💡Alright, so I've got this problem about an arithmetic sequence. It says that the sum of the first n terms is S_n. They give me that a_3 is 12, S_12 is greater than 0, and S_13 is less than 0. I need to find the range of the common difference d and the value of n where the sum S_n is maximized.First, let's recall what an arithmetic sequence is. It's a sequence where each term increases by a constant difference, which is d. So, the nth term can be written as a_n = a_1 + (n-1)d, where a_1 is the first term.Given that a_3 = 12, I can write that as:a_3 = a_1 + 2d = 12So, a_1 = 12 - 2d. That gives me the first term in terms of d.Now, the sum of the first n terms of an arithmetic sequence is given by:S_n = (n/2)(2a_1 + (n - 1)d)Alternatively, it can also be written as:S_n = n(a_1 + a_n)/2They give me S_12 > 0 and S_13 < 0. So, I need to write expressions for S_12 and S_13 and set up inequalities.Let's compute S_12 first:S_12 = (12/2)(2a_1 + 11d) = 6(2a_1 + 11d)We already know a_1 = 12 - 2d, so substituting that in:S_12 = 6(2*(12 - 2d) + 11d) = 6(24 - 4d + 11d) = 6(24 + 7d)So, S_12 = 6*(24 + 7d) > 0Dividing both sides by 6:24 + 7d > 0So, 7d > -24Therefore, d > -24/7Okay, so d must be greater than -24/7.Now, let's compute S_13:S_13 = (13/2)(2a_1 + 12d) = (13/2)(2*(12 - 2d) + 12d) = (13/2)(24 - 4d + 12d) = (13/2)(24 + 8d)So, S_13 = (13/2)*(24 + 8d) < 0Multiplying both sides by 2/13 (which is positive, so inequality sign doesn't change):24 + 8d < 0So, 8d < -24Therefore, d < -3So, from S_12 > 0, we have d > -24/7, and from S_13 < 0, we have d < -3.Therefore, the range of d is (-24/7, -3). Let me check if that makes sense.-24/7 is approximately -3.4286. So, d is between approximately -3.4286 and -3. So, d is negative, which means the sequence is decreasing.Now, moving on to part (2): Find the value of n when S_n is maximized.In an arithmetic sequence, the sum S_n is a quadratic function of n. It will have a maximum or minimum depending on the common difference d.Since d is negative here, the terms are decreasing. So, the sequence starts positive and eventually becomes negative. The sum S_n will increase to a certain point and then start decreasing once the negative terms dominate.To find the maximum sum, we need to find the point where the terms change from positive to negative. The maximum sum occurs just before the terms become negative.So, we need to find the smallest n such that a_n is still positive, and a_{n+1} is negative.Given that a_n = a_1 + (n-1)d, we can set up the inequality:a_n > 0 and a_{n+1} < 0So, let's write that:a_n = a_1 + (n - 1)d > 0a_{n+1} = a_1 + n*d < 0We know a_1 = 12 - 2d, so substituting:12 - 2d + (n - 1)d > 012 - 2d + n*d - d > 012 - 3d + n*d > 0Similarly, for a_{n+1}:12 - 2d + n*d < 0So, we have two inequalities:12 - 3d + n*d > 012 - 2d + n*d < 0Let me write them as:(12 - 3d) + n*d > 0(12 - 2d) + n*d < 0Let me denote x = n*d for simplicity.Then:12 - 3d + x > 012 - 2d + x < 0Subtracting the first inequality from the second:(12 - 2d + x) - (12 - 3d + x) < 0 - 0(12 - 2d + x - 12 + 3d - x) < 0d < 0Which is consistent because d is negative.But this doesn't directly help. Maybe another approach.Alternatively, let's solve for n in both inequalities.From the first inequality:12 - 3d + n*d > 0n*d > 3d - 12Since d is negative, dividing both sides by d reverses the inequality:n < (3d - 12)/dn < 3 - 12/dFrom the second inequality:12 - 2d + n*d < 0n*d < -12 + 2dAgain, d is negative, so dividing both sides by d reverses the inequality:n > (-12 + 2d)/dn > (-12)/d + 2So, combining both:(-12)/d + 2 < n < 3 - 12/dBut n must be an integer, so we need to find integer n in this interval.Alternatively, perhaps using the expressions for S_12 and S_13.We know that S_12 > 0 and S_13 < 0.So, S_12 is positive, S_13 is negative, which suggests that the maximum sum occurs somewhere between n=12 and n=13. But since n must be an integer, the maximum occurs at n=12 or n=13.But wait, S_12 is positive and S_13 is negative, so the maximum must be at n=12?Wait, but let's think about the behavior of S_n.Since the terms are decreasing, the sum will increase as long as the terms are positive, and start decreasing once the terms become negative.So, the maximum sum occurs at the point where the terms switch from positive to negative.Given that S_12 is positive and S_13 is negative, that suggests that the 13th term is negative, so the 12th term is still positive.Therefore, the maximum sum occurs at n=12.Wait, but let's verify.Alternatively, let's find when a_n changes sign.We have a_n = 12 - 2d + (n - 1)d = 12 - 2d + dn - d = 12 - 3d + dnSet a_n = 0:12 - 3d + dn = 0dn = 3d - 12n = (3d - 12)/d = 3 - 12/dSo, n = 3 - 12/dSince d is between -24/7 and -3, let's compute n.Let me compute 12/d:Since d is negative, 12/d is negative.So, n = 3 - 12/dGiven d > -24/7, so 12/d > 12/(-24/7) = -12*(7/24) = -7/2 = -3.5Similarly, d < -3, so 12/d > 12/(-3) = -4So, 12/d is between -4 and -3.5Therefore, n = 3 - (something between -4 and -3.5) = 3 + (something between 3.5 and 4)So, n is between 3 + 3.5 = 6.5 and 3 + 4 = 7So, n is between 6.5 and 7Since n must be an integer, the term where a_n changes sign is between 6.5 and 7, meaning that a_7 is the first negative term.Therefore, the maximum sum occurs at n=6, because up to n=6, all terms are positive, and starting from n=7, terms become negative, which would decrease the sum.Wait, but earlier I thought it was n=12 because S_12 is positive and S_13 is