💡Okay, so I have this equation to solve: 4 raised to the power of (x + 6) equals 64 raised to the power of x. Hmm, that is, 4^(x + 6) = 64^x. Alright, let me think about how to approach this.First, I remember that when dealing with exponential equations, it's helpful if I can express both sides with the same base. That way, I can set the exponents equal to each other and solve for x. So, I need to see if 4 and 64 can be expressed as powers of the same number.I know that 4 is 2 squared, so 4 = 2^2. What about 64? Let me think. 64 is a power of 2 as well. 2^6 is 64 because 2*2*2*2*2*2 = 64. So, 64 is 2^6. Alternatively, I also know that 64 is 4 cubed because 4*4*4 = 64. So, 64 = 4^3.Wait, so 64 can be written as 4^3. That might be useful because the other side of the equation is already in terms of 4. So, if I rewrite 64^x as (4^3)^x, that would be 4^(3x). So, the equation becomes 4^(x + 6) = 4^(3x).Now, since both sides have the same base, 4, I can set the exponents equal to each other. That's a rule I remember: if a^b = a^c, then b = c, provided that a is not 1 or 0. So, applying that here, I can write:x + 6 = 3x.Okay, now I have a linear equation to solve for x. Let me subtract x from both sides to get the x terms on one side.x + 6 - x = 3x - x.Simplifying that, I get:6 = 2x.Now, to solve for x, I need to isolate it. So, I'll divide both sides by 2.6 / 2 = 2x / 2.Which simplifies to:3 = x.So, x equals 3. Let me check if that makes sense by plugging it back into the original equation.Original equation: 4^(x + 6) = 64^x.Plugging in x = 3:Left side: 4^(3 + 6) = 4^9.Right side: 64^3.Now, let's compute both sides.4^9: 4 is 2^2, so 4^9 = (2^2)^9 = 2^(2*9) = 2^18.64^3: 64 is 2^6, so 64^3 = (2^6)^3 = 2^(6*3) = 2^18.Yes, both sides equal 2^18, so 4^9 equals 64^3. Therefore, x = 3 is indeed the correct solution.I think that's all. It wasn't too bad once I expressed both sides with the same base. It's important to remember that trick when dealing with exponential equations.

💡First, I need to determine Jessica's total monthly expenses from last year. She paid 1000 for rent, 200 for food, and 100 for car insurance. Adding these together gives her total monthly expenses.Next, I'll calculate the new monthly expenses for this year. Her rent increased by 30%, so I'll find 30% of 1000 and add it to the original rent amount. Her car insurance tripled, so I'll multiply the original car insurance cost by 3. For food, I'll denote the percentage increase as x% and express the new food cost accordingly.Then, I'll set up an equation for the total monthly expenses this year, which should equal last year's monthly expenses plus the additional 600 per month (since the total increase over the year is 7200, which is 600 per month).Finally, I'll solve the equation to find the value of x, which represents the percentage increase in Jessica's food costs.

💡Okay, so I have this function f(x) = sin(5π/4 - x) - cos(π/4 + x). There are two parts to the problem. First, I need to find the interval where f(x) is monotonically increasing. Second, given some cosine values involving angles α and β, I need to find f(β). Let me tackle each part step by step.Starting with the first part: finding where f(x) is increasing. To do this, I remember that a function is increasing where its derivative is positive. So, I need to find f'(x) and then determine where it's positive.First, let me simplify f(x) if possible. The function is f(x) = sin(5π/4 - x) - cos(π/4 + x). Maybe I can use some trigonometric identities to simplify this expression.Looking at sin(5π/4 - x), I recall that sin(A - B) = sinA cosB - cosA sinB. Similarly, cos(π/4 + x) can be expanded using cos(A + B) = cosA cosB - sinA sinB. Let me try expanding both terms.So, sin(5π/4 - x) = sin(5π/4)cosx - cos(5π/4)sinx. Similarly, cos(π/4 + x) = cos(π/4)cosx - sin(π/4)sinx.Now, I need to compute sin(5π/4) and cos(5π/4). 