💡Okay, so I have this problem where I'm given a sequence defined recursively. It starts with S₁ equals a, and then each subsequent term is 2 divided by the previous term. So S₂ is 2 over S₁, S₃ is 2 over S₂, and so on, all the way up to S₂₀₁₈. I need to express S₂₀₁₈ in terms of a. Hmm, that seems a bit involved, but maybe there's a pattern here.Let me write down the first few terms to see if I can spot a pattern. - S₁ = a- S₂ = 2 / S₁ = 2 / a- S₃ = 2 / S₂ = 2 / (2 / a) = a- S₄ = 2 / S₃ = 2 / a- S₅ = 2 / S₄ = 2 / (2 / a) = aWait a minute, this seems to be alternating between a and 2/a. So S₁ is a, S₂ is 2/a, S₃ is a, S₄ is 2/a, and so on. So it's like a cycle of two terms: a, 2/a, a, 2/a, etc.If that's the case, then the sequence is periodic with period 2. That means every odd term is a, and every even term is 2/a. So if I can figure out whether 2018 is odd or even, I can determine what S₂₀₁₈ is.Well, 2018 divided by 2 is 1009, which is an integer, so 2018 is even. That means S₂₀₁₈ should be equal to 2/a.But wait, let me double-check. Maybe I made a mistake in assuming the pattern continues indefinitely. Let me calculate a few more terms to see if the pattern holds.- S₅ = a- S₆ = 2 / S₅ = 2 / a- S₇ = 2 / S₆ = a- S₈ = 2 / S₇ = 2 / aYeah, it's definitely alternating. So every even term is 2/a, and every odd term is a. Since 2018 is even, S₂₀₁₈ is 2/a.I think that makes sense. The sequence alternates between a and 2/a, so depending on whether the term number is odd or even, it takes one of those two values. Since 2018 is even, it's 2/a.Just to make sure, let me think about why this pattern occurs. Starting from S₁ = a, each term is defined as 2 divided by the previous term. So S₂ is 2/a, then S₃ is 2 divided by S₂, which is 2/(2/a) = a. Then S₄ is 2 divided by S₃, which is 2/a again. So it's a repeating cycle of two terms.Therefore, for any term number n, if n is odd, it's a, and if n is even, it's 2/a. Since 2018 is even, S₂₀₁₈ is 2/a.I don't see any mistakes in this reasoning. The pattern holds consistently, and the term number being even determines the value. So I'm confident that S₂₀₁₈ is 2/a.

💡First, calculate the mass of each ball based on its number:1. For ( x = 1 ): mass ( = (1^2 - 5 cdot 1 + 30) = 1 - 5 + 30 = 26 ) grams.2. For ( x = 2 ): mass ( = (2^2 - 5 cdot 2 + 30) = 4 - 10 + 30 = 24 ) grams.3. For ( x = 3 ): mass ( = (3^2 - 5 cdot 3 + 30) = 9 - 15 + 30 = 24 ) grams.4. For ( x = 4 ): mass ( = (4^2 - 5 cdot 4 + 30) = 16 - 20 + 30 = 26 ) grams.5. For ( x = 5 ): mass ( = (5^2 - 5 cdot 5 + 30) = 25 - 25 + 30 = 30 ) grams.Identify pairs of balls with the same mass:- Balls 1 and 4 have a mass of 26 grams.- Balls 2 and 3 have a mass of 24 grams.Calculate the total number of ways to draw two balls from five:[binom{5}{2} = frac{5!}{2!(5-2)!} = 10]Determine the number of favorable outcomes (pairs with the same mass):- There are 2 favorable pairs.Calculate the probability:[P(text{same mass}) = frac{text{Number of favorable outcomes}}{text{Total possible outcomes}} = frac{2}{10} = frac{1}{5}]

💡First, I need to determine how many years into the future Ron will be four times as old as Maurice. Let's denote this number of years as ( x ).Currently, Ron is 43 years old, and Maurice is 7 years old. In ( x ) years, Ron's age will be ( 43 + x ) and Maurice's age will be ( 7 + x ).According to the problem, Ron's age at that future time will be four times Maurice's age. This gives us the equation:[43 + x = 4 times (7 + x)]Next, I'll solve this equation for ( x ). Expanding the right side:[43 + x = 28 + 4x]Subtracting ( 28 ) from both sides:[15 + x = 4x]Subtracting ( x ) from both sides:[15 = 3x]Finally, dividing both sides by 3:[x = 5]Therefore, in 5 years, Ron will be four times as old as Maurice.

