💡Okay, so I need to find the distance from the point (0, 3, -1) to the line that passes through the points (1, 2, 1) and (2, 4, 0). Hmm, I remember that the distance from a point to a line in 3D space can be found using a specific formula, but I'm a bit fuzzy on the exact steps. Let me try to recall and work through it step by step.First, I think I need the direction vector of the line. The line passes through two points: (1, 2, 1) and (2, 4, 0). To find the direction vector, I can subtract the coordinates of the first point from the second point. So, subtracting (1, 2, 1) from (2, 4, 0) gives me (2-1, 4-2, 0-1), which is (1, 2, -1). So, the direction vector **d** is (1, 2, -1).Next, I need a vector from a point on the line to the point (0, 3, -1). Let's take the point (1, 2, 1) on the line. The vector from (1, 2, 1) to (0, 3, -1) would be (0-1, 3-2, -1-1), which is (-1, 1, -2). Let me call this vector **v**.Now, I remember that the distance from the point to the line can be found using the cross product of **v** and **d**, divided by the magnitude of **d**. The formula is something like |**v** × **d**| / |**d**|. Let me write that down:Distance = |**v** × **d**| / |**d**|So, first, I need to compute the cross product **v** × **d**. Let's compute that.**v** is (-1, 1, -2) and **d** is (1, 2, -1).The cross product is calculated as:|i   j   k||-1   1  -2||1    2  -1|Which is i*(1*(-1) - (-2)*2) - j*(-1*(-1) - (-2)*1) + k*(-1*2 - 1*1)Calculating each component:i: (1*(-1) - (-2)*2) = (-1 + 4) = 3j: - [(-1*(-1) - (-2)*1)] = - [(1 + 2)] = -3k: (-1*2 - 1*1) = (-2 -1) = -3So, the cross product **v** × **d** is (3, -3, -3).Now, I need the magnitude of this cross product vector. The magnitude is sqrt(3^2 + (-3)^2 + (-3)^2) = sqrt(9 + 9 + 9) = sqrt(27) = 3*sqrt(3).Next, I need the magnitude of the direction vector **d**. **d** is (1, 2, -1), so its magnitude is sqrt(1^2 + 2^2 + (-1)^2) = sqrt(1 + 4 + 1) = sqrt(6).So, putting it all together, the distance is |**v** × **d**| / |**d**| = (3*sqrt(3)) / sqrt(6).I can simplify this by rationalizing the denominator:(3*sqrt(3)) / sqrt(6) = (3*sqrt(3)) / (sqrt(6)) * (sqrt(6)/sqrt(6)) = (3*sqrt(18)) / 6Simplify sqrt(18) to 3*sqrt(2):(3*3*sqrt(2)) / 6 = (9*sqrt(2)) / 6 = (3*sqrt(2)) / 2.Wait, that seems a bit different from my initial thought. Let me double-check my steps to make sure I didn't make a mistake.First, direction vector **d** is (1, 2, -1) – correct.Vector **v** from (1,2,1) to (0,3,-1) is (-1, 1, -2) – correct.Cross product calculation:i: (1*(-1) - (-2)*2) = (-1 + 4) = 3 – correct.j: - [(-1*(-1) - (-2)*1)] = - [(1 + 2)] = -3 – correct.k: (-1*2 - 1*1) = (-2 -1) = -3 – correct.So, cross product is (3, -3, -3) – correct.Magnitude of cross product: sqrt(9 + 9 + 9) = sqrt(27) = 3*sqrt(3) – correct.Magnitude of **d**: sqrt(1 + 4 + 1) = sqrt(6) – correct.So, distance is 3*sqrt(3)/sqrt(6). Simplify:sqrt(3)/sqrt(6) = sqrt(3/6) = sqrt(1/2) = 1/sqrt(2). So, 3*(1/sqrt(2)) = 3/sqrt(2). Rationalizing, that's 3*sqrt(2)/2.Ah, so that's the same as before. So, my initial calculation was correct.Alternatively, another way to find the distance is to parametrize the line and then find the value of the parameter t that minimizes the distance from the point to the line. Let me try that method to verify.Parametrize the line as:x = 1 + ty = 2 + 2tz = 1 - tSo, any