💡Okay, so I have this function f(x) = √2 sin(x + π/4), and I need to figure out on which interval it's monotonically increasing when x is between 0 and π. Hmm, okay. Let me think about how to approach this.First, I remember that to determine where a function is increasing or decreasing, I should look at its derivative. If the derivative is positive, the function is increasing; if it's negative, the function is decreasing. So, I need to find the derivative of f(x).Alright, f(x) = √2 sin(x + π/4). The derivative of sin(u) with respect to x is cos(u) times the derivative of u with respect to x. So, applying that here, the derivative f’(x) should be √2 cos(x + π/4) times the derivative of (x + π/4) with respect to x. The derivative of (x + π/4) is just 1, right? So, f’(x) = √2 cos(x + π/4).Okay, so f’(x) = √2 cos(x + π/4). Now, I need to find where this derivative is positive because that's where the function is increasing. So, I need to solve the inequality √2 cos(x + π/4) > 0.Since √2 is a positive constant, I can ignore it for the inequality, so it simplifies to cos(x + π/4) > 0. Now, I need to find the values of x in [0, π] where cos(x + π/4) is positive.I remember that cosine is positive in the first and fourth quadrants, which corresponds to angles between -π/2 to π/2, and so on periodically. But since x is between 0 and π, let's see what x + π/4 looks like.If x is 0, then x + π/4 is π/4. If x is π, then x + π/4 is π + π/4 = 5π/4. So, x + π/4 ranges from π/4 to 5π/4 as x goes from 0 to π.Now, cosine is positive in the first and fourth quadrants, which are angles between -π/2 to π/2 and 3π/2 to 5π/2, but since our angle x + π/4 is between π/4 and 5π/4, let's see where cosine is positive in that interval.Cosine is positive from π/4 up to π/2, because after π/2, cosine starts to decrease and becomes negative. Wait, no, actually, cosine is positive from -π/2 to π/2, so in the interval from π/4 to π/2, cosine is still positive. But after π/2, cosine becomes negative until 3π/2.Wait, but our interval is from π/4 to 5π/4. So, cosine is positive from π/4 to π/2, and then negative from π/2 to 3π/2, but our upper limit is 5π/4, which is less than 3π/2 (which is 4.712...), so up to 5π/4, cosine is negative.Wait, let me think again. Cosine is positive in the first and fourth quadrants, so between -π/2 to π/2, and then again between 3π/2 to 5π/2, etc. But in our case, x + π/4 goes from π/4 to 5π/4, which is from 45 degrees to 225 degrees.So, in this interval, cosine is positive from π/4 to π/2, which is 45 degrees to 90 degrees, and then negative from π/2 to 5π/4, which is 90 degrees to 225 degrees.Therefore, cos(x + π/4) > 0 when x + π/4 is in [π/4, π/2], which translates back to x being in [0, π/4].Wait, let me check that. If x + π/4 is in [π/4, π/2], then x is in [0, π/4]. So, yes, that makes sense.So, f’(x) is positive when x is between 0 and π/4, meaning the function f(x) is increasing on that interval.Let me just double-check by plugging in some values. If x is 0, f’(0) = √2 cos(π/4) = √2*(√2/2) = 1, which is positive. If x is π/4, f’(π/4) = √2 cos(π/2) = √2*0 = 0. So, at π/4, the derivative is zero, which is the point where the function stops increasing and starts decreasing.If I pick a value beyond π/4, say x = π/2, then f’(π/2) = √2 cos(3π/4) = √2*(-√2/2) = -1, which is negative, confirming that the function is decreasing there.So, putting it all together, the function f(x) is increasing on the interval [0, π/4].**Final Answer**The function ( f(x) ) is monotonically increasing on the interval (boxed{[0, frac{pi}{4}]}).

