💡Okay, so I have this problem here where I need to figure out which function is both even and increasing on the interval from negative infinity to zero. The options are A, B, C, and D. Let me take this step by step because I want to make sure I understand each part correctly.First, I remember that an even function satisfies the condition f(-x) = f(x) for all x in its domain. So, for each function, I need to check if replacing x with -x gives me the same function back. If it does, then it's even. If not, it's either odd or neither.Next, I need to check if the function is increasing on the interval (-∞, 0). To determine if a function is increasing, I can look at its derivative. If the derivative is positive on that interval, the function is increasing there. If the derivative is negative, it's decreasing.Alright, let's go through each option one by one.**Option A: f(x) = 2^x - 2^{-x}**First, check if it's even. Let's compute f(-x):f(-x) = 2^{-x} - 2^{x} = -(2^x - 2^{-x}) = -f(x)So, f(-x) = -f(x), which means this function is odd, not even. Therefore, Option A is out.**Option B: f(x) = x^2 - 3**Check if it's even:f(-x) = (-x)^2 - 3 = x^2 - 3 = f(x)So, yes, it's even. Now, check if it's increasing on (-∞, 0). Let's find the derivative:f'(x) = 2xNow, for x < 0, 2x is negative. So, the derivative is negative on (-∞, 0), which means the function is decreasing there, not increasing. So, Option B doesn't satisfy both conditions.**Option C: f(x) = -2 ln |x|**First, check if it's even:f(-x) = -2 ln |-x| = -2 ln |x| = f(x)So, it's even. Now, check if it's increasing on (-∞, 0). Let's find the derivative:f'(x) = derivative of -2 ln |x| with respect to x.Since x is negative in (-∞, 0), |x| = -x, so ln |x| = ln(-x). Therefore,f'(x) = -2 * (1/(-x)) * (-1) = -2 * (1/x)Wait, let me double-check that. The derivative of ln |x| is 1/x, regardless of the sign of x. So, f'(x) = -2 * (1/x).But x is negative in (-∞, 0), so 1/x is negative. Therefore, f'(x) = -2 * (negative) = positive.So, the derivative is positive on (-∞, 0), which means the function is increasing there. Therefore, Option C is both even and increasing on (-∞, 0).**Option D: f(x) = x cos(3x)**Check if it's even:f(-x) = (-x) cos(-3x) = -x cos(3x) = -f(x)So, f(-x) = -f(x), which means it's odd, not even. Therefore, Option D is out.So, after checking all options, only Option C satisfies both conditions: it's even and increasing on (-∞, 0).**Final Answer**boxed{C}

💡Alright, so I have this function ( f(x) = frac{1}{2}x^2 - (2a + 2)x + (2a + 1)ln x ). I need to do two things: first, discuss its monotonicity, and second, find the range of a positive real number ( lambda ) such that for any ( a ) in the interval ( left[frac{1}{2}, 2right] ) and any ( x_1, x_2 ) in ( [1, 2] ) with ( x_1 neq x_2 ), the inequality ( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| ) holds.Starting with part (1), discussing the monotonicity of ( f(x) ). Monotonicity is about where the function is increasing or decreasing, which is determined by the first derivative. So, I need to compute ( f'(x) ).Calculating the derivative:( f'(x) = frac{d}{dx}left( frac{1}{2}x^2 - (2a + 2)x + (2a + 1)ln x right) )Breaking it down term by term:- The derivative of ( frac{1}{2}x^2 ) is ( x ).- The derivative of ( -(2a + 2)x ) is ( -(2a + 2) ).- The derivative of ( (2a + 1)ln x ) is ( frac{2a + 1}{x} ).Putting it all together:( f'(x) = x - (2a + 2) + frac{2a + 1}{x} )Hmm, that simplifies to:( f'(x) = x - 2a - 2 + frac{2a + 1}{x} )I can combine the terms by getting a common denominator:( f'(x) = frac{x^2 - (2a + 2)x + (2a + 1)}{x} )Let me factor the numerator:Looking at ( x^2 - (2a + 2)x + (2a + 1) ), can I factor this quadratic?Let me try to factor it:Looking for two numbers that multiply to ( 2a + 1 ) and add up to ( -(2a + 2) ).Wait, that might not be straightforward. Alternatively, I can use the quadratic formula to find the roots.Quadratic equation: ( x^2 - (2a + 2)x + (2a + 1) = 0 )Using the quadratic formula:( x = frac{(2a + 2) pm