💡Okay, so I have this geometry problem here. Let me try to visualize it step by step. First, there's a circle with center O, and AB is a fixed diameter. So, AB is a straight line passing through the center O, right? Now, C is any point on the circle. From point C, a tangent line CT is drawn. I remember that a tangent to a circle is perpendicular to the radius at the point of contact. So, CT is perpendicular to OC. Then, D is the point diametrically opposite to C. That means CD is another diameter of the circle, passing through O as well. So, CD is perpendicular to AB if C is on the semicircle above AB, right? Wait, no, not necessarily. It depends on where C is. If C is somewhere on the circle, D will be directly opposite to it, so CD is a diameter, but its orientation depends on where C is.Now, as C moves over one semicircle, the line segment DT intersects the circle again at a point E. So, DT is a line from D to T, and since T is the point of tangency from C, which is on the circle, DT must be a secant line that intersects the circle again at E. The question is asking about the relationship of point E relative to AB as C moves. The options are whether E is always equidistant from A and B, mirrors C across AB, remains stationary, or varies randomly.Let me try to draw this in my mind. If I fix AB as the horizontal diameter, and O as the center, then C can be any point on the upper semicircle. The tangent at C, CT, will be perpendicular to OC. So, if OC is at some angle θ from AB, then CT will be at θ + 90 degrees.Point D is diametrically opposite to C, so it's on the lower semicircle. So, if C is at (r cos θ, r sin θ), then D is at (r cos (θ + π), r sin (θ + π)) which simplifies to (-r cos θ, -r sin θ).Now, the tangent line CT. The equation of the tangent at C can be written as x cos θ + y sin θ = r. Since T is the point of tangency, which is C itself, but wait, no. Wait, CT is the tangent line, so T is a point outside the circle from which the tangent is drawn to the circle at C. Wait, no, the tangent line is at C, so T is just a point on the tangent line. Wait, maybe I'm confusing something.Wait, the problem says "a tangent line CT is drawn to the circle." So, CT is the tangent at C, meaning T is just a point on the tangent line, but not necessarily on the circle. Wait, but if CT is the tangent at C, then T is the point of tangency, which is C itself. That can't be, because then DT would be DC, which is a diameter, and it would intersect the circle again at D, but D is already on the circle. Hmm, maybe I'm misunderstanding.Wait, perhaps CT is a tangent from point C to another point T on the circle. But that doesn't make sense because the tangent at C is just a line touching the circle at C. So, maybe T is a point outside the circle from which the tangent CT is drawn to the circle at C. So, T is an external point, and CT is the tangent from T to the circle at C.Wait, that makes more sense. So, T is an external point, and CT is tangent to the circle at C. So, CT is perpendicular to OC. So, T lies somewhere outside the circle such that CT is tangent at C.So, now, D is diametrically opposite to C, so CD is a diameter. Then, DT is a line from D to T, which is outside the circle, and this line DT intersects the circle again at E. So, E is another intersection point of DT with the circle.Now, as C moves over one semicircle, E moves in some way relative to AB. The question is asking about the relationship of E relative to AB as C moves.Let me try to analyze this. Maybe I can use some coordinate geometry to model this.Let me set up a coordinate system with O at (0,0), AB as the x-axis, so A is (-r,0) and B is (r,0), where r is the radius of the circle. Let me assume r = 1 for simplicity, so the circle has equation x² + y² = 1.Let point C be (cos θ, sin θ) on the upper semicircle, so θ ranges from 0 to π. Then, point D, diametrically opposite to C, is (-cos θ, -sin θ).Now, the tangent at C is given by the equation x cos θ + y sin θ = 1. So, point T is a point on this tangent line. But since CT is tangent at C, T must lie on this line. However, T is an external point from which the tangent is drawn to the circle at C. So, T can be any point on the tangent line, but in this case, since we're drawing CT as the tangent, T is just a point on the tangent line beyond C.Wait, but in the problem, it's just mentioned that CT is a tangent line, so T is the point of tangency, which is C itself. But that can't be because then DT would be DC, which