x+3 in [-5,5].So, 2x +3 >= -5 and 2x +3 <=5.Solve for x:First inequality: 2x +3 >= -5 => 2x >= -8 => x >= -4.Second inequality: 2x +3 <=5 => 2x <=2 => x <=1.Therefore, x in [-4,1].So, the solution set is all real numbers between -4 and 1, inclusive.But the answer choices are sets of specific numbers, not intervals. So, perhaps the question is asking for the boundary points where f(2x+3)=5, which are x=-4 and x=1.Therefore, the solution set is {-4,1}, which is option C.Wait, but in my earlier approach, I considered solving the inequality by substituting 2x+3 into f and solving for x, which gave me x in [-4,1]. But the answer choices are specific numbers, so I think the intended answer is the boundary points where f(2x+3)=5, which are x=-4 and x=1.Therefore, the correct answer is option C: {-4,1}.