sqrt{(2a + 2)^2 - 4 cdot 1 cdot (2a + 1)}}{2} )Compute discriminant ( D ):( D = (2a + 2)^2 - 4(2a + 1) )Expanding:( D = 4a^2 + 8a + 4 - 8a - 4 = 4a^2 )So,( x = frac{2a + 2 pm sqrt{4a^2}}{2} = frac{2a + 2 pm 2a}{2} )Simplify:First root:( x = frac{2a + 2 + 2a}{2} = frac{4a + 2}{2} = 2a + 1 )Second root:( x = frac{2a + 2 - 2a}{2} = frac{2}{2} = 1 )So, the numerator factors as ( (x - 1)(x - (2a + 1)) ). Therefore,( f'(x) = frac{(x - 1)(x - (2a + 1))}{x} )Since ( x > 0 ) (because of the ( ln x ) term in the original function), the sign of ( f'(x) ) depends on the numerator ( (x - 1)(x - (2a + 1)) ).Now, to discuss the monotonicity, I need to analyze where ( f'(x) ) is positive or negative. The critical points are at ( x = 1 ) and ( x = 2a + 1 ).The behavior of ( f'(x) ) depends on the relative positions of 1 and ( 2a + 1 ). So, let's consider different cases based on the value of ( a ).Case 1: ( 2a + 1 < 1 )This implies ( 2a < 0 ) or ( a < 0 ).In this case, the critical points are ordered as ( 2a + 1 < 1 ). So, the intervals to consider are ( (0, 2a + 1) ), ( (2a + 1, 1) ), and ( (1, infty) ).Testing the sign of ( f'(x) ) in each interval:- For ( x in (0, 2a + 1) ), both ( (x - 1) ) and ( (x - (2a + 1)) ) are negative, so their product is positive. Thus, ( f'(x) > 0 ).- For ( x in (2a + 1, 1) ), ( (x - 1) ) is negative and ( (x - (2a + 1)) ) is positive, so their product is negative. Thus, ( f'(x) < 0 ).- For ( x in (1, infty) ), both ( (x - 1) ) and ( (x - (2a + 1)) ) are positive, so their product is positive. Thus, ( f'(x) > 0 ).Case 2: ( 2a + 1 = 1 )This implies ( a = 0 ).Here, both critical points coincide at ( x = 1 ). The derivative becomes:( f'(x) = frac{(x - 1)^2}{x} )Since ( (x - 1)^2 ) is always non-negative and ( x > 0 ), ( f'(x) geq 0 ) for all ( x > 0 ). Thus, the function is non-decreasing on ( (0, infty) ).Case 3: ( 2a + 1 > 1 )This implies ( a > 0 ).The critical points are ordered as ( 1 < 2a + 1 ). So, the intervals are ( (0, 1) ), ( (1, 2a + 1) ), and ( (2a + 1, infty) ).Testing the sign of ( f'(x) ):- For ( x in (0, 1) ), ( (x - 1) ) is negative and ( (x - (2a + 1)) ) is negative, so their product is positive. Thus, ( f'(x) > 0 ).- For ( x in (1, 2a + 1) ), ( (x - 1) ) is positive and ( (x - (2a + 1)) ) is negative, so their product is negative. Thus, ( f'(x) < 0 ).- For ( x in (2a + 1, infty) ), both ( (x - 1) ) and ( (x - (2a + 1)) ) are positive, so their product is positive. Thus, ( f'(x) > 0 ).So, summarizing:- If ( a < 0 ), ( f(x) ) is increasing on ( (0, 2a + 1) ), decreasing on ( (2a + 1, 1) ), and increasing again on ( (1, infty) ).- If ( a = 0 ), ( f(x) ) is non-decreasing on ( (0, infty) ).- If ( a > 0 ), ( f(x) ) is increasing on ( (0, 1) ), decreasing on ( (1, 2a + 1) ), and increasing again on ( (2a + 1, infty) ).That's part (1). Now, moving on to part (2). The problem states:"For any ( a in left[frac{1}{2}, 2right] ), ( x_1, x_2 in [1, 2] ) (( x_1 neq x_2 )), there always exists ( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| ), find the range of the positive real number ( lambda )."So, we need to find the smallest ( lambda ) such that this inequality holds for all ( a ) in ( [frac{1}{2}, 2] ) and all ( x_1, x_2 ) in ( [1, 2] ).First, let's interpret this inequality. It resembles the definition of Lipschitz continuity, where the function ( f ) is Lipschitz continuous with respect to the function ( frac{1}{x} ). The Lipschitz constant here is ( lambda ).Alternatively, we can think of it as:( left| frac{f(x_1) - f(x_2)}{frac{1}{x_1} - frac{1}{x_2}} right| < lambda )Which suggests that the maximum value of the left-hand side over all ( x_1, x_2 ) in ( [1, 2] ) and ( a ) in ( [frac{1}{2}, 2] ) should be less than ( lambda ). Therefore, ( lambda ) must be greater than or equal to the supremum of the left-hand side.To find this supremum, we can consider the Mean Value Theorem. For some ( c ) between ( x_1 ) and ( x_2 ), we have:( |f(x_1) - f(x_2)| = |f'(c)| |x_1 - x_2| )Similarly, ( left|frac{1}{x_1} - frac{1}{x_2}right| = left| -frac{1}{x_1 