💡First, I observe that ( P ) and ( R ) are conjugates, as are ( Q ) and ( S ). This means that when I multiply ( P ) and ( R ), I can use the difference of squares formula to simplify the expression.[P times R = (sqrt{2024} + sqrt{2023})(sqrt{2024} - sqrt{2023}) = (sqrt{2024})^2 - (sqrt{2023})^2 = 2024 - 2023 = 1]Similarly, multiplying ( Q ) and ( S ) also results in 1:[Q times S = (-sqrt{2024} - sqrt{2023})(-sqrt{2024} + sqrt{2023}) = (sqrt{2024})^2 - (sqrt{2023})^2 = 2024 - 2023 = 1]Finally, multiplying these two results together gives:[PQRS = (P times R) times (Q times S) = 1 times 1 = 1]

💡Okay, so I have this problem here where the line y = 2x + 1 is tangent to the curve y = x³ + a x + b at the point (1, 3). I need to find the value of b. Hmm, let me think about how to approach this.First, I remember that for a line to be tangent to a curve at a particular point, two conditions must be satisfied. The first condition is that the point lies on both the line and the curve. The second condition is that the slope of the line is equal to the derivative of the curve at that point. So, I need to use both of these conditions to find the values of a and b.Let me start by verifying that the point (1, 3) lies on both the line and the curve. For the line y = 2x + 1, if I plug in x = 1, I get y = 2(1) + 1 = 3. So, yes, the point (1, 3) is on the line. Now, let's check the curve y = x³ + a x + b. Plugging in x = 1, we get y = (1)³ + a(1) + b = 1 + a + b. Since the point is (1, 3), this must equal 3. So, I can write the equation:1 + a + b = 3Simplifying this, I get:a + b = 2 ...(1)Okay, that's one equation. Now, moving on to the second condition: the slope of the tangent line must equal the derivative of the curve at that point. The slope of the line y = 2x + 1 is 2. So, I need to find the derivative of the curve y = x³ + a x + b and set it equal to 2 at x = 1.Let's compute the derivative of the curve. The derivative of y with respect to x is:dy/dx = 3x² + aSo, at x = 1, the derivative is:dy/dx = 3(1)² + a = 3 + aSince this must equal the slope of the tangent line, which is 2, I can write:3 + a = 2Solving for a, I subtract 3 from both sides:a = 2 - 3a = -1Okay, so a is -1. Now, I can plug this back into equation (1) to find b. From equation (1):a + b = 2-1 + b = 2Adding 1 to both sides:b = 2 + 1b = 3So, b is 3. Let me just double-check my work to make sure I didn't make any mistakes.First, plugging a = -1 and b = 3 into the curve equation:y = x³ + (-1)x + 3 = x³ - x + 3At x = 1, y = 1 - 1 + 3 = 3, which matches the point (1, 3). Good.Now, the derivative is dy/dx = 3x² - 1. At x = 1, dy/dx = 3(1)² - 1 = 3 - 1 = 2, which matches the slope of the tangent line. Perfect.So, everything checks out. The value of b is indeed 3.**Final Answer**The value of ( b ) is boxed{3}.

