💡Okay, so I need to find the maximum value of the function f(x) = x³ - 3x on the interval [-3, 3]. Alright, let's see. I remember that to find the maximum or minimum values of a function on a closed interval, I should check both the critical points inside the interval and the endpoints.First, I need to find the critical points. Critical points occur where the derivative is zero or undefined. Since f(x) is a polynomial, its derivative will exist everywhere, so I just need to find where the derivative is zero.Let me compute the derivative of f(x). The derivative of x³ is 3x², and the derivative of -3x is -3. So, f'(x) = 3x² - 3. Now, I need to set this equal to zero and solve for x.So, 3x² - 3 = 0. Let's factor out the 3: 3(x² - 1) = 0. That simplifies to x² - 1 = 0. Solving for x, I get x² = 1, so x = ±1. Therefore, the critical points are at x = 1 and x = -1.Now, I need to evaluate the function f(x) at these critical points and also at the endpoints of the interval, which are x = -3 and x = 3. The maximum value will be the largest of these values.Let's start with x = -3. Plugging into f(x): f(-3) = (-3)³ - 3*(-3) = -27 + 9 = -18.Next, x = -1. f(-1) = (-1)³ - 3*(-1) = -1 + 3 = 2.Then, x = 1. f(1) = (1)³ - 3*(1) = 1 - 3 = -2.Finally, x = 3. f(3) = (3)³ - 3*(3) = 27 - 9 = 18.So, the function values at the critical points and endpoints are: f(-3) = -18, f(-1) = 2, f(1) = -2, and f(3) = 18.Looking at these values, the largest one is 18 at x = 3. Therefore, the maximum value of f(x) on the interval [-3, 3] is 18.Wait, let me double-check my calculations to make sure I didn't make any mistakes. For f(-3): (-3)³ is -27, and -3*(-3) is +9, so -27 + 9 is indeed -18. For f(-1): (-1)³ is -1, and -3*(-1) is +3, so -1 + 3 is 2. For f(1): 1³ is 1, and -3*(1) is -3, so 1 - 3 is -2. For f(3): 3³ is 27, and -3*(3) is -9, so 27 - 9 is 18. All calculations seem correct.So, yes, the maximum value is 18 at x = 3. That corresponds to option D.

💡Okay, so I have this vector problem here. Let me try to understand what it's asking. We have three vectors: α, β, and γ. The conditions given are:1. The magnitude of α is 1, so |α| = 1.2. The magnitude of (α - β) is equal to the magnitude of β, so |α - β| = |β|.3. The dot product of (α - γ) and (β - γ) is zero, so (α - γ) · (β - γ) = 0.We need to find, for every determined β, the maximum and minimum values of |γ|, which are m and n respectively. Then, for any β, we have to find the minimum value of (m - n). Hmm, okay. Let me break this down step by step.First, since |α| = 1, I can represent α as a vector of length 1. Without loss of generality, I can place α along the x-axis in a coordinate system. So, let's say α = (1, 0). That should simplify things.Next, the condition |α - β| = |β|. Let me write β as (x, y). Then, |α - β| is the distance between α and β, which is sqrt[(1 - x)^2 + (0 - y)^2]. And |β| is sqrt[x^2 + y^2]. So, setting them equal:sqrt[(1 - x)^2 + y^2] = sqrt[x^2 + y^2]Let me square both sides to eliminate the square roots:(1 - x)^2 + y^2 = x^2 + y^2Simplify:1 - 2x + x^2 + y^2 = x^2 + y^2Subtract x^2 + y^2 from both sides:1 - 2x = 0So, 1 = 2x => x = 1/2.Therefore, β must lie on the vertical line x = 1/2. So, β can be represented as (1/2, y), where y is some real number.Okay, so β is (1/2, y). Now, let's move on to the third condition: (α - γ) · (β - γ) = 0.Let me denote γ as (a, b). Then, α - γ is (1 - a, 0 - b) = (1 - a, -b). Similarly, β - γ is (1/2 - a, y - b).Their dot product is:(1 - a)(1/2 - a) + (-b)(y - b) = 0Let me expand this:(1 - a)(1/2 - a) = (1)(1/2) - (1)(a) - (a)(1/2) + a^2 = 1/2 - a - a/2 + a^2 = 1/2 - (3a/2) + a^2And (-b)(y - b) = -b*y + b^2So, putting it all together:1/2 - (3a/2) + a^2 - b*y + b^2 = 0Let me rearrange terms:a^2 - (3a/2) + b^2 - b*y + 1/2 = 0Hmm, this looks like a quadratic equation in terms of a and b. Maybe I can complete the squares to figure out what kind of geometric shape this represents.Let me group the a terms and the b terms:(a^2 - (3a/2)) + (b^2 - b*y) + 1/2 = 0Completing the square for a:a^2 - (3a/2) = a^2 - (3/2)a + (9/16) - (9/16) = (a - 3/4)^2 - 9/16Similarly, for b:b^2 - b*y = b^2 - y*b + (y^2)/4 - (y^2)/4 = (b - y/2)^2 - y^2/4Substituting back into the equation:(a - 3/4)^2 - 9/16 + (b - y/2)^2 - y^2/4 + 1/2 = 0Combine constants:-9/16 - y^2/4 + 1/2 = (-9/16 + 8/16) - y^2/4 = (-1/16) - y^2/4So, the equation becomes:(a - 3/4)^2 + (b - y/2)^2 - 1/16 - y^2/4 = 0Bring the constants to the other side:(a - 3/4)^2 + (b - y/2)^2 = 1/16 + y^2/4Let me write the right-hand side as:(1/16) + (y^2)/4 = (1 + 4y^2)/16 = (4y^2 + 1)/16So, the equation is:(a - 3/4)^2 + (b - y/2)^2 = (sqrt(4y^2 + 1)/4)^2Wait, because sqrt(4y^2 + 1)/4 squared is (4y^2 + 1)/16, which matches the right-hand side.So, this equation represents a circle with center at (3/4, y/2) and radius sqrt(4y^2 + 1)/4.Therefore, γ lies on this circle. Now, we need to find the maximum and minimum values of |γ|, which is the distance from the origin to the point (a, b).So, |γ| = sqrt(a^2 + b^2). We need to find the maximum and minimum of this function on the circle defined above.To find the maximum and minimum distances from the origin to points on the circle, we can use the fact that the maximum distance is the distance from the origin to the center of the circle plus the radius, and the minimum distance is the distance from the origin to the center minus the radius.So, let's compute the distance from the origin to the center of the circle.Center is at (3/4, y/2). So, distance is sqrt[(3/4)^2 + (y/2)^2] = sqrt(9/16 + y^2/4).Radius is sqrt(4y^2 + 1)/4.Therefore, maximum |γ|, which is m, is:sqrt(9/16 + y^2/4) + sqrt(4y^2 + 1)/4Similarly, minimum |γ|, which is n, is:sqrt(9/16 + y^2/4) - sqrt(4y^2 + 1)/4Therefore, m - n is:[sqrt(9/16 + y^2/4) + sqrt(4y^2 + 1)/4] - [sqrt(9/16 + y^2/4) - sqrt(4y^2 + 1)/4] = 2 * [sqrt(4y^2 + 1)/4] = sqrt(4y^2 + 1)/2So, m - n = sqrt(4y^2 + 1)/2Now, we need to find the minimum value of m - n as y varies over all real numbers.So, the expression is sqrt(4y^2 + 1)/2. Let's denote this as f(y) = sqrt(4y^2 + 1)/2.We need to find the minimum of f(y). Since sqrt(4y^2 + 1) is a function that is minimized when y = 0.At y = 0, f(0) = sqrt(0 + 1)/2 = 1/2.As |y| increases, f(y) increases because sqrt(4y^2 + 1) increases.Therefore, the minimum value of f(y) is 1/2.Hence, the minimum value of m - n is 1/2.**Final Answer**boxed{dfrac{1}{2}}

💡Okay, so I have this problem here: the function f(x) = x² - 2a x + 3 is monotonic on the interval [2, 3]. I need to find the range of values for a. Hmm, okay, let's break this down step by step.First, I remember that a function is monotonic on an interval if it's either entirely non-increasing or non-decreasing throughout that interval. For quadratic functions like this one, which is a parabola, the function will be monotonic on an interval if that interval lies entirely on one side of the vertex of the parabola. So, I think I need to find the vertex of this quadratic function and then ensure that the interval [2, 3] doesn't straddle the vertex.The general form of a quadratic function is f(x) = ax² + bx + c, and its vertex is at x = -b/(2a). In this case, though, the function is f(x) = x² - 2a x + 3. So, comparing to the general form, the coefficient of x² is 1, the coefficient of x is -2a, and the constant term is 3.So, the vertex x-coordinate would be at x = -(-2a)/(2*1) = 2a/2 = a. Okay, so the vertex is at x = a. That means the axis of symmetry for this parabola is the vertical line x = a.Since the coefficient of x² is positive (it's 1), the parabola opens upwards. That means the function decreases to the left of the vertex and increases to the right of the vertex. So, to the left of x = a, the function is decreasing, and to the right of x = a, it's increasing.Now, for the function to be monotonic on the interval [2, 3], the entire interval must lie either entirely to the left of the vertex or entirely to the right of the vertex. If the interval [2, 3] includes the vertex x = a, then the function would first decrease and then increase (or vice versa) within that interval, making it non-monotonic.So, to ensure monotonicity, the vertex x = a must lie outside the interval [2, 3]. That means either a ≤ 2 or a ≥ 3. If a is less than or equal to 2, then the entire interval [2, 3] is to the right of the vertex, and since the parabola opens upwards, the function will be increasing on [2, 3]. Similarly, if a is greater than or equal to 3, the entire interval [2, 3] is