answer:# Solution: Part (1) - Monotonicity of f(x) Given f'(x) = e^{x} + a, we analyze the monotonicity based on the value of a: - **When a geq 0**: f'(x) = e^{x} + a > 0 quad text{for all } x in mathbb{R} This implies that f(x) is monotonically increasing on mathbb{R}. - **When a < 0**: Setting f'(x) = 0 gives us x = ln(-a). Therefore, we have: - On (-infty, ln(-a)), f'(x) < 0 and f(x) is monotonically decreasing. - On (ln(-a), +infty), f'(x) > 0 and f(x) is monotonically increasing. In conclusion, f(x) is monotonically increasing on mathbb{R} for a geq 0, and for a < 0, f(x) is monotonically decreasing on (-infty, ln(-a)) and monotonically increasing on (ln(-a), +infty). Part (2) - Range of b Given f'(1) = 0, we find: e + a = 0 implies a = -e Substituting a = -e into the inequality f(x) geq bx - 1, we get: e^{x} - ex - 1 geq bx - 1 This simplifies to: e^{x} geq (b + e)x quad text{for all } x in (0, +infty) Let g(x) = frac{e^{x}}{x} for x > 0, and find its derivative: g'(x) = frac{(x-1)e^{x}}{x^{2}} Setting g'(x) = 0 gives x = 1. Analyzing g(x): - On (0,1), g'(x) < 0 and g(x) is monotonically decreasing. - On (1,+infty), g'(x) > 0 and g(x) is monotonically increasing. The minimum value of g(x) is g(1) = e, implying: b + e leq e implies b leq 0 Therefore, the range of b is boxed{(-infty, 0]}. Part (3) - Range of a - **When a geq 0**: f(x) is monotonically increasing on mathbb{R}, so it has at most 1 zero. - **When a < 0**: For f(x) to have two distinct zeros, we require f(ln(-a)) < 0, leading to: -a + aln(-a) - 1 < 0 Let u(a) = -a + aln(-a) - 1 for a < 0. Finding its derivative: u'(a) = ln(-a) Setting u'(a) = 0 gives a = -1. Analyzing u(a): - On (-infty, -1), u'(a) > 0 and u(a) is monotonically increasing. - On (-1, 0), u'(a) < 0 and u(a) is monotonically decreasing. The maximum value of u(a) is u(-1) = 0, implying a < 0. Therefore, the range of a is boxed{(-infty, 0)}.

answer:The probability of an egg weighing not less than 30 grams includes the probabilities of it weighing exactly 30 grams up to any weight above 30 grams. Since the total probability must sum up to 1, and we know the probability of it weighing less than 30 grams is 0.3, we can calculate the probability of it not weighing less than 30 grams by subtracting from 1. So, the probability of an egg weighing not less than 30 grams is 1 - 0.3 = 0.7. Therefore, the correct answer is boxed{D}.

answer:First, let's handle the pairs with books of different genres: - There are three genres, and I can choose any two genres in binom{3}{2} = 3 ways. - For each picked genre, I have 4 books, so I can choose one book from each of the two chosen genres in 4 times 4 = 16 ways per genre pair. This gives us 3 times 16 = 48 combinations of books from different genres. Next, let's handle the pairs within the mystery genre: - I can select any two out of the four mystery books in binom{4}{2} = 6 ways. Adding the two parts together gives a total of 48 + 6 = 54 possible pairs. boxed{54}

answer:To determine the range of values for (m) that satisfies the inequality [ cos^2 theta + 2m sin theta - 2m - 2 < 0 ] for (0 leqslant theta leqslant frac{pi}{2}), we start by analyzing and transforming the inequality. **Step 1**: Consider the boundary condition ( theta = frac{pi}{2} ). - At ( theta = frac{pi}{2} ), we have ( cos theta = 0 ) and ( sin theta = 1 ). - Substituting these values into the inequality, we get: [ cos^2 left( frac{pi}{2} right) + 2m sin left( frac{pi}{2} right) - 2m - 2 < 0 ] [ 0 + 2m cdot 1 - 2m - 2 < 0 ] [ -2 < 0 ] - This inequality holds true for any (m), so the next step is to consider ( theta ) in the interval (left[0, frac{pi}{2}right)). **Step 2**: Substitute ( sin theta ) with ( t ) where ( t in [0,1) ) and note that ( cos^2 theta = 1 - sin^2 theta = 1 - t^2 ). - Transform the inequality: [ 1 - t^2 + 2m t - 2m - 2 < 0 ] **Step 3**: Rewrite the transformed inequality: [ 1 - t^2 + 2mt - 2m - 2 < 0 ] [ Rightarrow 2m t - 2m < t^2 - 1 + 2 ] [ Rightarrow 2m(t - 1) < t^2 - 1 ] [ Rightarrow m < frac{t^2 - 1}{2(t - 1)} ] **Step 4**: Simplify the right-hand side: [ m < frac{t^2 - 1}{2(t - 1)} = -frac{(1 - t^2)}{2(1 - t)} = -frac{(1 + t)(1 - t)}{2(1 - t)} ] [ m < -frac{1 + t}{2} ] **Step 5**: Deduce the minimum value of the function ( f(x) = x + frac{2}{x} ) in the domain (0 < 1-t leq 1). - We already know that for ( t in [0,1) ), this function is decreasing and the minimum value occurs when ( 1-t = 1 ). [ 1 + frac{2}{1} = 3 ] Therefore, ( (1 - t) + frac{2}{1 - t} geq 3 ). **Step 6**: Substitute the inequality back into the expression for ( m ): [ m > -frac{1}{2}(3) + 1 ] [ m > -frac{3}{2} + 1 ] [ m > -frac{1}{2} ] # Conclusion: The value of ( m ) must be greater than ( -frac{1}{2} ). Thus, the range of ( m ) is: [ boxed{m > -frac{1}{2}} ]