💡Okay, so I have this geometry problem here: In a triangle, one side is three times smaller than the sum of the other two sides. I need to prove that the angle opposite to this side is the smallest angle in the triangle. Hmm, let me try to break this down step by step.First, let me make sure I understand the problem correctly. We have a triangle with three sides, let's call them a, b, and c. One of these sides is three times smaller than the sum of the other two. So, if I take one side, say a, then a is three times smaller than b + c. That means 3a = b + c. Is that right? Yeah, because if a is three times smaller than b + c, then b + c must be three times a. So, 3a = b + c.Now, I need to prove that the angle opposite to side a is the smallest angle in the triangle. In a triangle, the smallest angle is opposite the shortest side. So, if I can show that side a is the shortest side, then the angle opposite to it will automatically be the smallest angle. That seems like a good approach.Alright, so let's assume that a is the side that is three times smaller than the sum of the other two sides. So, 3a = b + c. I need to show that a is less than both b and c. If I can prove that a < b and a < c, then a is indeed the shortest side, and the angle opposite to it will be the smallest.Let me start by expressing b and c in terms of a. From 3a = b + c, I can write b = 3a - c and c = 3a - b. Hmm, not sure if that helps directly. Maybe I should use the triangle inequality theorem, which states that the sum of any two sides of a triangle must be greater than the third side.So, applying the triangle inequality to our triangle:1. a + b > c2. a + c > b3. b + c > aWe already know that b + c = 3a, so substituting into the third inequality:3a > aWhich simplifies to 2a > 0. Well, that's always true since side lengths are positive. So, that doesn't give me new information.Let me look at the first inequality: a + b > c. Since b + c = 3a, I can express c as 3a - b. Substituting into the inequality:a + b > 3a - bSimplify:a + b > 3a - bBring like terms together:a + b + b > 3aa + 2b > 3aSubtract a from both sides:2b > 2aDivide both sides by 2:b > aSo, that tells me that b is greater than a. Good, that's one side.Now, let's look at the second inequality: a + c > b. Similarly, since b + c = 3a, I can express b as 3a - c. Substituting into the inequality:a + c > 3a - cSimplify:a + c > 3a - cBring like terms together:a + c + c > 3aa + 2c > 3aSubtract a from both sides:2c > 2aDivide both sides by 2:c > aSo, that tells me that c is also greater than a. Therefore, both b and c are greater than a, which means a is the shortest side.Since a is the shortest side, the angle opposite to it must be the smallest angle in the triangle. That's because in any triangle, the smallest angle is always opposite the shortest side, and the largest angle is opposite the longest side.Wait, let me make sure I didn't skip any steps. I used the triangle inequality to show that both b and c are greater than a, which makes a the shortest side. Then, by the properties of triangles, the angle opposite the shortest side is the smallest angle. So, the angle opposite side a is indeed the smallest angle in the triangle.Is there another way to approach this? Maybe using the Law of Cosines? Let me try that for verification.The Law of Cosines states that for any triangle with sides a, b, c opposite angles A, B, C respectively:a² = b² + c² - 2bc cos ASimilarly,b² = a² + c² - 2ac cos Bc² = a² + b² - 2ab cos CSince I want to compare the angles, maybe I can express the cosines of the angles in terms of the sides.Given that 3a = b + c, I can express b and c in terms of a. Let's say b = 3a - c. Hmm, but that might complicate things. Alternatively, maybe