💡Okay, so I have this geometry problem about a prism (ABC-A'B'C'). The base is an equilateral triangle with side length 2. The lateral edge (AA') forms a 45-degree angle with the edges (AB) and (AC) of the base. Also, point (A') is equidistant from the planes (ABC) and (BB'C'C). I need to find the length of (A'A).Alright, let me try to visualize this prism. Since the base is an equilateral triangle, all sides are equal, and all angles are 60 degrees. The prism has two congruent bases connected by rectangles. So, (ABC) is the base, and (A'B'C') is the top face.Now, the lateral edge (AA') forms a 45-degree angle with edges (AB) and (AC). Hmm, that means if I look at the edge (AA'), it's slanting at 45 degrees relative to both (AB) and (AC). Since (AB) and (AC) are edges of the base, which is an equilateral triangle, they are both of length 2.I think I should draw a diagram to help me understand the spatial relationships here. But since I can't draw, I'll try to imagine it. Point (A') is above point (A), connected by the lateral edge (AA'). The angle between (AA') and (AB) is 45 degrees, and the same with (AC).Wait, if (AA') makes a 45-degree angle with both (AB) and (AC), then the projection of (AA') onto the base (ABC) must form some specific triangle. Maybe a right triangle?Let me recall that in a prism, the lateral edges are perpendicular to the base if it's a right prism. But here, it's not necessarily a right prism because (AA') is at a 45-degree angle. So, this is an oblique prism.Since (AA') is at 45 degrees to both (AB) and (AC), the projection of (AA') onto the base should be such that it makes equal angles with (AB) and (AC). Given that (AB) and (AC) are sides of an equilateral triangle, they are at 60 degrees to each other.Hmm, maybe I can use some trigonometry here. If I consider the triangle formed by (AA'), its projection onto the base, and the height from (A') to the base, I can relate the angles.Let me denote the length of (AA') as (h). Since it's at a 45-degree angle with (AB) and (AC), the projection of (AA') onto the base should be equal to (h cos(45^circ)). But wait, the projection onto the base is a vector, not just a scalar length. So, maybe I need to consider the components of this projection along (AB) and (AC).Given that the projection makes equal angles with (AB) and (AC), and since (AB) and (AC) are at 60 degrees to each other, the projection vector should lie along the angle bisector of (BAC). In an equilateral triangle, the angle bisector is also the median and the altitude.So, the projection of (AA') onto the base is along the altitude from (A) to (BC). The length of this altitude is (h_{base} = sqrt{3}), as calculated from the side length 2.Now, if the projection of (AA') onto the base is along this altitude, then the length of the projection is (h cos(45^circ)). But this projection must also relate to the height of the prism.Wait, but the prism's height is the perpendicular distance between the bases. However, since (AA') is not perpendicular, the height of the prism is actually the vertical component of (AA'). So, if (AA') is at 45 degrees, then the vertical component is (h sin(45^circ)).But I also know that point (A') is equidistant from the planes (ABC) and (BB'C'C). The plane (ABC) is the base, and the plane (BB'C'C) is one of the lateral faces.So, the distance from (A') to (ABC) is the height of the prism, which is (h sin(45^circ)). The distance from (A') to (BB'C'C) is a bit trickier. Since (BB'C'C) is a rectangle, the distance from (A') to this plane would be the perpendicular distance from (A') to the plane.To find this distance, I might need to use the formula for the distance from a point to a plane in 3D space. But I need coordinates for that.Let me set up a coordinate system. Let me place point (A) at the origin ((0, 0, 0)). Since the base is an equilateral triangle with side length 2, I can place point (B) at ((2, 0, 0)). Point (C) would then be at ((1, sqrt{3}, 0)).Now, point (A') is somewhere above (A), but not directly above since the prism is oblique. Let me denote the coordinates of (A') as ((x, y, z)). Since (AA') forms a 45-degree angle with both (AB) and (AC), I can use the dot product to express this.The vector (AA') is ((x, y, z)). The vector (AB) is ((2, 0, 0)), and the vector (AC) is ((1, sqrt{3}, 0)).The angle between (AA') and (AB) is 45 degrees, so:[cos(45^circ) = frac{(x, y, z) cdot (2, 0, 0)}{| (x, y, z) | cdot | (2, 0, 0) |}]Similarly, the angle between (AA') and (AC) is 45 degrees:[cos(45^circ) = frac{(x, y, z) cdot (1, sqrt{3}, 0)}{| (x, y, z) | cdot | (1, sqrt{3}, 0) |}]Let me compute these dot products.First, for (AB):[frac{2x + 0 + 0}{sqrt{x^2 + y^2 + z^2} cdot 2} = frac{sqrt{2}}{2}]Simplify:[frac{2x}{2 sqrt{x^2 + y^2 + z^2}} = frac{sqrt{2}}{2}][frac{x}{sqrt{x^2 + y^2 + z^2}} = frac{sqrt{2}}{2}]Square both sides:[frac{x^2}{x^2 + y^2 + z^2} = frac{1}{2}][2x^2 = x^2 + y^2 + z^2][x^2 = y^2 + z^2]Similarly, for (AC):[frac{x + y sqrt{3} + 0}{sqrt{x^2 + y^2 + z^2} cdot 2} = frac{sqrt{2}}{2}]Simplify:[frac{x + y sqrt{3}}{2 sqrt{x^2 + y^2 + z^2}} = frac{sqrt{2}}{2}]Multiply both sides by 2:[frac{x + y sqrt{3}}{sqrt{x^2 + y^2 + z^2}} = sqrt{2}]Square both sides:[frac{(x + y sqrt{3})^2}{x^2 + y^2 + z^2} = 2]Expand the numerator:[frac{x^2 + 2xy sqrt{3} + 3y^2}{x^2 + y^2 + z^2} = 2]From the first equation, we have (x^2 = y^2 + z^2). Let's substitute (x^2) in the second equation:[frac{(y^2 + z^2) + 2xy sqrt{3} + 3y^2}{(y^2 + z^2) + y^2 + z^2} = 2]Simplify numerator and denominator:Numerator: (y^2 + z^2 + 2xy sqrt{3} + 3y^2 = 4y^2 + z^2 + 2xy sqrt{3})Denominator: (2y^2 + 2z^2)So:[frac{4y^2 + z^2 + 2xy sqrt{3}}{2y^2 + 2z^2} = 2]Multiply both sides by denominator:[4y^2 + z^2 + 2xy sqrt{3} = 4y^2 + 4z^2]Simplify:[z^2 + 2xy sqrt{3} = 4z^2][2xy sqrt{3} = 3z^2]From the first equation, (x^2 = y^2 + z^2), so (z^2 = x^2 - y^2). Substitute into the above equation:[2xy sqrt{3} = 3(x^2 - y^2)]Let me rearrange this:[3x^2 - 3y^2 - 2xy sqrt{3} = 0]This is a quadratic in terms of (x) and (y). Maybe I can express (x) in terms of (y) or vice versa.Let me assume (x = ky), where (k) is a constant. Then:[3(k y)^2 - 3y^2 - 2(k y)y sqrt{3} = 0]Simplify:[3k^2 y^2 - 3y^2 - 2k y^2 