💡Alright, so I have this problem about complex numbers and geometry. Let me try to break it down step by step. First, the problem says that z is a complex number with |z| = 1, which means z lies on the unit circle in the complex plane. The conjugate of z is denoted as (overline{z}). We have points A at (-1, 0) and B at (0, -1). The function given is (f(z) = |(z + 1)(overline{z} - i)|), and we need to find the shape formed by points Z, A, and B when f(z) is maximized.Okay, so since |z| = 1, z can be written in polar form as (z = costheta + isintheta). That makes sense because any point on the unit circle can be represented that way. The conjugate of z, (overline{z}), would then be (costheta - isintheta).Now, let's substitute z and (overline{z}) into the function f(z):(f(z) = |(costheta + isintheta + 1)(costheta - isintheta - i)|)Hmm, that looks a bit complicated. Maybe I can simplify each part separately.First, let's look at (z + 1):(z + 1 = costheta + isintheta + 1 = (1 + costheta) + isintheta)Okay, that's straightforward. Now, (overline{z} - i):(overline{z} - i = costheta - isintheta - i = costheta - i(sintheta + 1))So, now we have:(f(z) = |(1 + costheta + isintheta)(costheta - i(sintheta + 1))|)To find the modulus of the product, I can multiply the two complex numbers and then take the modulus. Let me compute the product first.Multiplying the two complex numbers:((1 + costheta + isintheta)(costheta - i(sintheta + 1)))Let me distribute this multiplication:First, multiply (1 + costheta) with (costheta):((1 + costheta)costheta = costheta + cos^2theta)Next, multiply (1 + costheta) with (-i(sintheta + 1)):((1 + costheta)(-i)(sintheta + 1) = -i(1 + costheta)(sintheta + 1))Similarly, multiply (isintheta) with (costheta):(isintheta cdot costheta = isinthetacostheta)And multiply (isintheta) with (-i(sintheta + 1)):(isintheta cdot (-i)(sintheta + 1) = -i^2sintheta(sintheta + 1))Since (i^2 = -1), this becomes:(sintheta(sintheta + 1))Okay, so putting it all together:The real parts are:(costheta + cos^2theta + sintheta(sintheta + 1))And the imaginary parts are:(-i(1 + costheta)(sintheta + 1) + isinthetacostheta)Let me simplify the real parts first:(costheta + cos^2theta + sin^2theta + sintheta)I know that (cos^2theta + sin^2theta = 1), so this simplifies to:(costheta + 1 + sintheta)So the real part is (1 + costheta + sintheta).Now, the imaginary parts:(-i(1 + costheta)(sintheta + 1) + isinthetacostheta)Let me factor out the i:(i[-(1 + costheta)(sintheta + 1) + sinthetacostheta])Let me expand (-(1 + costheta)(sintheta + 1)):(-( sintheta + 1 + costhetasintheta + costheta ))So, the expression inside the brackets becomes:(-sintheta - 1 - costhetasintheta - costheta + sinthetacostheta)Wait, the (-costhetasintheta) and (+sinthetacostheta) cancel each other out. So we're left with:(-sintheta - 1 - costheta)Therefore, the imaginary part is:(i(-sintheta - 1 - costheta))So, putting it all together, the product is:((1 + costheta + sintheta) + i(-sintheta - 1 - costheta))Which can be written as:((1 + costheta + sintheta) - i(1 + sintheta + costheta))Now, to find the modulus of this complex number, we use the formula ( |a + ib| = sqrt{a^2 + b^2} ).So, let's compute the modulus squared first:((1 + costheta + sintheta)^2 + (1 + sintheta + costheta)^2)Wait, both terms are the same, so this is:(2 times (1 + costheta + sintheta)^2)Therefore, the modulus is:(sqrt{2} times |1 + costheta + sintheta|)Since modulus is always non-negative, we can drop the absolute value:(f(z) = sqrt{2} times (1 + costheta + sintheta))But wait, (1 + costheta + sintheta) can be rewritten using a