Solutions to O Level Add Math paper 04. Find the value of k for which the coefficient of x in the expansion of 6 kx x is 860. [] The question is looking for the x term in the expansion of kx and x 6 r We should use the term expression, kx kx r 6 6r and x x r r 0 r. 6 6. r 0 r Solution : Given that 6 When r When r kx x, we have r kx kx r r 0 r r r r k x r 0 r, the coefficient of the x term, 4 k! k 40k 6 6 r 0 r, the coefficient of the x term, 6 6 6 4! 6 6r x x r 40 Therefore, 40 40k 860 k 8 k By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 Alternative solution: 6 Let y kx x a a x a x 860 x Let x 0, 0 d y 4 0 k kx 0 x a 6 860 x d y 60 k kx 0 x 6860 4 k kx 6 x a a x 860x 60 k 0 6860 k 4 k 4 7 k 8 k In Summary: Binomial Theorem is a simple topic that will continue to feature in the future examination. Each year, there will be one question on this. By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04. The acute angles A and B are such that tan B 8 calculator, find the exact value of A and tan B. Without using a sin A. [] 0 A, B 90 and 0 A B 80. To find the exact value of trigo ratio of sin A, we know that A is an acute angle, in quadrant I, its value must be positive. Since tana B 8, we can deduce that angle A B is in quadrant I too. Therefore 0 A B 90. Solution : Given that tana B 8, tan A tan B 8 tan Atan B tan A 8 tan A For tan A, tan A 8 4 tan A tan A 6 Therefore, sin A 6 6 A Alternative solution: From the ratios of tan A and tan (A+B), we know that 0 A, B, A B 90 PT PR 8 PQR ~ TSR, ST 6 ST 6 sin ST A PT 6 0 0 6 6 6 P S R T A B Q By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 In Summary: Exact value refers to non-rounding of the answer. Students should be well aware of doing mathematics without calculator. By KL Ang, Jan 0 Page 4
Solutions to O Level Add Math paper 04. A particle moves along the curve y in such a way that the y-coordinate of the x particle is increasing at a constant rate of 0.0 units per second. Find the y-coordinate of the particle at the instant that the x-coordinate of the particle is increasing at 0. units per second. [] Rate of change. Given that 0. 0, when 0., we have dt dt dt dt 0.0 0., the connected rate of change. Take note that the question is asking for the y-coordinate. Solution : Given that y x x ------ () () = (), 0.0 0. ------ () 4 x x x 4 y 4 In Summary: Typically question on rate of change is straight forward. The only issue is the finding of y-coordinate, instead of the usual value of x. By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 4. Express x x x as the sum of partial fractions. [6] The numerator is of degree, the denominator is of degree. This is a proper fraction. Solution : Let x x x A x B C x x x Axx Bx Cx When x, 4 C When 0 C 4 x, B B When x, A B A C Alternative solution: Let x A B C x x x x x. x 4 x x x x x x Axx Bx Cx x 4x 4 A Cx B Ax B B A C 4. x 4 x x x x x In Summary: Make sure the given fraction is a proper fraction. Must also check the correctness of the decompositions with any value of x. By KL Ang, Jan 0 Page 6
Solutions to O Level Add Math paper 04. An experiment in Physics to find the focal length, f m, of a lens, requires the student to place an object at a distance, u m, from the lens and to record the distance, v m, at which the image is seen on the other side of the lens. The table below shows some results. It is known that u, v and f are related by the equation error was made in recording one of the values of v. (i) u 0.0 0.00 0.0 0.00 v 0.60 0.99 0.6 0.0 u v. It is believed that an f Plot v against u and hence determine which value of v, in the table above, is the incorrect recording. [] (ii) Draw the straight line graph and use it to estimate a value of v to replace the incorrect recording of v found in part (i). [] (iii) Estimate the value of f. [] We were told that one of the data of v is wrong. By comparing the u and v, we suspect that the first point, 0.60 might be the one. Then find, graphically the suitable value of v. Solution: u 6.67.00 4.00. u 0.0 0.00 0.0 0.00 v 0.60 0.99 0.6 0.0.66.4.80 4.98 v (i) From the graph, it shows that v = 0.6 is probably the incorrect recording. (ii) From the graph, the estimate value of 4. 0. v is 0. v (iii) The v -axis intercept is 8.. By KL Ang, Jan 0 Page 7
Solutions to O Level Add Math paper 04 v f u Therefore, 8.. Hence f = 0. f In Summary: Graph plotting is a very time consuming activity. It is usually best to be done as the last question of the paper. By KL Ang, Jan 0 Page 8
Solutions to O Level Add Math paper 04 By KL Ang, Jan 0 Page 9
Solutions to O Level Add Math paper 04 cosec sec tan 6. (i) Prove that tan. [4] cosec sec tan (ii) Find, in radians, the acute angle for which cot. [] Part (i) has numerous ways to do. Part (ii) is for 0 Solution: (i) cosec sec tan sec tan cosec sec tan sec tan sec tan cosec sec tan sec tan cosec sin cos cos sin sin cos sin sin sin tan cos cosec sec tan tan tan tan tan Since 0, tan (ii) Given that cot. By KL Ang, Jan 0 Page 0