is a diameter, and it would intersect the circle again at D, but D is already on the circle. So, perhaps T is another point on the tangent line, not necessarily C.Wait, maybe I'm overcomplicating. Let me think differently. If CT is the tangent at C, then T is the point of tangency, which is C. So, DT is the line from D to C, which is the diameter CD. But then, CD intersects the circle again at D, which is already on the circle. So, that would mean E is D, but that's not possible because E is another intersection point. So, perhaps T is not C, but another point on the tangent line.Wait, perhaps T is a point outside the circle such that CT is tangent to the circle at C. So, T is an external point, and CT is the tangent from T to the circle at C. So, T lies on the tangent line at C, but outside the circle.So, in that case, the coordinates of T can be found. The tangent line at C is x cos θ + y sin θ = 1. So, any point T on this line can be represented as (cos θ + t sin θ, sin θ - t cos θ) for some parameter t. Because the direction vector of the tangent line is perpendicular to the radius, so it's (-sin θ, cos θ). So, moving along the tangent line from C, we can write T as C plus t times the direction vector.But since T is an external point, t must be such that T is outside the circle. So, for t > 0, T is on one side, and for t < 0, on the other. But in our case, since we're drawing the tangent from C, perhaps T is on the side away from the center.Wait, but actually, the tangent line at C is fixed, so T is just a point on that line. But in the problem, it's just mentioned that CT is a tangent line, so perhaps T is just a point on the tangent line, but not necessarily related to the circle beyond being the point of tangency.Wait, maybe I'm overcomplicating. Let me think about the line DT. Since D is (-cos θ, -sin θ), and T is a point on the tangent line at C, which is x cos θ + y sin θ = 1.So, the line DT connects D to T, and we need to find where else it intersects the circle, which is point E.So, let me parametrize the line DT. Let me write the parametric equations for DT. Let me denote T as (x_t, y_t), which lies on the tangent line x cos θ + y sin θ = 1.So, the parametric equations for DT can be written as:x = -cos θ + t(x_t + cos θ)y = -sin θ + t(y_t + sin θ)where t is a parameter. When t=0, we are at D, and when t=1, we are at T.We need to find the other intersection point E of DT with the circle x² + y² = 1.So, substituting the parametric equations into the circle equation:(-cos θ + t(x_t + cos θ))² + (-sin θ + t(y_t + sin θ))² = 1Expanding this:cos² θ - 2 t cos θ (x_t + cos θ) + t² (x_t + cos θ)² + sin² θ - 2 t sin θ (y_t + sin θ) + t² (y_t + sin θ)² = 1Simplify:(cos² θ + sin² θ) - 2 t [cos θ (x_t + cos θ) + sin θ (y_t + sin θ)] + t² [(x_t + cos θ)² + (y_t + sin θ)²] = 1Since cos² θ + sin² θ = 1, this simplifies to:1 - 2 t [cos θ x_t + cos² θ + sin θ y_t + sin² θ] + t² [(x_t + cos θ)² + (y_t + sin θ)²] = 1Subtract 1 from both sides:-2 t [cos θ x_t + sin θ y_t + (cos² θ + sin² θ)] + t² [(x_t + cos θ)² + (y_t + sin θ)²] = 0Again, cos² θ + sin² θ = 1, so:-2 t [cos θ x_t + sin θ y_t + 1] + t² [(x_t + cos θ)² + (y_t + sin θ)²] = 0Factor out t:t [ -2 (cos θ x_t + sin θ y_t + 1) + t ( (x_t + cos θ)² + (y_t + sin θ)² ) ] = 0So, the solutions are t=0, which corresponds to point D, and the other solution is when:-2 (cos θ x_t + sin θ y_t + 1) + t ( (x_t + cos θ)² + (y_t + sin θ)² ) = 0Solving for t:t = [2 (cos θ x_t + sin θ y_t + 1)] / [ (x_t + cos θ)² + (y_t + sin θ)² ]So, the parameter t for point E is this value.Now, since T lies on the tangent line at C, which is x cos θ + y sin θ = 1, we have:x_t cos θ + y_t sin θ = 1So, cos θ x_t + sin θ y_t = 1Therefore, substituting back into the expression for t:t = [2 (1 + 1)] / [ (x_t + cos θ)² + (y_t + sin θ)² ] = 4 / [ (x_t + cos θ)² + (y_t + sin θ)² ]Now, let's compute the denominator:(x_t + cos θ)² + (y_t + sin θ)² = x_t² + 2 x_t cos θ + cos² θ + y_t² + 2 y_t sin θ + sin² θ= (x_t² + y_t²) + 2 (x_t cos θ + y_t sin θ) + (cos² θ + sin² θ)= (x_t² + y_t²) + 2 (1) + 1= x_t² + y_t² + 3So, t = 4 / (x_t² + y_t² + 3)Now, we need to find the coordinates of E, which is given by:x = -cos θ + t (x_t + cos θ)y = -sin θ + t (y_t + sin θ)Substituting t:x = -cos θ + [4 / (x_t² + y_t² + 3)] (x_t + cos θ)y = -sin θ + [4 / (x_t² + y_t² + 3)] (y_t + sin θ)Now, this seems complicated, but maybe we can find a relationship between E and C.Wait, but I'm not sure if this is the best approach. Maybe there's a