negative. But according to this, the maximum occurs at n=6.Hmm, there's a contradiction here. Let me think again.Wait, S_n is the sum of the first n terms. The sum can still be increasing even if some terms are negative, as long as the added term is positive. But once the terms become negative, adding them will start decreasing the sum.But in this case, the terms start becoming negative at n=7, so the sum S_n will start decreasing from n=7 onwards.But S_12 is still positive, and S_13 is negative. So, the maximum must be somewhere between n=12 and n=13, but since n=13 is negative, the maximum is at n=12.Wait, but that contradicts the earlier conclusion that a_7 is negative, so the sum should start decreasing at n=7.Wait, perhaps I made a mistake in interpreting the sum.Let me think differently. The sum S_n is a quadratic function in n. The vertex of the parabola will give the maximum or minimum.Since d is negative, the coefficient of n^2 in S_n is (d/2), which is negative. So, the parabola opens downward, meaning the vertex is the maximum point.The vertex occurs at n = -b/(2a), where the quadratic is in the form an^2 + bn + c.Let me write S_n in terms of n.S_n = (n/2)(2a_1 + (n - 1)d) = (n/2)(2*(12 - 2d) + dn - d) = (n/2)(24 - 4d + dn - d) = (n/2)(24 - 5d + dn)So, S_n = (d/2)n^2 + (24 - 5d)/2 * nSo, in the form an^2 + bn + c, where a = d/2, b = (24 - 5d)/2The vertex is at n = -b/(2a) = -[(24 - 5d)/2] / [2*(d/2)] = -[(24 - 5d)/2] / d = -(24 - 5d)/(2d)Simplify:n = (5d - 24)/(2d) = (5d - 24)/(2d) = (5/2) - 12/dSo, n = 5/2 - 12/dGiven that d is between -24/7 and -3, let's compute n.First, compute 12/d:Since d is negative, 12/d is negative.Given d > -24/7, so 12/d > 12/(-24/7) = -12*(7/24) = -7/2 = -3.5Similarly, d < -3, so 12/d > 12/(-3) = -4So, 12/d is between -4 and -3.5Therefore, -12/d is between 3.5 and 4So, n = 5/2 - 12/d = 2.5 + (something between 3.5 and 4)So, n is between 2.5 + 3.5 = 6 and 2.5 + 4 = 6.5Therefore, n is between 6 and 6.5Since n must be an integer, the maximum occurs at n=6 or n=7.But wait, since the vertex is at n≈6.25, which is between 6 and 7, the maximum sum occurs at n=6 because the sum is increasing up to n=6 and starts decreasing from n=7.Wait, but earlier I thought that S_12 is positive and S_13 is negative, which suggests that the maximum is at n=12. But according to this, the maximum is at n=6.This is conflicting. Let me check my calculations.Wait, perhaps I made a mistake in the expression for S_n.Let me re-derive S_n.Given a_1 = 12 - 2dS_n = n/2 [2a_1 + (n - 1)d] = n/2 [2*(12 - 2d) + dn - d] = n/2 [24 - 4d + dn - d] = n/2 [24 - 5d + dn]So, S_n = (d/2)n^2 + (24 - 5d)/2 * nYes, that's correct.So, the vertex is at n = -b/(2a) = -[(24 - 5d)/2] / [2*(d/2)] = -[(24 - 5d)/2] / d = -(24 - 5d)/(2d) = (5d - 24)/(2d) = 5/2 - 12/dYes, that's correct.Given d is between -24/7 and -3, let's compute 12/d:For d = -24/7, 12/d = 12/(-24/7) = -7/2 = -3.5For d = -3, 12/d = -4So, 12/d is between -4 and -3.5, so -12/d is between 3.5 and 4Thus, n = 5/2 - 12/d = 2.5 + (something between 3.5 and 4) = between 6 and 6.5So, n is between 6 and 6.5, meaning the maximum occurs at n=6.But wait, S_12 is positive and S_13 is negative. How come the maximum is at n=6?Let me compute S_6 and S_7.Compute S_6:S_6 = (6/2)(2a_1 + 5d) = 3*(2*(12 - 2d) + 5d) = 3*(24 - 4d + 5d) = 3*(24 + d) = 72 + 3dSimilarly, S_7 = (7/2)(2a_1 + 6d) = (7/2)*(24 - 4d + 6d) = (7/2)*(24 + 2d) = 7*(12 + d) = 84 + 7dNow, let's see if S_7 > S_6:S_7 - S_6 = (84 + 7d) - (72 + 3d) = 12 + 4dGiven d > -24/7 ≈ -3.4286So, 4d > -24/7 *4 ≈ -13.714Thus, 12 + 4d > 12 -13.714 ≈ -1.714But since d < -3, 4d < -12Thus, 12 + 4d < 0Therefore, S_7 - S_6 < 0, so S_7 < S_6Therefore, the maximum occurs at n=6.Wait, but S_12 is positive and S_13 is negative. So, S_12 > 0 and S_13 < 0.But if the maximum is at n=6, then S_6 is the maximum, and S_n decreases after that.But S_12 is still positive, which is after n=6. So, how come S_12 is positive?Wait, maybe the sum starts decreasing after n=6, but it's still positive up to n=12.So, the maximum is at n=6, but the sum remains positive until n=12, and then becomes negative at n=13.That makes sense.So, the maximum sum occurs at n=6, and then the sum decreases, but remains positive until n=12, and then becomes negative at n=13.Therefore, the answer to part (2) is n=6.Wait, but let me confirm with specific values.Suppose d = -3. Let's see:a_1 = 12 - 2*(-3) = 12 + 6 = 18Then, a_n = 18 + (n-1)*(-3) = 18 -3n +3 = 21 -3nSo, a_3 = 21 -9 = 12, which matches.Compute S_12:S_12 = (12/2)(2*18 + 11*(-3)) = 6*(36 -33) = 6*3 = 18 >0S_13 = (13/2)(2*18 + 12*(-3)) = (13/2)*(36 -36) = 0 <0? Wait, S_13=0, but the problem says S_13 <0. So, d cannot be exactly -3.Similarly, if d approaches -3 from above, say d=-3.1:a_1 = 12 -2*(-3.1)=12+6.2=18.2a_n=18.2 + (n-1)*(-3.1)=18.2 -3.1n +3.1=21.3 -3.1nCompute S_12:S_12=(12/2)(2*18.2 +11*(-3.1))=6*(36.4 -34.1)=6*(2.3)=13.8>0S_13=(13/2)(2*18.2 +12*(-3.1))=(13/2)*(36.4 -37.2)=(13/2)*(-0.8)=13*(-0.4)=-5.2<0So, with d=-3.1, S_12=13.8>0, S_13=-5.2<0Now, compute S_6 and S_7:S_6=(6/2)(2*18.2 +5*(-3.1))=3*(36.4 -15.5)=3*(20.9)=62.7S_7=(7/2)(2*18.2 +6*(-3.1))=(7/2)*(36.4 -18.6)=(7/2)*(17.8)=7*8.9=62.3So, S_6=62.7, S_7=62.3, so S_7 < S_6, confirming that the maximum is at n=6.Similarly, let's take d=-24/7≈-3.4286a_1=12 -2*(-24/7)=12 +48/7≈12+6.857≈18.857a_n=18.857 + (n-1)*(-24/7)=18.857 -24/7 n +24/7≈18.857 +3.4286 -3.4286n≈22.2857 -3.4286nCompute S_6:S_6=(6/2)(2*18.857 +5*(-24/7))=3*(37.714 -17.142)=3*(20.572)=61.716S_7=(7/2)(2*18.857 +6*(-24/7))=(7/2)*(37.714 -20.571)=(7/2)*(17.143)=7*8.5715≈60So, S_6≈61.716, S_7≈60, so again S_7 < S_6, confirming maximum at n=6.Therefore, despite S_12 being positive, the maximum occurs at n=6 because the terms start becoming negative at n=7, causing the sum to decrease after that point.So, the answers are:(1) The common difference d is between -24/7 and -3, so d ∈ (-24/7, -3)(2) The maximum sum occurs at n=6