to the left of the vertex, and the function will be decreasing on [2, 3].Wait, let me double-check that. If a is less than or equal to 2, then the vertex is at x = a, which is less than or equal to 2. So, the interval [2, 3] starts at x = 2, which is to the right of the vertex. Since the parabola opens upwards, to the right of the vertex, the function is increasing. So, on [2, 3], the function would be increasing, which is monotonic.Similarly, if a is greater than or equal to 3, the vertex is at x = a, which is greater than or equal to 3. So, the interval [2, 3] ends at x = 3, which is to the left of the vertex. Since the parabola opens upwards, to the left of the vertex, the function is decreasing. So, on [2, 3], the function would be decreasing, which is also monotonic.Therefore, the function f(x) is monotonic on [2, 3] if and only if a is less than or equal to 2 or greater than or equal to 3. So, the range of values for a is all real numbers a such that a ≤ 2 or a ≥ 3.Let me think if there's another way to approach this problem, maybe using calculus. I know that if a function is differentiable, then it's monotonic on an interval if its derivative doesn't change sign on that interval. The derivative of f(x) is f'(x) = 2x - 2a. For the function to be monotonic on [2, 3], the derivative should either be non-negative throughout the interval or non-positive throughout the interval.So, let's compute the derivative: f'(x) = 2x - 2a. For f'(x) to be non-negative on [2, 3], we need 2x - 2a ≥ 0 for all x in [2, 3]. Similarly, for f'(x) to be non-positive on [2, 3], we need 2x - 2a ≤ 0 for all x in [2, 3].Let's consider the non-negative case first. 2x - 2a ≥ 0 ⇒ x - a ≥ 0 ⇒ x ≥ a. So, for this inequality to hold for all x in [2, 3], the smallest x in the interval, which is 2, must satisfy 2 ≥ a. Therefore, a ≤ 2.Now, for the non-positive case: 2x - 2a ≤ 0 ⇒ x - a ≤ 0 ⇒ x ≤ a. So, for this inequality to hold for all x in [2, 3], the largest x in the interval, which is 3, must satisfy 3 ≤ a. Therefore, a ≥ 3.So, combining both cases, we get that a must be less than or equal to 2 or greater than or equal to 3, which matches what I found earlier using the vertex approach.Just to be thorough, let's test a value in each interval to ensure that the function is indeed monotonic.First, let's take a = 1, which is less than 2. Then f(x) = x² - 2*1*x + 3 = x² - 2x + 3. The derivative is f'(x) = 2x - 2. On [2, 3], f'(2) = 4 - 2 = 2 > 0, and f'(3) = 6 - 2 = 4 > 0. So, the derivative is positive throughout, meaning the function is increasing on [2, 3], hence monotonic.Next, let's take a = 4, which is greater than 3. Then f(x) = x² - 2*4*x + 3 = x² - 8x + 3. The derivative is f'(x) = 2x - 8. On [2, 3], f'(2) = 4 - 8 = -4 < 0, and f'(3) = 6 - 8 = -2 < 0. So, the derivative is negative throughout, meaning the function is decreasing on [2, 3], hence monotonic.Now, let's check a value inside the interval [2, 3], say a = 2.5. Then f(x) = x² - 2*2.5*x + 3 = x² - 5x + 3. The derivative is f'(x) = 2x - 5. On [2, 3], f'(2) = 4 - 5 = -1 < 0, and f'(3) = 6 - 5 = 1 > 0. So, the derivative changes sign from negative to positive within the interval, meaning the function first decreases and then increases. Therefore, it's not monotonic on [2, 3], which confirms that a cannot be between 2 and 3.Similarly, if a is exactly 2, then f'(x) = 2x - 4. On [2, 3], f'(2) = 0, and f'(3) = 2. So, the derivative starts at 0 and increases to 2, which means the function is non-decreasing on [2, 3], hence monotonic. Similarly, if a is exactly 3, f'(x) = 2x - 6. On [2, 3], f'(2) = -2 < 0, and f'(3) = 0. So, the derivative starts negative and ends at 0, meaning the function is non-increasing on [2, 3], hence monotonic.Therefore, the critical points a = 2 and a = 3 are included in the range, as the function is still monotonic at these points.In summary, by analyzing the vertex of the quadratic function and ensuring that the interval [2, 3] does not contain the vertex, and by verifying using calculus with the derivative, I've determined that the function f(x) = x² - 2a x + 3 is monotonic on [2, 3] if and only if a is less than or equal to 2 or greater than or equal to 3. This gives the range of values for a as (-∞, 2] ∪ [3, ∞).