point on the line can be written as (1 + t, 2 + 2t, 1 - t).The distance squared from (0, 3, -1) to this point is:(1 + t - 0)^2 + (2 + 2t - 3)^2 + (1 - t - (-1))^2Simplify each term:(1 + t)^2 + (-1 + 2t)^2 + (2 - t)^2Expanding each:(1 + 2t + t^2) + (1 - 4t + 4t^2) + (4 - 4t + t^2)Combine like terms:1 + 2t + t^2 + 1 - 4t + 4t^2 + 4 - 4t + t^2Adding constants: 1 + 1 + 4 = 6Adding t terms: 2t -4t -4t = -6tAdding t^2 terms: t^2 + 4t^2 + t^2 = 6t^2So, distance squared is 6t^2 -6t +6.To find the minimum distance, take derivative with respect to t and set to zero.d/dt (6t^2 -6t +6) = 12t -6 = 0So, 12t -6 = 0 => t = 6/12 = 1/2.So, t = 1/2.Now, plug t = 1/2 into the distance squared:6*(1/2)^2 -6*(1/2) +6 = 6*(1/4) -3 +6 = 1.5 -3 +6 = 4.5So, distance is sqrt(4.5) = sqrt(9/2) = 3/sqrt(2) = 3*sqrt(2)/2.Same result as before. So, that confirms the distance is 3*sqrt(2)/2.Alternatively, I could have used the formula for the distance from a point to a line in 3D, which is |(P2 - P1) × (P1 - P0)| / |P2 - P1|, where P0 is the point, and P1 and P2 are points on the line.Let me define P0 as (0,3,-1), P1 as (1,2,1), and P2 as (2,4,0).Compute vector P2 - P1: (2-1, 4-2, 0-1) = (1,2,-1) – same as direction vector **d**.Compute vector P1 - P0: (1-0, 2-3, 1-(-1)) = (1,-1,2).Wait, hold on, in the formula, it's (P2 - P1) × (P0 - P1). Wait, let me check.Actually, the formula is |(P0 - P1) × (P2 - P1)| / |P2 - P1|.So, vector P0 - P1 is (0-1, 3-2, -1-1) = (-1,1,-2), which is the same as vector **v** earlier.So, cross product is (-1,1,-2) × (1,2,-1) = same as before, which is (3,-3,-3). The magnitude is 3*sqrt(3), and |P2 - P1| is sqrt(6). So, distance is 3*sqrt(3)/sqrt(6) = 3*sqrt(2)/2.Same result. So, all methods lead to the same answer.I think I've verified it sufficiently. So, the distance is 3*sqrt(2)/2.**Final Answer**The distance from the point (0, 3, -1) to the line is boxed{dfrac{3sqrt{2}}{2}}.

💡Okay, so I have this problem here where the sum of the first n terms of a sequence {a_n} is given by S_n = n² + 2n. I need to find the sum of the first n terms of another sequence, which is {1/(a_n a_{n+1})}. The options are given as A, B, C, D, so I need to figure out which one is correct.First, let me recall that the sum of the first n terms of a sequence is S_n. So, to find the nth term of the sequence {a_n}, I can use the formula a_n = S_n - S_{n-1} for n ≥ 2. For n = 1, a_1 is just S_1.Let me compute a_1 first. When n = 1, S_1 = 1² + 2*1 = 1 + 2 = 3. So, a_1 = 3.Now, for n ≥ 2, a_n = S_n - S_{n-1}. Let's compute that:S_n = n² + 2nS_{n-1} = (n - 1)² + 2(n - 1) = n² - 2n + 1 + 2n - 2 = n² - 1So, a_n = S_n - S_{n-1} = (n² + 2n) - (n² - 1) = n² + 2n - n² + 1 = 2n + 1Wait, that seems straightforward. So, a_n = 2n + 1 for n ≥ 1. Let me check for n = 1: 2*1 + 1 = 3, which matches a_1. So, that works.So, the general term of the sequence {a_n} is a_n = 2n + 1.Now, I need to find the sum of the first n terms of the sequence {1/(a_n a_{n+1})}. Let's denote this new sequence as b_n = 1/(a_n a_{n+1}).So, b_n = 1/[(2n + 1)(2(n + 1) + 1)] = 1/[(2n + 1)(2n + 3)]Hmm, so each term is 1 divided by the product of two consecutive odd numbers. This looks like it might be a telescoping series, so maybe I can express it as a difference of two fractions.I remember that for expressions like 1/(k(k + m)), we can use partial