x_2} (x_1 - x_2) right| = frac{|x_1 - x_2|}{x_1 x_2} )Therefore, the inequality becomes:( |f'(c)| |x_1 - x_2| < lambda frac{|x_1 - x_2|}{x_1 x_2} )We can cancel ( |x_1 - x_2| ) from both sides (since ( x_1 neq x_2 )):( |f'(c)| < lambda frac{1}{x_1 x_2} )But ( x_1, x_2 in [1, 2] ), so ( x_1 x_2 in [1, 4] ). Therefore, ( frac{1}{x_1 x_2} in left[frac{1}{4}, 1right] ).But this seems a bit tangled. Maybe another approach is better. Let's consider the function ( g(x) = f(x) - lambda cdot frac{1}{x} ). Then, the inequality ( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| ) can be rewritten as:( |g(x_1) - g(x_2)| < 0 )Wait, no. Let me think again.Wait, the original inequality is:( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| )Let me rearrange this:( frac{|f(x_1) - f(x_2)|}{left|frac{1}{x_1} - frac{1}{x_2}right|} < lambda )So, ( lambda ) must be greater than the maximum value of ( frac{|f(x_1) - f(x_2)|}{left|frac{1}{x_1} - frac{1}{x_2}right|} ) over all ( x_1, x_2 in [1, 2] ) and ( a in [frac{1}{2}, 2] ).Alternatively, using the Mean Value Theorem, as I started earlier, we have:( |f(x_1) - f(x_2)| = |f'(c)| |x_1 - x_2| )and( left|frac{1}{x_1} - frac{1}{x_2}right| = frac{|x_1 - x_2|}{x_1 x_2} )Therefore, the inequality becomes:( |f'(c)| |x_1 - x_2| < lambda frac{|x_1 - x_2|}{x_1 x_2} )Canceling ( |x_1 - x_2| ):( |f'(c)| < lambda frac{1}{x_1 x_2} )So, ( lambda > |f'(c)| x_1 x_2 )But ( c ) is between ( x_1 ) and ( x_2 ), so ( c in [1, 2] ). Also, ( x_1, x_2 in [1, 2] ), so ( x_1 x_2 in [1, 4] ).But this seems a bit indirect. Maybe another approach is better. Let's consider the function ( g(x) = f(x) - lambda cdot frac{1}{x} ). Then, the original inequality can be rewritten as:( |g(x_1) - g(x_2)| < 0 )Wait, no. Let me think again.Wait, if I rearrange the original inequality:( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| )This can be written as:( |f(x_1) - lambda cdot frac{1}{x_1} - (f(x_2) - lambda cdot frac{1}{x_2})| < 0 )But this implies that ( g(x_1) - g(x_2) ) is less than zero in absolute value, which is always true since absolute value is non-negative. That doesn't help.Wait, maybe I should consider the function ( g(x) = f(x) - lambda cdot frac{1}{x} ). Then, the inequality ( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| ) can be rewritten as:( |g(x_1) - g(x_2) + lambda left( frac{1}{x_1} - frac{1}{x_2} right)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| )But this seems more complicated. Maybe a better approach is to consider the function ( h(x) = f(x) + lambda cdot frac{1}{x} ) and analyze its monotonicity.Wait, perhaps a better idea is to consider the function ( f(x) ) and relate it to ( frac{1}{x} ). Let's think about the inequality:( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| )This can be interpreted as the function ( f(x) ) being Lipschitz continuous with respect to ( frac{1}{x} ) with constant ( lambda ).To find the smallest such ( lambda ), we can consider the maximum of ( frac{|f(x_1) - f(x_2)|}{|frac{1}{x_1} - frac{1}{x_2}|} ) over all ( x_1, x_2 in [1, 2] ) and ( a in [frac{1}{2}, 2] ).Alternatively, using the Mean Value Theorem, as before, we can express this ratio as ( |f'(c)| cdot frac{1}{left| frac{1}{x_1} - frac{1}{x_2} right| / |x_1 - x_2| } ). Wait, no, that might not be the right way.Wait, let's express the ratio:( frac{|f(x_1) - f(x_2)|}{|frac{1}{x_1} - frac{1}{x_2}|} = frac{|f'(c)| |x_1 - x_2|}{left| frac{x_2 - x_1}{x_1 x_2} right|} = |f'(c)| cdot frac{|x_1 - x_2| cdot x_1 x_2}{|x_1 - x_2|} = |f'(c)| cdot x_1 x_2 )So, the ratio simplifies to ( |f'(c)| cdot x_1 x_2 ), where ( c ) is between ( x_1 ) and ( x_2 ).Therefore, the inequality ( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| ) becomes:( |f'(c)| cdot x_1 x_2 < lambda )Thus, ( lambda ) must be greater than ( |f'(c)| cdot x_1 x_2 ) for all ( c in [1, 2] ) and ( x_1, x_2 in [1, 2] ).But since ( x_1, x_2 in [1, 2] ), ( x_1 x_2 in [1, 4] ). Therefore, the maximum value of ( |f'(c)| cdot x_1 