💡Alright, so I've got this problem here about the function ( f(x) = e^x - a sin x - 1 ) where ( a ) is a real number. There are two parts to this problem. Let me try to tackle them one by one.**Part (1): When ( a = 1 ), discuss the monotonicity of the function ( g(x) = frac{f(x)}{e^x} ) on ( left( -frac{pi}{2}, frac{3pi}{2} right) ).**Okay, so first, let's substitute ( a = 1 ) into ( f(x) ). That gives us:[ f(x) = e^x - sin x - 1 ]Then, ( g(x) ) is defined as ( frac{f(x)}{e^x} ), so let's write that out:[ g(x) = frac{e^x - sin x - 1}{e^x} ]Hmm, maybe I can simplify this expression a bit. Let's split the fraction:[ g(x) = frac{e^x}{e^x} - frac{sin x}{e^x} - frac{1}{e^x} ]Simplifying each term:[ g(x) = 1 - e^{-x} sin x - e^{-x} ]So,[ g(x) = 1 - e^{-x} (sin x + 1) ]Alright, now to discuss the monotonicity of ( g(x) ), I need to find its derivative ( g'(x) ) and analyze its sign.Let's compute ( g'(x) ):First, the derivative of 1 is 0. Then, for the second term, ( -e^{-x} (sin x + 1) ), we'll need to use the product rule.Let me denote ( u = -e^{-x} ) and ( v = sin x + 1 ).So, ( u' = e^{-x} ) (since the derivative of ( e^{-x} ) is ( -e^{-x} ), but there's a negative sign in front, so it becomes positive) and ( v' = cos x ).Applying the product rule:[ g'(x) = u'v + uv' = e^{-x} (sin x + 1) + (-e^{-x}) cos x ]Simplify this:[ g'(x) = e^{-x} (sin x + 1 - cos x) ]So,[ g'(x) = e^{-x} (sin x - cos x + 1) ]Now, since ( e^{-x} ) is always positive for all real ( x ), the sign of ( g'(x) ) depends on the expression ( sin x - cos x + 1 ).Let me denote ( h(x) = sin x - cos x + 1 ). So, ( g'(x) = e^{-x} h(x) ).So, the sign of ( g'(x) ) is the same as the sign of ( h(x) ).Therefore, I need to analyze ( h(x) = sin x - cos x + 1 ) over the interval ( left( -frac{pi}{2}, frac{3pi}{2} right) ).Hmm, maybe I can rewrite ( h(x) ) in a more manageable form. Let's see:[ h(x) = sin x - cos x + 1 ]I recall that ( sin x - cos x ) can be written as ( sqrt{2} sin left( x - frac{pi}{4} right) ). Let me verify that:Using the identity ( A sin x + B cos x = C sin(x + phi) ), where ( C = sqrt{A^2 + B^2} ) and ( phi = arctan left( frac{B}{A} right) ).In this case, ( A = 1 ) and ( B = -1 ), so:[ sin x - cos x = sqrt{1^2 + (-1)^2} sin left( x + phi right) ][ = sqrt{2} sin left( x - frac{pi}{4} right) ]Yes, that's correct because ( phi = arctan left( frac{-1}{1} right) = -frac{pi}{4} ).So, substituting back:[ h(x) = sqrt{2} sin left( x - frac{pi}{4} right) + 1 ]Therefore,[ h(x) = sqrt{2} sin left( x - frac{pi}{4} right) + 1 ]Now, let's analyze ( h(x) ) over ( left( -frac{pi}{2}, frac{3pi}{2} right) ).First, let's find the critical points where ( h(x) = 0 ):[ sqrt{2} sin left( x - frac{pi}{4} right) + 1 = 0 ][ sin left( x - frac{pi}{4} right) = -frac{1}{sqrt{2}} ]The solutions to this equation are:[ x - frac{pi}{4} = frac{5pi}{4} + 2pi n quad text{or} quad x - frac{pi}{4} = frac{7pi}{4} + 2pi n quad text{for integer } n ]So,[ x = frac{5pi}{4} + frac{pi}{4} + 2pi n = frac{3pi}{2} + 2pi n ][ x = frac{7pi}{4} + frac{pi}{4} + 2pi n = 2pi + 2pi n ]But our interval is ( left( -frac{pi}{2}, frac{3pi}{2} right) ). Let's check if these critical points lie within this interval.For ( n = 0 ):- ( x = frac{3pi}{2} ) is the upper bound of the interval.- ( x = 2pi ) is outside the interval.For ( n = -1 ):- ( x = frac{3pi}{2} - 2pi = -frac{pi}{2} ), which is the lower bound.- ( x = 2pi - 2pi = 0 ), which is inside the interval.Wait, so ( x = 0 ) is a critical point?Let me verify:At ( x = 0 ):[ h(0) = sqrt{2} sin left( -frac{pi}{4} right) + 1 = sqrt{2} left( -frac{sqrt{2}}{2} right) + 1 = -1 + 1 = 0 ]Yes, so ( x = 0 ) is a critical point.Similarly, at ( x = frac{3pi}{2} ):[ hleft( frac{3pi}{2} right) = sqrt{2} sin left( frac{3pi}{2} - frac{pi}{4} right) + 1 = sqrt{2} sin left( frac{5pi}{4} right) + 1 = sqrt{2} left( -frac{sqrt{2}}{2} right) + 1 = -1 + 1 = 0 ]So, ( x = frac{3pi}{2} ) is also a critical point.But since ( x = frac{3pi}{2} ) is the endpoint of the interval, we can consider it as a boundary.So, within the interval ( left( -frac{pi}{2}, frac{3pi}{2} right) ), the critical points are at ( x = 0 ) and ( x = frac{3pi}{2} ).Now, let's analyze the sign of ( h(x) ) in different sub-intervals:1. ( left( -frac{pi}{2}, 0 right) )2. ( left( 0, frac{3pi}{2} right) )Let's pick test points in each interval.**Interval 1: ( left( -frac{pi}{2}, 0 right) )**Let's choose ( x = -frac{pi}{4} ).Compute ( hleft( -frac{pi}{4} right) ):[ hleft( -frac{pi}{4} right) = sqrt{2} sin left( -frac{pi}{4} - frac{pi}{4} right) + 1 = sqrt{2} sin left( -frac{pi}{2} right) + 1 = sqrt{2} (-1) + 1 = -sqrt{2} + 1 approx -1.414 + 1 = -0.414 ]So, ( h(x) ) is negative in this interval.