💡Okay, so I have this problem where I need to find the intersection point of two lines, l₁ and l₂. The equations given are l₁: 3x + 4y - 2 = 0 and l₂: 2x + y + 2 = 0. Hmm, I remember that to find the intersection point of two lines, I need to solve these two equations simultaneously. That means finding the values of x and y that satisfy both equations at the same time.Alright, let me write down the equations again to make it clearer:1. 3x + 4y - 2 = 02. 2x + y + 2 = 0I think the best way to solve this is by using the substitution method. That means I'll solve one of the equations for one variable and then substitute that into the other equation. Looking at equation 2, it seems simpler because the coefficients are smaller, so maybe I can solve for y in terms of x.Starting with equation 2:2x + y + 2 = 0I can subtract 2x and 2 from both sides to isolate y:y = -2x - 2Okay, so now I have y expressed in terms of x. That means I can take this expression for y and plug it into equation 1 to find the value of x.Substituting y = -2x - 2 into equation 1:3x + 4(-2x - 2) - 2 = 0Let me simplify this step by step. First, distribute the 4 into the parentheses:3x + (-8x - 8) - 2 = 0Now, combine like terms:3x - 8x - 8 - 2 = 0Simplify the x terms and the constants:-5x - 10 = 0Now, solve for x:-5x = 10x = 10 / (-5)x = -2Alright, so x is -2. Now, I need to find the corresponding y value. I can plug this x value back into the expression I found for y earlier, which was y = -2x - 2.Substituting x = -2:y = -2(-2) - 2y = 4 - 2y = 2So, y is 2. Therefore, the intersection point of the two lines is at (-2, 2).Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting with equation 2, solving for y:2x + y + 2 = 0y = -2x - 2That looks correct. Then substituting into equation 1:3x + 4(-2x - 2) - 2 = 03x - 8x - 8 - 2 = 0-5x - 10 = 0-5x = 10x = -2Yes, that seems right. Then plugging x = -2 back into y:y = -2(-2) - 2 = 4 - 2 = 2Everything checks out. So, the intersection point is indeed (-2, 2). I think I've got it!

💡Okay, so I need to find the range of the function ( g(x) = frac{sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9}{sin x - 1} ) where ( sin x neq 1 ). Hmm, let me try to break this down step by step.First, I notice that the function involves both ( sin x ) and ( cos x ). Maybe I can express everything in terms of ( sin x ) to simplify it. I remember that ( cos^2 x = 1 - sin^2 x ), so I can substitute that into the equation.Let me rewrite the numerator:( sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9 )Substituting ( cos^2 x ) with ( 1 - sin^2 x ):( sin^3 x + 6 sin^2 x + sin x + 3(1 - sin^2 x) - 9 )Let me expand that:( sin^3 x + 6 sin^2 x + sin x + 3 - 3 sin^2 x - 9 )Now, combine like terms:- ( sin^3 x ) remains as is.- ( 6 sin^2 x - 3 sin^2 x = 3 sin^2 x )- ( sin x ) remains as is.- ( 3 - 9 = -6 )So the numerator simplifies to:( sin^3 x + 3 sin^2 x + sin x - 6 )So now, the function ( g(x) ) becomes:( frac{sin^3 x + 3 sin^2 x + sin x - 6}{sin x - 1} )Hmm, maybe I can factor the numerator. Let me try to factor ( sin^3 x + 3 sin^2 x + sin x - 6 ). I can try to factor by grouping or look for rational roots.Let me consider ( sin x ) as a variable, say ( y ). So the numerator is ( y^3 + 3y^2 + y - 6 ). Let me try to factor this.Testing ( y = 1 ): ( 1 + 3 + 1 - 6 = -1 neq 0 ).Testing ( y = 2 ): ( 8 + 12 + 2 - 6 = 16 neq 0 ).Testing ( y = -2 ): ( -8 + 12 - 2 - 6 = -4 neq 0 ).Testing ( y = -3 ): ( -27 + 27 - 3 - 6 = -9 neq 0 ).Hmm, none of these are roots. Maybe I made a mistake earlier in simplifying the numerator. Let me double-check.Original numerator after substitution:( sin^3 x + 6 sin^2 x + sin x + 3(1 - sin^2 x) - 9 )Expanding:( sin^3 x + 6 sin^2 x + sin x + 3 - 3 sin^2 x - 9 )Combining like terms:( sin^3 x + (6 sin^2 x - 3 sin^2 x) + sin x + (3 - 9) )Which is:( sin^3 x + 3 sin^2 x + sin x - 6 )Okay, that seems correct. Maybe I can factor it another way. Let me try to factor ( y^3 + 3y^2 + y - 6 ).Alternatively, perhaps I can perform polynomial division since the denominator is ( y - 1 ). Let me divide ( y^3 + 3y^2 + y - 6 ) by ( y - 1 ).Using synthetic division:Coefficients: 1 (y^3), 3 (y^2), 1 (y), -6 (constant)Divide by ( y - 1 ), so test root at y = 1.Bring down 1.Multiply 1 by 1: 1, add to next coefficient: 3 + 1 = 4.Multiply 4 by 1: 4, add to next coefficient: 1 + 4 = 5.Multiply 5 by 1: 5, add to last coefficient: -6 + 5 = -1.So the division gives a quotient of ( y^2 + 4y + 5 ) with a remainder of -1. Wait, but the original function has denominator ( sin x - 1 ), so if I have a remainder, that would mean the numerator isn't divisible by ( sin x - 1 ). But earlier, I thought it was. Maybe I made a mistake.Wait, let me check the synthetic division again.Dividing ( y^3 + 3y^2 + y - 6 ) by ( y - 1 ):Bring down 