💡Okay, so I have this problem about a parabola and a line intersecting it. The parabola is given by the equation ( y^2 = 4x ), and its focus is ( F ). There's a line ( l ) with a slope of 1 that passes through the focus ( F ) and intersects the parabola at points ( P ) and ( Q ). I need to find the value of ( frac{1}{|PF|} + frac{1}{|QF|} ).First, I remember that for a parabola in the form ( y^2 = 4ax ), the focus is at ( (a, 0) ). In this case, comparing ( y^2 = 4x ) to the standard form, ( 4a = 4 ), so ( a = 1 ). That means the focus ( F ) is at ( (1, 0) ).Next, the line ( l ) has a slope of 1 and passes through ( F(1, 0) ). The general equation of a line with slope ( m ) passing through a point ( (x_1, y_1) ) is ( y - y_1 = m(x - x_1) ). Plugging in the values, the equation of line ( l ) is ( y - 0 = 1 cdot (x - 1) ), which simplifies to ( y = x - 1 ).Now, I need to find the points where this line intersects the parabola. To do that, I'll substitute ( y = x - 1 ) into the equation of the parabola ( y^2 = 4x ). Substituting gives:[(x - 1)^2 = 4x]Expanding the left side:[x^2 - 2x + 1 = 4x]Bringing all terms to one side to form a quadratic equation:[x^2 - 6x + 1 = 0]So, the quadratic equation is ( x^2 - 6x + 1 = 0 ). Let me denote the roots of this equation as ( x_P ) and ( x_Q ), which correspond to the x-coordinates of points ( P ) and ( Q ).From the quadratic equation ( ax^2 + bx + c = 0 ), the sum of the roots is ( -b/a ) and the product is ( c/a ). Here, ( a = 1 ), ( b = -6 ), and ( c = 1 ). Therefore:- Sum of roots: ( x_P + x_Q = -(-6)/1 = 6 )- Product of roots: ( x_P x_Q = 1/1 = 1 )Now, I need to find the distances ( |PF| ) and ( |QF| ). Since ( F ) is at ( (1, 0) ), and points ( P ) and ( Q ) lie on the parabola, their coordinates are ( (x_P, y_P) ) and ( (x_Q, y_Q) ) respectively, where ( y_P = x_P - 1 ) and ( y_Q = x_Q - 1 ).The distance from a point ( (x, y) ) to ( F(1, 0) ) can be found using the distance formula:[|PF| = sqrt{(x_P - 1)^2 + (y_P - 0)^2}][|QF| = sqrt{(x_Q - 1)^2 + (y_Q - 0)^2}]But since ( y_P = x_P - 1 ) and ( y_Q = x_Q - 1 ), substituting these into the distance formula gives:[|PF| = sqrt{(x_P - 1)^2 + (x_P - 1)^2} = sqrt{2(x_P - 1)^2} = sqrt{2} |x_P - 1|][|QF| = sqrt{(x_Q - 1)^2 + (x_Q - 1)^2} = sqrt{2(x_Q - 1)^2} = sqrt{2} |x_Q - 1|]However, since the line ( l ) intersects the parabola at two points, and the parabola is symmetric about the x-axis, I can assume that both ( x_P ) and ( x_Q ) are greater than 1 because the focus is at (1,0) and the line with slope 1 will intersect the parabola on either side of the focus. Therefore, ( x_P - 1 ) and ( x_Q - 1 ) are positive, so the absolute value can be removed:[|PF| = sqrt{2}(x_P - 1)][|QF| = sqrt{2}(x_Q - 1)]Wait, hold on. I think I made a mistake here. The distance from ( P ) to ( F ) isn't necessarily ( sqrt{2}(x_P - 1) ). Let me double-check that.Given ( y_P = x_P - 1 ), so ( y_P = x_P - 1 ). Then, the distance ( |PF| ) is:[|PF| = sqrt{(x_P - 1)^2 + (y_P)^2} = sqrt{(x_P - 1)^2 + (x_P - 1)^2} = sqrt{2(x_P - 1)^2} = sqrt{2}|x_P - 1|]Since ( x_P ) is a point on the parabola, and the line intersects the parabola at two points, one on either side of the focus. Wait, actually, the parabola ( y^2 = 4x ) opens to the right, so the focus is at (1,0). The line with slope 1 passing through (1,0) would intersect the parabola at two points, both to the right of the vertex, which is at (0,0). So, ( x_P ) and ( x_Q ) are both greater than 1, meaning ( x_P - 