💡Alright, so I have this triangle ABC, and I need to find two things: first, the value of sin A, and second, the area of the triangle when b is 11. Let me try to figure this out step by step.First, the problem says that 4a equals the square root of 5 times c. So, 4a = √5 c. That means a is related to c by that equation. I can write that as a = (√5 / 4) * c. Maybe I'll need that later.Also, it's given that cos C is 3/5. Since cos C is positive, angle C must be acute, right? Because cosine is positive in the first and fourth quadrants, but in a triangle, all angles are between 0 and 180 degrees, so it's definitely acute.I remember that in a triangle, the sum of angles is 180 degrees, so A + B + C = 180. But I don't know B yet, so maybe I can find sin C first because I know cos C.Using the Pythagorean identity, sin²C + cos²C = 1. So, sin²C = 1 - cos²C. Plugging in cos C = 3/5, sin²C = 1 - (9/25) = 16/25. So sin C is 4/5 because it's positive in an acute angle.Now, I need to find sin A. Maybe I can use the Law of Sines here. The Law of Sines says that a/sin A = c/sin C. I have a relation between a and c, so let's plug that in.From 4a = √5 c, I can write a = (√5 / 4)c. So, plugging into the Law of Sines: (√5 / 4)c / sin A = c / (4/5). Let me write that out:(√5 / 4)c / sin A = c / (4/5)Hmm, the c's cancel out, so that leaves me with (√5 / 4) / sin A = 1 / (4/5). Simplify the right side: 1 divided by (4/5) is 5/4. So now I have:(√5 / 4) / sin A = 5/4Multiply both sides by sin A:√5 / 4 = (5/4) sin ANow, multiply both sides by 4:√5 = 5 sin ADivide both sides by 5:sin A = √5 / 5Okay, so that's part I done. Sin A is √5 over 5.Now, part II: If b = 11, find the area of triangle ABC.I need to find the area. I remember the formula for the area of a triangle is (1/2)ab sin C, but I need to figure out what a and b are.Wait, I know b is 11, but I don't know a or c. But I have a relation between a and c: 4a = √5 c. So, if I can find either a or c, I can find the other.Maybe I can use the Law of Sines again. The Law of Sines says that a/sin A = b/sin B = c/sin C.I know sin A is √5 / 5, sin C is 4/5, and b is 11. But I don't know sin B. Maybe I can find angle B first.Since A + B + C = 180, I can find angle B if I know angles A and C. But I don't know angle A or angle C in degrees, but I know their sines and cosines.Wait, I know cos C is 3/5, so angle C is arccos(3/5). Let me calculate that. Arccos(3/5) is approximately 53.13 degrees. Similarly, sin A is √5 / 5, so angle A is arcsin(√5 / 5). Let me calculate that as well. Arcsin(√5 / 5) is approximately 26.565 degrees.So, angle B would be 180 - 53.13 - 26.565, which is approximately 100.305 degrees. Hmm, that seems a bit messy, but maybe I can find sin B using the sine of the sum formula.Alternatively, since I know angles A and C, I can find angle B as 180 - A - C, and then sin B would be sin(180 - A - C) which is sin(A + C). Using the sine addition formula:sin(A + C) = sin A cos C + cos A sin CI know sin A is √5 / 5, cos C is 3/5, cos A I can find using the Pythagorean identity, and sin C is 4/5.So, let's find cos A. Since sin A is √5 / 5, cos²A = 1 - sin²A = 1 - (5/25) = 20/25 = 4/5. So cos A is 2√5 / 5.Now, plug into sin(A + C):sin A cos C + cos A sin C = (√5 / 5)(3/5) + (2√5 / 5)(4/5)Calculate each term:First term: (√5 / 5)(3/5) = 3√5 / 25Second term: (2√5 / 5)(4/5) = 8√5 / 25Add them together: 3√5 / 25 + 8√5 / 25 = 11√5 / 25So sin B is 11√5 / 25.Now, going back to the Law of Sines: a / sin A = b / sin BWe can write a = (b sin A) / sin BPlugging in the values:a = (11 * (√5 / 5)) / (11√5 / 25)Simplify numerator: 11√5 / 5Denominator: 11√5 / 25So, dividing these: (11√5 / 5) * (25 / 11√5) = (25 / 5) = 5So, a is 5.Now, since 4a = √5 c, and a is 5, then 4*5 = √5 c => 20 = √5 c => c = 20 / √5 = 4√5So, c is 4√5.Now, to find the area, I can use the formula (1/2)ab sin C. I have a = 5, b = 11, and sin C = 4/5.So, area = (1/2) * 5 * 11 * (4/5)Simplify:(1/2) * 5 * 11 * (4/5) = (1/2) * 11 * 4 = (1/2) * 44 = 22So, the area is 22.Wait, let me double-check that. If I use another formula, like (1/2)bc sin A, does it give the same result?b = 11, c = 4√5, sin A = √5 / 5So, area = (1/2) * 11 * 4√5 * (√5 / 5) = (1/2) * 11 * 4 * (5) / 5 = (1/2) * 11 * 4 = 22Yes, same result. So, that seems consistent.So, I think I did everything correctly. I found sin A using the Law of Sines and the given ratio, then found the sides using the Law of Sines again once I had all the sines of the angles, and then calculated the area with two different formulas to confirm.**Final Answer**Ⅰ. boxed{dfrac{sqrt{5}}{5}} Ⅱ. boxed{22}

💡Okay, so I have this problem: Given a real number a, if the function y = e^x + a x has a local maximum point greater than zero, then I need to find the range of a. Hmm, alright, let's break this down step by step.First, I remember that to find local maxima or minima of a function, we need to look at its critical points. Critical points occur where the first derivative is zero or undefined. Since this function is y = e^x + a x, it's differentiable everywhere because both e^x and a x are smooth functions. So, the critical points will be where the derivative equals zero.Let me compute the derivative of y with respect to x. The derivative of e^x is e^x, and the derivative of a x is a. So, y' = e^x + a. That seems straightforward.Now, to find the critical points, I set y' equal to zero:e^x + a = 0Solving for x, I get:e^x = -aTaking the natural logarithm of both sides, since e^x is always positive, -a must also be positive. So, -a > 0 implies that a < 0. That's an important point—so a has to be negative for there to be a real solution for x.So, x = ln(-a). Wait, hold on. If e^x = -a, then x = ln(-a). But ln is only defined for positive arguments, so -a must be positive, which again tells me a is negative.Now, the problem states that the local maximum point is greater than zero. So, I think this means that the x-coordinate of the critical point is greater than zero. So, x > 0.Given that x = ln(-a), and we want x > 0, let's see what that implies for a.Since ln(-a) > 0, and the natural logarithm function ln(z) is greater than zero when z > 1. So, ln(-a) > 0 implies that -a > 1, because ln(z) > 0 when z > 1.Therefore, -a > 1 implies that a < -1.So, putting it all together, for the function y = e^x + a x to have a local maximum at a