5π/4 is in the third quadrant where both sine and cosine are negative. The reference angle is π/4, so sin(5π/4) = -√2/2 and cos(5π/4) = -√2/2.Similarly, cos(π/4) = √2/2 and sin(π/4) = √2/2.So plugging these values in:sin(5π/4 - x) = (-√2/2)cosx - (-√2/2)sinx = (-√2/2)cosx + (√2/2)sinx.Similarly, cos(π/4 + x) = (√2/2)cosx - (√2/2)sinx.So f(x) becomes:f(x) = [(-√2/2)cosx + (√2/2)sinx] - [(√2/2)cosx - (√2/2)sinx]Let me distribute the negative sign in front of the second term:= (-√2/2 cosx + √2/2 sinx) - √2/2 cosx + √2/2 sinxNow, combine like terms:For cosx terms: (-√2/2 - √2/2) cosx = (-√2) cosxFor sinx terms: (√2/2 + √2/2) sinx = √2 sinxSo f(x) simplifies to:f(x) = √2 sinx - √2 cosxI can factor out √2:f(x) = √2 (sinx - cosx)Hmm, maybe I can write this as a single sine function using the identity A sinx + B cosx = C sin(x + φ). Let me recall that identity.In general, A sinx + B cosx = √(A² + B²) sin(x + φ), where φ = arctan(B/A) or something like that. Wait, actually, it's φ = arctan(B/A) if A is positive, but I might have to adjust based on the signs.In this case, f(x) = √2 (sinx - cosx). So, A = 1, B = -1. So, the amplitude would be √(1² + (-1)²) = √2. So, f(x) can be written as √2 * √2 sin(x - π/4). Wait, let me check that.Wait, more accurately, sinx - cosx can be written as √2 sin(x - π/4). Let me verify that:Using the identity sin(x - π/4) = sinx cos(π/4) - cosx sin(π/4) = (sinx - cosx)(√2/2). So, multiplying both sides by √2, we get √2 sin(x - π/4) = sinx - cosx. Perfect, so f(x) = √2 (sinx - cosx) = √2 * √2 sin(x - π/4) = 2 sin(x - π/4). Wait, no, hold on. Wait, if sin(x - π/4) = (sinx - cosx)/√2, then sinx - cosx = √2 sin(x - π/4). So, f(x) = √2 * √2 sin(x - π/4) = 2 sin(x - π/4). Wait, that seems too big. Wait, no, f(x) is √2 (sinx - cosx) which is √2 * √2 sin(x - π/4) = 2 sin(x - π/4). Hmm, but let me double-check.Wait, let me compute sin(x - π/4):sin(x - π/4) = sinx cos(π/4) - cosx sin(π/4) = (sinx - cosx)(√2/2). So, sinx - cosx = √2 sin(x - π/4). Therefore, f(x) = √2 * √2 sin(x - π/4) = 2 sin(x - π/4). Wait, that seems correct.But wait, in the original function, f(x) = sin(5π/4 - x) - cos(π/4 + x). After simplifying, I got f(x) = 2 sin(x - π/4). Hmm, that seems a bit different. Let me check my steps again.Wait, when I expanded sin(5π/4 - x), I got (-√2/2)cosx + (√2/2)sinx, which is correct. Then, cos(π/4 + x) is (√2/2)cosx - (√2/2)sinx. So, when I subtract that, it becomes:(-√2/2 cosx + √2/2 sinx) - (√2/2 cosx - √2/2 sinx) = (-√2/2 - √2/2) cosx + (√2/2 + √2/2) sinx = (-√2) cosx + √2 sinx.So, f(x) = √2 sinx - √2 cosx = √2 (sinx - cosx). Then, as I did before, sinx - cosx = √2 sin(x - π/4). So, f(x) = √2 * √2 sin(x - π/4) = 2 sin(x - π/4). Okay, that seems correct.Wait, but in the initial problem, f(x) is given as sin(5π/4 - x) - cos(π/4 + x). So, after simplifying, it's 2 sin(x - π/4). Hmm, that's a much simpler expression. So, f(x) = 2 sin(x - π/4).Now, to find where f(x) is increasing, I need to find where its derivative is positive. So, let's compute f'(x):f'(x) = d/dx [2 sin(x - π/4)] = 2 cos(x - π/4).So, f'(x) = 2 cos(x - π/4). We need to find where f'(x) > 0, which is where cos(x - π/4) > 0.The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, cos(x - π/4) > 0 when:-π/2 + 2πk < x - π/4 < π/2 + 2πk.Let me solve for x:Add π/4 to all parts:-π/2 + π/4 + 2πk < x < π/2 + π/4 + 2πk.Simplify the constants:-π/2 + π/4 = -π/4, and π/2 + π/4 = 3π/4.So, the intervals where f'(x) > 0 are:(-π/4 + 2πk, 3π/4 + 2πk) for any integer k.Therefore, f(x) is monotonically increasing on the intervals (-π/4 + 2πk, 3π/4 + 2πk), where k is any integer.Wait, but the