answer:Given the problem, we need to determine natural numbers n such that n^{2} + 492 is a perfect square. We can formulate this as: [ n^{2} + 492 = k^{2} ] for some natural number k > n. Rewriting the equation, we get: [ k^2 - n^2 = 492 ] We can factor this as: [ (k - n)(k + n) = 492 ] Next, note that 492 can be factored into its prime components: [ 492 = 2^2 cdot 3 cdot 41 ] Since both k - n and k + n should be integers and must have the same parity, they must both be even. The pairs of factors of 492 that satisfy this condition are (2, 246) and (6, 82). We will examine these pairs: Case 1: k - n = 2 and k + n = 246 Solving this system of equations: [ k - n = 2 ] [ k + n = 246 ] Adding these equations: [ (k - n) + (k + n) = 2k = 248 ] [ k = frac{248}{2} = 124 ] Subtracting the equations: [ (k + n) - (k - n) = 2n = 244 ] [ n = frac{244}{2} = 122 ] Case 2: k - n = 6 and k + n = 82 Solving this system of equations: [ k - n = 6 ] [ k + n = 82 ] Adding these equations: [ (k - n) + (k + n) = 2k = 88 ] [ k = frac{88}{2} = 44 ] Subtracting the equations: [ (k + n) - (k - n) = 2n = 76 ] [ n = frac{76}{2} = 38 ] Therefore, the natural numbers n that satisfy the given conditions are: [ boxed{n = 122 text{ and } n = 38} ] We have verified both pairs, and they both satisfy the original equation: ( n^2 + 492 ) is a perfect square for ( n = 122 ) and ( n = 38 ).

answer:Since f(x) is an odd function defined on [-2, 2], we have f(0) = 0. When x in (0, 2], f(x) = 2^x - 1 in (0, 3], thus when x in [-2, 2], f(x) in [-3, 3], If for every x_1 in [-2, 2], there exists an x_2 in [-2, 2] such that g(x_2) = f(x_1), then it is equivalent to g(x)_{text{max}} geq 3 and g(x)_{text{min}} leq -3, Since g(x) = x^2 - 2x + m = (x - 1)^2 + m - 1, for x in [-2, 2], thus g(x)_{text{max}} = g(-2) = 8 + m, g(x)_{text{min}} = g(1) = m - 1, then satisfying 8 + m geq 3 and m - 1 leq -3, solving gives m geq -5 and m leq -2, therefore, -5 leq m leq -2, hence the answer is: boxed{[-5, -2]} **Analysis:** Determine the range of the function f(x), and based on the condition, the relationship between the maximum and minimum values of the two functions can be established to reach the conclusion.

answer:Let's denote the cost price of the mangoes per kg as ( C ). According to the problem, selling the mangoes at Rs. 12.35 per kg would give the fruit seller a profit of 5%. This means that the selling price at a 5% profit is 105% of the cost price. So, we can write the following equation: ( 1.05C = 12.35 ) Now, we can solve for ( C ): ( C = frac{12.35}{1.05} ) ( C = 11.76 ) (approximately) Now we know the cost price of the mangoes per kg is Rs. 11.76. The fruit seller originally sold the mangoes at Rs. 10 per kg, which is less than the cost price, indicating a loss. The loss per kg can be calculated as: ( text{Loss per kg} = C - text{Original Selling Price} ) ( text{Loss per kg} = 11.76 - 10 ) ( text{Loss per kg} = 1.76 ) Now, to find the percentage of loss, we use the formula: ( text{Percentage of Loss} = left( frac{text{Loss per kg}}{C} right) times 100 ) ( text{Percentage of Loss} = left( frac{1.76}{11.76} right) times 100 ) ( text{Percentage of Loss} = left( frac{1.76}{11.76} right) times 100 ) ( text{Percentage of Loss} = 0.1496 times 100 ) ( text{Percentage of Loss} = 14.96 % ) Therefore, the percentage of loss at the original selling price is approximately boxed{14.96%} .

answer:1. **Lagrange's identity** relays a relationship between the sums of squares and the sum of products of two sequences. Our goal is to demonstrate Lagrange's identity: left(sum_{i=1}^{n} a_{i}^{2}right) cdot left(sum_{i=1}^{n} b_{i}^{2}right) = left(sum_{i=1}^{n} a_{i} b_{i}right)^{2} + sum_{1 leq i < j leq n}left(a_{i} b_{j} - a_{j} b_{i}right)^{2}. 2. We begin by expanding the left-hand side and right-hand side of Lagrange's identity separately: **Left-Hand Side**: [ left( sum_{i=1}^n a_i^2 right) cdot left( sum_{i=1}^n b_i^2 right) = sum_{i=1}^n sum_{j=1}^n a_i^2 b_j^2. ] **Right-Hand Side** consists of two terms: - The square of the sum of products: [ left( sum_{i=1}^n a_i b_i right)^2 = sum_{i=1}^n sum_{j=1}^n a_i b_i a_j b_j. ] - The sum of squares of differences: [ sum_{1 leq i < j leq n} left( a_i b_j - a_j b_i right)^2. ] 3. Expand the right-hand side of the equation as described below: - For the first term: [ left( sum_{i=1}^n a_i b_i right)^2 = sum_{i=1}^n sum_{j=1}^n a_i b_i a_j b_j. ] - For the second term, consider: [ sum_{1 leq i < j leq n} left( a_i b_j - a_j b_i right)^2 = sum_{1 leq i < j leq n} left( a_i^2 b_j^2 + a_j^2 b_i^2 - 2a_i b_i a_j b_j right). ] Let's adjust the terms of the full double sum: [ frac{1}{2} sum_{i=1}^n sum_{j=1}^n left(a_i b_j - a_j b_i right)^2 = frac{1}{2} sum_{i=1}^n sum_{j=1}^n left( a_i^2 b_j^2 + a_j^2 b_i^2 - 2a_i b_i a_j b_j right). ] 4. Subtract the first right-hand side term from the left-hand side: [ left(sum_{i=1}^{n} a_{i}^{2}right) cdot left(sum_{i=1}^{n} b_{i}^{2}right) - left( sum_{i=1}^{n} a_{i} b_{i} right)^{2} = frac{1}{2} sum_{i=1}^{n} sum_{j=1}^{n} left( a_i^2 b_j^2 + a_j^2 b_i^2 - 2a_i b_i a_j b_j right). ] Combining terms, we see: [ = frac{1}{2} sum_{i=1}^{n} sum_{j=1}^{n} left(a_i b_j - a_j b_i right)^2. ] Thus, we have verified: [ left(sum_{i=1}^{n} a_{i}^{2}right) cdot left(sum_{i=1}^{n} b_{i}^{2}right) = left(sum_{i=1}^{n} a_{i} b_{i}right)^{2} + sum_{1 leqslant i < j leqslant n}left(a_{i} b_{j} - a_{j} b_{i}right)^{2}. ] 5. **Proof of Cauchy-Schwarz Inequality**: Since each term left(a_{i} b_{j} - a_{j} b_{i}right)^{2} is non-negative, we conclude: [ sum_{1 leqslant i < j leqslant n}left(a_i b_j - a_j b_i right)^2 geq 0. ] -> Hence: [ left( sum_{i=1}^n a_i^2 right) left( sum_{i=1}^n b_i^2 right) geq left( sum_{i=1}^n a_i b_i right)^2. ] Conclusion: [ boxed{left( sum_{i=1}^n a_i^2 right) left( sum_{i=1}^n b_i^2 right) geq left( sum_{i=1}^n a_i b_i right)^2}. ] This is the Cauchy-Schwarz inequality.