point where x > 0, the parameter a must be less than -1.Wait, let me double-check this reasoning. So, we found that the critical point is at x = ln(-a). For this x to be greater than zero, ln(-a) > 0, which requires that -a > 1, hence a < -1. That seems correct.But just to be thorough, let's verify if this critical point is indeed a maximum. To do that, I can use the second derivative test.Compute the second derivative of y. The first derivative was y' = e^x + a, so the second derivative y'' is the derivative of y', which is e^x.Since e^x is always positive for any real x, the second derivative y'' is positive. That means that any critical point is a local minimum, not a maximum. Wait, that contradicts the problem statement which says there's a local maximum.Hold on, that's confusing. If the second derivative is always positive, then the function is concave upward everywhere, meaning any critical point is a local minimum. So, how can there be a local maximum?Wait, maybe I made a mistake in interpreting the problem. Let me read it again: "if the function y = e^x + a x has a local maximum point greater than zero, then the range of a is ______."Hmm, but according to my calculations, the function only has local minima, not maxima. That suggests that perhaps there's no local maximum unless... unless the function behaves differently at infinity or something.Wait, let's think about the behavior of y as x approaches infinity and negative infinity.As x approaches positive infinity, e^x grows exponentially, and a x grows linearly. Since e^x dominates, y approaches positive infinity.As x approaches negative infinity, e^x approaches zero, and a x approaches negative infinity if a is positive, or positive infinity if a is negative. Wait, but a is negative in our case because we found a < -1.So, as x approaches negative infinity, a x approaches positive infinity because a is negative and x is negative, so negative times negative is positive. So, y approaches positive infinity as x approaches negative infinity as well.Wait, so y tends to positive infinity as x approaches both positive and negative infinity. So, the function has a minimum somewhere in between.But the problem is talking about a local maximum. Hmm, that seems contradictory because if the function goes to positive infinity on both ends, it can only have a local minimum, not a maximum.Wait, maybe the problem is referring to a local maximum in the sense of a relative maximum, but given the function's behavior, it's impossible because it's always going to positive infinity on both sides. So, perhaps there's a misunderstanding here.Wait, perhaps I misapplied the second derivative test. Let me double-check.The first derivative is y' = e^x + a. Setting that equal to zero gives e^x = -a, so x = ln(-a). Then, the second derivative is y'' = e^x, which is always positive. So, that critical point is indeed a local minimum.Therefore, the function y = e^x + a x has only one critical point, which is a local minimum, not a maximum. So, how can the problem state that there's a local maximum? Maybe I misread the problem.Wait, let me read it again: "if the function y = e^x + a x has a local maximum point greater than zero, then the range of a is ______."Hmm, perhaps the problem is referring to a local maximum in the sense of a relative maximum, but given the function's behavior, it's impossible because it's always going to positive infinity on both ends. So, perhaps the problem is incorrectly stated, or maybe I'm missing something.Alternatively, maybe the problem is referring to a local maximum in the derivative function, but that doesn't make much sense. Or perhaps it's a translation issue, given that the original problem might not be in English.Wait, let me think differently. Maybe the function y = e^x + a x has a local maximum in terms of its graph, but given that the function tends to infinity on both ends, the only critical point is a minimum. So, perhaps the problem is incorrectly stated, or maybe I'm misunderstanding what is meant by "local maximum point greater than zero."Wait, perhaps "local maximum point greater than zero" refers to the y-value of the maximum being greater than zero, not the x-coordinate. That could make sense. Let me consider that possibility.So, if the function has a local maximum at some point x = c, and the y-value at that point is greater than zero, then we need to find the range of a such that y(c) > 0.But earlier, we saw that the function only has a local minimum, not a maximum. So, that seems contradictory again.Wait, maybe the function can have both a local maximum and a local minimum depending on the value of a. Let me check that.Wait, the first derivative is y' = e^x + a. Setting that equal to zero gives e^x = -a. So, if a is negative, we have a critical point at x = ln(-a). If a is positive, then e^x = -a has no real solution because e^x is always positive, so y' is always positive, meaning the function is increasing everywhere, so no critical points.Wait, so for a positive, the function is always increasing, so no local maxima or minima. For a negative, we have one critical point at x = ln(-a), which is a local minimum because the second derivative is positive.Therefore, the function y = e^x + a x can only have a local minimum when a is negative, and no local maxima at all. So, the problem statement might be incorrect, or perhaps I'm misinterpreting it.Alternatively, maybe the problem is referring to a local maximum in the sense of a relative maximum, but given the function's behavior, it's impossible because it's always going to positive infinity on both ends. So, perhaps the problem is incorrectly stated, or maybe I'm missing something.Wait, perhaps the problem is referring to a local maximum in the derivative function, but that doesn't make much sense. Or maybe it's a translation issue, given that the original problem might not be in English.Alternatively, maybe the function is y = e^x + a x, and we're supposed to find when it has a local maximum at x > 0, but given that the function only has a local minimum, perhaps the problem is asking for when the local minimum is greater than zero in terms of y-value.Wait, that could make sense. So, maybe the problem is asking for when the local minimum y-value is greater than zero. Let me consider that.So, if the local minimum is