question says "Find the interval where f(x) is monotonically increasing." It doesn't specify a particular interval, so I think it's expecting the general form. So, the answer is that f(x) is increasing on intervals of the form (-π/4 + 2πk, 3π/4 + 2πk) for any integer k.But let me double-check my steps to make sure I didn't make a mistake. Starting from f(x) = sin(5π/4 - x) - cos(π/4 + x), I expanded using trigonometric identities and simplified to f(x) = 2 sin(x - π/4). Then, taking the derivative, f'(x) = 2 cos(x - π/4). Setting f'(x) > 0 gives cos(x - π/4) > 0, which occurs when x - π/4 is in (-π/2 + 2πk, π/2 + 2πk). Solving for x gives x in (-π/4 + 2πk, 3π/4 + 2πk). That seems correct.Now, moving on to the second part: Given cos(α - β) = 3/5 and cos(α + β) = -3/5, where 0 < α < β ≤ π/2, find f(β).First, let's note that f(x) = 2 sin(x - π/4). So, f(β) = 2 sin(β - π/4).But to find f(β), I need to find sin(β - π/4). Alternatively, I can express f(β) in terms of α and β using the given information.Given cos(α - β) = 3/5 and cos(α + β) = -3/5. Let me write down these equations:1. cos(α - β) = 3/52. cos(α + β) = -3/5Since 0 < α < β ≤ π/2, let's analyze the possible values of α and β.First, α - β is negative because α < β, so α - β is in (-π/2, 0). Similarly, α + β is in (0, π) because both α and β are positive and less than or equal to π/2.Given cos(α - β) = 3/5, which is positive, so α - β is in the fourth quadrant, which makes sense since it's negative. Similarly, cos(α + β) = -3/5, which is negative, so α + β is in the second quadrant.Now, let's find sin(α - β) and sin(α + β) using the Pythagorean identity sin²θ + cos²θ = 1.For α - β:sin²(α - β) = 1 - cos²(α - β) = 1 - (9/25) = 16/25.Since α - β is in the fourth quadrant, sin(α - β) is negative. So, sin(α - β) = -4/5.For α + β:sin²(α + β) = 1 - cos²(α + β) = 1 - (9/25) = 16/25.Since α + β is in the second quadrant, sin(α + β) is positive. So, sin(α + β) = 4/5.Now, I can use these to find expressions involving α and β. Let me think about how to relate this to f(β) = 2 sin(β - π/4).Alternatively, perhaps I can find β first. Let me try to find α and β.We have two equations:1. cos(α - β) = 3/52. cos(α + β) = -3/5Let me denote θ = α - β and φ = α + β. Then, we have:cosθ = 3/5, cosφ = -3/5.We can find θ and φ, then solve for α and β.But since θ = α - β and φ = α + β, we can solve for α and β in terms of θ and φ.Adding θ and φ:θ + φ = (α - β) + (α + β) = 2α ⇒ α = (θ + φ)/2.Subtracting θ from φ:φ - θ = (α + β) - (α - β) = 2β ⇒ β = (φ - θ)/2.So, if I can find θ and φ, I can find α and β.Given cosθ = 3/5 and cosφ = -3/5.We can find θ and φ using arccos.θ = arccos(3/5). Since θ = α - β is negative, θ = -arccos(3/5).Similarly, φ = arccos(-3/5). Since φ is in the second quadrant, φ = π - arccos(3/5).Wait, let me clarify:cosθ = 3/5, θ is in the fourth quadrant, so θ = -arccos(3/5).cosφ = -3/5, φ is in the second quadrant, so φ = π - arccos(3/5).Let me denote arccos(3/5) as γ, where γ is in the first quadrant. So, γ = arccos(3/5).Then, θ = -γ, and φ = π - γ.So, θ = -γ, φ = π - γ.Now, α = (θ + φ)/2 = (-γ + π - γ)/2 = (π - 2γ)/2 = π/2 - γ.Similarly, β = (φ - θ)/2 = (π - γ - (-γ))/2 = (π)/2.Wait, β = π/2? But β is given to be ≤ π/2, so β = π/2.Wait, let me check:β = (φ - θ)/2 = [ (π - γ) - (-γ) ] / 2 = (π - γ + γ)/2 = π/2.So, β = π/2.But wait, if β = π/2, then α = π/2 - γ.Given that α < β, and β = π/2, then α must be less than π/2, which is consistent with the given condition 0 < α < β ≤ π/2.So, β = π/2.Now, let's compute f(β) = f(π/2) = 2 sin(π/2 - π/4) = 2 sin(π/4) = 2*(√2/2) = √2.Wait, that seems straightforward. Let me verify.Given β = π/2, then f(β) = 2 sin(π/2 - π/4) = 2 sin(π/4) = 2*(√2/2) = √2.Alternatively, since f(x) = 2 sin(x - π/4), plugging in x = π/2 gives 2 sin(π/2 - π/4) = 2 sin(π/4) = √2.So, f(β) = √2.But let me make sure I didn't make any mistakes in finding β.We had:θ = α - β = -γ, where γ = arccos(3/5).φ = α + β = π - γ.Then, solving for α and β:α = (θ + φ)/2 = (-γ + π - γ)/2 = (π - 2γ)/2 = π/2 - γ.β = (φ - θ)/2 = (π - γ - (-γ))/2 = π/2.So, β = π/2.Therefore, f(β) = f(π/2) = √2.Yes, that seems correct.So, summarizing:1. The function f(x) is monotonically increasing on intervals (-π/4 + 2πk, 3π/4 + 2πk) for any integer k.2. Given the conditions, f(β) = √2.I think that's the solution.