x_2 ) would be when ( x_1 x_2 ) is maximized, which is 4, and ( |f'(c)| ) is maximized.Thus, ( lambda ) must be at least ( 4 cdot max_{c in [1, 2]} |f'(c)| ).But let's compute ( f'(c) ):From part (1), ( f'(x) = frac{(x - 1)(x - (2a + 1))}{x} ).Given that ( a in [frac{1}{2}, 2] ), ( 2a + 1 in [2, 5] ). So, ( 2a + 1 geq 2 ), which is greater than 1.From part (1), when ( a > 0 ), ( f(x) ) is increasing on ( (0, 1) ), decreasing on ( (1, 2a + 1) ), and increasing again on ( (2a + 1, infty) ).But since ( x in [1, 2] ) and ( 2a + 1 geq 2 ), the interval ( [1, 2] ) is within the decreasing interval ( (1, 2a + 1) ). Therefore, on ( [1, 2] ), ( f(x) ) is decreasing.Therefore, ( f'(x) leq 0 ) on ( [1, 2] ). So, ( |f'(x)| = -f'(x) ).Thus, ( |f'(c)| = -f'(c) = -left( frac{(c - 1)(c - (2a + 1))}{c} right) ).But since ( c in [1, 2] ) and ( 2a + 1 geq 2 ), ( c - (2a + 1) leq 0 ). Therefore, ( (c - 1) geq 0 ) (since ( c geq 1 )) and ( (c - (2a + 1)) leq 0 ). So, their product is ( leq 0 ). Therefore, ( f'(c) leq 0 ), so ( |f'(c)| = -f'(c) = frac{(1 - c)(c - (2a + 1))}{c} ).But ( 1 - c leq 0 ) and ( c - (2a + 1) leq 0 ), so their product is ( geq 0 ). Therefore, ( |f'(c)| = frac{(c - 1)(2a + 1 - c)}{c} ).So, ( |f'(c)| = frac{(c - 1)(2a + 1 - c)}{c} ).We need to find the maximum of ( |f'(c)| cdot x_1 x_2 ). But since ( x_1, x_2 in [1, 2] ), ( x_1 x_2 in [1, 4] ). So, the maximum occurs when ( x_1 x_2 = 4 ) and ( |f'(c)| ) is maximized.Therefore, ( lambda geq 4 cdot max_{c in [1, 2]} |f'(c)| ).So, let's compute ( max_{c in [1, 2]} |f'(c)| ).Given ( |f'(c)| = frac{(c - 1)(2a + 1 - c)}{c} ).Let me denote ( c in [1, 2] ) and ( a in [frac{1}{2}, 2] ).So, ( 2a + 1 in [2, 5] ). Therefore, ( 2a + 1 - c geq 2 - 2 = 0 ) since ( c leq 2 ).Thus, ( |f'(c)| = frac{(c - 1)(2a + 1 - c)}{c} ).Let me denote ( t = c ), so ( t in [1, 2] ).Then, ( |f'(t)| = frac{(t - 1)(2a + 1 - t)}{t} ).We can write this as:( |f'(t)| = frac{(t - 1)(2a + 1 - t)}{t} )Let me denote ( s = t - 1 ), so ( s in [0, 1] ).Then, ( |f'(t)| = frac{s (2a + 1 - (1 + s))}{1 + s} = frac{s (2a - s)}{1 + s} )So, ( |f'(t)| = frac{s (2a - s)}{1 + s} ), where ( s in [0, 1] ) and ( a in [frac{1}{2}, 2] ).We need to find the maximum of this expression over ( s in [0, 1] ) and ( a in [frac{1}{2}, 2] ).Let me fix ( a ) and find the maximum over ( s ).For a fixed ( a ), the expression is:( frac{s (2a - s)}{1 + s} )Let me denote ( g(s) = frac{s (2a - s)}{1 + s} ).To find the maximum of ( g(s) ) over ( s in [0, 1] ), we can take the derivative with respect to ( s ) and set it to zero.Compute ( g'(s) ):Using quotient rule:( g'(s) = frac{(2a - s) + s(-1)(1 + s) - s(2a - s)(1)}{(1 + s)^2} )Wait, let me compute it step by step.Let me write ( g(s) = frac{s(2a - s)}{1 + s} ).Let me denote numerator ( N = s(2a - s) = 2a s - s^2 ).Denominator ( D = 1 + s ).Then, ( g'(s) = frac{N' D - N D'}{D^2} ).Compute ( N' = 2a - 2s ).Compute ( D' = 1 ).Thus,( g'(s) = frac{(2a - 2s)(1 + s) - (2a s - s^2)(1)}{(1 + s)^2} )Simplify numerator:First term: ( (2a - 2s)(1 + s) = 2a(1 + s) - 2s(1 + s) = 2a + 2a s - 2s - 2s^2 )Second term: ( -(2a s - s^2) = -2a s + s^2 )Combine both terms:( 2a + 2a s - 2s - 2s^2 - 2a s + s^2 )Simplify:- ( 2a )- ( 2a s - 2a s = 0 )- ( -2s )- ( -2s^2 + s^2 = -s^2 )So, numerator simplifies to:( 2a - 2s - s^2 )Therefore,( g'(s) = frac{2a - 2s - s^2}{(1 + s)^2} )Set ( g'(s) = 0 ):( 2a - 2s - s^2 = 0 )Rearranged:( s^2 + 2s - 2a = 0 )Solving for ( s ):( s = frac{-2 pm sqrt{4 + 8a}}{2} = frac{-2 pm 2sqrt{1 + 2a}}{2} = -1 pm sqrt{1 + 2a} )Since ( s geq 0 ), we discard the negative root:( s = -1 + sqrt{1 + 2a} )We need ( s in [0, 1] ). So, check if ( -1 + sqrt{1 + 2a} leq 1 ):( sqrt{1 + 2a} leq 2 )Square both sides:( 1 + 2a leq 4 )( 2a leq 3 )( a leq frac{3}{2} )Given ( a in [frac{1}{2}, 2] ), so for ( a leq frac{3}{2} ), the critical point ( s = -1 + sqrt{1 + 2a} ) is within ( [0, 1] ). For ( a > frac{3}{2} ), the critical point ( s ) would be greater than 1, so the maximum occurs at ( s = 1 ).Therefore, for ( a leq frac{3}{2} ), the