💡First, I need to understand the problem. I have two vectors, (overrightarrow{a} = (3, -1)) and (overrightarrow{b} = (-1, 2)). I need to find the coordinates of the vector (-3overrightarrow{a} - 2overrightarrow{b}).To do this, I'll start by scaling each vector according to the coefficients given.For (-3overrightarrow{a}):[-3overrightarrow{a} = -3 times (3, -1) = (-9, 3)]Next, for (-2overrightarrow{b}):[-2overrightarrow{b} = -2 times (-1, 2) = (2, -4)]Now, I'll add the results of these two scaled vectors:[(-9, 3) + (2, -4) = (-9 + 2, 3 - 4) = (-7, -1)]So, the coordinates of (-3overrightarrow{a} - 2overrightarrow{b}) are ((-7, -1)).Looking at the options provided:- A: ((7, 1))- B: ((-7, -1))- C: ((-7, 1))- D: ((7, -1))The correct answer is option B: ((-7, -1)).

💡Okay, so I have this problem: I need to find all real numbers ( b ) such that the inequality ( |x^2 + bx + 2b| leq 3 ) has exactly one solution in ( x ). Hmm, that sounds a bit tricky, but let me try to break it down step by step.First, I know that the expression inside the absolute value is a quadratic function. Let me denote it as ( f(x) = x^2 + bx + 2b ). So, the inequality becomes ( |f(x)| leq 3 ). This means that ( f(x) ) must lie between -3 and 3, inclusive. So, ( -3 leq f(x) leq 3 ).Now, the problem states that this inequality should have exactly one solution for ( x ). That means there should be only one value of ( x ) where ( f(x) ) is within the range of -3 to 3. Hmm, but since ( f(x) ) is a quadratic function, which is a parabola, it usually has a minimum or maximum point. Since the coefficient of ( x^2 ) is positive (1), it opens upwards, so it has a minimum point.So, the graph of ( f(x) ) is a parabola opening upwards. The inequality ( |f(x)| leq 3 ) would correspond to the region between the lines ( y = 3 ) and ( y = -3 ). For this inequality to have exactly one solution, the parabola must touch one of these lines exactly once. But since the parabola opens upwards, its minimum value is the lowest point on the graph. So, if the minimum value of ( f(x) ) is equal to 3, then the parabola just touches the line ( y = 3 ) at its vertex, and nowhere else does it enter the region ( |f(x)| leq 3 ). Wait, but if the minimum is 3, then ( f(x) ) is always greater than or equal to 3, so the inequality ( |f(x)| leq 3 ) would only be satisfied at the vertex where ( f(x) = 3 ). That would give exactly one solution.But hold on, what if the minimum is less than 3? Then the parabola would dip below 3, and depending on how low it goes, it might cross the line ( y = -3 ) or not. If it crosses ( y = -3 ), then there could be two points where ( f(x) = -3 ), which would mean two solutions for the inequality ( |f(x)| leq 3 ). But the problem says exactly one solution, so that can't happen. So, maybe the minimum has to be exactly 3 so that it only touches ( y = 3 ) once.Alternatively, if the minimum is exactly -3, then the parabola would touch ( y = -3 ) at its vertex, and since it opens upwards, it would cross ( y = 3 ) at two points. That would give two solutions for ( |f(x)| leq 3 ), which is not what we want. So, the minimum must be exactly 3 to have only one solution.Wait, let me think again. If the minimum is 3, then ( f(x) geq 3 ) for all ( x ), so the inequality ( |f(x)| leq 3 ) would only be satisfied when ( f(x) = 3 ), which is at the vertex. So, that would be exactly one solution. If the minimum is less than 3, say 2, then ( f(x) ) would dip down to 2, which is still above -3, so ( |f(x)| leq 3 ) would be satisfied for all ( x ) where ( f(x) ) is between -3 and 3. But since the parabola is opening upwards, it would cross ( y = 3 ) at two points, giving two solutions. So, that's not what we want.Alternatively, if the minimum is exactly -3, then ( f(x) ) would touch ( y = -3 ) at the vertex, and since it's opening upwards, it would cross ( y = 3 ) at two points, giving two solutions again. So, that's not good either.Therefore, the only way to have exactly one solution is if the minimum of ( f(x) ) is exactly 3, so that the parabola just touches ( y = 3 ) at one point and doesn't go below it, meaning ( |f(x)| leq 3 ) is only satisfied at that one point.So, I need to find the value(s) of ( b ) such that the minimum of ( f(x) = x^2 + bx + 2b ) is equal to 3.To find the minimum of a quadratic function ( ax^2 + bx + c ), the vertex occurs at ( x = -frac{b}{2a} ). In this case, ( a = 1 ), so the vertex is at ( x = -frac{b}{2} ).Let me compute the value of ( f(x) ) at this vertex:( fleft(-frac{b}{2}right) = left(-frac{b}{2}right)^2 + bleft(-frac{b}{2}right) + 2b )Simplify this:( fleft(-frac{b}{2}right) = frac{b^2}{4} - frac{b^2}{2} + 2b )Combine like terms:( frac{b^2}{4} - frac{2b^2}{4} + 2b = -frac{b^2}{4} + 2b )So, the minimum value is ( -frac{b^2}{4} + 2b ). We want this to be equal to 3:( -frac{b^2}{4} + 2b = 3 )Let me