Active Maths 4 Book 1: Additional Questions and Solutions

Active Maths 4 Book 1: Additional Questions and Solutions

Question 1 (a) Find the vertex of y = f(x) where f(x) = x 6x 7, stating whether it is a minimum or a maximum. (b) Find where the graph of y = f(x) crosses the axes and hence sketch the graph. Question The line l has equation y = x and the curve C has equation y = (x + )(x 4). (a) Sketch the line l and the curve C on the same axes, showing the coordinates of the x- and y-intercepts. (b) Show that the x-coordinates of the points of intersection of l and C satisfy the equation x 4x 5 = 0. (c) Hence, or otherwise, find the points of intersection of l and C. Question The height of a roller coaster h metres above ground level is given by the equation 1 x h = 100 ( 4 100 11 90 x + 4x ) where x is the distance in metres along the track. The track is 97.5 m long and goes underground at some points. A ride photo of the roller coaster can only be taken when h is stationary. (a) Find dh dx (b) Find d h dx The ride starts at x = 0. (c) Verify that this belongs to a stationary point. (d) At which other distances along the track can ride photos be taken? It is easier to take a photo when the ride is at a minimum point because the track is more accessible. (e) At what distance along the track should the ride photo be taken? Question 4 (a) Rewrite the following equation in the form f(x) = 0, where f(x) is of the form f(x) = ax + bx + cx + d: (x 1)(x + x + 1) = x 17 (b) Use the Factor Theorem to show that (x + ) is a factor of f(x). (c) Express f(x) in the form f(x) = (x + )(x + px + q), where p and q are constants. (d) Show that f(x) = 0 has only one real root. (e) Consider the equation y = f(x). (i) Find dy dx. (ii) Find the tangent to the curve y = f(x) at x = 1. (iii) Show that f(x) is increasing when x 4x > 0. (iv) Solve the quadratic inequality x 4x > 0. ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

Question 5 (a) A curve, C, has equation y = x + x + 6. Show that C has a stationary point at (0.5, 6.5). (b) The diagram shows the graphs of C and the line L, given by the y equation y = x +. L (i) Determine the coordinates of the points of intersection of C and L. (ii) Hence find the area of the shaded region bound by C and L. Question 6 The distance travelled by an object dropped from a height close to the earth s surface is closely approximated by the function d(t) = 1 gt C x where t is the time in seconds that have elapsed since the object was dropped, and g is the constant acceleration of the object due to gravity. (a) A stone is dropped from a height of 50 metres. (i) Find in terms of g the distance travelled by the stone in seconds. (ii) How long does it take the stone to fall a distance of g metres? (b) One method for determining the depth of a well is to drop a stone into it and then measure the time it takes until the splash is heard. (i) If d is the depth of the well and t 1 is the time it takes the stone to hit the water, write d in terms of g and t 1. (ii) Given that the speed of sound in air is approximately 0 m/s and t is the time taken for the sound to travel back up the well, write d in terms of t. (iii) T, the total time elapsed between dropping the stone and hearing the splash, can be written as T = pd + q d Find the value of p and the value of q. (iv) How deep is the well, if the total time elapsed is 4 seconds? Let g = 9.8 m/s. Question 7 The McCarthy family bought a house from the Moran family for 89,400. In lieu of a 0% down payment, the Moran s accepted a 10% down payment at the time of the sale and a promissory note from the McCarthy s for the remaining 10%, due in 4 years. The McCarthy s also agreed to make monthly interest payments to the Moran s at 11% per annum simple interest until the note expires. The McCarthy s obtained a 0 year mortgage, for the remaining 80% of the purchase price, from their bank at an APR of 4.55%. The bank in turn paid the sellers the remaining 80% of the purchase price, less a sales commission of % of the sales price paid to the sellers and buyers estate agents. Note: A promissory note is a financial instrument, in which one party promises in writing to pay a determinate sum of money to the other (the payee), either at a fixed or determinable future time or on demand of the payee, under specific terms. ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