maximum of ( g(s) ) occurs at ( s = -1 + sqrt{1 + 2a} ). For ( a > frac{3}{2} ), the maximum occurs at ( s = 1 ).Now, let's compute ( g(s) ) at these critical points and endpoints.First, for ( a leq frac{3}{2} ):( s = -1 + sqrt{1 + 2a} )Compute ( g(s) = frac{s(2a - s)}{1 + s} )Substitute ( s = -1 + sqrt{1 + 2a} ):Let me denote ( sqrt{1 + 2a} = t ), so ( t geq sqrt{1 + 2 cdot frac{1}{2}} = sqrt{2} approx 1.414 ), and ( t leq sqrt{1 + 2 cdot frac{3}{2}} = sqrt{4} = 2 ).Then, ( s = t - 1 ).So,( g(s) = frac{(t - 1)(2a - (t - 1))}{1 + (t - 1)} = frac{(t - 1)(2a - t + 1)}{t} )But ( 2a = t^2 - 1 ) because ( t = sqrt{1 + 2a} Rightarrow t^2 = 1 + 2a Rightarrow 2a = t^2 - 1 ).Substitute ( 2a = t^2 - 1 ):( g(s) = frac{(t - 1)(t^2 - 1 - t + 1)}{t} = frac{(t - 1)(t^2 - t)}{t} )Simplify numerator:( (t - 1)(t^2 - t) = (t - 1)t(t - 1) = t(t - 1)^2 )Thus,( g(s) = frac{t(t - 1)^2}{t} = (t - 1)^2 )So, ( g(s) = (t - 1)^2 = (sqrt{1 + 2a} - 1)^2 )Therefore, for ( a leq frac{3}{2} ), the maximum ( g(s) ) is ( (sqrt{1 + 2a} - 1)^2 ).For ( a > frac{3}{2} ), the maximum occurs at ( s = 1 ):Compute ( g(1) = frac{1(2a - 1)}{1 + 1} = frac{2a - 1}{2} )So, summarizing:- For ( a in [frac{1}{2}, frac{3}{2}] ), ( max_{s} g(s) = (sqrt{1 + 2a} - 1)^2 )- For ( a in (frac{3}{2}, 2] ), ( max_{s} g(s) = frac{2a - 1}{2} )Now, we need to find the maximum of ( |f'(c)| = g(s) ) over ( a in [frac{1}{2}, 2] ).So, let's analyze both cases.First, for ( a in [frac{1}{2}, frac{3}{2}] ):Compute ( (sqrt{1 + 2a} - 1)^2 ).Let me denote ( h(a) = (sqrt{1 + 2a} - 1)^2 ).Compute derivative of ( h(a) ):( h(a) = (sqrt{1 + 2a} - 1)^2 )Let ( u = sqrt{1 + 2a} ), so ( h(a) = (u - 1)^2 ).Then, ( du/da = frac{1}{2sqrt{1 + 2a}} cdot 2 = frac{1}{sqrt{1 + 2a}} ).Thus,( dh/da = 2(u - 1) cdot du/da = 2(sqrt{1 + 2a} - 1) cdot frac{1}{sqrt{1 + 2a}} )Simplify:( dh/da = 2 left( 1 - frac{1}{sqrt{1 + 2a}} right) )Since ( a geq frac{1}{2} ), ( sqrt{1 + 2a} geq sqrt{2} > 1 ), so ( dh/da > 0 ). Therefore, ( h(a) ) is increasing on ( [frac{1}{2}, frac{3}{2}] ).Thus, the maximum in this interval occurs at ( a = frac{3}{2} ):( h(frac{3}{2}) = (sqrt{1 + 2 cdot frac{3}{2}} - 1)^2 = (sqrt{4} - 1)^2 = (2 - 1)^2 = 1 )For ( a in (frac{3}{2}, 2] ):Compute ( frac{2a - 1}{2} ).This is a linear function increasing with ( a ). Therefore, its maximum occurs at ( a = 2 ):( frac{2 cdot 2 - 1}{2} = frac{4 - 1}{2} = frac{3}{2} = 1.5 )Comparing the maximums from both intervals:- For ( a in [frac{1}{2}, frac{3}{2}] ), the maximum ( |f'(c)| ) is 1.- For ( a in (frac{3}{2}, 2] ), the maximum ( |f'(c)| ) is 1.5.Therefore, the overall maximum of ( |f'(c)| ) over ( a in [frac{1}{2}, 2] ) is 1.5.But wait, let's verify this because earlier we had ( |f'(c)| = (sqrt{1 + 2a} - 1)^2 ) for ( a leq frac{3}{2} ), which at ( a = frac{3}{2} ) is 1, and for ( a > frac{3}{2} ), it's ( frac{2a - 1}{2} ), which at ( a = 2 ) is 1.5.So, the maximum ( |f'(c)| ) is 1.5.But wait, let's compute ( |f'(c)| ) at ( a = 2 ) and ( c = 1 ):( f'(1) = frac{(1 - 1)(1 - (2 cdot 2 + 1))}{1} = 0 )At ( c = 2 ):( f'(2) = frac{(2 - 1)(2 - (2 cdot 2 + 1))}{2} = frac{1 cdot (2 - 5)}{2} = frac{-3}{2} ), so ( |f'(2)| = frac{3}{2} = 1.5 ).Yes, that matches.Therefore, the maximum ( |f'(c)| ) over ( c in [1, 2] ) and ( a in [frac{1}{2}, 2] ) is 1.5.But earlier, I thought ( lambda geq 4 cdot max |f'(c)| ), but that might not be correct because ( x_1 x_2 ) can be up to 4, but ( |f'(c)| ) is 1.5. So, the maximum of ( |f'(c)| cdot x_1 x_2 ) is ( 1.5 cdot 4 = 6 ).Wait, but actually, ( |f'(c)| cdot x_1 x_2 leq |f'(c)| cdot 4 ). Since ( |f'(c)| leq 1.5 ), then ( |f'(c)| cdot 4 leq 6 ).But wait, is this the correct approach? Because ( c ) is between ( x_1 ) and ( x_2 ), and ( x_1 x_2 ) is not necessarily independent of ( c ). So, perhaps we need a different approach.Alternatively, let's consider the function ( g(x) = f(x) - lambda cdot frac{1}{x} ). We need ( |g(x_1) - g(x_2)| < 0 ), but that doesn't make sense because absolute value is non-negative. Maybe I misapplied this.Wait, going back to