fractions. Let me try that here.Let me write 1/[(2n + 1)(2n + 3)] as A/(2n + 1) + B/(2n + 3). To find A and B, I'll set up the equation:1 = A(2n + 3) + B(2n + 1)Expanding the right side:1 = (2A + 2B)n + (3A + B)Since this must hold for all n, the coefficients of n and the constant term must be equal on both sides. So:2A + 2B = 0 (coefficient of n)3A + B = 1 (constant term)Let me solve this system of equations. From the first equation:2A + 2B = 0 ⇒ A + B = 0 ⇒ B = -ASubstitute B = -A into the second equation:3A + (-A) = 1 ⇒ 2A = 1 ⇒ A = 1/2Then, B = -1/2So, the partial fractions decomposition is:1/[(2n + 1)(2n + 3)] = (1/2)/(2n + 1) - (1/2)/(2n + 3)Therefore, b_n = (1/2)[1/(2n + 1) - 1/(2n + 3)]Now, I need to find the sum of the first n terms of {b_n}, which is:Sum = Σ (from k=1 to n) [ (1/2)(1/(2k + 1) - 1/(2k + 3)) ]I can factor out the 1/2:Sum = (1/2) Σ (from k=1 to n) [1/(2k + 1) - 1/(2k + 3)]Let me write out the terms to see if it telescopes:For k=1: 1/3 - 1/5For k=2: 1/5 - 1/7For k=3: 1/7 - 1/9...For k=n: 1/(2n + 1) - 1/(2n + 3)When I add all these up, most terms should cancel out. Let's see:Sum = (1/2)[ (1/3 - 1/5) + (1/5 - 1/7) + (1/7 - 1/9) + ... + (1/(2n + 1) - 1/(2n + 3)) ]Looking at the terms inside the brackets:- The -1/5 cancels with the +1/5- The -1/7 cancels with the +1/7- This pattern continues, and all the intermediate terms cancel out.So, what's left is the first term of the first expression and the last term of the last expression:Sum = (1/2)[1/3 - 1/(2n + 3)]Simplify this:Sum = (1/2)(1/3 - 1/(2n + 3)) = (1/2)[ ( (2n + 3) - 3 ) / (3(2n + 3)) ) ]Wait, let me compute the difference inside:1/3 - 1/(2n + 3) = (2n + 3 - 3)/[3(2n + 3)] = (2n)/[3(2n + 3)]So, Sum = (1/2)(2n / [3(2n + 3)]) = (1/2)*(2n) / [3(2n + 3)] = n / [3(2n + 3)]So, the sum is n divided by [3 times (2n + 3)], which is option A.Wait, let me double-check my steps to make sure I didn't make a mistake.1. Found a_n correctly: a_n = 2n + 1.2. Expressed 1/(a_n a_{n+1}) as partial fractions: (1/2)[1/(2n+1) - 1/(2n+3)]. That seems correct.3. Summed from k=1 to n: telescoping series, so most terms cancel, leaving (1/2)(1/3 - 1/(2n+3)). That seems right.4. Simplified the expression: 1/3 - 1/(2n + 3) = (2n)/(3(2n + 3)). Then multiplied by 1/2, getting n/(3(2n + 3)). That seems correct.So, the answer should be option A: n / [3(2n + 3)].I think that's solid. I don't see any errors in the reasoning.**Final Answer**boxed{A}

💡First, I need to determine the probability that car Z will win in the race. I know the probabilities for cars X and Y winning, and the combined probability for any of the three cars winning.I'll start by converting the given probabilities to decimals for easier calculation. Car X has a probability of 1/4, which is 0.25, and car Y has a probability of 1/8, which is 0.125.Next, I'll add the probabilities of cars X and Y winning to find the combined probability of either car X or Y winning:0.25 (X) + 0.125 (Y) = 0.375.The total probability that one of the three cars (X, Y, or Z) will win is given as 0.4583333333333333. To find the probability of car Z winning, I'll subtract the combined probability of cars X and Y from this total:0.4583333333333333 - 0.375 = 0.0833333333333333.Finally, I'll convert the decimal probability of car Z winning back to a fraction. 0.0833333333333333 is equivalent to 1/12.