**Interval 2: ( left( 0, frac{3pi}{2} right) )**Let's choose ( x = frac{pi}{2} ).Compute ( hleft( frac{pi}{2} right) ):[ hleft( frac{pi}{2} right) = sqrt{2} sin left( frac{pi}{2} - frac{pi}{4} right) + 1 = sqrt{2} sin left( frac{pi}{4} right) + 1 = sqrt{2} left( frac{sqrt{2}}{2} right) + 1 = 1 + 1 = 2 ]So, ( h(x) ) is positive in this interval.Wait, but let me check another point to be sure. Let's pick ( x = pi ):[ h(pi) = sqrt{2} sin left( pi - frac{pi}{4} right) + 1 = sqrt{2} sin left( frac{3pi}{4} right) + 1 = sqrt{2} left( frac{sqrt{2}}{2} right) + 1 = 1 + 1 = 2 ]Still positive.And at ( x = frac{3pi}{2} ), as we saw earlier, ( h(x) = 0 ).So, summarizing:- For ( x in left( -frac{pi}{2}, 0 right) ), ( h(x) < 0 ) ⇒ ( g'(x) < 0 ) ⇒ ( g(x) ) is decreasing.- For ( x in left( 0, frac{3pi}{2} right) ), ( h(x) > 0 ) ⇒ ( g'(x) > 0 ) ⇒ ( g(x) ) is increasing.Therefore, the function ( g(x) ) is decreasing on ( left( -frac{pi}{2}, 0 right) ) and increasing on ( left( 0, frac{3pi}{2} right) ).**Part (2): When ( a = -3 ), prove that for all ( x in (0, +infty) ), ( f(x) < e^x + x + 1 - 2e^{-2x} ).**Alright, so now ( a = -3 ), so ( f(x) = e^x - (-3) sin x - 1 = e^x + 3 sin x - 1 ).We need to show that:[ e^x + 3 sin x - 1 < e^x + x + 1 - 2e^{-2x} ]Let me subtract ( e^x ) from both sides to simplify:[ 3 sin x - 1 < x + 1 - 2e^{-2x} ]Bring all terms to one side:[ 3 sin x - 1 - x - 1 + 2e^{-2x} < 0 ][ 3 sin x - x - 2 + 2e^{-2x} < 0 ]So, we need to show:[ 3 sin x - x - 2 + 2e^{-2x} < 0 quad text{for all } x > 0 ]Hmm, let me rearrange this:[ 3 sin x - x - 2 < -2e^{-2x} ]Multiply both sides by ( e^{2x} ) (which is positive, so inequality sign remains the same):[ e^{2x}(3 sin x - x - 2) < -2 ]So, define a function ( F(x) = e^{2x}(3 sin x - x - 2) ). We need to show that ( F(x) < -2 ) for all ( x > 0 ).Let me compute the derivative of ( F(x) ) to analyze its behavior.First, ( F(x) = e^{2x}(3 sin x - x - 2) ).Using the product rule:[ F'(x) = frac{d}{dx} [e^{2x}] cdot (3 sin x - x - 2) + e^{2x} cdot frac{d}{dx} [3 sin x - x - 2] ][ F'(x) = 2e^{2x}(3 sin x - x - 2) + e^{2x}(3 cos x - 1) ][ F'(x) = e^{2x} [2(3 sin x - x - 2) + (3 cos x - 1)] ][ F'(x) = e^{2x} [6 sin x - 2x - 4 + 3 cos x - 1] ][ F'(x) = e^{2x} (6 sin x - 2x + 3 cos x - 5) ]So,[ F'(x) = e^{2x} (6 sin x - 2x + 3 cos x - 5) ]Now, since ( e^{2x} > 0 ) for all ( x ), the sign of ( F'(x) ) depends on the expression ( 6 sin x - 2x + 3 cos x - 5 ).Let me denote this expression as ( G(x) = 6 sin x - 2x + 3 cos x - 5 ).We need to analyze ( G(x) ) for ( x > 0 ).Hmm, perhaps I can find an upper bound for ( G(x) ) to show that it's always negative, which would imply ( F'(x) < 0 ), meaning ( F(x) ) is decreasing.Let me try to bound ( G(x) ).First, note that ( sin x ) and ( cos x ) are bounded between -1 and 1.So, ( 6 sin x leq 6 ) and ( 3 cos x leq 3 ).But that might not be tight enough.Alternatively, perhaps I can use some inequalities.Wait, let's consider the term ( -2x ). For ( x > 0 ), this is negative and becomes more negative as ( x ) increases.Also, ( 6 sin x + 3 cos x ) can be written as ( R sin(x + phi) ), where ( R = sqrt{6^2 + 3^2} = sqrt{45} = 3sqrt{5} approx 6.708 ).So, ( 6 sin x + 3 cos x = 3sqrt{5} sin(x + phi) ), where ( phi = arctanleft( frac{3}{6} right) = arctanleft( frac{1}{2} right) approx 0.4636 ) radians.Therefore,[ G(x) = 3sqrt{5} sin(x + phi) - 2x - 5 ]Now, the maximum value of ( 3sqrt{5} sin(x + phi) ) is ( 3sqrt{5} approx 6.708 ).So,[ G(x) leq 6.708 - 2x - 5 = 1.708 - 2x ]For ( x > 0 ), ( 1.708 - 2x ) is decreasing and becomes negative when ( x > 0.854 ).But we need to consider all ( x > 0 ), including ( x ) close to 0.Wait, let's evaluate ( G(x) ) at ( x = 0 ):[ G(0) = 6 cdot 0 - 0 + 3 cdot 1 - 5 = 3 - 5 = -2 ]So, at ( x = 0 ), ( G(0) = -2 ).Now, let's check the derivative of ( G(x) ):[ G'(x) = 6 cos x - 2 - 3 sin x ]Hmm, not sure if that helps directly.Alternatively, perhaps I can use the fact that ( sin x leq x ) for ( x geq 0 ). Wait, actually, ( sin x leq x ) for ( x geq 0 ), but here we have ( 6 sin x ). Hmm, not sure.Alternatively, let's consider that ( 6 sin x - 2x + 3 cos x - 5 leq 6 cdot 1 - 2x + 3 cdot 1 - 5 = 6 - 2x + 3 - 5 = 4 - 2x ).So,[ G(x) leq 4 - 2x ]For ( x > 2 ), this is negative, but for ( 0 < x < 2 ), it's positive or negative?At ( x = 1 ), ( 4 - 2(1) = 2 > 0 ).But we need a tighter bound.Wait, perhaps I can use the inequality ( sin x leq x ) for ( x geq 0 ).So,[ 6 sin x leq 6x ][ 3 cos x leq 3 ]Thus,[ G(x) = 6 sin x - 2x + 3 cos x - 5 leq 6x - 2x + 3 - 5 = 4x - 2 ]So,[ G(x) leq 4x - 2 ]For ( x > 0.5 ), this is positive, but for ( x < 0.5 ), it's negative.Hmm, not helpful for all ( x > 0 ).Wait, perhaps another approach. Let's consider the function ( G(x) = 6 sin x - 2x + 3 cos x - 5 ).Let me compute its maximum value.To find the maximum, set ( G'(x) = 0 ):[ G'(x) = 6 cos x - 2 - 3 sin x = 0 ][ 6 cos x - 3 sin x = 2 ]This is a bit complicated, but perhaps I can write it as:[ 6 cos x - 3 sin x = 2 ]Divide both sides by 3:[ 2 cos x - sin x = frac{2}{3} ]Let me write this as:[ 2 cos x - sin x = frac{2}{3} ]This is a linear combination of sine and cosine. Let me write it as ( R cos(x + phi) = frac{2}{3} ), where ( R = sqrt{2^2 + (-1)^2} = sqrt{5} approx 2.236 ).So,[ sqrt{5} cos(x + phi) = frac{2}{3} ][ cos(x + phi) = frac{2}{3 sqrt{5}} approx frac{2}{6.708} approx 0.298 ]So, there are solutions for ( x ), meaning that ( G(x) ) has critical points.But this might not be the most straightforward way.Alternatively, perhaps I can consider that ( G(x) ) is always negative for ( x > 0 ).Wait, at ( x = 0 ), ( G(0) = -2 ).Let me compute ( G(x) ) at ( x = pi/2 ):[ G(pi/2) = 6 cdot 1 - 2 cdot frac{pi}{2} + 3 cdot 0 - 5 = 6 - pi - 5 = 1 - pi approx -2.1416 ]Negative.At ( x = pi ):[ G(pi) = 6 cdot 0 - 2pi + 