sqrt{3} = 0]Factor out (y^2):[y^2 (3k^2 - 3 - 2k sqrt{3}) = 0]Since (y) can't be zero (otherwise, (x) would be zero, and (A') would be at the origin, which doesn't make sense), we have:[3k^2 - 3 - 2k sqrt{3} = 0]This is a quadratic equation in (k):[3k^2 - 2sqrt{3}k - 3 = 0]Let me solve for (k):Using quadratic formula:[k = frac{2sqrt{3} pm sqrt{(2sqrt{3})^2 - 4 cdot 3 cdot (-3)}}{2 cdot 3}]Calculate discriminant:[(2sqrt{3})^2 - 4 cdot 3 cdot (-3) = 12 + 36 = 48]So,[k = frac{2sqrt{3} pm sqrt{48}}{6} = frac{2sqrt{3} pm 4sqrt{3}}{6}]Thus,First solution:[k = frac{2sqrt{3} + 4sqrt{3}}{6} = frac{6sqrt{3}}{6} = sqrt{3}]Second solution:[k = frac{2sqrt{3} - 4sqrt{3}}{6} = frac{-2sqrt{3}}{6} = -frac{sqrt{3}}{3}]So, (k = sqrt{3}) or (k = -frac{sqrt{3}}{3}).Let me consider both cases.Case 1: (k = sqrt{3}), so (x = sqrt{3} y).From (x^2 = y^2 + z^2):[(3 y^2) = y^2 + z^2 implies z^2 = 2 y^2 implies z = sqrt{2} y]Case 2: (k = -frac{sqrt{3}}{3}), so (x = -frac{sqrt{3}}{3} y).From (x^2 = y^2 + z^2):[left( frac{3}{9} y^2 right) = y^2 + z^2 implies frac{1}{3} y^2 = y^2 + z^2 implies z^2 = -frac{2}{3} y^2]This is impossible since (z^2) can't be negative. So, Case 2 is invalid.Therefore, only Case 1 is valid: (x = sqrt{3} y) and (z = sqrt{2} y).Now, let me express coordinates of (A') as ((sqrt{3} y, y, sqrt{2} y)).Now, I need to find the value of (y). For that, I can use the fact that point (A') is equidistant from planes (ABC) and (BB'C'C).First, the distance from (A') to plane (ABC). Since plane (ABC) is the base, which is the xy-plane in my coordinate system. The distance from a point ((x, y, z)) to the xy-plane is simply (|z|). So, distance from (A') to (ABC) is (|sqrt{2} y|).Next, the distance from (A') to plane (BB'C'C). Plane (BB'C'C) is a rectangle. Let me find the equation of this plane.Points (B), (B'), (C'), and (C) lie on this plane. Coordinates:- (B = (2, 0, 0))- (C = (1, sqrt{3}, 0))- (B') is the top of (B), so since (AA') is ((sqrt{3} y, y, sqrt{2} y)), assuming the prism is such that all lateral edges are equal and similarly oriented, but actually, in a prism, the lateral edges are all congruent and parallel. So, (BB') should be equal and parallel to (AA'). Therefore, (B') is (B + (x, y, z)), but wait, no, in a prism, the translation vector is the same for all vertices.Wait, actually, in a prism, each vertex is translated by the same vector to get the top face. So, if (A') is (A + vec{v}), then (B') is (B + vec{v}), and (C') is (C + vec{v}).But in this case, (A') is not directly above (A), so (vec{v}) is the vector from (A) to (A'), which is ((sqrt{3} y, y, sqrt{2} y)). Therefore, (B') is (B + vec{v} = (2 + sqrt{3} y, 0 + y, 0 + sqrt{2} y)), and (C') is (C + vec{v} = (1 + sqrt{3} y, sqrt{3} + y, 0 + sqrt{2} y)).So, plane (BB'C'C) has points (B = (2, 0, 0)), (B' = (2 + sqrt{3} y, y, sqrt{2} y)), (C' = (1 + sqrt{3} y, sqrt{3} + y, sqrt{2} y)), and (C = (1, sqrt{3}, 0)).To find the equation of plane (BB'C'C), I can use three points: (B), (B'), and (C).First, find two vectors on the plane:Vector (BB' = B' - B = (sqrt{3} y, y, sqrt{2} y))Vector (BC = C - B = (-1, sqrt{3}, 0))Now, the normal vector to the plane is the cross product of (BB') and (BC).Compute cross product:Let me denote (BB' = (sqrt{3} y, y, sqrt{2} y)) and (BC = (-1, sqrt{3}, 0)).Cross product (BB' times BC):[begin{vmatrix}mathbf{i} & mathbf{j} & mathbf{k} sqrt{3} y & y & sqrt{2} y -1 & sqrt{3} & 0 end{vmatrix}]Calculate determinant:[mathbf{i} (y cdot 0 - sqrt{2} y cdot sqrt{3}) - mathbf{j} (sqrt{3} y cdot 0 - sqrt{2} y cdot (-1)) + mathbf{k} (sqrt{3} y cdot sqrt{3} - y cdot (-1))]Simplify each component:- i-component: (0 - sqrt{6} y = -sqrt{6} y)- j-component: (- (0 + sqrt{2} y) = -sqrt{2} y)- k-component: (3 y + y = 4 y)So, the normal vector is ((- sqrt{6} y, - sqrt{2} y, 4 y)). We can factor out (y):[y (- sqrt{6}, - sqrt{2}, 4)]So, the normal vector is proportional to ((- sqrt{6}, - sqrt{2}, 4)).The equation of the plane can be written as:[- sqrt{6} (x - 2) - sqrt{2} (y - 0) + 4 (z - 0) = 0]Simplify:[- sqrt{6} x + 2 sqrt{6} - sqrt{2} y + 4 z = 0]Bring constants to the other side:[- sqrt{6} x - sqrt{2} y + 4 z = -2 sqrt{6}]So, the equation of plane (BB'C'C) is:[- sqrt{6} x - sqrt{2} y + 4 z = -2 sqrt{6}]Now, the distance from point (A') ((sqrt{3} y, y, sqrt{2} y)) to this plane is given by the formula:[text{Distance} = frac{| - sqrt{6} (sqrt{3} y) - sqrt{2} (y) + 4 (sqrt{2} y) + 2 sqrt{6} |}{sqrt{ (sqrt{6})^2 + (sqrt{2})^2 + 4^2 }}]Simplify numerator:First, compute each term:- ( - sqrt{6} (sqrt{3} y) = - sqrt{18} y = - 3 sqrt{2} y )- ( - sqrt{2} y )- ( 4 (sqrt{2} y) = 4 sqrt{2} y )- ( + 2 sqrt{6} )Combine like terms:- Terms with ( sqrt{2} y ): ( -3 sqrt{2} y - sqrt{2} y + 4 sqrt{2} y = 0 )- Constant term: ( + 2 sqrt{6} )So, numerator is ( | 0 + 2 sqrt{6} | = 2 sqrt{6} )Denominator:[sqrt{ (sqrt{6})^2 + (sqrt{2})^2 + 4^2 } = sqrt{6 + 2 + 16} = sqrt{24} = 2 sqrt{6}]Therefore, distance is:[frac{2 sqrt{6}}{2 sqrt{6}} = 1]So, the distance from (A') to plane (BB'C'C) is 1.Earlier, we found that the distance from (A') to plane (ABC) is (|sqrt{2} y|).Given that (A') is equidistant from both planes, we have:[|sqrt{2} y| = 1]Since (y) is a coordinate, it can be positive or negative, but distance is positive, so:[sqrt{2} |y| = 1 implies |y| = frac{1}{sqrt{2}} implies y = pm frac{sqrt{2}}{2}]Since the prism is above the base, we can assume (y) is positive. So, (y = frac{sqrt{2}}{2}).Now, recall that (A') has coordinates ((sqrt{3} y, y, sqrt{2} y)). Plugging in (y = frac{sqrt{2}}{2}):- (x = sqrt{3} cdot frac{sqrt{2}}{2} = frac{sqrt{6}}{2})- (y = frac{sqrt{2}}{2})- (z = sqrt{2} cdot frac{sqrt{2}}{2} = frac{2}{2} = 1)So, coordinates of (A') are (left( frac{sqrt{6}}{2}, frac{sqrt{2}}{2}, 1 right)).Now, the length (AA') is the distance from (A(0,0,0)) to (A'left( frac{sqrt{6}}{2}, frac{sqrt{2}}{2}, 1 right)):[AA' = sqrt{ left( frac{sqrt{6}}{2} - 0 right)^2 + left( frac{sqrt{2}}{2} - 0 right)^2 + (1 - 0)^2 }]Calculate each term:- ( left( frac{sqrt{6}}{2} right)^2 = frac{6}{4} = frac{3}{2} )- ( left( frac{sqrt{2}}{2} right)^2 = frac{2}{4} = frac{1}{2} )- (1^2 = 1)Sum:[frac{3}{2} + frac{1}{2} + 1 = 2 + 1 = 3]So,[AA' = sqrt{3}]Therefore, the length of (A'A) is (sqrt{3}).

