answer:1. **Define the problem and the given inequality:** We are given the inequality: [ frac{2}{sqrt{3}}tan^{-1}left(frac{2(b^2 - a^2)}{(a^2+2)(b^2+2)}right)leq int limits_a^b frac{(x^2+1)(x^2+x+1)}{(x^3 + x^2 + 1) (x^3 + x + 1)}dxleq frac{4}{sqrt{3}}tan^{-1}left(frac{(b-a)sqrt{3}}{a+b+2(1+ab)}right) ] We need to prove both parts of the inequality. 2. **Prove the right-hand side (RHS) inequality:** Denote the double inequality as ( A leq B leq C ), where: - ( A = frac{2}{sqrt{3}}tan^{-1}left(frac{2(b^2 - a^2)}{(a^2+2)(b^2+2)}right) ) - ( B = int limits_a^b frac{(x^2+1)(x^2+x+1)}{(x^3 + x^2 + 1) (x^3 + x + 1)}dx ) - ( C = frac{4}{sqrt{3}}tan^{-1}left(frac{(b-a)sqrt{3}}{a+b+2(1+ab)}right) ) 3. **Consider the function ( C - B ) and its derivative:** Fix ( a > 0 ) and consider ( C - B ) as a function of ( b ). We need to show that ( C(b) - B(b) ) is non-decreasing in ( b ). 4. **Compute the derivative ( C'(b) - B'(b) ):** [ C'(b) - B'(b) = frac{(b-1)^2(b^4+2b^3+b^2+2b+1)}{(b^2+b+1)(b^3+b+1)(b^3+b^2+1)} ] Since the numerator ((b-1)^2(b^4+2b^3+b^2+2b+1)) is always non-negative and the denominator ((b^2+b+1)(b^3+b+1)(b^3+b^2+1)) is always positive for ( b > 0 ), we have: [ C'(b) - B'(b) geq 0 ] This implies that ( C(b) - B(b) ) is non-decreasing in ( b ). 5. **Evaluate ( C(a) - B(a) ):** [ C(a) - B(a) = 0 ] Since ( C(b) - B(b) ) is non-decreasing and ( C(a) - B(a) = 0 ), it follows that: [ C(b) - B(b) geq 0 quad text{for all} quad b geq a ] Therefore, ( B leq C ). 6. **Prove the left-hand side (LHS) inequality:** The proof for the LHS inequality follows a similar approach. We need to show that ( B - A ) is non-decreasing in ( b ). 7. **Consider the function ( B - A ) and its derivative:** Fix ( a > 0 ) and consider ( B - A ) as a function of ( b ). We need to show that ( B(b) - A(b) ) is non-decreasing in ( b ). 8. **Compute the derivative ( B'(b) - A'(b) ):** [ B'(b) - A'(b) = frac{(b-1)^2(b^4+2b^3+b^2+2b+1)}{(b^2+b+1)(b^3+b+1)(b^3+b^2+1)} ] Since the numerator ((b-1)^2(b^4+2b^3+b^2+2b+1)) is always non-negative and the denominator ((b^2+b+1)(b^3+b+1)(b^3+b^2+1)) is always positive for ( b > 0 ), we have: [ B'(b) - A'(b) geq 0 ] This implies that ( B(b) - A(b) ) is non-decreasing in ( b ). 9. **Evaluate ( B(a) - A(a) ):** [ B(a) - A(a) = 0 ] Since ( B(b) - A(b) ) is non-decreasing and ( B(a) - A(a) = 0 ), it follows that: [ B(b) - A(b) geq 0 quad text{for all} quad b geq a ] Therefore, ( A leq B ). 10. **Combine the results:** Since we have shown both ( A leq B ) and ( B leq C ), the double inequality is proven. (blacksquare)

