answer：Okay, so I have this problem here: I need to find the range of the function ( f(x) = cosleft(frac{pi}{3} - xright) + sinleft(frac{pi}{2} + xright) ). The options are A: ([-2, 2]), B: ([- sqrt{3}, sqrt{3}]), C: ([-1, 1]), and D: ([- frac{sqrt{3}}{2}, frac{sqrt{3}}{2}]). Hmm, let's see how to approach this.First, I remember that to find the range of a trigonometric function, it's often helpful to simplify the expression using trigonometric identities. Maybe I can rewrite the function in a way that combines the sine and cosine terms into a single trigonometric function. That might make it easier to determine the maximum and minimum values.Looking at the function, I see two terms: ( cosleft(frac{pi}{3} - xright) ) and ( sinleft(frac{pi}{2} + xright) ). I think I can use the angle subtraction and addition formulas here.Starting with ( cosleft(frac{pi}{3} - xright) ), I recall the identity ( cos(A - B) = cos A cos B + sin A sin B ). Applying this, I get:[cosleft(frac{pi}{3} - xright) = cosleft(frac{pi}{3}right)cos x + sinleft(frac{pi}{3}right)sin x]I know that ( cosleft(frac{pi}{3}right) = frac{1}{2} ) and ( sinleft(frac{pi}{3}right) = frac{sqrt{3}}{2} ), so substituting these values in:[cosleft(frac{pi}{3} - xright) = frac{1}{2}cos x + frac{sqrt{3}}{2}sin x]Now, moving on to the second term, ( sinleft(frac{pi}{2} + xright) ). I remember the identity ( sin(A + B) = sin A cos B + cos A sin B ). Applying this:[sinleft(frac{pi}{2} + xright) = sinleft(frac{pi}{2}right)cos x + cosleft(frac{pi}{2}right)sin x]I know that ( sinleft(frac{pi}{2}right) = 1 ) and ( cosleft(frac{pi}{2}right) = 0 ), so substituting these:[sinleft(frac{pi}{2} + xright) = 1 cdot cos x + 0 cdot sin x = cos x]So now, substituting both simplified terms back into the original function:[f(x) = left(frac{1}{2}cos x + frac{sqrt{3}}{2}sin xright) + cos x]Combine like terms:[f(x) = frac{1}{2}cos x + cos x + frac{sqrt{3}}{2}sin x = frac{3}{2}cos x + frac{sqrt{3}}{2}sin x]Okay, so now the function is simplified to ( f(x) = frac{3}{2}cos x + frac{sqrt{3}}{2}sin x ). This looks like a linear combination of sine and cosine, which I remember can be written in the form ( R sin(x + phi) ) or ( R cos(x + phi) ), where ( R ) is the amplitude and ( phi ) is the phase shift.The formula for combining ( a sin x + b cos x ) into a single sine function is:[a sin x + b cos x = R sin(x + phi)]where ( R = sqrt{a^2 + b^2} ) and ( phi = arctanleft(frac{b}{a}right) ).In our case, ( a = frac{sqrt{3}}{2} ) and ( b = frac{3}{2} ). Let me compute ( R ):[R = sqrt{left(frac{sqrt{3}}{2}right)^2 + left(frac{3}{2}right)^2} = sqrt{frac{3}{4} + frac{9}{4}} = sqrt{frac{12}{4}} = sqrt{3}]So, ( R = sqrt{3} ). That means the amplitude of the function is ( sqrt{3} ).Now, the function can be rewritten as:[f(x) = sqrt{3} sin(x + phi)]where ( phi ) is some phase shift. However, since we're interested in the range of the function, the phase shift doesn't affect the range—it only affects where the maximum and minimum occur. The range of ( sin ) function is always ([-1, 1]), so multiplying by ( sqrt{3} ) scales this to ([- sqrt{3}, sqrt{3}]).Therefore, the range of ( f(x) ) is ([- sqrt{3}, sqrt{3}]), which corresponds to option B.Wait, let me double-check my steps to make sure I didn't make a mistake. I used the angle subtraction and addition formulas correctly, substituted the known values of sine and cosine for the specific angles, combined like terms, and then used the amplitude formula for a linear combination of sine and cosine. It all seems to add up. So, yeah, I think B is the correct answer.**Final Answer**boxed{B}