I can express everything in terms of a.Wait, since 3a = b + c, I can write b = 3a - c. Then, substitute into the Law of Cosines for angle A:a² = (3a - c)² + c² - 2(3a - c)c cos ALet me expand that:a² = 9a² - 6ac + c² + c² - 2(3a - c)c cos ASimplify:a² = 9a² - 6ac + 2c² - 2(3a - c)c cos AHmm, this seems messy. Maybe there's a better way. Alternatively, since I already know that a is the shortest side, and thus angle A is the smallest angle, maybe I don't need to go into the Law of Cosines.But just to be thorough, let's see if we can compare angles using the Law of Sines. The Law of Sines states that:a / sin A = b / sin B = c / sin C = 2RWhere R is the radius of the circumscribed circle.Since a is the shortest side, sin A must be the smallest sine value among the three angles. Because the sine function is increasing in the interval [0, π], the smallest sine corresponds to the smallest angle.Therefore, angle A is the smallest angle.Okay, that seems consistent with what I found earlier.Wait, but I need to make sure that all angles are less than 90 degrees or not. Because if one angle is obtuse, the sine function still holds, but the relationship between sides and angles might be different.But in this case, since a is the shortest side, and the sum of the other two sides is 3a, which is greater than a, the triangle is valid, and all angles are less than 180 degrees. So, the Law of Sines applies.Therefore, since a is the shortest side, angle A is the smallest angle.I think I've covered different approaches to confirm that angle opposite side a is indeed the smallest angle. I started with the triangle inequality to show that a is the shortest side, then used the Law of Sines to confirm that the angle opposite the shortest side is the smallest angle.Is there anything else I need to consider? Maybe edge cases? For example, what if the triangle is degenerate? But since 3a = b + c, and in a degenerate triangle, the sum of two sides equals the third, which would make the triangle have zero area. But in our case, 3a = b + c, which is greater than a, so it's a valid triangle.Also, what if a is equal to b or c? But from the inequalities, we saw that b > a and c > a, so a cannot be equal to b or c. Therefore, a is strictly less than both b and c.So, I think I've covered all bases here. The key steps were using the triangle inequality to show that a is the shortest side and then applying the fundamental property of triangles that the smallest angle is opposite the shortest side.

💡First, I need to determine the number of months A kept his oxen in the pasture. Let's denote this number as ( x ).Next, I'll calculate the cost each person incurs based on the number of oxen and the duration they used the pasture. - **A's cost**: ( 10 ) oxen for ( x ) months.- **B's cost**: ( 12 ) oxen for ( 5 ) months.- **C's cost**: ( 15 ) oxen for ( 3 ) months.The total cost of the pasture is ( 175 ) Rs, so I can set up the equation:[10x + 12 times 5 + 15 times 3 = 175]Simplifying the equation:[10x + 60 + 45 = 175][10x + 105 = 175]Subtracting ( 105 ) from both sides:[10x = 70]Finally, solving for ( x ):[x = frac{70}{10} = 7]Therefore, A kept his oxen in the pasture for ( 7 ) months.