the original inequality:( |f(x_1) - f(x_2)| < lambda left|frac{1}{x_1} - frac{1}{x_2}right| )This can be rewritten as:( left| frac{f(x_1) - f(x_2)}{frac{1}{x_1} - frac{1}{x_2}} right| < lambda )Which is similar to the definition of the derivative, but with respect to ( frac{1}{x} ).Alternatively, consider the function ( h(x) = f(x) + lambda cdot frac{1}{x} ). Then, the inequality becomes:( |h(x_1) - h(x_2)| < 0 )Which again doesn't make sense because absolute value is non-negative.Wait, perhaps I should think in terms of the function ( f(x) ) being Lipschitz with respect to ( frac{1}{x} ). The Lipschitz constant ( lambda ) must satisfy:( |f(x_1) - f(x_2)| leq lambda left| frac{1}{x_1} - frac{1}{x_2} right| )To find the smallest such ( lambda ), we can consider the maximum of ( frac{|f(x_1) - f(x_2)|}{|frac{1}{x_1} - frac{1}{x_2}|} ) over all ( x_1, x_2 in [1, 2] ) and ( a in [frac{1}{2}, 2] ).Using the Mean Value Theorem, as before, this ratio is equal to ( |f'(c)| cdot x_1 x_2 ) for some ( c ) between ( x_1 ) and ( x_2 ).Therefore, the maximum value of this ratio is the maximum of ( |f'(c)| cdot x_1 x_2 ).Given that ( x_1, x_2 in [1, 2] ), ( x_1 x_2 in [1, 4] ). So, the maximum occurs when ( x_1 x_2 ) is maximized, i.e., 4, and ( |f'(c)| ) is maximized.From earlier, the maximum ( |f'(c)| ) is 1.5. Therefore, the maximum ratio is ( 1.5 cdot 4 = 6 ).Thus, ( lambda ) must be greater than or equal to 6.But let's verify this. Suppose ( lambda = 6 ). Then, for any ( x_1, x_2 in [1, 2] ) and ( a in [frac{1}{2}, 2] ), we have:( |f(x_1) - f(x_2)| < 6 left| frac{1}{x_1} - frac{1}{x_2} right| )Is this true?From the earlier analysis, the maximum ratio ( frac{|f(x_1) - f(x_2)|}{|frac{1}{x_1} - frac{1}{x_2}|} ) is 6, so ( lambda ) must be at least 6. Therefore, the range of ( lambda ) is ( [6, infty) ).But let's double-check. Suppose ( a = 2 ), ( x_1 = 1 ), ( x_2 = 2 ).Compute ( |f(1) - f(2)| ):First, compute ( f(1) ):( f(1) = frac{1}{2}(1)^2 - (2 cdot 2 + 2)(1) + (2 cdot 2 + 1)ln 1 = frac{1}{2} - 6 + 5 cdot 0 = frac{1}{2} - 6 = -frac{11}{2} )Compute ( f(2) ):( f(2) = frac{1}{2}(4) - (4 + 2)(2) + 5 ln 2 = 2 - 12 + 5 ln 2 = -10 + 5 ln 2 )Thus, ( |f(1) - f(2)| = | -frac{11}{2} - (-10 + 5 ln 2) | = | -frac{11}{2} + 10 - 5 ln 2 | = | frac{9}{2} - 5 ln 2 | )Compute ( frac{9}{2} approx 4.5 ), ( 5 ln 2 approx 3.4658 ), so ( 4.5 - 3.4658 approx 1.0342 )Compute ( left| frac{1}{1} - frac{1}{2} right| = frac{1}{2} )Thus, the ratio ( frac{|f(1) - f(2)|}{|frac{1}{1} - frac{1}{2}|} = frac{1.0342}{0.5} approx 2.0684 ), which is less than 6.Wait, but earlier we thought the maximum ratio is 6. So, perhaps my earlier conclusion was incorrect.Wait, maybe I need to consider the maximum of ( |f'(c)| cdot x_1 x_2 ), but when ( x_1 x_2 ) is 4, ( c ) must be somewhere in ( [1, 2] ), but ( |f'(c)| ) is 1.5 at ( c = 2 ). So, ( 1.5 cdot 4 = 6 ).But in the example above, the ratio is only about 2.0684. So, perhaps the maximum occurs elsewhere.Wait, let's consider ( x_1 = 1 ) and ( x_2 ) approaching 1. Then, the ratio ( frac{|f(x_1) - f(x_2)|}{|frac{1}{x_1} - frac{1}{x_2}|} ) approaches ( |f'(1)| cdot x_1 x_2 ).But ( f'(1) = 0 ), so the ratio approaches 0.Wait, maybe the maximum occurs when ( x_1 ) and ( x_2 ) are such that ( c ) is at the point where ( |f'(c)| ) is maximized, and ( x_1 x_2 ) is maximized.But ( x_1 x_2 ) is maximized when ( x_1 = x_2 = 2 ), but ( x_1 neq x_2 ), so approaching 2 from below.Wait, perhaps the maximum ratio occurs when ( x_1 ) and ( x_2 ) are such that ( c ) is at the point where ( |f'(c)| ) is maximized, and ( x_1 x_2 ) is as large as possible.But since ( x_1, x_2 in [1, 2] ), the maximum ( x_1 x_2 ) is 4 when ( x_1 = x_2 = 2 ), but ( x_1 neq x_2 ), so we can approach 4 by taking ( x_1 = 2 ) and ( x_2 ) approaching 2.But in that case, ( c ) approaches 2, and ( |f'(c)| ) approaches 1.5. So, the ratio approaches ( 1.5 cdot 4 = 6 ).Therefore, the supremum of the ratio is 6, so ( lambda ) must be at least 6.Thus, the range of ( lambda ) is ( [6, infty) ).