💡Alright, so I've got these four problems to solve, all involving square roots. I remember that square roots can sometimes be tricky, especially when you have to multiply, add, or subtract them. But I think if I take it step by step, I can handle each problem one by one.Starting with the first problem: √18 × √6. Hmm, I recall that when you multiply two square roots, you can combine them under a single square root. So, √18 × √6 should be equal to √(18×6). Let me calculate that: 18 times 6 is 108. So, it's √108. Now, I need to simplify √108. I know that 108 can be broken down into factors where one is a perfect square. Let's see, 108 divided by 36 is 3, and 36 is a perfect square. So, √108 is the same as √(36×3), which is √36 × √3. √36 is 6, so this simplifies to 6√3. Okay, that seems straightforward.Moving on to the second problem: √8 - √2 + 2√(1/2). This one looks a bit more complicated because there are three terms, and one of them has a square root of a fraction. Let me break it down. First, simplify each term individually. Starting with √8: 8 can be broken down into 4×2, and 4 is a perfect square. So, √8 is √(4×2) which is √4 × √2, which is 2√2. Next, the term √2 is already simplified, so I'll leave that as it is. Now, the third term is 2√(1/2). I think I can simplify √(1/2) first. √(1/2) is the same as √1 divided by √2, which is 1/√2. But having a square root in the denominator isn't usually preferred, so I can rationalize it by multiplying numerator and denominator by √2. That gives me √2/2. So, √(1/2) is √2/2. Therefore, 2√(1/2) is 2 × (√2/2), which simplifies to √2. Now, putting it all together: √8 - √2 + 2√(1/2) becomes 2√2 - √2 + √2. Combining like terms: 2√2 - √2 is √2, and then adding another √2 gives me 2√2. Wait, that doesn't seem right. Let me double-check. 2√2 minus √2 is indeed √2, and then adding √2 should give me 2√2. Yeah, that makes sense.Now, the third problem: √12 × (√9 / 3) ÷ (√3 / 3). Okay, this one involves multiplication and division of square roots. Let me simplify each part first. √12 can be simplified since 12 is 4×3, and 4 is a perfect square. So, √12 is √4 × √3, which is 2√3. Next, √9 is straightforward; √9 is 3. So, √9 divided by 3 is 3/3, which is 1. Then, √3 divided by 3 is √3/3. So, putting it all together: √12 × (√9 / 3) ÷ (√3 / 3) becomes 2√3 × 1 ÷ (√3 / 3). Dividing by a fraction is the same as multiplying by its reciprocal, so this becomes 2√3 × 1 × 3/√3. Simplifying that, 2√3 × 3/√3. The √3 in the numerator and denominator cancel out, leaving 2 × 3, which is 6. So, the answer is 6.Finally, the fourth problem: (√7 + √5) × (√7 - √5). This looks like a product of a sum and difference, which I remember follows the difference of squares formula: (a + b)(a - b) = a² - b². Applying that here, (√7)² - (√5)². Squaring √7 gives me 7, and squaring √5 gives me 5. So, 7 - 5 is 2. That was pretty straightforward.Wait, let me just go back and make sure I didn't make any mistakes. For the second problem, I had √8 - √2 + 2√(1/2). I simplified √8 to 2√2, and 2√(1/2) to √2. So, 2√2 - √2 + √2. Combining them: 2√2 - √2 is √2, and then adding another √2 gives me 2√2. Yeah, that seems correct. Okay, I think I've got all the answers now.