at x = ln(-a), and we want y(ln(-a)) > 0.Let's compute y at x = ln(-a):y(ln(-a)) = e^{ln(-a)} + a * ln(-a) = (-a) + a * ln(-a)So, y(ln(-a)) = -a + a ln(-a)We want this to be greater than zero:-a + a ln(-a) > 0Factor out a:a(-1 + ln(-a)) > 0But we know that a < 0, so let's write a = -b where b > 0.Then, the inequality becomes:(-b)(-1 + ln(b)) > 0Which simplifies to:b(-1 + ln(b)) > 0Since b > 0, the sign of the expression depends on (-1 + ln(b)).So, we have:-1 + ln(b) > 0Which implies:ln(b) > 1Therefore:b > e^1 = eSince b = -a, and a < 0, we have:-a > eWhich implies:a < -eSo, the range of a is (-∞, -e)Wait, that's different from my initial conclusion of a < -1. So, perhaps the problem is asking for when the local minimum y-value is greater than zero, which would require a < -e.But the problem statement says "has a local maximum point greater than zero," which is confusing because the function only has a local minimum. So, maybe the problem is misstated, or perhaps I'm misinterpreting it.Alternatively, maybe the problem is referring to a local maximum in terms of the derivative, but that doesn't make sense because the derivative is e^x + a, which is always increasing since e^x is increasing. So, the derivative can have at most one critical point, which is a minimum.Wait, perhaps the problem is referring to the function having a local maximum at some point, but given the function's behavior, it's impossible. So, maybe the answer is that no such a exists, but that seems unlikely.Alternatively, perhaps I made a mistake in my earlier reasoning. Let me go back.We have y = e^x + a xy' = e^x + aSet y' = 0: e^x = -aSo, x = ln(-a), which requires a < 0.Now, the second derivative y'' = e^x > 0, so it's a local minimum.So, the function has a local minimum at x = ln(-a). So, the problem is asking for when this local minimum is greater than zero in terms of x-coordinate, which is ln(-a) > 0, implying -a > 1, so a < -1.Alternatively, if the problem is asking for when the y-value at the local minimum is greater than zero, then we have:y(ln(-a)) = -a + a ln(-a) > 0Which, as I computed earlier, leads to a < -e.But the problem says "has a local maximum point greater than zero," which is confusing because the function only has a local minimum. So, perhaps the problem is incorrectly stated, or perhaps I'm misinterpreting it.Alternatively, maybe the problem is referring to the function having a local maximum at some point x > 0, but given that the function only has a local minimum, perhaps the problem is asking for when the local minimum is at x > 0, which would require a < -1.So, perhaps the answer is a < -1, which is (-∞, -1).But earlier, I thought it might be a < -e if considering the y-value. But the problem specifically mentions "a local maximum point greater than zero," which is unclear.Wait, perhaps "greater than zero" refers to the x-coordinate of the maximum point. But since the function only has a local minimum, maybe the problem is incorrectly stated, or perhaps it's a translation issue.Alternatively, maybe the problem is referring to the function having a local maximum in terms of its concavity, but that doesn't make sense either.Wait, perhaps the function y = e^x + a x can have a local maximum if a is such that the function changes concavity, but that's not the case here because y'' = e^x is always positive, so the function is always concave upward.Hmm, this is confusing. Let me try to summarize:- The function y = e^x + a x has a critical point at x = ln(-a) when a < 0.- This critical point is a local minimum because y'' = e^x > 0.- Therefore, the function cannot have a local maximum.So, the problem statement might be incorrect, or perhaps I'm misinterpreting it.Alternatively, maybe the problem is referring to the function having a local maximum at some point x > 0, but given that the function only has a local minimum, perhaps the problem is asking for when the local minimum is at x > 0, which would require a < -1.Alternatively, if the problem is asking for when the function has a local maximum at some x > 0, which is impossible because the function only has a local minimum, then the answer would be that no such a exists.But that seems unlikely because the problem is asking for a range of a.Alternatively, perhaps the problem is referring to the function having a local maximum in terms of its derivative, but that doesn't make sense because the derivative is e^x + a, which is always increasing.Wait, maybe the problem is referring to the function y = e^x + a x having a local maximum in terms of its graph, but given that the function tends to infinity on both ends, it's impossible.Alternatively, perhaps the problem is referring to a local maximum in terms of the parameter a, but that doesn't make sense either.Wait, perhaps the problem is misstated, and it should say "local minimum" instead of "local maximum." If that's the case, then the range of a would be a < -1, as we found earlier.Alternatively, if the problem is correctly stated, then perhaps the answer is that no such a exists because the function cannot have a local maximum.But given that the problem is asking for a range of a, I think it's more likely that there's a misstatement, and it should be a local minimum. So, assuming that, the range of a would be a < -1.Alternatively, if the problem is correctly stated, and it's referring to the y-value of the local maximum being greater than zero, but since there is no local maximum, perhaps the problem is incorrectly stated.In conclusion, given the function y = e^x + a x, it only has a local minimum at x = ln(-a) when a < 0. Therefore, if the problem is asking for when this local minimum is at x > 0, then a < -1. If it's asking for when the y-value at the local minimum is greater than zero, then a < -e. However, since the problem mentions a local maximum, which doesn't exist for this function, I think the problem might be misstated. But assuming it's referring to the local minimum, the range of a is (-∞, -1).Alternatively, if the problem is correctly stated and refers to a local maximum, then perhaps the answer is that no such a exists, but that seems unlikely given the problem's phrasing.So, to sum up, I think the problem might have a typo, and it should be a local minimum. In that case, the range of a is a < -1. Therefore, the range of a is (-∞, -1).