answer：Alright, so I've got this vector problem here with four options, A through D. I need to figure out which one is correct. Let me take it step by step.First, let's understand what each option is saying.Option A: If the magnitude of the sum of vectors a and b is equal to the magnitude of a minus the magnitude of b, then a and b are perpendicular. Hmm, okay. So, |a + b| = |a| - |b| implies a perpendicular to b. I'm not sure about that. I remember that when vectors are perpendicular, their dot product is zero, and the magnitude of their sum is calculated using the Pythagorean theorem. But does |a + b| = |a| - |b| necessarily mean they are perpendicular? Maybe not. Let me think of an example.Suppose a is (3, 0) and b is (-1, 0). Then a + b is (2, 0), so |a + b| is 2. |a| is 3, |b| is 1, so |a| - |b| is 2. So in this case, |a + b| equals |a| - |b|, but a and b are not perpendicular; they're actually colinear but pointing in opposite directions. So, this would mean that Option A is incorrect because the vectors don't have to be perpendicular for |a + b| to equal |a| - |b|.Option B: If a and b are perpendicular, then |a + b| equals |a| - |b|. Wait, if they are perpendicular, then the magnitude of their sum should be sqrt(|a|² + |b|²), right? Because when vectors are perpendicular, the Pythagorean theorem applies. So, unless one of the vectors has zero magnitude, |a + b| would be greater than both |a| and |b|. Therefore, |a + b| can't be |a| - |b| unless one of them is zero, but the problem states both are non-zero vectors. So, Option B is also incorrect.Option C: If |a + b| equals |a| - |b|, then there exists a real number λ such that a = λb. Hmm, okay. So, if the magnitude of the sum is |a| - |b|, does that mean they are scalar multiples of each other? Let's see. From my earlier example, when a was (3, 0) and b was (-1, 0), |a + b| was 2, which is |a| - |b|, and indeed, a was a scalar multiple of b (λ = -3). So, in that case, it worked. Is this always true?Let me think. If |a + b| = |a| - |b|, then the vectors must be pointing in exactly opposite directions. Because if they weren't, the magnitude of the sum would be larger. So, that would mean one is a scalar multiple of the other with a negative scalar. So, yes, there must exist a real number λ such that a = λb. So, Option C seems correct.Option D: If there exists a real number λ such that a = λb, then |a + b| = |a| - |b|. Hmm, so if a and b are scalar multiples, does that mean their sum's magnitude is |a| - |b|? Wait, let's test this with an example. Suppose a is (2, 0) and b is (1, 0). Then a = 2b. So, |a + b| is |(3, 0)| = 3. |a| is 2, |b| is 1, so |a| - |b| is 1. But 3 is not equal to 1. So, in this case, even though a is a scalar multiple of b, |a + b| is not equal to |a| - |b|. Therefore, Option D is incorrect.So, summarizing:- Option A: Incorrect, because the vectors don't have to be perpendicular.- Option B: Incorrect, because when vectors are perpendicular, |a + b| is larger than both |a| and |b|.- Option C: Correct, because if |a + b| equals |a| - |b|, the vectors must be colinear and pointing in opposite directions.- Option D: Incorrect, because even if vectors are colinear, their sum's magnitude depends on the direction; it's only when they are in opposite directions that |a + b| equals |a| - |b|.Therefore, the correct answer is Option C.**Final Answer**boxed{C}