💡Alright, so I have this problem where I need to prove that if a function ( f ) satisfies the equation[ f(x+1) + f(x-1) = sqrt{2} f(x) ]for all real numbers ( x ), then the function ( f ) is periodic. Hmm, okay. I remember that a periodic function repeats its values at regular intervals, called periods. So, I need to show that there exists some positive number ( T ) such that ( f(x + T) = f(x) ) for all ( x ).First, let me write down the given equation again:[ f(x+1) + f(x-1) = sqrt{2} f(x) ]This looks like a functional equation that relates the values of the function at points ( x+1 ), ( x ), and ( x-1 ). Maybe I can use this equation to express ( f(x+1) ) or ( f(x-1) ) in terms of ( f(x) ) and then iterate this process to find a pattern or a recurrence relation.Let me try to express ( f(x+1) ) from the equation:[ f(x+1) = sqrt{2} f(x) - f(x-1) ]Okay, so ( f(x+1) ) is expressed in terms of ( f(x) ) and ( f(x-1) ). Maybe I can use this recursively to find expressions for ( f(x+2) ), ( f(x+3) ), and so on.Let's compute ( f(x+2) ). Using the same equation, but replacing ( x ) with ( x+1 ):[ f((x+1)+1) + f((x+1)-1) = sqrt{2} f(x+1) ][ f(x+2) + f(x) = sqrt{2} f(x+1) ]From this, we can solve for ( f(x+2) ):[ f(x+2) = sqrt{2} f(x+1) - f(x) ]But we already have an expression for ( f(x+1) ):[ f(x+1) = sqrt{2} f(x) - f(x-1) ]So, substitute this into the expression for ( f(x+2) ):[ f(x+2) = sqrt{2} (sqrt{2} f(x) - f(x-1)) - f(x) ][ f(x+2) = 2 f(x) - sqrt{2} f(x-1) - f(x) ][ f(x+2) = f(x) - sqrt{2} f(x-1) ]Hmm, interesting. So, ( f(x+2) ) is expressed in terms of ( f(x) ) and ( f(x-1) ). Maybe I can continue this process to find ( f(x+3) ), ( f(x+4) ), etc.Let's compute ( f(x+3) ). Again, using the original equation, replace ( x ) with ( x+2 ):[ f((x+2)+1) + f((x+2)-1) = sqrt{2} f(x+2) ][ f(x+3) + f(x+1) = sqrt{2} f(x+2) ]We have expressions for both ( f(x+1) ) and ( f(x+2) ):[ f(x+1) = sqrt{2} f(x) - f(x-1) ][ f(x+2) = f(x) - sqrt{2} f(x-1) ]Substitute these into the equation for ( f(x+3) ):[ f(x+3) + (sqrt{2} f(x) - f(x-1)) = sqrt{2} (f(x) - sqrt{2} f(x-1)) ][ f(x+3) + sqrt{2} f(x) - f(x-1) = sqrt{2} f(x) - 2 f(x-1) ]Subtract ( sqrt{2} f(x) - f(x-1) ) from both sides:[ f(x+3) = -2 f(x-1) + f(x-1) ][ f(x+3) = -f(x-1) ]Wait, that's a nice simplification. So, ( f(x+3) = -f(x-1) ). Let me note that down.Now, let's compute ( f(x+4) ). Replace ( x ) with ( x+3 ) in the original equation:[ f((x+3)+1) + f((x+3)-1) = sqrt{2} f(x+3) ][ f(x+4) + f(x+2) = sqrt{2} f(x+3) ]We have expressions for ( f(x+2) ) and ( f(x+3) ):[ f(x+2) = f(x) - sqrt{2} f(x-1) ][ f(x+3) = -f(x-1) ]Substitute these into the equation for ( f(x+4) ):[ f(x+4) + (f(x) - sqrt{2} f(x-1)) = sqrt{2} (-f(x-1)) ][ f(x+4) + f(x) - sqrt{2} f(x-1) = -sqrt{2} f(x-1) ]Subtract ( f(x) - sqrt{2} f(x-1) ) from both sides:[ f(x+4) = -sqrt{2} f(x-1) - f(x) + sqrt{2} f(x-1) ][ f(x+4) = -f(x) ]Oh, that's a significant result. So, ( f(x+4) = -f(x) ). Let's keep that in mind.Now, let's compute ( f(x+8) ). If ( f(x+4) = -f(x) ), then replacing ( x ) with ( x+4 ):[ f((x+4)+4) = -f(x+4) ][ f(x+8) = -(-f(x)) ][ f(x+8) = f(x) ]So, we've shown that ( f(x+8) = f(x) ) for all ( x ). This means that the function ( f ) repeats its values every 8 units. Therefore, ( f ) is periodic with period 8.Wait, but is 8 the fundamental period? Or could there be a smaller period? Let's check if a smaller period exists.Suppose there exists a period ( T ) such that ( T < 8 ) and ( f(x+T) = f(x) ) for all ( x ). From our previous results, we have ( f(x+4) = -f(x) ). If ( T ) were a period, then ( f(x+4) = f(x) ) would imply ( -f(x) = f(x) ), which would mean ( f(x) = 0 ) for all ( x ). But the problem doesn't specify that ( f ) is the zero function, so we can't assume that. Therefore, 8 is indeed the fundamental period.So, putting it all together, by repeatedly applying the given functional equation, we derived that ( f(x+8) = f(x) ), proving that ( f ) is periodic with period 8.**Final Answer**The function ( f ) is periodic with period 8, so the final answer is boxed{8}.