more geometric way to see this.Let me think about the properties of the circle and the tangent. Since CT is tangent at C, and D is diametrically opposite to C, then CD is a diameter. So, angle CTD is a right angle because the angle subtended by a diameter is a right angle. Wait, no, that's for points on the circle. Since T is outside the circle, maybe not.Wait, actually, since CT is tangent at C, and CD is a diameter, then angle CTD is equal to angle CBD because of some properties. Hmm, not sure.Alternatively, maybe we can use inversion or some other transformation, but that might be too advanced.Wait, another approach: Since CT is tangent at C, then CT is perpendicular to OC. So, OC is perpendicular to CT.Also, since D is diametrically opposite to C, then OD = OC, and D is (-cos θ, -sin θ).Now, let's consider triangle CTD. Since CT is tangent at C, and CD is a diameter, maybe there's some reflection property.Wait, if I reflect point T over AB, what happens? Since AB is the x-axis, reflecting over AB would invert the y-coordinate.But I'm not sure if that helps.Wait, let me think about the line DT. Since D is (-cos θ, -sin θ) and T is on the tangent line at C, which is x cos θ + y sin θ = 1.So, the line DT connects (-cos θ, -sin θ) to T, which is on x cos θ + y sin θ = 1.We need to find where else this line intersects the circle, which is point E.Wait, maybe instead of using coordinates, I can use some geometric properties.Since CT is tangent at C, then CT is perpendicular to OC. So, triangle OCT is right-angled at C.Also, since D is diametrically opposite to C, then CD is a diameter, so angle CED is a right angle if E is on the circle. Wait, no, because E is another intersection point, so maybe not.Wait, let me think about the power of point D with respect to the circle. The power of D is equal to the square of the tangent from D to the circle. But since D is on the circle, the power is zero, which doesn't help.Wait, but D is on the circle, so any line through D will intersect the circle again at another point, which is E.Wait, since CT is tangent at C, and D is diametrically opposite to C, then maybe there's a relationship between E and C.Wait, let me consider the reflection of C over AB. If I reflect C over AB, which is the x-axis, then the reflection point C' would be (cos θ, -sin θ). Now, is E equal to C'? Let's see.If E is the reflection of C over AB, then E would be (cos θ, -sin θ). But D is (-cos θ, -sin θ). So, if E is (cos θ, -sin θ), then the line DE would connect (-cos θ, -sin θ) to (cos θ, -sin θ), which is a horizontal line along y = -sin θ, intersecting the circle at E.But in our case, E is the intersection of DT with the circle again. So, unless T is chosen such that DT passes through (cos θ, -sin θ), which is the reflection of C, then E would be the reflection.But is that the case?Wait, let's see. If E is the reflection of C over AB, then E is (cos θ, -sin θ). So, the line DE connects (-cos θ, -sin θ) to (cos θ, -sin θ), which is a horizontal line. So, the equation of DE is y = -sin θ.Now, does this line intersect the tangent line CT at point T?The tangent line CT is x cos θ + y sin θ = 1. So, substituting y = -sin θ into this equation:x cos θ + (-sin θ) sin θ = 1x cos θ - sin² θ = 1x cos θ = 1 + sin² θx = (1 + sin² θ)/cos θSo, the point T would be at ((1 + sin² θ)/cos θ, -sin θ)But is this consistent with CT being tangent at C?Wait, point C is (cos θ, sin θ). The tangent at C is x cos θ + y sin θ = 1. So, point T is on this tangent line, and we found that when DE is the horizontal line y = -sin θ, T is at ((1 + sin² θ)/cos θ, -sin θ).Now, let's check if CT is indeed tangent at C. The line CT connects C (cos θ, sin θ) to T ((1 + sin² θ)/cos θ, -sin θ). Let's find the slope of CT.Slope m = [ -sin θ - sin θ ] / [ (1 + sin² θ)/cos θ - cos θ ]= [ -2 sin θ ] / [ (1 + sin² θ - cos² θ)/cos θ ]= [ -2 sin θ ] / [ (1 + sin² θ - (1 - sin² θ))/cos θ ]= [ -2 sin θ ] / [ (2 sin² θ)/cos θ ]= [ -2 sin θ ] * [ cos θ / (2 sin² θ) ]= -cos θ / sin θSo, the slope of CT is -cot θ.But the slope of the tangent at C is -cot θ because the tangent line is x cos θ + y sin θ = 1, which can be rewritten as y = (-cos θ / sin θ) x + 1/sin θ, so the slope is -cot θ. So, yes, CT has the correct slope.Therefore, when E is the reflection of C over AB, the line DT intersects the circle again at E, which is the reflection of C over AB.Therefore, as C moves over one semicircle, E mirrors C across AB.So, the answer should be B) E mirrors C across AB.