💡Okay, so I have this problem about a geometric sequence. All the terms are positive, which is good to know. The sum of the first three terms, S₃, is 14, and the third term, a₃, is 8. I need to find the sixth term, a₆. The options are 16, 32, 64, or 128. Hmm, let's break this down step by step.First, let me recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, usually denoted as r. So, if the first term is a₁, then the second term is a₁ * r, the third term is a₁ * r², and so on. In general, the nth term is aₙ = a₁ * r^(n-1).Given that, I know that a₃ = 8. Using the formula for the nth term, that would be a₃ = a₁ * r² = 8. So, I can write that as equation (1): a₁ * r² = 8.Also, the sum of the first three terms, S₃, is 14. The sum of the first n terms of a geometric sequence is given by Sₙ = a₁ * (1 - rⁿ) / (1 - r) when r ≠ 1. So, for n = 3, S₃ = a₁ * (1 - r³) / (1 - r) = 14. Let's write that as equation (2): a₁ * (1 - r³) / (1 - r) = 14.Now, I have two equations:1. a₁ * r² = 82. a₁ * (1 - r³) / (1 - r) = 14I need to solve these two equations to find r and a₁, and then use them to find a₆.Let me see if I can express a₁ from equation (1) and substitute it into equation (2). From equation (1), a₁ = 8 / r². So, substituting into equation (2):(8 / r²) * (1 - r³) / (1 - r) = 14Let me simplify this equation. First, let's write it out:(8 / r²) * (1 - r³) / (1 - r) = 14I can factor the numerator (1 - r³). Remember that 1 - r³ factors into (1 - r)(1 + r + r²). So, substituting that in:(8 / r²) * [(1 - r)(1 + r + r²)] / (1 - r) = 14Notice that (1 - r) cancels out in the numerator and denominator:(8 / r²) * (1 + r + r²) = 14So, now we have:8 * (1 + r + r²) / r² = 14Let me write this as:8(1 + r + r²) / r² = 14To make it easier, let's multiply both sides by r² to eliminate the denominator:8(1 + r + r²) = 14r²Expanding the left side:8 + 8r + 8r² = 14r²Now, let's bring all terms to one side to form a quadratic equation:8 + 8r + 8r² - 14r² = 0Simplify the terms:8 + 8r - 6r² = 0Let me rearrange the terms in descending order of powers of r:-6r² + 8r + 8 = 0It's usually easier to work with positive coefficients, so let's multiply the entire equation by -1:6r² - 8r - 8 = 0Now, we have a quadratic equation: 6r² - 8r - 8 = 0To solve for r, we can use the quadratic formula. The quadratic formula is r = [ -b ± sqrt(b² - 4ac) ] / (2a), where a = 6, b = -8, and c = -8.Plugging in the values:r = [ -(-8) ± sqrt( (-8)² - 4 * 6 * (-8) ) ] / (2 * 6)r = [ 8 ± sqrt(64 + 192) ] / 12r = [ 8 ± sqrt(256) ] / 12r = [ 8 ± 16 ] / 12So, we have two possible solutions:1. r = (8 + 16) / 12 = 24 / 12 = 22. r = (8 - 16) / 12 = (-8) / 12 = -2/3But wait, the problem states that all terms of the geometric sequence are positive. If r were negative, then the terms would alternate in sign. Since all terms are positive, r must be positive. Therefore, we discard r = -2/3 and take r = 2.Now that we have r = 2, we can find a₁ using equation (1): a₁ * r² = 8Plugging in r = 2:a₁ * (2)² = 8a₁ * 4 = 8a₁ = 8 / 4a₁ = 2So, the first term a₁ is 2, and the common ratio r is 2.Now, we need to find a₆, the sixth term. Using the formula for the nth term of a geometric sequence:aₙ = a₁ * r^(n-1)So, a₆ = a₁ * r^(6-1) = a₁ * r^5Plugging in the known values:a₆ = 2 * (2)^5Calculate 2^5:2^5 = 32So, a₆ = 2 * 32 = 64Therefore, the sixth term is 64, which corresponds to option C.Let me double-check my calculations to make sure I didn't make any mistakes.Starting with a₁ = 2, r = 2:a₁ = 2a₂ = a₁ * r = 2 * 2 = 4a₃ = a₂ * r = 4 * 2 = 8 (which matches the given a₃ = 8)Sum of first three terms: 2 + 4 + 8 = 14 (which matches S₃ = 14)So, everything checks out. Therefore, a₆ = 64 is indeed the correct answer.

💡Okay, so I have this geometry problem here about half-planes sharing a common edge and the sum of dihedral angles around that edge. Hmm, let me try to visualize this. First, I know that a half-plane is like a flat surface that extends infinitely in one direction, right? So if there are multiple half-planes sharing a common edge, they must all meet along that edge. Now, the question is about the dihedral angles between each pair of consecutive half-planes. Dihedral angles are the angles between two planes, measured along their line of intersection, which in this case is the common edge ( l ).I remember that in three-dimensional space, when multiple planes meet along a common line, the sum of the dihedral angles around that line should relate to the full angle around a point in space. But wait, in three dimensions, the full angle around a point isn't just ( 360^circ ) like in two dimensions. Or is it?Let me think. If I have a point in space, the total solid angle around it is ( 4pi ) steradians. But dihedral angles are different; they're the angles between two planes, not solid angles. So maybe the sum of dihedral angles around a common edge is related to the full angle in a different way.Wait, if I take a plane that's perpendicular to the common edge ( l ), then the intersection of each half-plane with this perpendicular plane would give me lines radiating from the point where the perpendicular plane meets ( l ). So, in this perpendicular plane, the dihedral angles between the half-planes would correspond to the angles between these lines.Ah, so in this perpendicular plane, the dihedral angles become planar angles around a point. And I know that the sum of angles around a point in a plane is ( 360^circ ) or ( 2pi ) radians. So does that mean the sum of the dihedral angles is also ( 2pi )?But wait, dihedral angles are measured in three dimensions, so is it the same as the planar angles? Or is there a factor I'm missing? Let me double-check. If I have two half-planes meeting along edge ( l ), the dihedral angle between them is the angle you'd measure if you look along the edge ( l ). So, if I have multiple such dihedral angles around ( l ), their sum should account for the full "circle" around ( l ).But in three dimensions, a full circle around a line isn't just ( 360^circ ). Or is it? Wait, no, because when you look along the line ( l ), you're effectively in a two-dimensional space, so the angles around ( l ) should still add up to ( 360^circ ). So, even though we're in three dimensions, the dihedral angles around a common edge should sum up to ( 2pi ) radians.Let me think of a simple case. If there are two half-planes meeting along ( l ), the dihedral angle between them would be ( pi ) radians or ( 180^circ ). If there are three half-planes, each pair forming a dihedral angle of ( 2pi/3 ) radians, their sum would be ( 2pi ). Similarly, for ( n ) half-planes, each dihedral angle would be ( 2pi/n ), and their sum would be ( 2pi ).So, regardless of the number of half-planes, as long as they share a common edge and span the entire space around that edge, the sum of their dihedral angles should be ( 2pi ) radians.I think that makes sense. The key idea is that by intersecting with a perpendicular plane, we reduce the problem to planar angles around a point, whose sum is known to be ( 2pi ). Therefore, the sum of the dihedral angles must also be ( 2pi ).**Final Answer**The sum of the dihedral angles is boxed{2pi}.

