answer:To analyze the function f(x)=aln x+frac{b}{x}+frac{c}{{x}^{2}} where aneq 0 and determine the conditions under which it has both a maximum and a minimum value, we follow these steps: 1. **Domain of the Function**: The function is defined for x > 0, so its domain is left(0,+infty right). 2. **First Derivative**: To find the critical points, we calculate the first derivative of f(x): [ {f'}left(xright)=frac{a}{x}-frac{b}{{x}^{2}}-frac{2c}{{x}^{3}}=frac{a{x}^{2}-bx-2c}{{x}^{3}} ] 3. **Critical Points**: The critical points occur when {f'}left(xright)=0, which simplifies to the quadratic equation ax^{2}-bx-2c=0. For the function to have both a maximum and a minimum, this equation must have two positive roots, denoted as x_{1} and x_{2}. 4. **Sum and Product of Roots**: From the quadratic equation, we can express the sum and product of the roots as: [ x_{1}+x_{2}=frac{b}{a} quad text{and} quad x_{1}x_{2}=frac{-2c}{a} ] For both roots to be positive, we require x_{1}+x_{2} > 0 and x_{1}x_{2} > 0, leading to frac{b}{a} > 0 and frac{-2c}{a} > 0. 5. **Discriminant**: The discriminant of the quadratic equation, Delta =b^{2}+8ac, must be greater than 0 for the equation to have two distinct real roots. Therefore, we have Delta =b^{2}+8ac > 0. 6. **Analysis of Conditions**: - From frac{b}{a} > 0, we conclude that ab > 0. - From frac{-2c}{a} > 0, we conclude that ac < 0. - Combining these, we get abcdot ac=a^{2}bc < 0, which implies bc < 0. Therefore, the correct statements that must be true for the function to have both a maximum and a minimum value are: - ab > 0 (B) - ac < 0 (D) - Delta = b^{2}+8ac > 0 (C) Hence, the correct answer is boxed{BCD}.

answer:**Step 1: Understanding the problem** We are given a line (7x + 11y = 77) and we need to determine the number of lattice points within the triangle formed by this line and the coordinate axes. A **lattice point** is a point ((x, y)) where both (x) and (y) are integers. **Step 2: Finding intercepts** Let's find the intercepts of the given line on the coordinate axes. For the x-intercept: [ 7x + 11 cdot 0 = 77 implies x = frac{77}{7} = 11 ] So the x-intercept is ((11, 0)). For the y-intercept: [ 7 cdot 0 + 11y = 77 implies y = frac{77}{11} = 7 ] So the y-intercept is ((0, 7)). **Step 3: Counting interior lattice points using Pick's Theorem** Pick's Theorem states that for any simple polygon with vertices on lattice points, the area (A) of the polygon is given by: [ A = I + frac{B}{2} - 1 ] where (I) is the number of interior lattice points and (B) is the number of boundary lattice points. First, we calculate the area of the triangle with vertices at ((0,0)), ((11,0)), and ((0,7)): [ A = frac{1}{2} times 11 times 7 = 38.5 ] **Step 4: Counting boundary lattice points** We need to count the number of lattice points on the boundary of the triangle: 1. Points on the x-axis from ((0,0)) to ((11,0)): - There are 12 points (including both endpoints). 2. Points on the y-axis from ((0,0)) to ((0,7)): - There are 8 points (including both endpoints). 3. Points on the line segment ((11,0)) to ((0,7)) on (7x + 11y = 77): To find the lattice points on this line segment, we note: [ (x, y) = left(frac{77 - 11y}{7}, yright) ] (y) goes from 0 to 7, but not all will yield integers for (x). Checking each (y) from 0 to 7: - ((11, 0)) - ((4, 3)) - ((0, 7)) So there are 3 points. Sum of all boundary points: [ 12 + 8 + 3 - 3 = 20 quad text{(subtract 3 to avoid double-counting the vertices)} ] Thus, (B = 20). **Step 5: Applying Pick's Theorem** Using Pick's Theorem: [ 38.5 = I + frac{20}{2} - 1 implies 38.5 = I + 10 - 1 implies 38.5 = I + 9 ] Solve for (I): [ I = 38.5 - 9 = 29.5 ] **Step 6: Concluding Number of Lattice Points** Including boundary lattice points, the total number of lattice points: [ I + B = 29 + 20 ] [ I + B = 49 ] Therefore, the total number of lattice points is: [ boxed{49} ]

answer:1. **Geometric Configuration**: - AB is the diameter of a circle with center O. - E is the diametrically opposite point to A on the circle, making AE the another diameter. - CD is a chord such that angle CDE = 60^circ. 2. **Triangle Configuration**: - Triangle CDE is formed where angle CDE = 60^circ. By properties of the circle, angle COE = 180^circ - angle CDE = 120^circ. - With C moving over the semicircle, angle COE remains as 120^circ. 3. **Angle Bisector Intersection & Circle Behavior**: - Let P be the intersection of the bisector of angle OCD with the circle. - Since angle CDE = 60^circ, irrespective of C's position, the bisector of angle OCD continually bisects the smaller arc ED. - Due to the constant angle, as C traces the semicircle, the nature of angle OCD changes, thus influencing the bisector to variably intersect the circle. 4. **Conclusion**: - The position where the bisector of angle OCD cuts the circle varies as C moves over the semicircle, influenced by the continually changing angular relationship within triangle OCD. text{varies} The final answer is boxed{textbf{(C) Varies}}

answer:First, we find the divisors of each number: - Divisors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60 - Divisors of 18: 1, 2, 3, 6, 9, 18 Now, we find the common divisors from both sets: Common divisors of 60 and 18 are 1, 2, 3, 6. Sum the common divisors: [ 1 + 2 + 3 + 6 = boxed{12} ]