answer:(1) Since f(x)=sin{x}, The derivative f'(x)=cos{x}. The slope of the tangent line at the point Pleft(dfrac{pi}{4}, fleft(dfrac{pi}{4}right)right) is k=f'left(dfrac{pi}{4}right) = cos{dfrac{pi}{4}}=dfrac{sqrt{2}}{2}. Since fleft(dfrac{pi}{4}right)=sin{dfrac{pi}{4}}=dfrac{sqrt{2}}{2}, the coordinates of the tangent point are left(dfrac{pi}{4}, dfrac{sqrt{2}}{2}right). Therefore, the equation of the tangent line is y-dfrac{sqrt{2}}{2}=dfrac{sqrt{2}}{2}left(x-dfrac{pi}{4}right), which simplifies to x-sqrt{2}y+1-dfrac{pi}{4}=0. boxed{y=frac{sqrt{2}}{2}x+frac{1}{2}-frac{sqrt{2}pi}{8}} (2) The derivative of g(x) is g'(x)=m-dfrac{3}{6}x^2=m-dfrac{1}{2}x^2. Setting the derivative to zero gives x^2=2m. For mleq 0, g'(x)<0, and g(x) is monotonically decreasing over all x. For m>0, the condition g'(x)leq 0 implies g(x) is monotonically decreasing when x^2geq 2m. Thus, the intervals of monotonous decrease are (-infty, -sqrt{2m}] or [sqrt{2m}, +infty). boxed{text{Decreasing intervals: }(-infty, -sqrt{2m}]cup[sqrt{2m}, +infty)} (3) When m=1, the function g(x) becomes g(x)=x-dfrac{x^3}{6}. Define the function h(x)=g(x)+dfrac{x^3}{6}-f(x)=x-sin{x}. The derivative of h(x) is h'(x)=1-cos{x}. For x>0, cos{x}leq 1. Hence h'(x)=1-cos{x}geq 0, which implies the function h(x) is monotonically increasing. When x=0, h(0)=0. Therefore, for x>0, h(x)>0, which means that boxed{f(x)<g(x)+dfrac{x^3}{6}} text{ is true}.