3 cdot (-1) - 5 = -2pi - 3 - 5 approx -6.283 - 8 = -14.283 ]Negative.At ( x = pi/4 ):[ G(pi/4) = 6 cdot frac{sqrt{2}}{2} - 2 cdot frac{pi}{4} + 3 cdot frac{sqrt{2}}{2} - 5 ][ = 3sqrt{2} - frac{pi}{2} + frac{3sqrt{2}}{2} - 5 ][ = frac{9sqrt{2}}{2} - frac{pi}{2} - 5 ]Approximately:[ frac{9 cdot 1.414}{2} - frac{3.1416}{2} - 5 ][ approx frac{12.726}{2} - 1.5708 - 5 ][ approx 6.363 - 1.5708 - 5 approx -0.2078 ]Still negative.At ( x = pi/6 ):[ G(pi/6) = 6 cdot frac{1}{2} - 2 cdot frac{pi}{6} + 3 cdot frac{sqrt{3}}{2} - 5 ][ = 3 - frac{pi}{3} + frac{3sqrt{3}}{2} - 5 ]Approximately:[ 3 - 1.0472 + 2.598 - 5 ][ approx 3 - 1.0472 + 2.598 - 5 approx -0.4492 ]Negative.Hmm, so at several points, ( G(x) ) is negative. Maybe ( G(x) ) is always negative for ( x > 0 ).Wait, but earlier, when I tried to bound ( G(x) ), I saw that for small ( x ), ( G(x) ) could be positive or negative.Wait, let me check ( x = 0.1 ):[ G(0.1) = 6 sin(0.1) - 2(0.1) + 3 cos(0.1) - 5 ]Approximately:[ 6(0.0998) - 0.2 + 3(0.995) - 5 ][ approx 0.5988 - 0.2 + 2.985 - 5 ][ approx 0.5988 - 0.2 = 0.3988; 0.3988 + 2.985 = 3.3838; 3.3838 - 5 = -1.6162 ]Negative.At ( x = 0.5 ):[ G(0.5) = 6 sin(0.5) - 2(0.5) + 3 cos(0.5) - 5 ]Approximately:[ 6(0.4794) - 1 + 3(0.8776) - 5 ][ approx 2.8764 - 1 + 2.6328 - 5 ][ approx 2.8764 - 1 = 1.8764; 1.8764 + 2.6328 = 4.5092; 4.5092 - 5 = -0.4908 ]Still negative.Wait, so maybe ( G(x) ) is always negative for ( x > 0 ). If that's the case, then ( F'(x) = e^{2x} G(x) < 0 ) for all ( x > 0 ), meaning ( F(x) ) is decreasing on ( (0, +infty) ).Since ( F(x) ) is decreasing, and as ( x to 0^+ ), let's compute the limit:[ lim_{x to 0^+} F(x) = lim_{x to 0} e^{2x}(3 sin x - x - 2) ][ = 1 cdot (0 - 0 - 2) = -2 ]So, ( F(x) ) approaches -2 as ( x to 0^+ ).Since ( F(x) ) is decreasing, for ( x > 0 ), ( F(x) < F(0) = -2 ).Therefore,[ F(x) = e^{2x}(3 sin x - x - 2) < -2 ]Which is exactly what we needed to show.Thus, the inequality ( f(x) < e^x + x + 1 - 2e^{-2x} ) holds for all ( x in (0, +infty) ).**Final Answer**(1) (boxed{text{g(x) is decreasing on } left( -frac{pi}{2}, 0 right) text{ and increasing on } left( 0, frac{3pi}{2} right)})(2) (boxed{f(x) < e^{x} + x + 1 - 2e^{-2x} text{ for all } x in (0, +infty)})

💡Alright, so I have this problem: I need to find the largest prime number of the form (2^n + n^2 - 1) that's less than 100, and (n) itself has to be a prime number. Hmm, okay. Let me try to figure this out step by step.First, I know that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, (n) has to be a prime number, and when I plug it into the expression (2^n + n^2 - 1), the result also needs to be a prime number less than 100.Let me list out the prime numbers first because (n) has to be prime. The prime numbers less than, say, 10 are 2, 3, 5, 7, 11, 13, etc. But since the result (2^n + n^2 - 1) has to be less than 100, I probably don't need to go too high with (n). Let me test with smaller primes first.Starting with (n = 2):(2^2 + 2^2 - 1 = 4 + 4 - 1 = 7). Okay, 7 is a prime number, so that's good. But is it the largest? Let's check the next prime.Next, (n = 3):(2^3 + 3^2 - 1 = 8 + 9 - 1 = 16). Hmm, 16 is not a prime number because it's divisible by 2, 4, and 8. So, that doesn't work.Moving on to (n = 5):(2^5 + 5^2 - 1 = 32 + 25 - 1 = 56). 56 is also not a prime number; it's divisible by 2, 4, 7, 8, etc. So, that's out.Next prime is (n = 7):(2^7 + 7^2 - 1 = 128 + 49 - 1 = 176). Wait, 176 is way above 100, so that's too big. Plus, even if it were under 100, 176 is not a prime number. So, that's no good.What about (n = 11)? Let me check:(2^{11} + 11^2 - 1 = 2048 + 121 - 1 = 2168). That's way over 100, so definitely not needed.Wait a second, maybe I went too far with (n = 7). Let me double-check the calculations for (n = 7). (2^7) is 128, (7^2) is 49, so 128 + 49 is 177, minus 1 is 176. Yeah, that's correct. So, 176 is too big.So, up to (n = 7), the only prime result we got was 7 when (n = 2). But wait, is there a prime number between (n = 2) and (n = 7) that I might have missed?Let me see, primes less than 7 are 2, 3, 5. I've already checked those. So, no, I haven't missed any.But just to be thorough, let me check (n = 13) just in case, even though it's likely over 100:(2^{13} + 13^2 - 1 = 8192 + 169 - 1 = 8360). Yep, way over 100.So, it seems like the only prime number of the form (2^n + n^2 - 1) that's less than 100 is 7 when (n = 2). But wait, the question asks for the largest such prime. Is there a possibility that a higher (n) could result in a prime number less than 100?Let me think. Maybe I should check (n = 1), but 1 isn't a prime number, so that's out. What about (n = 0)? No, 0 isn't prime either. So, the next prime after 2 is 3, which we've already checked.Wait, maybe I made a mistake with (n = 3). Let me recalculate:(2^3 = 8), (3^2 = 9), so 8 + 9 = 17, minus 1 is 16. Yeah, that's correct. 16 isn't prime.Hmm, so maybe 7 is indeed the only prime number that fits the criteria. But just to be absolutely sure, let me check (n = 11) again:(2^{11} = 2048), which is way over 100, so no need to consider that.Wait, what about (n = 5)? (2^5 = 32), (5^2 = 25), so 32 + 25 = 57, minus 1 is 56. 56 isn't prime. Okay, that's correct.So, after checking all primes (n) where (2^n + n^2 - 1) is less than 100, the only prime result I get is 7 when (n = 2). Therefore, 7 must be the largest prime number of that form under 100.But hold on, is there a chance that a higher (n) could give a prime number just below 100? Let me see. For example, (n = 7) gives 176, which is over 100. What about (n = 5)? 56 is under 100 but not prime. (n = 3) gives 16, which is under 100 but not prime. So, no, there's no prime result between 7 and 100 for this form.Therefore, I'm confident that the largest prime number of the form (2^n + n^2 - 1) that is less than 100, where (n) is also a prime number, is 7.