(a) Find the McCarthy s down payment and the amount they borrowed from their bank. (b) What amount did the McCarthy s borrow from the Moran family? (c) Find the McCarthy s monthly interest-only payment to the Moran s. (d) Find the Moran family s total income from all aspects of the down payment. (e) How much did the Moran s receive from the McCarthy s bank? (f) Find the Moran s total income from all aspects of the sale. (g) Calculate to the nearest euro the McCarthy s monthly repayments. Question 8 A is the closed interval [ 5, 5] The function f is defined on A by: A graph of the function f is shown. f: A R: x 1 x 8 7 6 5 4 A 1 B C 0 D 7 6 5 4 1 0 1 4 5 6 7 1 (i) Find the co-ordinates of A, B, C and D (ii) State whether f is injective. Give a reason for your answer. (iii) State whether f is surjective. Give a reason for your answer. (iv) For what positive value of x A is f(x) a minimum? (v) Find the minimum value of f(x). (vi) On the plane above, graph the function 1 g(x) = (x ) by transforming the graph of f(x). (vii) Use long division and factoring to show that the function h(x) = x + 4x + 5 x + x + 1 4 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

can be written as h(x) = + (x + 1) Then, on the plane above, graph h by transforming the graph of f(x). (viii) Use integration to find the area of the region bounded by the graphs of f(x), g(x), the x and y axes and the line x =. Question 9 A bungee jumper plummets from a high bridge to the river below and then bounces back over and over again. At time t seconds after her jump, her height H (in metres) above the river is given by H(t) = 100 + 75e pt ( cos π 4 t ) where p. (i) If at t = 1s, the jumper s height above the river is 100 5 75 m, find the value of p. (ii) Find her height, to two decimal places, at the times indicated in the table. e t 0 1 4 6 8 1 H(t) (iii) If the jumper s height can also be represented by the function L(t) = 100 + 75e pt sin ( π 4 t + q ) where q, find the smallest positive value of q. (iv) Find H (t), the derivative of H(t). (v) Using the fact that a sin q + b cos q = a + b sin (q + a), where a = sin 1 ( write H (t) in the form a, where a, b and a. b e pt sin ( π 4 t + a ) b a + b ) ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 5

(vi) The graphs of H(t) and H (t) in the domain 0 t 10 are shown below. 00 H(t) 180 160 140 10 100 80 60 40 0 0 H'(t) 0 0 40 60 C 10 0 0 40 50 60 70 80 90 100 110 10 t (a) From the graph estimate the shortest distance between the jumper and the river. (b) Find the co-ordinates of the point C and hence find to two decimal places the shortest distance between the jumper and the river. (c) Find the percentage error in your estimate from part (a). (d) Use the graph to estimate lim (H(t)) as t 6 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

Solutions Question 1 f(x) = x 6x 7 (a) Function is a quadratic and the coefficient of x is positive, therefore vertex will be a minimum. To find the vertex, we need to complete the square f(x) = (x x) 7 = (x x + 1) 7 = (x 1) 10 To find x-value of vertex, put x 1 = 0 Vertex is (1, 10) Minimum x = 1 (b) f(x) crosses y-axis at x = 0 which gives y = 7 f(x) crosses x-axis at y = 0 (0, 7) x 6x 7 = 0 Using the completed square form: (x 1) 10 = 0 (x 1) = 10 x 1 = ± 10 x = 1 ± 10 ( 1 + 10 ) (, 0, 1 10 ), 0 14 1 10 8 6 y 4 1 10/ 0 1 + 10/ x.5 1.5 1 0.5 0 0.5 1 1.5.5.5 4 4.5 4 6 8 10 (1, 10) ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 7

Question (a) 5 0 5 0 15 10 y 5 0 (1.5, 0) (, 0) (4, 0) 8 6 4 0 4 6 5 (0, ) 8 10 x 10 (0, 8) (b) At points of intersection: x = (x + )(x 4) x = x x 8 0 = x 4x 5 Shown (c) x 4x 5 = 0 (x 5)(x + 1) = 0 x = 5, x = 1 x = 5 y = (5) y = 7 (5, 7) x = 1 y = ( 1) y = 5 ( 1, 5) Question 1 x h = 100 ( 4 100 11 90 x + 4x ) (a) dh dx = 1 100 ( x 00 11 0 x + 8x ) (b) d h dx = 1 100 ( x 100 11 15 ) x + 8 (c) For x = 0 to belong to a stationary point, dh must = 0 when x = 0. dx Substitute x = 0 into dh dx : 1 100 ( 0 00 11 0 (0) + 8(0) ) = 0, so x = 0 is at a stationary point. (d) Put dh = 0 & solve for x: dx 1 100 ( x 00 11 0 x + 8x ) = 0 8 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

