answer:To calculate the cost of the alternative plan for Darnell and compare it with his current plan, we proceed as follows: 1. **Calculate the cost for texting on the alternative plan:** - Darnell sends 60 texts per month. - The alternative plan charges 1 per 30 texts. - Therefore, the cost for texting is calculated as 1 times frac{60}{30} = 1 times 2 = 2. 2. **Calculate the cost for calling on the alternative plan:** - Darnell spends 60 minutes on the phone per month. - The alternative plan charges 3 per 20 minutes. - Therefore, the cost for calling is calculated as 3 times frac{60}{20} = 3 times 3 = 9. 3. **Calculate the total cost on the alternative plan:** - The total cost is the sum of the texting and calling costs: 2 + 9 = 11. 4. **Compare the cost with his current plan:** - Darnell currently pays 12 per month. - On the alternative plan, he would pay 11 per month. - Therefore, Darnell would pay 12 - 11 = 1 less on the alternative plan. Hence, Darnell would pay boxed{1} less on the alternative plan.

answer:If A={-1,0,1} and set B has only two elements, then B={1,2}; (|x|+1=1), x=0, (|x|=2), x=±1, therefore A={0}, {1}, {-1}, {0,1}, {0,-1}, {1,-1}, {0,1,-1}, a total of 7. So the answer is boxed{{1,2}}, boxed{7}. Directly according to the definition of the mapping, the conclusion can be drawn. This question tests the definition of mapping and the student's calculation ability, which is relatively basic.

answer:This problem involves comprehensive application of functions, operation of exponential functions, properties of absolute value functions, and the concept of classification and discussion. It is a moderate-difficulty problem. According to the given condition, the function y=|f(x)| is monotonically increasing. We can solve for the range of the real number a by discussing the properties of the function and the absolute value function. Solution steps: 1. For any x_1, x_2in[1,2], and x_1 < x_2, according to the given condition [|f(x_1)|-|f(x_2)|](x_1-x_2) > 0, the function y=|f(x)| is monotonically increasing. 2. When ageqslant 0, f(x) is an increasing function on [1,2], then f(1)geqslant 0, solving for a yields: 0leqslant aleqslant frac{e^{2}}{2}. 3. When a < 0, |f(x)|=f(x), let frac{e^{x}}{2}=- frac{a}{e^{x}}, solving for x yields: x=ln ;sqrt{-2a}. 4. Since the monotonically increasing interval of the absolute value function is [ln ;sqrt{-2a},+infty), therefore, ln ;sqrt{-2a}leqslant 1, solving for a yields: -;frac{e^{2}}{2}leqslant a < 0. 5. Combining the above results, the range of a is left[-;frac{e^{2}}{2}, frac{e^{2}}{2}right]. Therefore, the correct answer is option B: boxed{left[left.- frac{e^{2}}{2}, frac{e^{2}}{2} right.right]}.

answer:To compute A(2, 1), we follow the definition of the Ackermann function A(m, n) step by step, using the rules provided: 1. Starting with A(2, 1), we apply rule (3) since m > 0 and n > 0, which gives us: [A(2, 1) = A(1, A(2, 0))] 2. Next, to find A(2, 0), we apply rule (2) because m > 0 and n = 0, resulting in: [A(2, 0) = A(1, 1)] Thus, A(2, 1) = A(1, A(1, 1)). 3. To compute A(1, 1), we use rule (3), which requires us to find A(1, 0): [A(1, 1) = A(0, A(1, 0))] 4. For A(1, 0), applying rule (2) gives us: [A(1, 0) = A(0, 1)] Therefore, A(1, 1) = A(0, A(0, 1)). 5. Now, we need to compute A(0, 1). Using rule (1), we get: [A(0, 1) = 2] So, A(1, 1) = A(0, 2). 6. Applying rule (1) again for A(0, 2), we find: [A(0, 2) = 3] Thus, A(1, 1) = 3, and A(2, 1) = A(1, 3). 7. To compute A(1, 3), we apply rule (3), which requires us to find A(1, 2): [A(1, 3) = A(0, A(1, 2))] 8. For A(1, 2), applying rule (3) again, we need A(1, 1), which we already found to be 3. So: [A(1, 2) = A(0, A(1, 1)) = A(0, 3)] 9. Finally, applying rule (1) for A(0, 3), we get: [A(0, 3) = 4] Therefore, A(2, 1) = A(0, 4). 10. Applying rule (1) one last time for A(0, 4), we find: [A(0, 4) = 5] Thus, the final answer is boxed{5}.