💡Okay, so I have this problem here about pegs on a board, spaced 1 unit apart both horizontally and vertically. There's a rubber band stretched over four pegs forming a quadrilateral, and I need to find its area. The answer choices are from 4 to 6, so it's multiple choice, which is helpful.First, I should probably visualize the figure they're talking about. Since it's an Asymptote figure, I can imagine a grid with pegs at integer coordinates. The rubber band is stretched over four pegs, forming a quadrilateral. The Asymptote code draws lines between (0,1), (1,3), (4,1), and (3,0), connecting them in order. So, the quadrilateral has vertices at these four points.Let me write down the coordinates of the four vertices to make it clearer:1. (0,1)2. (1,3)3. (4,1)4. (3,0)Hmm, okay. So, these are the four points that form the quadrilateral. I need to find the area of this shape. Since it's a quadrilateral, one way to find the area is to use the shoelace formula, which is a mathematical algorithm to determine the area of a polygon when given the coordinates of its vertices.But before jumping into that, maybe I can try to visualize the shape or break it down into simpler shapes whose areas I can calculate and then add or subtract accordingly.Let me plot these points mentally:- Starting at (0,1), which is on the left side, one unit up from the origin.- Then moving to (1,3), which is one unit to the right and two units up.- Next, to (4,1), which is three units to the right and two units down.- Then to (3,0), which is one unit to the left and one unit down.- Finally, back to (0,1).So, this quadrilateral seems to have two sides that are slanting upwards and two that are slanting downwards. It might be a convex quadrilateral, but I'm not entirely sure. Maybe it's a trapezoid or some kind of kite? Hmm, not sure.Alternatively, perhaps using the shoelace formula is the most straightforward method here. I remember the shoelace formula works for any polygon given their coordinates, so that might be a good approach.Let me recall the shoelace formula. For a polygon with vertices (x1, y1), (x2, y2), ..., (xn, yn), the area is given by:Area = |(x1y2 + x2y3 + ... + xn y1) - (y1x2 + y2x3 + ... + ynx1)| / 2So, I need to apply this formula to the four points given.Let me list the coordinates again in order:1. (0,1)2. (1,3)3. (4,1)4. (3,0)And then back to the first point to complete the cycle:5. (0,1)So, applying the shoelace formula, I need to compute two sums:Sum1 = (x1y2) + (x2y3) + (x3y4) + (x4y5)Sum2 = (y1x2) + (y2x3) + (y3x4) + (y4x5)Then, subtract Sum2 from Sum1 and take the absolute value, then divide by 2.Let me compute Sum1 first:x1y2 = 0 * 3 = 0x2y3 = 1 * 1 = 1x3y4 = 4 * 0 = 0x4y5 = 3 * 1 = 3So, Sum1 = 0 + 1 + 0 + 3 = 4Now, Sum2:y1x2 = 1 * 1 = 1y2x3 = 3 * 4 = 12y3x4 = 1 * 3 = 3y4x5 = 0 * 0 = 0So, Sum2 = 1 + 12 + 3 + 0 = 16Now, subtract Sum2 from Sum1:4 - 16 = -12Take the absolute value: |-12| = 12Divide by 2: 12 / 2 = 6So, according to the shoelace formula, the area is 6 square units. Hmm, that's one of the answer choices, option E.Wait, but before I conclude, let me double-check my calculations because sometimes it's easy to make a mistake with the coordinates.Let me write down the coordinates again:1. (0,1)2. (1,3)3. (4,1)4. (3,0)5. (0,1)Compute Sum1:x1y2: 0*3 = 0x2y3: 1*1 = 1x3y4: 4*0 = 0x4y5: 3*1 = 3Sum1: 0 + 1 + 0 + 3 = 4Sum2:y1x2: 1*1 = 1y2x3: 3*4 = 12y3x4: 1*3 = 3y4x5: 0*0 = 0Sum2: 1 + 12 + 3 + 0 = 16Difference: 4 - 16 = -12Absolute value: 12Area: 12 / 2 = 6Okay, that seems consistent. So, the area is 6.Alternatively, maybe I can try another method to confirm, like dividing the quadrilateral into triangles and calculating their areas separately.Let me see. If I draw a diagonal from (0,1) to (4,1), that would split the quadrilateral into two triangles: one with vertices (0,1), (1,3), (4,1), and another with vertices (0,1), (4,1), (3,0).Wait, but actually, that's not quite right because the quadrilateral is (0,1), (1,3), (4,1), (3,0). So, if I draw a diagonal from (1,3) to (3,0), that might split it into two triangles.Alternatively, maybe it's better to use the shoelace formula as I did before because it's straightforward.But just to make sure, let me try another approach. Maybe using vectors or coordinate geometry.Alternatively, I can use the area formula for a trapezoid, but I need to check if this quadrilateral is a trapezoid. A trapezoid has at least one pair of parallel sides.Looking at the sides:From (0,1) to (1,3): slope is (3-1)/(1-0) = 2/1 = 2From (1,3) to (4,1): slope is (1-3)/(4-1) = (-2)/3From (4,1) to (3,0): slope is (0-1)/(3-4) = (-1)/(-1) = 1From (3,0) to (0,1): slope is (1-0)/(0-3) = 1/(-3) = -1/3So, none of these slopes are equal, so there are no parallel sides. Therefore, it's not a trapezoid. So, that approach won't work.Alternatively, maybe I can use the vector cross product method. For a polygon, the area can be calculated by summing the cross products of consecutive vectors.But that's essentially the shoelace formula, so I think I'm back to the same method.Alternatively, maybe