💡Alright, so I have this function here: y equals one half sin 2x tan x plus 2 sin x tan(x over 2). I need to find the range of this function. Hmm, okay. Let me try to break this down step by step.First, I remember that the range of a function is the set of all possible output values (y-values) it can produce. So, I need to figure out what values y can take on as x varies over its domain. But before I can find the range, I should probably simplify the function as much as possible. Maybe I can use some trigonometric identities to make this expression simpler.Looking at the function, I see sin 2x, which I know is equal to 2 sin x cos x. That's one identity. Then there's tan x, which is sin x over cos x. And there's also tan(x/2), which I think has an identity too. Let me recall: tan(x/2) is equal to (1 - cos x) over sin x. Yeah, that sounds right.So, let me rewrite the function using these identities. Starting with the first term: one half sin 2x tan x. Since sin 2x is 2 sin x cos x, and tan x is sin x over cos x, substituting these in, we get:(1/2) * (2 sin x cos x) * (sin x / cos x)Let me compute that. The 2 and the 1/2 cancel out, so we have sin x cos x * sin x / cos x. The cos x in the numerator and denominator also cancel out, leaving us with sin squared x.Okay, so the first term simplifies to sin²x. That wasn't too bad.Now, the second term is 2 sin x tan(x/2). Using the identity for tan(x/2), which is (1 - cos x)/sin x, substituting that in, we get:2 sin x * (1 - cos x)/sin xHere, the sin x in the numerator and denominator cancel out, leaving us with 2*(1 - cos x). So, the second term simplifies to 2 - 2 cos x.Putting it all together, the original function simplifies to:y = sin²x + 2 - 2 cos xHmm, okay. That's a simpler expression. Now, maybe I can express this in terms of cos x only, using the Pythagorean identity. I know that sin²x is equal to 1 - cos²x. So, substituting that in:y = (1 - cos²x) + 2 - 2 cos xSimplifying this, we get:y = 1 - cos²x + 2 - 2 cos xCombine like terms:y = 3 - cos²x - 2 cos xHmm, this looks like a quadratic in terms of cos x. Maybe I can rewrite it as such. Let me set u = cos x, then the equation becomes:y = 3 - u² - 2uWhich can be rewritten as:y = -u² - 2u + 3Or, rearranged:y = - (u² + 2u) + 3I can complete the square for the quadratic in u. Let's see:u² + 2u = (u + 1)² - 1So, substituting back:y = - [(u + 1)² - 1] + 3Which simplifies to:y = - (u + 1)² + 1 + 3y = - (u + 1)² + 4So, now we have:y = - (cos x + 1)² + 4Wait, that's interesting. So, y is equal to negative (cos x + 1) squared plus 4. Now, since (cos x + 1) squared is always non-negative, the term being subtracted is non-negative. Therefore, y will be less than or equal to 4.But what about the minimum value? Since (cos x + 1) squared can get as large as... Well, cos x ranges from -1 to 1. So, cos x + 1 ranges from 0 to 2. Therefore, (cos x + 1) squared ranges from 0 to 4.So, when (cos x + 1) squared is 0, y is 4. When it's 4, y is 0. So, y ranges from 0 to 4. But wait, does it actually reach 4?Wait, when is (cos x + 1) squared equal to 0? That happens when cos x + 1 = 0, which is when cos x = -1. So, x is pi, 3pi, etc. Let's check y at x = pi.At x = pi, sin 2x is sin 2pi, which is 0. tan x is tan pi, which is 0. So, the first term is 0. The second term is 2 sin x tan(x/2). At x = pi, sin x is 0, so the second term is also 0. So, y is 0 at x = pi. So, y can be 0.Similarly, when (cos x + 1) squared is 4, that would require cos x + 1 = 2, so cos x = 1. That happens at x = 0, 2pi, etc. Let's check y at x = 0.At x = 0, sin 2x is 0, tan x is 0, so the first term is 0. The second term is 2 sin x tan(x/2). At x = 0, sin x is 0, so the second term is also 0. So, y is 0 at x = 0 as well. Wait, that's not 4. Hmm, that's confusing.Wait, according to our earlier equation, y = - (cos x + 1)^2 + 4. So, when (cos x + 1)^2 is 0, y is 4. But when cos x = -1, (cos x + 1)^2 is 0, so y is 4. But when I plug x = pi into the original function, I get y = 0. That seems contradictory.Wait, maybe I made a mistake in simplifying the function. Let me go back and check.Original function: y = (1/2) sin 2x tan x + 2 sin x tan(x/2)First term: (1/2) sin 2x tan xsin 2x is 2 sin x cos x, tan x is sin x / cos xSo, (1/2)*(2 sin x cos x)*(sin x / cos x) = (1/2)*2 sin²x = sin²xSecond term: 2 sin x tan(x/2)tan(x/2) is (1 - cos x)/sin xSo, 2 sin x * (1 - cos x)/sin x = 2(1 - cos x)So, y = sin²x + 