💡To determine the largest possible sum of the digits in a 24-hour digital watch display, we need to consider both the hours and minutes sections.1. **Hours Section**: The hours range from 00 to 23. The maximum digit sum for the hours occurs at 19, which is (1 + 9 = 10).2. **Minutes Section**: The minutes range from 00 to 59. The maximum digit sum for the minutes occurs at 59, which is (5 + 9 = 14).3. **Total Sum**: Adding the maximum sums from both sections gives (10 + 14 = 24).Thus, the largest possible sum of the digits in the display is (boxed{textbf{(C)} 24}).

💡Alright, so I've got this problem here: If 5 knicks equal 3 knacks and 2 knacks equal 7 knocks, how many knicks are equal to 49 knocks? Hmm, okay, let's try to wrap my head around this.First off, I need to understand the relationships between these units: knicks, knacks, and knocks. It seems like they're all different units, and I need to convert from knocks to knicks using knacks as an intermediary. So, it's like a chain of conversions: knocks to knacks, and then knacks to knicks.Let's start with what's given:1. 5 knicks = 3 knacks2. 2 knacks = 7 knocksI need to find out how many knicks are equivalent to 49 knocks. So, essentially, I need to convert 49 knocks into knacks first, and then convert those knacks into knicks.Starting with the second equation: 2 knacks = 7 knocks. I can use this to find out how many knacks are equivalent to 1 knock. If 2 knacks equal 7 knocks, then 1 knock equals 2/7 knacks. That makes sense because if I divide both sides by 7, I get 1 knock = (2/7) knacks.Now, if 1 knock is 2/7 knacks, then 49 knocks would be 49 times that. So, 49 knocks * (2/7 knacks/knock). Let's calculate that: 49 * (2/7) = (49/7) * 2 = 7 * 2 = 14 knacks. Okay, so 49 knocks equal 14 knacks.Now, I need to convert knacks into knicks using the first equation: 5 knicks = 3 knacks. I can rearrange this to find out how many knicks are in 1 knack. If 3 knacks equal 5 knicks, then 1 knack equals 5/3 knicks. That seems right because if I divide both sides by 3, I get 1 knack = (5/3) knicks.So, if 1 knack is 5/3 knicks, then 14 knacks would be 14 times that. Let's do the math: 14 * (5/3) = (14*5)/3 = 70/3 knicks. Hmm, 70 divided by 3 is approximately 23.333... So, that's 23 and 1/3 knicks.Wait, let me double-check my calculations to make sure I didn't make a mistake. Starting from 49 knocks, I converted to knacks by multiplying by 2/7, which gave me 14 knacks. Then, I converted knacks to knicks by multiplying by 5/3, resulting in 70/3 knicks, which is indeed 23 and 1/3 knicks. That seems consistent.Is there another way to approach this problem? Maybe by finding a direct conversion factor from knocks to knicks without going through knacks? Let's see.From the given equations:5 knicks = 3 knacks2 knacks = 7 knocksI can combine these two equations to find a relationship between knicks and knocks. If 2 knacks equal 7 knocks, then 1 knack equals 7/2 knocks. Similarly, from the first equation, 1 knack equals 5/3 knicks.So, if 1 knack is 5/3 knicks and also 7/2 knocks, then I can set up a proportion:5/3 knicks = 7/2 knocksTo find out how many knicks equal 1 knock, I can divide both sides by 7/2:(5/3 knicks) / (7/2 knocks) = 1 knockDividing by a fraction is the same as multiplying by its reciprocal, so:5/3 * 2/7 knicks per knock = (10/21) knicks per knockSo, 1 knock equals 10/21 knicks. Therefore, 49 knocks would be:49 knocks * (10/21 knicks/knock) = (49*10)/21 knicks = 490/21 knicksSimplifying 490 divided by 21: 21*23 = 483, so 490 - 483 = 7, which is 23 and 7/21, which simplifies to 23 and 1/3 knicks.Okay, so both methods give me the same result: 23 and 1/3 knicks. That makes me more confident that the answer is correct.Just to recap:- Converted knocks to knacks: 49 knocks * (2/7 knacks/knock) = 14 knacks- Converted knacks to knicks: 14 knacks * (5/3 knicks/knack) = 70/3 knicks = 23 1/3 knicksAlternatively:- Found a direct conversion factor: 1 knock = 10/21 knicks- Applied it to 49 knocks: 49 * (10/21) = 70/3 knicks = 23 1/3 knicksBoth approaches lead to the same answer, so I think I'm good.I guess the key here was setting up the conversion factors correctly and ensuring that the units cancel out appropriately. It's like a unit analysis problem, where you multiply by fractions that represent the relationships between the units until you end up with the desired unit.Another thing to note is that when dealing with ratios, it's essential to keep track of which unit is in the numerator and which is in the denominator to ensure that the units cancel correctly. For example, when converting knocks to knacks, I used the ratio of knacks per knock, which allowed the knock units to cancel out, leaving me with knacks. Then, converting knacks to knicks required the ratio of knicks per knack, canceling out the knacks and leaving me with knicks.It's also helpful to simplify fractions whenever possible to make the calculations easier. For instance, when I had 49 knocks multiplied by 2/7 knacks per knock, simplifying 49/7 to 7 made the multiplication straightforward.Overall, this problem was a good exercise in applying conversion factors and unit analysis to solve for an unknown quantity. It reinforces the importance of carefully setting up the relationships between units and systematically applying the conversions step by step.