💡Alright, so I have this problem here where I need to define a function ( f(x) ) on ( mathbb{R} ) that satisfies the differential inequality ( f'(x) > 1 - f(x) ). Then, I have to determine which of the given options A, B, C, or D is correct based on this condition. First, let me try to understand the inequality ( f'(x) > 1 - f(x) ). This is a first-order linear differential inequality. I remember that for differential equations of the form ( f'(x) + P(x)f(x) = Q(x) ), we can use integrating factors to solve them. Maybe I can apply a similar approach here.Let me rewrite the inequality:( f'(x) + f(x) > 1 )So, comparing this to the standard linear differential equation ( f'(x) + P(x)f(x) = Q(x) ), I can see that ( P(x) = 1 ) and ( Q(x) = 1 ). The integrating factor ( mu(x) ) would be ( e^{int P(x) dx} = e^{int 1 dx} = e^{x} ).Multiplying both sides of the inequality by the integrating factor ( e^{x} ):( e^{x}f'(x) + e^{x}f(x) > e^{x} )Notice that the left side is the derivative of ( e^{x}f(x) ). So, this simplifies to:( frac{d}{dx} [e^{x}f(x)] > e^{x} )Now, let's integrate both sides from some point ( a ) to ( x ). But since the problem doesn't specify initial conditions, I might need another approach. Maybe I can define a new function to capture this behavior.Let me define ( g(x) = e^{x}f(x) - e^{x} ). Then, let's compute its derivative:( g'(x) = e^{x}f(x) + e^{x}f'(x) - e^{x} )Simplify this:( g'(x) = e^{x}f(x) + e^{x}f'(x) - e^{x} = e^{x}(f(x) + f'(x) - 1) )From the original inequality ( f'(x) > 1 - f(x) ), we can rearrange it to ( f'(x) + f(x) > 1 ). Therefore, ( f(x) + f'(x) - 1 > 0 ).Substituting back into ( g'(x) ):( g'(x) = e^{x}(f(x) + f'(x) - 1) > 0 )Since ( e^{x} ) is always positive, ( g'(x) > 0 ) implies that ( g(x) ) is an increasing function on ( mathbb{R} ).Now, let's analyze the options given:**Option A: ( ef(1) - e > e^{2}f(2) - e^{2} )**This can be rewritten as:( e(f(1) - 1) > e^{2}(f(2) - 1) )Dividing both sides by ( e ) (since ( e > 0 )):( f(1) - 1 > e(f(2) - 1) )But since ( g(x) = e^{x}(f(x) - 1) ) is increasing, ( g(1) < g(2) ). Therefore:( e^{1}(f(1) - 1) < e^{2}(f(2) - 1) )Which implies:( e(f(1) - 1) < e^{2}(f(2) - 1) )So, the inequality in Option A is reversed. Hence, Option A is incorrect.**Option B: ( e^{2015}f(2015) - e^{2015} > e^{2016}f(2016) - e^{2016} )**Similarly, this can be written as:( e^{2015}(f(2015) - 1) > e^{2016}(f(2016) - 1) )Dividing both sides by ( e^{2015} ):( f(2015) - 1 > e(f(2016) - 1) )Again, since ( g(x) ) is increasing, ( g(2015) < g(2016) ):( e^{2015}(f(2015) - 1) < e^{2016}(f(2016) - 1) )Which implies:( f(2015) - 1 < e(f(2016) - 1) )So, the inequality in Option B is also reversed. Hence, Option B is incorrect.**Option C: ( e^{2}f(2) + e^{2} > e f(1) + e )**Let me rearrange this:( e^{2}(f(2) + 1) > e(f(1) + 1) )Dividing both sides by ( e ):( e(f(2) + 1) > f(1) + 1 )Hmm, this seems a bit different. Maybe I need to define another function to analyze this. Let me consider ( h(x) = e^{x}f(x) + e^{x} ). Then, compute its derivative:( h'(x) = e^{x}f(x) + e^{x}f'(x) + e^{x} = e^{x}(f(x) + f'(x) + 1) )From the original inequality ( f'(x) > 1 - f(x) ), adding ( f(x) ) to both sides gives ( f'(x) + f(x) > 1 ). Therefore, ( f(x) + f'(x) + 1 > 2 ). So, ( h'(x) = e^{x}(f(x) + f'(x) + 1) > 2e^{x} > 0 ).Thus, ( h(x) ) is also an increasing function. Therefore, ( h(2) > h(1) ).Which means:( e^{2}f(2) + e^{2} > e f(1) + e )So, Option C is correct.**Option D: ( e^{2016}f(2016) + e^{2016} < e^{2015}f(2015) + e^{2015} )**Similarly, this can be written as:( e^{2016}(f(2016) + 1) < e^{2015}(f(2015) + 1) )Dividing both sides by ( e^{2015} ):( e(f(2016) + 1) < f(2015) + 1 )But since ( h(x) ) is increasing, ( h(2016) > h(2015) ):( e^{2016}(f(2016) + 1) > e^{2015}(f(2015) + 1) )Which implies:( e(f(2016) + 1) > f(2015) + 1 )So, the inequality in Option D is reversed. Hence, Option D is incorrect.Therefore, after analyzing all the options, the correct one is Option C.**Final Answer**boxed{C}