x 00 11 0 x + 8x = 0 x 110x + 400x = 0 x(x 110x + 400) = 0 x(x 80)(x 0) = 0 x = 0, x = 80, x = 0 So ride photos can be taken at x = 0 m and at x = 80 m. (e) At x = 0, d h dx = 1 100 ( (0) 100 11 15 ) (0) + 8 = 0.05, (< 0 Maximum) At x = 80, d h dx = 1 100 ( (80) 100 11 15 ) (80) + 8 0.1 (> 0 Minimum) The ride photo should be taken 80 m along the track. Question 4 (a) (x 1)(x + x + 1) = x 17 x + x + x x x 1 = x 17 x x + 16 = 0 (b) If (x + ) is a factor then, by the Factor Theorem, f( ) will be equal to zero. Using the equation you found in part (a): f( ) = ( ) ( ) + 16 = 8 (4) + 16 = 0 (x + ) is a factor of f(x). (c) Factorise the cubic by equating coefficients: x x + 16 = (x + )(px + qx + r) = px + (q + p)x + (r + q)x + r Equating coefficients: p = 1, r = 16 r = 8 q + p = q + = q = 4. So x x + 16 = (x + )(x 4x + 8) (d) f(x) = 0 has at least one real root, x =, as shown in part (b). As shown in part (c), f(x) = (x + )(x 4x + 8), so if f(x) = 0 then (x + )(x 4x + 8) = 0, and so any other roots of f(x) = 0 will be the roots of (x 4x + 8) = 0. The discriminant of (x 4x + 8) is: ( 4) (4 1 8) = 16 (e) 16 < 0, so (x 4x + 8) = 0 has no real roots Therefore f(x) = (x + )(x 4x + 8) only has one real root, x = (i) dy dx = f ʹ(x) = x 4x. (ii) The gradient of the tangent is given by: m = f ʹ(1) = (1) 4(1) = 1 When x = 1, y = (1) (1) + 16 = 15 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 9

Now use y y 1 = m(x x 1 ) to find the equation of the tangent: y 15 = (x 1) y = x + 16 (iii) f(x) is increasing where f ʹ(x) > 0, i.e. when x 4x > 0. (iv) The graph of y = x 4x is drawn below. 0 4 x 4x = 0 x(x 4) = 0, so the graph crosses the x-axis at x = 0 and x = 4. The graph is positive (i.e. above the x-axis) when x < 0 and when x > 4. Therefore, x 4x > 0 when x < 0 and when x > 4. Question 5 (a) dy dx = x + 1 Stationary points occur when dy dx = 0, x + 1 = 0 x = 0.5 x = 0.5 y = (0.5) + 0.5 + 6 = 0.5 + 0.5 + 6 = 6.5 So there is a stationary point at (0.5, 6.5). (b) (i) Finding the points of intersection: (ii) x + x + 6 = x + x = 4 x = ± Use the equation for L to find the corresponding y-values: When x =, y = + = 4 When x =, y = + = 0 So the coordinates of intersection are (, 4) and (, 0). y L (, 4) (, 0) x C 10 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