💡Alright, so I've got this problem about an arithmetic sequence. Let me try to wrap my head around it step by step. The problem says:"Given a non-zero arithmetic sequence {a_n}, it satisfies a₃ - a₇² + a₁₁ = 0. The sum of the first 13 terms S₁₃ is ______."Okay, so first, let's recall what an arithmetic sequence is. An arithmetic sequence is a sequence of numbers where the difference between consecutive terms is constant. This constant difference is called the common difference, usually denoted by 'd'. The general formula for the nth term of an arithmetic sequence is:a_n = a₁ + (n - 1)dwhere a₁ is the first term.Now, the problem mentions a₃, a₇, and a₁₁. Let me write down expressions for each of these using the general formula.a₃ = a₁ + 2d a₇ = a₁ + 6d a₁₁ = a₁ + 10dSo, substituting these into the given equation:a₃ - a₇² + a₁₁ = 0 => (a₁ + 2d) - (a₁ + 6d)² + (a₁ + 10d) = 0Hmm, that looks a bit complicated. Let me simplify this step by step.First, let's expand the squared term:(a₁ + 6d)² = a₁² + 12a₁d + 36d²So, substituting back into the equation:(a₁ + 2d) - (a₁² + 12a₁d + 36d²) + (a₁ + 10d) = 0Now, let's combine like terms:First, the a₁ terms: a₁ + a₁ = 2a₁ Then the d terms: 2d + 10d = 12d Then the squared terms: -a₁² -12a₁d -36d²So putting it all together:2a₁ + 12d - a₁² - 12a₁d - 36d² = 0Hmm, this is a quadratic equation in terms of a₁ and d. It looks a bit messy. Maybe there's a smarter way to approach this problem without getting bogged down in expanding everything.Wait a minute, I remember that in an arithmetic sequence, the terms equidistant from the ends have a special relationship. Specifically, the sum of terms equidistant from the ends is constant and equal to twice the middle term. For example, in a sequence with an odd number of terms, the middle term is the average of the sequence.In this case, a₃ and a₁₁ are equidistant from a₇. Let me check:a₃ is the 3rd term, a₇ is the 7th term, and a₁₁ is the 11th term. So, a₃ is 4 terms before a₇, and a₁₁ is 4 terms after a₇. So, they are symmetric around a₇.Therefore, the sum of a₃ and a₁₁ should be equal to 2a₇.Let me write that down:a₃ + a₁₁ = 2a₇So, substituting back into the original equation:a₃ - a₇² + a₁₁ = 0 => (a₃ + a₁₁) - a₇² = 0 => 2a₇ - a₇² = 0Oh, that's much simpler! So, we have:2a₇ - a₇² = 0Let's factor this equation:a₇(2 - a₇) = 0So, the solutions are:a₇ = 0 or a₇ = 2But the problem states that it's a non-zero arithmetic sequence. Hmm, does that mean all terms are non-zero, or just that the sequence isn't the zero sequence? Well, if a₇ were zero, then depending on the common difference, other terms could be non-zero. But the problem says "non-zero arithmetic sequence," which might imply that all terms are non-zero. So, a₇ can't be zero because that would make the 7th term zero, which is not allowed.Therefore, we discard a₇ = 0 and take the other solution:a₇ = 2Alright, so now we know that the 7th term is 2. Now, the question asks for the sum of the first 13 terms, S₁₃.I remember that the sum of the first n terms of an arithmetic sequence is given by:S_n = n/2 * (a₁ + a_n)Alternatively, it can also be written as:S_n = n * a_avgwhere a_avg is the average of the first and last terms.Since we're dealing with the first 13 terms, n = 13. So, S₁₃ = 13/2 * (a₁ + a₁₃)But wait, we don't know a₁ or a₁₃. Hmm, but maybe we can express a₁₃ in terms of a₇.Since a₇ is the 7th term, a₁₃ is the 13th term, which is 6 terms after a₇. So, a₁₃ = a₇ + 6dSimilarly, a₁ is the first term, which is 6 terms before a₇. So, a₁ = a₇ - 6dTherefore, a₁ + a₁₃ = (a₇ - 6d) + (a₇ + 6d) = 2a₇So, the sum S₁₃ becomes:S₁₃ = 13/2 * (2a₇) = 13 * a₇Since we found that a₇ = 2, substituting that in:S₁₃ = 13 * 2 = 26So, the sum of the first 13 terms is 26.Wait, let me verify that again to make sure I didn't make a mistake. So, we started with the given equation, used the property of arithmetic sequences that a₃ + a₁₁ = 2a₇, substituted into the equation to find a₇, then used the sum formula which simplified nicely because the average term is a₇, so the sum is just 13 times a₇. That makes sense because in an arithmetic sequence, the average of the first and last term is equal to the middle term when the number of terms is odd, which 13 is. So, yeah, that seems correct.I think that's solid. I don't see any mistakes in the logic or calculations. So, I'm confident that the sum of the first 13 terms is 26.**Final Answer**The sum of the first 13 terms is boxed{26}.