1 ) and ( x_Q - 1 ) are positive, so ( |PF| = sqrt{2}(x_P - 1) ) and ( |QF| = sqrt{2}(x_Q - 1) ).But actually, I think I might be overcomplicating this. There's a property of parabolas that the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix. For the parabola ( y^2 = 4x ), the directrix is the line ( x = -1 ). So, the distance from ( P(x_P, y_P) ) to the focus ( F(1,0) ) is equal to the distance from ( P ) to the directrix, which is ( x_P + 1 ). Therefore, ( |PF| = x_P + 1 ). Similarly, ( |QF| = x_Q + 1 ).Wait, that's a much simpler way to find ( |PF| ) and ( |QF| )! I should have remembered that property. So, instead of using the distance formula, I can directly say that ( |PF| = x_P + 1 ) and ( |QF| = x_Q + 1 ).So, now, I have:- ( |PF| = x_P + 1 )- ( |QF| = x_Q + 1 )I need to compute ( frac{1}{|PF|} + frac{1}{|QF|} ), which is:[frac{1}{x_P + 1} + frac{1}{x_Q + 1}]To compute this, I can combine the fractions:[frac{(x_Q + 1) + (x_P + 1)}{(x_P + 1)(x_Q + 1)} = frac{x_P + x_Q + 2}{(x_P + 1)(x_Q + 1)}]From earlier, I know that ( x_P + x_Q = 6 ) and ( x_P x_Q = 1 ). Let's compute the numerator and the denominator separately.Numerator:[x_P + x_Q + 2 = 6 + 2 = 8]Denominator:[(x_P + 1)(x_Q + 1) = x_P x_Q + x_P + x_Q + 1 = 1 + 6 + 1 = 8]Therefore, the expression simplifies to:[frac{8}{8} = 1]So, the value of ( frac{1}{|PF|} + frac{1}{|QF|} ) is 1.Wait, let me just verify that I didn't make any mistakes in my calculations. I used the property that the distance from a point on the parabola to the focus is equal to the distance to the directrix, which is ( x + 1 ) for this parabola. Then, I found the sum and product of the roots from the quadratic equation, which gave me ( x_P + x_Q = 6 ) and ( x_P x_Q = 1 ). Plugging these into the numerator and denominator, I got both equal to 8, leading to the result 1. That seems correct.I think another way to approach this is by using parametric equations or calculus, but since I got a straightforward answer using the properties of the parabola and quadratic equations, I feel confident that the answer is 1.**Final Answer**boxed{C}

💡To find the area of trapezoid ( PQRS ) with ( PQ parallel RS ), we start by analyzing the given areas of triangles ( PQT ) and ( PRT ). 1. **Identify Similar Triangles:** - Triangles ( PQT ) and ( PRT ) share a common vertex at ( P ) and their bases ( QT ) and ( RT ) lie on the same line ( QS ). Therefore, they are similar triangles.2. **Determine the Ratio of Areas:** - The ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides. Given that ( [PQT] = 60 ) and ( [PRT] = 30 ), the ratio of their areas is ( frac{30}{60} = frac{1}{2} ). Thus, the ratio of their corresponding sides is ( sqrt{frac{1}{2}} = frac{1}{sqrt{2}} ).3. **Calculate the Area of Triangle ( QST ):** - Since triangles ( PRT ) and ( QST ) are similar and share the same height from ( T ) to ( RS ), the ratio of their areas is the same as the ratio of their bases. Using the ratio ( frac{1}{2} ), the area of triangle ( QST ) is ( frac{1}{2} times 30 = 15 ) square units.4. **Sum the Areas:** - The trapezoid ( PQRS ) is composed of triangles ( PQT ), ( PRT ), and ( QST ). Adding their areas together: ( 60 + 30 + 15 = 105 ) square units.However, upon reviewing the calculation, it's clear that the initial approach missed considering the contribution of triangle ( QST ) correctly. To ensure accuracy, let's re-examine the relationships between the triangles and apply the correct proportional reasoning.