rearrange this equation:( -frac{b^2}{4} + 2b - 3 = 0 )Multiply both sides by -4 to eliminate the fraction:( b^2 - 8b + 12 = 0 )Now, we have a quadratic equation in terms of ( b ):( b^2 - 8b + 12 = 0 )Let me try to factor this:Looking for two numbers that multiply to 12 and add up to -8. Hmm, -6 and -2.So, ( (b - 6)(b - 2) = 0 )Therefore, the solutions are ( b = 6 ) and ( b = 2 ).Wait, let me check if these values actually result in the minimum being 3.First, for ( b = 6 ):( f(x) = x^2 + 6x + 12 )The vertex is at ( x = -frac{6}{2} = -3 )Compute ( f(-3) = (-3)^2 + 6*(-3) + 12 = 9 - 18 + 12 = 3 ). So, yes, the minimum is 3.For ( b = 2 ):( f(x) = x^2 + 2x + 4 )The vertex is at ( x = -frac{2}{2} = -1 )Compute ( f(-1) = (-1)^2 + 2*(-1) + 4 = 1 - 2 + 4 = 3 ). So, the minimum is also 3.Therefore, both ( b = 6 ) and ( b = 2 ) satisfy the condition that the minimum of ( f(x) ) is 3, which means the inequality ( |f(x)| leq 3 ) has exactly one solution at the vertex.Wait, but let me think again. If the minimum is 3, then ( f(x) geq 3 ) for all ( x ), so ( |f(x)| leq 3 ) would only be satisfied when ( f(x) = 3 ), which is exactly at the vertex. So, that's one solution. If the minimum were less than 3, say 2, then ( f(x) ) would dip below 3, and since it's a parabola opening upwards, it would cross ( y = 3 ) at two points, giving two solutions. Similarly, if the minimum were below -3, it would cross both ( y = 3 ) and ( y = -3 ), giving more solutions. So, indeed, the only way to have exactly one solution is when the minimum is exactly 3.Therefore, the values of ( b ) are 2 and 6.Wait, but let me just double-check if there's any other scenario where the inequality could have exactly one solution. For example, if the parabola is tangent to both ( y = 3 ) and ( y = -3 ), but that would require the minimum to be -3 and the maximum to be 3, which isn't possible because the parabola only has a minimum or maximum, not both. Since this parabola opens upwards, it only has a minimum, so it can't be tangent to both lines. Therefore, the only way is when the minimum is exactly 3.So, I think I've covered all the cases, and the conclusion is that ( b = 2 ) and ( b = 6 ) are the only real numbers that satisfy the condition.

💡Okay, so I have this problem where a piece of string fits exactly around the perimeter of a rectangle. The rectangle has a length that's twice its width, and the area of the rectangle is 288. I need to find the area of the largest circle that can be formed from this string, rounded to the nearest whole number. Hmm, let me break this down step by step.First, let me visualize the rectangle. It has a length and a width, and the length is twice the width. So, if I let the width be some value, say, 'w', then the length would be '2w'. That makes sense because the problem says the length is twice the width.Now, the area of the rectangle is given as 288. I remember that the area of a rectangle is calculated by multiplying its length by its width. So, in this case, it would be:Area = length × width288 = 2w × wSimplifying that, I get:288 = 2w²To find the value of 'w', I can divide both sides by 2:288 ÷ 2 = w²144 = w²Now, to find 'w', I take the square root of both sides:w = √144w = 12Okay, so the width of the rectangle is 12 units. Since the length is twice the width, the length would be:Length = 2w = 2 × 12 = 24Alright, so now I know the dimensions of the rectangle: 12 units in width and 24 units in length.Next, I need to find the perimeter of the rectangle because the string fits exactly around it. The perimeter of a rectangle is calculated by adding up all its sides, which is 2 times the length plus 2 times the width. So,Perimeter = 2 × length + 2 × widthPerimeter = 2 × 24 + 2 × 12Calculating that:Perimeter = 48 + 24 = 72So, the length of the string is 72 units.Now, this string is going to be formed into a circle. The circumference of the circle will be equal to the length of the string, which is 72 units. I need to find the area of this circle. To find the area, I first need to find the radius of the circle.I remember that the circumference of a circle is given by the formula:Circumference = 2 × π × radiusSo, plugging in the circumference we have:72 = 2 × π × rI can solve for 'r' by dividing both sides by 2π:r = 72 / (2π) = 36 / πSo, the radius of the circle is 36 divided by π.Now, to find the area of the circle, I use the formula:Area = π × radius²Substituting the radius we found:Area = π × (36 / π)²Let me compute that step by step. First, square the radius:(36 / π)² = (36²) / (π²) = 1296 / π²Now, multiply by π:Area = π × (1296 / π²) = 1296 / πSo, the area is 1296 divided by π. To get a numerical value, I can approximate π as 3.1416.Calculating that:Area ≈ 1296 / 3.1416 ≈ 412.53Rounding this to the nearest whole number, I get 413.Wait, let me double-check my calculations to make sure I