trigonometric identity. Let me recall that:(costheta + sintheta = sqrt{2}sinleft(theta + frac{pi}{4}right))Yes, that's correct. Because:(costheta + sintheta = sqrt{2}sinleft(theta + frac{pi}{4}right))So, substituting back, we have:(f(z) = sqrt{2} times left(1 + sqrt{2}sinleft(theta + frac{pi}{4}right)right))Which simplifies to:(f(z) = sqrt{2} + 2sinleft(theta + frac{pi}{4}right))Wait, no, hold on. Let me check that again.If (costheta + sintheta = sqrt{2}sinleft(theta + frac{pi}{4}right)), then:(1 + costheta + sintheta = 1 + sqrt{2}sinleft(theta + frac{pi}{4}right))Therefore, (f(z) = sqrt{2} times left(1 + sqrt{2}sinleft(theta + frac{pi}{4}right)right))So, (f(z) = sqrt{2} + 2sinleft(theta + frac{pi}{4}right))Wait, that doesn't seem right. Let me compute it step by step.We have:(f(z) = sqrt{2} times (1 + costheta + sintheta))But (1 + costheta + sintheta = 1 + sqrt{2}sinleft(theta + frac{pi}{4}right))So, substituting:(f(z) = sqrt{2} times left(1 + sqrt{2}sinleft(theta + frac{pi}{4}right)right))Which is:(f(z) = sqrt{2} + 2sinleft(theta + frac{pi}{4}right))Yes, that's correct.Now, to find the maximum value of f(z), we need to maximize this expression. The maximum value of (sinleft(theta + frac{pi}{4}right)) is 1, so the maximum value of f(z) is:(f(z)_{text{max}} = sqrt{2} + 2 times 1 = sqrt{2} + 2)So, the maximum value is (2 + sqrt{2}).Now, we need to find the value of (theta) where this maximum occurs. Since (sinleft(theta + frac{pi}{4}right) = 1), we have:(theta + frac{pi}{4} = frac{pi}{2} + 2kpi), where k is an integer.Therefore, (theta = frac{pi}{2} - frac{pi}{4} + 2kpi = frac{pi}{4} + 2kpi).So, (theta = frac{pi}{4}) is the angle where the maximum occurs. Therefore, the point Z is at:(z = cosfrac{pi}{4} + isinfrac{pi}{4} = frac{sqrt{2}}{2} + ifrac{sqrt{2}}{2})So, the coordinates of Z are (left(frac{sqrt{2}}{2}, frac{sqrt{2}}{2}right)).Now, we have points A(-1, 0), B(0, -1), and Z(left(frac{sqrt{2}}{2}, frac{sqrt{2}}{2}right)). We need to determine the shape of triangle ZAB.To find the shape, let's compute the distances between each pair of points.First, distance between Z and A:( |ZA| = sqrt{left(frac{sqrt{2}}{2} - (-1)right)^2 + left(frac{sqrt{2}}{2} - 0right)^2} )Simplify:( |ZA| = sqrt{left(frac{sqrt{2}}{2} + 1right)^2 + left(frac{sqrt{2}}{2}right)^2} )Compute each term:(left(frac{sqrt{2}}{2} + 1right)^2 = left(1 + frac{sqrt{2}}{2}right)^2 = 1 + sqrt{2} + frac{2}{4} = 1 + sqrt{2} + frac{1}{2} = frac{3}{2} + sqrt{2})(left(frac{sqrt{2}}{2}right)^2 = frac{2}{4} = frac{1}{2})So, |ZA| squared is:(frac{3}{2} + sqrt{2} + frac{1}{2} = 2 + sqrt{2})Therefore, |ZA| = (sqrt{2 + sqrt{2}})Similarly, let's compute |ZB|:( |ZB| = sqrt{left(frac{sqrt{2}}{2} - 0right)^2 + left(frac{sqrt{2}}{2} - (-1)right)^2} )Simplify:( |ZB| = sqrt{left(frac{sqrt{2}}{2}right)^2 + left(frac{sqrt{2}}{2} + 1right)^2} )Which is the same as |ZA|, so |ZB| squared is also 2 + (sqrt{2}), hence |ZB| = (sqrt{2 + sqrt{2}})Now, let's compute |AB|:Points A(-1, 0) and B(0, -1):( |AB| = sqrt{(0 - (-1))^2 + (-1 - 0)^2} = sqrt{1 + 1} = sqrt{2} )So, |AB| = (sqrt{2})Now, let's summarize the distances:- |ZA| = |ZB| = (sqrt{2 + sqrt{2}})- |AB| = (sqrt{2})So, triangle ZAB has two sides equal (|ZA| = |ZB|) and one side different. Therefore, it is an isosceles triangle.But wait, let's check if it's a right-angled triangle as well. For that, we can use the Pythagorean theorem.If triangle ZAB is right-angled, then the sum of squares of two sides should equal the square of the third side.Compute |ZA|² + |ZB|²:( (2 + sqrt{2}) + (2 + sqrt{2}) = 4 + 2sqrt{2} )Compare with |AB|²:( (sqrt{2})^2 = 2 )4 + 2√2 ≠ 