💡Okay, so I have this function f(x) = 2x³ - a x² + 1, and I need to find the range of values for a such that the function has exactly two zeros in the interval [1/2, 2]. Hmm, okay, let's break this down step by step.First, I remember that to find the zeros of a function, I can set f(x) equal to zero and solve for x. But since this is a cubic equation, it might have up to three real roots. The question is specifically asking for exactly two zeros in the interval [1/2, 2]. So, I need to figure out the conditions on a that make this happen.I also recall that for a function to have exactly two zeros in an interval, it must cross the x-axis twice within that interval. This usually implies that the function has a local maximum and a local minimum within the interval, and that the function changes direction (from increasing to decreasing or vice versa) in such a way that it crosses the x-axis twice.So, maybe I should start by finding the derivative of f(x) to locate its critical points. The derivative, f'(x), will tell me where the function is increasing or decreasing, which can help identify local maxima and minima.Calculating the derivative:f'(x) = d/dx [2x³ - a x² + 1] = 6x² - 2a x.Setting the derivative equal to zero to find critical points:6x² - 2a x = 0Factor out 2x:2x(3x - a) = 0So, x = 0 or x = a/3.Okay, so the critical points are at x = 0 and x = a/3. Since our interval of interest is [1/2, 2], x = 0 is outside this interval, so the only critical point inside [1/2, 2] is x = a/3, provided that a/3 is within [1/2, 2].So, for x = a/3 to be in [1/2, 2], we need 1/2 ≤ a/3 ≤ 2. Multiplying all parts by 3, that gives 3/2 ≤ a ≤ 6. So, if a is between 3/2 and 6, the critical point x = a/3 is inside our interval. Otherwise, it's outside.Now, let's consider the behavior of f(x) on the interval [1/2, 2]. Since f'(x) is a quadratic, it can change signs. Let's analyze the sign of f'(x) around the critical points.For x < 0: Let's pick x = -1 (though this is outside our interval, just for understanding). Plugging into f'(x): 6*(-1)² - 2a*(-1) = 6 + 2a. Since a is a real number, this is positive. So, f(x) is increasing for x < 0.Between x = 0 and x = a/3: Let's pick x = a/6, which is between 0 and a/3. Plugging into f'(x): 6*(a/6)² - 2a*(a/6) = 6*(a²/36) - 2a²/6 = a²/6 - a²/3 = -a²/6. This is negative, so f(x) is decreasing in this interval.For x > a/3: Let's pick x = a/2, which is greater than a/3 (assuming a > 0). Plugging into f'(x): 6*(a/2)² - 2a*(a/2) = 6*(a²/4) - a² = (3a²/2) - a² = a²/2. This is positive, so f(x) is increasing for x > a/3.So, summarizing:- For x < 0: f(x) is increasing.- For 0 < x < a/3: f(x) is decreasing.- For x > a/3: f(x) is increasing.But since our interval is [1/2, 2], and assuming a/3 is within this interval (i.e., 3/2 ≤ a ≤ 6), then on [1/2, a/3], f(x) is decreasing, and on [a/3, 2], f(x) is increasing.Therefore, the function has a local minimum at x = a/3 within [1/2, 2]. So, the function decreases from x = 1/2 to x = a/3, reaching a minimum, then increases from x = a/3 to x = 2.For the function to have exactly two zeros in [1/2, 2], it must cross the x-axis once while decreasing and once while increasing. That means the function must be positive at x = 1/2, negative at the local minimum x = a/3, and positive again at x = 2. Alternatively, it could be negative at x = 1/2, positive at the local minimum, and negative at x = 2, but since the leading coefficient is positive (2), as x approaches infinity, f(x) tends to infinity, so it's more likely that the function starts positive at x = 1/2, dips below zero at the minimum, and then comes back up to positive at x = 2.Wait, but let's check the value at x = 1/2 and x = 2.Calculating f(1/2):f(1/2) = 2*(1/2)³ - a*(1/2)² + 1 = 2*(1/8) - a*(1/4) + 1 = 1/4 - a/4 + 1 = (1 + 1/4) - a/4 = 5/4 - a/4.Similarly, f(2):f(2) = 2*(8) - a*(4) + 1 = 16 - 4a + 1 = 17 - 4a.So, f(1/2) = 5/4 - a/4, and f(2) = 17 - 4a.We need the function to cross the x-axis twice in [1/2, 2]. So, for that, the function must start at f(1/2), go down to a minimum at x = a/3, and then come back up to f(2). So, for two zeros, f(1/2) and f(2) must be positive, and the minimum at x = a/3 must be negative.Therefore, the conditions are:1. f(1/2) > 0: 5/4 - a/4 > 0 ⇒ 5 - a > 0 ⇒ a < 5.2. f(2) > 0: 17 - 4a > 0 ⇒ 17 > 4a ⇒ a < 17/4 = 4.25.3. f(a/3) < 0: The function at the local minimum is negative.So, combining conditions 1 and 2, a must be less than 4.25.But also, we need the local minimum at x = a/3 to be negative. Let's compute f(a/3):f(a/3) = 2*(a/3)³ - a*(a/3)² + 1 = 2*(a³/27) - a*(a²/9) + 1 = (2a³)/27 - (a³)/9 + 1.Simplify this:(2a³)/27 - (3a³)/27 + 1 = (-a³)/27 + 1.So, f(a/3) = 1 - (a³)/27.We need this to be less than zero:1 - (a³)/27 < 0 ⇒ (a³)/27 > 1 ⇒ a³ > 27 ⇒ a > 3.Wait, hold on, that can't be right because 3³ is 27, so a³ > 27 implies a > 3. But earlier, from condition 2, a must be less than 17/4, which is 4.25. So, combining all these, a must be greater than 3 and less than 4.25.But wait, earlier I thought that a must be between 3/2 and 6 for the critical point x = a/3 to be in [1/2, 2]. So, if a > 3, then a/3 > 1, which is still within [1/2, 2] as long as a ≤ 6. But since a < 4.25, which is less than 6, so x = a/3 is still within [1, 4.25/3 ≈ 1.416], which is within [1/2, 2].Wait, but hold on, if a > 3, then a/3 > 1, so the critical point is in [1, 2], which is still within [1/2, 2]. So, that's fine.But let me double-check the calculation for f(a/3):f(a/3) = 2*(a/3)^3 - a*(a/3)^2 + 1= 2*(a³/27) - a*(a²/9) + 1= (2a³)/27 - (a³)/9 + 1= (2a³ - 3a³)/27 + 1= (-a³)/27 + 1.Yes, that's correct. So, f(a/3) = 1 - (a³)/27.So, setting this less than zero:1 - (a³)/27 < 0 ⇒ (a³)/27 > 1 ⇒ a³ > 27 ⇒ a > 3.So, a must be greater than 3.But earlier, from f(1/2) > 0, we had a < 5, and from f(2) > 0, a < 17/4 = 4.25. So, combining these, a must be greater than 3 and less than 4.25.But wait, is that all? Let me think.We also need to ensure that the function actually crosses the x-axis twice in [1/2, 2]. So, f(1/2) > 0, f(a/3) < 0, and f(2) > 0. So, the function starts positive at x = 1/2, dips below zero at x = a/3, and then comes back up to positive at x = 2. Therefore, it must cross the x-axis once between 1/2 and a/3, and once between a/3 and 2. So, that gives exactly two zeros in [1/2, 2].But wait, what if the function doesn't cross the x-axis at all? Or crosses only