answer：Okay, so I have this problem about a trapezoid ABCD where the bases AB and CD are 8 and 17 units long, respectively. The legs of the trapezoid are extended beyond points A and B to meet at a point E. I need to find the ratio of the area of triangle EAB to the area of trapezoid ABCD. Hmm, okay, let me try to visualize this.First, I know that in a trapezoid, the two bases are parallel. So, AB is parallel to CD, with lengths 8 and 17. The legs are AD and BC, which are the non-parallel sides. When these legs are extended beyond A and B, they meet at point E. So, point E is formed by extending the legs until they intersect.I remember that when two sides of a trapezoid are extended, the triangles formed are similar. Specifically, triangle EAB and triangle EDC should be similar because they are both formed by the same point E and the two bases AB and CD. Since AB and CD are parallel, the corresponding angles at E should be equal, making the triangles similar by the AA (Angle-Angle) similarity criterion.So, if triangles EAB and EDC are similar, the ratio of their corresponding sides should be equal. The ratio of AB to CD is 8 to 17. Therefore, the ratio of similarity between triangle EAB and triangle EDC is 8:17.Now, since the triangles are similar, the ratio of their areas is the square of the ratio of their corresponding sides. So, the area ratio of triangle EAB to triangle EDC should be (8/17)^2, which is 64/289. That means the area of triangle EAB is 64/289 times the area of triangle EDC.But wait, the problem is asking for the ratio of the area of triangle EAB to the area of trapezoid ABCD, not triangle EDC. So, I need to relate the area of the trapezoid to the areas of these triangles.I recall that the area of the trapezoid ABCD can be found by subtracting the area of triangle EAB from the area of triangle EDC. In other words, [ABCD] = [EDC] - [EAB]. That makes sense because trapezoid ABCD is the part of triangle EDC that's left after removing triangle EAB.Let me denote the area of triangle EAB as A. Then, the area of triangle EDC would be A + [ABCD]. But from the similarity ratio, we know that [EDC] = (17/8)^2 * [EAB] = (289/64) * A. So, substituting back, we have:[EDC] = (289/64) * ABut [EDC] is also equal to A + [ABCD]So, setting them equal:(289/64) * A = A + [ABCD]Let me solve for [ABCD]:(289/64) * A - A = [ABCD](289/64 - 64/64) * A = [ABCD](225/64) * A = [ABCD]So, [ABCD] = (225/64) * ATherefore, the ratio of [EAB] to [ABCD] is A / (225/64 * A) = 64/225.Wait, so the ratio is 64/225. That seems straightforward, but let me double-check my steps.1. Identified that triangles EAB and EDC are similar with a side ratio of 8:17.2. Therefore, area ratio is (8/17)^2 = 64/289.3. Expressed [EDC] as [EAB] + [ABCD].4. Substituted [EDC] = (289/64)*[EAB] into the equation.5. Solved for [ABCD] in terms of [EAB], getting [ABCD] = (225/64)*[EAB].6. Therefore, the ratio [EAB]/[ABCD] is 64/225.Yes, that seems correct. I don't think I made any calculation errors. So, the ratio is 64/225.**Final Answer**The ratio of the area of triangle ( EAB ) to the area of trapezoid ( ABCD ) is boxed{dfrac{64}{225}}.

answer：First, I need to calculate the compound interest (C.I.) on Rs. 6000 for 2 years at 15% per annum using the formula:[ C.I. = P left(1 + frac{r}{100}right)^t - P ]where ( P = 6000 ), ( r = 15% ), and ( t = 2 ) years.Next, I'll compute the compound interest:[ C.I. = 6000 left(1 + 0.15right)^2 - 6000 ][ C.I. = 6000 times 1.3225 - 6000 ][ C.I. = 7935 - 6000 ][ C.I. = 1935 ]The simple interest (S.I.) is half of this compound interest:[ S.I. = frac{1935}{2} = 967.5 ]Using the simple interest formula:[ S.I. = frac{P times r times t}{100} ]where ( S.I. = 967.5 ), ( r = 12% ), and ( t = 4 ) years, I'll solve for ( P ):[ 967.5 = frac{P times 12 times 4}{100} ][ 967.5 = frac{48P}{100} ][ 967.5 = 0.48P ][ P = frac{967.5}{0.48} ][ P = 2015.625 ]Therefore, the sum placed on simple interest is Rs. 2015.625.

