💡Alright, so I have this problem where I need to convert the rectangular coordinates (4, -4√3, 4) to spherical coordinates. I remember that spherical coordinates are represented as (ρ, θ, φ), where ρ is the radius, θ is the azimuthal angle in the xy-plane, and φ is the polar angle from the positive z-axis. First, I need to recall the formulas to convert from rectangular (x, y, z) to spherical coordinates. From what I remember, the formulas are:1. ρ = √(x² + y² + z²)2. θ = arctan(y / x)3. φ = arccos(z / ρ)Okay, so let me start by calculating ρ. Plugging in the values:ρ = √(4² + (-4√3)² + 4²)Calculating each term:4² = 16(-4√3)² = (-4)² * (√3)² = 16 * 3 = 484² = 16So, adding them up: 16 + 48 + 16 = 80Therefore, ρ = √80. Simplifying √80, since 80 = 16 * 5, so √80 = 4√5. Got that, ρ is 4√5.Next, let's find φ. The formula is φ = arccos(z / ρ). Here, z is 4, and ρ is 4√5.So, φ = arccos(4 / (4√5)) = arccos(1 / √5)I can leave it as arccos(1/√5) for now, or maybe express it in terms of inverse trigonometric functions, but I think it's fine as is for now.Now, the tricky part is finding θ. The formula is θ = arctan(y / x). Here, y is -4√3 and x is 4.So, θ = arctan((-4√3) / 4) = arctan(-√3)Hmm, arctan(-√3). I remember that arctan(√3) is π/3 because tan(π/3) = √3. But since we have a negative value, arctan(-√3) would be -π/3. However, θ is supposed to be between 0 and 2π, so I need to adjust this negative angle to get it within the correct range.Since the point (4, -4√3, 4) is in the fourth quadrant of the xy-plane (because x is positive and y is negative), θ should be in the fourth quadrant. The reference angle is π/3, so θ = 2π - π/3 = 5π/3.Let me double-check that. If θ is 5π/3, then:cos(5π/3) = cos(π/3) = 1/2sin(5π/3) = -sin(π/3) = -√3/2Which matches the signs of x and y in our original coordinates. So, θ = 5π/3 is correct.Putting it all together, the spherical coordinates are (4√5, 5π/3, arccos(1/√5)).Wait, should I express φ in a different form? Let me see. arccos(1/√5) is fine, but sometimes it's expressed using arcsin or arctan. Let me verify if there's another way to write it.Alternatively, since φ is the angle from the positive z-axis, and we have z = 4, and ρ = 4√5, we can also find φ using the relationship:sin(φ) = √(x² + y²) / ρCalculating √(x² + y²):√(4² + (-4√3)²) = √(16 + 48) = √64 = 8So, sin(φ) = 8 / (4√5) = 2 / √5Therefore, φ = arcsin(2 / √5)But since φ is between 0 and π, both arccos(1/√5) and arcsin(2/√5) are valid expressions for φ. However, arccos(1/√5) is probably the more straightforward expression here because it directly uses z and ρ.Just to be thorough, let me check if arccos(1/√5) is equal to arcsin(2/√5). Using the identity that sin(φ) = √(1 - cos²(φ)):If cos(φ) = 1/√5, then sin(φ) = √(1 - (1/5)) = √(4/5) = 2/√5Yes, that's consistent. So both expressions are correct, but arccos(1/√5) is simpler in this context.Alright, so I think I've got all the components:- ρ = 4√5- θ = 5π/3- φ = arccos(1/√5)I should probably rationalize the denominator for 1/√5 to make it look neater, but in terms of exact value, it's fine as is. Alternatively, I can write it as √5/5, but since it's inside the arccos, either form is acceptable.Let me recap:1. Calculated ρ correctly by taking the square root of the sum of squares of x, y, z.2. Found θ by taking arctan(y/x), adjusted for the correct quadrant.3. Determined φ using arccos(z/ρ), which gave me arccos(1/√5).Everything seems to check out. I don't think I made any calculation errors, but let me quickly verify each step:- ρ: √(16 + 48 + 16) = √80 = 4√5. Correct.- θ: arctan(-√3) = -π/3, adjusted to 5π/3. Correct.- φ: arccos(4 / (4√5)) = arccos(1/√5). Correct.Yes, all steps are correct. So, I'm confident that the spherical coordinates are (4√5, 5π/3, arccos(1/√5)).**Final Answer**The spherical coordinates are boxed{left(4sqrt{5}, frac{5pi}{3}, cos^{-1}left(frac{1}{sqrt{5}}right)right)}.