💡Okay, let's see. I have this problem about a quadratic function, f(x) = x² + ax + b, where a and b are real numbers. The range of this function is given as [0, +∞), which means the function can take any value from 0 upwards but never goes below 0. That tells me something about the vertex of the parabola. Since the range starts at 0, the vertex must be at the minimum point, right? So the vertex is at (h, 0) for some h.Now, the quadratic equation is f(x) = x² + ax + b. I remember that the vertex form of a quadratic is f(x) = (x - h)² + k, where (h, k) is the vertex. In this case, since the range is [0, +∞), k must be 0. So, f(x) can be written as (x - h)². Expanding this, we get x² - 2hx + h². Comparing this with the given quadratic x² + ax + b, we can see that a must be -2h and b must be h². So, a = -2h and b = h².But I don't know what h is yet. Maybe I can find it using the other part of the problem. The problem says that the solution set of the inequality f(x) < c is (m, m + 6). That means when we solve x² + ax + b < c, the solution is all x between m and m + 6. So, the inequality x² + ax + b - c < 0 has roots at m and m + 6.Let me write that down: x² + ax + (b - c) < 0. The roots of the equation x² + ax + (b - c) = 0 are m and m + 6. Since the quadratic opens upwards (the coefficient of x² is positive), the inequality x² + ax + (b - c) < 0 is satisfied between the two roots. So, the distance between the roots is (m + 6) - m = 6. That means the difference between the roots is 6.I remember that for a quadratic equation ax² + bx + c = 0, the difference between the roots is sqrt(D)/|a|, where D is the discriminant. But in our case, the quadratic is x² + ax + (b - c) = 0, so a is 1, b is a, and c is (b - c). Wait, that might be confusing because the coefficients are named a, b, c, but in the quadratic equation, they are usually a, b, c. Maybe I should use different notation.Let me denote the quadratic as x² + px + q = 0, where p = a and q = b - c. Then, the roots are m and m + 6. The sum of the roots is m + (m + 6) = 2m + 6, which should be equal to -p (from Vieta's formula). So, 2m + 6 = -p. Also, the product of the roots is m(m + 6) = m² + 6m, which should be equal to q. So, m² + 6m = q = b - c.But I also know that the original quadratic f(x) = x² + ax + b has a range of [0, +∞). That means its discriminant must be zero because it just touches the x-axis at the vertex. The discriminant of f(x) is a² - 4b = 0. So, a² = 4b.Earlier, I found that a = -2h and b = h². So, substituting into a² = 4b, we get (-2h)² = 4h², which is 4h² = 4h². That checks out, but it doesn't give me new information.Going back to the inequality, the quadratic x² + ax + (b - c) has roots m and m + 6. The difference between the roots is 6, so the discriminant of this quadratic must be (6)² = 36. The discriminant D is p² - 4q, which is a² - 4(b - c) = 36.But from earlier, I know that a² = 4b. So, substituting that into the discriminant, we get 4b - 4(b - c) = 36. Let's simplify that: 4b - 4b + 4c = 36, which simplifies to 4c = 36. Therefore, c = 9.Wait, let me double-check that. If the discriminant is 36, then a² - 4(b - c) = 36. Since a² = 4b, substituting gives 4b - 4b + 4c = 36, so 4c = 36, c = 9. Yeah, that seems right.So, the value of c is 9.