answer：First, I need to identify the formula for calculating the area of a rhombus when the lengths of its diagonals are known. The formula is:Area = (d₁ × d₂) / 2where d₁ and d₂ are the lengths of the diagonals.Next, I'll substitute the given values into the formula. The diagonals are 12 cm and 10 cm.Area = (12 cm × 10 cm) / 2Now, I'll perform the multiplication:12 cm × 10 cm = 120 cm²Finally, I'll divide by 2 to find the area:120 cm² / 2 = 60 cm²So, the area of the rhombus is 60 square centimeters.

answer：Okay, so I have this problem about four points on a parabola, and I need to find the minimum area of a triangle formed by some intersections and the focus of the parabola. Let me try to break this down step by step.First, the parabola given is ( y^2 = 4x ). I remember that the standard form of a parabola is ( y^2 = 4ax ), so in this case, ( a = 1 ). That means the focus ( F ) is at ( (1, 0) ). Got that.There are four points ( A_1, A_2, A_3, A_4 ) on the parabola. The problem says that ( A_1, F, A_3 ) are collinear, and ( A_2, F, A_4 ) are collinear. So, lines ( A_1A_3 ) and ( A_2A_4 ) both pass through the focus ( F ).I need to find the intersection points ( M ) and ( N ). ( M ) is where lines ( A_1A_2 ) and ( A_3A_4 ) meet, and ( N ) is where lines ( A_1A_4 ) and ( A_2A_3 ) meet. Then, I need to find the area of triangle ( FMN ) and its minimum value.Hmm, okay. Maybe parametrizing the points on the parabola would help. For a parabola ( y^2 = 4x ), a parametric representation is ( (at^2, 2at) ). Since ( a = 1 ), it's ( (t^2, 2t) ). Wait, but in the problem, the points are given as ( A_1, A_2, A_3, A_4 ). Maybe I can use parameters ( t_1, t_2, t_3, t_4 ) for these points.So, let me denote:- ( A_1 = (t_1^2, 2t_1) )- ( A_2 = (t_2^2, 2t_2) )- ( A_3 = (t_3^2, 2t_3) )- ( A_4 = (t_4^2, 2t_4) )But wait, the problem mentions the points are on ( y^2 = 4x ), so maybe scaling is different. Let me check: for ( y^2 = 4x ), the parametric equations are ( x = at^2 ), ( y = 2at ). So, if ( a = 1 ), it's ( x = t^2 ), ( y = 2t ). So, yes, ( A_1 = (t_1^2, 2t_1) ), etc.But in the problem, the points are given as ( A_1, A_2, A_3, A_4 ), so maybe I can assign parameters ( t_1, t_2, t_3, t_4 ) accordingly.Given that ( A_1, F, A_3 ) are collinear, so the line ( A_1A_3 ) passes through ( F(1, 0) ). Similarly, the line ( A_2A_4 ) passes through ( F(1, 0) ).So, maybe I can write the equations of these lines and use the fact that they pass through ( F ) to find relations between the parameters.Let me recall that the equation of a line through two points ( (x_1, y_1) ) and ( (x_2, y_2) ) is ( (y - y_1) = m(x - x_1) ), where ( m = frac{y_2 - y_1}{x_2 - x_1} ).Alternatively, for a parabola, the equation of the chord joining two points ( t_1 ) and ( t_2 ) is ( 2x - (t_1 + t_2)y + 2t_1 t_2 = 0 ). Wait, is that correct?Let me derive it. Let me take two points on the parabola ( y^2 = 4x ): ( (t_1^2, 2t_1) ) and ( (t_2^2, 2t_2) ). The slope ( m ) is ( frac{2t_2 - 2t_1}{t_2^2 - t_1^2} = frac{2(t_2 - t_1)}{(t_2 - t_1)(t_2 + t_1)} } = frac{2}{t_1 + t_2} ).So, the equation of the line is ( y - 2t_1 = frac{2}{t_1 + t_2}(x - t_1^2) ).Multiplying both sides by ( t_1 + t_2 ):( (t_1 + t_2)(y - 2t_1) = 2(x - t_1^2) )Expanding:( (t_1 + t_2)y - 2t_1(t_1 + t_2) = 2x - 2t_1^2 )Simplify:( (t_1 + t_2)y - 2t_1^2 - 2t_1 t_2 = 2x - 2t_1^2 )Subtract ( 2x ) from both sides and add ( 2t_1^2 ):( (t_1 + t_2)y - 2t_1 t_2 = 2x )Rearranged:( 2x - (t_1 + t_2)y + 2t_1 t_2 = 0 )Yes, that's correct.So, the equation of chord ( A_1A_3 ) is ( 2x - (t_1 + t_3)y + 2t_1 t_3 = 0 ).Since this line passes through ( F(1, 0) ), substituting ( x = 1 ), ( y = 0 ):( 2(1) - (t_1 + t_3)(0) + 2t_1 t_3 = 0 )Simplify:( 2 + 2t_1 t_3 = 0 )So, ( 2t_1 t_3 = -2 ) => ( t_1 t_3 = -1 )Therefore, ( t_3 = -1/t_1 )Similarly, for the line ( A_2A_4 ) passing through ( F(1, 0) ), the equation is ( 2x - (t_2 + t_4)y + 2t_2 t_4 = 0 ).Substituting ( F(1, 0) ):( 2(1) - (t_2 + t_4)(0) + 2t_2 t_4 = 0 )Simplify:( 2 + 2t_2 t_4 = 0 ) => ( t_2 t_4 = -1 ) => ( t_4 = -1/t_2 )Okay, so now we have relations:- ( t_3 = -1/t_1 )- ( t_4 = -1/t_2 )So, we can express all four points in terms of ( t_1 ) and ( t_2 ):- ( A_1 = (t_1^2, 2t_1) )- ( A_2 = (t_2^2, 2t_2) )- ( A_3 = ( ( -1/t_1 )^2, 2(-1/t_1) ) = (1/t_1^2, -2/t_1) )- ( A_4 = ( ( -1/t_2 )^2, 2(-1/t_2) ) = (1/t_2^2, -2/t_2) )Alright, now I need to find the intersection points ( M ) and ( N ).First, let's find ( M ), which is the intersection of lines ( A_1A_2 ) and ( A_3A_4 ).Let me find the equations of these two lines.Starting with line ( A_1A_2 ). Using the chord equation again:( 2x - (t_1 + t_2)y + 2t_1 t_2 = 0 )Similarly, line ( A_3A_4 ) has points ( A_3 = (1/t_1^2, -2/t_1) ) and ( A_4 = (1/t_2^2, -2/t_2) ). So, using the chord equation for these two points:Let me compute the equation of line ( A_3A_4 ). Using the chord equation formula:( 2x - (t_3 + t_4)y + 2t_3 t_4 = 0 )But ( t_3 = -1/t_1 ), ( t_4 = -1/t_2 ), so:( 2x - ( (-1/t_1) + (-1/t_2) ) y + 2*(-1/t_1)*(-1/t_2) = 0 )Simplify:( 2x - ( - (1/t_1 + 1/t_2) ) y + 2*(1/(t_1 t_2)) = 0 )Which is:( 2x + (1/t_1 + 1/t_2)y + 2/(t_1 t_2) = 0 )Multiply through by ( t_1 t_2 ) to eliminate denominators:( 2 t_1 t_2 x + (t_2 + t_1) y + 2 = 0 )So, equation of line ( A_3A_4 ) is:( 2 t_1 t_2 x + (t_1 + t_2) y + 2 = 0 )Now, we have the equations of lines ( A_1A_2 ) and ( A_3A_4 ):1. ( 2x - (t_1 + t_2)y + 2t_1 t_2 = 0 )2. ( 2 t_1 t_2 x + (t_1 + t_2) y + 2 = 0 )We need to solve these two equations to find point ( M ).Let me write them as:1. ( 2x - (t_1 + t_2)y = -2 t_1 t_2 )2. ( 2 t_1 t_2 x + (t_1 + t_2) y = -2 )Let me denote ( S = t_1 + t_2 ) and ( P = t_1 t_2 ) to simplify the equations.So, equations become:1. ( 2x - S y = -2P )2. ( 2P x + S y = -2 )Now, let's solve for ( x ) and ( y ).From equation 1: ( 2x - S y = -2P )From equation 2: ( 2P x + S y = -2 )Let me add the two equations:( 2x - S y + 2P x + S y = -2P - 2 )Simplify:( (2 + 2P) x = -2(P + 1) )So,( x = frac{ -2(P + 1) }{ 2(1 + P) } = frac{ -2(P + 1) }{ 2(P + 1) } = -1 )So, ( x = -1 ). Now, substitute back into equation 1 to find ( y ).From equation 1:( 2*(-1) - S y = -2P )Simplify:( -2 - S y = -2P )So,( -S y = -2P + 2 )Multiply both sides by (-1):( S y = 2P - 2 )Thus,( y = frac{2P - 2}{S} = frac{2(P - 1)}{S} )Since ( P = t_1 t_2 ) and ( S = t_1 + t_2 ), we have:( y = frac{2(t_1 t_2 - 1)}{t_1 + t_2} )Therefore, point ( M ) is ( (-1, frac{2(t_1 t_2 - 1)}{t_1 + t_2}) )Now, let's find point ( N ), which is the intersection of lines ( A_1A_4 ) and ( A_2A_3 ).First, find the equations of these lines.Starting with line ( A_1A_4 ). Points ( A_1 = (t_1^2, 2t_1) ) and ( A_4 = (1/t_2^2, -2/t_2) ).Using the chord equation formula again:( 2x - (t_1 + t_4)y + 2 t_1 t_4 = 0 )But ( t_4 = -1/t_2 ), so:( 2x - (t_1 - 1/t_2)y + 2 t_1 (-1/t_2) = 0 )Simplify:( 2x - (t_1 - 1/t_2)y - 2 t_1 / t_2 = 0 )Alternatively, maybe it's better to compute the equation using the two-point form.Slope of ( A_1A_4 ):( m = frac{ -2/t_2 - 2t_1 }{ 1/t_2^2 - t_1^2 } )Simplify numerator:( -2/t_2 - 2t_1 = -2( t_1 + 1/t_2 ) )Denominator:( 1/t_2^2 - t_1^2 = (1 - t_1^2 t_2^2)/t_2^2 )So, slope ( m = frac{ -2( t_1 + 1/t_2 ) }{ (1 - t_1^2 t_2^2)/t_2^2 } = frac{ -2 t_2^2 ( t_1 + 1/t_2 ) }{ 1 - t_1^2 t_2^2 } )Simplify numerator:( -2 t_2^2 ( t_1 + 1/t_2 ) = -2 t_2^2 t_1 - 2 t_2 )Denominator:( 1 - t_1^2 t_2^2 )So, slope ( m = frac{ -2 t_2^2 t_1 - 2 t_2 }{ 1 - t_1^2 t_2^2 } )This seems complicated. Maybe using the chord equation is better.Wait, another approach: since ( A_1A_4 ) is a chord between ( t_1 ) and ( t_4 = -1/t_2 ), so the equation is:( 2x - (t_1 + t_4)y + 2 t_1 t_4 = 0 )Substitute ( t_4 = -1/t_2 ):( 2x - (t_1 - 1/t_2)y + 2 t_1 (-1/t_2) = 0 )Simplify:( 2x - (t_1 - 1/t_2)y - 2 t_1 / t_2 = 0 )Multiply through by ( t_2 ) to eliminate denominators:( 2 t_2 x - (t_1 t_2 - 1) y - 2 t_1 = 0 )Similarly, for line ( A_2A_3 ), which connects ( A_2 = (t_2^2, 2t_2) ) and ( A_3 = (1/t_1^2, -2/t_1) ). Using the chord equation:( 2x - (t_2 + t_3)y + 2 t_2 t_3 = 0 )But ( t_3 = -1/t_1 ), so:( 2x - (t_2 - 1/t_1)y + 2 t_2 (-1/t_1) = 0 )Simplify:( 2x - (t_2 - 1/t_1)y - 2 t_2 / t_1 = 0 )Multiply through by ( t_1 ):( 2 t_1 x - (t_1 t_2 - 1) y - 2 t_2 = 0 )So, now we have the equations of lines ( A_1A_4 ) and ( A_2A_3 ):1. ( 2 t_2 x - (t_1 t_2 - 1) y - 2 t_1 = 0 )2. ( 2 t_1 x - (t_1 t_2 - 1) y - 2 t_2 = 0 )Let me denote ( Q = t_1 t_2 - 1 ) to simplify.So, equations become:1. ( 2 t_2 x - Q y - 2 t_1 = 0 )2. ( 2 t_1 x - Q y - 2 t_2 = 0 )Subtract equation 2 from equation 1:( 2 t_2 x - Q y - 2 t_1 - (2 t_1 x - Q y - 2 t_2) = 0 )Simplify:( 2 t_2 x - 2 t_1 x - 2 t_1 + 2 t_2 = 0 )Factor:( 2x(t_2 - t_1) - 2(t_1 - t_2) = 0 )Factor out 2:( 2 [x(t_2 - t_1) - (t_1 - t_2)] = 0 )Simplify inside:( x(t_2 - t_1) + (t_2 - t_1) = 0 )Factor:( (t_2 - t_1)(x + 1) = 0 )Since ( t_2 neq t_1 ) (as all points are distinct), we have ( x + 1 = 0 ) => ( x = -1 )So, ( x = -1 ). Substitute back into equation 1 to find ( y ).From equation 1:( 2 t_2 (-1) - Q y - 2 t_1 = 0 )Simplify:( -2 t_2 - Q y - 2 t_1 = 0 )So,( -Q y = 2 t_1 + 2 t_2 )Thus,( y = - frac{2(t_1 + t_2)}{Q} = - frac{2(t_1 + t_2)}{t_1 t_2 - 1} )Therefore, point ( N ) is ( (-1, - frac{2(t_1 + t_2)}{t_1 t_2 - 1}) )So, now we have both points ( M ) and ( N ):- ( M = (-1, frac{2(t_1 t_2 - 1)}{t_1 + t_2}) )- ( N = (-1, - frac{2(t_1 + t_2)}{t_1 t_2 - 1}) )Now, we need to find the area of triangle ( FMN ), where ( F = (1, 0) ).First, let's note that both ( M ) and ( N ) lie on the vertical line ( x = -1 ). So, the line ( MN ) is vertical, and the base of the triangle ( FMN ) can be considered as the segment ( MN ), and the height would be the horizontal distance from ( F ) to this line, which is ( |1 - (-1)| = 2 ).Wait, actually, in coordinate geometry, the area of a triangle can be calculated using the determinant formula. But since ( M ) and ( N ) are on the same vertical line, the area can be found as ( frac{1}{2} times |MN| times |FK| ), where ( K ) is the foot of the perpendicular from ( F ) to line ( MN ). Since ( MN ) is vertical at ( x = -1 ), the distance from ( F(1, 0) ) to ( MN ) is ( |1 - (-1)| = 2 ).So, the area ( S ) is ( frac{1}{2} times |MN| times 2 = |MN| ).But let me confirm this. Alternatively, using coordinates:Points:- ( F = (1, 0) )- ( M = (-1, y_1) )- ( N = (-1, y_2) )The area can be calculated using the formula:( S = frac{1}{2} | (x_F(y_M - y_N) + x_M(y_N - y_F) + x_N(y_F - y_M)) | )Substituting:( S = frac{1}{2} | 1(y_1 - y_2) + (-1)(y_2 - 0) + (-1)(0 - y_1) | )Simplify:( S = frac{1}{2} | (y_1 - y_2) - y_2 + y_1 | = frac{1}{2} | 2 y_1 - 2 y_2 | = | y_1 - y_2 | )So, the area is ( | y_1 - y_2 | ). Since ( y_1 = frac{2(t_1 t_2 - 1)}{t_1 + t_2} ) and ( y_2 = - frac{2(t_1 + t_2)}{t_1 t_2 - 1} ), let's