2, so it's not a right-angled triangle.Alternatively, check |ZA|² + |AB|²:( (2 + sqrt{2}) + 2 = 4 + sqrt{2} )Compare with |ZB|²:( 2 + sqrt{2} )Not equal.Similarly, |ZB|² + |AB|² = same as above.So, it's not a right-angled triangle. Therefore, it's just an isosceles triangle.Wait, but let me think again. Maybe I made a mistake in calculating the distances.Wait, |ZA| squared was 2 + √2, and |AB| squared was 2. So, 2 + √2 is approximately 3.414, and 2 is 2. So, 2 + √2 is not equal to 2, so the triangle is not right-angled.Therefore, the triangle is isosceles but not right-angled. So, the answer should be D: Isosceles triangle.But wait, let me check the options again:A: Equilateral triangleB: Right-angled triangleC: Isosceles right-angled triangleD: Isosceles triangleSo, since it's isosceles but not right-angled, D is the correct answer.But wait, let me think again. Maybe I made a mistake in computing the distances.Wait, when I computed |ZA| squared, I had:(left(frac{sqrt{2}}{2} + 1right)^2 + left(frac{sqrt{2}}{2}right)^2)Which is:(left(1 + frac{sqrt{2}}{2}right)^2 + left(frac{sqrt{2}}{2}right)^2)Expanding (left(1 + frac{sqrt{2}}{2}right)^2):(1 + sqrt{2} + frac{2}{4} = 1 + sqrt{2} + frac{1}{2} = frac{3}{2} + sqrt{2})Adding (left(frac{sqrt{2}}{2}right)^2 = frac{1}{2}):Total |ZA| squared = (frac{3}{2} + sqrt{2} + frac{1}{2} = 2 + sqrt{2})Yes, that's correct.Similarly, |ZB| squared is the same.|AB| squared is 2.So, the sides are (sqrt{2 + sqrt{2}}), (sqrt{2 + sqrt{2}}), and (sqrt{2}).So, two sides are equal, one is different. Therefore, it's an isosceles triangle.But let me check if it's a right-angled isosceles triangle. For that, the sides should satisfy |ZA|² = |ZB|² = 2 + √2, and |AB|² = 2.If it were a right-angled isosceles triangle, then |ZA|² + |ZB|² should equal |AB|², but that's not the case here. Alternatively, if |ZA|² = |ZB|² = |AB|², but that's not true either.Wait, no, in a right-angled isosceles triangle, the legs are equal, and the hypotenuse is √2 times the legs. So, if |ZA| = |ZB|, then |AB| should be |ZA| * √2.But |ZA| = (sqrt{2 + sqrt{2}}), so |AB| should be (sqrt{2 + sqrt{2}} * sqrt{2} = sqrt{2*(2 + sqrt{2})} = sqrt{4 + 2sqrt{2}}). But |AB| is √2, which is not equal to √(4 + 2√2). Therefore, it's not a right-angled isosceles triangle.Therefore, the triangle is just isosceles.So, the answer is D: Isosceles triangle.Wait, but let me think again. Maybe I made a mistake in interpreting the function f(z). Let me go back to the function:(f(z) = |(z + 1)(overline{z} - i)|)I expanded it correctly, but let me double-check the modulus calculation.We had:(f(z) = |(1 + costheta + sintheta) - i(1 + sintheta + costheta)|)So, the modulus is:(sqrt{(1 + costheta + sintheta)^2 + (1 + sintheta + costheta)^2})Which is:(sqrt{2*(1 + costheta + sintheta)^2})So, (f(z) = sqrt{2} * |1 + costheta + sintheta|)Since 1 + cosθ + sinθ is always positive (because |1 + cosθ + sinθ| ≥ 1 - √2 ≈ -0.414, but since we're taking modulus, it's non-negative), so f(z) = √2*(1 + cosθ + sinθ)Then, as I did before, express 1 + cosθ + sinθ as 1 + √2 sin(θ + π/4)So, f(z) = √2*(1 + √2 sin(θ + π/4)) = √2 + 2 sin(θ + π/4)Wait, that's not correct. Let me compute it correctly.Wait, 1 + cosθ + sinθ = 1 + √2 sin(θ + π/4)Therefore, f(z) = √2*(1 + √2 sin(θ + π/4)) = √2 + 2 sin(θ + π/4)Yes, that's correct.So, the maximum value of f(z) is when sin(θ + π/4) = 1, so f(z) = √2 + 2*1 = 2 + √2And θ + π/4 = π/2 + 2kπ, so θ = π/4 + 2kπSo, z = cos(π/4) + i sin(π/4) = √2/2 + i√2/2So, point Z is (√2/2, √2/2)Then, distances:|ZA|: distance from (-1,0) to (√2/2, √2/2)Compute:√[(√2/2 + 1)^2 + (√2/2 - 