💡Okay, so I have this geometry problem here. Let me try to figure it out step by step. We have triangle ABC with sides AB = 5 and AC = 7. M is the midpoint of side BC, and the length of AM is 4. We need to find the length of BC. The answer choices are given, so that might help if I get stuck or want to check my work.First, let me visualize the triangle. ABC is a triangle where AB is 5 units, AC is 7 units, and M is the midpoint of BC. So, BM equals MC. Since M is the midpoint, BC is twice the length of BM or MC. If I can find BM, I can just double it to get BC.I remember there's a theorem related to the length of a median in a triangle. I think it's called Apollonius's Theorem. Let me recall what it says. Apollonius's Theorem states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. Wait, that might be a bit confusing. Let me write it down to clarify. For triangle ABC, with median AM, Apollonius's Theorem would state:AB² + AC² = 2AM² + 2BM²Yes, that seems right. So, plugging in the values we have:AB is 5, so AB² is 25.AC is 7, so AC² is 49.AM is 4, so AM² is 16.BM is half of BC, which we can denote as x. So, BM² is x².Putting it all into the equation:25 + 49 = 2*(16) + 2*(x²)Let me compute the left side first: 25 + 49 is 74.On the right side: 2*16 is 32, and 2*x² is 2x². So, the equation becomes:74 = 32 + 2x²Now, subtract 32 from both sides to isolate the term with x²:74 - 32 = 2x²42 = 2x²Divide both sides by 2:21 = x²So, x is the square root of 21. But x is BM, which is half of BC. Therefore, BC is 2 times x, which is 2*sqrt(21).Let me double-check my steps to make sure I didn't make a mistake.1. I used Apollonius's Theorem correctly: AB² + AC² = 2AM² + 2BM².2. Plugged in the given lengths: 25 + 49 = 2*16 + 2x².3. Calculated the left side: 74.4. Calculated the right side: 32 + 2x².5. Subtracted 32 from 74 to get 42 = 2x².6. Divided by 2 to get x² = 21.7. Took the square root to find x = sqrt(21).8. Remembered that BC is twice BM, so BC = 2*sqrt(21).Everything seems to check out. I don't see any errors in my calculations or reasoning. So, I think the length of BC is 2 times the square root of 21.**Final Answer**The length of side BC is boxed{2sqrt{21}}.

💡First, determine the number of green M&Ms remaining after Carter eats 12 of them:20 green M&Ms - 12 eaten = 8 green M&Ms.Next, calculate the total number of M&Ms in the jar after the changes. Since the probability of picking a green M&M is 25%, and there are 8 green M&Ms:Total M&Ms = 8 green M&Ms / 0.25 = 32 M&Ms.Subtract the known quantities of green and yellow M&Ms to find the number of red M&Ms remaining:32 total M&Ms - 8 green M&Ms - 14 yellow M&Ms = 10 red M&Ms.Determine how many red M&Ms the sister ate:20 initial red M&Ms - 10 remaining red M&Ms = 10 red M&Ms eaten.Finally, calculate the ratio of red M&Ms eaten to the initial number:10 eaten : 20 initial = 1:2.