💡Alright, so I have this problem about a triangle ABC, and I need to figure out what kind of triangle it is based on the given conditions. The options are right-angled, equilateral, isosceles, or isosceles right-angled. Let me try to break this down step by step.First, the problem gives me two equations:1. ((a + b + c)(b + c - a) = 3abc)2. (sin A = 2 sin B cos C)I need to analyze these equations to determine the type of triangle ABC is.Starting with the first equation: ((a + b + c)(b + c - a) = 3abc). Hmm, this looks a bit complicated, but maybe I can simplify it. Let me expand the left side:[(a + b + c)(b + c - a) = (b + c)^2 - a^2]Expanding ((b + c)^2):[(b + c)^2 = b^2 + 2bc + c^2]So, subtracting (a^2):[(b^2 + 2bc + c^2) - a^2 = 3abc]Which simplifies to:[b^2 + 2bc + c^2 - a^2 = 3abc]Hmm, I remember from the Law of Cosines that (a^2 = b^2 + c^2 - 2bc cos A). Maybe I can substitute that into the equation.So, replacing (a^2) in the equation:[b^2 + 2bc + c^2 - (b^2 + c^2 - 2bc cos A) = 3abc]Simplifying inside the parentheses:[b^2 + 2bc + c^2 - b^2 - c^2 + 2bc cos A = 3abc]The (b^2) and (c^2) terms cancel out:[2bc + 2bc cos A = 3abc]Factor out (2bc):[2bc(1 + cos A) = 3abc]Divide both sides by (bc) (assuming (b) and (c) are not zero, which they can't be in a triangle):[2(1 + cos A) = 3a]Wait, that seems a bit odd. I have (2(1 + cos A) = 3a). But (a) is a side length, and the left side is a scalar multiple of a trigonometric function. Maybe I made a mistake in substituting the Law of Cosines.Let me double-check that step. I had:[b^2 + 2bc + c^2 - a^2 = 3abc]And I substituted (a^2 = b^2 + c^2 - 2bc cos A), so:[b^2 + 2bc + c^2 - (b^2 + c^2 - 2bc cos A) = 3abc]Which simplifies to:[2bc + 2bc cos A = 3abc]Yes, that seems correct. So, factoring out (2bc):[2bc(1 + cos A) = 3abc]Divide both sides by (bc):[2(1 + cos A) = 3a]Wait, but (a) is a length, and the left side is a number. This doesn't make sense dimensionally. Did I make a mistake earlier?Let me go back to the original equation:[(a + b + c)(b + c - a) = 3abc]Maybe instead of expanding, I can use substitution or another identity. Alternatively, perhaps assuming the triangle is equilateral might simplify things.If the triangle is equilateral, then (a = b = c). Let's test that.Let (a = b = c). Then the left side becomes:[(a + a + a)(a + a - a) = (3a)(a) = 3a^2]The right side is:[3abc = 3a cdot a cdot a = 3a^3]So, (3a^2 = 3a^3) implies (a^2 = a^3), which means (a = 1) (assuming (a neq 0)). So, in an equilateral triangle with side length 1, this equation holds. But does it hold for any equilateral triangle? If (a = 2), then left side is (3 cdot 2^2 = 12), and right side is (3 cdot 2^3 = 24), which doesn't hold. Hmm, so maybe the triangle isn't necessarily equilateral, or perhaps I need to consider another approach.Let me think about the second equation: (sin A = 2 sin B cos C). Maybe I can use trigonometric identities or the Law of Sines here.I know that in any triangle, the sum of angles is (180^circ), so (A + B + C = 180^circ). Also, from the Law of Sines, (frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R), where (R) is the circumradius.Maybe I can express (sin A) and (sin B) in terms of sides.But before that, let me recall that (sin A = 2 sin B cos C). I wonder if there's a trigonometric identity that can help here.I remember that (sin(A) = sin(B + C)), since (A = 180^circ - B - C). And (sin(B + C) = sin B cos C + cos B sin C). So,[sin A = sin(B + C) = sin B cos C + cos B sin C]But the given equation is (sin A = 2 sin B cos C). So,[sin B cos C + cos B sin C = 2 sin B cos C]Subtracting (sin B cos C) from both sides:[cos B sin C = sin B cos C]Which simplifies to:[frac{sin C}{cos C} = frac{sin B}{cos B}]So,[tan C = tan B]Since (B) and (C) are angles in a triangle, they must be between (0^circ) and (180^circ), and since (tan) is periodic with period (180^circ), the only way (tan C = tan B) is if (C = B) or (C = B + 180^circ). But since both angles are less than (180^circ), (C = B).So, angle (B = C), which means the triangle is isosceles with sides (b = c).Okay, so from the second equation, we've deduced that the triangle is isosceles with (b = c).Now, going back to the first equation: ((a + b + c)(b + c - a) = 3abc). Since we