4038/02 October/November 2008
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1 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version 1.1 ADDITIONAL MATHEMATICS Paper Suggested Solutions 1. Topic: Exponential Functions (i) Given: V =10000e -pt When the man bought the motorcycle, t = 0. value of motorcycle when bought = 10000e 0 (ii) When t = 1, v = 4000: 4000 = 10000e 0.4 = e -1p ln 0.4 = 1p ln e ln 0.4 -p (1) p = V = 10000e t. (1) = $ /0 October/November (18) Sub t = 18 into (1): Value after 18 months = 10000e 59.8 $50 ( s.f.) (iii) Sub v = 1000 into (1): 1000 = 10000e t Note: V < $1000 when t = = e t months, since the e pt curve describes the depreciation of value ln 0.1 = t ln e over time. V t = ln months 0 months (nearest month) t. Topic: Quadratic Equations (Sum & Product of Roots) x 4x + = 0 a =, b = 4, c = Sum of roots: α + β = b a = 4 = Product of roots: αβ = c = a New sum of roots: (α + ) + (β + ) = α + β + 4 =[(α + β) αβ] + 4 = () + 4 = 5 New product of roots: (α + )(β + ) = α β + α + β + 4 =(αβ) + (α + β ) + 4 = (αβ) + [(α + β) αβ] + 4 = [4 ] + 4 = = 8 1 = 4 4 equation with roots α +, β + : x 5x + = 0 4 4x 0x + = 0 Sub α + β =, αα = Useful expression: α + β = (α + β) αα Sub α + β =, αα = x (Sum of roots)x + (Product of roots) = 0 age of motorcycle when expected value is $1000 is 0 months / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 1 / 8
2 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version 1.1. Topic: Trigonometry (Trigonometric Identities & Equations) (i) tan A + cot A = cosec A L.H.S.: sin A cos A (ii) tan A + cot A = + cos A sin A = sin A+cos A cos A sin A = = = From (i): cosec A = sin A = sin A = 1 sin A cos A 1 sin A cos A sin A = cosec A = R.H.S. (Proved) Basic angle α = sin A 41.81, , , , 18.19, , A = 0.9, 69.1, 00.9, 49.1 sin A + cos A = 1 sin A = sin A cos A 90 A 18., 498. A 69.1, S T α A α C 70 α A 60 0 A 70 A 41.8, A 0.9, Topic: Logarithms (i) + log (x 7) = log (x ) log (x ) log (x 7) = (ii) log 5 y log y 5 = log 5 y = log 5 y Let log 5 y = x. 1 log x = x 7 x = x 7 x = 9(x 7) x = 7x 6 60 = 5x x =.4 x 1 = x x 1 = x (x + 1)(x 1) = 0 x = 1 or 1 log 5 y = 1 or 1 y = 5 1 or 5 1 y = 1 or 5 5 Change of Base of Log: log a b = 1 log b a Quotient Law: m log a m log a n = log a n y = a x x = log a y / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. / 8
3 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topic: Polynomials (Factor Theorem & Remainder Theorem) Factor Theorem: (i) f(x) = (x x +1) (x + 1) (x ) = (x 6x +) (x f(a) = 0 (x a) is a factor of f(x) x ) = x 4 x - 4x 6x + 6x + 1x + x x 4 = x 4 8x + 4x + 10x 4 (ii) f(x) = 0 Check: f( 1) = ( 1) 4 8( 1) + 4( 1) + 10( 1) 4 = 0 (x x + 1) (x + 1) (x ) = 0 No. of real roots = 4 (iii) f(x) = (x x + 1) (x + 1) (x ) x = ± 9 4(1)(1) or 1 or = ± 5, 1, 6. Topic: Geometric Proofs (i) (ii) When x = 1, f 1 = [1 + 1] 4 = 1 4 = Remainder = Remainder Theorem: f(x) divided by (x a) remainder is f(a) AE = CE (Tangents from a common external point E) AO = CO (Radius of circle) OAE = OCE = 90 (Radius Tangent) AEO CEO (SAS) Let AOE = θ. COE = θ ( AEO CEO) AOC = θ ABC = θ ( at centre = at circumference) EO // DB ( AOE = ABC = θ - corresponding s) By Midpoint Theorem, E is the mid-pt of AD / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. / 8
4 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topic: Trigonometry (Trigonometric Functions) (i) Amplitude = 4 60 (ii) Period = = 180 cycles (iii) Minimum point occurs when cos x = 1 x = cos -1 ( 1) x = 180 x = 90 y = 4 ( 1) = 6 y = a cos bx + c: Amplitude = a Period = 60 b coordinates of the minimum point of the curve is (90, 6) (v) (iv) When y = 0, 4cos x - = 0 4cos x = cos x = 1 Basic angle α = 60 x = 60, 00 x = 0, S T A α α 0 C α = 60 0 x x 60 x = 60 x = 0 x = 00 x = 150 (vi) coordinates where curve meets the x-axis are (0, 0)and (150, 0) / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 4 / 8