0)^2]= √[(1 + √2/2)^2 + (√2/2)^2]= √[1 + √2 + (2/4) + (2/4)]= √[1 + √2 + 1/2 + 1/2]= √[2 + √2]Similarly, |ZB|: distance from (0,-1) to (√2/2, √2/2)= √[(√2/2 - 0)^2 + (√2/2 + 1)^2]= √[(√2/2)^2 + (1 + √2/2)^2]Same as above, so √[2 + √2]|AB|: distance from (-1,0) to (0,-1)= √[(0 - (-1))^2 + (-1 - 0)^2] = √[1 + 1] = √2So, sides are √[2 + √2], √[2 + √2], and √2So, two sides equal, one different. Therefore, isosceles triangle.But wait, let me check if it's a right-angled isosceles triangle. For that, the sides should satisfy |ZA|² + |ZB|² = |AB|² or something similar.But |ZA|² = 2 + √2, |ZB|² = 2 + √2, |AB|² = 2So, |ZA|² + |ZB|² = 4 + 2√2, which is not equal to |AB|².Alternatively, |ZA|² + |AB|² = (2 + √2) + 2 = 4 + √2, which is not equal to |ZB|².Similarly, |ZB|² + |AB|² = same as above.Therefore, it's not a right-angled triangle.Hence, the triangle is isosceles but not right-angled.Therefore, the answer is D: Isosceles triangle.

💡Alright, let's tackle this problem step by step. So, we have two conditions here: condition p, which is ( a leqslant 1 ), and condition q, which is ( |a| leqslant 1 ). The question is asking us to determine the relationship between these two conditions, specifically whether p is a sufficient condition, a necessary condition, both, or neither for q.First, let's recall what sufficient and necessary conditions mean. A sufficient condition means that if p is true, then q must also be true. In other words, p being true guarantees that q is true. On the other hand, a necessary condition means that if q is true, then p must also be true. So, p is necessary for q if q can't be true without p being true.Now, let's look at the given conditions. Condition p is ( a leqslant 1 ). This means that a can be any number less than or equal to 1, which could be negative, zero, or positive up to 1. Condition q is ( |a| leqslant 1 ), which translates to ( -1 leqslant a leqslant 1 ). So, q restricts a to be between -1 and 1, inclusive.Let's analyze whether p is a sufficient condition for q. If p is true, meaning ( a leqslant 1 ), does that necessarily mean q is true? Well, if a is less than or equal to 1, a could still be greater than 1 in the negative direction, right? For example, if a is -2, then p is true because -2 is less than or equal to 1, but q is false because the absolute value of -2 is 2, which is greater than 1. So, p being true doesn't guarantee that q is true. Therefore, p is not a sufficient condition for q.Next, let's check if p is a necessary condition for q. If q is true, meaning ( |a| leqslant 1 ), does that mean p must be true? Well, if ( |a| leqslant 1 ), then a is between -1 and 1. So, a is definitely less than or equal to 1. Therefore, if q is true, p must also be true. This means that p is a necessary condition for q.To summarize, p is necessary for q because whenever q is true, p must also be true. However, p is not sufficient for q because p being true doesn't necessarily make q true. So, p is a necessary but not sufficient condition for q.I think that makes sense. Let me just double-check with an example. Suppose a = 0.5. Then, both p and q are true. If a = -0.5, again, both p and q are true. If a = 1, both are true. But if a = -1, p is true, and q is also true. Now, if a = 2, p is false, and q is false. If a = -2, p is true, but q is false. So, in the case where a = -2, p is true, but q is false, which confirms that p is not sufficient. And in all cases where q is true, p is also true, confirming that p is necessary.Yeah, I think that solidifies the conclusion. So, p is a necessary but not sufficient condition for q.