didn't make a mistake. So, 36 divided by π is approximately 11.459. Squaring that gives approximately 131.33. Then, multiplying by π gives approximately 412.53, which rounds to 413. Hmm, but in the initial problem, the answer was given as 412. Did I do something wrong?Let me recalculate:36 / π ≈ 36 / 3.1416 ≈ 11.45911.459 squared is approximately 131.33131.33 multiplied by π is approximately 131.33 × 3.1416 ≈ 412.53So, 412.53 rounds to 413. But the initial answer was 412. Maybe I should check if the perimeter was calculated correctly.Perimeter of the rectangle: 2 × (length + width) = 2 × (24 + 12) = 2 × 36 = 72. That seems correct.Circumference of the circle: 72 = 2πr, so r = 36/π. That's correct.Area: πr² = π × (36/π)² = π × 1296/π² = 1296/π ≈ 412.53. So, 412.53 is approximately 413 when rounded to the nearest whole number.Wait, but 412.53 is closer to 413, right? Because 0.53 is more than 0.5, so it should round up. So, maybe the initial answer was incorrect, and the correct answer is 413.Alternatively, perhaps I should use a more precise value of π. Let me try using π ≈ 3.1415926535.So, 1296 / π ≈ 1296 / 3.1415926535 ≈ 412.5316So, 412.5316 is approximately 412.53, which is still closer to 413. So, I think the correct rounded answer is 413.But in the initial problem, the answer was given as 412. Maybe there was a miscalculation there. Let me check again.Wait, perhaps I made a mistake in calculating the area. Let me go back.Area = π × (36/π)² = π × (1296/π²) = 1296/πYes, that's correct. So, 1296 divided by π is approximately 412.53, which is 413 when rounded.Hmm, maybe the initial answer was rounded differently, or perhaps I made a mistake in the perimeter calculation. Let me verify the perimeter again.The rectangle has length 24 and width 12, so perimeter is 2*(24 + 12) = 2*36 = 72. That's correct.Circumference of the circle is 72, so radius is 72/(2π) = 36/π. Correct.Area is π*(36/π)^2 = π*(1296/π²) = 1296/π ≈ 412.53. So, 413.Wait, maybe the initial problem used a different approximation for π, like 22/7, which is approximately 3.142857.Let me try that.1296 / (22/7) = 1296 * (7/22) = (1296 * 7) / 221296 * 7 = 90729072 / 22 = 412.3636...So, 412.3636 is approximately 412.36, which would round to 412.Ah, so depending on the approximation of π, the result can be 412 or 413. If using 22/7, it's approximately 412.36, which rounds to 412. If using 3.1416, it's approximately 412.53, which rounds to 413.So, perhaps the initial answer used 22/7 for π, resulting in 412. But using a more precise value of π, it's closer to 413.I think, in most cases, using π ≈ 3.1416 is more accurate, so the area should be approximately 413. However, since the problem asks to round to the nearest whole number, and 412.53 is closer to 413, I think the correct answer is 413.But to be thorough, let me check with π ≈ 3.141592653589793.1296 / π ≈ 1296 / 3.141592653589793 ≈ 412.5316So, 412.5316 is approximately 412.53, which is 413 when rounded to the nearest whole number.Therefore, I think the correct answer is 413.Wait, but the initial problem's answer was 412. Maybe I should check if I made a mistake in the area formula.Area = πr², where r = 36/π.So, Area = π*(36/π)² = π*(1296/π²) = 1296/π. That's correct.So, 1296 divided by π is approximately 412.53, which is 413.I think the initial answer might have been using a different approximation or rounding method. But based on standard rounding rules, 412.53 rounds up to 413.So, I think the correct answer is 413.But just to be absolutely sure, let me calculate 1296 divided by π with more precision.Using a calculator, 1296 ÷ π ≈ 412.5316.Yes, that's approximately 412.53, which is 413 when rounded to the nearest whole number.Therefore, I conclude that the area of the largest circle that can be formed from the string is approximately 413.

💡Okay, so I have this function f(x) = e^x - (1/2)a x² + (a - e)x, where x is greater than or equal to 0, and e is approximately 2.71828. There are two parts to this problem. The first part is to find the minimum value of f(x) when a = 0. The second part is to determine the number of monotonic intervals of f(x) when 1 < a < e.Starting with part (1): When a = 0, the function simplifies. Let me substitute a = 0 into f(x). So, f(x) becomes e^x - (1/2)*0*x² + (0 - e)x, which simplifies to e^x - e x. So, f(x) = e^x - e x for x ≥ 0.To find the minimum value of this function, I need to find its critical points by taking the derivative and setting it equal to zero. The derivative of f(x) with respect to x is f'(x) = d/dx [e^x - e x] = e^x - e.Setting f'(x) equal to zero: e^x - e = 0. Solving for x, we get e^x = e, which implies x = 1 because ln(e) = 1.Now, I need to check if this critical point is a minimum. I can do this by using the second derivative test. The second derivative of f(x) is f''(x) = d/dx [e^x - e] = e^x. At x = 