Solutions to O Level Add Math paper 04 7. y A B y x y x C O The diagram shows a trapezium OABC in which OA is parallel to CB and O is the origin. The side AB is parallel to the x-axis and the diagonal AC is parallel to the y-axis. The side OA has equation y x and the side OC has equation y x. The x-coordinate of A is h. (i) Express the coordinates of A, B and C in terms of h. [] x (ii) In the case where h = 4, find the area of the trapezium OABC. [] The first challenge is the find the x-coordinate of C. There are numerous ways to find the area. Without knowing (i), we can still solve for area. Solution: (i) Ah h,, C h, h, h y x h Equation of BC, When y h, h h x h 7 h x 7 x h 4 By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 7 B h, h 4 (ii) When 4 h, A 4,8, B 7,8, C 4, the area of OABC 0 0 4 8 7 8 4 0 0 4 6 46 88 the area of OABC is square units In Summary: This question is very straight forward. Finding area of a quadrilateral is a regular affair in this exam. By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 8. It is given that f(x) is such that f (x) sin 4x cosx. Given also that f 0, show that f'' x 4f x cos4x. [7] d This is the only non-routine question in this paper. Firstly, f ' x f. Secondly d f '' x f d y. Let y fx, sin 4x cosx. 4y cos4x. Solution: Let y fx, given that sin 4x cosx and when x, y 0. y sin 4x cosx cos4x sin x y C where C is a constant. 4 When x, y 0. cos sin 0 C 4 0 C 4 C 4 cos4x sin y x 4 4 d y 4cos4x sin x d y cos4x sin x 4y 4cos4x sin x 4 4 4 d y 4y 4cos4x sin x cos4x sin x d y 4y cos4x By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 In Summary: The best part of this question is that if you make any mistake along the way, you will know it in the end and let you work backward to correct the error. By KL Ang, Jan 0 Page 4
Solutions to O Level Add Math paper 04 9. The equation of a curve is y ax x a where a is a constant. (i) In the case where a =, find the set of values of x for which the curve lies completely above the line y = 9. [] (ii) In the case where a = 4, show that the line y x is a tangent to the curve. [] (iii) Find the other value of a for which the line y x is tangent to the curve. [] (i) solve x x 9. (ii) To show y 4x x and y x has a repeated root. (iii) To find a for y ax x a tangent to y x. Solution: (i) Given that y ax x a, when a, y x x For y 9, we get x x 9 x x 0 x x 4 0 From the graph, x 4 or x 4 O x x : x 4or x (ii) When a 4, the curve is y 4x x. The intersection of the line and the curve is 4x x x 4x 4x 0 the discriminant of the quadratic equation, 4 4 4 0 Hence the line y x is tangent to the curve. By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 (iii) Let x ax x a ax 4x a 0 Set the discriminant to zero, 4 4 a a 0 6 a 4a 4 a a 0 4 a a 0 a 4 or a a is the other value. 0 In Summary: (i) needs answer in Set notation. (ii) is to show discriminant is zero. (iii) is to let discriminant to be zero. By KL Ang, Jan 0 Page 6
Solutions to O Level Add Math paper 04 0. x m L m x m 60 60 A gardener uses 0 m of fencing to enclose a plot of the shape shown above. The shape consists of two equilateral triangles of side x m and a rectangle with sides x m and l m. 0x 4 (i) Show that the area of the plot is x m. [] (ii) Given that x can vary, find the value of x for which the area of the plot is stationary. [4] (iii) Explain why this value of x gives the gardener the largest area possible. [] (i) This is about surd conjugates, simplifying surd. (ii) is to find the turning point and (iii) is to determine the nature of this turning point. Solution: (i) The perimeter, L 4x 0 L 6 x ---------- () The area, A Lx A Substitute () into (4), x xx Lx ---------- (4) A 6 xx x 0 x A 4 x By KL Ang, Jan 0 Page 7
Solutions to O Level Add Math paper 04 (ii) da Let 0, da 0 4 x da 6 4 x 0 6 4 6 x 4 x x x 64 4 4 64 4 64 x x 4 (iii) Since 4 0, the coefficient of the x term is negative, hence equation A 0 x 4 x has a maximum turning point at 4 x. In Summary: Simplifying surds and Properties of quadratic curve. By KL Ang, Jan 0 Page 8
Solutions to O Level Add Math paper 04. The point P lies on the curve y x ln x. The tangent to the curve at P is parallel to the line y x. (i) Find the coordinates of P. [] The normal to the curve y x ln x at P meets the line y x at point Q. (ii) Show that the x-coordinate of Q is k e, where k is a constant to be found. [] Gradient of the curve at the point of tangent is the same as the gradient of the line. Forming equation of the normal to find the intersection. Solution: (i) The gradient of point P is. y x ln x The gradient of the curve, d y ln x x x ln x Therefore, ln x ln x x e y e P e, e (ii) Let The gradient of the normal at P is. The equation of the normal at P is, y e x e y x e x x x e x 6 e e By KL Ang, Jan 0 Page 9
x e Solutions to O Level Add Math paper 04 In Summary: A simple routine question on coordinate geometry. The handling of e may give problem to some students. By KL Ang, Jan 0 Page 0
Solutions to O Level Add Math paper 04. A curve has the equation x 4 y. (i) Explain why the lowest point on the curve has coordinates,4. [] (ii) Find the coordinates of the points at which the curve intersects the x-axis. [] (iii) Sketch the graph of x 4 y. [] (iv) Using your graph, state the number of solutions to each of the following equations. (a) 4 7 x [] (b) 4 x [] (c) 4 0 x [] Sketching Quadratic curve and modulus quadratic curve. Solution: (i) Given that x 4 x 0 x 4 4 y, y 4, y has a minimum value of 4. When x, y 4. (ii) When y 0, 0 x 4 x 4 x x x or x By KL Ang, Jan 0 Page
Solutions to O Level Add Math paper 04 (iii) the graph of y x 4 y y 7 4 y O x y (iv) (a) x 4 7 (b) x 4 (c) x 4, solutions., 4 solutions., no solution. In Summary: Questions on Curve sketching will continue to appear in the future papers. By KL Ang, Jan 0 Page