💡First, I need to determine the total number of ways to choose 3 pieces of silverware from the drawer. There are 6 forks, 8 spoons, and 6 knives, making a total of 20 pieces of silverware. The number of ways to choose 3 pieces from 20 is calculated using the combination formula:[dbinom{20}{3} = frac{20 times 19 times 18}{3 times 2 times 1} = 1140]Next, I'll calculate the number of favorable outcomes, which is selecting one fork, one spoon, and one knife. - The number of ways to choose one fork from 6 is 6.- The number of ways to choose one spoon from 8 is 8.- The number of ways to choose one knife from 6 is 6.Multiplying these together gives the total number of favorable outcomes:[6 times 8 times 6 = 288]Finally, the probability of selecting one fork, one spoon, and one knife is the ratio of favorable outcomes to the total number of possible outcomes:[frac{288}{1140} = boxed{frac{24}{95}}]

💡First, I need to convert the base 6 number (524_6) to its base 10 equivalent. To do this, I'll break down each digit and multiply it by (6) raised to the power of its position.Starting with the rightmost digit:- The digit 4 is in the (6^0) place, so it contributes (4 times 1 = 4).- The digit 2 is in the (6^1) place, contributing (2 times 6 = 12).- The digit 5 is in the (6^2) place, contributing (5 times 36 = 180).Adding these together: (180 + 12 + 4 = 196). So, (524_6 = 196_{10}).Next, I need to express (196) as (2cd_{10}), where (c) and (d) are single base-10 digits. Dividing (196) by 2 gives (98), so (c = 9) and (d = 8).Finally, I'll calculate (frac{c cdot d}{12}):[frac{9 times 8}{12} = frac{72}{12} = 6]

💡Okay, so I have this problem about an arithmetic sequence. Let me see if I can figure it out step by step. First, the problem says that {a_n} is an arithmetic sequence, and it gives me that a₁ + a₇ + a₁₃ equals 4. I need to find the value of a₂ + a₁₂. The options are A: 2, B: 1, C: 3/2, D: 8/3.Hmm, arithmetic sequences. I remember that in an arithmetic sequence, each term is obtained by adding a common difference to the previous term. So, the nth term can be expressed as a_n = a₁ + (n-1)d, where d is the common difference.Let me write down what I know:a₁ + a₇ + a₁₃ = 4.I need to express a₇ and a₁₃ in terms of a₁ and d.So, a₇ = a₁ + (7-1)d = a₁ + 6d.Similarly, a₁₃ = a₁ + (13-1)d = a₁ + 12d.So, substituting back into the equation:a₁ + (a₁ + 6d) + (a₁ + 12d) = 4.Let me simplify that:a₁ + a₁ + 6d + a₁ + 12d = 4.Combine like terms:3a₁ + 18d = 4.Hmm, okay. So, 3a₁ + 18d = 4. Maybe I can factor out a 3:3(a₁ + 6d) = 4.Divide both sides by 3:a₁ + 6d = 4/3.Wait a minute, a₁ + 6d is actually a₇, right? Because a₇ = a₁ + 6d. So, that means a₇ = 4/3.Alright, so a₇ is 4/3. Now, the question is asking for a₂ + a₁₂.Let me write expressions for a₂ and a₁₂.a₂ = a₁ + (2-1)d = a₁ + d.a₁₂ = a₁ + (12-1)d = a₁ + 11d.So, a₂ + a₁₂ = (a₁ + d) + (a₁ + 11d) = 2a₁ + 12d.Hmm, 2a₁ + 12d. Let me see if I can relate this to something I already know.I know that 3a₁ + 18d = 4 from earlier. So, 2a₁ + 12d is two-thirds of that equation, right?Because 2a₁ + 12d is (2/3)*(3a₁ + 18d) = (2/3)*4 = 8/3.Wait, is that correct? Let me check.If I take 3a₁ + 18d = 4, then multiplying both sides by (2/3):(2/3)*(3a₁ + 18d) = (2/3)*4.Which simplifies to 2a₁ + 12d = 