1, f''(1) = e^1 = e, which is positive. Since the second derivative is positive, the function is concave up at x = 1, indicating a local minimum.Therefore, the minimum value of f(x) occurs at x = 1. Plugging x = 1 back into f(x): f(1) = e^1 - e*1 = e - e = 0. So, the minimum value is 0.Moving on to part (2): We need to determine the number of monotonic intervals of f(x) when 1 < a < e. Monotonic intervals refer to intervals where the function is either entirely increasing or entirely decreasing. To find these, we'll analyze the first derivative of f(x).First, let's compute the first derivative f'(x). The function is f(x) = e^x - (1/2)a x² + (a - e)x. So, f'(x) = d/dx [e^x] - d/dx [(1/2)a x²] + d/dx [(a - e)x] = e^x - a x + (a - e).So, f'(x) = e^x - a x + (a - e). To find the critical points, set f'(x) = 0: e^x - a x + (a - e) = 0.This equation might be tricky to solve analytically, so perhaps we can analyze the behavior of f'(x) to determine the number of critical points, which in turn will tell us the number of monotonic intervals.Let me define g(x) = f'(x) = e^x - a x + (a - e). To analyze g(x), let's compute its derivative, g'(x): g'(x) = d/dx [e^x - a x + (a - e)] = e^x - a.So, g'(x) = e^x - a. Setting g'(x) = 0 gives e^x - a = 0, which implies x = ln(a). So, the critical point of g(x) is at x = ln(a).Now, since 1 < a < e, ln(a) will be between 0 and 1 because ln(1) = 0 and ln(e) = 1. So, x = ln(a) is within the interval (0,1).Let's analyze the behavior of g(x):1. For x < ln(a): Since x < ln(a), e^x < a, so g'(x) = e^x - a < 0. Therefore, g(x) is decreasing on [0, ln(a)).2. For x > ln(a): Since x > ln(a), e^x > a, so g'(x) = e^x - a > 0. Therefore, g(x) is increasing on (ln(a), ∞).So, g(x) has a minimum at x = ln(a). Let's compute the value of g(x) at this minimum: g(ln(a)) = e^{ln(a)} - a ln(a) + (a - e) = a - a ln(a) + a - e = 2a - a ln(a) - e.Now, let's denote this minimum value as h(a) = 2a - a ln(a) - e. We need to analyze the sign of h(a) because it will determine the number of critical points of f'(x), which in turn affects the number of monotonic intervals.Let's analyze h(a):h(a) = 2a - a ln(a) - e.We can factor out an a: h(a) = a(2 - ln(a)) - e.We need to find when h(a) is positive, negative, or zero because:- If h(a) < 0, then g(x) = f'(x) crosses the x-axis twice, meaning f'(x) has two critical points, leading to three monotonic intervals for f(x).- If h(a) = 0, then g(x) touches the x-axis at one point, meaning f'(x) has one critical point, leading to two monotonic intervals for f(x).- If h(a) > 0, then g(x) does not cross the x-axis, meaning f'(x) has no critical points, so f(x) is either always increasing or always decreasing, leading to one monotonic interval.But wait, in our case, since g(x) has a minimum at x = ln(a), if h(a) < 0, then g(x) will cross the x-axis twice, implying two critical points for f'(x). If h(a) = 0, it touches the x-axis once, implying one critical point. If h(a) > 0, it doesn't cross, implying no critical points.However, let's think about the behavior of g(x) as x approaches 0 and as x approaches infinity.As x approaches 0 from the right:g(0) = e^0 - a*0 + (a - e) = 1 + (a - e) = a - e + 1.Since 1 < a < e, a - e is negative, but a - e + 1 is a - (e - 1). Since e ≈ 2.718, e - 1 ≈ 1.718. So, a is between 1 and e, so a - (e - 1) is between 1 - 1.718 = -0.718 and e - 1.718 ≈ 1. So, g(0) can be negative or positive depending on a.Wait, let's compute g(0):g(0) = e^0 - a*0 + (a - e) = 1 + 0 + (a - e) = a - e + 1.So, g(0) = a - e + 1.Since 1 < a < e, let's compute the range of g(0):When a = 1, g(0) = 1 - e + 1 = 2 - e ≈ 2 - 2.718 ≈ -0.718.When a = e, g(0) = e - e + 1 = 1.So, as a increases from 1 to e, g(0) increases from approximately -0.718 to 1.Therefore, there exists a value a0 in (1, e) such that g(0) = 0 when a = a0.Solving for a0: a0 - e + 1 = 0 ⇒ a0 = e - 1 ≈ 2.718 - 1 ≈ 1.718.So, when a = e - 1, g(0) = 0.Therefore, for a < e - 1, g(0) < 0.For a > e - 1, g(0) > 0.Now, let's consider the behavior of g(x) as x approaches infinity:As x → ∞, e^x dominates, so g(x) = e^x - a x + (a - e) → ∞.So, regardless of a, g(x) tends to infinity as x increases.Now, let's analyze h(a) = 2a - a ln(a) - e.We need to find when h(a) is positive, negative, or zero.Let's compute h(a) at a = e - 1:h(e - 1) = 2(e - 1) - (e - 1) ln(e - 1) - e.Simplify:= 2e - 2 - (e - 1) ln(e - 1) - e= e - 2 - (e - 1) ln(e - 1)We can compute this numerically:e ≈ 2.718, so e - 1 ≈ 1.718.ln(e - 1) ≈ ln(1.718) ≈ 0.542.So,h(e - 1) ≈ 2.718 - 2 - (1.718)(0.542) ≈ 0.718 - 0.931 ≈ -0.213.So, h(e - 1) ≈ -0.213 < 0.At a = e:h(e) = 2e - e ln(e) - e = 2e - e*1 - e = 2e - e - e = 0.So, h(e) = 0.At a = 1:h(1) = 2*1 - 1*ln(1) - e = 2 - 0 - e ≈ 2 - 2.718 ≈ -0.718.So, h(1) ≈ -0.718.So, h(a) starts at h(1) ≈ -0.718, increases to h(e - 1) ≈ -0.213, and then increases further to h(e) = 0.Wait, but