answer:Given the problem: [ text{Given } frac{x^{2}}{1+x^{2}}+frac{y^{2}}{1+y^{2}}+frac{z^{2}}{1+z^{2}}=2. ] We need to prove that: [ frac{x}{1+x^{2}}+frac{y}{1+y^{2}}+frac{z}{1+z^{2}} leq sqrt{2}. ] We start with the given equation: # Proof 1: 1. By subtracting from 1 inside the fractions, we can write: [ frac{x^{2}}{1+x^{2}}+frac{1}{1+x^{2}} = 1, ] [ frac{y^{2}}{1+y^{2}}+frac{1}{1+y^{2}} = 1, ] [ frac{z^{2}}{1+z^{2}}+frac{1}{1+z^{2}} = 1, ] 2. Summing these: [ frac{x^{2}}{1+x^{2}} + frac{y^{2}}{1+y^{2}} + frac{z^{2}}{1+z^{2}} + frac{1}{1+x^{2}} + frac{1}{1+y^{2}} + frac{1}{1+z^{2}} = 1 + 1 + 1. ] Which simplifies to: [ 2 + left(frac{1}{1+x^{2}} + frac{1}{1+y^{2}} + frac{1}{1+z^{2}}right) = 2. ] Hence, we obtain: [ frac{1}{1+x^{2}} + frac{1}{1+y^{2}} + frac{1}{1+z^{2}} = 1. ] 3. Let ( k ) be a constant and add the constant ( k ) into the inequality: [ 2 + k = frac{x^{2}}{1+x^{2}} + frac{y^{2}}{1+y^{2}} + frac{z^{2}}{1+z^{2}} + k left(frac{1}{1+x^{2}} + frac{1}{1+y^{2}} + frac{1}{1+z^{2}}right). ] This can be rewritten as: [ 2 + k = left(frac{x^{2}}{1+x^{2}} + frac{k}{1+x^{2}}right) + left(frac{y^{2}}{1+y^{2}} + frac{k}{1+y^{2}}right) + left(frac{z^{2}}{1+z^{2}} + frac{k}{1+z^{2}}right). ] 4. Using the Cauchy-Schwarz inequality on the RHS: [ 2 + k geq 2sqrt{k} left(frac{x}{1+x^{2}} + frac{y}{1+y^{2}} + frac{z}{1+z^{2}}right). ] 5. The equality condition of Cauchy-Schwarz implies ( x^2 = y^2 = z^2 = k ). By substituting ( x^2 = y^2 = z^2 = k, ): 6. The value of ( k ) simplifies to: [ k = 2. ] 7. Therefore, we have: [ frac{x}{1+x^{2}} + frac{y}{1+y^{2}} + frac{z}{1+z^{2}} leq frac{k + 2}{2sqrt{k}} = sqrt{2}. ] # Conclusion: [ boxed{frac{x}{1+x^{2}} + frac{y}{1+y^{2}} + frac{z}{1+z^{2}} leq sqrt{2}} ] This completes the proof.

answer:Let's assume that A initially invests Rs. x. Since A doubles his capital after 6 months, A's investment for the first 6 months is Rs. x and for the next 6 months is Rs. 2x. B invests Rs. 4500 for the entire year. To find out how much A initially invested, we need to equate A's and B's investment over the year since they are supposed to divide the profit in a 1:1 ratio. A's investment for the year can be calculated as the average of his investment over the 12 months, which is (x for the first 6 months + 2x for the next 6 months) / 2. A's average investment for the year = (x + 2x) / 2 = (3x) / 2 Since B's investment is constant throughout the year, B's average investment for the year is Rs. 4500. According to the problem, A's and B's investments should be equal for them to share the profit in a 1:1 ratio. So, we can set up the equation: (3x) / 2 = 4500 Now, we solve for x: 3x = 4500 * 2 3x = 9000 x = 9000 / 3 x = 3000 Therefore, A initially invested Rs. boxed{3000} .

answer:To solve for the range of values for omega given that the function f(x) = cos omega x - 1 has exactly 3 zeros in the interval [0, 2pi], we proceed as follows: 1. **Identify the Condition for Zeros**: For f(x) = cos omega x - 1 to have zeros, we need cos omega x = 1. This is because the equation becomes 0 when cos omega x = 1. 2. **Period of the Function**: The period of the cosine function, cos omega x, is frac{2pi}{omega}. This means that the function repeats its values every frac{2pi}{omega} units along the x-axis. 3. **Determine the Number of Periods in [0, 2pi]**: To have exactly 3 zeros within the interval [0, 2pi], the interval must contain exactly two full periods of the function. This is because the cosine function has one zero per half period when shifted down by 1. Therefore, we set up the inequality to find the range of omega that satisfies this condition: [ 2 cdot frac{2pi}{omega} leq 2pi < 3 cdot frac{2pi}{omega} ] Simplifying this inequality: [ 2 leq omega < 3 ] 4. **Conclusion**: The range of values for omega that ensures the function f(x) = cos omega x - 1 has exactly 3 zeros in the interval [0, 2pi] is [2, 3). Therefore, the final answer is boxed{[2, 3)}.

answer:Since a < 0, then 1-a > 1, 1+a < 1, thus f(1-a)geqslant f(1+a) is equivalent to -(1-a)geqslant (1+a)^{2}+2a, which means a^{2}+3a+2leqslant 0, we get -2leqslant aleqslant -1, hence, the range of the real number a is boxed{[-2,-1]}, so the correct choice is: boxed{text{B}} By determining the range of 1-a and 1+a based on the condition, and combining the expression of the piecewise function for transformation and solution, we can solve the problem. This question mainly examines the solution of inequalities. Determining the range of variables 1-a and 1+a based on the expression of the piecewise function is the key to solving this problem.