2 - 2 cos xThen, sin²x = 1 - cos²xSo, y = 1 - cos²x + 2 - 2 cos x = 3 - cos²x - 2 cos xThen, completing the square:y = - (cos x + 1)^2 + 4Wait, that seems correct. So, when cos x = -1, y = 4. But when I plug x = pi into the original function, I get y = 0. So, that must mean that at x = pi, the function is 0, but according to the simplified equation, it should be 4. That's a problem.Wait, maybe I made a mistake in simplifying the function. Let me check the original function again.Original function: y = (1/2) sin 2x tan x + 2 sin x tan(x/2)At x = pi, sin 2x = sin 2pi = 0, tan x = tan pi = 0, so first term is 0.Second term: 2 sin x tan(x/2). At x = pi, sin x = 0, so tan(x/2) is tan(pi/2), which is undefined. Wait, tan(pi/2) is undefined, so actually, at x = pi, tan(x/2) is undefined. So, the function isn't defined at x = pi. That's an important point.So, x cannot be pi, because tan(x/2) would be undefined. Similarly, x cannot be any odd multiple of pi, because tan(x/2) would be undefined. So, the domain of the function excludes points where x is an odd multiple of pi.Therefore, when cos x approaches -1, x approaches pi, but x cannot be pi. So, (cos x + 1)^2 approaches 0, but never actually reaches 0. Therefore, y approaches 4, but never actually reaches 4.Similarly, when cos x approaches 1, x approaches 0, and y approaches 0. But at x = 0, tan(x/2) is 0, so y is 0. So, y can be 0, but it can't be 4.Wait, so y can be 0, but it can't be 4. So, the range is from 0 up to, but not including, 4.But wait, let me check another point. Let's take x = pi/2. Then, sin 2x is sin pi, which is 0. tan x is tan(pi/2), which is undefined. So, x cannot be pi/2 either.Wait, so the domain of the function is all real numbers except where tan x or tan(x/2) is undefined. So, x cannot be pi/2 + k pi, or pi + 2k pi, etc. So, the function is defined on intervals between these points.But regardless, when we simplified the function, we got y = - (cos x + 1)^2 + 4. Since cos x ranges from -1 to 1, but excluding the points where cos x = -1 (since x cannot be pi, etc.), so (cos x + 1)^2 ranges from just above 0 to 4.Therefore, y = - (cos x + 1)^2 + 4 ranges from just below 4 down to 0.But wait, when cos x approaches 1, (cos x + 1)^2 approaches 4, so y approaches 0. When cos x approaches -1, (cos x + 1)^2 approaches 0, so y approaches 4.But at x = 0, cos x = 1, and y = 0. So, y can be 0. But when cos x approaches -1, y approaches 4, but never actually reaches 4 because x cannot be pi.So, the range of y is [0, 4), meaning y can be 0 up to, but not including, 4.Wait, but earlier, when I plugged in x = pi/2, the function was undefined. So, does that affect the range? Let me think.Actually, the function is defined for all x except where tan x or tan(x/2) is undefined. So, between those undefined points, the function is continuous and can take on all values between 0 and 4.But wait, let me check another value. Let's take x = pi/3. Then, cos x = 0.5. So, (cos x + 1)^2 = (1.5)^2 = 2.25. So, y = -2.25 + 4 = 1.75. So, y = 1.75 is attainable.Similarly, take x = pi/4. cos x = sqrt(2)/2 ≈ 0.707. So, (cos x + 1)^2 ≈ (1.707)^2 ≈ 2.918. So, y ≈ -2.918 + 4 ≈ 1.082.Wait, but what about when x approaches pi? Let's say x approaches pi from the left. Then, cos x approaches -1 from above. So, (cos x + 1)^2 approaches 0. So, y approaches 4 from below.Similarly, when x approaches 0, cos x approaches 1, so (cos x + 1)^2 approaches 4, so y approaches 0.Therefore, the function y can take on all values from 0 up to, but not including, 4.But wait, when x is pi/2, the function is undefined, but near pi/2, what happens? Let's take x approaching pi/2 from below. Then, tan x approaches infinity, but in the first term, we have sin 2x tan x. sin 2x at pi/2 is sin pi, which is 0. So, the first term is 0. The second term is 2 sin x tan(x/2). At x approaching pi/2, sin x approaches 1, and tan(x/2) approaches tan(pi/4) = 1. So, the second term approaches 2*1*1 = 2. So, near x = pi/2, y approaches 2.Similarly, approaching pi/2 from above, tan x approaches negative infinity, but sin 2x is sin(pi + something), which is negative. So, sin 2x tan x would be negative times negative, which is positive. But sin 2x is approaching 0, so the first term is still approaching 0. The second term is 2 sin x tan(x/2). At x approaching pi/2 from above, sin x is still approaching 1, and tan(x/2) approaches tan(pi/4) = 1. So, the second