💡First, I need to determine all possible sums when rolling a cube twice. The smallest possible sum is 2 (1+1), and the largest is 12 (6+6). Next, I identify which of these sums are prime numbers. The prime numbers between 2 and 12 are 2, 3, 5, 7, and 11.For each prime sum, I list all possible combinations of rolls that result in that sum:- Sum of 2: (1,1)- Sum of 3: (1,2), (2,1)- Sum of 5: (1,4), (4,1), (2,3), (3,2)- Sum of 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3)- Sum of 11: (5,6), (6,5)Adding up all these combinations gives a total of 15 favorable outcomes.Since each roll of the cube has 6 possible outcomes, there are a total of 6 * 6 = 36 possible combinations when rolling the cube twice.Finally, the probability that the sum of two rolls is prime is the number of favorable outcomes divided by the total number of possible outcomes:e = 15 / 36 = 5 / 12

💡First, I need to determine the fraction of yellow and blue flowers in the garden. It's given that nine tenths of the flowers are yellow, so:[text{Fraction of yellow flowers} = frac{9}{10}]The remaining flowers are blue:[text{Fraction of blue flowers} = frac{1}{10}]Next, I'll find out how many of these yellow and blue flowers are daisies.For the yellow flowers, half are tulips, so the other half must be daisies:[text{Fraction of yellow daisies} = frac{9}{10} times frac{1}{2} = frac{9}{20}]For the blue flowers, four fifths are daisies:[text{Fraction of blue daisies} = frac{1}{10} times frac{4}{5} = frac{4}{50} = frac{2}{25}]Now, I'll add the fractions of yellow and blue daisies to find the total fraction of daisies:[text{Total fraction of daisies} = frac{9}{20} + frac{2}{25} = frac{45}{100} + frac{8}{100} = frac{53}{100}]Converting this fraction to a percentage:[frac{53}{100} times 100% = 53%]Therefore, the percent of the flowers that are daisies is 53%.

💡Okay, so I have this problem about oblique coordinates. Hmm, oblique coordinates are a bit different from the usual Cartesian coordinates because the axes aren't perpendicular. In this case, the angle between the x and y axes is 60 degrees. I need to solve two parts: first, find the magnitude of vector OP when P has coordinates (3, -2), and second, find the equation of a circle centered at O with radius 2 in this oblique coordinate system.Starting with part (1). The coordinates of point P are given as (3, -2), which means that the vector OP is 3 times the unit vector e1 plus (-2) times the unit vector e2. So, OP = 3e1 - 2e2. To find the magnitude of OP, I need to compute |OP|.I remember that the magnitude of a vector can be found using the dot product. Specifically, |OP|² = OP · OP. So, let's compute that.OP · OP = (3e1 - 2e2) · (3e1 - 2e2). Expanding this, I get 3e1 · 3e1 + 3e1 · (-2e2) + (-2e2) · 3e1 + (-2e2) · (-2e2). Simplifying each term:- 3e1 · 3e1 = 9(e1 · e1) = 9|e1|². Since e1 is a unit vector, |e1| = 1, so this term is 9.- 3e1 · (-2e2) = -6(e1 · e2). Similarly, (-2e2) · 3e1 = -6(e2 · e1). So combined, these two terms give -12(e1 · e2).- (-2e2) · (-2e2) = 4(e2 · e2) = 4|e2|². Again, since e2 is a unit vector, this is 4.So putting it all together: |OP|² = 9 - 12(e1 · e2) + 4. That simplifies to |OP|² = 13 - 12(e1 · e2).Now, I need to find e1 · e2. Since e1 and e2 are unit vectors with an angle of 60 degrees between them, the dot product is |e1||e2|cos(theta) = 1*1*cos(60°) = 0.5.So substituting back in: |OP|² = 13 - 12*(0.5) = 13 - 6 = 7. Therefore, |OP| = sqrt(7). That seems straightforward.Moving on to part (2). I need to find the equation of a circle with center O and radius 2 in this oblique coordinate system. In Cartesian coordinates, the equation would be x² + y² = r², but since the axes are oblique, the equation will be different.In general, for an oblique coordinate system, the equation of a circle can be found by considering the magnitude of the vector OP. So, |OP| = 2. Using the same approach as in part (1), |OP|² = 4.Expressing OP as x e1 + y e2, then |OP|² = (x e1 + y e2) · (x e1 + y e2). Expanding this, we get x²(e1 · e1) + 2xy(e1 · e2) + y²(e2 · e2). Since e1 and e2 are unit vectors, e1 · e1 = 1 and e2 · e2 = 1. The cross term is 2xy(e1 · e2) = 2xy*(0.5) = xy.So putting it all together: |OP|² = x² + xy + y². And since |OP| = 2, we have x² + xy + y² = 4.Wait, let me double-check that. If the angle is 60 degrees, then the dot product is 0.5, so the cross term is 2xy*(0.5) = xy. So yes, the equation becomes x² + xy + y² = 4. That makes sense.I think that's it. So, for part (1), the magnitude is sqrt(7), and for part (2), the equation is x² + xy + y² = 4.