once? So, we need to ensure that the function actually crosses the x-axis twice. So, the conditions f(1/2) > 0, f(a/3) < 0, and f(2) > 0 are sufficient for two crossings.But let's also check the endpoints. If f(1/2) is positive, f(a/3) is negative, and f(2) is positive, then by the Intermediate Value Theorem, there must be a zero between 1/2 and a/3, and another zero between a/3 and 2. So, that's two zeros.But we also need to ensure that there are no more zeros outside these intervals. Since the function is increasing before x = a/3 and decreasing after, but wait, no, actually, in our interval [1/2, 2], the function is decreasing from 1/2 to a/3 and increasing from a/3 to 2. So, it can only have two zeros in this interval.But wait, what if a is such that f(1/2) is zero or f(2) is zero? Then, we might have a zero at the endpoint, which could affect the count.But the problem says "exactly two zeros in the interval [1/2, 2]." So, if f(1/2) = 0 or f(2) = 0, that would be a zero at the endpoint, which is still counted as a zero in the interval. So, we need to be careful about that.But in our earlier conditions, we have f(1/2) > 0 and f(2) > 0. So, if a is such that f(1/2) = 0, that would be a = 5, because f(1/2) = 5/4 - a/4 = 0 ⇒ a = 5. Similarly, f(2) = 0 ⇒ a = 17/4 = 4.25.So, if a = 5, f(1/2) = 0, and if a = 4.25, f(2) = 0. So, in those cases, the function would have a zero at the endpoint. But the problem says "exactly two zeros in the interval [1/2, 2]." So, if a = 5, f(1/2) = 0, and then f(a/3) < 0, so the function would cross from positive to negative at x = a/3, and then back to positive at x = 2. So, that would be two zeros: one at x = 1/2 and one between a/3 and 2. Similarly, if a = 4.25, f(2) = 0, and the function would cross from positive to negative at x = a/3, and then back to zero at x = 2. So, that would be two zeros: one between 1/2 and a/3, and one at x = 2.But wait, in the case where a = 5, f(1/2) = 0, and f(a/3) = f(5/3). Let's compute f(5/3):f(5/3) = 2*(125/27) - 5*(25/9) + 1 = 250/27 - 125/9 + 1 = 250/27 - 375/27 + 27/27 = (250 - 375 + 27)/27 = (-98)/27 ≈ -3.63 < 0.So, at a = 5, f(1/2) = 0, f(5/3) < 0, and f(2) = 17 - 20 = -3 < 0. Wait, that's not right. Wait, f(2) when a = 5 is 17 - 20 = -3, which is negative. So, f(2) is negative when a = 5. But earlier, we had f(2) > 0 when a < 17/4 ≈ 4.25. So, when a = 5, f(2) is negative.Wait, this is conflicting with my earlier conclusion. Let me recast this.Wait, f(2) = 17 - 4a. So, when a = 5, f(2) = 17 - 20 = -3, which is negative. So, at a = 5, f(1/2) = 0, f(5/3) < 0, and f(2) < 0. So, the function starts at zero at x = 1/2, goes down to a minimum at x = 5/3, and then continues decreasing to x = 2, which is also negative. So, in this case, the function only has one zero at x = 1/2, because it doesn't come back up to positive at x = 2. So, that would mean only one zero in the interval, which contradicts our earlier thought.Wait, so perhaps my initial conditions were incorrect. Let me re-examine.If a = 5, f(1/2) = 0, f(5/3) < 0, and f(2) = -3 < 0. So, the function is zero at x = 1/2, negative at x = 5/3, and negative at x = 2. So, it doesn't cross the x-axis again after x = 1/2. So, only one zero in the interval.Similarly, if a = 4.25, f(2) = 0, and f(a/3) = f(17/12). Let's compute f(17/12):f(17/12) = 2*(17/12)^3 - (17/4)*(17/12)^2 + 1.This seems complicated, but let's compute it step by step.First, (17/12)^2 = 289/144.(17/12)^3 = (17^3)/(12^3) = 4913/1728.So, f(17/12) = 2*(4913/1728) - (17/4)*(289/144) + 1.Compute each term:2*(4913/1728) = 9826/1728 ≈ 5.687.(17/4)*(289/144) = (17*289)/(4*144) = (4913)/576 ≈ 8.525.So, f(17/12) ≈ 5.687 - 8.525 + 1 ≈ -1.838 < 0.So, f(17/12) < 0. Therefore, at a = 17/4, f(2) = 0, and f(a/3) < 0. So, the function starts at f(1/2) = 5/4 - (17/4)/4 = 5/4 - 17/16 = (20/16 - 17/16) = 3/16 > 0, goes down to f(17/12) < 0, and then comes back up to f(2) = 0. So, in this case, the function crosses the x-axis once between 1/2 and 17/12, and once at x = 2. So, that's two zeros in the interval.Therefore, when a = 17/4, we have exactly two zeros: one in (1/2, 17/12) and one at x = 2.Similarly, when a = 5, f(1/2) = 0, f(5/3) < 0, and f(2) < 0, so only one zero in the interval.Therefore, our earlier conclusion that a must be greater than 3 and less than 17/4 is correct, but we need to include a = 17/4 because at that point, the function has a zero at x = 2, which is still within the interval.But wait, what about a = 3? Let's check a = 3.f(a/3) = f(1) = 2*(1)^3 - 3*(1)^2 + 1 = 2 - 3 + 1 = 0.So, at a = 3, f(1) = 0. So, the function has a zero at x = 1, which is within [1/2, 2]. Also, f(1/2) = 5/4 - 3/4 = 2/4 = 1/2 > 0, and f(2) = 17 - 12 = 5 > 0. So, the function starts positive at x = 1/2, dips to zero at x = 1, and then goes back up to positive at x = 2. So, does that count as two zeros? Well, at x = 1, it's a zero, but it's a point where the function touches the x-axis and turns around. So, is that considered one zero or two?In terms of multiplicity, if the function just touches the x-axis, it's a repeated root, so it's a zero of multiplicity two. But in terms of distinct zeros, it's just one zero. So, in the interval [1/2, 2], the function has one zero at x = 1, but it's a double root. So, depending on the interpretation, it might be considered as one zero or two zeros.But the problem says "exactly two zeros." So, if we consider multiplicity, then at a = 3, the function has a double root at x = 1, so that's two zeros counting multiplicity. But if we consider distinct zeros, it's only one. So, this is a bit ambiguous.But in calculus, when we talk about zeros, we usually mean distinct zeros unless specified otherwise. So, at a = 3, the function has a double root, so it's technically one zero with multiplicity two. So, if the problem counts multiplicity, then a = 3 would be included, but if not, it's excluded.But let's see what happens just above a = 3. For a slightly greater than 3, say a = 3.1, f(a/3) = 1 - (3.1)^3 / 27 ≈ 1 - 30.076 / 27 ≈ 1 - 1.113 ≈ -0.113 < 0. So, the function dips below zero, creating two distinct zeros. So, for a > 3, we have two distinct zeros. At a = 3, we have a double zero. So, depending on the problem's requirement, if they want exactly two distinct zeros, then a must be greater than 3. If they allow for a double zero as two zeros, then a can be equal to 3.But the problem says "exactly two