term approaches 2 again.So, near pi/2, y approaches 2. So, y can take on values around 2, but not exactly at pi/2.Wait, so does that mean that the function can take on all values between 0 and 4, excluding 4? Or are there gaps?Wait, let's think about the expression y = - (cos x + 1)^2 + 4. Since cos x ranges from -1 to 1, excluding points where x is an odd multiple of pi, (cos x + 1)^2 ranges from just above 0 to 4. Therefore, y ranges from just below 4 down to 0.But is y continuous over its domain? Since the function is undefined at certain points, but between those points, it's continuous. So, the range should be all real numbers between 0 and 4, excluding 4.Wait, but when x approaches pi, y approaches 4, but never reaches it. So, the maximum value y approaches is 4, but doesn't include it. The minimum value is 0, which is achieved at x = 0.Therefore, the range of y is [0, 4).But wait, earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, let me clarify. At x = 0, cos x = 1, so (cos x + 1)^2 = 4, so y = -4 + 4 = 0. So, y = 0 is attainable at x = 0.At x approaching pi, y approaches 4, but doesn't reach it because x cannot be pi.So, the range is all real numbers y such that 0 ≤ y < 4.Therefore, the range of the function is [0, 4).But wait, earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, let me double-check. At x = 0, sin 2x = 0, tan x = 0, so first term is 0. The second term is 2 sin x tan(x/2). At x = 0, sin x = 0, so the second term is 0. So, y = 0 + 0 = 0. So, y can be 0.At x = pi/2, the function is undefined, but near x = pi/2, y approaches 2.At x approaching pi, y approaches 4, but doesn't reach it.So, the function can take on all values from 0 up to, but not including, 4.Therefore, the range is [0, 4).Wait, but earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, I'm getting confused. Let me summarize:- The function simplifies to y = - (cos x + 1)^2 + 4.- cos x ranges from -1 to 1, but x cannot be pi, 3pi, etc., where cos x = -1.- Therefore, (cos x + 1)^2 ranges from just above 0 to 4.- So, y ranges from just below 4 down to 0.- At x = 0, y = 0.- As x approaches pi, y approaches 4, but never reaches it.Therefore, the range of y is [0, 4).But wait, earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, I think I'm mixing up the points. At x = 0, y = 0, which is valid because x = 0 is in the domain. At x = pi, y would be undefined because tan(x/2) is undefined there. So, y can be 0, but cannot be 4.Therefore, the range is [0, 4).Wait, but let me check another point. Let's take x = pi/3. Then, cos x = 0.5, so (cos x + 1)^2 = (1.5)^2 = 2.25. So, y = -2.25 + 4 = 1.75. So, y = 1.75 is attainable.Similarly, take x = pi/4. cos x = sqrt(2)/2 ≈ 0.707. So, (cos x + 1)^2 ≈ (1.707)^2 ≈ 2.918. So, y ≈ -2.918 + 4 ≈ 1.082.Wait, but what about when x approaches pi/2? Let's say x approaches pi/2 from below. Then, tan x approaches infinity, but sin 2x approaches sin pi = 0. So, the first term is 0. The second term is 2 sin x tan(x/2). At x approaching pi/2, sin x approaches 1, and tan(x/2) approaches tan(pi/4) = 1. So, the second term approaches 2*1*1 = 2. So, y approaches 2.Similarly, approaching pi/2 from above, tan x approaches negative infinity, but sin 2x approaches 0, so the first term is still 0. The second term is 2 sin x tan(x/2). At x approaching pi/2 from above, sin x approaches 1, and tan(x/2) approaches tan(pi/4) = 1. So, the second term approaches 2 again.So, near pi/2, y approaches 2. So, y can take on values around 2, but not exactly at pi/2.Therefore, the function is continuous on its domain, which is all real numbers except where tan x or tan(x/2) is undefined. Between those points, the function can take on all values between 0 and 4, excluding 4.Therefore, the range of y is [0, 4).Wait, but earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, I think I'm overcomplicating this. The key is that y = - (cos x + 1)^2 + 4, and since (cos x + 1)^2 is always non-negative and can approach 0 but never actually be 0 (because cos x cannot be -1 due to the domain restrictions), y can approach 4 but never reach it. However, y can reach 0 when cos x = 1, which happens at x = 0, 2pi, etc., which are in the domain.Therefore, the range of y is [0, 4).But wait, let me check another point. Let's take x = pi/6. Then, cos x = sqrt(3)/2 ≈ 0.866. So, (cos x + 1)^2 ≈ (1.866)^2 ≈ 3.482. So, y ≈ -3.482 + 4 ≈ 0.518.Similarly, take x = pi/12. cos