💡Okay, so I have this inequality: ( x^2 - 2x + 5 geq a^2 - 3a ), and it's supposed to hold for any real number ( x ). I need to find the range of the real number ( a ). Hmm, let me think about how to approach this.First, the inequality involves a quadratic expression in ( x ) on the left side. Since it has to hold for any real ( x ), that means the left side must always be greater than or equal to the right side, regardless of what ( x ) is. So, maybe I should analyze the quadratic expression ( x^2 - 2x + 5 ) to find its minimum value. If I can find the minimum value of the left side, then I can set that minimum to be greater than or equal to ( a^2 - 3a ). That should give me an inequality to solve for ( a ).Alright, let's rewrite the quadratic expression ( x^2 - 2x + 5 ) in a form that makes it easier to find the minimum. I remember that completing the square is a good method for this. Let me try that.Starting with ( x^2 - 2x + 5 ), I can complete the square for the ( x ) terms. The coefficient of ( x ) is -2, so half of that is -1, and squaring it gives 1. So, I can rewrite the expression as:( (x^2 - 2x + 1) + 4 )Which simplifies to:( (x - 1)^2 + 4 )Okay, so ( x^2 - 2x + 5 ) is equivalent to ( (x - 1)^2 + 4 ). Since ( (x - 1)^2 ) is a square term, it's always non-negative for any real ( x ). The smallest value it can take is 0, which happens when ( x = 1 ). Therefore, the minimum value of the entire expression ( (x - 1)^2 + 4 ) is 4.So, the left side of the inequality ( x^2 - 2x + 5 ) has a minimum value of 4. Since the inequality must hold for all real ( x ), the right side ( a^2 - 3a ) must be less than or equal to this minimum value. Otherwise, there would be some ( x ) for which the inequality doesn't hold.Therefore, we can set up the inequality:( a^2 - 3a leq 4 )Now, let's solve this quadratic inequality for ( a ). First, I'll bring all terms to one side to set the inequality to zero:( a^2 - 3a - 4 leq 0 )Now, I need to solve ( a^2 - 3a - 4 leq 0 ). To do this, I'll first find the roots of the quadratic equation ( a^2 - 3a - 4 = 0 ). I can use the quadratic formula:( a = frac{3 pm sqrt{9 + 16}}{2} )Calculating the discriminant:( sqrt{25} = 5 )So, the roots are:( a = frac{3 + 5}{2} = 4 ) and ( a = frac{3 - 5}{2} = -1 )These roots divide the real number line into three intervals:1. ( (-infty, -1) )2. ( (-1, 4) )3. ( (4, infty) )To determine where the quadratic expression ( a^2 - 3a - 4 ) is less than or equal to zero, I can test each interval.First, let's consider the interval ( (-infty, -1) ). Let's pick a test point, say ( a = -2 ):( (-2)^2 - 3(-2) - 4 = 4 + 6 - 4 = 6 ), which is positive. So, the expression is positive in this interval.Next, the interval ( (-1, 4) ). Let's choose ( a = 0 ):( 0^2 - 3(0) - 4 = -4 ), which is negative. So, the expression is negative in this interval.Finally, the interval ( (4, infty) ). Let's take ( a = 5 ):( 5^2 - 3(5) - 4 = 25 - 15 - 4 = 6 ), which is positive. So, the expression is positive in this interval.Since we are looking for where ( a^2 - 3a - 4 leq 0 ), the solution is the interval where the expression is negative or zero, which is ( [-1, 4] ).Therefore, the range of the real number ( a ) is ( [-1, 4] ).Let me double-check my steps to make sure I didn't make any mistakes. I completed the square correctly, found the minimum value, set up the inequality, solved the quadratic equation, tested the intervals, and concluded the range. Everything seems to add up. So, I think the correct answer is ( [-1, 4] ), which is option A.

💡Okay, so I have this function ( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 ) defined on the interval ([-2018, 0) cup (0, 2018]). I need to find its maximum value ( M ) and minimum value ( N ), and then compute ( M + N ). The answer choices are 3, 2, 1, and 0. Hmm, let's try to figure this out step by step.First, I should probably simplify the function to make it easier to work with. Let me expand the numerator:( (2^x + 1)^2 = 2^{2x} + 2 cdot 2^x + 1 ).So, substituting back into the function, we get:( f(x) = frac{2^{2x} + 2 cdot 2^x + 1}{2^x x} + 1 ).Now, let's break this fraction into separate terms:( f(x) = frac{2^{2x}}{2^x x} + frac{2 cdot 2^x}{2^x x} + frac{1}{2^x x} + 1 ).Simplifying each term:1. ( frac{2^{2x}}{2^x