let's check the derivative of h(a) to see how it behaves.h(a) = 2a - a ln(a) - e.Compute h'(a):h'(a) = 2 - [ln(a) + a*(1/a)] - 0 = 2 - ln(a) - 1 = 1 - ln(a).So, h'(a) = 1 - ln(a).Set h'(a) = 0: 1 - ln(a) = 0 ⇒ ln(a) = 1 ⇒ a = e.So, h(a) has a critical point at a = e, which is a maximum because h'(a) changes from positive to negative at a = e.Wait, let's see:For a < e, ln(a) < 1, so h'(a) = 1 - ln(a) > 0. So, h(a) is increasing on (1, e).At a = e, h(a) reaches a maximum of 0.So, h(a) increases from a = 1 to a = e, starting at h(1) ≈ -0.718, increasing to h(e) = 0.Therefore, h(a) is always negative for a in (1, e), except at a = e where it is zero.Wait, but earlier I thought h(e - 1) ≈ -0.213, which is still negative, and h(e) = 0.So, h(a) is negative for all a in (1, e), and zero at a = e.Therefore, the minimum value of g(x) = f'(x) is h(a) = 2a - a ln(a) - e < 0 for 1 < a < e.This means that g(x) = f'(x) has two roots: one in (0, ln(a)) and another in (ln(a), ∞). Because g(x) is decreasing on [0, ln(a)) and increasing on (ln(a), ∞), and since the minimum value is negative, it must cross the x-axis twice.Wait, but let's think about the behavior at x = 0 and x approaching infinity.At x = 0, g(0) = a - e + 1.As we saw earlier, when a < e - 1, g(0) < 0.When a = e - 1, g(0) = 0.When a > e - 1, g(0) > 0.So, for a < e - 1, g(0) < 0, and since g(x) approaches infinity as x increases, and it has a minimum at x = ln(a) where g(ln(a)) < 0, then g(x) must cross the x-axis once in (ln(a), ∞). But wait, since g(x) is decreasing from x = 0 to x = ln(a), and g(0) < 0, and the minimum at x = ln(a) is also < 0, then g(x) remains negative on [0, ln(a)), and then increases from x = ln(a) onwards, crossing the x-axis once in (ln(a), ∞). So, only one critical point for f'(x), meaning f(x) has two monotonic intervals.Wait, that contradicts my earlier thought. Let me clarify.If a < e - 1, then g(0) < 0, and since g(x) is decreasing from x = 0 to x = ln(a), and the minimum at x = ln(a) is also < 0, then g(x) remains negative on [0, ln(a)). Then, for x > ln(a), g(x) increases from g(ln(a)) < 0 to infinity, so it must cross the x-axis once in (ln(a), ∞). Therefore, f'(x) = 0 has only one solution in (ln(a), ∞). So, f(x) has one critical point, meaning two monotonic intervals: decreasing on [0, x1) and increasing on (x1, ∞).If a = e - 1, then g(0) = 0, and the minimum at x = ln(a) is h(a) ≈ -0.213 < 0. So, g(x) starts at 0, decreases to a minimum below zero, then increases back to infinity. Therefore, f'(x) = 0 has two solutions: one at x = 0 and another in (ln(a), ∞). But x = 0 is the boundary, so f(x) is decreasing on [0, x1) and increasing on (x1, ∞). So, still two monotonic intervals.If a > e - 1, then g(0) > 0. So, g(x) starts positive at x = 0, decreases to a minimum at x = ln(a) where g(ln(a)) < 0, then increases to infinity. Therefore, g(x) must cross the x-axis twice: once in (0, ln(a)) and once in (ln(a), ∞). Therefore, f'(x) = 0 has two solutions, x1 and x2, with x1 < ln(a) < x2. Therefore, f(x) has three monotonic intervals: increasing on [0, x1), decreasing on (x1, x2), and increasing on (x2, ∞).Wait, but earlier I thought h(a) is negative for all a in (1, e), which is correct. So, the minimum of g(x) is negative, but depending on g(0), which is a - e + 1, we have different scenarios.So, summarizing:- For 1 < a < e - 1: g(0) < 0, so f'(x) has one critical point in (ln(a), ∞). Therefore, f(x) has two monotonic intervals.- For a = e - 1: g(0) = 0, so f'(x) has a critical point at x = 0 and another in (ln(a), ∞). But since x = 0 is the boundary, f(x) is decreasing on [0, x1) and increasing on (x1, ∞). So, still two monotonic intervals.- For e - 1 < a < e: g(0) > 0, so f'(x) has two critical points: one in (0, ln(a)) and another in (ln(a), ∞). Therefore, f(x) has three monotonic intervals.Therefore, the number of monotonic intervals of f(x) when 1 < a < e is:- Two intervals when 1 < a ≤ e - 1.- Three intervals when e - 1 < a < e.But the problem states 1 < a < e, so we need to specify the number of intervals based on the value of a within that range.So, the answer is:When 1 < a ≤ e - 1, f(x) has two monotonic intervals.When e - 1 < a < e, f(x) has three monotonic intervals.Therefore, the number of monotonic intervals depends on whether a is less than or equal to e - 1 or greater than e - 1 within the interval (1, e).To confirm, let's consider specific values:- Let a = 1.5, which is less than e - 1 ≈ 1.718. Then, g(0) = 1.5 - e + 1 ≈ 2.5 - 2.718 ≈ -0.218 < 0. So, f'(x) has one critical point in (ln(1.5), ∞). Therefore, f(x) has two monotonic intervals.- Let a = 2, which is greater than e - 1. Then, g(0) = 2 - e + 1 ≈ 3 - 2.718 ≈ 0.282 > 0. So, f'(x) has two critical points: one in (0, ln(2)) and another in (ln(2), ∞). Therefore, f(x) has three monotonic intervals.Thus, the conclusion is correct.