compute ( | y_1 - y_2 | ).Compute ( y_1 - y_2 ):( frac{2(t_1 t_2 - 1)}{t_1 + t_2} - left( - frac{2(t_1 + t_2)}{t_1 t_2 - 1} right) = frac{2(t_1 t_2 - 1)}{t_1 + t_2} + frac{2(t_1 + t_2)}{t_1 t_2 - 1} )Let me denote ( A = t_1 t_2 ) and ( B = t_1 + t_2 ) for simplicity.So, expression becomes:( frac{2(A - 1)}{B} + frac{2B}{A - 1} )Combine the terms:( frac{2(A - 1)^2 + 2B^2}{B(A - 1)} )Expand numerator:( 2(A^2 - 2A + 1) + 2B^2 = 2A^2 - 4A + 2 + 2B^2 )So, numerator is ( 2A^2 - 4A + 2 + 2B^2 ), denominator is ( B(A - 1) ).But ( A = t_1 t_2 ) and ( B = t_1 + t_2 ). Also, note that ( (t_1 + t_2)^2 = t_1^2 + 2 t_1 t_2 + t_2^2 ), so ( B^2 = t_1^2 + 2A + t_2^2 ).But I don't know if that helps. Alternatively, maybe express in terms of ( A ) and ( B ):Wait, let me see if I can relate ( A ) and ( B ). Since ( t_3 = -1/t_1 ) and ( t_4 = -1/t_2 ), perhaps there's some symmetry or relation we can exploit.Alternatively, maybe set ( t_1 = k ) and ( t_2 = m ), so ( A = k m ), ( B = k + m ). Then, the expression becomes:Numerator: ( 2(k m)^2 - 4(k m) + 2 + 2(k + m)^2 )Denominator: ( (k + m)(k m - 1) )But this seems messy. Maybe another approach.Wait, let me compute ( y_1 y_2 ):( y_1 y_2 = frac{2(t_1 t_2 - 1)}{t_1 + t_2} times left( - frac{2(t_1 + t_2)}{t_1 t_2 - 1} right) = -4 )So, ( y_1 y_2 = -4 ). Interesting, so the product of ( y_1 ) and ( y_2 ) is constant, equal to -4.Given that, we can express ( y_2 = -4 / y_1 ). Therefore, ( | y_1 - y_2 | = | y_1 + 4 / y_1 | ).So, the area ( S = | y_1 + 4 / y_1 | ). Since area is positive, we can consider ( S = | y_1 + 4 / y_1 | ).But ( y_1 ) can be positive or negative. Let me assume ( y_1 > 0 ), then ( y_2 = -4 / y_1 < 0 ). So, ( | y_1 - y_2 | = | y_1 + 4 / y_1 | = y_1 + 4 / y_1 ).Similarly, if ( y_1 < 0 ), then ( y_2 = -4 / y_1 > 0 ), so ( | y_1 - y_2 | = | y_1 + 4 / y_1 | = | y_1 + 4 / y_1 | ). But since ( y_1 ) is negative, ( y_1 + 4 / y_1 ) is negative, so absolute value would be ( - y_1 - 4 / y_1 ), which is the same as ( | y_1 | + | 4 / y_1 | ).In either case, ( S = | y_1 | + | 4 / y_1 | ). Let me denote ( | y_1 | = a ), then ( S = a + 4 / a ).We need to find the minimum value of ( S = a + 4 / a ), where ( a > 0 ).This is a standard optimization problem. The function ( f(a) = a + 4 / a ) has its minimum at ( a = sqrt{4} = 2 ), and the minimum value is ( 2 + 4 / 2 = 4 ).Therefore, the minimum area is 4.But let me verify this with the actual coordinates.If ( S = | y_1 - y_2 | = | y_1 + 4 / y_1 | ), and the minimum of ( a + 4 / a ) is 4 when ( a = 2 ), then yes, the minimum area is 4.So, the minimum area of triangle ( FMN ) is 4.**Final Answer**The minimum value of the area of ( triangle FMN ) is (boxed{4}).