1.Multiply 1 by 1: 1, add to 3: 4.Multiply 4 by 1: 4, add to 1: 5.Multiply 5 by 1: 5, add to -6: -1.So the remainder is indeed -1, not zero. That means ( y - 1 ) is not a factor of the numerator. But in the original function, the denominator is ( sin x - 1 ), so if the numerator isn't divisible by it, we can't cancel it out. That complicates things.Wait, maybe I made a mistake in simplifying the numerator. Let me go back.Original numerator:( sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9 )Express ( cos^2 x ) as ( 1 - sin^2 x ):( sin^3 x + 6 sin^2 x + sin x + 3(1 - sin^2 x) - 9 )Expanding:( sin^3 x + 6 sin^2 x + sin x + 3 - 3 sin^2 x - 9 )Combine like terms:( sin^3 x + (6 sin^2 x - 3 sin^2 x) + sin x + (3 - 9) )Which is:( sin^3 x + 3 sin^2 x + sin x - 6 )Hmm, that seems correct. So the numerator is ( y^3 + 3y^2 + y - 6 ), which doesn't factor nicely with ( y - 1 ). So maybe I need another approach.Alternatively, perhaps I can write the numerator as ( (y - 1)(something) + remainder ). Since the denominator is ( y - 1 ), maybe I can express the function as:( g(x) = frac{(y - 1)(quadratic) + remainder}{y - 1} )Which would be ( quadratic + frac{remainder}{y - 1} ). But since the remainder is -1, as we saw earlier, that would mean:( g(x) = y^2 + 4y + 5 - frac{1}{y - 1} )But that seems more complicated. Maybe I can proceed differently.Alternatively, perhaps I can consider the function ( g(x) ) as a function of ( y = sin x ), with ( y in [-1, 1] ) and ( y neq 1 ). So let me define ( h(y) = frac{y^3 + 3y^2 + y - 6}{y - 1} ).Let me compute ( h(y) ) for ( y in [-1, 1) ).But since ( y - 1 ) is in the denominator, and ( y neq 1 ), I need to handle this carefully.Alternatively, perhaps I can perform polynomial division correctly. Let me try dividing ( y^3 + 3y^2 + y - 6 ) by ( y - 1 ).Using long division:Divide ( y^3 ) by ( y ): ( y^2 ). Multiply ( y - 1 ) by ( y^2 ): ( y^3 - y^2 ). Subtract from the numerator:( (y^3 + 3y^2 + y - 6) - (y^3 - y^2) = 4y^2 + y - 6 ).Now, divide ( 4y^2 ) by ( y ): ( 4y ). Multiply ( y - 1 ) by ( 4y ): ( 4y^2 - 4y ). Subtract:( (4y^2 + y - 6) - (4y^2 - 4y) = 5y - 6 ).Now, divide ( 5y ) by ( y ): 5. Multiply ( y - 1 ) by 5: ( 5y - 5 ). Subtract:( (5y - 6) - (5y - 5) = -1 ).So the division gives ( y^2 + 4y + 5 ) with a remainder of -1. Therefore, the function can be written as:( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} )Hmm, that seems correct. So ( g(x) = h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ), where ( y = sin x ) and ( y in [-1, 1) ).Now, I need to find the range of ( h(y) ) for ( y in [-1, 1) ).Let me analyze each term:1. ( y^2 + 4y + 5 ): This is a quadratic function. Let me find its vertex. The vertex occurs at ( y = -b/(2a) = -4/(2*1) = -2 ). Since the coefficient of ( y^2 ) is positive, the parabola opens upwards. At ( y = -2 ), the minimum value is ( (-2)^2 + 4*(-2) + 5 = 4 - 8 + 5 = 1 ). But since ( y in [-1, 1) ), the minimum in this interval occurs at ( y = -1 ):( (-1)^2 + 4*(-1) + 5 = 1 - 4 + 5 = 2 ).At ( y = 1 ), it would be ( 1 + 4 + 5 = 10 ), but since ( y ) approaches 1 from the left, the value approaches 10.2. ( -frac{1}{y - 1} ): Let me analyze this term. As ( y ) approaches 1 from below, ( y - 1 ) approaches 0 from the negative side, so ( frac{1}{y - 1} ) approaches negative infinity, and thus ( -frac{1}{y - 1} ) approaches positive infinity. At ( y = -1 ), ( -frac{1}{-1 - 1} = -frac{1}{-2} = 0.5 ).So combining both terms:At ( y = -1 ):( h(-1) = 2 + 0.5 = 2.5 ).As ( y ) approaches 1 from below:( y^2 + 4y + 5 ) approaches 10, and ( -frac{1}{y - 1} ) approaches positive infinity, so ( h(y) ) approaches positive infinity.But wait, that contradicts the earlier thought that the range was [2,12). Maybe I made a mistake.Wait, let me compute ( h(y) ) at ( y = -1 ):( h(-1) = (-1)^2 + 4*(-1) + 5 - frac{1}{-1 - 1} = 1 - 4 + 5 - frac{1}{-2} = 2 + 0.5 = 2.5 ).Wait, but earlier, when I thought the function simplified to ( (y + 2)(y + 3) ), which would be ( y^2 + 5y + 6 ), but that seems different from ( y^2 + 4y + 5 - frac{1}{y - 1} ). So perhaps I made a mistake earlier in factoring.Wait, let me go back. Initially, I thought the numerator was ( (sin x - 1)(sin x + 2)(sin x + 3) ), but that must be incorrect because when I performed the division, I got a remainder. So that initial factoring was wrong.So, the correct expression is ( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ).Therefore, I need to find the range of ( h(y) ) for ( y in [-1, 1) ).Let me consider ( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ).To find the range, I can analyze the behavior of ( h(y) ) as ( y ) varies from -1 to 1.First, at ( y = -1 ):( h(-1) = (-1)^2 + 4*(-1) + 5 - frac{1}{-1 - 1} = 1 - 4 + 5 - frac{1}{-2} = 2 + 0.5 = 2.5 ).As ( y ) increases from -1 towards 1, let's see how each term behaves:- ( y^2 + 4y + 5 ) is a quadratic that increases as ( y ) increases from -1 to 1 because the vertex is at ( y = -2 ), which is outside our interval. So in our interval, it's increasing.