zeros," without specifying distinctness. So, it's safer to include a = 3, because it's a zero of multiplicity two, hence two zeros.Wait, but earlier, when a = 3, f(1/2) = 1/2 > 0, f(1) = 0, and f(2) = 5 > 0. So, the function touches the x-axis at x = 1 and doesn't cross it again. So, in terms of crossing, it's only one crossing, but in terms of zeros, it's two zeros (counting multiplicity). So, this is a bit tricky.But in the context of the problem, I think they are referring to distinct zeros. Because if they included multiplicity, they would probably specify. So, to have exactly two distinct zeros in the interval, a must be greater than 3, because at a = 3, it's a double root, so only one distinct zero.But wait, let's think again. If a = 3, the function has a double root at x = 1, so it's tangent to the x-axis there. So, in the interval [1/2, 2], it's just touching the x-axis at x = 1, but not crossing it again. So, in terms of crossings, it's only one crossing, but in terms of zeros, it's two zeros (with multiplicity). So, depending on the interpretation, the answer could be a > 3 or a ≥ 3.But let's check the problem statement again: "find the range of values for a such that the function has exactly two zeros in the interval [1/2, 2]." It doesn't specify distinct zeros, so maybe it includes multiplicity. So, in that case, a = 3 would be included because it has two zeros (counting multiplicity). But wait, actually, at a = 3, the function has a double root, so it's one zero with multiplicity two. So, in terms of the number of zeros, it's still one zero, but with multiplicity two. So, I think the problem is referring to distinct zeros, so a must be greater than 3.But to be thorough, let's consider both cases.Case 1: Exactly two distinct zeros in [1/2, 2]. Then, a must be greater than 3 and less than 17/4.Case 2: Exactly two zeros, counting multiplicity. Then, a can be equal to 3 as well, because it has a double root. But in that case, the function only crosses the x-axis once, but has two zeros when counting multiplicity.But in most problems like this, unless specified, they refer to distinct zeros. So, I think the correct range is a > 3 and a < 17/4.But let's also check what happens when a approaches 17/4 from below.As a approaches 17/4 from below, f(2) approaches zero from above, and f(a/3) approaches 1 - ( (17/4)^3 ) / 27. Let's compute that:(17/4)^3 = 4913 / 64.So, 4913 / 64 divided by 27 is 4913 / (64*27) = 4913 / 1728 ≈ 2.84.So, 1 - 2.84 ≈ -1.84 < 0.So, as a approaches 17/4 from below, f(a/3) approaches approximately -1.84, which is still negative. So, the function still has two zeros in the interval.At a = 17/4, f(2) = 0, and f(a/3) < 0. So, the function has a zero at x = 2 and another zero between 1/2 and a/3. So, that's two zeros.Therefore, including a = 17/4, the function has exactly two zeros in the interval.So, putting it all together, the range of a is 3 < a ≤ 17/4.But wait, earlier I thought a must be greater than 3, but when a = 3, the function has a double root, which is technically two zeros (with multiplicity). So, if the problem counts multiplicity, then a can be equal to 3. But if not, then a must be greater than 3.But let's check the problem statement again: "exactly two zeros in the interval [1/2, 2]." It doesn't specify distinct, so maybe it's safer to include a = 3.But wait, at a = 3, the function has a double root at x = 1, so it's one zero with multiplicity two. So, in terms of the number of zeros, it's still one zero, but with multiplicity two. So, if the problem counts multiplicity, then it's two zeros; otherwise, it's one.But in most cases, unless specified, "exactly two zeros" refers to distinct zeros. So, I think the correct range is a > 3 and a ≤ 17/4.But let's also consider the case when a is less than 3. For example, a = 2.At a = 2, f(a/3) = f(2/3) = 2*(8/27) - 2*(4/9) + 1 = 16/27 - 8/9 + 1 = 16/27 - 24/27 + 27/27 = (16 - 24 + 27)/27 = 19/27 > 0.So, f(a/3) > 0. Therefore, the function is decreasing from x = 1/2 to x = 2/3, reaching a minimum at x = 2/3, which is still positive, and then increasing from x = 2/3 to x = 2. So, the function is always positive in [1/2, 2], hence no zeros.Similarly, if a is between 3/2 and 3, say a = 2.5, f(a/3) = f(5/6) = 2*(125/216) - 2.5*(25/36) + 1.Compute:2*(125/216) = 250/216 ≈ 1.157.2.5*(25/36) = 62.5/36 ≈ 1.736.So, f(5/6) ≈ 1.157 - 1.736 + 1 ≈ 0.421 > 0.So, f(a/3) > 0, meaning the function doesn't dip below zero, so no zeros in the interval.Therefore, for a ≤ 3, the function doesn't have two zeros in the interval. So, the lower bound is a > 3.But wait, earlier I thought that for the critical point x = a/3 to be in [1/2, 2], a must be ≥ 3/2. But if a is between 3/2 and 3, the function's minimum is still positive, so no zeros. So, only when a > 3 does the function dip below zero, creating two zeros.Therefore, the range of a is (3, 17/4].But wait, let's check a = 3.5, which is between 3 and 17/4 (which is 4.25).f(1/2) = 5/4 - 3.5/4 = (5 - 3.5)/4 = 1.5/4 = 0.375 > 0.f(a/3) = f(3.5/3) = f(7/6) ≈ 1 - (3.5)^3 / 27 ≈ 1 - 42.875 / 27 ≈ 1 - 1.587 ≈ -0.587 < 0.f(2) = 17 - 4*3.5 = 17 - 14 = 3 > 0.So, the function starts positive at x = 1/2, dips below zero at x = 7/6, and comes back up to positive at x = 2. So, two zeros in the interval.Similarly, at a = 4, f(a/3) = f(4/3) = 1 - (64)/27 ≈ 1 - 2.37 ≈ -1.37 < 0, and f(2) = 17 - 16 = 1 > 0. So, two zeros.At a = 17/4 = 4.25, f(2) = 0, and f(a/3) = f(17/12) ≈ -1.838 < 0, so two zeros: one between 1/2 and 17/12, and one at x = 2.Therefore, the range of a is 3 < a ≤ 17/4.But wait, earlier I thought that the critical point x = a/3 must be in [1/2, 2], which requires a ≥ 3/2. But in reality, for a > 3, x = a/3 is in [1, 2], which is within [1/2, 2]. So, the lower bound is a > 3, not a ≥ 3/2.But wait, if a is between 3/2 and 3, x = a/3 is in [1/2, 1], but f(a/3) > 0, so the function doesn't dip below zero, hence no zeros in the interval.Therefore, the correct range is a > 3 and a ≤ 17/4.So, the range of a is (3, 17/4].But let's also check a = 3. Let me compute f(x) at a = 3:f(x) = 2x³ - 3x² + 1.Let's find its zeros.We know that x = 1 is a root because f(1) = 2 - 3 + 1 = 0.Let's factor it:f(x) = (x - 1)(2x² - x - 1).Now, factor 2x² - x - 1:Looking for two numbers m and n such that m * n = -2 and m + n = -1.Wait, 2x² - x - 1 = (2x + 1)(x - 1).Yes, because (2x + 1)(x - 1) = 2x² - 2x + x - 1 = 