answer：Okay, so I have this equation: (2x^3 - 7x^2 + 7x + p = 0). It says that it has three distinct roots, and these roots form a geometric progression. I need to find (p) and solve the equation. Hmm, okay, let's break this down step by step.First, I remember that a geometric progression means each term is multiplied by a common ratio. So, if the roots are in a geometric progression, I can denote them as (a), (ar), and (ar^2), where (a) is the first term and (r) is the common ratio. Since they are distinct, (r) can't be 1 or -1, I think.Now, since these are the roots of the cubic equation, I can use Vieta's formulas, which relate the coefficients of the polynomial to the sums and products of its roots. For a cubic equation (ax^3 + bx^2 + cx + d = 0), the sum of the roots is (-b/a), the sum of the products of the roots taken two at a time is (c/a), and the product of the roots is (-d/a).Applying this to my equation, (2x^3 - 7x^2 + 7x + p = 0), the coefficients are (a = 2), (b = -7), (c = 7), and (d = p). So, let's write down the Vieta's formulas for this equation:1. Sum of roots: (a + ar + ar^2 = -b/a = 7/2)2. Sum of products two at a time: (a cdot ar + a cdot ar^2 + ar cdot ar^2 = c/a = 7/2)3. Product of roots: (a cdot ar cdot ar^2 = -d/a = -p/2)Okay, so let me write these equations out:1. (a(1 + r + r^2) = 7/2)2. (a^2(r + r^2 + r^3) = 7/2)3. (a^3 r^3 = -p/2)Hmm, so I have three equations with two unknowns: (a) and (r). Maybe I can solve for (a) and (r) first, and then find (p).Looking at equations 1 and 2, both equal to 7/2. Maybe I can divide equation 2 by equation 1 to eliminate the constants.So, equation 2 divided by equation 1:(frac{a^2(r + r^2 + r^3)}{a(1 + r + r^2)} = frac{7/2}{7/2})Simplifying, this becomes:(frac{a(r + r^2 + r^3)}{1 + r + r^2} = 1)So,(a(r + r^2 + r^3) = 1 + r + r^2)Hmm, that's interesting. Let's factor (r) from the numerator on the left side:(a r (1 + r + r^2) = 1 + r + r^2)Assuming (1 + r + r^2 neq 0) (which it isn't because the roots are distinct and real, so (r) can't be a root of (1 + r + r^2 = 0)), we can divide both sides by (1 + r + r^2):(a r = 1)So, (a = 1/r). That's a useful relation. Now, let's substitute (a = 1/r) back into equation 1.Equation 1: (a(1 + r + r^2) = 7/2)Substituting (a = 1/r):(frac{1}{r}(1 + r + r^2) = 7/2)Simplify the left side:(frac{1 + r + r^2}{r} = 7/2)Which is:(frac{1}{r} + 1 + r = 7/2)Let me write this as:(r + 1 + frac{1}{r} = 7/2)Hmm, that's a bit tricky. Maybe I can multiply both sides by (r) to eliminate the denominator:(r(r) + r(1) + r(frac{1}{r}) = frac{7}{2} r)Simplify:(r^2 + r + 1 = frac{7}{2} r)Bring all terms to one side:(r^2 + r + 1 - frac{7}{2} r = 0)Combine like terms:(r^2 - frac{5}{2} r + 1 = 0)Multiply both sides by 2 to eliminate the fraction:(2r^2 - 5r + 2 = 0)Now, I have a quadratic equation in terms of (r). Let's solve for (r) using the quadratic formula:(r = frac{5 pm sqrt{(-5)^2 - 4 cdot 2 cdot 2}}{2 cdot 2})Calculate discriminant:(sqrt{25 - 16} = sqrt{9} = 3)So,(r = frac{5 pm 3}{4})Which gives two solutions:1. (r = frac{5 + 3}{4} = frac{8}{4} = 2)2. (r = frac{5 - 3}{4} = frac{2}{4} = frac{1}{2})So, (r = 2) or (r = frac{1}{2}).Since (a = 1/r), let's find (a) for each case.Case 1: (r = 2)(a = 1/2)So, the roots are:1. (a = 1/2)2. (ar = (1/2) cdot 2 = 1)3. (ar^2 = (1/2) cdot 4 = 2)Case 2: (r = 1/2)(a = 1/(1/2) = 2)So, the roots are:1. (a = 2)2. (ar = 2 cdot (1/2) = 1)3. (ar^2 = 2 cdot (1/4) = 1/2)So, in both cases, the roots are (1/2), (1), and (2). That makes sense because reversing the order just changes the common ratio from (2) to (1/2), but the roots themselves remain the same.Now, let's find (p). From equation 3:(a^3 r^3 = -p/2)Let's compute (a^3 r^3) for both cases.Case 