now know (b = c), let's substitute (c = b) into the equation.So, replacing (c) with (b):[(a + b + b)(b + b - a) = 3ab cdot b]Simplify:[(a + 2b)(2b - a) = 3ab^2]Let me compute the left side:[(a + 2b)(2b - a) = (2b + a)(2b - a) = (2b)^2 - a^2 = 4b^2 - a^2]So,[4b^2 - a^2 = 3ab^2]Let me rearrange this:[4b^2 - 3ab^2 - a^2 = 0]Factor out (b^2) from the first two terms:[b^2(4 - 3a) - a^2 = 0]Hmm, not sure if that's helpful. Maybe I can express (a) in terms of (b). Let's treat this as a quadratic in (a):[-a^2 - 3ab^2 + 4b^2 = 0]Multiply both sides by -1:[a^2 + 3ab^2 - 4b^2 = 0]This is a quadratic equation in terms of (a):[a^2 + 3b^2 a - 4b^2 = 0]Let me write it as:[a^2 + (3b^2)a - 4b^2 = 0]Using the quadratic formula to solve for (a):[a = frac{ -3b^2 pm sqrt{(3b^2)^2 + 16b^2} }{2}]Simplify inside the square root:[(3b^2)^2 + 16b^2 = 9b^4 + 16b^2]So,[a = frac{ -3b^2 pm sqrt{9b^4 + 16b^2} }{2}]Factor out (b^2) inside the square root:[sqrt{b^2(9b^2 + 16)} = b sqrt{9b^2 + 16}]So,[a = frac{ -3b^2 pm b sqrt{9b^2 + 16} }{2}]Since side lengths are positive, we discard the negative solution:[a = frac{ -3b^2 + b sqrt{9b^2 + 16} }{2}]Hmm, this seems complicated. Maybe there's a simpler approach.Wait, earlier I assumed (b = c), so the triangle is isosceles with sides (b = c). Maybe I can use the Law of Cosines for angle (A), since in an isosceles triangle, angles opposite equal sides are equal.So, in triangle ABC, since (b = c), angles (B) and (C) are equal. Let me denote (B = C = theta), so angle (A = 180^circ - 2theta).From the Law of Cosines:[a^2 = b^2 + c^2 - 2bc cos A]But (b = c), so:[a^2 = 2b^2 - 2b^2 cos A]Which simplifies to:[a^2 = 2b^2(1 - cos A)]But (A = 180^circ - 2theta), so (cos A = cos(180^circ - 2theta) = -cos 2theta).So,[a^2 = 2b^2(1 - (-cos 2theta)) = 2b^2(1 + cos 2theta)]Using the double-angle identity, (1 + cos 2theta = 2cos^2 theta), so:[a^2 = 2b^2 cdot 2cos^2 theta = 4b^2 cos^2 theta]Thus,[a = 2b cos theta]But (theta = B = C), so (cos theta = cos B). Therefore,[a = 2b cos B]Now, going back to the first equation, which after substitution became:[4b^2 - a^2 = 3ab^2]But from above, (a = 2b cos B), so let's substitute that into the equation:[4b^2 - (2b cos B)^2 = 3 cdot 2b cos B cdot b^2]Simplify:[4b^2 - 4b^2 cos^2 B = 6b^3 cos B]Divide both sides by (b^2) (assuming (b neq 0)):[4 - 4 cos^2 B = 6b cos B]Let me rearrange this:[-4 cos^2 B - 6b cos B + 4 = 0]Multiply both sides by -1:[4 cos^2 B + 6b cos B - 4 = 0]This is a quadratic equation in terms of (cos B):[4 cos^2 B + 6b cos B - 4 = 0]Let me denote (x = cos B), then:[4x^2 + 6b x - 4 = 0]Using the quadratic formula:[x = frac{ -6b pm sqrt{(6b)^2 + 64} }{8}]Simplify inside the square root:[36b^2 + 64]So,[x = frac{ -6b pm sqrt{36b^2 + 64} }{8}]Since (x = cos B) must be between -1 and 1, and (b > 0), we take the positive root:[x = frac{ -6b + sqrt{36b^2 + 64} }{8}]This seems complicated. Maybe there's a different approach.Wait, earlier I had (a = 2b cos B). Let me use that in the equation (4b^2 - a^2 = 3ab^2):[4b^2 - (2b cos B)^2 = 3 cdot 2b cos B cdot b^2]Which simplifies to:[4b^2 - 4b^2 cos^2 B = 6b^3 cos B]Divide both sides by (b^2):[4 - 4 cos^2 B = 6b cos B]Let me rearrange:[4 cos^2 B + 6b cos B - 4 = 0]This is the same quadratic as before. Maybe instead of solving for (cos B), I can express (b) in terms of (cos B).Let me denote (x = cos B), then:[4x^2 + 6b x - 4 = 0]Solving for (b):[6b x = -4x^2 + 4]So,[b = frac{ -4x^2 + 4 }{6x } = frac{4(1 - x^2)}{6x} = frac{2(1 - x^2)}{3x}]But (1 - x^2 = sin^2 B), so:[b = frac{2 sin^2 B}{3x} = frac{2 sin^2 B}{3 cos B}]But from the Law of Sines, (frac{a}{sin A} = frac{b}{sin B}), and since (A = 180^circ - 2B), (sin A = sin(2B) = 2 sin B cos B).So,[frac{a}{2 sin B cos B} = frac{b}{sin B}]Simplify:[frac{a}{2 cos