💡First, I need to determine how much the dealer initially marked up the articles above the cost price. The dealer offers a cash discount of 20%, which means the selling price after the discount is 80% of the marked price.Next, the dealer still makes a 20% profit on the cost price after applying the cash discount. This implies that the selling price after the discount is 120% of the cost price.By setting up the equation 0.8 * Marked Price = 1.2 * Cost Price, I can solve for the Marked Price in terms of the Cost Price. This reveals that the Marked Price is 150% of the Cost Price, indicating a 50% markup.Finally, the additional allowance of selling 10 articles for the price of 9 does not affect the initial markup percentage, as it is already accounted for in the profit calculation.

💡First, I recognize that the square root of a negative number involves imaginary numbers. However, the problem specifies that √(-a) is a rational number, which means it must be a real number. This implies that -a must be a perfect square because the square root of a perfect square is rational.Since -a is a perfect square, it follows that a is the negative of a perfect square. Therefore, a is the opposite of a perfect square.Looking at the options provided:- A: Rational number – This is true, but it doesn't fully capture the specific relationship.- B: The opposite of a perfect square – This accurately describes a based on the reasoning.- C: Perfect square – This is incorrect because a is the opposite of a perfect square.- D: Negative real number – While a is indeed a negative real number, this option is too broad and doesn't specifically address the relationship to perfect squares.Thus, the most precise and correct answer is B: The opposite of a perfect square.

💡Okay, so I have this equation: (x^4 - x^2 - 2 = 0). Hmm, it's a quartic equation because the highest power of x is 4. Quartic equations can be tricky, but maybe I can simplify it somehow.Let me think. I remember that sometimes you can substitute a part of the equation with a new variable to make it easier. In this case, I see that (x^4) is ((x^2)^2), so maybe I can let (y = x^2). That way, the equation becomes (y^2 - y - 2 = 0). That looks like a quadratic equation in terms of y, which I know how to solve!Alright, so now I have (y^2 - y - 2 = 0). To solve this quadratic equation, I can use the quadratic formula: (y = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = -1), and (c = -2).Plugging in those values: (y = frac{-(-1) pm sqrt{(-1)^2 - 4(1)(-2)}}{2(1)}). Simplifying that: (y = frac{1 pm sqrt{1 + 8}}{2}), which is (y = frac{1 pm sqrt{9}}{2}). So, (y = frac{1 pm 3}{2}).That gives me two solutions for y: (y = frac{1 + 3}{2} = 2) and (y = frac{1 - 3}{2} = -1). Okay, so y can be 2 or -1.But remember, (y = x^2), so I need to solve for x. Let's start with (y = 2): (x^2 = 2). Taking the square root of both sides gives (x = sqrt{2}) or (x = -sqrt{2}).Now, for (y = -1): (x^2 = -1). Wait a minute, the square of a real number can't be negative. So, this equation doesn't have real solutions. That means the only real solutions come from (y = 2).So, putting it all together, the real solutions to the original equation (x^4 - x^2 - 2 = 0) are (x = sqrt{2}) and (x = -sqrt{2}).I think that's it! I don't see any mistakes in my steps. I substituted to make it a quadratic, solved for y, and then found x from y. The negative y didn't give any real solutions, so only the positive y worked out.