1: (a = 1/2), (r = 2)((1/2)^3 cdot 2^3 = (1/8) cdot 8 = 1)So,(1 = -p/2)Multiply both sides by 2:(2 = -p)Thus, (p = -2)Case 2: (a = 2), (r = 1/2)(2^3 cdot (1/2)^3 = 8 cdot (1/8) = 1)Same result:(1 = -p/2)So, again, (p = -2)Therefore, (p = -2).To confirm, let's plug (p = -2) back into the original equation and check if the roots are indeed (1/2), (1), and (2).The equation becomes:(2x^3 - 7x^2 + 7x - 2 = 0)Let's test (x = 1/2):(2(1/2)^3 - 7(1/2)^2 + 7(1/2) - 2)Calculate each term:(2(1/8) = 1/4)(-7(1/4) = -7/4)(7(1/2) = 7/2)(-2)Add them up:(1/4 - 7/4 + 7/2 - 2)Convert all to quarters:(1/4 - 7/4 + 14/4 - 8/4 = (1 - 7 + 14 - 8)/4 = 0/4 = 0)Good, (x = 1/2) is a root.Now, test (x = 1):(2(1)^3 - 7(1)^2 + 7(1) - 2 = 2 - 7 + 7 - 2 = 0)Perfect, (x = 1) is a root.Lastly, test (x = 2):(2(8) - 7(4) + 7(2) - 2 = 16 - 28 + 14 - 2 = 0)Yes, (x = 2) is also a root.So, everything checks out. Therefore, (p = -2) and the roots are (1/2), (1), and (2).**Final Answer**The value of (p) is (boxed{-2}), and the roots of the equation are (x = frac{1}{2}), (x = 1), and (x = 2).

answer：Okay, so I have this problem here: Given f(x) = e^x + sin x, then f'(x) = (     ). The options are A: ln x + cos x, B: ln x - cos x, C: e^x + cos x, D: e^x - cos x. Hmm, I need to find the derivative of f(x). Let me think about how to approach this.First, I remember that to find the derivative of a function, I can use basic differentiation rules. The function f(x) is composed of two parts: e^x and sin x. So, I should find the derivatives of each part separately and then add them together.Starting with the first term, e^x. I recall that the derivative of e^x is e^x itself. That's one of those special functions where the derivative is the same as the original function. So, the derivative of e^x is e^x. That part seems straightforward.Now, moving on to the second term, sin x. I think the derivative of sin x is cos x. Let me double-check that. Yes, from what I remember, the derivative of sin x with respect to x is indeed cos x. So, the derivative of sin x is cos x.Putting it all together, the derivative of f(x) should be the sum of the derivatives of each term. So, f'(x) = derivative of e^x plus derivative of sin x, which is e^x + cos x.Looking back at the options, option C is e^x + cos x, which matches what I just calculated. Option A is ln x + cos x, but ln x is the natural logarithm, and I don't see a logarithm in the original function, so that doesn't seem right. Option B is ln x - cos x, which also includes ln x, so that's probably not correct either. Option D is e^x - cos x, which has the right first term but the wrong sign for the second term. Since the derivative of sin x is cos x, not negative cos x, option D is incorrect.Wait, hold on a second. Let me make sure I didn't mix up any signs. The derivative of sin x is cos x, right? So, it should be positive cos x. If it were the derivative of cos x, that would be negative sin x. But here, we're dealing with sin x, so the derivative is positive cos x. Therefore, option C is correct.I think I might have confused myself earlier if I thought about the derivative of cos x, but no, in this case, it's sin x, so it's definitely cos x. So, f'(x) = e^x + cos x, which is option C.Just to recap, f(x) = e^x + sin x. The derivative of e^x is e^x, and the derivative of sin x is cos x. Adding them together gives e^x + cos x, which is option C. So, I feel confident that option C is the correct answer.I don't see any mistakes in my reasoning. I considered each term separately, applied the correct differentiation rules, and checked the signs. Everything seems to add up. I think I'm ready to select option C as the correct derivative of f(x).**Final Answer**boxed{C}