💡Okay, so I've got this problem here with two quadratic equations that share a common root. The equations are:1. ( x^2 + 2019a x + b = 0 )2. ( x^2 + 2019b x + a = 0 )And I'm told that ( a neq b ). I need to figure out what the common root could be. Hmm, let's think about how to approach this.First, since both equations have a common root, let's call that root ( r ). That means if I plug ( r ) into both equations, they should both equal zero. So, substituting ( r ) into the first equation gives:( r^2 + 2019a r + b = 0 ) ...(1)And substituting ( r ) into the second equation gives:( r^2 + 2019b r + a = 0 ) ...(2)Now, if both of these equations equal zero, I can set them equal to each other because they both equal zero. So, subtracting equation (2) from equation (1):( (r^2 + 2019a r + b) - (r^2 + 2019b r + a) = 0 )Simplifying this, the ( r^2 ) terms cancel out:( 2019a r + b - 2019b r - a = 0 )Let me factor out the common terms. I see ( 2019r ) is common in the first two terms, and then ( b - a ) is the remaining part:( 2019r(a - b) + (b - a) = 0 )Wait, ( (b - a) ) is the same as ( -(a - b) ), so I can rewrite this as:( 2019r(a - b) - (a - b) = 0 )Now, factor out ( (a - b) ):( (a - b)(2019r - 1) = 0 )Hmm, so this equation is the product of two terms equal to zero. That means either ( (a - b) = 0 ) or ( (2019r - 1) = 0 ). But wait, the problem says ( a neq b ), so ( (a - b) ) can't be zero. Therefore, the other factor must be zero:( 2019r - 1 = 0 )Solving for ( r ):( 2019r = 1 )( r = frac{1}{2019} )So, the common root must be ( frac{1}{2019} ). Let me just double-check my steps to make sure I didn't make a mistake.1. I set ( r ) as the common root and substituted it into both equations.2. Subtracted the two equations to eliminate ( r^2 ).3. Factored out ( (a - b) ) and recognized that since ( a neq b ), the other factor must be zero.4. Solved for ( r ) and got ( frac{1}{2019} ).Everything seems to check out. I don't see any errors in my reasoning. So, yeah, the common root is definitely ( frac{1}{2019} ).**Final Answer**The common root is boxed{dfrac{1}{2019}}.

💡Okay, so I have this function f(x) = ax³ - 3x² + 1, and I need to find the range of the real number a such that the function has exactly one root x₀, and x₀ is less than 0. Hmm, okay. Let me try to figure this out step by step.First, I remember that for a cubic function like this, the number of real roots can vary depending on the coefficients. Since it's a cubic, it can have one or three real roots. But in this case, we want exactly one real root, and that root has to be negative. So, I need to figure out the conditions on a that make this happen.I think I should start by analyzing the function's behavior. Maybe taking the derivative to find critical points would help. Let's compute f'(x):f'(x) = d/dx [ax³ - 3x² + 1] = 3ax² - 6x.Okay, so f'(x) = 3ax² - 6x. To find critical points, set f'(x) = 0:3ax² - 6x = 03x(ax - 2) = 0So, the critical points are at x = 0 and x = 2/a. Interesting. So, depending on the value of a, the critical points will be in different places.Now, since we're interested in the number of real roots, especially exactly one negative root, I should consider the behavior of the function as x approaches positive and negative infinity.For a cubic function, as x approaches positive infinity, the term with the highest degree (ax³) dominates. So, if a is positive, f(x) tends to positive infinity as x approaches positive infinity, and negative infinity as x approaches negative infinity. If a is negative, it's the opposite: f(x) tends to negative infinity as x approaches positive infinity and positive infinity as x approaches negative infinity.Wait, so if a is positive, the function goes from negative infinity to positive infinity, which means it must cross the x-axis at least once. If a is negative, it goes from positive infinity to negative infinity, so it must cross the x-axis at least once as well. But we want exactly one root, so maybe the function doesn't have any local maxima or minima that cross the x-axis, which would cause more roots.So, perhaps the function should be monotonic? But wait, since it's a cubic, it can't be entirely monotonic unless the derivative doesn't change sign. But in our case, the derivative is a quadratic, so it can have two critical points, which are x = 0 and x = 2/a.So, if a ≠ 0, the function has two critical points. If a = 0, then f(x) becomes a quadratic: f(x) = -3x² + 1, which has two real roots, so that's not what we want. So, a can't be zero.So, let's consider a ≠ 0. Now, depending on the