💡Alright, so I have this function f(x) = sin x cos x - sin²x + 1/2, and I need to figure out where it's monotonically increasing. Then, there's this triangle problem where I have to find the range of f(B) given some condition. Let me tackle part (I) first.Starting with f(x) = sin x cos x - sin²x + 1/2. Hmm, I remember that sin x cos x can be rewritten using a double-angle identity. Specifically, sin(2x) = 2 sin x cos x, so sin x cos x = (1/2) sin(2x). That might simplify things.So, substituting that in, f(x) becomes (1/2) sin(2x) - sin²x + 1/2. Now, looking at the sin²x term, I recall that sin²x can be expressed using another identity: sin²x = (1 - cos(2x))/2. Let me substitute that in as well.So, replacing sin²x, we have f(x) = (1/2) sin(2x) - (1 - cos(2x))/2 + 1/2. Let me simplify this step by step.First, expand the terms: (1/2) sin(2x) - 1/2 + (cos(2x))/2 + 1/2. The -1/2 and +1/2 cancel each other out, so we're left with (1/2) sin(2x) + (1/2) cos(2x). That looks like a combination of sine and cosine terms with the same argument, 2x. Maybe I can write this as a single sine function using the amplitude-phase form. The general form is A sin(θ + φ), where A is the amplitude and φ is the phase shift.To find A, we use the formula A = sqrt((1/2)^2 + (1/2)^2) = sqrt(1/4 + 1/4) = sqrt(1/2) = √2/2. Next, to find φ, we know that tan φ = (coefficient of cos)/(coefficient of sin) = (1/2)/(1/2) = 1. So φ = arctan(1) = π/4. Therefore, f(x) can be rewritten as (√2/2) sin(2x + π/4). Now, to find where f(x) is monotonically increasing, we need to look at its derivative. The derivative of f(x) with respect to x is f’(x) = (√2/2) * 2 cos(2x + π/4) = √2 cos(2x + π/4).We need to find the intervals where f’(x) > 0, which means cos(2x + π/4) > 0. The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k. So, setting up the inequality:-π/2 + 2πk < 2x + π/4 < π/2 + 2πk.Let me solve for x:Subtract π/4 from all parts:-π/2 - π/4 + 2πk < 2x < π/2 - π/4 + 2πk.Simplify the left side: -3π/4 + 2πk < 2x < π/4 + 2πk.Divide all parts by 2:-3π/8 + πk < x < π/8 + πk.So, the function f(x) is monotonically increasing on the intervals (-3π/8 + πk, π/8 + πk) for any integer k.Alright, that seems solid. Now, moving on to part (II).We have a triangle ABC with sides a, b, c opposite angles A, B, C respectively. The given condition is b cos 2A = b cos A - a sin B, and 0 < A < π/2. We need to find the range of f(B).First, let me write down the given equation:b cos 2A = b cos A - a sin B.I can rearrange this equation:b cos 2A - b cos A + a sin B = 0.Factor out b:b (cos 2A - cos A) + a sin B = 0.Hmm, maybe using the Law of Sines here would help. The Law of Sines states that a/sin A = b/sin B = c/sin C = 2R, where R is the circumradius. So, a = 2R sin A and b = 2R sin B.Let me substitute a and b in terms of sin A and sin B.So, b = 2R sin B, a = 2R sin A.Substituting into the equation:2R sin B (cos 2A - cos A) + 2R sin A sin B = 0.Factor out 2R sin B:2R sin B [ (cos 2A - cos A) + sin A ] = 0.Since 2R sin B is not zero (because in a triangle, angles are between 0 and