- ( -frac{1}{y - 1} ) also increases as ( y ) increases because as ( y ) approaches 1 from below, ( y - 1 ) approaches 0 from the negative side, so ( frac{1}{y - 1} ) approaches negative infinity, and thus ( -frac{1}{y - 1} ) approaches positive infinity.Therefore, both terms are increasing as ( y ) increases from -1 to 1. So ( h(y) ) is increasing on ( [-1, 1) ).At ( y = -1 ), ( h(-1) = 2.5 ).As ( y ) approaches 1 from the left, ( h(y) ) approaches infinity because the term ( -frac{1}{y - 1} ) dominates and goes to infinity.Wait, but that contradicts the initial thought that the range was [2,12). So perhaps I made a mistake in the initial approach.Wait, let me double-check the initial substitution and simplification.Original function:( g(x) = frac{sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9}{sin x - 1} )Express ( cos^2 x = 1 - sin^2 x ):Numerator becomes:( sin^3 x + 6 sin^2 x + sin x + 3(1 - sin^2 x) - 9 )Expanding:( sin^3 x + 6 sin^2 x + sin x + 3 - 3 sin^2 x - 9 )Combine like terms:( sin^3 x + (6 sin^2 x - 3 sin^2 x) + sin x + (3 - 9) )Which is:( sin^3 x + 3 sin^2 x + sin x - 6 )So that's correct.Then, trying to factor ( sin^3 x + 3 sin^2 x + sin x - 6 ), I thought it was ( (sin x - 1)(sin x + 2)(sin x + 3) ), but when I expanded that, it would be:( (sin x - 1)(sin x + 2)(sin x + 3) )First multiply ( (sin x + 2)(sin x + 3) ):( sin^2 x + 5 sin x + 6 )Then multiply by ( (sin x - 1) ):( sin^3 x + 5 sin^2 x + 6 sin x - sin^2 x - 5 sin x - 6 )Combine like terms:( sin^3 x + (5 sin^2 x - sin^2 x) + (6 sin x - 5 sin x) - 6 )Which is:( sin^3 x + 4 sin^2 x + sin x - 6 )Wait, that's different from our numerator, which was ( sin^3 x + 3 sin^2 x + sin x - 6 ). So my initial factoring was incorrect. Therefore, the numerator is not ( (sin x - 1)(sin x + 2)(sin x + 3) ), but rather something else.Therefore, my earlier conclusion that ( g(x) = (sin x + 2)(sin x + 3) ) was wrong. That was a mistake.So, going back, the correct expression after division is ( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ).Now, to find the range of ( h(y) ) for ( y in [-1, 1) ).Let me consider the behavior of ( h(y) ):1. As ( y ) approaches 1 from below, ( -frac{1}{y - 1} ) approaches positive infinity, so ( h(y) ) approaches infinity.2. At ( y = -1 ), ( h(-1) = (-1)^2 + 4*(-1) + 5 - frac{1}{-1 - 1} = 1 - 4 + 5 - frac{1}{-2} = 2 + 0.5 = 2.5 ).3. Let me check the derivative to see if there's a minimum or maximum in the interval.Compute ( h'(y) ):( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} )So,( h'(y) = 2y + 4 - frac{d}{dy}left( frac{1}{y - 1} right) )The derivative of ( frac{1}{y - 1} ) is ( -frac{1}{(y - 1)^2} ), so:( h'(y) = 2y + 4 - left( -frac{1}{(y - 1)^2} right) = 2y + 4 + frac{1}{(y - 1)^2} )Since ( y in [-1, 1) ), ( y - 1 ) is negative, so ( (y - 1)^2 ) is positive, and ( frac{1}{(y - 1)^2} ) is positive. Therefore, ( h'(y) = 2y + 4 + text{positive} ).Let me check if ( h'(y) ) is always positive in the interval.At ( y = -1 ):( h'(-1) = 2*(-1) + 4 + frac{1}{(-1 - 1)^2} = -2 + 4 + frac{1}{4} = 2 + 0.25 = 2.25 > 0 ).At ( y = 0 ):( h'(0) = 0 + 4 + frac{1}{(0 - 1)^2} = 4 + 1 = 5 > 0 ).At ( y ) approaching 1 from below:( 2y + 4 ) approaches ( 2*1 + 4 = 6 ), and ( frac{1}{(y - 1)^2} ) approaches infinity. So ( h'(y) ) approaches infinity.Therefore, ( h'(y) > 0 ) for all ( y in [-1, 1) ). This means ( h(y) ) is strictly increasing on this interval.Since ( h(y) ) is strictly increasing, its minimum value is at ( y = -1 ), which is 2.5, and as ( y ) approaches 1, ( h(y) ) approaches infinity.Wait, but earlier, when I thought the function simplified to ( (y + 2)(y + 3) ), I concluded the range was [2,12). But that was incorrect because the factoring was wrong. So now, with the correct expression, the range should be [2.5, ∞).But wait, let me double-check the value at ( y = -1 ):( h(-1) = (-1)^2 + 4*(-1) + 5 - frac{1}{-1 - 1} = 1 - 4 + 5 - frac{1}{-2} = 2 + 0.5 = 2.5 ).Yes, that's correct.As ( y ) approaches 1, ( h(y) ) approaches infinity.Therefore, the range of ( g(x) ) is [2.5, ∞).But wait, the initial problem statement says "as ( x ) ranges over all real numbers such that ( sin x neq 1 )". So ( y ) can approach 1 but never equals 1, so ( h(y) ) can get arbitrarily large but never actually reaches infinity. So the range is [2.5, ∞).But wait, let me check if there's any other critical points or if the function could have a lower minimum elsewhere.Since ( h(y) ) is strictly increasing on [-1,1), the minimum is at y=-1, which is 2.5, and it increases without bound as y approaches 1.Therefore, the range