answer：Alright, so I have this problem about the function ( f(x) = e^x + a sin x + b ). It has two parts, and I need to solve both. Let me take them one at a time.Starting with part (1): When ( a = 1 ) and ( x in [0, +infty) ), the function ( f(x) geq 0 ) always holds. I need to find the range of ( b ).Okay, so ( f(x) = e^x + sin x + b ). The problem says that this function is always non-negative for all ( x ) starting from 0 to infinity. So, I need to ensure that ( e^x + sin x + b geq 0 ) for all ( x geq 0 ).Hmm, I think I should analyze the behavior of ( f(x) ) to find the minimum value it can take. If I can find the minimum value of ( f(x) ) on ( [0, +infty) ), then setting that minimum to be greater than or equal to zero will give me the condition on ( b ).To find the minimum, I can take the derivative of ( f(x) ) and find its critical points. So, let me compute ( f'(x) ):( f'(x) = frac{d}{dx} (e^x + sin x + b) = e^x + cos x ).Now, I need to find where ( f'(x) = 0 ):( e^x + cos x = 0 ).But wait, ( e^x ) is always positive for all real ( x ), and ( cos x ) oscillates between -1 and 1. So, ( e^x + cos x ) will always be greater than or equal to ( e^x - 1 ). Since ( e^x ) is at least 1 when ( x geq 0 ), ( e^x - 1 geq 0 ). Therefore, ( f'(x) geq 0 ) for all ( x geq 0 ).This means that ( f(x) ) is monotonically increasing on ( [0, +infty) ). So, the minimum value of ( f(x) ) occurs at ( x = 0 ).Calculating ( f(0) ):( f(0) = e^0 + sin 0 + b = 1 + 0 + b = 1 + b ).Since ( f(x) geq 0 ) for all ( x geq 0 ), the minimum value ( 1 + b ) must be greater than or equal to 0:( 1 + b geq 0 Rightarrow b geq -1 ).So, for part (1), the range of ( b ) is ( b geq -1 ).Moving on to part (2): It says that the tangent line of ( f(x) ) at ( x = 0 ) is ( x - y - 1 = 0 ). I need to find the range of the real number ( m ) such that the equation ( f(x) = frac{m - 2x}{x} ) has exactly two solutions.First, let me parse the given information. The tangent line at ( x = 0 ) is given by ( x - y - 1 = 0 ). I can rewrite this equation in slope-intercept form to find the slope and the point of tangency.Rewriting ( x - y - 1 = 0 ):( -y = -x + 1 Rightarrow y = x - 1 ).So, the tangent line is ( y = x - 1 ). This means that at ( x = 0 ), the function ( f(x) ) has the same value as the tangent line, and their derivatives are equal.Let me compute ( f(0) ) and ( f'(0) ):( f(0) = e^0 + a sin 0 + b = 1 + 0 + b = 1 + b ).( f'(x) = e^x + a cos x ), so ( f'(0) = e^0 + a cos 0 = 1 + a times 1 = 1 + a ).Now, the tangent line at ( x = 0 ) is ( y = x - 1 ), so the slope of the tangent line is 1. Therefore, ( f'(0) = 1 ):( 1 + a = 1 Rightarrow a = 0 ).Also, the point ( (0, f(0)) ) lies on the tangent line ( y = x - 1 ). Plugging ( x = 0 ) into the tangent line equation:( y = 0 - 1 = -1 ).Therefore, ( f(0) = -1 ). But we have ( f(0) = 1 + b ), so:( 1 + b = -1 Rightarrow b = -2 ).So, now we know ( a = 0 ) and ( b = -2 ), so the function is ( f(x) = e^x + 0 times sin x - 2 = e^x - 2 ).Now, the equation given is ( f(x) = frac{m - 2x}{x} ). Let me write that out:( e^x - 2 = frac{m - 2x}{x} ).I need to solve this equation for ( x ) and find the values of ( m ) such that there are exactly two solutions.First, let me simplify the equation:( e^x - 2 = frac{m - 2x}{x} ).Multiply both sides by ( x ) to eliminate the denominator (assuming ( x neq 0 )):( x(e^x - 2) = m - 2x ).Let me expand the left side:( x e^x - 2x = m - 2x ).Now, I can add ( 2x ) to both sides to simplify:( x e^x = m ).So, the equation simplifies to ( x e^x = m ).Therefore, the problem reduces to finding the number of solutions to ( x e^x = m ). We need exactly two solutions for ( x ).Let me analyze the function ( g(x) = x e^x ). I need to find the values of ( m ) for which the horizontal line ( y = m ) intersects the graph of ( g(x) ) exactly twice.First, let's find the critical points of ( g(x) ) to understand its behavior.Compute the derivative ( g'(x) ):( g'(x) = frac{d}{dx} (x e^x) = e^x + x e^x = e^x (1 + x) ).Set ( g'(x) = 0 ):( e^x (1 + x) = 0 ).Since ( e^x ) is never zero, the critical point occurs when ( 1 + x = 0 Rightarrow x = -1 ).So, ( x = -1 ) is the critical point. Let's analyze the behavior of ( g(x) ) around this point.For ( x < -1 ), let's pick ( x = -2 ):( g'(-2) = e^{-2} (1 - 2) = e^{-2} (-1) < 0 ). So, ( g(x) ) is decreasing on ( (-infty, -1) ).For ( x > -1 ), let's pick ( x = 0 ):( g'(0) = e^0 (1 + 0) = 1 > 0 ). So, ( g(x) ) is increasing on ( (-1, +infty) ).Therefore, ( x = -1 ) is a local minimum.Compute ( g(-1) ):( g(-1) = (-1) e^{-1} = -frac{1}{e} ).Now, let's analyze the limits:As ( x to -infty ), ( x e^x ) approaches 0 because ( e^x ) decays to zero faster than ( x ) goes to negative infinity. So, ( lim_{x to -infty} g(x) = 0 ).As ( x to +infty ), ( x e^x ) goes to infinity because both ( x ) and ( e^x ) grow without bound.Also, at ( x = 0 ), ( g(0) = 0 times e^0 = 0 ).So, putting this all together, the graph of ( g(x) = x e^x ) has the following behavior:- As ( x to -infty ), ( g(x) to 0 ) from below (since ( x ) is negative and ( e^x ) is positive, so ( g(x) ) is negative approaching zero).- It decreases until ( x = -1 ), reaching a minimum of ( -1/e ).- Then, it increases from ( x = -1 ) onwards, passing through ( (0, 0) ) and going to infinity as ( x to +infty ).So, the graph has a "U" shape on the right side of ( x = -1 ) and approaches zero from below on the left side.Now, the equation ( g(x) = m ) will have:- Two solutions when ( m ) is between the minimum value ( -1/e ) and zero.- One solution when ( m = -1/e ) or ( m geq 0 ).- No solution when ( m < -1/e ).But wait, let me verify that.Wait, when ( m = -1/e ), the equation ( g(x) = m ) has exactly one solution at ( x = -1 ).When ( m > 0 ), the equation ( g(x) = m ) has exactly one solution on ( x > 0 ) because ( g(x) ) is increasing there and goes to infinity.When ( m = 0 ), the equation ( g(x) = 0 ) has solutions at ( x = 0 ) and as ( x to -infty ), but actually, ( g(x) ) approaches zero from below as ( x to -infty ), so ( g(x) = 0 ) only at ( x = 0 ). So, only one solution.Wait, that contradicts my earlier thought. Let me think again.Wait, ( g(x) = x e^x ). When ( x = 0 ), ( g(x) = 0 ). For ( x > 0 ), ( g(x) ) increases from 0 to infinity. For ( x < 0 ), ( g(x) ) approaches 0 from below as ( x to -infty ) and has a minimum at ( x = -1 ) of ( -1/e ).So, for ( m > 0 ), the equation ( g(x) = m ) has exactly one solution in ( x > 0 ).For ( m = 0 ), the equation ( g(x) = 0 ) has exactly one solution at ( x = 0 ).For ( -1/e < m < 0 ), the equation ( g(x) = m ) has two solutions: one in ( x < -1 ) and another in ( -1 < x < 0 ).For ( m = -1/e ), the equation has exactly one solution at ( x = -1 ).For ( m < -1/e ), the equation has no solutions