💡Alright, so I have this problem about converting a polar equation to a rectangular one and then finding the minimum distance from a point on an ellipse to that line. Let me try to figure this out step by step.First, part (1) asks to convert the polar equation of line ( l ) into a rectangular coordinate equation. The given equation is ( rho sinleft(theta - frac{pi}{4}right) = 3sqrt{2} ). Hmm, I remember that in polar coordinates, ( rho ) is the distance from the origin, and ( theta ) is the angle from the positive x-axis. I think the formula for converting polar equations to rectangular involves using ( x = rho cos theta ) and ( y = rho sin theta ). But this equation has ( sin(theta - frac{pi}{4}) ), which is a bit tricky. Maybe I can use the sine subtraction formula to expand it. The sine of a difference is ( sin(A - B) = sin A cos B - cos A sin B ). So, applying that here:( sinleft(theta - frac{pi}{4}right) = sin theta cos frac{pi}{4} - cos theta sin frac{pi}{4} ).I know that ( cos frac{pi}{4} = sin frac{pi}{4} = frac{sqrt{2}}{2} ). So substituting that in:( sinleft(theta - frac{pi}{4}right) = sin theta cdot frac{sqrt{2}}{2} - cos theta cdot frac{sqrt{2}}{2} ).So the original equation becomes:( rho left( frac{sqrt{2}}{2} sin theta - frac{sqrt{2}}{2} cos theta right) = 3sqrt{2} ).Let me factor out ( frac{sqrt{2}}{2} ):( frac{sqrt{2}}{2} rho (sin theta - cos theta) = 3sqrt{2} ).Now, multiplying both sides by ( frac{2}{sqrt{2}} ) to simplify:( rho (sin theta - cos theta) = 6 ).But ( rho sin theta ) is ( y ) and ( rho cos theta ) is ( x ), so substituting those in:( y - x = 6 ).Wait, that seems too straightforward. Let me double-check. So if I have ( rho sin(theta - frac{pi}{4}) = 3sqrt{2} ), expanding it gives me ( frac{sqrt{2}}{2} (y - x) = 3sqrt{2} ). Multiplying both sides by ( sqrt{2} ) gives ( frac{1}{2}(y - x) = 3 ), so ( y - x = 6 ). Yeah, that seems right. So the rectangular equation is ( y - x = 6 ), or rearranged as ( x - y + 6 = 0 ). I think that's correct.Moving on to part (2). We have an ellipse ( C: frac{x^2}{16} + frac{y^2}{9} = 1 ). So this is an ellipse centered at the origin, with semi-major axis 4 along the x-axis and semi-minor axis 3 along the y-axis. The question is to find the minimum distance from a point ( P ) on this ellipse to the line ( l ) we found earlier, which is ( x - y + 6 = 0 ).I remember that the distance from a point ( (x_0, y_0) ) to the line ( ax + by + c = 0 ) is given by ( frac{|ax_0 + by_0 + c|}{sqrt{a^2 + b^2}} ). So in this case, the line is ( x - y + 6 = 0 ), so ( a = 1 ), ( b = -1 ), and ( c = 6 ). Therefore, the distance from ( P(x, y) ) to the line is ( frac{|x - y + 6|}{sqrt{1 + 1}} = frac{|x - y + 6|}{sqrt{2}} ).Since ( P ) is on the ellipse, we can parameterize it using the standard parametric equations for an ellipse. For an ellipse ( frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ), the parametric equations are ( x = a cos alpha ) and ( y = b sin alpha ), where ( alpha ) is the parameter. So here, ( a = 4 ) and ( b = 3 ), so ( x = 4 cos alpha ) and ( y = 3 sin alpha ).Substituting these into the distance formula, we get:( d(alpha) = frac{|4 cos alpha - 3 sin alpha + 6|}{sqrt{2}} ).Now, to find the minimum distance, we need to minimize this expression with respect to ( alpha ). The expression inside the absolute value is ( 4 cos alpha - 3 sin alpha + 6 ). Let me denote ( A = 4 cos alpha - 3 sin alpha ), so the expression becomes ( |A + 6| ).I recall that expressions of the form ( A cos alpha + B sin alpha ) can be rewritten as ( R cos(alpha - phi) ), where ( R = sqrt{A^2 + B^2} ) and ( phi = arctanleft(frac{B}{A}right) ). Wait, actually, in this case, it's ( 4 cos alpha - 3 sin alpha ), so it's similar to ( A cos alpha + B sin alpha ), but with a negative sign on the sine term. So, I can write it as ( R cos(alpha + phi) ), perhaps?Let me compute ( R ). Here, ( A = 4 ) and ( B = -3 ), so ( R = sqrt{4^2 + (-3)^2} = sqrt{16 + 9} = sqrt{25} = 5 ). So, ( 4 cos alpha - 3 sin alpha = 5 cos(alpha + phi) ), where ( phi ) is such that ( cos phi = frac{4}{5} ) and ( sin