5 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topic: Applications of Differentiation & Integration (Maxima & Minima, Area Under Curve) (i) y = x ax + b = dx x a From the diagram, minimum point is (, 0) Sub = 0 and x = into : dx dx () a = 0 1 a = 0 a = 1 Sub (, 0) into y: a = 1, b = 16 0 = 1() + b 0 = b b = 16 (ii) From (i), y = x 1x + 16 = dx x 1 At maximum point, = dx x 1 = 0 (x 4) = 0 (x + )(x ) = 0 x = or (rejected max. point occurs when x < 0 in the diagram) Sub x = into y: Check: d y = 6x dx y = ( ) 1 ( ) + 16 Sub x = into d y dx : d = y dx = 6( ) = 1 < 0 = (, ) is a maximum point coordinates of maximum point is (, ). (iii) Area of shaded region = (x 1x + 16) dx 0 = x4 1x x 0 = x4 4 6x + 16x 0 = [ ()4 4 6() + 16()] [ (0)4 4 6(0) + 16(0)] = = 1 units / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 5 / 8
6 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topic: Further Trigonometric Identities (R-Formula) (i) From the diagram, OAD = θ (corresponding s) sin θ = OD 4 OD = 4 sin θ ABC = BAC = ( θ) = θ cos θ = BC BC = cos θ L = OD BC = 4 sin θ cos θ (Shown) (ii) 4 sin θ cos θ = R sin (θ α) = R (sinθ cos α cos θ sin α) = R cos α sinθ R sin α cos θ Comparing coefficients, 4 = R cos α. (1) = R sin α. () () : sin α = (1) cos α 4 tan α = 1 α 6.6 Addition Formula: sin (A B) = sin A cos B cos A sin B (1) + () : R cos α + R sin α = 4 + R (cos α + sin α) = R = 0 R = 0 or 0 (rejected) L = 0 sin (θ 6.6 ) 90 θ S A θ φ φ 0 (iii) When L =, 0 sin (θ 6.6 ) = 70 sin (θ 6.6 ) = φ Basic angle φ 4.1 θ , θ 68.7, (rejected θ is acute) θ 68.7 ( sig. fig.) T C / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 6 / 8
7 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topics: Coordinate Geometry, Integration, Applications of Differentiation (Rate of Change) (i) = 6 dx (x 1) Sub x = into : dx Gradient of curve at P(, 9) = 6 = Gradient of normal = Equation of normal to the curve at P: y 9 = (x ) y = x y = x + 1. (1) At Q(0, y), Gradients m 1 and m of two lines m 1 m = 1 Equation of line with gradient m and point (x 1, y 1 ): (y y 1 ) = m(x x 1 ) (ii) y = dx (x 1) 6 = 6(x 1) dx = 6(x 1) 1 ( 1) = (x 1) c + c y =. () x 1 Since P(, 9) lies on the curve, sub x =, y = 9 into (): 9 = () 1 + c 9 = 1 + c c = 10 y = + 10 x 1 (aa + (aa + b) n b)n+1 dx = + c a(n + 1) Sub x = 0 into (1): y = (0) + 1 = 1 Q = (0, 1) At R(x, 0), Sub y = 0 into (1): x + 1 = 0 x = 1 x = 8 R = (8, 0) Mid-point of QR = 0+8, 1+0 = (4, 6) Midpoint of (x 1, y 1 ) and (x, y ): x 1 + x, y 1 + y (iii) x-coordinate increases at 0.0 units per second dx dt = 0.0 Rate of change of y-coordinate: = dx dt dx dt 6 = 0.0 (x 1) At P(, 9), sub x = into : dt = dt (() 1) = 0.0 units per second Chain Rule: = dx dt dx dt y-coordinate increases at 0.0 units per second / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 7 / 8
8 GCE O Level October/November 008 Suggested Solutions Additional Mathematics (408/0) version Topics: Coordinate Geometry, Circles (i) Centre of C 1, O = (0, 0) Radius of C 1, OP = (0 8) + (0 + 6) = 10 units Equation of C 1 : (x 0) + (y 0) = 10 (ii) Center of C, Q = Midpoint of OP = 0+8, 0+( 6) = (4, ) x + y = 100 Radius of C, QP = (4 8) + ( + 6) = 5 units Equation of C : (x 4) + (y + ) = 5 x 8x y + 6y + 9 = 5 x + y 8x + 6y = 0 Length of Line Segment = (x 1 x ) + (y 1 y ) Equation of circle with centre (a, b) and radius r: (x a) + (y b) = r Midpoint of (x 1, y 1 ) and (x, y ): x 1 + x, y 1 + y (iii) Gradient of OP = = 4 Gradient of AB = 4 Using Q(4, ) from (ii): Equation of AB: y + = 4 (x 4) y = 4 x 5. (1) A and B also lie on C 1 with equation obtained in (i): x + y = () Sub (1) into (): x x = 100 x + 16x 00 x + 65 = x 00 x + 65 = x 00 x 75 = 0 x 8x 11 = 0 x = 8 ± 64 4(1)( 11) (1) = 8 ± 108 = 8 ±6 = 4 ± Gradients m 1 and m of two lines m 1 m = 1 Equation of line with gradient m and point (x 1, y 1 ): (y y 1 ) = m(x x 1 ) x-coordinates of A and B are 4 + and 4 respectively / missloi@exampaper.com.sg facebook.com/jossstickstuition twitter.com/missloi Copyright 008 ϕ exampaper.com.sg. All rights reserved. 8 / 8
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