answer:Since line l intersects with plane alpha and line l is not perpendicular to plane alpha, we know:- l is perpendicular to a set of parallel lines within alpha, hence option A is incorrect;- There are infinitely many lines within alpha that are perpendicular to l, and these lines are parallel to each other, hence option B is incorrect;- There are no lines within alpha that are parallel to l, hence option C is incorrect;- There are infinitely many lines within alpha that are skew to l, hence option D is correct.Therefore, the correct answer is boxed{text{D}}.

answer:See task 110964.## Answer1352 / 15Send comment

answer:Let's rewrite the solution step by step, following the given instructions: Part (1)Given a=1, the function becomes f(x) = x^2e^{1-x} - (x-1). We are interested in finding the extreme values of f(x) in the interval (frac{3}{4}, 2).1. First Derivative: To find the extreme values, we calculate the first derivative of f(x): [ f'(x) = (2x - x^2)e^{1-x} - 1 = frac{(2x - x^2) - e^{x-1}}{e^{x-1}}. ]2. Define h(x): Let h(x) = (2x - x^2) - e^{x-1}, with its domain being mathbb{R}.3. Derivative of h(x): The derivative of h(x) is h'(x) = 2 - 2x - e^{x-1}. It's clear that h'(x) is decreasing on (frac{3}{4}, 2).4. Sign of h(x): Since h(frac{3}{4}) = frac{1}{2} - frac{1}{sqrt[4]{e}} 0 for x in (frac{3}{4}, 1) and f'(x) 0, so a > -1.3. Sum of Roots: Since x_1 + x_2 = 2 and x_1 < x_2, we have x_1 < 1.4. Given Inequality: We have the inequality x_1g(x_1) leq lambda f'(x_1).5. Substituting f'(x): After substituting and simplifying, we get ({2-x_1})({2x_1}){e^{1-x}}leq lambda[{({2x_1-x_1^2}){e^{1-x_{1}}}+({2x_1-x_1^2})}].6. Solving for lambda: The inequality simplifies to {x_1}[{2{e^{1-x_1}}-lambda({{e^{1-x_1}}+1})}]leq 0. For x_1 in (0, 1), we find that lambda geq frac{2e}{e+1}.7. Monotonicity of k(x): The function k(x) = frac{2{e^{1-x}}}{{e^{1-x}+1}} is monotonically decreasing on mathbb{R}. Thus, for x_1 in (0, 1), lambda geq frac{2e}{e+1}.Therefore, the value of lambda is boxed{frac{2e}{e+1}}.

answer:To solve: int _{ 0 }^{ 2 }f(x)dx equals int _{ 0 }^{ 1 }f(x)dx+ int _{ 1 }^{ 2 }f(x)dx = int _{ 0 }^{ 1 }x^{2}dx+ int _{ 1 }^{ 2 }(2-x)dx = frac {1}{3}x^{3}|_{0}^{1}+(2x- frac {1}{2}x^{2})|_{1}^{2} = (frac {1}{3}-0)+(2- frac {3}{2}) = frac {5}{6} Therefore, the answer is: boxed{frac {5}{6}}

answer:To find the greatest common divisor (GCD) of 840 and 1,764, we can use the Euclidean algorithm, which involves a series of division steps.Let's denote the larger number 1,764 by a and the smaller number 840 by b.1. Divide a by b and find the remainder: a = 1,764, quad b = 840 1,764 div 840 = 2 text{ remainder } 84 Rightarrow text{remainder } r_1 = 1,764 - 2 cdot 840 = 84 2. Replace a with b and b with r_1, and repeat: a = 840, quad b = 84 840 div 84 = 10 text{ remainder } 0 Since the remainder is 0, we have found that 84 is a divisor of 840.We could continue the algorithm, but since we've reached a remainder of 0, we stop here and conclude that 84 is the GCD of 840 and 1,764. This is because 84 is the non-zero remainder that divided the previous divisor (840) exactly.Hence, the greatest common divisor (GCD) of 840 and 1,764 is boxed{84}.

answer:Notice,begin{aligned}n^{2} & =left(m^{2}-4right)left(m^{2}+12 m+32right)+4 & =(m+2)(m-2)(m+4)(m+8)+4 & =left(m^{2}+6 m+8right)left(m^{2}+6 m-16right)+4 .end{aligned}Let t=m^{2}+6 m-4.Then n^{2}=(t+12)(t-12)+4, which means(t+n)(t-n)=140 text {. }Assume n is a non-negative integer. Clearly, t+n and t-n have the same parity.Since t+n geqslant t-n, we havet+n=-2,-10,14,70 text {. }Correspondingly, t-n=-70,-14,10,2.Thus, (t, n)=(-36,34),(-12,2),(12,2),(36,34) text {. }Therefore, m^{2}+6 m-4= pm 36, pm 12.Solving these, we get m=4,-10,-8,2,-2,-4.Hence, there are 6 points P that satisfy the conditions.