of ( g(x) ) is [2.5, ∞).But wait, let me check if I made a mistake in the initial substitution.Wait, the original function was:( g(x) = frac{sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9}{sin x - 1} )After substitution, I got:( frac{sin^3 x + 3 sin^2 x + sin x - 6}{sin x - 1} )Then, performing polynomial division, I got:( y^2 + 4y + 5 - frac{1}{y - 1} )Which is correct.So, the function ( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ) is indeed the correct expression.Thus, the range is [2.5, ∞).But wait, in the initial problem, the user wrote:"Find the range of the function[ g(x) = frac{sin^3 x + 6 sin^2 x + sin x + 3 cos^2 x - 9}{sin x - 1}, ]as ( x ) ranges over all real numbers such that ( sin x neq 1 )."And the initial assistant response concluded the range was [2,12), but that was based on incorrect factoring.So, to correct that, the range should be [2.5, ∞).But let me double-check by plugging in some values.At ( y = -1 ):( h(-1) = 2.5 ).At ( y = 0 ):( h(0) = 0 + 0 + 5 - frac{1}{-1} = 5 + 1 = 6 ).At ( y = 0.5 ):( h(0.5) = 0.25 + 2 + 5 - frac{1}{-0.5} = 7.25 + 2 = 9.25 ).As ( y ) approaches 1, ( h(y) ) approaches infinity.So, the function starts at 2.5 when ( y = -1 ) and increases without bound as ( y ) approaches 1.Therefore, the range of ( g(x) ) is [2.5, ∞).But wait, 2.5 is 5/2, so in boxed form, it would be ( boxed{[frac{5}{2}, infty)} ).But let me confirm once more.Alternatively, perhaps I can express ( h(y) ) differently.Given ( h(y) = y^2 + 4y + 5 - frac{1}{y - 1} ), and knowing that ( y in [-1,1) ), let me see if I can find the minimum value.Since ( h(y) ) is strictly increasing, the minimum is at ( y = -1 ), which is 2.5, and it increases to infinity as ( y ) approaches 1.Therefore, the range is ( [frac{5}{2}, infty) ).Yes, that seems correct.

💡Okay, so I have this problem where I need to find the sum of all zero points of the function f[g(x)]. The functions f(x) and g(x) are both piecewise functions, which means they have different expressions depending on whether x is positive or negative. Let me try to break this down step by step.First, let me write down the functions again to make sure I have them right.f(x) is defined as:- 2^(x - 2) - 1, when x is greater than or equal to 0- x + 2, when x is less than 0g(x) is defined as:- x² - 2x - 1, when x is greater than or equal to 0- 1/x, when x is less than 0So, f[g(x)] means I need to substitute g(x) into f(x). That is, wherever there's an x in f(x), I replace it with g(x). But since both f and g are piecewise, I need to consider different cases based on the value of x.Let me think about how to approach this. Since g(x) is piecewise, its expression depends on whether x is non-negative or negative. Similarly, f(x) is also piecewise, so the expression for f[g(x)] will depend on whether g(x) is non-negative or negative. Therefore, I need to figure out when g(x) is non-negative and when it's negative.Let me first analyze g(x):For x >= 0, g(x) = x² - 2x - 1. This is a quadratic function. Let me find its roots to see where it's positive or negative.The quadratic equation x² - 2x - 1 = 0 can be solved using the quadratic formula:x = [2 ± sqrt(4 + 4)] / 2 = [2 ± sqrt(8)] / 2 = [2 ± 2*sqrt(2)] / 2 = 1 ± sqrt(2)So, the roots are x = 1 + sqrt(2) and x = 1 - sqrt(2). Since sqrt(2) is approximately 1.414, 1 - sqrt(2) is approximately -0.414. But since we're considering x >= 0, the relevant root is x = 1 + sqrt(2) ≈ 2.414.Therefore, for x >= 0, g(x) is negative between x = 0 and x = 1 + sqrt(2), and positive when x > 1 + sqrt(2).For x < 0, g(x) = 1/x. Since x is negative, 1/x is also negative. So, for x < 0, g(x) is always negative.Now, let's consider f[g(x)].Since f(x) is defined differently based on whether its input is non-negative or negative, I need to consider the sign of g(x) to determine which expression of f to use.So, breaking it down:1. When x >= 1 + sqrt(2): g(x) is positive, so f[g(x)] = 2^(g(x) - 2) - 12. When 0 <= x < 1 + sqrt(2): g(x) is negative, so f[g(x)] = g(x) + 23. When x < 0: g(x) is negative, so f[g(x)] = g(x) + 2Wait, hold on. For x < 0, g(x) is 1/x, which is negative, so f[g(x)] uses the x < 0 case of f, which is x + 2. So, f[g(x)] = g(x) + 2.Similarly, for 0 <= x < 1 + sqrt(2), g(x) is negative, so f[g(x)] is g(x) + 2.And for x >= 1 + sqrt(2), g(x) is positive, so f[g(x)] is 2^(g(x) - 2) - 1.So, in summary:f[g(x)] = { 2^(g(x) - 2) - 1, when x >= 1 + sqrt(2) g(x) + 2, when 0 <= x < 1 + sqrt(2) g(x) + 2, when x < 0}But since g(x) is different in each interval, let's substitute the expressions accordingly.So:1. For x >= 1 + sqrt(2): g(x) = x² - 2x - 1 So, f[g(x)] = 2^(x² - 2x - 1 - 2) - 1 = 2^(x² - 2x - 3) - 12. For 0 <= x < 1 + sqrt(2): g(x) = x² - 2x - 1 So, f[g(x)] = (x² - 2x - 1) + 2 = x² - 2x + 13. For x < 0: g(x) = 1/x So, f[g(x)] = (1/x) + 2Now, I need to find the zero points of f[g(x)], which means solving f[g(x)] = 0 in each interval.Let's handle each case separately.