B} = b]But earlier, I had (a = 2b cos B), which is consistent because:[a = 2b cos B implies b = frac{a}{2 cos B}]So, substituting (b = frac{2 sin^2 B}{3 cos B}) into (b = frac{a}{2 cos B}):[frac{2 sin^2 B}{3 cos B} = frac{a}{2 cos B}]Multiply both sides by (2 cos B):[frac{4 sin^2 B}{3} = a]But from earlier, (a = 2b cos B), and (b = frac{2 sin^2 B}{3 cos B}), so:[a = 2 cdot frac{2 sin^2 B}{3 cos B} cdot cos B = frac{4 sin^2 B}{3}]Which matches the previous result. So, (a = frac{4 sin^2 B}{3}).But I'm not sure if this is leading me anywhere. Maybe I should consider specific cases or look for another relationship.Wait, from the second equation, we have (sin A = 2 sin B cos C). But since (B = C), this becomes:[sin A = 2 sin B cos B = sin 2B]But (A = 180^circ - 2B), so:[sin(180^circ - 2B) = sin 2B]Which is true because (sin(180^circ - x) = sin x). So, this equation is always satisfied when (B = C), which we already deduced.So, the second equation doesn't give us new information beyond confirming that (B = C).Going back to the first equation, perhaps I can use the fact that (a = 2b cos B) and substitute into the equation (4b^2 - a^2 = 3ab^2):[4b^2 - (2b cos B)^2 = 3 cdot 2b cos B cdot b^2]Simplify:[4b^2 - 4b^2 cos^2 B = 6b^3 cos B]Divide by (b^2):[4 - 4 cos^2 B = 6b cos B]Let me denote (x = cos B), then:[4 - 4x^2 = 6b x]From earlier, we have (a = 2b x), and from the Law of Sines, (a = 2b x), which is consistent.But I still have two variables here: (b) and (x). Maybe I can express (b) in terms of (x) and substitute.From (4 - 4x^2 = 6b x), solve for (b):[b = frac{4 - 4x^2}{6x} = frac{2(1 - x^2)}{3x}]But (1 - x^2 = sin^2 B), so:[b = frac{2 sin^2 B}{3x}]And from (a = 2b x), we have:[a = 2 cdot frac{2 sin^2 B}{3x} cdot x = frac{4 sin^2 B}{3}]So, (a = frac{4 sin^2 B}{3}).But I also know from the Law of Sines that:[frac{a}{sin A} = frac{b}{sin B}]Since (A = 180^circ - 2B), (sin A = sin 2B = 2 sin B cos B). So,[frac{a}{2 sin B cos B} = frac{b}{sin B}]Simplify:[frac{a}{2 cos B} = b]But from earlier, (b = frac{2 sin^2 B}{3x}), and (x = cos B), so:[frac{a}{2 cos B} = frac{2 sin^2 B}{3 cos B}]Multiply both sides by (2 cos B):[a = frac{4 sin^2 B}{3}]Which matches our previous result. So, everything is consistent, but I'm not getting a specific value for (B) or the sides.Maybe I need to consider specific values. Let's assume that the triangle is equilateral, so all angles are (60^circ). Then, (A = B = C = 60^circ).Check the first equation:[(a + b + c)(b + c - a) = 3abc]If (a = b = c), then:[(3a)(a) = 3a cdot a cdot a implies 3a^2 = 3a^3 implies a^2 = a^3 implies a = 1]So, for an equilateral triangle with side length 1, the equation holds. But if (a = 2), then:[(6)(2) = 3 cdot 2 cdot 2 cdot 2 implies 12 = 24]Which is false. So, the first equation only holds for an equilateral triangle with side length 1, not in general. Therefore, the triangle isn't necessarily equilateral.Wait, but earlier from the second equation, we deduced that (B = C), so the triangle is isosceles. So, it's at least isosceles. But the first equation might impose additional constraints.Let me try to find the specific angles. From the first equation, after substitution, we have:[4 - 4 cos^2 B = 6b cos B]But (b = frac{2 sin^2 B}{3 cos B}), so substitute:[4 - 4 cos^2 B = 6 cdot frac{2 sin^2 B}{3 cos B} cdot cos B]Simplify:[4 - 4 cos^2 B = 4 sin^2 B]But (sin^2 B = 1 - cos^2 B), so:[4 - 4 cos^2 B = 4(1 - cos^2 B)]Simplify:[4 - 4 cos^2 B = 4 - 4 cos^2 B]Which is an identity, meaning it holds for any (B). So, the first equation doesn't give us any new information beyond what we already have from the second equation, which is that the triangle is isosceles.Therefore, combining both conditions, the triangle must be isosceles. But let me check if it could be a