phi = frac{3}{5} ). Wait, let me verify that.The formula is ( A cos alpha + B sin alpha = R cos(alpha - phi) ), where ( R = sqrt{A^2 + B^2} ), ( cos phi = frac{A}{R} ), and ( sin phi = frac{B}{R} ). But in our case, it's ( 4 cos alpha - 3 sin alpha ), which can be written as ( 4 cos alpha + (-3) sin alpha ). So, ( A = 4 ), ( B = -3 ), so ( R = 5 ), ( cos phi = frac{4}{5} ), ( sin phi = frac{-3}{5} ). Therefore, ( 4 cos alpha - 3 sin alpha = 5 cos(alpha + phi) ), where ( phi = arctanleft(frac{B}{A}right) = arctanleft(frac{-3}{4}right) ). So, ( phi ) is in the fourth quadrant, but since cosine is even and sine is odd, it might be easier to represent it as ( 5 cos(alpha + phi) ).Therefore, the expression ( 4 cos alpha - 3 sin alpha + 6 ) becomes ( 5 cos(alpha + phi) + 6 ). So, the distance is ( frac{|5 cos(alpha + phi) + 6|}{sqrt{2}} ).Now, to find the minimum distance, we need to minimize ( |5 cos(alpha + phi) + 6| ). Since the absolute value is always non-negative, the minimum occurs when ( 5 cos(alpha + phi) + 6 ) is as small as possible. The range of ( cos(alpha + phi) ) is between -1 and 1, so ( 5 cos(alpha + phi) ) ranges from -5 to 5. Therefore, ( 5 cos(alpha + phi) + 6 ) ranges from 1 to 11. So, the minimum value inside the absolute value is 1, and the maximum is 11.But wait, since we have ( |5 cos(alpha + phi) + 6| ), the minimum occurs when ( 5 cos(alpha + phi) + 6 ) is closest to zero. However, since ( 5 cos(alpha + phi) + 6 ) is always positive (because the minimum is 1), the minimum distance is when ( 5 cos(alpha + phi) + 6 ) is minimized, which is 1. Therefore, the minimum distance is ( frac{1}{sqrt{2}} ), which simplifies to ( frac{sqrt{2}}{2} ).Wait, let me double-check that. If ( 5 cos(alpha + phi) + 6 ) is minimized at 1, then the distance is ( frac{1}{sqrt{2}} ). But is that correct? Because the distance formula is ( frac{|expression|}{sqrt{2}} ), and the expression is minimized at 1, so yes, the minimum distance is ( frac{1}{sqrt{2}} ), which is ( frac{sqrt{2}}{2} ).Alternatively, maybe I can think of it as the distance from the ellipse to the line. The line is ( x - y + 6 = 0 ), and the ellipse is ( frac{x^2}{16} + frac{y^2}{9} = 1 ). To find the minimum distance, another approach is to use Lagrange multipliers, but that might be more complicated. Alternatively, I can parametrize the ellipse and then find the minimum distance as I did before.Wait, another thought: the minimum distance from the ellipse to the line is the same as the minimum value of the distance function over all points on the ellipse. So, by parametrizing the ellipse and expressing the distance in terms of a single variable ( alpha ), I can then find the minimum by taking the derivative and setting it to zero. But since I already expressed the distance as ( frac{|5 cos(alpha + phi) + 6|}{sqrt{2}} ), and since ( 5 cos(alpha + phi) + 6 ) has a minimum value of 1, the minimum distance is indeed ( frac{sqrt{2}}{2} ).Wait, but let me think again. If ( 5 cos(alpha + phi) + 6 ) can be as low as 1, then the distance is ( frac{1}{sqrt{2}} ). But is there a point on the ellipse where this occurs? Because sometimes, the minimal value might not be attainable due to the constraints of the ellipse.Wait, no, because the ellipse is a closed curve, and the distance function is continuous, so the minimum must be attained at some point on the ellipse. Therefore, the minimal distance is indeed ( frac{sqrt{2}}{2} ).Wait, but let me verify with another approach. Suppose I use Lagrange multipliers. Let me set up the function to minimize: ( f(x, y) = frac{|x - y + 6|}{sqrt{2}} ), subject to the constraint ( frac{x^2}{16} + frac{y^2}{9} = 1 ).But since the absolute value complicates things, maybe I can square the distance to make it easier. So, minimize ( (x - y + 6)^2 ) subject to ( frac{x^2}{16} + frac{y^2}{9} = 1 ).Using Lagrange multipliers, set up the gradient of ( f ) equal to lambda times the gradient of the constraint.Let me define ( f(x, y) = (x - y + 6)^2 ) and ( g(x, y) = frac{x^2}{16} + frac{y^2}{9} - 1 = 0 ).Compute gradients:( nabla f = [2(x - y + 6), -2(x - y + 6)] ).