answer:# Solução: 1. **Definir os preços iniciais**: - Suponhamos que uma dúzia de ovos custava ( R 1,00 ). - Portanto, 10 maçãs também custavam ( R 1,00 ). 2. **Calcular os novos preços após a variação percentual**: - O preço dos ovos diminuiu em ( 10 % ): [ text{Novo preço dos ovos} = 1,00 - 0,10 times 1,00 = 1,00 - 0,10 = R 0,90 ] - O preço das maçãs aumentou ( 2 % ): [ text{Novo preço das maçãs} = 1,00 + 0,02 times 1,00 = 1,00 + 0,02 = R 1,02 ] 3. **Calcular o total gasto antes e depois da variação**: - Inicialmente, o total gasto na compra de uma dúzia de ovos e 10 maçãs era: [ text{Total inicial} = 1,00 + 1,00 = R 2,00 ] - Agora, o total gasto com os novos preços é: [ text{Total novo} = 0,90 + 1,02 = R 1,92 ] 4. **Determinar o aumento de preço absoluto e percentual**: - O aumento absoluto no gasto total é: [ text{Aumento absoluto} = 1,92 - 2,00 = -R 0,08 ] Isso mostra uma redução no gasto, o que contradiz a suposição de que haveria um aumento. Revisando a interpretação inicial para os sinais de mudança, devemos identificar corretamente as porcentagens fornecidas. Notando um possível erro inicial no primeiro passo dos cálculos acima, especialmente devido à Variação de custo. Devemos reitrerar partes corretas do conteúdo original: O preço de uma dúzia de ovos subiu em 10%, portanto: [ text{Novo preço dos ovos} = 1,00 times (1 + 0,10) = 1,00 times 1,10 = R 1,10 ] O preço das maçãs aumentou em 2%, então: [ text{Novo preço das maçãs} = 1,00 times 1,02 = R 1,02 ] Sabe-se que o cálculo correto resulta na seguinte expressão: [ 1,02 + 1,10 = 2.08, e assim a_2 para a nova soma total ] 5. **Verificar a soma das partes e reafirmação da correção**: Portanto, 6. **Conclusão**: [ text{Aumento percentual} = frac{2,08-2,00}{2,00} times 100 = frac{0,08}{2,00} times 100 = 0,04 times 100 = 4 % ] A opção correta é ( boxed{4 %} ).

answer:Step 1: Count the total number of integers in the specified range: from 1 to 19 inclusive, there are 19 integers. So, there are (binom{19}{2} = 171) ways to choose two of them without regard to order. Step 2: Count the ways to select pairs with an odd sum. An odd sum occurs when one integer is odd and the other is even. - Odd integers between 1 and 19: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 (10 odd integers) - Even integers between 1 and 19: 2, 4, 6, 8, 10, 12, 14, 16, 18 (9 even integers) - Choose one odd and one even integer: (10 times 9 = 90) Step 3: Check which of these pairs have a product greater than 100. - The pairs to be checked are: ((1, x), (3, x), (5, x), (7, x), (9, x), (11, x), (13, x), (15, x), (17, x), (19, x)) for all even (x). - Evaluating ((odd, even)) pairs that satisfy (odd times even > 100): - Minimum product pairs: ((5, 21), (7, 15), (9, 12), (11, 10), (13, 8), (15, 7), (17, 6), (19, 6)) - Valid pairs after ensuring they are within the range: ((9, 12), (11, 10), (13, 8) = 3) pairs only (as others exceed the range or do not meet product criterion). So, the number of desirable pairs is 3. The probability is therefore (dfrac{3}{171}). Conclusion: The probability that the sum of the chosen integers is odd and their product is greater than 100 is (boxed{dfrac{1}{57}}).

answer:Given that the triangles are similar in a sequence as stated (PQR sim XYZ and XYZ sim GHI), we utilize transitive property to identify PQR sim GHI. We have the proportion: [ frac{PQ}{HI} = frac{QR}{GH} ] Given: - PQ = 5 - HI = 30 - QR = 15 Calculating GH using transitive similarity: [ frac{5}{30} = frac{15}{GH} Rightarrow GH = frac{15 times 30}{5} = 90 ] Now, we connect this with triangle XYZ: [ frac{QR}{GH} = frac{XY}{YZ} ] Substituting known values: [ frac{15}{90} = frac{XY}{15} Rightarrow XY = frac{15^2}{90} = boxed{2.5} ]