x} = frac{2^x}{x} ).2. ( frac{2 cdot 2^x}{2^x x} = frac{2}{x} ).3. ( frac{1}{2^x x} ) remains as it is.So, putting it all together:( f(x) = frac{2^x}{x} + frac{2}{x} + frac{1}{2^x x} + 1 ).Hmm, that looks a bit complicated, but maybe I can factor out ( frac{1}{x} ):( f(x) = frac{2^x + 2 + frac{1}{2^x}}{x} + 1 ).Wait, that might not be too helpful. Maybe I should consider the behavior of the function as ( x ) approaches 0 and as ( x ) becomes large (either positive or negative).Let's consider the limits:1. As ( x to 0^+ ) (approaching 0 from the positive side): - ( 2^x to 1 ) - So, the numerator ( (2^x + 1)^2 to (1 + 1)^2 = 4 ) - The denominator ( 2^x x to 1 cdot 0 = 0 ) - Therefore, ( frac{4}{0} ) tends to ( +infty ) - Adding 1, the function tends to ( +infty )2. As ( x to 0^- ) (approaching 0 from the negative side): - ( 2^x to 1 ) as well because ( 2^x ) is continuous. - The numerator is still 4. - The denominator ( 2^x x to 1 cdot 0^- = 0^- ) - So, ( frac{4}{0^-} ) tends to ( -infty ) - Adding 1, the function tends to ( -infty )So, near 0, the function goes to positive infinity on the right and negative infinity on the left. That suggests that the function has no maximum or minimum in the entire interval because it can get arbitrarily large in both directions. But wait, the interval is ([-2018, 0) cup (0, 2018]), so it's bounded. Hmm, maybe I need to look for critical points where the function could attain a maximum or minimum.To find critical points, I should take the derivative of ( f(x) ) and set it equal to zero. Let's compute ( f'(x) ).First, let me write ( f(x) ) again:( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 ).Let me denote ( g(x) = frac{(2^x + 1)^2}{2^x x} ), so ( f(x) = g(x) + 1 ). Therefore, ( f'(x) = g'(x) ).Compute ( g'(x) ):Let me write ( g(x) = frac{(2^x + 1)^2}{2^x x} ). Let me denote ( u = (2^x + 1)^2 ) and ( v = 2^x x ). Then, ( g(x) = frac{u}{v} ), so ( g'(x) = frac{u'v - uv'}{v^2} ).First, compute ( u = (2^x + 1)^2 ). Then, ( u' = 2(2^x + 1) cdot ln 2 cdot 2^x ).Next, compute ( v = 2^x x ). Then, ( v' = 2^x ln 2 cdot x + 2^x cdot 1 = 2^x (x ln 2 + 1) ).Now, plug into ( g'(x) ):( g'(x) = frac{[2(2^x + 1) cdot ln 2 cdot 2^x] cdot [2^x x] - [(2^x + 1)^2] cdot [2^x (x ln 2 + 1)]}{(2^x x)^2} ).This looks quite messy. Maybe I can factor out some terms to simplify.Let me factor out ( 2^x ) from numerator and denominator:Numerator:( 2(2^x + 1) cdot ln 2 cdot 2^x cdot 2^x x - (2^x + 1)^2 cdot 2^x (x ln 2 + 1) )Factor out ( 2^x (2^x + 1) ):( 2^x (2^x + 1) [2 ln 2 cdot 2^x x - (2^x + 1)(x ln 2 + 1)] )Denominator:( (2^x x)^2 = 2^{2x} x^2 )So, ( g'(x) = frac{2^x (2^x + 1) [2 ln 2 cdot 2^x x - (2^x + 1)(x ln 2 + 1)]}{2^{2x} x^2} )Simplify:( g'(x) = frac{(2^x + 1) [2 ln 2 cdot 2^x x - (2^x + 1)(x ln 2 + 1)]}{2^x x^2} )This is still complicated, but maybe I can set the numerator equal to zero to find critical points.So, set the numerator equal to zero:( (2^x + 1) [2 ln 2 cdot 2^x x - (2^x + 1)(x ln 2 + 1)] = 0 )Since ( 2^x + 1 ) is always positive (as ( 2^x > 0 ) for all real x), the term ( (2^x + 1) ) can never be zero. Therefore, the critical points occur when:( 2 ln 2 cdot 2^x x - (2^x + 1)(x ln 2 + 1) = 0 )Let me denote ( y = 2^x ). Then, ( y > 0 ) for all real x.Substituting ( y ) into the equation:( 2 ln 2 cdot y x - (y + 1)(x ln 2 + 1) = 0 )Let me expand the second term:( (y + 1)(x ln 2 + 1) = y x ln 2 + y + x ln 2 + 1 )So, substituting back:( 2 ln 2 cdot y x - (y x ln 2 + y + x ln 2 + 1) = 0 )Simplify:( 2 ln 2 cdot y x - y x ln 2 - y - x ln 2 - 1 = 0 )Combine like terms:( (2 ln 2 cdot y x - y x ln 2) - y - x ln 2 - 1 = 0 )Which simplifies to:( ln 2 cdot y x - y - x ln 2 - 1 = 0 )Factor terms:Let me factor ( y ) from the first two terms and ( x