answer:1. Consider point ( P(x, y) ) on the ellipse ( C_1 ) with foci ( F_1(-c, 0) ) and ( F_2(c, 0) ). The ellipse can be described by the equation [ frac{x^2}{a^2} + frac{y^2}{b^2} = 1 ] where ( c = sqrt{a^2 - b^2} ). 2. The position vectors from point ( P ) to the foci ( F_1 ) and ( F_2 ) are: [ overrightarrow{P F_1} = (-c - x, -y), quad overrightarrow{P F_2} = (c - x, -y). ] 3. The dot product ( overrightarrow{P F_1} cdot overrightarrow{P F_2} ) can be computed as follows: [ overrightarrow{P F_1} cdot overrightarrow{P F_2} = (-c - x)(c - x) + (-y)(-y) ] Simplifying this, we get: [ overrightarrow{P F_1} cdot overrightarrow{P F_2} = -c^2 - cx + cx + x^2 + y^2 = x^2 + y^2 - c^2. ] 4. Substituting ( y^2 = b^2 - frac{b^2 x^2}{a^2} ) (from the ellipse equation), we have: [ overrightarrow{P F_1} cdot overrightarrow{P F_2} = x^2 + b^2 - frac{b^2 x^2}{a^2} - c^2. ] Simplify this expression: [ overrightarrow{P F_1} cdot overrightarrow{P F_2} = x^2 left( 1 - frac{b^2}{a^2} right) + b^2 - c^2 = frac{c^2}{a^2} x^2 + b^2 - c^2. ] 5. To find the range of ( overrightarrow{P F_1} cdot overrightarrow{P F_2} ), we evaluate it from ( x = 0 ) to ( x = pm a ): [ overrightarrow{P F_1} cdot overrightarrow{P F_2} = frac{c^2}{a^2} cdot a^2 + b^2 - c^2 = c^2 + b^2 - c^2 = b^2. ] Given that ( c^2 leq b^2 leq 3c^2 ), we have: [ c^2 leq b^2 leq 3c^2. ] 6. From the relationships between ( a^2, b^2, ) and ( c^2 ): [ c^2 leq a^2 - c^2 leq 3c^2 implies frac{1}{4} leq frac{c^2}{a^2} leq frac{1}{2}, ] hence: [ frac{1}{4} leq e^2 leq frac{1}{2} implies frac{1}{2} leq e leq frac{sqrt{2}}{2}. ] For part (2): 1. When ( e = frac{1}{2} ), we have ( a = 2c ) and ( b = sqrt{3} c ). Therefore, the hyperbola ( C_2 ) has the equation [ frac{x^2}{c^2} - frac{y^2}{3c^2} = 1, ] and vertex ( A(2c, 0) ). 2. Let ( B(x_0, y_0) ) be a point on ( C_2 ) in the first quadrant, satisfying [ frac{x_0^2}{c^2} - frac{y_0^2}{3c^2} = 1. ] 3. When ( AB ) is perpendicular to the x-axis, we have ( x_0 = 2c ) and ( y_0 = 3c ). Then: [ tan angle BF_1A = frac{3c}{3c} = 1 implies angle BF_1A = frac{pi}{4}. ] Thus: [ angle BAF_1 = frac{pi}{2} = 2 cdot frac{pi}{4} implies lambda = 2. ] 4. For ( x_0 neq 2c ): [ tan angle BAF_1 = frac{-y_0}{x_0 - a} = frac{-y_0}{x_0 - 2c}, ] and [ tan angle BF_1A = frac{y_0}{x_0 + c}. ] 5. Using trigonometric relationships: [ tan 2angle BF_1A = frac{2tan angle BF_1A}{1 - tan^2 angle BF_1A} = frac{2 cdot frac{y_0}{x_0 + c}}{1 - left(frac{y_0}{x_0 + c}right)^2}. ] 6. Since ( y_0^2 = 3c^2 left( frac{x_0^2}{c^2} - 1 right) = 3(x_0^2 - c^2) ), we get: [ tan 2angle BF_1A = frac{2y_0 cdot (x_0 + c)}{(x_0 + c)^2 - 3(x_0^2 - c^2)} = frac{-y_0}{x_0 - 2c} = tan angle BAF_1. ] 7. As ( 2angle BF_1A ) and ( angle BAF_1 ) lie in ( (0, frac{pi}{2}) cup (frac{pi}{2}, pi) ): [ 2 angle BF_1A = angle BAF_1 implies lambda = 2. ] Conclusion: There exists a constant ( lambda = 2 ) such that ( angle BAF_1 = lambda angle BF_1A ) holds. [ boxed{2} ]

answer:The simplified expression will contain terms for every monomial of the form (x^a y^b z^c w^d), where (a, b, c,) and (d) are non-negative integers, (a) is even, and (a+b+c+d=2008). 1. There are 1005 even values of (a) with (0 leq a leq 2008). 2. For each such value of (a), (b) can take any integer values between 0 and (2008-a), and the values of (c) and (d) are then uniquely determined as (2008-a-b-c) and (b+c), respectively. 3. Thus, the number of terms in the simplified expression is: [ (2009-0) + (2009-2) + cdots + (2009-2008) = 2009 + 2007 + cdots + 1 ] This sum is the sum of the first 1005 odd positive integers, which is ((1005)^2 = 1,010,025). Conclusion: The number of terms in the simplified expression is (boxed{1,010,025}).

answer:To find the height of the room, we can use the Pythagorean theorem in three dimensions. The longest pole that can be placed in the room will be the diagonal of the room, which can be calculated using the lengths of the room, the breadth of the room, and the height of the room. Let's denote the height of the room as ( h ). The diagonal of the room ( d ) can be calculated using the formula: [ d = sqrt{l^2 + b^2 + h^2} ] where ( l ) is the length of the room, ( b ) is the breadth of the room, and ( h ) is the height of the room. Given that the longest pole is 17 m long, we have: [ d = 17 ] The length ( l ) of the room is 12 m, and the breadth ( b ) of the room is 8 m. Plugging these values into the formula, we get: [ 17 = sqrt{12^2 + 8^2 + h^2} ] Squaring both sides to eliminate the square root, we get: [ 17^2 = 12^2 + 8^2 + h^2 ] [ 289 = 144 + 64 + h^2 ] [ 289 = 208 + h^2 ] Subtracting 208 from both sides, we get: [ h^2 = 289 - 208 ] [ h^2 = 81 ] Taking the square root of both sides, we find: [ h = sqrt{81} ] [ h = 9 ] Therefore, the height of the room is boxed{9} meters.