💡Alright, so I have this identity to prove: [(ax + by + cz)^2 + (bx + cy + az)^2 + (cx + ay + bz)^2 = (cx + by + az)^2 + (bx + ay + cz)^2 + (ax + cy + bz)^2]Hmm, okay. At first glance, it looks like both sides are sums of squares of linear combinations of x, y, and z with coefficients a, b, and c. The left side has the terms (ax + by + cz), (bx + cy + az), and (cx + ay + bz), while the right side has (cx + by + az), (bx + ay + cz), and (ax + cy + bz). I notice that the coefficients a, b, and c are being permuted in some way in each term on both sides. Maybe if I can show that both sides simplify to the same expression, the identity will hold. Let me start by expanding both sides and see if they end up being the same. First, let's expand the left-hand side (LHS):1. Expand ((ax + by + cz)^2): [ a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2acxz + 2bcyz ]2. Expand ((bx + cy + az)^2): [ b^2x^2 + c^2y^2 + a^2z^2 + 2bcxy + 2abxz + 2cayz ]3. Expand ((cx + ay + bz)^2): [ c^2x^2 + a^2y^2 + b^2z^2 + 2caxy + 2bcxz + 2abyz ]Now, let's add all these expanded terms together:[(a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2acxz + 2bcyz) + (b^2x^2 + c^2y^2 + a^2z^2 + 2bcxy + 2abxz + 2cayz) + (c^2x^2 + a^2y^2 + b^2z^2 + 2caxy + 2bcxz + 2abyz)]Let's combine like terms:- For (x^2): (a^2x^2 + b^2x^2 + c^2x^2 = (a^2 + b^2 + c^2)x^2)- For (y^2): (b^2y^2 + c^2y^2 + a^2y^2 = (a^2 + b^2 + c^2)y^2)- For (z^2): (c^2z^2 + a^2z^2 + b^2z^2 = (a^2 + b^2 + c^2)z^2)Now, for the cross terms:- For (xy): (2abxy + 2bcxy + 2caxy = 2(ab + bc + ca)xy)- For (xz): (2acxz + 2abxz + 2bcxz = 2(ab + bc + ca)xz)- For (yz): (2bcyz + 2cayz + 2abyz = 2(ab + bc + ca)yz)So, combining all these, the LHS simplifies to:[(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) + 2(ab + bc + ca)(xy + xz + yz)]Okay, that's the LHS. Now, let's do the same for the right-hand side (RHS):1. Expand ((cx + by + az)^2): [ c^2x^2 + b^2y^2 + a^2z^2 + 2bcxy + 2acxz + 2abyz ]2. Expand ((bx + ay + cz)^2): [ b^2x^2 + a^2y^2 + c^2z^2 + 2abxy + 2bcxz + 2acyz ]3. Expand ((ax + cy + bz)^2): [ a^2x^2 + c^2y^2 + b^2z^2 + 2acxy + 2abxz + 2bcyz ]Now, add these expanded terms together:[(c^2x^2 + b^2y^2 + a^2z^2 + 2bcxy + 2acxz + 2abyz) + (b^2x^2 + a^2y^2 + c^2z^2 + 2abxy + 2bcxz + 2acyz) + (a^2x^2 + c^2y^2 + b^2z^2 + 2acxy + 2abxz + 2bcyz)]Again, let's combine like terms:- For (x^2): (c^2x^2 + b^2x^2 + a^2x^2 = (a^2 + b^2 + c^2)x^2)- For (y^2): (b^2y^2 + a^2y^2 + c^2y^2 = (a^2 + b^2 + c^2)y^2)- For (z^2): (a^2z^2 + c^2z^2 + b^2z^2 = (a^2 + b^2 + c^2)z^2)Now, for the cross terms:- For (xy): (2bcxy + 2abxy + 2acxy = 2(ab + bc + ca)xy)- For (xz): (2acxz + 2bcxz + 2abxz = 2(ab + bc + ca)xz)- For (yz): (2abyz + 2acyz + 2bcyz = 2(ab + bc + ca)yz)So, combining all these, the RHS simplifies to:[(a^2 + b^2 + c^2)(x^2 + y^2 + z^2) + 2(ab + bc + ca)(xy + xz + yz)]Wait a minute, that's exactly the same as the LHS! So both sides simplify to the same expression, which means the identity holds true.I think that's it. By expanding both sides and simplifying, I've shown that they are equal. It seems like the way the coefficients a, b, and c are permuted in each term doesn't affect the overall sum because when you expand and combine like terms, the symmetric nature of the expressions ensures equality.Just to double-check, maybe I can plug in some specific values for a, b, c, x, y, z and see if both sides are equal.Let's say a = 1, b = 2, c = 3, x = 1, y = 1, z = 1.Compute LHS:[(1*1 + 2*1 + 3*1)^2 + (2*1 + 3*1 + 1*1)^2 + (3*1 + 1*1 + 2*1)^2][(1 + 2 + 3)^2 + (2 + 3 + 1)^2 + (3 + 1 + 2)^2][6^2 + 6^2 + 6^2 = 36 + 36 + 36 = 108]Compute RHS:[(3*1 + 2*1 + 1*1)^2 + (2*1 + 1*1 + 3*1)^2 + (1*1 + 3*1 + 2*1)^2][(3 + 2 + 1)^2 + (2 + 1 + 3)^2 + (1 + 3 + 2)^2][6^2 + 6^2 + 6^2 = 36 + 36 + 36 = 108]They are equal. Let's try another set of values to be sure.Let a = 2, b = 3, c = 4, x = 2, y = 3, z = 4.Compute LHS:[(2*2 + 3*3 + 4*4)^2 + (3*2 + 4*3 + 2*4)^2 + (4*2 + 2*3 + 3*4)^2][(4 + 9 + 16)^2 + (6 + 12 + 8)^2 + (8 + 6 + 12)^2][29^2 + 26^2 + 26^2 = 841 + 676 + 676 = 2193]Compute RHS:[(4*2 + 3*3 + 2*4)^2 + (3*2 + 2*3 + 4*4)^2 + (2*2 + 4*3 + 3*4)^2][(8 + 9 + 8)^2 + (6 + 6 + 16)^2 + (4 + 12 + 12)^2][25^2 + 28^2 + 28^2 = 625 + 784 + 784 = 2193]Again, both sides are equal. This gives me more confidence that the identity is indeed correct.I think I've covered both the algebraic expansion and numerical verification. It seems solid. Maybe there's a more elegant way to see why this identity holds without expanding everything, perhaps by recognizing some symmetry or using properties of symmetric functions, but for now, the expansion method works.

💡First, I need to determine the cost price of the bicycle for person A. Person A sells the bicycle to person B at a 25% profit. This means that the selling price for A, which becomes the cost price for B, is 125% of A's cost price.Next, person B sells the bicycle to person C at a 25% profit. The selling price for B, which is the final selling price, is given as Rs. 225. This selling price is 125% of B's cost price.I can set up the equation using the final selling price and work backwards to find A's cost price. By substituting the relationships between the cost prices and selling prices, I can solve for A's cost price.Finally, by performing the calculations, I find that the cost price of the bicycle for A is Rs. 144.

