answer:To negate the given proposition "forall x gt 0, 2^{x} gt 0", we follow the rules of logical negations. The negation of a universal quantifier (forall) is an existential quantifier (exists), and the negation of the inequality > is leqslant. Therefore, the negation of the given proposition is: - Original Proposition: forall x gt 0, 2^{x} gt 0 - Negation: exists x gt 0, 2^{x} leqslant 0 This matches with option A: exists x gt 0, 2^{x} leqslant 0. Thus, the correct answer is boxed{text{A}}.

answer:First, we use the identity cos^2 frac{theta}{2} = frac{1 + cos theta}{2} and simplify the expression: [ cos^2 frac{theta}{2} cdot (1 - sin theta) = frac{1 + cos theta}{2} cdot (1 - sin theta). ] Then, substitute sin theta = 2 sin frac{theta}{2} cos frac{theta}{2}: [ 1 - sin theta = 1 - 2 sin frac{theta}{2} cos frac{theta}{2}. ] Thus the original expression becomes: [ f(theta) = frac{1 + cos theta}{2} cdot left(1 - 2 sin frac{theta}{2} cos frac{theta}{2}right) = frac{(1 + cos theta)(1 - sin theta)}{2}. ] We have that 1 + cos theta = 2 cos^2 frac{theta}{2} and 1 - sin theta = 1 - 2sin frac{theta}{2} cos frac{theta}{2}, so [ f(theta) = cos^2 frac{theta}{2} cdot (1 - 2sin frac{theta}{2} cos frac{theta}{2}). ] Now maximize f(theta) from 0 to frac{pi}{2}: The function simplifies further since sin theta = 2 sin frac{theta}{2} cos frac{theta}{2}, [ f(theta) = cos^2 frac{theta}{2} (1 - sin theta) = cos^2 frac{theta}{2} left(1 - 2sin frac{theta}{2} cos frac{theta}{2}right). ] Solving for theta = 0 (as sin theta and cos theta values are straightforward here), [ f(0) = cos^2 0 cdot (1 - 0) = 1. ] As cos^2 frac{theta}{2} and the term 1 - 2sin frac{theta}{2} cos frac{theta}{2} are always non-negative and decrease with increasing theta within [0, frac{pi}{2}], the expression attains its maximum at theta = 0 yielding boxed{1}.

answer:Let's break down the problem and solution step-by-step. **Problem Statement:** Identify all triplets of natural numbers ((a, b, c)) such that the product of any two numbers in the triplet, increased by 1, is divisible by twice the third number. **Given:** For triplet ((a, b, c)): 1. ((a cdot b) + 1) is divisible by (2c) 2. ((b cdot c) + 1) is divisible by (2a) 3. ((c cdot a) + 1) is divisible by (2b) **Solution:** Let's denote the three natural numbers as (a), (b), and (c). We want to find (a, b, c in mathbb{N}) (natural numbers) that satisfy the following three conditions: 1. ((a cdot b) + 1 equiv 0 pmod{2c}) 2. ((b cdot c) + 1 equiv 0 pmod{2a}) 3. ((c cdot a) + 1 equiv 0 pmod{2b}) This can also be written as: 1. ((a cdot b) + 1 = k cdot 2c) for some integer (k) 2. ((b cdot c) + 1 = m cdot 2a) for some integer (m) 3. ((c cdot a) + 1 = n cdot 2b) for some integer (n) We are given the hint that all three numbers (a, b, c) should be mutually prime and odd. This hints at the compactness and simplicity of their relationship. We start by checking small odd numbers: **Step-by-step Calculation:** 1. Let's first assume (a = b = c = 1): - For (a = b = c = 1): - ( (1 cdot 1) + 1 = 2 ) is divisible by (2 cdot 1 = 2). - ( (1 cdot 1) + 1 = 2 ) is divisible by (2 cdot 1 = 2). - ( (1 cdot 1) + 1 = 2 ) is divisible by (2 cdot 1 = 2). All conditions are satisfied. 2. Can there be any other triplets? - If any of (a), (b), or (c) were greater than 1, say (a > 1), (b > 1) or (c > 1), then ((a cdot b) + 1) would not necessarily yield divisibility by (2c) without additional constraints. Conclusively, the only natural numbers that satisfy such compact mutually prime and additional divisibility constraints are (a = 1), (b = 1), and (c = 1). # Conclusion: Hence, the only valid triplet of natural numbers that satisfy the given conditions is ((1, 1, 1)). [ boxed{(1, 1, 1)} ]

answer:Let's denote the radian measure of the sector's central angle as alpha. The area of the sector can be calculated as S = frac{1}{2}alpha r^2 = frac{1}{2}alpha times 1^2 = 1. Solving for alpha, we find that alpha = 2. Therefore, the answer is boxed{2}.