because ( g(x) ) never goes below ( -1/e ).Therefore, to have exactly two solutions, ( m ) must satisfy ( -1/e < m < 0 ).So, the range of ( m ) is ( (-1/e, 0) ).But wait, let me double-check.If ( m ) is between ( -1/e ) and 0, then ( g(x) = m ) will have two solutions: one on ( (-infty, -1) ) and another on ( (-1, 0) ). That makes sense because ( g(x) ) is decreasing on ( (-infty, -1) ) and increasing on ( (-1, +infty) ). So, for ( m ) between the minimum and zero, the horizontal line will intersect the graph twice.Yes, that seems correct.So, putting it all together, for part (2), the range of ( m ) is ( -1/e < m < 0 ).**Final Answer**- (1) (boxed{b geqslant -1})- (2) (boxed{-dfrac{1}{e} < m < 0})

answer：Okay, so I have this problem here where I need to find the sum of f(n) from n=1 to n=2048. The function f(n) is defined as log base 8 of n if that logarithm is rational, otherwise it's 0. Hmm, interesting. Let me try to break this down step by step.First, I need to understand when log₈(n) is rational. Remember, a rational number is a fraction where both the numerator and the denominator are integers. So, log₈(n) being rational means that it can be expressed as a fraction a/b where a and b are integers and b is not zero.I know that logarithms can be tricky, but maybe I can express n in terms of powers of 8 to make this easier. If I let n = 8^k, where k is a rational number, then log₈(n) would just be k. But wait, n has to be a positive integer, so 8^k must also be an integer. That means k has to be such that 8^k is an integer. Since 8 is 2^3, maybe I can express k as a fraction with denominator 3. Let me think about that.If I write k as m/3, where m is an integer, then 8^(m/3) = (2^3)^(m/3) = 2^m. So, n would be 2^m. That makes sense because 2^m is definitely an integer for integer m. Therefore, log₈(n) is rational if and only if n is a power of 2. So, n can be 2^0, 2^1, 2^2, ..., up to 2^k where 2^k ≤ 2048.Now, I need to figure out the maximum value of k such that 2^k ≤ 2048. Let's calculate that. 2^10 is 1024, and 2^11 is 2048. So, k can range from 0 to 11. That means n can be 1, 2, 4, 8, ..., up to 2048.For each of these n, f(n) is log₈(n). Let's express log₈(n) in terms of m. Since n = 2^m, log₈(2^m) = m * log₈(2). I know that log₈(2) is 1/3 because 8^(1/3) = 2. So, log₈(2^m) = m/3. Therefore, f(n) = m/3 for n = 2^m.Now, I need to sum f(n) for all n from 1 to 2048. But since f(n) is zero unless n is a power of 2, I only need to sum f(n) for n = 2^0, 2^1, ..., 2^11. Each of these corresponds to m from 0 to 11, so the sum becomes the sum of m/3 for m from 0 to 11.Let me write that out:Sum = (0/3) + (1/3) + (2/3) + (3/3) + ... + (11/3)This is the same as (1/3) times the sum of m from 0 to 11.The sum of the first k integers starting from 0 is (k)(k+1)/2. Here, k is 11, so the sum is (11)(12)/2 = 66.Therefore, the total sum is (1/3) * 66 = 22.Wait, let me double-check my steps to make sure I didn't make a mistake. I started by recognizing that log₈(n) is rational only when n is a power of 2, which makes sense because 8 is 2^3. Then, I expressed n as 2^m and found that log₈(n) = m/3. I identified the range of m from 0 to 11 because 2^11 is 2048. Then, I summed m/3 from m=0 to m=11, which is (1/3) times the sum of m from 0 to 11. The sum of m from 0 to 11 is indeed 66, so 66 divided by 3 is 22. That seems correct.I don't see any errors in my reasoning, so I think the answer is 22.