( nabla g = [frac{2x}{16}, frac{2y}{9}] = [frac{x}{8}, frac{2y}{9}] ).Set ( nabla f = lambda nabla g ):So,1. ( 2(x - y + 6) = lambda cdot frac{x}{8} ).2. ( -2(x - y + 6) = lambda cdot frac{2y}{9} ).Let me denote ( k = x - y + 6 ). Then, equation 1 becomes ( 2k = frac{lambda x}{8} ), so ( lambda = frac{16k}{x} ).Equation 2 becomes ( -2k = frac{2 lambda y}{9} ), so ( lambda = frac{-9k}{y} ).Therefore, ( frac{16k}{x} = frac{-9k}{y} ).Assuming ( k neq 0 ) (since if ( k = 0 ), the distance would be zero, but that's not possible because the line doesn't intersect the ellipse, as we'll see later), we can divide both sides by ( k ):( frac{16}{x} = frac{-9}{y} ).Cross-multiplying:( 16y = -9x ).So, ( y = -frac{9}{16}x ).Now, substitute ( y = -frac{9}{16}x ) into the ellipse equation:( frac{x^2}{16} + frac{(-frac{9}{16}x)^2}{9} = 1 ).Simplify:( frac{x^2}{16} + frac{frac{81}{256}x^2}{9} = 1 ).Simplify the second term:( frac{81}{256 cdot 9}x^2 = frac{9}{256}x^2 ).So, the equation becomes:( frac{x^2}{16} + frac{9x^2}{256} = 1 ).Convert to a common denominator:( frac{16x^2}{256} + frac{9x^2}{256} = 1 ).Combine terms:( frac{25x^2}{256} = 1 ).So, ( x^2 = frac{256}{25} ), which gives ( x = pm frac{16}{5} ).Therefore, ( x = frac{16}{5} ) or ( x = -frac{16}{5} ).Corresponding ( y ) values:For ( x = frac{16}{5} ), ( y = -frac{9}{16} cdot frac{16}{5} = -frac{9}{5} ).For ( x = -frac{16}{5} ), ( y = -frac{9}{16} cdot (-frac{16}{5}) = frac{9}{5} ).So, the points are ( (frac{16}{5}, -frac{9}{5}) ) and ( (-frac{16}{5}, frac{9}{5}) ).Now, compute the distance from these points to the line ( x - y + 6 = 0 ).First, for ( (frac{16}{5}, -frac{9}{5}) ):Distance ( d = frac{|frac{16}{5} - (-frac{9}{5}) + 6|}{sqrt{2}} = frac{|frac{16}{5} + frac{9}{5} + 6|}{sqrt{2}} = frac{|frac{25}{5} + 6|}{sqrt{2}} = frac{|5 + 6|}{sqrt{2}} = frac{11}{sqrt{2}} ).For ( (-frac{16}{5}, frac{9}{5}) ):Distance ( d = frac{|-frac{16}{5} - frac{9}{5} + 6|}{sqrt{2}} = frac{|-frac{25}{5} + 6|}{sqrt{2}} = frac{|-5 + 6|}{sqrt{2}} = frac{1}{sqrt{2}} ).So, the distances are ( frac{11}{sqrt{2}} ) and ( frac{1}{sqrt{2}} ). Therefore, the minimum distance is ( frac{1}{sqrt{2}} ), which simplifies to ( frac{sqrt{2}}{2} ).Wait, but earlier when I used the parametrization method, I also got ( frac{sqrt{2}}{2} ). So, both methods agree. Therefore, the minimum distance is indeed ( frac{sqrt{2}}{2} ).But just to be thorough, let me check if the line ( x - y + 6 = 0 ) intersects the ellipse. If it does, then the minimum distance would be zero, but since we found a positive minimum distance, it must not intersect.To check for intersection, solve the system:1. ( x - y + 6 = 0 ) => ( y = x + 6 ).2. ( frac{x^2}{16} + frac{y^2}{9} = 1 ).Substitute ( y = x + 6 ) into the ellipse equation:( frac{x^2}{16} + frac{(x + 6)^2}{9} = 1 ).Multiply through by 144 (the least common multiple of 16 and 9):( 9x^2 + 16(x^2 + 12x + 36) = 144 ).Expand:( 9x^2 + 16x^2 + 192x + 576 = 144 ).Combine like terms:( 25x^2 + 192x + 576 - 144 = 0 ).Simplify:( 25x^2 + 192x + 432 = 0 ).Compute the discriminant:( D = 192^2 - 4 cdot 25 cdot 432 ).Calculate:( 192^2 = 36864 ).( 4 cdot 25 cdot 432 = 100 cdot 432 = 43200 ).So, ( D = 36864 - 43200 = -6336 ).Since the discriminant is negative, there are no real solutions, meaning the line does not intersect the ellipse. Therefore, the minimum distance is indeed ( frac{sqrt{2}}{2} ).So, summarizing:1. The rectangular equation of line ( l ) is ( x - y + 6 = 0 ).2. The minimum distance from a point ( P ) on the ellipse ( C ) to line ( l ) is ( frac{sqrt{2}}{2} ).

💡First, I need to determine the current number of books Jason and Mary have together.Jason currently has 18 books. He plans to buy 8 more books next month, so he will have a total of 18 + 8 = 26 books.Mary initially has 42 books but lost 6, so she now has 42 - 6 = 36 books. She will borrow 5 more books next week, bringing her total to 36 + 5 = 41 books.Adding Jason's and Mary's current books together: 18 + 36 = 54 books.Next month, after both have acquired more books, their total will be 26 + 41 = 67 books.