2x² - x - 1.So, f(x) = (x - 1)(2x + 1)(x - 1) = (x - 1)^2 (2x + 1).So, the roots are x = 1 (double root) and x = -1/2.So, in the interval [1/2, 2], the only zero is x = 1, which is a double root. So, counting multiplicity, it's two zeros, but as distinct zeros, it's one.Therefore, if the problem counts multiplicity, then a = 3 is included, giving exactly two zeros (counting multiplicity). If not, it's excluded.But since the problem says "exactly two zeros," without specifying, it's safer to assume they mean distinct zeros. So, a must be greater than 3.Therefore, the range is 3 < a ≤ 17/4.But let's also check a = 17/4:f(x) = 2x³ - (17/4)x² + 1.We can check if x = 2 is a root:f(2) = 16 - (17/4)*4 + 1 = 16 - 17 + 1 = 0.So, x = 2 is a root. Let's factor it:f(x) = (x - 2)(2x² + (something)x + something).But perhaps it's easier to compute f(a/3) when a = 17/4:a/3 = 17/12 ≈ 1.4167.f(17/12) = 2*(17/12)^3 - (17/4)*(17/12)^2 + 1.As computed earlier, this is approximately -1.838 < 0.So, the function starts at f(1/2) = 5/4 - (17/4)/4 = 5/4 - 17/16 = (20/16 - 17/16) = 3/16 > 0, dips below zero at x = 17/12, and then comes back up to zero at x = 2. So, two zeros: one between 1/2 and 17/12, and one at x = 2.Therefore, a = 17/4 is included.So, putting it all together, the range of a is 3 < a ≤ 17/4.But wait, earlier I thought the lower bound was 3/2, but that's only for the critical point x = a/3 to be in [1/2, 2]. However, for the function to have two zeros, the critical point must not only be in the interval but also the function must dip below zero there. So, the actual lower bound is a > 3.Therefore, the range of a is (3, 17/4].But let's also check a = 3. Let me compute f(x) at a = 3:f(x) = 2x³ - 3x² + 1.As before, f(1) = 0, and it's a double root. So, in the interval [1/2, 2], the function touches the x-axis at x = 1 but doesn't cross it again. So, only one distinct zero. Therefore, a = 3 is excluded if we're considering distinct zeros.So, the final answer is a must be greater than 3 and less than or equal to 17/4.Therefore, the range of a is (3, 17/4].But wait, in the initial problem, the user wrote:"Given the function f(x)=2x^{3}-ax^{2}+1, find the range of values for a such that the function has exactly two zeros in the interval [1/2,2]."So, the answer is a must be greater than 3 and less than or equal to 17/4.But let me check the calculations again to ensure I didn't make any mistakes.We have:1. f(1/2) = 5/4 - a/4 > 0 ⇒ a < 5.2. f(2) = 17 - 4a > 0 ⇒ a < 17/4.3. f(a/3) = 1 - (a³)/27 < 0 ⇒ a > 3.So, combining these, a must be greater than 3 and less than 17/4.Therefore, the range is (3, 17/4].But wait, 17/4 is 4.25, which is less than 5, so the stricter condition is a < 17/4.So, the range is 3 < a ≤ 17/4.Therefore, the final answer is (3, 17/4].But in the initial problem, the user's solution had a range of (3/2, 17/4]. But after careful consideration, I think the correct range is (3, 17/4].Wait, but in the initial problem, the user's solution had a range of (3/2, 17/4], but after my analysis, I think it's (3, 17/4].But let me check again.When a = 3/2 = 1.5, f(a/3) = f(0.5) = 2*(0.125) - 1.5*(0.25) + 1 = 0.25 - 0.375 + 1 = 0.875 > 0.So, the function doesn't dip below zero, so no zeros in the interval.When a = 3, f(a/3) = 0, so it's a double root.When a > 3, f(a/3) < 0, so two zeros.When a = 17/4, f(2) = 0, so two zeros.When a > 17/4, f(2) < 0, so only one zero in the interval.Therefore, the correct range is 3 < a ≤ 17/4.So, the initial user's solution had a lower bound of 3/2, which is incorrect because for a between 3/2 and 3, the function doesn't dip below zero, hence no zeros in the interval.Therefore, the correct range is (3, 17/4].

💡Okay, so I have this inequality: (x + y - 6 < 0). I need to figure out which side of the line (x + y - 6 = 0) this inequality represents. The options are upper right, upper left, lower right, and lower left. Hmm, okay, let's break this down step by step.First, I think I should understand what the line (x + y - 6 = 0) looks like. Maybe I can rewrite it in a more familiar form, like slope-intercept form. So, if I solve for y, I get:(y = -x + 6)Alright, so this is a straight line with a slope of -1 and a y-intercept at (0, 6). That makes sense. It should be a diagonal line going from the upper left to the lower right of the coordinate plane.Now, the inequality is (x + y - 6 < 0), which simplifies to (x + y < 6). I remember that inequalities like this represent a region on one side of the line. But how do I figure out which side?I think the best way is to pick a test point that's not on the line and see if it satisfies the inequality. Usually, the origin (0,0) is a good choice because it's easy to plug in. Let's try that:Plugging in (0,0) into (x + y < 6):(0 + 0 < 6)(0 < 6)That's true, so the origin is part of the region represented by the inequality. But where is the origin located relative to the line (y = -x + 6)? The origin is at (0,0), which is below the line because when x is 0, y is 6 on the line, and 0 is less than 6. So, the region that satisfies the inequality is below the line.But wait, the options are about left lower side, upper left, etc. So, being below the line would correspond to the lower part. Since the line goes from upper left to lower right, the lower side would be towards the lower left and lower right. But how do I determine whether it's the lower left or lower right?Maybe I should consider the direction of the inequality. The inequality is (x + y < 6), which is less than, so it's the region below the line. Now, if I think about the line (y = -x + 6), it's decreasing from left to right. So, below the line would be the area that is to the left and below the line, which would be the lower left side.Alternatively, I could pick another test point to confirm. Let's say I pick the point (1,1). Plugging into the inequality:(1 + 1 < 6)(2 < 6)That's true, so (1,1) is also in the region. Where is (1,1) located? It's to the lower left of the line because if I plug x=1 into the line equation, y would be 5, so (1,1) is below that.What if I pick a point to the right, like (7,0)? Plugging into the inequality:(7 + 0 < 6)(7 < 6)That's false, so (7,0) is not in the region. So, the region is not extending to the lower right beyond the line.Therefore, combining these observations, the region represented by (x + y - 6 < 0) is on the lower left side of the line.