**Final Answer**(1) boxed{sqrt{7}} (2) boxed{x^2 + xy + y^2 = 4}

💡Alright, so I have this quadratic function f(x) = x² + 4x + m, where m is a real constant. The problem says that its graph intersects the coordinate axes at three points, and there's a circle C that passes through these three points. I need to find the range of m for part (I) and then prove that circle C passes through a fixed point independent of m, and find that point for part (II).Starting with part (I). The quadratic function is a parabola. It intersects the coordinate axes, which are the x-axis and y-axis. Since it's a quadratic, it can intersect the x-axis at two points or one point (if it's tangent) or no points. But since it's given that it intersects the coordinate axes at three points, that must mean it intersects the x-axis at two distinct points and the y-axis at one point.So, first, let's find the y-intercept. The y-intercept occurs when x = 0. Plugging in x = 0 into f(x), we get f(0) = 0 + 0 + m = m. So, the y-intercept is at (0, m).Next, the x-intercepts. These occur where f(x) = 0, so solving x² + 4x + m = 0. For a quadratic equation ax² + bx + c = 0, the discriminant is D = b² - 4ac. For real and distinct roots, the discriminant must be positive.In this case, a = 1, b = 4, c = m. So, discriminant D = 16 - 4*1*m = 16 - 4m. For two distinct real roots, D > 0, so 16 - 4m > 0. Solving this, 16 > 4m => 4 > m => m < 4.But also, since the graph intersects the y-axis at (0, m), m cannot be zero because if m were zero, the y-intercept would be at the origin, which is also on the x-axis. So, in that case, the graph would intersect the x-axis at two points (including the origin) and the y-axis at the origin, but that would be only two distinct points, not three. Therefore, m cannot be zero. So, m ≠ 0.Putting it together, m must be less than 4 and not equal to zero. So, the range of m is m ∈ (-∞, 0) ∪ (0, 4).Wait, let me double-check that. If m = 0, the quadratic becomes x² + 4x = 0, which factors to x(x + 4) = 0, so roots at x = 0 and x = -4. So, the graph would intersect the x-axis at (0, 0) and (-4, 0), and the y-axis at (0, 0). So, all three points coincide at (0, 0) and (-4, 0). So, actually, it's only two distinct points: (0, 0) and (-4, 0). So, yes, m cannot be zero because we need three distinct points. So, m must be less than 4 and not equal to zero.Therefore, the range of m is m < 4 and m ≠ 0. So, in interval notation, that's (-∞, 0) ∪ (0, 4).Okay, that seems solid for part (I). Now, moving on to part (II). I need to prove that the circle C passes through a fixed point independent of m and find that point.First, let's recall that a circle passing through three non-collinear points is uniquely determined. Here, the three points are the two x-intercepts and the y-intercept of the parabola. Let's denote these points as A, B, and C.Let me denote the points:- A: (0, m) – the y-intercept.- B: (x₁, 0) – one x-intercept.