ln 2 ) from the next two:( y (ln 2 cdot x - 1) - (x ln 2 + 1) = 0 )Hmm, interesting. Let me write it as:( y (ln 2 cdot x - 1) = x ln 2 + 1 )But ( y = 2^x ), so:( 2^x (ln 2 cdot x - 1) = x ln 2 + 1 )This equation seems transcendental, meaning it might not have a closed-form solution. Maybe I can analyze it graphically or look for specific solutions.Let me test x = 1:Left side: ( 2^1 (ln 2 cdot 1 - 1) = 2 (ln 2 - 1) approx 2 (0.6931 - 1) = 2 (-0.3069) = -0.6138 )Right side: ( 1 cdot ln 2 + 1 approx 0.6931 + 1 = 1.6931 )Not equal.x = 2:Left: ( 4 (ln 2 cdot 2 - 1) = 4 (1.3862 - 1) = 4 (0.3862) ≈ 1.5448 )Right: ( 2 ln 2 + 1 ≈ 1.3862 + 1 = 2.3862 )Not equal.x = 0.5:Left: ( sqrt{2} (ln 2 cdot 0.5 - 1) ≈ 1.4142 (0.3466 - 1) ≈ 1.4142 (-0.6534) ≈ -0.925 )Right: ( 0.5 ln 2 + 1 ≈ 0.3466 + 1 = 1.3466 )Not equal.x = -1:Left: ( 2^{-1} (ln 2 cdot (-1) - 1) = 0.5 (-0.6931 - 1) = 0.5 (-1.6931) ≈ -0.8466 )Right: ( (-1) ln 2 + 1 ≈ -0.6931 + 1 = 0.3069 )Not equal.Hmm, seems like it's not easy to find a solution by plugging in. Maybe there's a symmetry or substitution I can use.Wait, let me consider the substitution ( t = x ln 2 ). Then, ( 2^x = e^{x ln 2} = e^t ). Let me rewrite the equation:( e^t (t - 1) = t + 1 )So,( e^t (t - 1) - t - 1 = 0 )Let me denote ( h(t) = e^t (t - 1) - t - 1 ). I need to find t such that ( h(t) = 0 ).Compute h(0):( h(0) = e^0 (-1) - 0 - 1 = -1 - 1 = -2 )h(1):( h(1) = e^1 (0) - 1 - 1 = 0 - 2 = -2 )h(2):( h(2) = e^2 (1) - 2 - 1 ≈ 7.389 - 3 ≈ 4.389 )So, h(2) is positive. Since h(1) = -2 and h(2) ≈ 4.389, by Intermediate Value Theorem, there is a root between t=1 and t=2.Similarly, h(-1):( h(-1) = e^{-1} (-2) - (-1) - 1 ≈ 0.3679 (-2) + 1 - 1 ≈ -0.7358 )h(-2):( h(-2) = e^{-2} (-3) - (-2) - 1 ≈ 0.1353 (-3) + 2 - 1 ≈ -0.4059 + 1 ≈ 0.5941 )So, h(-2) is positive. Since h(-1) ≈ -0.7358 and h(-2) ≈ 0.5941, there is a root between t=-2 and t=-1.Therefore, there are two critical points: one for t between -2 and -1, and another for t between 1 and 2. Translating back to x:Since ( t = x ln 2 ), so:1. For t between -2 and -1: x between ( -2 / ln 2 ) and ( -1 / ln 2 ). Numerically, ( ln 2 ≈ 0.6931 ), so ( -2 / 0.6931 ≈ -2.885 ) and ( -1 / 0.6931 ≈ -1.4427 ). So, x is between approximately -2.885 and -1.4427.2. For t between 1 and 2: x between ( 1 / 0.6931 ≈ 1.4427 ) and ( 2 / 0.6931 ≈ 2.885 ).So, we have critical points in the intervals (-2.885, -1.4427) and (1.4427, 2.885). Since our interval is [-2018, 0) ∪ (0, 2018], these critical points are within the domain.Now, to find whether these critical points correspond to maxima or minima, I would need to analyze the second derivative or use test points around these intervals. However, this seems quite involved. Maybe there's a smarter way.Wait, looking back at the function ( f(x) ), perhaps it's symmetric in some way. Let me check:If I substitute ( x ) with ( -x ), what happens?Compute ( f(-x) ):( f(-x) = frac{(2^{-x} + 1)^2}{2^{-x} (-x)} + 1 )Simplify:( f(-x) = frac{(2^{-x} + 1)^2}{2^{-x} (-x)} + 1 = -frac{(2^{-x} + 1)^2}{2^{-x} x} + 1 )Multiply numerator and denominator by ( 2^x ):( f(-x) = -frac{(1 + 2^x)^2}{x} + 1 )Compare this to ( f(x) ):( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 )Hmm, not exactly symmetric, but there's a relationship. Let me see:Let me denote ( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 ).Then, ( f(-x) = -frac{(2^x + 1)^2}{x} + 1 ).Wait, so ( f(-x) = -2^x f(x) + 1 )?Wait, let me compute:( f(-x) = -frac{(2^{-x} + 1)^2}{2^{-x} x} + 1 = -frac{(1 + 2^x)^2}{x} + 1 ).But ( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 ).So, ( f(-x) = -frac{(2^x + 1)^2}{x} + 1 ).Let me express ( f(x) ) as:( f(x) = frac{(2^x + 1)^2}{2^x x} + 1 = frac{(2^x + 1)^2}{2^x x} + 1 ).So, if I denote ( A = frac{(2^x + 1)^2}{2^x x} ), then ( f(x) = A + 1 ) and ( f(-x) = -A + 1 ).Therefore, ( f(x) + f(-x) = (A + 1) + (-A + 1) = 2 ).Wow, that's interesting! So, for any x ≠ 0, ( f(x) + f(-x) = 2 ). That means the function is symmetric around the point (0,1). So, if ( f(a) = M ), then ( f(-a) = 2 - M ). Similarly, if ( f(b) = N ), then ( f(-b) = 2 - N ).But wait, since the function is defined on both sides of 0, and it's symmetric around (0,1), the maximum and minimum values should satisfy ( M + N = 2 ). Because if M is the maximum at some point x, then the corresponding point -x would have value 2 - M, which would be the minimum if M is the maximum.Alternatively, if M is the maximum, then the minimum N would be 2 - M, so M + N = M + (2 - M) = 2.Therefore, regardless of the specific values of M and N, their sum is 2.So, the answer should be 2, which is option B.**Final Answer**boxed{B}

💡Alright, so I have this math problem here: I need to find the range of values of ( x ) that make the expression ( frac{sqrt{1 - x}}{2 + x} ) meaningful. Hmm, okay, let's break this down step by step.First, I know that for an expression to be meaningful, especially one that involves a square root and a denominator, there are certain conditions that need to be satisfied. I remember that the expression inside a square root has to be non-negative because you can't take the square root of a negative number in the set of real numbers. Also, the denominator can't be zero because division by zero is undefined. So, I need to make sure both of these conditions are met.Let me start with the square root part: ( sqrt{1 - x} ). The expression inside the square root, which is ( 1 - x ), must be greater than or equal to zero. So, I can write that as:[ 1 - x geq 0 ]Now, solving for ( x ), I'll subtract 1 from both sides:[ -x geq -1 ]But wait, when I multiply or divide both sides of an inequality by a negative number, I have to remember to flip the inequality sign. So, if I multiply both sides by -1, the inequality becomes:[ x leq 1 ]Okay, so that gives me one condition: ( x ) has to be less than or equal to 1.Next, I need to look at the denominator: ( 2 + x ). The denominator can't be zero because division by zero is undefined. So, I set up the equation:[ 2 + x neq 0 ]Solving for ( x ), I subtract 2 from both sides:[ x neq -2 ]So, that's the second condition: ( x ) can't be equal to -2.Now, I need to combine these two conditions to find the range of ( x ) that satisfies both. The first condition says ( x leq 1 ), and the second condition says ( x neq -2 ). So, putting them together, ( x ) can be any real number less than or equal to 1, except for -2.Wait, let me double-check to make sure I didn't miss anything. The square root condition is straightforward, and the denominator condition is also clear. Is there any overlap or something I need to be careful about?Well, if ( x leq 1 ), then ( x ) can be any number up to 1, but it can't be -2. So, I think that's all there is to it. I don't see any other restrictions here. The numerator is a square root, which only imposes the non-negativity condition, and the denominator just can't be zero.So, to recap, the expression ( frac{sqrt{1 - x}}{2 + x} ) is meaningful when:1. ( 1 - x geq 0 ) which simplifies to ( x leq 1 ).2. ( 2 + x neq 0 ) which simplifies to ( x neq -2 ).Therefore, combining these two conditions, the range of ( x ) is all real numbers less than or equal to 1, excluding -2.I think that's it. I don't see any other constraints or potential mistakes in my reasoning. It seems pretty straightforward once you break it down into the two main conditions: the square root and the denominator.**Final Answer**The range of values of ( x ) is boxed{x leq 1 text{ and } x neq -2}.