8/3.Yes, that's correct. So, a₂ + a₁₂ = 8/3.Alternatively, I remember that in an arithmetic sequence, the sum of terms equidistant from the ends is constant. So, a₂ and a₁₂ are equidistant from a₇. Since a₂ is 5 terms before a₇ and a₁₂ is 5 terms after a₇, their sum should be twice a₇.So, a₂ + a₁₂ = 2a₇.Since a₇ is 4/3, then 2*(4/3) = 8/3.That's the same answer as before. So, that must be the correct one.Let me just recap to make sure I didn't make any mistakes.1. Expressed a₇ and a₁₃ in terms of a₁ and d.2. Substituted into the given equation to get 3a₁ + 18d = 4.3. Recognized that a₁ + 6d = 4/3, which is a₇.4. Expressed a₂ and a₁₂ in terms of a₁ and d.5. Found that a₂ + a₁₂ = 2a₁ + 12d.6. Noticed that 2a₁ + 12d is two-thirds of 3a₁ + 18d, which is 8/3.7. Alternatively, used the property that the sum of terms equidistant from the center is twice the middle term.Both methods gave me the same answer, so I'm pretty confident it's correct.**Final Answer**boxed{D}

💡Okay, so I have this function f(x) which is equal to x squared for all real numbers x. The problem is asking me to evaluate f(x + 3). Hmm, let me think about how to approach this.First, I know that f(x) = x². That means whenever I see f of something, I can replace that something with x and then square it. So, if I have f of (x + 3), I should replace x in the function with (x + 3). That would give me f(x + 3) = (x + 3)².Wait, but I need to make sure I'm doing this correctly. Let me double-check. The function f takes any input, squares it, and gives the result. So, if the input is (x + 3), then f(x + 3) should indeed be (x + 3) squared.Now, I should probably expand this expression to make it simpler. Expanding (x + 3)² means I need to multiply (x + 3) by itself. Let me write that out:(x + 3)(x + 3)Using the distributive property, I can expand this:First, multiply x by x, which is x².Then, x times 3 is 3x.Next, 3 times x is another 3x.Finally, 3 times 3 is 9.So, putting it all together:x² + 3x + 3x + 9Combine like terms:x² + 6x + 9Okay, so f(x + 3) simplifies to x² + 6x + 9. Let me make sure I didn't make any mistakes in the expansion. I can use the formula for the square of a binomial, which is (a + b)² = a² + 2ab + b². Applying that here, a is x and b is 3.So, (x)² + 2*(x)*(3) + (3)² = x² + 6x + 9. Yep, that matches what I got earlier. Good, I didn't make a mistake there.Just to be thorough, let me plug in a specific value for x to test both the original function and the transformed function. Let's say x = 1.Original function: f(1) = 1² = 1.Now, f(x + 3) when x = 1 is f(1 + 3) = f(4) = 4² = 16.Using the expanded form: f(x + 3) = x² + 6x + 9. Plugging in x = 1: 1² + 6*1 + 9 = 1 + 6 + 9 = 16. That matches, so it seems correct.Let me try another value, maybe x = 0.Original function: f(0) = 0² = 0.f(x + 3) when x = 0 is f(0 + 3) = f(3) = 3² = 9.Using the expanded form: 0² + 6*0 + 9 = 0 + 0 + 9 = 9. Perfect, that works too.How about a negative number? Let's take x = -2.Original function: f(-2) = (-2)² = 4.f(x + 3) when x = -2 is f(-2 + 3) = f(1) = 1² = 1.Using the expanded form: (-2)² + 6*(-2) + 9 = 4 - 12 + 9 = 1. That also checks out.Alright, so after expanding and testing with a few values, I'm confident that f(x + 3) is indeed x² + 6x + 9. I think I've got it right!