💡Okay, so I have this problem where I need to find the value of 'a' such that the minimum value of the function f(x) = |x + 1| + |2x + a| is 3. Hmm, okay, let's try to figure this out step by step.First, I remember that absolute value functions can be tricky because they have corners or points where the expression inside the absolute value changes sign. So, for f(x) = |x + 1| + |2x + a|, there are two critical points where the expressions inside the absolute values change their behavior. These points are when x + 1 = 0 and when 2x + a = 0. Solving these, I get x = -1 and x = -a/2. So, these are the points where the function might change its slope.Now, to find the minimum value of f(x), I should analyze the function in different intervals determined by these critical points. The critical points divide the real line into intervals, and in each interval, the function f(x) will be linear because the absolute value expressions will either be positive or negative, and their sum will be a linear function.So, let's consider the two critical points: x = -1 and x = -a/2. Depending on the value of 'a', these two points can be in different orders on the number line. That is, -a/2 could be to the left or right of -1. Therefore, I need to consider two cases:1. Case 1: -a/2 > -1, which implies that a < 2.2. Case 2: -a/2 < -1, which implies that a > 2.If a = 2, then both critical points coincide at x = -1, so that's a special case which I can handle separately.Let me start with Case 1: a < 2, so -a/2 > -1. This means that the critical points are ordered as x = -1 and then x = -a/2, with -a/2 being to the right of -1 on the number line.So, in this case, the function f(x) can be broken down into three intervals:- For x ≤ -1: Both x + 1 and 2x + a are negative or zero. Therefore, |x + 1| = -(x + 1) and |2x + a| = -(2x + a). So, f(x) = -(x + 1) - (2x + a) = -3x -1 -a.- For -1 < x < -a/2: Here, x + 1 is positive, but 2x + a is still negative. Therefore, |x + 1| = x + 1 and |2x + a| = -(2x + a). So, f(x) = (x + 1) - (2x + a) = -x + 1 - a.- For x ≥ -a/2: Both x + 1 and 2x + a are positive. Therefore, |x + 1| = x + 1 and |2x + a| = 2x + a. So, f(x) = (x + 1) + (2x + a) = 3x + 1 + a.Now, let's analyze the behavior of f(x) in each interval.For x ≤ -1: f(x) = -3x -1 -a. The coefficient of x is -3, which is negative, so the function is decreasing as x increases in this interval.For -1 < x < -a/2: f(x) = -x + 1 - a. The coefficient of x is -1, still negative, so the function continues to decrease as x increases in this interval.For x ≥ -a/2: f(x) = 3x + 1 + a. The coefficient of x is 3, which is positive, so the function starts increasing as x increases in this interval.Therefore, the function f(x) is decreasing up to x = -a/2 and then starts increasing. So, the minimum value occurs at x = -a/2.So, let's compute f(-a/2):f(-a/2) = |(-a/2) + 1| + |2*(-a/2) + a|.Simplify each term:First term: |(-a/2) + 1| = |1 - a/2|.Second term: |2*(-a/2) + a| = |-a + a| = |0| = 0.Wait, that's interesting. So, f(-a/2) = |1 - a/2| + 0 = |1 - a/2|.But we are given that the minimum value of f(x) is 3. So, |1 - a/2| = 3.Therefore, 1 - a/2 = 3 or 1 - a/2 = -3.Solving the first equation: 1 - a/2 = 3 => -a/2 = 2 => a = -4.Solving the second equation: 1 - a/2 = -3 => -a/2 = -4 => a = 8.But wait, in Case 1, we assumed that a < 2. So, a = -4 is valid because -4 < 2, but a = 8 is not because 8 > 2. So, in this case, a can only be -4.Now, let's move to Case 2: a > 2, so -a/2 < -1. This means that the critical points are ordered as x = -a/2 and then x = -1, with -a/2 being to the left of -1 on the number line.So, in this case, the function f(x) can be broken down into three intervals:- For x ≤ -a/2: Both x + 1 and 2x + a are negative or zero. Therefore, |x + 1| = -(x + 1) and |2x + a| = -(2x + a). So, f(x) = -(x + 1) - (2x + a) = -3x -1 -a.