💡Okay, so I have this problem where Jerry starts at 0 on the real number line and tosses a fair coin 10 times. Every time he gets heads, he moves 1 unit in the positive direction, and tails means he moves 1 unit in the negative direction. I need to find the probability that he reaches +6 at some point during these 10 tosses. The answer should be in the form of a fraction c/d, where c and d are coprime, and then find c + d. The options are A) 127, B) 557, C) 514, D) 657, E) 673.Alright, let me try to break this down step by step. First, I know that each coin toss is independent and has a 50-50 chance of being heads or tails. So, each move is either +1 or -1 with equal probability.Jerry starts at 0, and we want to know the probability that he reaches +6 at least once during his 10 moves. This is similar to a one-dimensional random walk problem where we want the probability that the walk reaches a certain point.I remember something called the reflection principle in probability, which is used to calculate such probabilities. The reflection principle helps in counting the number of paths that cross a certain boundary, which in this case is +6.Let me recall how the reflection principle works. If we want to find the number of paths that reach +6 at some point, we can consider the first time the path reaches +6 and then reflect the subsequent steps. This reflection helps in counting the number of paths that cross the boundary, which can then be used to find the probability.But before diving into that, maybe I should think about the total number of possible paths. Since each toss has 2 outcomes, the total number of paths after 10 tosses is 2^10, which is 1024. So, the denominator of our probability will be 1024.Now, the numerator is the number of paths that reach +6 at some point during the 10 tosses. To find this, I can use the reflection principle. The idea is to count the number of paths that reach +6 and then adjust for the overcounting.Let me denote the number of heads as H and tails as T. Each head gives +1, and each tail gives -1. So, after 10 tosses, the position is H - T. But since H + T = 10, the position is 2H - 10. So, if we want the position to be +6 at some point, we need 2H - 10 = 6, which implies H = 8. So, to reach +6, Jerry needs 8 heads and 2 tails.But wait, this is only the number of paths that end at +6 after 10 tosses. However, we want the number of paths that reach +6 at least once during the 10 tosses, not necessarily ending there. So, the number of paths that end at +6 is C(10,8) = 45. But this is just the number of paths that end at +6, not the ones that reach +6 at some point.So, I need a different approach. The reflection principle can help here. The reflection principle states that the number of paths that reach a certain level can be found by reflecting the paths that cross that level.Let me try to apply the reflection principle. The formula for the number of paths that reach +a in n steps is 2 * C(n, (n + a)/2) if n and a have the same parity. In our case, a = 6 and n = 10. Since 10 and 6 are both even, their sum is 16, which is even, so (n + a)/2 = (10 + 6)/2 = 8. So, the number of paths that reach +6 is 2 * C(10,8) = 2 * 45 = 90.Wait, but this seems too high because the total number of paths is 1024, and 90 is much less than that. Hmm, maybe I'm misapplying the reflection principle.Let me think again. The reflection principle is used to count the number of paths that cross a certain boundary, but in this case, we want the number of paths that reach +6 at least once. So, perhaps the formula is different.I recall that the number of paths that reach +a at least once in n steps is equal to the sum over k from 0 to floor((n - a)/2) of (-1)^k * C(n, (a + k)/2). But I'm not sure if that's correct.Alternatively, I remember that the probability of reaching +a in a symmetric random walk can be calculated using the formula:P = 2 * (C(n, (n + a)/2) / 2^n)But this is similar to what I did earlier, giving 2 * 45 / 1024 = 90 / 1024 = 45 / 512.Wait, that seems plausible. So, the probability is 45/512. Let me check if this makes sense.If Jerry needs to reach +6, which requires 8 heads and 2 tails, the number of paths that end at +6 is C(10,8) = 45. But the reflection principle says that the number of paths that reach +6 at some point is twice that, so 90. Therefore, the probability is 90 / 1024 = 45 / 512.Yes, that seems correct. So, the probability is 45/512. Now, we need to make sure that 45 and 512 are coprime. 45 factors into 3^2 * 5, and 512 is 2^9. They have no common factors, so the fraction is already in simplest terms.Therefore, c = 45 and d = 512, so c + d = 45 + 512 = 557.Looking back at the options, B is 557, so that should be the answer.Wait, just to make sure, let me think if there's another way to approach this. Maybe using states or recursion.Suppose we model this as a Markov chain where each state is the current position, from -10 to +10. We start at 0, and we want the probability of reaching +6 before step 10.But that might be more complicated. Alternatively, using the concept of Gambler's Ruin, where we calculate the probability of reaching a certain wealth before going bankrupt.In this case, the "wealth" is the position, and we want the probability of reaching +6 before reaching -infinity, but since we have a finite number of steps, it's a bit different.Alternatively, using the range of the walk. The maximum position reached during the walk is a key factor. The probability that the maximum position is at least 6.I think the reflection principle is the right way to go here, and the calculation seems correct. So, I think 557 is the right answer.