💡Okay, so I have this function: y equals 4 cosine squared x minus 4 times the square root of 3 times sine x cosine x minus 1. Hmm, that looks a bit complicated, but maybe I can simplify it. Let me write it down again to see it clearly:y = 4cos²x - 4√3 sinx cosx - 1Alright, so part (1) asks for the smallest positive period of the function. Hmm, periods. I remember that for trigonometric functions, the period is the length after which the function repeats itself. For cosine and sine functions, the basic period is 2π, but when there are coefficients inside the function, like cos(2x), the period changes. Specifically, if you have cos(kx), the period is 2π divided by k.Looking at my function, I see that it has cos²x and sinx cosx terms. Both of these can be rewritten using double-angle identities, which might make it easier to find the period. Let me recall the double-angle identities:cos²x can be written as (1 + cos2x)/2, and sinx cosx is equal to (sin2x)/2. Let me substitute these into the function.So, substituting cos²x:4cos²x = 4*(1 + cos2x)/2 = 2*(1 + cos2x) = 2 + 2cos2xSimilarly, the term -4√3 sinx cosx becomes:-4√3*(sin2x)/2 = -2√3 sin2xSo putting it all together, the function becomes:y = 2 + 2cos2x - 2√3 sin2x - 1Simplify that:2 - 1 is 1, so:y = 1 + 2cos2x - 2√3 sin2xHmm, that looks better. Now, I have a function in terms of cos2x and sin2x. Maybe I can combine these into a single sine or cosine function using the amplitude-phase form. The general form is A cosθ + B sinθ = C cos(θ - φ), where C is the amplitude and φ is the phase shift.Let me compute the amplitude C:C = sqrt(A² + B²) where A is the coefficient of cos2x and B is the coefficient of sin2x.Here, A = 2 and B = -2√3.So, C = sqrt(2² + (-2√3)²) = sqrt(4 + 12) = sqrt(16) = 4Okay, so the amplitude is 4. Now, let's find the phase shift φ.tanφ = B/A = (-2√3)/2 = -√3So, φ is the angle whose tangent is -√3. I know that tan(π/3) = √3, so tan(-π/3) = -√3. Therefore, φ = -π/3.So, we can write:2cos2x - 2√3 sin2x = 4 cos(2x - (-π/3)) = 4 cos(2x + π/3)Wait, hold on. Let me double-check that. The formula is A cosθ + B sinθ = C cos(θ - φ). So, in this case, it's 2cos2x - 2√3 sin2x = 4 cos(2x - φ). So, solving for φ:cosφ = A/C = 2/4 = 1/2sinφ = B/C = (-2√3)/4 = -√3/2So, φ is the angle where cosφ = 1/2 and sinφ = -√3/2. That corresponds to φ = -π/3 or φ = 5π/3. But since we're dealing with the phase shift, it's usually expressed as a positive angle, so φ = 5π/3. However, in the formula, it's θ - φ, so:2cos2x - 2√3 sin2x = 4 cos(2x - 5π/3)But 5π/3 is equivalent to -π/3 in terms of cosine because cosine is an even function. So, cos(2x - 5π/3) = cos(2x + π/3). So, either way, it's the same.Therefore, the function simplifies to:y = 1 + 4 cos(2x + π/3)Alternatively, I could have written it as a sine function, but cosine is fine.Now, the function is y = 1 + 4 cos(2x + π/3). So, to find the period, since the argument of the cosine is 2x + π/3, the coefficient of x is 2, so the period is 2π divided by 2, which is π. So, the period is π. That answers part (1).Moving on to part (2): Find the maximum value of the function and the corresponding x values.Since the function is y = 1 + 4 cos(2x + π/3), the maximum value occurs when cos(2x + π/3) is maximum, which is 1. So, the maximum value is 1 + 4*1 = 5. Wait, hold on, that contradicts the previous solution. Wait, in the initial solution, the function was rewritten as -4 sin(2x - π/6) + 2, which would have a maximum value of 6. Hmm, so which one is correct?Wait, let me check my steps again.Original function: y = 4cos²x - 4√3 sinx cosx -1I used the double-angle identities:cos²x = (1 + cos2x)/2, so 4cos²x = 2 + 2cos2xsinx cosx = (sin2x)/2, so -4√3 sinx cosx = -2√3 sin2xThus, y = 2 + 2cos2x - 2√3 sin2x -1 = 1 + 2cos2x - 2√3 sin2xThen, I combined 2cos2x - 2√3 sin2x into 4 cos(2x + π/3). So, y = 1 + 4 cos(2x + π/3)But in the initial solution, the function was rewritten as -4 sin(2x - π/6) + 2. Let me see if these are equivalent.Wait, perhaps I made a mistake in combining the terms. Let me try again.Given 2cos2x - 2√3 sin2x.Expressed as R cos(2x + φ).Compute R: sqrt(2² + (2√3)^2) = sqrt(4 + 12) = sqrt(16) = 4So, R = 4.Then, cosφ = 2/4 = 1/2, sinφ = -2√3 /4 = -√3/2So, φ is the angle where cosφ = 1/2 and sinφ = -√3/2, which is φ = -π/3.Therefore, 2cos2x - 2√3 sin2x = 4 cos(2x - (-π/3)) = 4 cos(2x + π/3)So, y = 1 + 4 cos(2x + π/3)Alternatively, if I express it as a sine function, it would be:4 cos(2x + π/3) = 4 sin(2x + π/3 + π/2) = 4 sin(2x + 5π/6)But that might complicate things. Alternatively, perhaps the initial solution expressed it as a sine function with a different phase shift.Wait, let me see the initial solution:They wrote it as -4 sin(2x - π/6) + 2.Let me check if 1 + 4 cos(2x + π/3) is equal to -4 sin(2x - π/6) + 2.Compute -4 sin(2x - π/6) + 2.Express sin(2x - π/6) as sin2x cosπ/6 - cos2x sinπ/6.Which is sin2x*(√3/2) - cos2x*(1/2)So, -4 sin(2x - π/6) = -4*(√3/2 sin2x - 1/2 cos2x) = -2√3 sin2x + 2 cos2xThus, -4 sin(2x - π/6) + 2 = 2 cos2x - 2√3 sin2x + 2Which is the same as 1 + 4 cos(2x + π/3) ?Wait, 2 cos2x - 2√3 sin2x + 2 is equal to y.But in my expression, y = 1 + 4 cos(2x + π/3). So, let's compute 1 + 4 cos(2x + π/3):cos(2x + π/3) = cos2x cosπ/3 - sin2x sinπ/3 = (1/2)cos2x - (√3/2) sin2xThus, 4 cos(2x + π/3) = 2 cos2x - 2√3 sin2xTherefore, y = 1 + 4 cos(2x + π/3) = 1 + 2 cos2x - 2√3 sin2xBut in the initial solution, they have y = -4 sin(2x - π/6) + 2, which is equal to 2 cos2x - 2√3 sin2x + 2. Wait, that's different from my expression.Wait, my expression is y = 1 + 2 cos2x - 2√3 sin2x, whereas the initial solution is y = 2 cos2x - 2√3 sin2x + 2, which is 2 + 2 cos2x - 2√3 sin2x.Wait, so there's a discrepancy here. Which one is correct?Wait, let's go back to the original function:y = 4cos²x - 4√3 sinx cosx -1Compute 4cos²x: 4*(1 + cos2x)/2 = 2 + 2cos2xCompute -4√3 sinx cosx: -4√3*(sin2x)/2 = -2√3 sin2xSo, y = 2 + 2cos2x - 2√3 sin2x -1 = 1 + 2cos2x - 2√3 sin2xSo, that's correct. So, the initial solution must have made a mistake in their simplification because they ended up with y = -4 sin(2x - π/6) + 2, which is 2 + 2cos2x - 2√3 sin2x, which is different from the correct expression.Therefore, my expression is correct: y = 1 + 4 cos(2x + π/3). So, the maximum value is 1 + 4*1 = 5, and the minimum is 1 - 4 = -3.But in the initial solution, they had the maximum value as 6, which is conflicting.Wait, perhaps I made a mistake in combining the terms.Wait, let me try another approach. Let me express 2cos2x - 2√3 sin2x as R sin(2x + φ). Maybe that's where the confusion is.Expressing as sine:R sin(2x + φ) = R sin2x cosφ + R cos2x sinφSet equal to 2cos2x - 2√3 sin2x.So, equate coefficients:R cosφ = -2√3R sinφ = 2So, R = sqrt( (-2√3)^2 + 2^2 ) = sqrt(12 + 4) = sqrt(16) = 4So, R = 4.Then, cosφ = -2√3 /4 = -√3 / 2sinφ = 2 /4 = 1/2So, φ is the angle where cosφ = -√3/2 and sinφ = 1/2, which is φ = 5π/6.Therefore, 2cos2x - 2√3 sin2x = 4 sin(2x + 5π/6)Thus, y = 1 + 4 sin(2x + 5π/6)So, that's another way to write it.So, in this case, the maximum value is 1 + 4*1 = 5, same as before.But in the initial solution, they had y = -4 sin(2x - π/6) + 2, which would have a maximum value of 2 + 4 = 6. So, that's conflicting.Wait, let's compute -4 sin(2x - π/6) + 2.As I did earlier, sin(2x - π/6) = sin2x cosπ/6 - cos2x sinπ/6 = (√3/2) sin2x - (1/2) cos2xThus, -4 sin(2x - π/6) = -4*(√3/2 sin2x - 1/2 cos2x) = -2√3 sin2x + 2 cos2xSo, -4 sin(2x - π/6) + 2 = 2 cos2x - 2√3 sin2x + 2But in our function, y = 1 + 2cos2x - 2√3 sin2xSo, the initial solution is incorrect because they have +2 instead of +1.Therefore, the correct expression is y = 1 + 4 cos(2x + π/3) or y = 1 + 4 sin(2x + 5π/6). Both are correct, just different phase shifts.So, the maximum value is 5, not 6. So, the initial solution had an error.Therefore, for part (2), the maximum value is 5, and it occurs when cos(2x + π/3) = 1.So, 2x + π/3 = 2πk, where k is an integer.Thus, 2x = -π/3 + 2πkSo, x = -π/6 + πkTherefore, the maximum value is 5, and occurs at x = -π/6 + πk, where k is integer.But in the initial solution, they had maximum value 6 at x = -π/6 + πk, which is conflicting.So, perhaps the initial solution had a mistake in their simplification.Moving on to part (3): Find the intervals of monotonic increase.Given that y = 1 + 4 cos(2x + π/3), the function is a cosine function with amplitude 4, shifted vertically by 1, and with a phase