💡Okay, so I have this problem with two parts. Let me try to understand it step by step. First, there are two fixed points, M at (0,1) and N at (1,2). There's a moving point P such that the ratio of its distance from M to its distance from N is √2. I need to find the equation of the trajectory of point P and identify what shape this graph is. Then, there's a line y = kx - 1 that intersects this trajectory at points A and B. For part (1), I think I need to use the distance formula. Let me denote point P as (x, y). The distance from P to M is √[(x - 0)^2 + (y - 1)^2], which simplifies to √(x² + (y - 1)²). Similarly, the distance from P to N is √[(x - 1)^2 + (y - 2)^2]. Given that the ratio of these distances is √2, so I can write:√(x² + (y - 1)²) / √[(x - 1)² + (y - 2)²] = √2To get rid of the square roots, I'll square both sides:(x² + (y - 1)²) / [(x - 1)² + (y - 2)²] = 2Multiplying both sides by the denominator:x² + (y - 1)² = 2[(x - 1)² + (y - 2)²]Let me expand both sides:Left side: x² + y² - 2y + 1Right side: 2[(x² - 2x + 1) + (y² - 4y + 4)] = 2x² - 4x + 2 + 2y² - 8y + 8Simplify the right side: 2x² + 2y² - 4x - 8y + 10Now, set left side equal to right side:x² + y² - 2y + 1 = 2x² + 2y² - 4x - 8y + 10Bring all terms to one side:x² + y² - 2y + 1 - 2x² - 2y² + 4x + 8y - 10 = 0Combine like terms:(-x²) + (-y²) + 4x + 6y - 9 = 0Multiply both sides by -1 to make it look nicer:x² + y² - 4x - 6y + 9 = 0Hmm, this looks like the equation of a circle. Let me complete the squares to write it in standard form.For x terms: x² - 4x. Take half of -4, which is -2, square it to get 4.For y terms: y² - 6y. Take half of -6, which is -3, square it to get 9.So, add and subtract these squares:(x² - 4x + 4) + (y² - 6y + 9) = -9 + 4 + 9Simplify:(x - 2)² + (y - 3)² = 4So, the equation is a circle with center at (2, 3) and radius 2. That answers part (1).Moving on to part (2), I need to find the range of values for k such that the line y = kx - 1 intersects the circle at two points A and B. To find the range of k, I think I need to ensure that the distance from the center of the circle to the line is less than the radius. If the distance is equal to the radius, the line is tangent, and if it's greater, there's no intersection.The formula for the distance from a point (h, m) to the line ax + by + c = 0 is |ah + bm + c| / √(a² + b²). First, write the line in standard form: y = kx - 1 becomes kx - y - 1 = 0.The center of the circle is (2, 3). Plugging into the distance formula:|k*2 - 1*3 - 1| / √(k² + 1) < 2Simplify numerator:|2k - 3 - 1| = |2k - 4|So, |2k - 4| / √(k² + 1) < 2Multiply both sides by √(k² + 1):|2k - 4| < 2√(k² + 1)Square both sides to eliminate the absolute value and square root:(2k - 4)² < 4(k² + 1)Expand left side:4k² - 16k + 16 < 4k² + 4Subtract 4k² from both sides:-16k + 16 < 4Subtract 16:-16k < -12Divide by -16, remembering to flip the inequality:k > 12/16 = 3/4So, the range of k is k > 3/4. That's part (2).Now, part (3) asks whether there exists a value of k