answer：Okay, so I need to find the equation of a hyperbola C. The problem says that the hyperbola has its center at the origin O, and its foci are on the y-axis. That means it's a vertical hyperbola. The eccentricity is given as √5 / 2. There's also a circle with radius 2 that's tangent to one of the asymptotes of the hyperbola, and the center of this circle is one of the foci of the hyperbola.First, let me recall some properties of hyperbolas. For a vertical hyperbola centered at the origin, the standard form is:(y² / a²) - (x² / b²) = 1Where:- a is the distance from the center to the vertices along the y-axis.- b is related to the conjugate axis.- The foci are located at (0, ±c), where c² = a² + b².- The eccentricity e is given by e = c / a.Given that the eccentricity e is √5 / 2, so:e = c / a = √5 / 2Which means c = (√5 / 2) * aAlso, since c² = a² + b², substituting c:( (√5 / 2) * a )² = a² + b²Simplify:(5/4) * a² = a² + b²Subtract a² from both sides:(5/4)a² - a² = b²Which simplifies to:(1/4)a² = b²So, b = (1/2)aAlright, so b is half of a.Now, the problem mentions a circle with radius 2 that's tangent to the asymptote of the hyperbola, and the center of this circle is one of the foci. So, the circle is centered at (0, c) or (0, -c), and it's tangent to one of the asymptotes.Let me recall that the asymptotes of a vertical hyperbola are given by:y = ±(a / b) xSince b = (1/2)a, then a / b = a / (a/2) = 2So, the asymptotes are y = ±2xSo, the equations of the asymptotes are y = 2x and y = -2x.Now, the circle is centered at (0, c) and has a radius of 2. It's tangent to one of these asymptotes. So, the distance from the center of the circle (0, c) to the asymptote must be equal to the radius, which is 2.The formula for the distance from a point (x0, y0) to the line Ax + By + C = 0 is |Ax0 + By0 + C| / sqrt(A² + B²).Let me write the asymptote in standard form. Let's take y = 2x. To write it as Ax + By + C = 0, we can rearrange it to 2x - y = 0.So, A = 2, B = -1, C = 0.The distance from (0, c) to this line is:|2*0 + (-1)*c + 0| / sqrt(2² + (-1)²) = | -c | / sqrt(4 + 1) = |c| / sqrt(5)Since distance can't be negative, it's c / sqrt(5). This distance is equal to the radius, which is 2.So:c / sqrt(5) = 2Multiply both sides by sqrt(5):c = 2 * sqrt(5)But earlier, we had c = (√5 / 2) * aSo, set them equal:(√5 / 2) * a = 2 * sqrt(5)Multiply both sides by 2:√5 * a = 4 * sqrt(5)Divide both sides by sqrt(5):a = 4So, a = 4.Since b = (1/2)a, then b = 2.Now, we can write the equation of the hyperbola.It's a vertical hyperbola, so:(y² / a²) - (x² / b²) = 1Substituting a = 4 and b = 2:(y² / 16) - (x² / 4) = 1Looking at the options:A: x²/16 - y²/4 = 1 → This is a horizontal hyperbola, so incorrect.B: y²/4 - x² = 1 → This has a = 2, but we found a = 4, so incorrect.C: y² - x²/4 = 1 → This is similar to B but without denominators under y², so incorrect.D: y²/16 - x²/4 = 1 → This matches our equation, so correct.Therefore, the answer is D.**Final Answer**boxed{D}