The area between L and C is equal to the area under C between x = and x = minus the area under L between x = and x =. The area under C between x = and x = is given by: ( x + x + 6)dx = [ x + x ] + 6x = ( + ) ( + 6() ( ) = ( 8 + 14 ) ( 8 10 ) = 56 + ( ) + 6( ) ) The area under L between x = and x = is equal to the area of a triangle of base 4 and height 4 [the vertices of the triangle are (, 0), (, 0) and (, 4).] So, the area beneath L is 1 4 4 = 8 The area between C and L is therefore: 56 8 = 56 4 = Question 6 (a) (i) d() = 1 g() = 9g metres. (ii) 1 gt = g t = 4 (b) (i) d = 1 g t 1 t = seconds. (ii) d = 0t (iii) t 1 = d g d t 1 = g (iv) t = d 0 1 0 d + g d = 4 Let d = x T = t 1 + t d T = g + d 0 T = g 1 d + 0 d 1 T = 0 d + g d p = 1 0 q = g d = x, 1 0 x + g x 4 = 0 0.000x + 0.451754x 4 = 0 x = 0.451754 ± (0.451754) 4(0.000)( 4) (0.000) x = 8.8 m. The well has a depth of 8.8 m. ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 11

Question 7 (a) Down payment: 89,400 10% = 8,940 (b) 8,940 8940 11% (c) Monthly Interest: = 56.95 1 (d) Total Income = 8,940 + 8,940 + 56.95 48 = 9501.60 (e) 80% of 89,400 = 11,50 % of 89,400 = 11,68 Moran's received 99,88 (f) Total Income: 99,88 + 95,01.60 = 94,851.60 (g) Monthly interest rate F = P(1 + i) t 104.55 = 100(1 + i) 1 (1 + i) 1 = 1.0455 i = 1 1.0455 1 = 0.00714819 i (1 + i) A = 1150 [ 40 (1 + i) 40 1 ] where i = 0.00714819 A = 1,96.7 Question 8 (i) A ( 1, 1) C (, 1 4 ) B (1, 1) D (, 1 4 ) (ii) f is not injective, as f( 1) = 1 = f(1) (iii) f is not surjective, as there does not exist y such that f(0) = y ( iv) 5 (v) f(5) = 1 5 = 1 5 1 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

(vi) 8 7 6 h(x) 5 4 g(x) (vii) A 1 B C D 0 7 6 5 4 1 0 1 4 5 6 7 x + x + 1 x + 4x + 5 x + 4x + h(x) = + x + x + 1 h(x) = + (x + 1) (viii) Area = 1 1 x dx = x dx 1 = [ 1 x ] 1 = [ 1 + 1 ] = 1 1 Question 9 (i) 100 + 75e 1p cos (π) = 100 75 e 5 75e 1p = 75 e 5 1p = 5 p = 1 0 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 1

(ii) t H(t) 0 175.00 1 150.45 100.00 4 8.60 6 100.00 8 150.7 1 58.84 (iii) 100 + 75e pt sin ( π 4 t + q ) = 100 + 75ept cos ( π 4 t ) sin ( π 4 t + q ) = cos ( π 4 t ) sin π 4 t cos q + sin q cos π 4 t = cos π 4 t q = π or sin q = 1 and cos q = 0 smallest value for which this is true is q = π (iv) H (t) = 75e pt ( π 4 sin ( π 4 t ) ) + cos ( π 4 t ) = 75 e t = 15 (v) H (t) = 15 0 ( π 4 sin ( π 4 t ) ) 15 4 e 4 e t 4 e 75 pept t 0 [ 5π sin ( π 4 t ) + cos ( π 4 t ) ] t 0 [ = 15 4 e t 0 = 15 5π 4 + 1 e t (vi) (a) 9 m. (b) 15 4 e t 0 [ e t 0 sin ( π 4 t + a ) = 0 sin ( π 4 t + a ) = 0 for t > 0 π 4 t + a = π π 4 t + 0.06576181 = π t =.919s 0 cos ( π 4 t ) 5π + 1 sin ( π 4 t + a )], where a = 1 ( sin 1 5π + 1 sin ( π 4 t + a ) 0 sin ( π 4 t + a ) 5π + 1 ] sin ( π 4 t + a ) = 0 C = (.919, 0) H(.919) = 100 + 75 e.919 = 8.471 m. 0 cos ( π 4.919 ) 5π + 1 ) = 0.06576181 14 ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS

(c) Error = 9 8.471 = 0.59 % Error = 0.59 8.471 100 = 1.8% ( d.p.) (d) lim H(t) = 100 t ACTIVE MATHS 4 BOOK 1: ADDITIONAL QUESTIONS AND SOLUTIONS 15