such that the dot product of vectors OA and OB equals 11. If it exists, find k; otherwise, explain why not.Vectors OA and OB are from the origin to points A and B. Their dot product is x1x2 + y1y2 = 11.Given that A and B lie on both the circle and the line y = kx - 1, I can find their coordinates in terms of k.Let me denote A as (x1, y1) and B as (x2, y2). Since they lie on the line, y1 = kx1 - 1 and y2 = kx2 - 1.So, the dot product OA · OB = x1x2 + y1y2 = x1x2 + (kx1 - 1)(kx2 - 1) = 11.Let me expand this:x1x2 + (k²x1x2 - kx1 - kx2 + 1) = 11Combine like terms:(1 + k²)x1x2 - k(x1 + x2) + 1 = 11So,(1 + k²)x1x2 - k(x1 + x2) - 10 = 0Now, I need expressions for x1 + x2 and x1x2. To find these, I can solve the system of equations consisting of the circle and the line.Substitute y = kx - 1 into the circle equation:(x - 2)² + (y - 3)² = 4Replace y with kx - 1:(x - 2)² + (kx - 1 - 3)² = 4Simplify:(x - 2)² + (kx - 4)² = 4Expand both terms:(x² - 4x + 4) + (k²x² - 8kx + 16) = 4Combine like terms:(1 + k²)x² + (-4 - 8k)x + (4 + 16) = 4Simplify:(1 + k²)x² - (4 + 8k)x + 20 = 4Subtract 4:(1 + k²)x² - (4 + 8k)x + 16 = 0So, quadratic in x: (1 + k²)x² - 4(1 + 2k)x + 16 = 0Let me write it as:(1 + k²)x² - 4(2k + 1)x + 16 = 0From this quadratic equation, the sum of roots x1 + x2 is [4(2k + 1)] / (1 + k²), and the product x1x2 is 16 / (1 + k²).So, x1 + x2 = [4(2k + 1)] / (1 + k²) and x1x2 = 16 / (1 + k²).Now, plug these into the equation we had earlier:(1 + k²)x1x2 - k(x1 + x2) - 10 = 0Substitute x1x2 and x1 + x2:(1 + k²)*(16 / (1 + k²)) - k*[4(2k + 1)/(1 + k²)] - 10 = 0Simplify term by term:First term: (1 + k²)*(16 / (1 + k²)) = 16Second term: -k*[4(2k + 1)/(1 + k²)] = -4k(2k + 1)/(1 + k²)Third term: -10So, equation becomes:16 - [4k(2k + 1)/(1 + k²)] - 10 = 0Simplify:6 - [4k(2k + 1)/(1 + k²)] = 0Bring the fraction to the other side:6 = [4k(2k + 1)/(1 + k²)]Multiply both sides by (1 + k²):6(1 + k²) = 4k(2k + 1)Expand both sides:6 + 6k² = 8k² + 4kBring all terms to one side:6 + 6k² - 8k² - 4k = 0Simplify:-2k² - 4k + 6 = 0Multiply both sides by -1:2k² + 4k - 6 = 0Divide by 2:k² + 2k - 3 = 0Factor:(k + 3)(k - 1) = 0So, k = -3 or k = 1But from part (2), we know that k must be greater than 3/4. So, k = -3 is less than 3/4, which is not allowed. Therefore, the only possible value is k = 1.Let me check if this works. If k = 1, then the line is y = x - 1. Let's find points A and B by solving the system:y = x - 1(x - 2)² + (y - 3)² = 4Substitute y:(x - 2)² + (x - 1 - 3)² = 4Simplify:(x - 2)² + (x - 4)² = 4Expand:(x² - 4x + 4) + (x² - 8x + 16) = 4Combine:2x² - 12x + 20 = 42x² - 12x + 16 = 0Divide by 2:x² - 6x + 8 = 0Factor:(x - 2)(x - 4) = 0So, x = 2 or x = 4Thus, points A and B are (2, 1) and (4, 3). Compute OA · OB:(2)(4) + (1)(3) = 8 + 3 = 11Yes, it works. So, k = 1 is the solution.**Final Answer**(1) The equation of the trajectory is boxed{(x - 2)^2 + (y - 3)^2 = 4}, which is a circle.(2) The range of values for ( k ) is boxed{k > dfrac{3}{4}}.(3) The value of ( k ) is boxed{1}.