π, so sin B ≠ 0), the expression inside the brackets must be zero:(cos 2A - cos A) + sin A = 0.So, cos 2A - cos A + sin A = 0.Let me recall that cos 2A can be written in terms of cos²A or sin²A. Let me use cos 2A = 1 - 2 sin²A.So, substituting:(1 - 2 sin²A) - cos A + sin A = 0.Simplify:1 - 2 sin²A - cos A + sin A = 0.Hmm, this seems a bit complicated. Maybe another identity for cos 2A would be better. Alternatively, perhaps factor the equation differently.Wait, another approach: Let me write cos 2A as cos²A - sin²A.So, cos 2A = cos²A - sin²A.Substituting back:cos²A - sin²A - cos A + sin A = 0.Let me rearrange terms:cos²A - cos A - sin²A + sin A = 0.Factor terms:cos A (cos A - 1) - sin A (sin A - 1) = 0.Hmm, not sure if that helps. Alternatively, group cos²A - sin²A and -cos A + sin A.So, (cos²A - sin²A) + (-cos A + sin A) = 0.Factor cos²A - sin²A as (cos A - sin A)(cos A + sin A).So, (cos A - sin A)(cos A + sin A) + (-cos A + sin A) = 0.Factor out (-cos A + sin A):Wait, (-cos A + sin A) = -(cos A - sin A). So, let me write it as:(cos A - sin A)(cos A + sin A) - (cos A - sin A) = 0.Factor out (cos A - sin A):(cos A - sin A)(cos A + sin A - 1) = 0.So, either cos A - sin A = 0 or cos A + sin A - 1 = 0.Case 1: cos A - sin A = 0.This implies cos A = sin A, so tan A = 1, which gives A = π/4.Case 2: cos A + sin A - 1 = 0.So, cos A + sin A = 1.Let me square both sides to see if this gives any solutions:(cos A + sin A)^2 = 1^2 => cos²A + 2 sin A cos A + sin²A = 1.Simplify: (cos²A + sin²A) + 2 sin A cos A = 1 => 1 + sin 2A = 1.So, sin 2A = 0.Which implies 2A = nπ, so A = nπ/2.But since 0 < A < π/2, the only possible solution is A = 0 or A = π/2, but both are excluded. So, no solution in this case.Therefore, the only solution is A = π/4.So, angle A is π/4. Now, in triangle ABC, the sum of angles is π, so A + B + C = π.Given A = π/4, so B + C = 3π/4.We need to find the range of f(B). Remember, f(x) = sin x cos x - sin²x + 1/2, which we simplified earlier to (√2/2) sin(2x + π/4).So, f(B) = (√2/2) sin(2B + π/4).We need to find the range of this function given that B is an angle in a triangle with A = π/4. So, B must satisfy 0 < B < 3π/4, because A = π/4 and A + B + C = π, so B can be from just above 0 to just below 3π/4.But let's think about the possible values of 2B + π/4.If B is in (0, 3π/4), then 2B is in (0, 3π/2), so 2B + π/4 is in (π/4, 7π/4).So, the argument of the sine function is between π/4 and 7π/4.The sine function reaches its maximum at π/2 and minimum at 3π/2.So, sin(θ) ranges from -1 to 1 as θ goes from π/4 to 7π/4.But wait, actually, sin(π/4) = √2/2, sin(π/2) = 1, sin(3π/2) = -1, and sin(7π/4) = -√2/2.So, the maximum value of sin(2B + π/4) is 1, and the minimum is -1.Therefore, f(B) = (√2/2) sin(2B + π/4) ranges from (√2/2)(-1) to (√2/2)(1), which is [-√2/2, √2/2].But wait, let me check if 2B + π/4 actually reaches π/2 and 3π/2.When does 2B + π/4 = π/2? Solving for B: 2B = π/2 - π/4 = π/4, so B = π/8.Similarly, 2B + π/4 = 3π/2: 2B = 3π/2 - π/4 = 5π/4, so B = 5π/8.But 5π/8 is less than 3π/4 (which is 6π/8), so yes, B