shift of -π/6 (since 2x + π/3 = 2(x + π/6)).The period is π, as found earlier.To find the intervals where the function is increasing, we can look at the derivative.Compute dy/dx:dy/dx = -8 sin(2x + π/3)The function is increasing when dy/dx > 0, so when sin(2x + π/3) < 0.So, sin(2x + π/3) < 0.Solve for x:2x + π/3 ∈ (π + 2πk, 2π + 2πk) for integer k.So, 2x + π/3 > π + 2πk and 2x + π/3 < 2π + 2πkSubtract π/3:2x > 2π/3 + 2πk and 2x < 5π/3 + 2πkDivide by 2:x > π/3 + πk and x < 5π/6 + πkTherefore, the function is increasing on the intervals (π/3 + πk, 5π/6 + πk) for integer k.But in the initial solution, they wrote the intervals as [π/3 + kπ, 5π/6 + kπ], which is correct, but they used square brackets, implying closed intervals. However, since the function is increasing on open intervals between the critical points, it's more accurate to use open intervals. However, since the function is continuous and differentiable everywhere, it's increasing on those closed intervals as well, because at the endpoints, the derivative is zero, so the function isn't decreasing there. So, both open and closed intervals are acceptable, but usually, when talking about increasing intervals, we consider open intervals where the derivative is positive. But in the initial solution, they used closed intervals, which is also acceptable.So, the intervals of monotonic increase are [π/3 + kπ, 5π/6 + kπ] for integer k.Finally, part (4): Find the axis of symmetry of the function.An axis of symmetry for a function is a vertical line x = a such that the function is symmetric about that line. For a sinusoidal function like y = A cos(Bx + C) + D, the axis of symmetry occurs at the maximum and minimum points, or at the midline crossings.But more precisely, for a cosine function, the axis of symmetry is at the peaks and troughs. However, since the function is periodic, there are infinitely many axes of symmetry spaced by half the period.Given that the function is y = 1 + 4 cos(2x + π/3), the general form is y = A cos(Bx + C) + D.The axis of symmetry occurs where the function reaches its maximum or minimum, which is at the peaks and troughs.But more accurately, for a cosine function, the axis of symmetry is at the points where the argument of the cosine is an integer multiple of π, i.e., where cos(2x + π/3) = ±1.So, solving 2x + π/3 = πk, where k is integer.Thus, 2x = -π/3 + πkx = -π/6 + (π/2)kTherefore, the axes of symmetry are at x = -π/6 + (π/2)k for integer k.Alternatively, since the function is periodic with period π, the axes of symmetry are spaced by π/2, starting from x = -π/6.So, the equation of the axis of symmetry is x = -π/6 + (π/2)k, which can also be written as x = π/3 + (π/2)k, since -π/6 + π/2 = π/3.Therefore, the axes of symmetry are x = π/3 + (π/2)k for integer k.In the initial solution, they wrote x = π/3 + (kπ)/2, which is the same as x = π/3 + (π/2)k, so that's correct.So, summarizing:(1) The smallest positive period is π.(2) The maximum value is 5, occurring at x = -π/6 + πk.(3) The function is increasing on the intervals [π/3 + kπ, 5π/6 + kπ].(4) The axes of symmetry are x = π/3 + (π/2)k.But wait, in the initial solution, for part (2), they had the maximum value as 6, which is incorrect. So, I need to correct that.Also, for part (4), the initial solution wrote x = π/3 + (kπ)/2, which is the same as x = π/3 + (π/2)k, so that's correct.So, in conclusion, the correct answers are:(1) π(2) Maximum value 5 at x = -π/6 + πk(3) Increasing on [π/3 + kπ, 5π/6 + kπ](4) Axes of symmetry at x = π/3 + (π/2)kBut in the initial solution, part (2) was incorrect, as they had maximum value 6. So, I need to make sure to correct that.Wait, let me double-check the maximum value.Given y = 1 + 4 cos(2x + π/3). The maximum value of cos is 1, so y_max = 1 + 4*1 = 5. The minimum is 1 - 4 = -3. So, yes, maximum is 5.Therefore, the initial solution had an error in part (2), giving maximum value as 6, which is incorrect.So, to sum up, my corrected answers are:(1) The smallest positive period is π.(2) The maximum value is 5, occurring at x = -π/6 + πk.(3) The function is increasing on the intervals [π/3 + kπ, 5π/6 + kπ].(4) The axes of symmetry are x = π/3 + (π/2)k.I think that's all.