- C: (x₂, 0) – the other x-intercept.We know from the quadratic equation that x₁ + x₂ = -b/a = -4, and x₁x₂ = c/a = m.So, x₁ and x₂ are the roots, so x₁ + x₂ = -4, x₁x₂ = m.Now, the circle passes through these three points. Let's denote the general equation of a circle as x² + y² + Dx + Ey + F = 0, where D, E, F are constants.Since points A, B, and C lie on this circle, substituting each into the circle's equation will give us equations to solve for D, E, F in terms of m.First, let's substitute point A: (0, m).Plugging into the circle equation: 0² + m² + D*0 + E*m + F = 0 => m² + E*m + F = 0. Let's call this Equation (1).Next, substitute point B: (x₁, 0).Plugging into the circle equation: x₁² + 0² + D*x₁ + E*0 + F = 0 => x₁² + D*x₁ + F = 0. Let's call this Equation (2).Similarly, substitute point C: (x₂, 0).Plugging into the circle equation: x₂² + 0² + D*x₂ + E*0 + F = 0 => x₂² + D*x₂ + F = 0. Let's call this Equation (3).So, now we have three equations:1. m² + E*m + F = 02. x₁² + D*x₁ + F = 03. x₂² + D*x₂ + F = 0We can subtract Equation (2) from Equation (3) to eliminate F:(x₂² - x₁²) + D*(x₂ - x₁) = 0Factor x₂² - x₁² as (x₂ - x₁)(x₂ + x₁):(x₂ - x₁)(x₂ + x₁) + D*(x₂ - x₁) = 0Factor out (x₂ - x₁):(x₂ - x₁)(x₂ + x₁ + D) = 0Since x₂ ≠ x₁ (because the quadratic has two distinct roots, as m < 4 and m ≠ 0), so x₂ - x₁ ≠ 0. Therefore, the other factor must be zero:x₂ + x₁ + D = 0But from earlier, x₁ + x₂ = -4, so:-4 + D = 0 => D = 4So, we found D = 4.Now, let's substitute D = 4 into Equation (2):x₁² + 4*x₁ + F = 0 => F = -x₁² - 4x₁Similarly, from Equation (3):x₂² + 4*x₂ + F = 0 => F = -x₂² - 4x₂Therefore, -x₁² - 4x₁ = -x₂² - 4x₂Simplify:x₂² - x₁² + 4x₂ - 4x₁ = 0Factor x₂² - x₁² as (x₂ - x₁)(x₂ + x₁) and 4(x₂ - x₁):(x₂ - x₁)(x₂ + x₁ + 4) = 0Again, since x₂ ≠ x₁, we have:x₂ + x₁ + 4 = 0But x₁ + x₂ = -4, so:-4 + 4 = 0 => 0 = 0Which is always true, so no new information here.Now, let's go back to Equation (1):m² + E*m + F = 0We have F in terms of x₁ or x₂. Let's express F as -x₁² - 4x₁.But from the quadratic equation, x₁² + 4x₁ + m = 0, so x₁² = -4x₁ - mTherefore, F = -(-4x₁ - m) - 4x₁ = 4x₁ + m - 4x₁ = mSo, F = mTherefore, Equation (1) becomes:m² + E*m + m = 0Simplify:m² + (E + 1)m = 0Factor:m(m + E + 1) = 0Since m ≠ 0 (from part (I)), we can divide both sides by m:m + E + 1 = 0 => E = -m - 1So, now we have D = 4, E = -m -1, F = m.Therefore, the equation of the circle is:x² + y² + 4x + (-m -1)y + m = 0Simplify:x² + y² + 4x - (m + 1)y + m = 0Now, let's write this as:x² + y² + 4x - y - m y + m = 0Group the terms with m:x² + y² + 4x - y + m(-y + 1) = 0So, the equation is:x² + y² + 4x - y + m(1 - y) = 0Now, since this equation must hold for all values of m in the range we found earlier, the coefficients of m must be zero for the equation to hold regardless of m. However, in this case, the equation is specific to each m, but we are to find a point that lies on the circle for all m.Alternatively, we can think of this as a family of circles parameterized by m, and we need to find a point (x, y) that satisfies the equation for all m.So, to find such a point, the equation must hold true regardless of the value of m. That means the coefficient of m must be zero, and the remaining equation must also be satisfied.So, set the coefficient of m to zero:1 - y = 0 => y = 1And the remaining equation:x² + y² + 4x - y = 0But since y = 1, substitute y = 1 into this equation:x² + (1)² + 4x - 1 = 0 => x² + 1 + 4x - 1 = 0 => x² + 4x = 0Factor:x(x + 4) = 0 => x = 0 or x = -4Therefore, the points are (0, 1) and (-4, 1).So, regardless of the value of m, the circle passes through these two points. Therefore, the fixed points are (0, 1) and (-4, 1).Wait, but the problem says "a fixed point," singular. So, maybe I need to check if both are fixed or if only one is considered.Looking back, the circle passes through three points: (0, m), (x₁, 0), and (x₂, 0). The circle's equation depends on m, but we found that regardless of m, it must pass through (0, 1) and (-4, 1). So, these are two fixed points.But the problem says "a fixed point," so maybe it's expecting one point. But from our derivation, there are two such points. Maybe both are acceptable, or perhaps the problem expects both.Alternatively, perhaps I made a miscalculation. Let me double-check.We had the circle equation:x² + y² + 4x - (m + 1)y + m = 0Which can be rewritten as:x² + y² + 4x - y - m y + m = 0Then, grouping terms with m:x² + y² + 4x - y + m(1 - y) = 0So, for this equation to hold for all m, the coefficient