**Case 1: x >= 1 + sqrt(2)**Equation: 2^(x² - 2x - 3) - 1 = 0Let me solve this equation.2^(x² - 2x - 3) - 1 = 0=> 2^(x² - 2x - 3) = 1Since 2^0 = 1, so the exponent must be 0:x² - 2x - 3 = 0Solving the quadratic equation:x² - 2x - 3 = 0Using quadratic formula:x = [2 ± sqrt(4 + 12)] / 2 = [2 ± sqrt(16)] / 2 = [2 ± 4] / 2So, x = (2 + 4)/2 = 6/2 = 3Or x = (2 - 4)/2 = (-2)/2 = -1But in this case, x >= 1 + sqrt(2) ≈ 2.414, so x = 3 is valid, but x = -1 is not in this interval, so we discard it.So, one solution is x = 3.**Case 2: 0 <= x < 1 + sqrt(2)**Equation: x² - 2x + 1 = 0This is a quadratic equation. Let me factor it:x² - 2x + 1 = (x - 1)^2 = 0So, x = 1 is a double root.Now, check if x = 1 is within the interval 0 <= x < 1 + sqrt(2). Since 1 < 2.414, yes, it is valid.So, another solution is x = 1.**Case 3: x < 0**Equation: (1/x) + 2 = 0Let me solve for x.1/x + 2 = 0=> 1/x = -2=> x = -1/2Check if x = -1/2 is in the interval x < 0. Yes, it is.So, another solution is x = -1/2.Now, compiling all the solutions:- x = 3- x = 1- x = -1/2So, the zero points are at x = -1/2, x = 1, and x = 3.The problem asks for the sum of all zero points. So, let's add them up:3 + 1 + (-1/2) = 4 - 1/2 = 3.5Expressed as a fraction, that's 7/2.So, the sum of all zero points is 7/2.Let me just double-check my work to make sure I didn't make any mistakes.In Case 1, solving 2^(x² - 2x - 3) - 1 = 0 led me to x² - 2x - 3 = 0, which gave x = 3 and x = -1. Discarded x = -1 because it's not in the interval. That seems correct.In Case 2, solving x² - 2x + 1 = 0 gave x = 1, which is valid. Correct.In Case 3, solving 1/x + 2 = 0 gave x = -1/2, which is valid. Correct.Adding them up: 3 + 1 - 1/2 = 3.5 or 7/2. Yep, that seems right.I think that's all. So, the sum is 7/2.**Final Answer**The sum of all zero points of the function ( f[g(x)] ) is boxed{dfrac{7}{2}}.

💡Okay, so I need to figure out how many integers between 200 and 250 have three different digits in increasing order. Hmm, let's break this down step by step.First, the numbers we're looking at are from 200 up to, but not including, 250. So, all these numbers are three-digit numbers where the first digit is 2. That means the number is in the form 2ab, where 'a' is the second digit and 'b' is the third digit.Now, the problem says that the digits must be in increasing order and all different. So, each digit must be larger than the one before it. That means 2 < a < b. Let me think about what values 'a' can take. Since the number is less than 250, the second digit 'a' can only be 0, 1, 2, 3, or 4. If 'a' were 5 or higher, the number would be 250 or more, which is outside our range. So, 'a' can be 0, 1, 2, 3, or 4.Now, for each possible value of 'a', I need to figure out how many valid 'b's there are such that 'b' is greater than 'a' and all digits are unique.Let's start with 'a' = 0. If 'a' is 0, then 'b' has to be greater than 0. Since 'b' is a single digit, it can be from 1 to 9. That gives us 9 possible values for 'b' (1, 2, 3, 4, 5, 6, 7, 8, 9). So, for 'a' = 0, there are 9 numbers.Next, 'a' = 1. Here, 'b' has to be greater than 1. So, 'b' can be 2 to 9. That's 8 numbers. Wait, but hold on, the number is 21b, and we need all digits to be different. Since the first digit is 2, and 'a' is 1, 'b' can't be 2 because that would repeat the first digit. So, actually, 'b' can be 3 to 9, which is 7 numbers. So, for 'a' = 1, there are 7 numbers.Moving on to 'a' = 2. Now, 'b' has to be greater than 2. So, 'b' can be 3 to 9. That's 7 numbers. But wait, the first digit is 2, and 'a' is also 2. So, we have two 2s, which means the digits aren't all different. Oh, that's a problem. So, actually, 'a' can't be 2 because that would mean the second digit is the same as the first digit, which violates the "different digits" condition. Hmm, so does that mean 'a' can't be 2? Wait, no, the digits just have to be different, so if 'a' is 2, the number would be 22b, which has repeating digits. So, 'a' can't be 2. Therefore, for 'a' = 2, there are 0 valid numbers.Wait, that contradicts my earlier thought. Let me clarify. The number is 2ab, and all digits must be different. So, if 'a' is 2, then the first and second digits are both 2, which is not allowed. So, 'a' can't be 2. Therefore, 'a' can only be 0, 1, 3, or 4.So, let's correct that. 