answer:To evaluate each option step-by-step: **Option A:** We apply the rule of exponents that states when multiplying powers with the same base, we add the exponents. Thus, [a^{4}cdot a^{3} = a^{4+3} = a^{7}] This matches the operation given in option A, making it correct. **Option B:** Using the rule that states when raising a power to another power, we multiply the exponents, we get: [(a^{2})^{3} = a^{2cdot3} = a^{6}] The operation given in option B suggests (a^{2})^{3} = a^{5}, which is incorrect based on our calculation. Therefore, option B is incorrect. **Option C:** Subtracting like terms, we have: [3a^{2} - a^{2} = (3-1)a^{2} = 2a^{2}] The operation in option C suggests 3a^{2} - a^{2} = 2, which is incorrect because the result is 2a^{2}, not 2. Therefore, option C is incorrect. **Option D:** Expanding the square of a binomial, we have: [(a-b)^{2} = (a-b)(a-b) = a^{2} - 2ab + b^{2}] The operation in option D suggests (a-b)^{2} = a^{2} - b^{2}, which is incorrect because the correct expansion includes the term -2ab. Therefore, option D is incorrect. Given the evaluations above, the only correct operation is in: [boxed{A}]

answer:For the triangle formed by three adjacent extremal points on the graph of the function y=Asinomega x (omega>0) to be a right-angled triangle with an area of 1, the triangle must be an isosceles right-angled triangle. Therefore, we have frac{1}{2} times 4A times 2A = 1, which implies A = frac{1}{2}. Furthermore, since the period is 4A=2, we have frac{2pi}{omega} = 2, which leads to omega = pi. Therefore, the correct choice is: boxed{text{C}}. This problem examines the graph and properties of the sine function and tests the students' computational skills. It is considered a medium-difficulty question.

answer:(Ⅰ) To prove this, we have the following since a_n and a_{n+1} are the roots of the equation x^2 - 2^n x + b_n = 0: begin{cases} a_n + a_{n+1} = 2^n b_n = a_n cdot a_{n+1} end{cases} Since frac{a_{n+1} - frac{1}{3} cdot 2^{n+1}}{a_n - frac{1}{3} cdot 2^n} = frac{2^n - a_n - frac{1}{3} cdot 2^{n+1}}{a_n - frac{1}{3} cdot 2^n} = frac{- left(a_n - frac{1}{3} cdot 2^nright)}{a_n - frac{1}{3} cdot 2^n} = -1 Thus, the sequence left{a_n - frac{1}{3} cdot 2^nright} is a geometric progression with the first term a_1 - frac{2}{3} = frac{1}{3} and common ratio -1. (Ⅱ) From the previous result, we have a_n - frac{1}{3}cdot 2^n = frac{1}{3}cdot (-1)^{n-1} which yields a_n = frac{1}{3} [2^n - (-1)^n] Therefore, S_n = sum_{k=1}^{n} a_k = frac{1}{3}(2 + 2^2 + 2^3 + cdots + 2^n) - frac{1}{3} [(-1) + (-1)^2 + cdots + (-1)^n] which simplifies to S_n = frac{1}{3} left[2^{n+1} - 2 - frac{(-1)^n - 1}{2}right] As a result, b_n = a_n cdot a_{n+1} = frac{1}{9} [2^n - (-1)^n] cdot [2^{n+1} - (-1)^{n+1}] = frac{1}{9} [2^{2n+1} - (-2)^n - 1] In order for b_n > lambda S_n to hold for all n in mathbb{N}^*, we need frac{1}{9}[2^{2n+1} - (-2)^n - 1] - frac{lambda}{3} left[2^{n+1} - 2 - frac{(-1)^n - 1}{2}right] > 0 For odd n, this becomes frac{1}{9}(2^{n+1} - 1)(2^n + 1) - frac{lambda}{3}(2^{n+1} - 1) > 0 Since 2^{n+1} - 1 > 0, we have lambda < frac{1}{3}(2^n + 1) for all odd n, with the minimum value of frac{1}{3}(2^n + 1) for odd n being 1. Hence lambda < 1. For even n, the inequality becomes frac{1}{9}(2^{n+1} + 1)(2^n - 1) - frac{2lambda}{3}(2^n - 1) > 0 Since 2^n - 1 > 0, we have lambda < frac{1}{6}(2^{n+1} + 1) for all even n, with the minimum value of frac{1}{6}(2^{n+1} + 1) for even n being frac{3}{2}. Therefore lambda < frac{3}{2}. However, we must satisfy both conditions simultaneously, so the appropriate range for lambda is the intersection of the two ranges, which is (-infty, 1). Therefore, there exists a constant lambda such that b_n > lambda S_n for all n in mathbb{N}^* when lambda in (-infty, 1), and the final answer is boxed{lambda in (-infty, 1)}

answer:To solve this problem, we start by expressing the coordinates of point A on the line l: y=2x as A(a,2a), where a>0 because A is in the first quadrant. Given that B is the point (5,0), the midpoint C of the diameter AB would have coordinates Cleft(frac{a+5}{2},aright) because the midpoint is calculated as the average of the x and y coordinates of A and B. The equation of the circle with diameter AB can be derived using the fact that a point P(x,y) lies on a circle if and only if it satisfies the equation (x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0, where (x_1,y_1) and (x_2,y_2) are the endpoints of the diameter. Substituting the coordinates of A and B into this formula, we get the equation of circle C as (x-5)(x-a)+y(y-2a)=0. To find the intersection of the circle with the line l: y=2x, we substitute y=2x into the circle's equation, leading to a system of equations: [ left{ begin{array}{l} y=2x (x-5)(x-a)+y(y-2a)=0 end{array} right. ] Solving this system gives us the coordinates of point D as D(1,2). Next, we calculate the vectors overrightarrow{AB} and overrightarrow{CD}. The vector overrightarrow{AB} is (5-a,-2a), obtained by subtracting the coordinates of A from B. Similarly, overrightarrow{CD} is calculated as (-frac{a+3}{2},2-a), by subtracting the coordinates of D from C. Given that overrightarrow{AB} cdot overrightarrow{CD} = 0, we set up the dot product equation: [ (5-a) times left(-frac{a+3}{2}right) + (-2a) times (2-a) = 0 ] Expanding and simplifying this equation gives us: [ frac{1}{2}(a^{2}-2a-15) + 2a^{2} - 4a = 0 ] Solving this quadratic equation for a, we find that a=3 or a=-1. However, since a>0 (because A is in the first quadrant), we conclude that a=3. This means the abscissa of point A is 3. Therefore, the correct answer is boxed{A}.