I can use the formula for the area of a quadrilateral using coordinates, which is similar to the shoelace formula.Wait, actually, the shoelace formula is the most reliable here because it directly uses the coordinates.Alternatively, maybe I can use the area formula for a polygon by dividing it into triangles.Let me try that.So, if I pick a point inside the quadrilateral and connect it to all four vertices, I can divide the quadrilateral into four triangles. But that might complicate things.Alternatively, I can split the quadrilateral into two triangles by drawing a diagonal.Let me try that.Let's split the quadrilateral into two triangles: Triangle 1 with vertices (0,1), (1,3), (4,1), and Triangle 2 with vertices (0,1), (4,1), (3,0).Wait, but actually, that's not the case because the quadrilateral is (0,1), (1,3), (4,1), (3,0). So, if I draw a diagonal from (1,3) to (3,0), I can split it into two triangles: Triangle A: (0,1), (1,3), (3,0), and Triangle B: (1,3), (4,1), (3,0).Wait, that might not cover the entire quadrilateral. Let me make sure.Alternatively, maybe I should split it from (0,1) to (4,1), which is a horizontal line. Then, Triangle 1 would be (0,1), (1,3), (4,1), and Triangle 2 would be (0,1), (4,1), (3,0).Let me calculate the area of each triangle.For Triangle 1: (0,1), (1,3), (4,1)Using the shoelace formula for triangles:Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2|Plugging in the coordinates:x1 = 0, y1 = 1x2 = 1, y2 = 3x3 = 4, y3 = 1So,Area = |0*(3 - 1) + 1*(1 - 1) + 4*(1 - 3)| / 2Simplify:= |0 + 1*0 + 4*(-2)| / 2= |0 + 0 - 8| / 2= | -8 | / 2= 8 / 2 = 4So, Area of Triangle 1 is 4.Now, Triangle 2: (0,1), (4,1), (3,0)Again, using the shoelace formula:x1 = 0, y1 = 1x2 = 4, y2 = 1x3 = 3, y3 = 0Area = |0*(1 - 0) + 4*(0 - 1) + 3*(1 - 1)| / 2Simplify:= |0 + 4*(-1) + 3*0| / 2= |0 - 4 + 0| / 2= | -4 | / 2= 4 / 2 = 2So, Area of Triangle 2 is 2.Therefore, total area of the quadrilateral is 4 + 2 = 6.Okay, that matches the shoelace formula result. So, that's reassuring.Alternatively, maybe I can use Pick's Theorem, which is another method to find the area of a polygon with integer coordinates.Pick's Theorem states that the area A of a simple polygon with integer coordinates is:A = I + B/2 - 1where I is the number of interior lattice points, and B is the number of boundary lattice points.So, to use Pick's Theorem, I need to count the number of interior points and boundary points of the quadrilateral.But since I don't have the actual figure in front of me, it's a bit tricky, but maybe I can estimate.Alternatively, since I already have the area as 6 from the shoelace formula, maybe I can check if Pick's Theorem gives the same result.But to do that, I need to count I and B.Alternatively, maybe it's better to stick with the shoelace formula since it's more straightforward in this case.Wait, but just to try, let me see.First, I need to count the number of boundary points, B.Boundary points are the lattice points on the edges of the quadrilateral.Each edge is a line segment between two vertices.So, let's consider each side:1. From (0,1) to (1,3): The change in x is 1, change in y is 2. So, the number of lattice points on this edge is gcd(1,2) + 1 = 1 + 1 = 2. But since the endpoints are already counted, the number of boundary points on this edge is 2, but excluding the endpoints, it's 0. Wait, no, actually, the formula is that the number of lattice points on a line segment between two points is gcd(delta_x, delta_y) + 1. So, for this edge, gcd(1,2) = 1, so number of lattice points is 2. But since we're counting boundary points, we include all points on the edges, so for each edge, the number of boundary points is gcd(delta_x, delta_y) + 1, but we have to be careful not to double-count the vertices.Wait, maybe it's better to calculate the total number of boundary points by summing the number of lattice points on each edge and then subtracting the 4 vertices which are counted twice.So, let's do that.For each edge:1. From (0,1) to (1,3): delta_x = 1, delta_y = 2. gcd(1,2) = 1. So, number of lattice points = 1 + 1 = 2.2. From (1,3) to (4,1): delta_x = 3, delta_y = -2. gcd(3,2) = 1. So, number of lattice points = 1 + 1 = 2.3. From (4,1) to (3,0): delta_x = -1, delta_y = -1. gcd(1,1) = 1. So, number of lattice points = 1 + 1 = 2.4. From (3,0) to (0,1): delta_x = -3, delta_y = 1. gcd(3,1) = 1. So, number of lattice points = 1 + 1 = 2.So, total boundary points on all edges: 2 + 2 + 2 + 2 = 8.But each vertex is shared by two edges, so we've counted each vertex twice. There are 4 vertices, so we need to subtract 4 to avoid double-counting.Therefore, total boundary points B = 8 - 4 = 4.Wait, that seems low. Let me think again.Wait, no, actually, the formula is that the number of boundary points on each edge is gcd(delta_x, delta_y) + 1, which includes the endpoints. So, when we sum all edges, we get the total number of boundary points, but each vertex is shared by two edges, so we've counted each vertex twice. Therefore, the total boundary points B is (sum of (gcd(delta_x, delta_y) + 1) for all edges) - number of vertices.So, in this case:Sum of (gcd + 1) for each edge:Edge 1: 1 + 1 = 2Edge 2: 1 + 1 = 2Edge 3: 1 + 1 = 2Edge 4: 1 + 1 = 2Total sum: 2 + 2 + 2 + 2 = 8Number of vertices: 4Therefore, total boundary points B = 8 - 4 = 4.Wait, that seems correct. So, B = 4.Now, according to Pick's Theorem, A = I + B/2 - 1.We know A is 6 from the shoelace formula, so:6 = I + 4/2 - 1Simplify:6 = I + 2 - 16 = I + 1Therefore, I = 5.So, there are 5 interior points.But without seeing the figure, it's hard to verify, but since the shoelace formula gave us 6, and Pick's Theorem corroborates it with I = 5 and B = 4, that seems consistent.Therefore, the area is indeed 6.Alternatively, another way to think about it is to use coordinate geometry to find the area.But I think I've already confirmed it sufficiently with two methods: shoelace formula and decomposition into triangles, both giving the same result.So, I'm confident that the area is 6.**Final Answer**boxed{E}