'a' can be 0, 1, 3, or 4.So, for 'a' = 0: 9 numbers.For 'a' = 1: 7 numbers.Now, 'a' = 3. Then, 'b' has to be greater than 3. So, 'b' can be 4 to 9. That's 6 numbers.For 'a' = 4. Then, 'b' has to be greater than 4. So, 'b' can be 5 to 9. That's 5 numbers.Wait, but earlier I thought 'a' could be 2, but now I realize it can't because of the repeating digit. So, 'a' can only be 0, 1, 3, 4.So, adding up the numbers:- 'a' = 0: 9- 'a' = 1: 7- 'a' = 3: 6- 'a' = 4: 5Total = 9 + 7 + 6 + 5 = 27.Wait, but earlier I thought 'a' could be 2, which would have given 7 numbers, but now I'm excluding it because of the repeating digit. So, is the total 27 or 34?Wait, let me double-check. If 'a' = 2 is allowed, then 'b' would have to be greater than 2, but since the first digit is 2, 'a' = 2 would make the first two digits both 2, which is not allowed. So, 'a' can't be 2. Therefore, the total is 27.But in the initial thought process, I considered 'a' = 2 and got 7 numbers, leading to a total of 34. But that was incorrect because 'a' = 2 would result in repeating digits.So, the correct total is 27.Wait, but let me think again. Maybe I made a mistake in excluding 'a' = 2. Let's see: the number is 2ab, with 2 < a < b, and all digits different.If 'a' = 2, then 2 < 2 < b, which is not possible because 2 is not less than 2. So, 'a' can't be 2. Therefore, 'a' can only be 0, 1, 3, 4.So, the total is 9 + 7 + 6 + 5 = 27.But wait, earlier I thought 'a' = 2 would give 7 numbers, but that's incorrect because 'a' can't be 2. So, the correct answer is 27.Wait, but let me check with an example. Let's list some numbers:For 'a' = 0: 201, 203, 204, 205, 206, 207, 208, 209. Wait, that's 8 numbers, not 9. Because 'b' has to be greater than 0, but also different from 2. Wait, no, 'b' just has to be greater than 'a', which is 0, and different from 2. So, 'b' can be 1, 3, 4, 5, 6, 7, 8, 9. That's 8 numbers, not 9. Hmm, so maybe my initial count for 'a' = 0 was wrong.Wait, if 'a' = 0, then 'b' has to be greater than 0 and different from 2. So, 'b' can be 1, 3, 4, 5, 6, 7, 8, 9. That's 8 numbers, not 9. So, I think I made a mistake there.Similarly, for 'a' = 1, 'b' has to be greater than 1 and different from 2. So, 'b' can be 3, 4, 5, 6, 7, 8, 9. That's 7 numbers.For 'a' = 3, 'b' has to be greater than 3 and different from 2. So, 'b' can be 4, 5, 6, 7, 8, 9. That's 6 numbers.For 'a' = 4, 'b' has to be greater than 4 and different from 2. So, 'b' can be 5, 6, 7, 8, 9. That's 5 numbers.So, adding up: 8 + 7 + 6 + 5 = 26.Wait, now I'm getting 26. Hmm, this is confusing.Wait, let's try to list all possible numbers:For 'a' = 0:- 201, 203, 204, 205, 206, 207, 208, 209. That's 8 numbers.For 'a' = 1:- 213, 214, 215, 216, 217, 218, 219. That's 7 numbers.For 'a' = 3:- 234, 235, 236, 237, 238, 239. That's 6 numbers.For 'a' = 4:- 245, 246, 247, 248, 249. That's 5 numbers.Total: 8 + 7 + 6 + 5 = 26.Wait, but earlier I thought it was 27. So, which is correct?Wait, maybe I missed a number when 'a' = 0. Let's see: 201, 203, 204, 205, 206, 207, 208, 209. That's 8 numbers. Is there another number? 202 is invalid because of repeating digits. 200 is invalid because 'a' is 0 and 'b' is 0, which is repeating. So, no, 8 is correct.Similarly, for 'a' = 1: 213, 214, 215, 216, 217, 218, 219. That's 7.For 'a' = 3: 234, 235, 236, 237, 238, 239. That's 6.For 'a' = 4: 245, 246, 247, 248, 249. That's 5.Total: 26.Wait, but earlier I thought it was 27. Maybe I double-counted somewhere.Alternatively, perhaps the initial assumption that 'a' can't be 2 is incorrect. Let's see: if 'a' = 2, then the number is 22b. But since the first digit is 2, and 'a' is also 2, that's repeating, so it's invalid. Therefore, 'a' can't be 2.So, the total is 26.Wait, but in the initial thought process, I thought it was 34, but that was because I included 'a' = 2, which is invalid.Wait, but let's think again. Maybe I'm missing some numbers.Wait, for 'a' = 0, 'b' can be 1, 3, 4, 5, 6, 7, 8, 9. That's 8 numbers.For 'a' = 1, 'b' can be 3, 4, 5, 6, 7, 8, 9. That's 7 numbers.For 'a' = 3, 'b' can be 4, 5, 6, 7, 8, 9. That's 6 numbers.For 'a' = 4, 'b' can be 5, 6, 7, 8, 9. That's 5 numbers.Total: 8 + 7 + 6 + 5 = 26.Wait, but let me check if there are any other possibilities. For example, if 'a' = 0, can 'b' be 2? No, because 'b' has to be greater than 'a' (which is 0), but also different from the first digit, which is 2. So, 'b' can't be 2. So, 'b' can be 1, 3, 4, 5, 6, 7, 8, 9.Similarly, for 'a' = 1, 'b' can't be 2, so 'b' is 3-9.For 'a' = 3, 'b' is 4-9.For 'a' = 4, 'b' is 5-9.So, 8 + 7 + 6 + 5 = 26.Wait, but I'm still confused because earlier I thought it was 34. Maybe the initial thought process was wrong.Alternatively, perhaps the problem allows 'a' to be equal to the first digit, but no, the digits must be different.Wait, let me think differently. Maybe I should consider all three digits being in increasing order, regardless of the first digit.So, the number is 2ab, with 2 < a < b, and all digits different.So, 2 < a < b.So, 'a' can be from 3 to 4, because if 'a' is 5, then the number would be 25b, which is 250 or more, which is outside