💡Alright, so I have this function f(x) = x e^{kx}, where k is not zero. I need to do three things with this function: find the tangent line at (0, f(0)), determine where it's increasing or decreasing, and figure out the range of k that makes it strictly increasing on (-1, 1). Let's take it step by step.Starting with part (I): finding the equation of the tangent line at (0, f(0)). I remember that the equation of a tangent line at a point (a, f(a)) is given by y = f(a) + f’(a)(x - a). So, in this case, a is 0, so I need to find f(0) and f’(0).Calculating f(0): f(0) = 0 * e^{k*0} = 0 * e^0 = 0 * 1 = 0. So, f(0) is 0. That means the point is (0, 0).Now, I need to find the derivative f’(x) to get the slope at x = 0. The function is f(x) = x e^{kx}, so I'll use the product rule for differentiation. The product rule says that if you have two functions multiplied together, their derivative is the derivative of the first times the second plus the first times the derivative of the second.Let’s let u = x and v = e^{kx}. Then, f(x) = u * v, so f’(x) = u’ * v + u * v’.First, find u’ and v’:- u = x, so u’ = 1.- v = e^{kx}, so v’ = k e^{kx} (since the derivative of e^{kx} is k e^{kx}).Putting it all together:f’(x) = u’ * v + u * v’ = 1 * e^{kx} + x * k e^{kx} = e^{kx} + kx e^{kx}.Simplify that: f’(x) = e^{kx}(1 + kx).Now, evaluate f’(0):f’(0) = e^{k*0}(1 + k*0) = e^0(1 + 0) = 1 * 1 = 1.So, the slope of the tangent line at x = 0 is 1. Therefore, the equation of the tangent line is y = f(0) + f’(0)(x - 0) = 0 + 1*(x - 0) = x. So, the tangent line is y = x.Alright, part (I) seems done.Moving on to part (II): determining the intervals of monotonicity for f(x). Monotonicity refers to where the function is increasing or decreasing. To find this, we look at the sign of the derivative f’(x).We already found f’(x) = e^{kx}(1 + kx). Since e^{kx} is always positive for any real x and k, the sign of f’(x) depends solely on the term (1 + kx).So, f’(x) > 0 when (1 + kx) > 0, and f’(x) < 0 when (1 + kx) < 0.Let’s solve for x in both cases.Case 1: (1 + kx) > 01 + kx > 0kx > -1x > -1/k (if k > 0)x < -1/k (if k < 0)Case 2: (1 + kx) < 01 + kx < 0kx < -1x < -1/k (if k > 0)x > -1/k (if k < 0)So, depending on the sign of k, the intervals where f’(x) is positive or negative change.Let’s break it down:If k > 0:- f’(x) > 0 when x > -1/k- f’(x) < 0 when x < -1/kIf k < 0:- f’(x) > 0 when x < -1/k- f’(x) < 0 when x > -1/kTherefore, the function f(x) is increasing on the interval where f’(x) > 0 and decreasing where f’(x) < 0.So, summarizing:- For k > 0: - Increasing on (-1/k, ∞) - Decreasing on (-∞, -1/k)- For k < 0: - Increasing on (-∞, -1/k) - Decreasing on (-1/k, ∞)That should cover part (II).Now, part (III): If f(x) is strictly increasing on (-1, 1), find the range of feasible values for k.From part (II), we know that f(x) is increasing when f’(x) > 0, which depends on the sign of (1 + kx). So, for f(x) to be strictly increasing on (-1, 1), we need f’(x) > 0 for all x in (-1, 1).Given f’(x) = e^{kx}(1 + kx), and e^{kx} is always positive, we need 1 + kx > 0 for all x in (-1, 1).Therefore, 1 + kx > 0 for all x in (-1, 1).We need to ensure that 1 + kx > 0 for x = -1 and x = 1, as these are the endpoints of the interval. If it's positive at both endpoints, and since it's a linear function in x (1 + kx is linear), it will be positive throughout the interval.So, let's check the conditions:1. At x = -1: 1 + k*(-1) > 0 => 1 - k > 0 => -k > -1 => k < 1.2. At x = 1: 1 + k*(1) > 0 => 1 + k > 0 => k > -1.So, combining these two inequalities:-1 < k < 1.But wait, we have to consider the sign of k because in part (II), the behavior changes depending on whether k is positive or