can reach 5π/8, so 2B + π/4 can reach 3π/2.Similarly, when B approaches 0, 2B + π/4 approaches π/4, and sin(π/4) = √2/2, so f(B) approaches (√2/2)(√2/2) = 1/2.Wait, but earlier I thought the range was from -√2/2 to √2/2, but when B approaches 0, f(B) approaches 1/2, which is greater than -√2/2. So, is the minimum actually -√2/2?Wait, let me think again. The function sin(theta) over theta in (π/4, 7π/4) does indeed reach -1 at 3π/2, which is within the interval. So, the minimum value is -1, and the maximum is 1. Therefore, f(B) ranges from -√2/2 to √2/2.But when B approaches 0, f(B) approaches (√2/2) sin(π/4) = (√2/2)(√2/2) = 1/2, which is within the range. Similarly, when B approaches 3π/4, 2B + π/4 approaches 3π/2 + π/4 = 7π/4, and sin(7π/4) = -√2/2, so f(B) approaches (√2/2)(-√2/2) = -1/2. Wait, that's conflicting with my earlier conclusion.Wait, hold on. Let me compute f(B) when B approaches 3π/4.2B + π/4 = 2*(3π/4) + π/4 = 3π/2 + π/4 = 7π/4.sin(7π/4) = -√2/2, so f(B) = (√2/2)*(-√2/2) = - (2/4) = -1/2.Similarly, when B approaches 0, 2B + π/4 approaches π/4, sin(π/4) = √2/2, so f(B) approaches (√2/2)*(√2/2) = 1/2.But earlier, I thought the range was from -√2/2 to √2/2, but actually, the function f(B) ranges between -1/2 and 1/2 because the maximum and minimum of sin(theta) are scaled by √2/2.Wait, no. Wait, f(B) = (√2/2) sin(theta), and sin(theta) ranges from -1 to 1, so f(B) ranges from -√2/2 to √2/2. But when B is in (0, 3π/4), theta = 2B + π/4 is in (π/4, 7π/4). So, sin(theta) does indeed reach -1 and 1 within this interval.Wait, when does sin(theta) = 1? At theta = π/2, which is when 2B + π/4 = π/2 => B = π/8.Similarly, sin(theta) = -1 at theta = 3π/2, which is when 2B + π/4 = 3π/2 => B = 5π/8.So, B can be π/8 and 5π/8, which are both within (0, 3π/4). Therefore, f(B) does reach √2/2 and -√2/2.But wait, when B approaches 0, f(B) approaches 1/2, which is less than √2/2 (~0.707). Similarly, when B approaches 3π/4, f(B) approaches -1/2, which is greater than -√2/2 (~-0.707). So, the function f(B) oscillates between -√2/2 and √2/2 as B varies from 0 to 3π/4, but the endpoints of B don't reach the extreme values of f(B). However, since B can take values where f(B) reaches those extremes, the range is indeed from -√2/2 to √2/2.Wait, but when B = π/8, f(B) = √2/2, and when B = 5π/8, f(B) = -√2/2. So, even though B can't be exactly 0 or 3π/4, it can get close enough that f(B) approaches 1/2 and -1/2, but the actual maximum and minimum of f(B) are √2/2 and -√2/2 because those are achieved at B = π/8 and B = 5π/8, which are valid angles within the triangle.So, the range of f(B) is [-√2/2, √2/2].I think that's it. So, summarizing:(I) The function f(x) is monotonically increasing on intervals (-3π/8 + kπ, π/8 + kπ) for any integer k.(II) The range of f(B) is [-√2/2, √2/2].**Final Answer**(I) The function is monotonically increasing on the intervals boxed{left[ kpi - frac{3pi}{8}, kpi + frac{pi}{8} right]} for any integer ( k ).(II) The range of values for ( f(B) ) is boxed{left[ -frac{sqrt{2}}{2}, frac{sqrt{2}}{2} right]}.