- For -a/2 < x < -1: Here, 2x + a is positive, but x + 1 is still negative. Therefore, |x + 1| = -(x + 1) and |2x + a| = 2x + a. So, f(x) = -(x + 1) + (2x + a) = x -1 + a.- For x ≥ -1: Both x + 1 and 2x + a are positive. Therefore, |x + 1| = x + 1 and |2x + a| = 2x + a. So, f(x) = (x + 1) + (2x + a) = 3x + 1 + a.Now, let's analyze the behavior of f(x) in each interval.For x ≤ -a/2: f(x) = -3x -1 -a. The coefficient of x is -3, which is negative, so the function is decreasing as x increases in this interval.For -a/2 < x < -1: f(x) = x -1 + a. The coefficient of x is 1, which is positive, so the function is increasing as x increases in this interval.For x ≥ -1: f(x) = 3x + 1 + a. The coefficient of x is 3, which is positive, so the function continues to increase as x increases in this interval.Therefore, the function f(x) is decreasing up to x = -a/2 and then starts increasing. So, the minimum value occurs at x = -a/2.Wait, that's the same conclusion as in Case 1. So, let's compute f(-a/2):f(-a/2) = |(-a/2) + 1| + |2*(-a/2) + a|.Simplify each term:First term: |(-a/2) + 1| = |1 - a/2|.Second term: |2*(-a/2) + a| = |-a + a| = |0| = 0.So, f(-a/2) = |1 - a/2| + 0 = |1 - a/2|.Again, since the minimum value is 3, |1 - a/2| = 3.So, 1 - a/2 = 3 or 1 - a/2 = -3.Solving the first equation: 1 - a/2 = 3 => -a/2 = 2 => a = -4.Solving the second equation: 1 - a/2 = -3 => -a/2 = -4 => a = 8.But in Case 2, we assumed that a > 2. So, a = 8 is valid because 8 > 2, but a = -4 is not because -4 < 2. So, in this case, a can only be 8.Now, let's consider the special case when a = 2. Then, both critical points coincide at x = -1.So, f(x) = |x + 1| + |2x + 2| = |x + 1| + 2|x + 1| = 3|x + 1|.So, f(x) = 3|x + 1|, which has a minimum value of 0 at x = -1. But the problem states that the minimum value is 3, so a = 2 is not a valid solution.Therefore, the possible values of 'a' are -4 and 8.Wait, but I should verify these solutions to make sure they work.Let's check a = -4.Then, f(x) = |x + 1| + |2x - 4|.The critical points are x = -1 and x = 2.Breaking f(x) into intervals:For x ≤ -1: f(x) = -3x -1 -(-4) = -3x + 3.For -1 < x < 2: f(x) = -x + 1 -(-4) = -x + 5.For x ≥ 2: f(x) = 3x +1 + (-4) = 3x -3.So, the function decreases until x = 2, then starts increasing. The minimum is at x = 2.f(2) = |2 + 1| + |4 - 4| = 3 + 0 = 3. Perfect.Now, check a = 8.Then, f(x) = |x + 1| + |2x + 8|.The critical points are x = -1 and x = -4.Breaking f(x) into intervals:For x ≤ -4: f(x) = -3x -1 -8 = -3x -9.For -4 < x < -1: f(x) = x -1 +8 = x +7.For x ≥ -1: f(x) = 3x +1 +8 = 3x +9.So, the function decreases until x = -4, then starts increasing. The minimum is at x = -4.f(-4) = |-4 + 1| + | -8 + 8| = | -3 | + |0| = 3 + 0 = 3. Perfect.So, both a = -4 and a = 8 give the function f(x) with a minimum value of 3.Therefore, the values of 'a' are -4 and 8.**Final Answer**The value of the real number ( a ) is boxed{-4} or boxed{8}.