💡Okay, so I have this problem here about a shape made up of three similar right-angled triangles. The smallest one has two sides of length 2. I need to find the area of the whole shape. Hmm, let me try to visualize this. Since all the triangles are similar, their sides are proportional. First, let me think about the smallest triangle. It's a right-angled triangle with two sides of length 2. So, that must mean it's an isosceles right-angled triangle because both legs are equal. In such a triangle, the hypotenuse can be found using the Pythagorean theorem. So, if each leg is 2, then the hypotenuse squared is 2 squared plus 2 squared, which is 4 + 4 = 8. Therefore, the hypotenuse is the square root of 8, which simplifies to 2 times the square root of 2. Okay, got that.Now, since all three triangles are similar, their sides are in proportion. Let me denote the smallest triangle as Triangle 1, the middle one as Triangle 2, and the largest one as Triangle 3. Since they are similar, the ratio of their corresponding sides is constant. Let me figure out what that ratio is.If Triangle 1 has legs of 2 and hypotenuse of 2√2, then Triangle 2 must have sides that are multiples of these. Let me assume that the ratio of the sides from Triangle 1 to Triangle 2 is some factor, say 'k'. Similarly, from Triangle 2 to Triangle 3, it would be another factor, maybe 'k' again if they are increasing by the same ratio each time. Wait, but the problem doesn't specify how the triangles are arranged, so I need to make sure I'm not assuming something incorrectly.Alternatively, maybe all three triangles are arranged in such a way that each subsequent triangle is scaled by a factor of √2. That would make sense because in similar right-angled triangles, especially isosceles ones, scaling by √2 often relates the sides. Let me test this idea.If Triangle 2 is scaled by √2 from Triangle 1, then its legs would be 2√2, and its hypotenuse would be 2√2 * √2 = 4. Similarly, Triangle 3 would be scaled by another √2 from Triangle 2, making its legs 4 and hypotenuse 4√2. Hmm, that seems plausible.Now, let me calculate the areas of each triangle. The area of a right-angled triangle is (base * height)/2. For Triangle 1, that's (2 * 2)/2 = 2. For Triangle 2, which has legs of 2√2, the area would be (2√2 * 2√2)/2. Let's compute that: 2√2 * 2√2 = 8, so 8/2 = 4. Okay, so the area of Triangle 2 is 4.For Triangle 3, with legs of 4, the area would be (4 * 4)/2 = 16/2 = 8. So, the areas are 2, 4, and 8. Adding them up gives 2 + 4 + 8 = 14. Hmm, 14 is one of the answer choices, option A.Wait, but let me double-check because sometimes when triangles are arranged together, their areas might overlap or something. But the problem says the shape is made up of three similar right-angled triangles, so I think it's just the sum of their areas. So, 2 + 4 + 8 should be correct, which is 14.Alternatively, maybe the scaling factor isn't √2 each time. Let me think again. If the triangles are similar, the ratio of their areas is the square of the ratio of their sides. So, if the sides are in a ratio of k, then the areas are in a ratio of k squared.Suppose the ratio from Triangle 1 to Triangle 2 is k, then the area ratio is k squared. Similarly, from Triangle 2 to Triangle 3, it's another k squared. So, the areas would be 2, 2k², and 2k⁴. But I don't know what k is yet.Wait, maybe I can find k by considering the arrangement of the triangles. If they are arranged such that the hypotenuse of one becomes the leg of the next, then the scaling factor would be √2. Because in an isosceles right triangle, the hypotenuse is √2 times the leg. So, if the hypotenuse of Triangle 1 is 2√2, that becomes the leg of Triangle 2. Therefore, Triangle 2 has legs of 2√2, and its hypotenuse would be 2√2 * √2 = 4. Then, Triangle 3 would have legs of 4, and hypotenuse of 4√2. So, that matches my earlier calculation.Therefore, the areas are 2, 4, and 8, adding up to 14. So, I think the answer is 14, which is option A.But wait, let me check the answer choices again. Option B is 12 + 12√2, which is about 24.97, which is much larger than 14. Option C is 28, which is also larger. Option D is 24 + 20√2, which is even bigger, and E is 56. So, 14 is the smallest option, but I want to make sure I didn't make a mistake.Wait, maybe I misinterpreted the arrangement of the triangles. If the triangles are arranged such that each subsequent triangle is built on the hypotenuse of the previous one, then the scaling factor might be different. Let me think about that.If Triangle 1 has legs of 2, hypotenuse 2√2. Then, Triangle 2 is built on that hypotenuse, so its legs would be 2√2 each, making its hypotenuse 2√2 * √2 = 4. Then, Triangle 3 is built on that hypotenuse, so its legs are 4 each, hypotenuse 4√2. So, that's the same as before. So, the areas are still 2, 4, 8, totaling 14.Alternatively, maybe the triangles are arranged differently, like one on top of the other, but I think the key is that they are similar and the scaling factor is consistent. So, I think 14 is correct.Wait, another thought: maybe the triangles are not all isosceles. The problem says they are similar right-angled triangles, but doesn't specify they are isosceles. So, maybe the smallest triangle has legs of 2 and 2, but the others have different leg lengths. Hmm, that complicates things.If the triangles are similar but not necessarily isosceles, then the ratio of sides could be different. Let me denote the sides of Triangle 1 as a = 2, b = 2, and hypotenuse c = 2√2. Then, Triangle 2 would have sides ka, kb, kc, and Triangle 3 would have sides k²a, k²b, k²c, assuming the scaling factor is k each time.But without knowing the arrangement, it's hard to determine k. However, if the triangles are arranged such that each hypotenuse becomes a leg of the next triangle, then the scaling factor k would be √2, as before. Because the hypotenuse of Triangle 1 is 2√2, which becomes the leg of Triangle 2, so k = √2.Therefore, Triangle 2 has legs 2√2 and 2√2, hypotenuse 4. Triangle 3 has legs 4 and 4, hypotenuse 4√2. So, areas are 2, 4, 8, total 14.Alternatively, if the triangles are arranged differently, maybe the scaling factor is different. But I think the most logical arrangement is each hypotenuse becoming a leg of the next triangle, leading to scaling by √2 each time.So, I think the total area is 14, which is option A.