💡Okay, so I have this function f(x) = x³ - 3x² - 9x + 1, and I need to find its intervals of monotonicity and its extrema. Hmm, okay, let's start by recalling what monotonicity means. If a function is increasing on an interval, its derivative is positive there, and if it's decreasing, the derivative is negative. So, to find where the function is increasing or decreasing, I need to find its derivative first.Alright, let's compute the derivative of f(x). The derivative of x³ is 3x², the derivative of -3x² is -6x, the derivative of -9x is -9, and the derivative of a constant like 1 is 0. So, putting it all together, f'(x) = 3x² - 6x - 9.Now, to find the critical points, I need to set the derivative equal to zero and solve for x. So, 3x² - 6x - 9 = 0. Hmm, this is a quadratic equation. Maybe I can factor it. Let's see: 3x² - 6x - 9. I can factor out a 3 first, so 3(x² - 2x - 3) = 0. Now, factoring x² - 2x - 3. Looking for two numbers that multiply to -3 and add to -2. Hmm, -3 and +1. So, (x - 3)(x + 1) = 0. Therefore, the critical points are x = 3 and x = -1.Okay, so I have critical points at x = -1 and x = 3. These points divide the real number line into three intervals: (-∞, -1), (-1, 3), and (3, ∞). To determine the sign of the derivative in each interval, I can pick a test point in each interval and plug it into f'(x).Let's start with the interval (-∞, -1). I'll pick x = -2. Plugging into f'(x): 3(-2)² - 6(-2) - 9 = 3(4) + 12 - 9 = 12 + 12 - 9 = 15. That's positive, so f'(x) is positive in (-∞, -1), meaning f(x) is increasing there.Next, the interval (-1, 3). Let's choose x = 0. Plugging into f'(x): 3(0)² - 6(0) - 9 = 0 - 0 - 9 = -9. That's negative, so f'(x) is negative in (-1, 3), meaning f(x) is decreasing there.Finally, the interval (3, ∞). Let's take x = 4. Plugging into f'(x): 3(4)² - 6(4) - 9 = 3(16) - 24 - 9 = 48 - 24 - 9 = 15. Positive again, so f'(x) is positive in (3, ∞), meaning f(x) is increasing there.So, summarizing the intervals of monotonicity:- Increasing on (-∞, -1)- Decreasing on (-1, 3)- Increasing on (3, ∞)Now, for the extrema. Since the function changes from increasing to decreasing at x = -1, there should be a local maximum there. Conversely, at x = 3, the function changes from decreasing to increasing, so there should be a local minimum.Let's find the corresponding y-values for these critical points by plugging them back into the original function f(x).First, for x = -1:f(-1) = (-1)³ - 3(-1)² - 9(-1) + 1 = -1 - 3 + 9 + 1 = 6.Next, for x = 3:f(3) = 3³ - 3(3)² - 9(3) + 1 = 27 - 27 - 27 + 1 = -26.So, the function has a local maximum at (-1, 6) and a local minimum at (3, -26).Wait, let me double-check my calculations to make sure I didn't make any mistakes. For f(-1): (-1)^3 is -1, -3*(-1)^2 is -3*1 which is -3, -9*(-1) is +9, and +1. So, -1 -3 +9 +1 is indeed 6. For f(3): 3^3 is 27, -3*(3)^2 is -27, -9*3 is -27, and +1. So, 27 -27 -27 +1 is -26. Yep, that's correct.So, putting it all together, the function f(x) = x³ - 3x² - 9x + 1 is increasing on (-∞, -1) and (3, ∞), decreasing on (-1, 3), with a local maximum at x = -1 (y = 6) and a local minimum at x = 3 (y = -26).