sign of a, the critical points will be in different positions. Let's analyze both cases: a > 0 and a < 0.Case 1: a > 0.In this case, the critical points are at x = 0 and x = 2/a, which is positive since a is positive. So, the function has a local maximum at x = 0 and a local minimum at x = 2/a.Let me compute f(0). f(0) = a*(0)³ - 3*(0)² + 1 = 1. So, the function has a local maximum at (0, 1). Then, it decreases until x = 2/a, where it has a local minimum, and then increases again.Now, since we want exactly one real root, and that root is negative, the function should not cross the x-axis after x = 0. Because if it crosses the x-axis after x = 0, we would have another root. So, the local minimum at x = 2/a must be above the x-axis. That way, the function only crosses the x-axis once, somewhere to the left of x = 0.So, let's compute f(2/a):f(2/a) = a*(2/a)³ - 3*(2/a)² + 1= a*(8/a³) - 3*(4/a²) + 1= 8/a² - 12/a² + 1= (-4)/a² + 1So, f(2/a) = 1 - 4/a².We want this local minimum to be above the x-axis, so:1 - 4/a² > 0=> 1 > 4/a²=> a² > 4=> |a| > 2But since we are in the case where a > 0, this simplifies to a > 2.Okay, so for a > 2, the function has a local minimum above the x-axis, meaning it only crosses the x-axis once, and since f(0) = 1 is positive, that root must be to the left of x = 0, i.e., negative.Case 2: a < 0.In this case, the critical points are at x = 0 and x = 2/a, which is negative since a is negative. So, the function has a local maximum at x = 0 and a local minimum at x = 2/a (which is negative).Again, f(0) = 1, which is positive. The function decreases from x = 0 to x = 2/a, then increases again. But since a is negative, as x approaches positive infinity, f(x) tends to negative infinity, and as x approaches negative infinity, f(x) tends to positive infinity.So, in this case, the function will cross the x-axis once to the right of x = 0 (since it goes from positive infinity to negative infinity) and once to the left of x = 2/a (since it goes from positive infinity to negative infinity). Wait, that would mean three real roots, right? Because it crosses once on the left, once between x = 2/a and x = 0, and once on the right? Hmm, maybe not.Wait, let me think again. If a is negative, the function tends to positive infinity as x approaches negative infinity and tends to negative infinity as x approaches positive infinity. So, it must cross the x-axis at least once on the positive side and once on the negative side. But depending on the local minimum, it might have three real roots.Wait, let's compute f(2/a) when a < 0.f(2/a) = 1 - 4/a², same as before.But since a is negative, a² is still positive, so 1 - 4/a² is the same as in the previous case.If a < 0, then |a| > 2 would mean a < -2. So, if a < -2, then f(2/a) = 1 - 4/a² > 0, similar to the a > 2 case.Wait, but in this case, the local minimum is at x = 2/a, which is negative. So, if f(2/a) > 0, then the function only crosses the x-axis once on the positive side, but we want it to have exactly one root, which is negative. So, this seems conflicting.Wait, if a < -2, then f(2/a) > 0, so the function has a local minimum above the x-axis at x = 2/a (negative). Then, since f(x) tends to positive infinity as x approaches negative infinity and tends to negative infinity as x approaches positive infinity, it must cross the x-axis once on the positive side and once on the negative side. But if the local minimum is above the x-axis, it might only cross once on the negative side?Wait, no. Let me sketch the graph mentally. If a < 0, the function tends to positive infinity as x approaches negative infinity, comes down to a local maximum at x = 0 (f(0) = 1), then decreases to a local minimum at x = 2/a (which is negative, since a is negative). If the local minimum is above the x-axis, then the function doesn't cross the x-axis between x = 2/a and x = 0, but it still goes to negative infinity as x approaches positive infinity, so it must cross the x-axis once on the positive side.But we want exactly one root, which is negative. So, if a < -2, the function crosses the x-axis once on the positive side and once on the negative side, which is two roots, which is not what we want. So, maybe a < 0 is not acceptable?Wait, but when a < 0, the function tends to positive infinity as x approaches negative infinity and tends to negative infinity as x approaches positive infinity. So, it must cross the x-axis at least once on the positive side. If the local minimum is above the x-axis, then it only crosses once on the positive side and once on the negative