answer:(1) Since f(0) = e^0cos 0 = 1, the point of tangency is (0, 1). The derivative of the function is f'(x) = e^xcos x - e^xsin x = e^x(cos x - sin x). Thus, f'(0) = 1, which means the slope of the tangent line is k = 1. Therefore, the equation of the tangent line is y = x + 1, or in standard form, boxed{x - y + 1 = 0}. (2) Let f'(x) = e^x(cos x - sin x) = 0. Since x in [0, frac{pi}{2}], the only solution is x = frac{pi}{4}. On the interval [0, frac{pi}{4}], f'(x) > 0, so f(x) is increasing. On the interval [frac{pi}{4}, frac{pi}{2}], f'(x) < 0, so f(x) is decreasing. Therefore, the maximum value of f(x) is f(frac{pi}{4}) = e^{frac{pi}{4}}cos{frac{pi}{4}} = boxed{frac{sqrt{2}}{2}e^{frac{pi}{4}}}. Since f(0) = 1 and f(frac{pi}{2}) = 0, the minimum value of f(x) is boxed{0}. Hence, the range of f(x) on [0, frac{pi}{2}] is boxed{[0, frac{sqrt{2}}{2}e^{frac{pi}{4}}]}.

answer:To prove the proposition by contradiction, we start by assuming the opposite of what we want to prove. We want to prove that in a right triangle ABC, at least one of the acute angles angle A and angle B is not greater than 45^{circ}. Therefore, the assumption for contradiction would be that both acute angles are greater than 45^{circ}. 1. Assume angle A > 45^{circ} and angle B > 45^{circ}. 2. In a right triangle, the sum of the angles is 180^{circ}. Since angle C is a right angle (90^{circ}), we have angle A + angle B = 90^{circ}. 3. If both angle A > 45^{circ} and angle B > 45^{circ}, then angle A + angle B > 90^{circ}, which contradicts the fact that angle A + angle B = 90^{circ}. 4. Therefore, our initial assumption must be false, and at least one of the acute angles angle A or angle B must not be greater than 45^{circ}. This logical process shows that to start proving the proposition by contradiction, we should assume both angle A > 45^{circ} and angle B > 45^{circ}. Hence, the correct choice is: boxed{text{A}}.

answer:Firstly, let's translate and address the set M: M = { x mid x^2 geq 1 }. This inequality x^2 geq 1 holds for x geq 1 or x leq -1, because any real number squared that is greater than or equal to one, can be positive or negative. Therefore, M can be redefined as: M = { x mid x geq 1 text{ or } x leq -1 }. Now, let's examine the set N: N = { y mid y = 3x^2 + 1, x in mathbb{R} }. For any real number x, the expression 3x^2 + 1 is always greater than or equal to 1, because 3x^2 is non-negative for all x in mathbb{R}, and adding 1 ensures the expression is at least 1. Hence, N is: N = { y mid y geq 1 }. To find M cap N, we are looking for elements that are common to both M and N. The intersection will be all x values that satisfy both conditions: x geq 1 or x leq -1, and at the same time, make 3x^2 + 1 greater than or equal to 1. Since any x that satisfies x geq 1 or x leq -1 will automatically make 3x^2 + 1 greater than or equal to 1, the intersection is simply the values of x that satisfy the original inequality x^2 geq 1. Therefore, the intersection M cap N is: M cap N = { x mid x geq 1 text{ or } x leq -1 }. Comparing this with the given options, we can see that option D corresponds to the intersection we have found. So the correct answer is: boxed{ text{D. } { x mid x geq 1 text{ or } x leq -1 } }

answer:To find out how many books Gwen had on each shelf, we need to divide the total number of books by the total number of shelves. Gwen has 5 shelves of mystery books and 3 shelves of picture books, which makes a total of 5 + 3 = 8 shelves. She has a total of 32 books. So, to find out how many books are on each shelf, we divide the total number of books by the total number of shelves: 32 books ÷ 8 shelves = 4 books per shelf. Therefore, Gwen had boxed{4} books on each shelf.