x ≈ 0.966. So, (cos x + 1)^2 ≈ (1.966)^2 ≈ 3.865. So, y ≈ -3.865 + 4 ≈ 0.135.So, y can take on values close to 0, but not exactly 0 except at x = 0.Wait, but at x = 0, y = 0. So, y can be 0, and can take on all values up to, but not including, 4.Therefore, the range is [0, 4).Wait, but earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, I think I'm getting confused because I'm mixing up the points where cos x = -1 and cos x = 1. At x = 0, cos x = 1, which is valid, and y = 0. At x = pi, cos x = -1, but x = pi is not in the domain because tan(x/2) is undefined there. So, y approaches 4 as x approaches pi, but never reaches it.Therefore, the range of y is [0, 4).But wait, let me think again. The function simplifies to y = - (cos x + 1)^2 + 4. Since (cos x + 1)^2 is always non-negative, y is always less than or equal to 4. However, because cos x cannot be -1 (as x cannot be pi, etc.), y cannot be 4. But y can be 0 when cos x = 1, which is allowed.Therefore, the range is [0, 4).But wait, earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.Wait, I think I'm stuck in a loop here. Let me try to visualize the function.If I plot y = - (cos x + 1)^2 + 4, it's a downward-opening parabola in terms of cos x. The maximum value would be at the vertex, which is when cos x + 1 = 0, i.e., cos x = -1. But since cos x = -1 is not in the domain, the function approaches 4 but never reaches it. The minimum value is when cos x = 1, which is y = 0, and that's attainable.Therefore, the range is [0, 4).Wait, but earlier, I thought y could be 0, but when I plugged in x = pi, I got y = 0, but x = pi is not in the domain. Wait, no, at x = 0, y = 0. So, y can be 0.I think I've convinced myself that the range is [0, 4). So, y can take on any value from 0 up to, but not including, 4.Therefore, the range of the function is [0, 4).

💡Okay, so I have this problem here: Given the function f(x) = x² + 2x f'(1), I need to find f'(1). Hmm, that seems a bit tricky at first glance because f'(1) is part of the function itself. I guess I need to figure out a way to solve for f'(1) by using differentiation.Alright, let me start by recalling how to find the derivative of a function. The function f(x) is given as x squared plus 2x times f'(1). So, f(x) = x² + 2x f'(1). I need to find f'(x), which is the derivative of f with respect to x.Let me write that down: f'(x) = derivative of x² plus derivative of 2x f'(1). The derivative of x² is straightforward, that's 2x. Now, what about the derivative of 2x f'(1)? Well, f'(1) is a constant because it's the derivative of f at a specific point, x=1. So, when I take the derivative of 2x times a constant, that's just 2 times that constant. So, the derivative of 2x f'(1) is 2 f'(1).Putting it all together, f'(x) = 2x + 2 f'(1). Okay, so now I have an expression for f'(x) in terms of x and f'(1). But I need to find f'(1), so maybe I can substitute x=1 into this equation.Let me do that. If I plug in x=1 into f'(x), I get f'(1) = 2*1 + 2 f'(1). Simplifying that, f'(1) = 2 + 2 f'(1). Hmm, now I have an equation where f'(1) is on both sides. I need to solve for f'(1).Let me rearrange the equation. If I subtract 2 f'(1) from both sides, I get f'(1) - 2 f'(1) = 2. That simplifies to -f'(1) = 2. So, if I multiply both sides by -1, I get f'(1) = -2.Wait, does that make sense? Let me check my steps again. I started with f(x) = x² + 2x f'(1). Took the derivative, got f'(x) = 2x + 2 f'(1). Plugged in x=1, so f'(1) = 2 + 2 f'(1). Then subtracted 2 f'(1) from both sides: f'(1) - 2 f'(1) = 2, which is -f'(1) = 2, so f'(1) = -2. Yeah, that seems consistent.I guess another way to think about it is that f'(1) is a constant, so when I differentiate, it doesn't change with x. So, by plugging in x=1, I can solve for that constant. It's like setting up an equation where the derivative at a specific point depends on itself, but through differentiation, I can isolate it.Let me also think about what the function f(x) looks like. It's a quadratic function because of the x² term, and the linear term is scaled by f'(1). Since f'(1) turned out to be -2, the function is f(x) = x² + 2x*(-2) = x² - 4x. Let me check the derivative of that: f'(x) = 2x - 4. Then, f'(1) would be 2*1 - 4 = -2, which matches what I found earlier. So, that checks out.I think I'm confident that f'(1) is indeed -2. I don't see any mistakes in my reasoning, and the check at the end confirms it.