answer：Alright, so I have this sequence problem here. It says that starting from the second number, each number is equal to the difference between 1 and the reciprocal of the number before it. They give me that the first number, a₁, is 2, and they want me to find a₂₀₁₅. Hmm, okay, let's break this down step by step.First, let me make sure I understand the rule correctly. Each term after the first is defined as 1 minus the reciprocal of the previous term. So, mathematically, that would be:aₙ = 1 - (1 / aₙ₋₁)Got it. So, starting with a₁ = 2, I can find a₂, then a₃, and so on, using this formula. Maybe if I compute the first few terms, I can see a pattern or something.Let's start:a₁ = 2a₂ = 1 - (1 / a₁) = 1 - (1 / 2) = 1 - 0.5 = 0.5Okay, so a₂ is 0.5. Now, let's find a₃:a₃ = 1 - (1 / a₂) = 1 - (1 / 0.5) = 1 - 2 = -1Hmm, a₃ is -1. Interesting. Now, a₄:a₄ = 1 - (1 / a₃) = 1 - (1 / (-1)) = 1 - (-1) = 1 + 1 = 2Wait a minute, a₄ is back to 2. That's the same as a₁. So, does this mean the sequence repeats every three terms? Let's check a₅ to be sure.a₅ = 1 - (1 / a₄) = 1 - (1 / 2) = 0.5Yep, that's the same as a₂. So, it seems like the sequence cycles every three terms: 2, 0.5, -1, 2, 0.5, -1, and so on.So, if the sequence repeats every three terms, then to find a₂₀₁₅, I can figure out where 2015 falls in the cycle. That is, I can divide 2015 by 3 and see what the remainder is. The remainder will tell me which term in the cycle a₂₀₁₅ corresponds to.Let me do that division. 2015 divided by 3. Let's see:3 goes into 20 six times (3*6=18), remainder 2.Bring down the 1: 21. 3 goes into 21 seven times exactly, remainder 0.Bring down the 5: 3 goes into 5 once, remainder 2.Wait, that seems a bit confusing. Maybe I should just compute 2015 ÷ 3.Alternatively, I can think of it as 3*671 = 2013, right? Because 3*700 is 2100, which is too big, so 3*600=1800, 3*70=210, so 600+70=670, 3*670=2010. Then 2015 - 2010 is 5, so 3*1=3, so 2015 is 3*671 + 2, because 671*3=2013, and 2015-2013=2.So, 2015 divided by 3 is 671 with a remainder of 2. That means a₂₀₁₅ is the same as a₂, which is 0.5.Wait, let me double-check that. If the cycle is 2, 0.5, -1, then the first term is 2 (remainder 1), the second term is 0.5 (remainder 2), the third term is -1 (remainder 0), and then it repeats. So, if the remainder is 1, it's 2; remainder 2, it's 0.5; and if there's no remainder, it's -1.Since 2015 divided by 3 gives a remainder of 2, that corresponds to the second term in the cycle, which is 0.5. So, a₂₀₁₅ is 0.5.Just to make sure I didn't make a mistake earlier, let me compute a few more terms to confirm the cycle.a₁ = 2a₂ = 1 - 1/2 = 0.5a₃ = 1 - 1/0.5 = 1 - 2 = -1a₄ = 1 - 1/(-1) = 1 + 1 = 2a₅ = 1 - 1/2 = 0.5a₆ = 1 - 1/0.5 = -1a₇ = 1 - 1/(-1) = 2Yep, it definitely cycles every three terms. So, my earlier conclusion seems solid.Therefore, since 2015 mod 3 is 2, a₂₀₁₅ is equal to a₂, which is 0.5.Just to wrap up, 0.5 is the same as 1/2, so I can write the answer as 1/2 in a box.**Final Answer**The value of ( a_{2015} ) is boxed{dfrac{1}{2}}.