negative.Let me think again. If k is positive, then f(x) is increasing for x > -1/k. So, to ensure that on (-1, 1), f(x) is increasing, we need -1/k <= -1, which would mean k <= 1. But since k is positive, it's 0 < k <= 1.Similarly, if k is negative, f(x) is increasing for x < -1/k. So, to ensure that on (-1, 1), f(x) is increasing, we need -1/k >= 1, which implies k >= -1. Since k is negative, it's -1 <= k < 0.Combining both cases, k must satisfy -1 <= k < 0 or 0 < k <= 1. But since k ≠ 0, we exclude 0.Therefore, the range of feasible values for k is [-1, 0) ∪ (0, 1].Wait, but earlier I had -1 < k < 1, but considering the behavior based on the sign of k, it's actually -1 <= k < 0 and 0 < k <= 1. So, including -1 and 1 but excluding 0.Yes, that makes sense because at k = -1 and k = 1, the critical point x = -1/k is at x = 1 and x = -1, respectively, which are the endpoints of the interval (-1, 1). So, at k = -1, the function is increasing on (-∞, 1), which includes (-1, 1). Similarly, at k = 1, the function is increasing on (-1, ∞), which includes (-1, 1).Therefore, the range is k ∈ [-1, 0) ∪ (0, 1].I think that's it. Let me just double-check.If k = -1, f’(x) = e^{-x}(1 - x). At x = -1, f’(-1) = e^{1}(1 - (-1)) = e*(2) > 0. At x = 1, f’(1) = e^{-1}(1 - 1) = 0. Wait, at x = 1, f’(1) = 0. So, is the function still strictly increasing on (-1, 1)?Strictly increasing means f’(x) > 0 for all x in (-1, 1). At x = 1, f’(1) = 0, which is not greater than 0. So, does that mean k = -1 is not acceptable?Hmm, that's a good point. Similarly, for k = 1, f’(x) = e^{x}(1 + x). At x = -1, f’(-1) = e^{-1}(1 - 1) = 0. So, again, at x = -1, f’(-1) = 0, which is not greater than 0.Therefore, strictly increasing requires f’(x) > 0 for all x in (-1, 1), not just non-negative. So, we need to exclude the cases where f’(x) = 0 at the endpoints.Therefore, k must satisfy -1 < k < 1, excluding k = -1 and k = 1.But wait, earlier reasoning suggested that at k = -1, the function is increasing on (-∞, 1), but at x = 1, the derivative is zero. So, on the open interval (-1, 1), does x = 1 count? Since it's an open interval, x = 1 is not included. Similarly, for k = 1, the derivative at x = -1 is zero, but x = -1 is not included in the open interval (-1, 1).Wait, the interval is (-1, 1), which is open, so it doesn't include the endpoints. Therefore, even though at k = -1, f’(1) = 0, since x = 1 is not in the interval, it's okay. Similarly, at k = 1, f’(-1) = 0, but x = -1 is not in the interval.Therefore, k = -1 and k = 1 are acceptable because the points where f’(x) = 0 are at the endpoints, which are excluded from the interval (-1, 1). So, on the open interval (-1, 1), f’(x) > 0 for k = -1 and k = 1.Wait, let me verify for k = -1:f’(x) = e^{-x}(1 - x). For x in (-1, 1), 1 - x is always positive because x < 1, so 1 - x > 0. Also, e^{-x} is always positive. Therefore, f’(x) > 0 for all x in (-1, 1). So, k = -1 is acceptable.Similarly, for k = 1:f’(x) = e^{x}(1 + x). For x in (-1, 1), 1 + x > 0 because x > -1, so 1 + x > 0. e^{x} is always positive. Therefore, f’(x) > 0 for all x in (-1, 1). So, k = 1 is acceptable.Therefore, including k = -1 and k = 1 is correct because the points where f’(x) = 0 are at the endpoints, which are not included in the interval (-1, 1). Hence, the range is k ∈ [-1, 0) ∪ (0, 1].So, to summarize:(I) The tangent line at (0, f(0)) is y = x.(II) The function is increasing on (-1/k, ∞) for k > 0 and on (-∞, -1/k) for k < 0. It is decreasing on (-∞, -1/k) for k > 0 and on (-1/k, ∞) for k < 0.(III) The range of k for which f(x) is strictly increasing on (-1, 1) is k ∈ [-1, 0) ∪ (0, 1].I think that covers all parts. I should probably write this up neatly.