💡Alright, so I'm trying to figure out this problem about a construction team doing a blasting operation. They need to make sure people evacuate to a safe area that's at least 50 meters away from the blasting point. There's a fuse involved that burns at a speed of 0.2 meters per second, and the blaster is leaving at a speed of 3 meters per second. The fuse is ignited in a direction away from the blasting point. I need to find out what condition the length of the fuse, which we'll call x meters, should satisfy. Okay, let's break this down. First, the fuse is burning away from the blasting point, so it's not burning towards where the blaster is going. The blaster is moving away from the blasting point at 3 meters per second, and the fuse is burning at 0.2 meters per second in the opposite direction. So, the fuse is burning away from the blaster, which might mean that the blaster has more time to get to safety because the fuse isn't burning towards them. But wait, the fuse is being ignited in a direction away from the blasting point, so it's burning away from the blaster. That might mean that the blaster doesn't have to worry about the fuse burning towards them, but they still need to make sure they're at least 50 meters away when the blast happens. So, the time it takes for the fuse to burn completely should be enough time for the blaster to get to the safe distance.Let me think about the time it takes for the fuse to burn. If the fuse is x meters long and it burns at 0.2 meters per second, then the time it takes for the fuse to burn completely is x divided by 0.2, which is x/0.2 seconds.Now, the blaster is moving away from the blasting point at 3 meters per second. So, the time it takes for the blaster to reach the safe area, which is 50 meters away, is 50 divided by 3, which is 50/3 seconds.But wait, the fuse is burning away from the blaster, so the blaster doesn't have to cover the entire 50 meters before the fuse burns out. Instead, the blaster only needs to cover the distance from the current position to the safe area, which is 50 meters, while the fuse is burning away. So, the time it takes for the blaster to reach the safe area should be less than or equal to the time it takes for the fuse to burn out.So, the time for the blaster to reach safety is 50/3 seconds, and the time for the fuse to burn out is x/0.2 seconds. To ensure safety, the fuse should burn out after the blaster has reached the safe area. Therefore, x/0.2 should be greater than or equal to 50/3.Wait, but the fuse is burning away from the blaster, so maybe the blaster doesn't have to cover the entire 50 meters before the fuse burns out. Instead, the fuse burning away might mean that the blaster has more time to get to safety because the fuse isn't burning towards them. So, perhaps the blaster only needs to cover the distance from the current position to the safe area, which is 50 meters, while the fuse is burning away.But I'm getting a bit confused here. Let me try to visualize it. The blaster is at the blasting point, and the fuse is ignited in a direction away from the blaster. So, the fuse is burning away from the blaster, which means the blaster doesn't have to worry about the fuse burning towards them. Therefore, the blaster can take their time to move away, and the fuse will burn out on its own.But the problem says that people must evacuate to a safe area at least 50 meters away. So, the blaster needs to be at least 50 meters away when the fuse burns out. So, the time it takes for the fuse to burn out should be enough time for the blaster to reach 50 meters away.So, the time for the fuse to burn out is x/0.2 seconds, and the time for the blaster to reach 50 meters is 50/3 seconds. To ensure safety, the fuse should burn out after the blaster has reached the safe area. Therefore, x/0.2 should be greater than or equal to 50/3.Wait, but if the fuse is burning away from the blaster, does that affect the distance the blaster needs to cover? Or is it just about the time it takes for the fuse to burn out versus the time it takes for the blaster to reach safety?I think it's just about the time. The fuse burning away doesn't directly affect the distance the blaster needs to cover, but it does affect the time they have to cover that distance. So, the blaster needs to cover 50 meters in the time it takes for the fuse to burn out.Therefore, the time for the blaster to reach safety is 50/3 seconds, and the time for the fuse to burn out is x/0.2 seconds. To ensure safety, the fuse should burn out after the blaster has reached the safe area. So, x/0.2 should be greater than or equal to 50/3.But wait, the problem says "at least 50 meters away," so the blaster needs to be at least 50 meters away when the fuse burns out. So, the time it takes for the fuse to burn out should be enough time for the blaster to reach 50 meters.Therefore, the inequality should be x/0.2 ≥ 50/3.But let me double-check. If the fuse is longer, it takes more time to burn out, giving the blaster more time to reach safety. If the fuse is shorter, it burns out faster, giving the blaster less time. So, to ensure the blaster has enough time, the fuse needs to be long enough that x/0.2 is greater than or equal to 50/3.Yes, that makes sense. So, the condition is x/0.2 ≥ 50/3.But let me write it in terms of x. So, x/0.2 ≥ 50/3.To make it simpler, I can multiply both sides by 0.2 to get x ≥ (50/3)*0.2.Calculating that, (50/3)*0.2 = (50*0.2)/3 = 10/3 ≈ 3.333 meters.So, the fuse needs to be at least approximately 3.333 meters long.But the problem asks for the inequality, not the numerical value. So, the inequality is x/0.2 ≥ 50/3.Alternatively, I can write it as x ≥ (50/3)*0.2, but the first form is probably better.Wait, but in the initial problem, it says "the fuse is ignited in a direction away from the blasting point." Does that affect the distance the blaster needs to cover? Or is it just about the time?I think it's just about the time because the fuse burning away doesn't reduce the distance the blaster needs to cover. The blaster still needs to cover 50 meters to be safe. The fuse burning away just means that the blaster doesn't have to worry about the fuse burning towards them, but they still need to reach 50 meters away.So, the time for the fuse to burn out is x/0.2, and the time for the blaster to reach 50 meters is 50/3. To ensure safety, x/0.2 should be greater than or equal to 50/3.Therefore, the inequality is x/0.2 ≥ 50/3.I think that's the correct condition.

💡Okay, so I need to figure out how many two-digit numbers have at least one digit that is a 4. Hmm, let's see. First, I should understand what a two-digit number is. A two-digit number ranges from 10 to 99, right? So there are 90 two-digit numbers in total.Now, the question is asking for numbers that have at least one digit as 4. That means either the tens digit is 4, or the ones digit is 4, or both. I think I can approach this by counting how many numbers have 4 in the tens place and how many have 4 in the ones place, and then subtract the overlap where both digits are 4 to avoid double-counting.Let me start with the tens digit. If the tens digit is 4, then the number looks like 4X, where X can be any digit from 0 to 9. So that gives me 10 numbers: 40, 41, 42, 43, 44, 45, 46, 47, 48, and 49.Next, for the ones digit being 4, the number looks like X4, where X can be any digit from 1 to 9 (since it's a two-digit number, the tens digit can't be 0). That gives me 9 numbers: 14, 24, 34, 44, 54, 64, 74, 84, and 94.Wait, I notice that the number 44 appears in both lists. So if I just add the two counts together, I'll be counting 44 twice. To get the correct total, I need to subtract the overlap once.So, the total number of two-digit numbers with at least one digit being 4 is 10 (from the tens place) plus 9 (from the ones place) minus 1 (for the overlap), which equals 18.Let me double-check to make sure I didn't miss anything. The tens place gives me 10 numbers, the ones place gives me 9, and subtracting the one overlap seems right. Yeah, that makes sense. So, I think the answer is 18.