side? Wait, no, because if the local minimum is above the x-axis, then it can't cross the x-axis between x = 2/a and x = 0, but it still has to come from positive infinity, go down to the local maximum at x = 0, which is 1, then go down to the local minimum at x = 2/a, which is above the x-axis, and then go to negative infinity. So, it must cross the x-axis once on the positive side and once on the negative side, making two roots. But we want exactly one root, which is negative. So, this is conflicting.Wait, maybe I made a mistake. Let me think again.If a < 0, then x = 2/a is negative. So, the function has a local maximum at x = 0 (f(0) = 1) and a local minimum at x = 2/a (which is negative). If the local minimum is above the x-axis, then the function only crosses the x-axis once on the positive side, because it goes from positive infinity to 1 at x = 0, then decreases to a local minimum above the x-axis, and then increases to negative infinity. Wait, no, it can't increase to negative infinity because a is negative, so as x approaches positive infinity, f(x) tends to negative infinity. So, the function must cross the x-axis once on the positive side. But on the negative side, it goes from positive infinity to the local minimum at x = 2/a, which is above the x-axis, so it doesn't cross the x-axis on the negative side. So, in this case, the function only crosses the x-axis once on the positive side, but we need it to cross exactly once on the negative side.Wait, so if a < 0, the function tends to positive infinity as x approaches negative infinity, comes down to a local maximum at x = 0 (f(0) = 1), then decreases to a local minimum at x = 2/a (which is negative). If the local minimum is above the x-axis, then the function doesn't cross the x-axis on the negative side, but it still must cross the x-axis once on the positive side because it goes to negative infinity as x approaches positive infinity. So, in this case, the function has exactly one real root on the positive side, which is not what we want. We want exactly one real root on the negative side.Therefore, if a < 0, the function will have one real root on the positive side and possibly another on the negative side, but if the local minimum is above the x-axis, it only has one real root on the positive side. But we need exactly one real root on the negative side, so a < 0 is not acceptable.Wait, maybe if the local minimum is below the x-axis, then the function would have three real roots: one on the negative side, one between x = 2/a and x = 0, and one on the positive side. But we want exactly one real root on the negative side, so that's not acceptable either.Therefore, perhaps a < 0 is not acceptable because it either results in two real roots (one positive, one negative) or three real roots, neither of which satisfies the condition of exactly one real root on the negative side.So, maybe only a > 2 is acceptable.Wait, let me confirm.When a > 2, the function has a local maximum at x = 0 (f(0) = 1) and a local minimum at x = 2/a (positive). If the local minimum is above the x-axis, then the function only crosses the x-axis once on the negative side because it goes from negative infinity to the local maximum at x = 0, which is positive, then decreases to a local minimum above the x-axis, and then increases to positive infinity. So, it only crosses the x-axis once on the negative side.Yes, that makes sense. So, when a > 2, the function has exactly one real root, which is negative.When a = 2, f(2/a) = f(1) = 1 - 4/4 = 0. So, the local minimum is exactly at the x-axis, meaning the function touches the x-axis at x = 1, which would be a repeated root. But the problem states exactly one root, so a = 2 is not acceptable because it results in a repeated root, making it technically two roots (one simple and one double). So, a must be greater than 2.When 0 < a < 2, f(2/a) = 1 - 4/a² < 0, so the local minimum is below the x-axis, meaning the function crosses the x-axis three times: once on the negative side, once between x = 0 and x = 2/a, and once on the positive side. So, that's three real roots, which is more than we want.When a = 0, as we saw earlier, the function becomes quadratic with two real roots, which is also not what we want.When a < 0, as discussed, the function either has two real roots (one positive, one negative) or three real roots, neither of which satisfies the condition.Therefore, the only acceptable case is when a > 2.So, the range of the real number a is (2, +∞).**Final Answer**The range of the real number ( a ) is boxed{(2, +infty)}.