our range.Wait, no, 'a' can be 0, 1, 2, 3, 4, but with 2 < a < b.Wait, no, 2 < a < b means 'a' has to be greater than 2. So, 'a' can be 3 or 4.Wait, that's a different approach. So, if 2 < a < b, then 'a' must be greater than 2, so 'a' can be 3 or 4.Then, for 'a' = 3, 'b' can be 4 to 9. That's 6 numbers.For 'a' = 4, 'b' can be 5 to 9. That's 5 numbers.Total: 6 + 5 = 11.Wait, that's even less. So, which approach is correct?Wait, the problem says "three different digits in increasing order." So, does that mean strictly increasing, or non-decreasing? I think it means strictly increasing, so each digit must be larger than the previous one.So, 2 < a < b.Therefore, 'a' must be greater than 2, so 'a' can be 3 or 4.So, for 'a' = 3, 'b' can be 4-9: 6 numbers.For 'a' = 4, 'b' can be 5-9: 5 numbers.Total: 11 numbers.Wait, but that contradicts the earlier counts. So, which is correct?Wait, let's think about what "three different digits in increasing order" means. It means that each digit is larger than the one before it. So, 2 < a < b.Therefore, 'a' must be greater than 2, so 'a' can be 3 or 4.So, for 'a' = 3, 'b' can be 4-9: 6 numbers.For 'a' = 4, 'b' can be 5-9: 5 numbers.Total: 11 numbers.But earlier, I considered 'a' = 0 and 'a' = 1, but if 2 < a < b, then 'a' has to be greater than 2, so 'a' can't be 0 or 1.Wait, that makes sense. So, the correct approach is that 'a' must be greater than 2, so 'a' can be 3 or 4.Therefore, the total is 11 numbers.Wait, but let's list them:For 'a' = 3:- 234, 235, 236, 237, 238, 239. That's 6 numbers.For 'a' = 4:- 245, 246, 247, 248, 249. That's 5 numbers.Total: 11 numbers.So, the answer should be 11.But wait, earlier I thought it was 26 or 27. So, which is correct?Wait, maybe the problem allows 'a' to be less than or equal to 2, as long as the digits are increasing. So, for example, 201 is 2 < 0 < 1, but that's not increasing because 2 > 0. So, that's not valid.Wait, no, 201: 2 > 0, so it's not increasing. So, 201 is invalid.Similarly, 213: 2 < 1 is false, so 213 is invalid.Wait, so actually, if 'a' is less than 2, then 2 > a, so the digits wouldn't be increasing. Therefore, 'a' must be greater than 2.Therefore, the only valid 'a's are 3 and 4, leading to 11 numbers.Wait, but let's think again. If 'a' is 0, then 2 > 0, so the digits aren't increasing. Similarly, if 'a' is 1, 2 > 1, so digits aren't increasing. Therefore, 'a' must be greater than 2, so 'a' can be 3 or 4.Therefore, the total is 11.But earlier, I thought it was 26 or 27, but that was because I considered 'a' = 0 and 'a' = 1, which don't satisfy the increasing order.So, the correct answer is 11.Wait, but let me check with the initial thought process. The initial thought process considered 'a' = 0, 1, 2, 3, 4, leading to 9 + 7 + 7 + 6 + 5 = 34, but that was incorrect because 'a' = 2 leads to repeating digits, and 'a' = 0 and 'a' = 1 don't satisfy the increasing order.Therefore, the correct answer is 11.Wait, but let me think again. Maybe the problem allows the digits to be in non-decreasing order, but the user specified "increasing order," which usually means strictly increasing.So, in that case, the answer is 11.But wait, let me check the initial problem statement again: "How many integers between 200 and 250 have three different digits in increasing order?"So, "increasing order" means each digit is larger than the previous one. Therefore, 2 < a < b.Therefore, 'a' must be greater than 2, so 'a' can be 3 or 4.Thus, the total is 11.Wait, but let me list all possible numbers to confirm:For 'a' = 3:- 234, 235, 236, 237, 238, 239. That's 6 numbers.For 'a' = 4:- 245, 246, 247, 248, 249. That's 5 numbers.Total: 11 numbers.Yes, that seems correct.Wait, but earlier I thought it was 26 or 27, but that was because I considered 'a' = 0 and 'a' = 1, which don't satisfy the increasing order.Therefore, the correct answer is 11.But wait, let me think again. Maybe the problem allows the digits to be in non-decreasing order, but the user specified "increasing order," which usually means strictly increasing.So, in that case, the answer is 11.But wait, let me check the initial thought process again. The initial thought process considered 'a' = 0, 1, 2, 3, 4, leading to 9 + 7 + 7 + 6 + 5 = 34, but that was incorrect because 'a' = 2 leads to repeating digits, and 'a' = 0 and 'a' = 1 don't satisfy the increasing order.Therefore, the correct answer is 11.Wait, but I'm still confused because the initial thought process was different. Maybe I need to clarify.In summary, the correct approach is:- The number is 2ab, with 2 < a < b, and all digits different.- Therefore, 'a' must be greater than 2, so 'a' can be 3 or 4.- For 'a' = 3, 'b' can be 4-9: 6 numbers.- For 'a' = 4, 'b' can be 5-9: 5 numbers.- Total: 11 numbers.Therefore, the answer is 11.