of m must be zero, so 1 - y = 0 => y = 1.Then, substituting y = 1 into the remaining equation:x² + 1 + 4x - 1 = 0 => x² + 4x = 0 => x(x + 4) = 0 => x = 0 or x = -4.So, indeed, the circle must pass through both (0, 1) and (-4, 1) regardless of m.Therefore, the fixed points are (0, 1) and (-4, 1).But the problem says "a fixed point," so maybe it's expecting both? Or perhaps it's a typo, and it's supposed to say "fixed points." Alternatively, maybe only one of them is considered, but from the derivation, both are fixed.Alternatively, perhaps the problem is considering that the circle passes through both, so both are fixed points.Therefore, the conclusion is that the circle passes through the fixed points (0, 1) and (-4, 1), independent of m.So, summarizing:(I) The range of m is m < 4 and m ≠ 0, so m ∈ (-∞, 0) ∪ (0, 4).(II) The circle passes through the fixed points (0, 1) and (-4, 1).

💡Okay, so I have this problem here about finding the monotonic increasing interval of the function y equals x cubed. The options are A: negative infinity to zero, B: zero to positive infinity, C: negative infinity to positive infinity, and D: negative infinity to zero union zero to positive infinity. Hmm, let me think. I remember that to find where a function is increasing or decreasing, we usually look at its derivative. The derivative tells us the slope of the function at any point, right? So if the derivative is positive, the function is increasing, and if it's negative, the function is decreasing.Alright, so let me find the derivative of y equals x cubed. The derivative of x cubed is 3x squared. Okay, so y prime equals 3x squared. Now, 3x squared is always non-negative because any real number squared is non-negative, and multiplying by 3 keeps it non-negative. So, y prime is greater than or equal to zero for all real numbers x.Wait, but does that mean the function is increasing everywhere? Because if the derivative is always non-negative, that suggests that the function is either increasing or staying constant. But in this case, y prime equals zero only when x is zero. So, at x equals zero, the slope is zero, but everywhere else, the slope is positive. So, does that mean the function is increasing on the entire real line? Because even though the slope is zero at x equals zero, it's still not decreasing anywhere. It's just momentarily flat at that point. So, I think the function is increasing on negative infinity to positive infinity. That would be option C.But let me double-check. Maybe I'm missing something. If I look at the graph of y equals x cubed, it's a continuous curve that passes through the origin. For negative x values, as x increases, y increases as well. Similarly, for positive x values, as x increases, y increases. So, it seems like the function is increasing throughout its entire domain.Wait, but what about around x equals zero? Since the derivative is zero there, is it still considered increasing? I think so because the function doesn't decrease on either side of zero. It just has a horizontal tangent at that point. So, the function is still increasing overall.Therefore, I think the correct answer is C: negative infinity to positive infinity. But just to be thorough, let me consider the other options. Option A is negative infinity to zero, which would only cover the left side of the graph. But the function is also increasing on the right side, so that can't be the full answer. Option B is zero to positive infinity, which is the right side, but again, it's only half of the function. Option D is the union of negative infinity to zero and zero to positive infinity, which is essentially the same as negative infinity to positive infinity, except it excludes the point at zero. But since the function is increasing at zero as well, because the derivative is zero there, I think it's still included in the increasing interval.So, yeah, I'm pretty confident that the answer is C. The function y equals x cubed is increasing on the entire real line.