💡Okay, so I have this problem where I need to find all polynomials ( W ) with real coefficients such that if ( x + y ) is a rational number, then ( W(x) + W(y) ) is rational. Hmm, interesting. Let me try to break this down step by step.First, I know that ( W ) is a polynomial, so it has a general form like ( W(x) = a_n x^n + a_{n-1} x^{n-1} + dots + a_1 x + a_0 ), where the coefficients ( a_i ) are real numbers. The problem is telling me that whenever ( x + y ) is rational, the sum ( W(x) + W(y) ) must also be rational. I think a good starting point is to consider specific cases or properties that such a polynomial must satisfy. Maybe I can fix some values or look for patterns. Let me try to fix a rational number ( q ) and consider ( x + y = q ). Then, for any ( x ), ( y = q - x ), so ( W(x) + W(q - x) ) must be rational.Wait, so if I define a function ( f(x) = W(x) + W(q - x) ), then ( f(x) ) must be rational for all ( x ). But ( f(x) ) is a polynomial because ( W ) is a polynomial, right? So ( f(x) ) is a polynomial that only takes rational values. But polynomials that take only rational values are quite special. In fact, if a polynomial with real coefficients takes only rational values at all rational inputs, then it must be a linear polynomial with rational coefficients. Hmm, is that always true? I'm not entirely sure, but I think it's something like that.Wait, actually, if a polynomial takes only rational values at all rational points, then it must have rational coefficients. But in this case, ( f(x) ) is a polynomial that takes rational values for all real ( x ), not just rational ( x ). That's a stronger condition. In fact, if a polynomial takes only rational values everywhere, then it must be a constant polynomial because a non-constant polynomial would take on infinitely many values, including irrational ones. But ( f(x) ) is not necessarily constant, but it's taking only rational values.Wait, no, that's not quite right. If a polynomial takes only rational values everywhere, it doesn't have to be constant. For example, ( f(x) = x ) takes rational values when ( x ) is rational, but it's not constant. But in our case, ( f(x) ) must take rational values for all real ( x ), which is a much stronger condition. In fact, the only polynomials that take rational values everywhere are constant polynomials because otherwise, they would take on irrational values as well.Wait, let me think again. If ( f(x) = W(x) + W(q - x) ) is a polynomial that is rational for all real ( x ), then it must be a constant polynomial. Because if it's not constant, it would take on infinitely many values, including irrational ones. So, ( f(x) ) must be a constant, say ( c ), which is rational.So, ( W(x) + W(q - x) = c ) for all ( x ). That seems like a functional equation. Maybe I can use this to find the form of ( W(x) ).Let me consider the case when ( q = 0 ). Then, ( W(x) + W(-x) = c ). Hmm, but ( q ) is any rational number, so maybe I can choose different ( q )s to get more information.Wait, actually, ( q ) is fixed once we choose it, but since the condition must hold for all rational ( q ), maybe I can vary ( q ) to get more constraints on ( W(x) ).Alternatively, maybe I can subtract two such equations for different ( q )s. Let me try that.Suppose I have two rational numbers, ( q_1 ) and ( q_2 ). Then, for each, I have:1. ( W(x) + W(q_1 - x) = c_1 )2. ( W(x) + W(q_2 - x) = c_2 )Subtracting these two equations, I get:( W(q_1 - x) - W(q_2 - x) = c_1 - c_2 )Let me make a substitution: let ( y = q_2 - x ). Then, ( x = q_2 - y ), and ( q_1 - x = q_1 - q_2 + y ). So, the equation becomes:( W(q_1 - q_2 + y) - W(y) = c_1 - c_2 )This suggests that the difference ( W(y + d) - W(y) ) is constant for all ( y ), where ( d = q_1 - q_2 ). But ( d ) is a rational number since ( q_1 ) and ( q_2 ) are rational. So, for any rational ( d ), ( W(y + d) - W(y) ) is constant. Wait, this seems like a property of linear functions. If ( W ) is linear, then ( W(y + d) - W(y) ) would be ( a d ), which is constant. But if ( W ) is of higher degree, say quadratic, then ( W(y + d) - W(y) ) would involve terms with ( y ), so it wouldn't be constant.Therefore, this suggests that ( W ) must be a linear polynomial. Let me test this hypothesis.Assume ( W(x) = a x + b ). Then, ( W(x) + W(y) = a(x + y) + 2b ). If ( x + y ) is rational, then ( a(x + y) + 2b ) must be rational. Since ( x + y ) is rational, let's denote it by ( q ). Then, ( W(x) + W(y) = a q + 2b ). For this to be rational for all rational ( q ), both ( a ) and ( b ) must be rational numbers. Because if ( a ) were irrational, then ( a q ) would be irrational for some rational ( q ), making ( a q + 2b ) irrational, which contradicts the condition.Therefore, ( W(x) ) must be a linear polynomial with rational coefficients. Wait, but the problem states that ( W ) has real coefficients. So, if ( a ) and ( b ) are rational, they are also real. So, that fits.But let me check if there are any other possibilities. Suppose ( W ) is a constant polynomial, say ( W(x) = c ). Then, ( W(x) + W(y) = 2c ), which is rational if ( c ) is rational. So, constant polynomials with rational constants are also solutions.But in the earlier analysis, when I considered ( W(x) + W(q - x) = c ), if ( W ) is constant, then ( c = 2W(0) ), which is rational if ( W(0) ) is rational. So, constant polynomials with rational constants are included in the linear case when ( a = 0 ).Therefore, the general solution is that ( W(x) ) is a linear polynomial with rational coefficients, i.e., ( W(x) = a x + b ) where ( a ) and ( b ) are rational numbers.Let me verify this with an example. Suppose ( W(x) = 2x + 3 ). Then, if ( x + y ) is rational, say ( q ), then ( W(x) + W(y) = 2x + 3 + 2y + 3 = 2(x + y) + 6 = 2q + 6 ), which is rational since ( q ) is rational.Another example: ( W(x) = sqrt{2}x + 1 ). Here, ( a = sqrt{2} ) is irrational. If ( x + y = 1 ), then ( W(x) + W(y) = sqrt{2}(x + y) + 2 = sqrt{2} + 2 ), which is irrational. So, this doesn't satisfy the condition, confirming that ( a ) must be rational.Similarly, if ( W(x) = x + sqrt{2} ), then ( W(x) + W(y) = x + y + 2sqrt{2} ). If ( x + y ) is rational, say ( q ), then ( W(x) + W(y) = q + 2sqrt{2} ), which is irrational. Hence, ( b ) must also be rational.Therefore, the only polynomials ( W ) that satisfy the given condition are linear polynomials with rational coefficients.**Final Answer**The polynomials ( W ) are all linear polynomials with rational coefficients. Thus, the solution is boxed{W(x) = ax + b} where ( a ) and ( b ) are rational numbers.

💡Okay, so I have this function f(x) = x e^x - a e^{2x}, and I need to find the range of values for the real number a such that the function has exactly two extreme points, x1 and x2, with x1 < x2. Hmm, let me think about how to approach this.First, I remember that extreme points of a function occur where the first derivative is zero or undefined. Since this function is differentiable everywhere (it's composed of exponential functions and polynomials, which are smooth), I just need to find where the derivative equals zero.So, let me compute the derivative of f(x). The function is f(x) = x e^x - a e^{2x}. To find f'(x), I'll differentiate each term separately.The derivative of x e^x is e^x + x e^x by the product rule. Then, the derivative of -a e^{2x} is -a * 2 e^{2x} by the chain rule. So putting it all together, f'(x) = e^x + x e^x - 2a e^{2x}.Hmm, I can factor out e^x from all terms. Let me try that:f'(x) = e^x (1 + x - 2a e^x).So, f'(x) = e^x (x + 1 - 2a e^x).Since e^x is always positive for any real x, the sign of f'(x) depends on the expression (x + 1 - 2a e^x). Therefore, to find the critical points, I need to solve the equation:x + 1 - 2a e^x = 0.Let me denote this equation as g(x) = x + 1 - 2a e^x = 0. So, the problem reduces to finding the number of real roots of g(x) = 0.The original function f(x) will have exactly two extreme points if and only if g(x) = 0 has exactly two distinct real roots. So, I need to find the values of a for which this equation has exactly two solutions.To analyze the number of solutions, I can study the function g(x) = x + 1 - 2a e^x. Let me compute its derivative to understand its behavior.g'(x) = derivative of x + 1 - 2a e^x = 1 - 2a e^x.So, g'(x) = 1 - 2a e^x.The critical points of g(x) occur where g'(x) = 0, so:1 - 2a e^x = 0 => 2a e^x = 1 => e^x = 1/(2a) => x = ln(1/(2a)).But wait, ln(1/(2a)) is only defined if 1/(2a) > 0, which implies that a > 0. So, if a ≤ 0, then g'(x) = 1 - 2a e^x is always positive because 2a e^x would be non-positive (since a ≤ 0), so 1 - (non-positive) is always positive. Therefore, if a ≤ 0, g(x) is strictly increasing everywhere.But if a > 0, then g(x) has a critical point at x = ln(1/(2a)). Let me analyze both cases.Case 1: a ≤ 0.As I noted, g'(x) = 1 - 2a e^x is always positive because 2a e^x ≤ 0. So, g(x) is strictly increasing on ℝ. Therefore, the equation g(x) = 0 can have at most one real root. But since we need exactly two extreme points, this case is not suitable. So, a must be greater than 0.Case 2: a > 0.In this case, g(x) has a critical point at x = ln(1/(2a)). Let me compute the value of g(x) at this critical point to determine whether it's a maximum or a minimum.Compute g(ln(1/(2a))):g(ln(1/(2a))) = ln(1/(2a)) + 1 - 2a e^{ln(1/(2a))}.Simplify e^{ln(1/(2a))} = 1/(2a). So,g(ln(1/(2a))) = ln(1/(2a)) + 1 - 2a*(1/(2a)) = ln(1/(2a)) + 1 - 1 = ln(1/(2a)).So, g(ln(1/(2a))) = ln(1/(2a)).Now, to have exactly two real roots, the function g(x) must cross the x-axis twice. Since g(x) is strictly increasing before x = ln(1/(2a)) and strictly decreasing after that, it will have a maximum at x = ln(1/(2a)). For the equation g(x) = 0 to have two real roots, the maximum value of g(x) must be positive. If the maximum is zero, it will have exactly one real root, and if the maximum is negative, it will have no real roots.Therefore, the condition is that g(ln(1/(2a))) > 0.From above, g(ln(1/(2a))) = ln(1/(2a)) > 0.So, ln(1/(2a)) > 0.This implies that 1/(2a) > e^0 = 1.Therefore, 1/(2a) > 1 => 2a < 1 => a < 1/2.But since we are in the case where a > 0, combining these, we have 0 < a < 1/2.Let me verify this conclusion.If a is between 0 and 1/2, then 1/(2a) > 1, so ln(1/(2a)) > 0, which makes the maximum of g(x) positive, hence g(x) = 0 has two real roots.If a = 1/2, then 1/(2a) = 1, so ln(1/(2a)) = ln(1) = 0. Then, g(0) = 0 + 1 - 2*(1/2)*e^0 = 1 - 1 = 0. So, the maximum is zero, which means g(x) touches the x-axis at x = 0, so only one real root.If a > 1/2, then 1/(2a) < 1, so ln(1/(2a)) < 0. Then, the maximum of g(x) is negative, so g(x) = 0 has no real roots.If a ≤ 0, as discussed earlier, g(x) is strictly increasing, so only one real root.Therefore, the range of a for which f(x) has exactly two extreme points is 0 < a < 1/2.I think that makes sense. Let me just recap:1. Compute f'(x) and set it to zero, leading to g(x) = x + 1 - 2a e^x = 0.2. Analyze g(x) by finding its derivative, which gives critical points.3. Determine the behavior of g(x) based on the value of a.4. Conclude that for two extreme points, a must be between 0 and 1/2.Yeah, that seems solid. I don't think I missed anything here.**Final Answer**The range of values for the real number ( a ) is boxed{left(0, dfrac{1}{2}right)}.