💡Okay, so I have this problem here where I need to find the coefficient of the term containing ( x^2 ) in the expansion of ( left(x - frac{3}{x}right)^n ). First, I need to figure out what ( n ) is because it's given as an integral. The integral is from 1 to ( e^6 ) of ( frac{1}{x} , dx ). Alright, let me start by solving that integral. The integral of ( frac{1}{x} ) with respect to ( x ) is ( ln|x| ), right? So evaluating from 1 to ( e^6 ), it should be ( ln(e^6) - ln(1) ). I remember that ( ln(e^6) ) simplifies because the natural logarithm and the exponential function are inverses. So ( ln(e^6) = 6 ). And ( ln(1) ) is 0 because ( e^0 = 1 ). So subtracting those, ( 6 - 0 = 6 ). Therefore, ( n = 6 ).Okay, so now I know that ( n = 6 ). The next part is to expand ( left(x - frac{3}{x}right)^6 ) and find the coefficient of the ( x^2 ) term. I remember that the binomial expansion of ( (a + b)^n ) is given by the sum from ( r = 0 ) to ( n ) of ( binom{n}{r} a^{n - r} b^r ). In this case, ( a = x ) and ( b = -frac{3}{x} ). So each term in the expansion will be ( binom{6}{r} x^{6 - r} left(-frac{3}{x}right)^r ).Let me write that out more clearly: each term is ( binom{6}{r} times x^{6 - r} times left(-frac{3}{x}right)^r ). Simplifying this, the ( x ) terms can be combined. So ( x^{6 - r} times left(frac{1}{x}right)^r = x^{6 - r - r} = x^{6 - 2r} ). And the constants are ( binom{6}{r} times (-3)^r ). Therefore, each term is ( binom{6}{r} times (-3)^r times x^{6 - 2r} ). I need the term where the exponent of ( x ) is 2. So I set ( 6 - 2r = 2 ) and solve for ( r ).Subtracting 6 from both sides gives ( -2r = -4 ), and dividing both sides by -2 gives ( r = 2 ). So the term where ( r = 2 ) will have ( x^2 ).Now, I need to find the coefficient for this term. Plugging ( r = 2 ) into the coefficient part: ( binom{6}{2} times (-3)^2 ).Calculating ( binom{6}{2} ), which is the number of combinations of 6 things taken 2 at a time. I remember that ( binom{6}{2} = frac{6!}{2!(6 - 2)!} = frac{6 times 5}{2 times 1} = 15 ).Then, ( (-3)^2 = 9 ). Multiplying these together, ( 15 times 9 = 135 ).So the coefficient of the ( x^2 ) term is 135. Let me just double-check my steps to make sure I didn't make a mistake.1. Calculated the integral correctly: ( int_{1}^{e^6} frac{1}{x} dx = 6 ). That seems right.2. Expanded ( left(x - frac{3}{x}right)^6 ) using binomial theorem. Yes, that's the standard approach.3. Simplified the exponents correctly: ( x^{6 - 2r} ). That makes sense because ( x^{6 - r} times x^{-r} = x^{6 - 2r} ).4. Solved for ( r ) when exponent is 2: ( 6 - 2r = 2 ) leads to ( r = 2 ). Correct.5. Calculated the binomial coefficient ( binom{6}{2} = 15 ). That's right.6. Calculated ( (-3)^2 = 9 ). Correct.7. Multiplied 15 and 9 to get 135. Yep, that's accurate.I think all the steps check out. So the coefficient should be 135.**Final Answer**boxed{135}