💡First, I need to compare the values of ( a = log_2 0.3 ), ( b = 3^{0.2} ), and ( c = 0.3^2 ).Starting with ( a = log_2 0.3 ), since ( 0.3 ) is less than ( 1 ), the logarithm will be negative. Specifically, ( log_2 0.3 ) is less than ( log_2 1 = 0 ).Next, for ( b = 3^{0.2} ), since the base ( 3 ) is greater than ( 1 ) and the exponent ( 0.2 ) is positive, ( b ) will be greater than ( 1 ).Finally, ( c = 0.3^2 ) is calculated by squaring ( 0.3 ), resulting in ( 0.09 ).Now, comparing the three values:- ( a ) is negative,- ( c ) is ( 0.09 ),- ( b ) is greater than ( 1 ).Thus, the order from smallest to largest is ( a < c < b ).

💡Okay, so I have this problem where I need to find the number of monomials (x^a y^b z^c) with non-zero coefficients in the expression ((x + y + z)^{2020} + (x - y - z)^{2020}). Hmm, that seems a bit complicated at first glance, but maybe I can break it down step by step.First, I remember that when you expand ((x + y + z)^n), you get a lot of terms, each of which is a product of powers of (x), (y), and (z). The coefficients are determined by the multinomial theorem. But in this case, we have two expressions: ((x + y + z)^{2020}) and ((x - y - z)^{2020}). When we add them together, some terms might cancel out or reinforce each other.Let me think about how to approach this. Maybe I can substitute (t = y + z) to simplify the expression. So, if I let (t = y + z), then the first term becomes ((x + t)^{2020}) and the second term becomes ((x - t)^{2020}). That seems manageable.Now, expanding both ((x + t)^{2020}) and ((x - t)^{2020}) using the binomial theorem. The binomial theorem says that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n-k} b^k). Applying this to both expressions:[(x + t)^{2020} = sum_{k=0}^{2020} binom{2020}{k} x^{2020 - k} t^k][(x - t)^{2020} = sum_{k=0}^{2020} binom{2020}{k} x^{2020 - k} (-t)^k]When I add these two expansions together, the terms with odd powers of (t) will cancel out because ((-t)^k) will be negative when (k) is odd. So, only the even powers of (t) will remain. That simplifies things a bit.So, adding them together:[(x + t)^{2020} + (x - t)^{2020} = 2 sum_{k text{ even}} binom{2020}{k} x^{2020 - k} t^k]Now, substituting back (t = y + z), we have:[2 sum_{k text{ even}} binom{2020}{k} x^{2020 - k} (y + z)^k]Next, I need to expand ((y + z)^k) for each even (k). Using the binomial theorem again:[(y + z)^k = sum_{m=0}^{k} binom{k}{m} y^m z^{k - m}]So, substituting this back into our expression, we get:[2 sum_{k text{ even}} binom{2020}{k} x^{2020 - k} sum_{m=0}^{k} binom{k}{m} y^m z^{k - m}]Now, combining the sums, each term in the expansion will be of the form:[2 binom{2020}{k} binom{k}{m} x^{2020 - k} y^m z^{k - m}]Here, (k) is even, and (m) ranges from 0 to (k). Each monomial is determined by the exponents (a = 2020 - k), (b = m), and (c = k - m). To find the number of distinct monomials, I need to count all possible combinations of (a), (b), and (c) such that (a + b + c = 2020) and the coefficient is non-zero. Since the coefficient is non-zero only when (k) is even, we can let (k = 2i) where (i) ranges from 0 to 1010 (since (2020 / 2 = 1010)).For each (i), (k = 2i), and then (m) can range from 0 to (2i). So, for each (i), there are (2i + 1) possible values of (m) (from 0 to (2i)). Therefore, the total number of monomials is the sum of (2i + 1) for (i) from 0 to 1010.This is an arithmetic series where the first term is 1 (when (i = 0)) and the last term is (2*1010 + 1 = 2021). The number of terms in this series is 1011 (from (i = 0) to (i = 1010)).The sum of the first (n) odd numbers is (n^2). Here, (n = 1011), so the total number of monomials is (1011^2).Calculating (1011^2):[1011^2 = (1000 + 11)^2 = 1000^2 + 2*1000*11 + 11^2 = 1,000,000 + 22,000 + 121 = 1,022,121]So, the number of monomials with non-zero coefficients is 1,022,121.**Final Answer**The number of monomials with non-zero coefficients is boxed{1022121}.