right-angled triangle as well.If the triangle is right-angled and isosceles, then it must have angles (45^circ, 45^circ, 90^circ). Let's see if this satisfies the first equation.Let (A = 90^circ), (B = C = 45^circ). Then, sides are in the ratio (1 : 1 : sqrt{2}). Let me assign (b = c = 1), then (a = sqrt{2}).Check the first equation:[(a + b + c)(b + c - a) = (sqrt{2} + 1 + 1)(1 + 1 - sqrt{2}) = (sqrt{2} + 2)(2 - sqrt{2})]Multiply:[(sqrt{2} + 2)(2 - sqrt{2}) = 2 cdot sqrt{2} - (sqrt{2})^2 + 4 - 2 sqrt{2} = 2sqrt{2} - 2 + 4 - 2sqrt{2} = 2]The right side is:[3abc = 3 cdot sqrt{2} cdot 1 cdot 1 = 3sqrt{2} approx 4.2426]Which is not equal to 2. So, the first equation doesn't hold for a right-angled isosceles triangle. Therefore, the triangle isn't a right-angled isosceles triangle.So, the triangle is isosceles, but not necessarily right-angled. Therefore, the answer is C: An isosceles triangle.

💡Okay, so I have this equation: ( x^2 + frac{y^2}{cos theta} = 4 ). I need to figure out which type of conic section this equation cannot represent. The options are Circle, Ellipse, Hyperbola, and Parabola. Hmm, let me think about this step by step.First, I remember that the general form of a conic section is ( Ax^2 + By^2 + Cx + Dy + E = 0 ). Depending on the coefficients ( A ) and ( B ), we can determine whether it's a circle, ellipse, hyperbola, or parabola. In this case, the equation is ( x^2 + frac{y^2}{cos theta} = 4 ). So, comparing it to the general form, ( A = 1 ), ( B = frac{1}{cos theta} ), and there are no linear terms or constants except for the 4 on the right side.Let me recall the conditions for each conic:1. **Circle**: This happens when ( A = B ). So, in our equation, ( 1 = frac{1}{cos theta} ). That would mean ( cos theta = 1 ). I know that ( cos theta = 1 ) when ( theta = 0 ) or ( 2pi ), etc. So, yes, it can represent a circle when ( theta ) is such that ( cos theta = 1 ).2. **Ellipse**: This occurs when ( A ) and ( B ) are both positive and ( A neq B ). In our case, ( A = 1 ) is positive, and ( B = frac{1}{cos theta} ) will be positive as long as ( cos theta > 0 ). Since ( cos theta ) ranges between -1 and 1, ( B ) will be positive when ( cos theta ) is positive. So, for ( 0 < cos theta < 1 ), ( B ) is greater than 1, making it an ellipse. Therefore, it can represent an ellipse.3. **Hyperbola**: This happens when one of ( A ) or ( B ) is positive and the other is negative. In our equation, ( A = 1 ) is always positive. So, ( B = frac{1}{cos theta} ) will be negative if ( cos theta ) is negative. Since ( cos theta ) can be negative (between -1 and 0), ( B ) can be negative, making the equation represent a hyperbola. So, it can also represent a hyperbola.4. **Parabola**: This is a bit trickier. A parabola occurs when either ( A = 0 ) or ( B = 0 ), but not both. In our equation, ( A = 1 ) is always non-zero. For ( B ) to be zero, ( frac{1}{cos theta} ) would have to be zero, which is impossible because ( cos theta ) can never be infinity. Therefore, ( B ) can never be zero. So, the equation cannot represent a parabola.Wait, let me double-check. The standard form of a parabola is either ( y^2 = 4ax ) or ( x^2 = 4ay ), where only one variable is squared. In our equation, both ( x^2 ) and ( y^2 ) are present, so it's not a parabola. Parabolas have only one squared term, so this equation can't represent a parabola.So, putting it all together, the equation can represent a circle when ( cos theta = 1 ), an ellipse when ( 0 < cos theta < 1 ), and a hyperbola when ( -1 < cos theta < 0 ). However, it cannot represent a parabola because it always has both ( x^2 ) and ( y^2 ) terms, which is characteristic of circles, ellipses, and hyperbolas, but not parabolas.Therefore, the answer should be D. Parabola.