💡First, I need to determine the base cost of the cell phone plan, which is a fixed 25 per month.Next, I'll calculate the cost for the text messages. Alex sent 150 text messages, and each text costs 0.10. So, the total cost for texts is 150 multiplied by 0.10, which equals 15.Then, I'll calculate the cost for the minutes talked. Alex talked for a total of 28 hours. Since the plan includes 25 hours, I'll convert both hours to minutes for easier calculation. 28 hours is 1680 minutes, and 25 hours is 1500 minutes. The extra minutes used are 1680 minus 1500, which equals 180 minutes.For the included 1500 minutes, the cost is 1500 minutes multiplied by 0.05 per minute, totaling 75. For the extra 180 minutes, the cost is 180 minutes multiplied by 0.15 per minute, totaling 27.Finally, I'll add up all these costs: 25 for the base plan, 15 for texts, 75 for included minutes, and 27 for extra minutes. The total amount Alex has to pay is 142.

💡Okay, so I have this equation to solve: √(1 + sin6) + √(1 - sin6). And I need to figure out which of the options A, B, C, or D it equals. The options are all in terms of sin3 or cos3, which makes me think I need to use some trigonometric identities to simplify the expression.First, I remember that expressions with square roots and sine or cosine can sometimes be simplified using identities. Maybe I can use the identity for sin of a double angle or something like that. Let me think... Oh, wait, there's an identity that relates 1 ± sinθ to something squared. Let me recall it.I think it's something like 1 + sinθ = (sin(θ/2) + cos(θ/2))². Let me check that. If I expand (sin(θ/2) + cos(θ/2))², I get sin²(θ/2) + 2sin(θ/2)cos(θ/2) + cos²(θ/2). Since sin²x + cos²x = 1, that simplifies to 1 + sinθ. Yes, that's correct! Similarly, 1 - sinθ would be (sin(θ/2) - cos(θ/2))².So, applying this to my problem, where θ is 6, I can rewrite the square roots:√(1 + sin6) = √[(sin3 + cos3)²] = |sin3 + cos3|Similarly, √(1 - sin6) = √[(sin3 - cos3)²] = |sin3 - cos3|So now, my expression becomes |sin3 + cos3| + |sin3 - cos3|Hmm, okay. Now, I need to figure out the signs inside the absolute values. Since 3 radians is approximately 171.9 degrees, which is in the second quadrant where sine is positive and cosine is negative. So, sin3 is positive, and cos3 is negative.Therefore, sin3 + cos3: sin3 is positive, cos3 is negative. Depending on their magnitudes, this could be positive or negative. Similarly, sin3 - cos3: sin3 is positive, and subtracting a negative would make it sin3 + |cos3|, which is definitely positive.Wait, let me calculate sin3 and cos3 numerically to get a better sense.Calculating sin3: 3 radians is about 171.9 degrees. Sin(3) ≈ 0.1411.Calculating cos3: Cos(3) ≈ -0.98999.So, sin3 ≈ 0.1411, cos3 ≈ -0.98999.Therefore, sin3 + cos3 ≈ 0.1411 - 0.98999 ≈ -0.84889, which is negative.And sin3 - cos3 ≈ 0.1411 - (-0.98999) ≈ 0.1411 + 0.98999 ≈ 1.1311, which is positive.So, |sin3 + cos3| = |-0.84889| = 0.84889|sin3 - cos3| = |1.1311| = 1.1311Adding them together: 0.84889 + 1.1311 ≈ 1.98, which is approximately 2.Wait, but the options are 2sin3, -2sin3, 2cos3, -2cos3. Let me compute those:2sin3 ≈ 2 * 0.1411 ≈ 0.2822-2sin3 ≈ -0.28222cos3 ≈ 2 * (-0.98999) ≈ -1.97998-2cos3 ≈ 1.97998Hmm, so my computed value is approximately 1.98, which is close to -2cos3, since -2cos3 ≈ 1.97998. So, that would be option D.But wait, let me think again. When I took the square roots, I had to consider the absolute values. Since sin3 + cos3 was negative, its absolute value is positive, and sin3 - cos3 was positive, so its absolute value is itself. So, adding them together, I got approximately 1.98, which is close to 2. But in the options, the closest is -2cos3, which is approximately 1.98. So, that seems to match.But let me try another approach to confirm. Maybe squaring the original expression.Let me denote S = √(1 + sin6) + √(1 - sin6). Then, S² = (1 + sin6) + (1 - sin6) + 2√{(1 + sin6)(1 - sin6)}.Simplifying, S² = 2 + 2√(1 - sin²6). But 1 - sin²6 is cos²6, so S² = 2 + 2|cos6|.Now, cos6: 6 radians is about 343.8 degrees, which is in the fourth quadrant where cosine is positive. So, cos6 is positive. Therefore, |cos6| = cos6.Thus, S² = 2 + 2cos6.Now, I can use the double-angle identity: cos6 = 2cos²3 - 1.So, substituting, S² = 2 + 2(2cos²3 - 1) = 2 + 4cos²3 - 2 = 4cos²3.Therefore, S² = 4cos²3, so S = ±2|cos3|.But since S is a sum of square roots, which are always non-negative, S must be non-negative. So, S = 2|cos3|.But earlier, I found that cos3 is negative, so |cos3| = -cos3. Therefore, S = 2*(-cos3) = -2cos3.So, that's option D.Wait, let me verify once more. If I compute -2cos3, cos3 is approximately -0.98999, so -2cos3 ≈ 1.97998, which is approximately 2, which matches my earlier calculation. So, that seems consistent.Therefore, the answer should be D: -2cos3.**Final Answer**boxed{D}