💡Alright, so I've got this problem here: Given that (3^a = 4^b = 5^c = 6), I need to find the value of (frac{1}{a} + frac{1}{b} + frac{1}{c}). Hmm, okay, let's break this down step by step.First off, all these expressions (3^a), (4^b), and (5^c) are equal to 6. That means (3^a = 6), (4^b = 6), and (5^c = 6). So, I can write them as separate equations:1. (3^a = 6)2. (4^b = 6)3. (5^c = 6)Now, I need to find (frac{1}{a} + frac{1}{b} + frac{1}{c}). To do that, I probably need to find the values of (a), (b), and (c) first, right? But since they're all expressed in terms of exponents, maybe I can use logarithms to solve for them.Let me recall that if I have an equation like (x^y = z), I can solve for (y) by taking the logarithm of both sides. Specifically, (y = frac{log z}{log x}). Yeah, that sounds familiar from logarithm properties.So, applying that to each equation:1. For (3^a = 6), taking the logarithm of both sides gives (a = frac{log 6}{log 3}).2. For (4^b = 6), similarly, (b = frac{log 6}{log 4}).3. For (5^c = 6), (c = frac{log 6}{log 5}).Okay, so now I have expressions for (a), (b), and (c). The next step is to find (frac{1}{a} + frac{1}{b} + frac{1}{c}). Let's compute each reciprocal:1. (frac{1}{a} = frac{log 3}{log 6})2. (frac{1}{b} = frac{log 4}{log 6})3. (frac{1}{c} = frac{log 5}{log 6})So, adding them up:[frac{1}{a} + frac{1}{b} + frac{1}{c} = frac{log 3}{log 6} + frac{log 4}{log 6} + frac{log 5}{log 6}]Since all the denominators are the same, I can combine the numerators:[frac{log 3 + log 4 + log 5}{log 6}]Hmm, I remember that the sum of logarithms is the logarithm of the product. So, (log 3 + log 4 + log 5 = log (3 times 4 times 5)). Let me calculate that product:(3 times 4 = 12), and (12 times 5 = 60). So, it's (log 60).Therefore, the expression simplifies to:[frac{log 60}{log 6}]Wait a second, that looks like a logarithm change of base formula. Specifically, (log_b a = frac{log a}{log b}). So, (frac{log 60}{log 6} = log_6 60).So, putting it all together, (frac{1}{a} + frac{1}{b} + frac{1}{c} = log_6 60).Let me just double-check to make sure I didn't make any mistakes. Starting from the beginning, I took the logarithm of each equation correctly, solved for (a), (b), and (c), then took their reciprocals and added them up. Combined the logarithms into a single log of the product, which is correct. Then recognized the change of base formula to express it as (log_6 60). Yeah, that seems right.I wonder if there's another way to approach this problem without using logarithms? Maybe by expressing 6 in terms of powers of 3, 4, and 5, but that seems more complicated. Logarithms seem to be the straightforward method here.Alternatively, could I express all the terms with the same base? For example, express 6 as (3^a), (4^b), and (5^c), but that's essentially what I did by taking logarithms. So, I think the logarithmic approach is the most efficient here.Just to recap:1. Given (3^a = 4^b = 5^c = 6).2. Solved for (a), (b), and (c) using logarithms.3. Took reciprocals and summed them.4. Combined the logarithms into a single term.5. Applied the change of base formula to get the final answer.Everything seems to check out. I don't see any errors in my reasoning, so I feel confident that (log_6 60) is the correct answer.**Final Answer**The value of (frac{1}{a} + frac{1}{b} + frac{1}{c}) is boxed{log_{6} 60}.