ADDITIONAL MATHEMATICS

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1 00-CE MTH HONG KONG EXMINTIONS UTHORITY HONG KONG CERTIFICTE OF EDUCTION EXMINTION 00 DDITIONL MTHEMTICS 8.0 am.00 am (½ hours) This paper must be answered in English. nswer LL questions in Section and any FOUR questions in Section.. Write your answers in the answer book provided. For Section, there is no need to start each question on a fresh page.. ll working must be clearly shown. 4. Unless otherwise specified, numerical answers must be eact. 5. In this paper, vectors may be represented by bold-type letters such as u, but candidates are epected to use appropriate symbols such as u r in their working. 6. The diagrams in the paper are not necessarily drawn to scale. 香港考試局保留版權 Hong Kong Eaminations uthority ll Rights Reserved CE- MTH

2 FORMULS FOR REFERENCE sin ( ± ) = sin cos ± cos sin cos ( ± ) = cos cos m sin sin tan ± tan tan ( ± ) = m tan tan sin cos = sin ( + ) + sin ( ) cos cos = cos ( + ) + cos ( ) + sin + sin = sin cos + sin sin = cos sin + cos + cos = cos cos + cos cos = sin sin sin sin = cos ( ) cos ( + ) * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * Section (6 marks) nswer LL questions in this section.. If n is a positive integer and the coefficient of in the epansion of n ( + ) + ( + ) n is 75, find the value(s) of n. (4 marks). Find the equation of the tangent to the curve C : y = ( ) 4 which is parallel to the line y = (4 marks) 4 +. Let sin y = 00. Find d y. (4 marks) d 00-CE- MTH 保留版權 ll Rights Reserved 00

3 4. Find 0 d. [Hint : Let = sinθ.] (4 marks) 5. P (, y) is a variable point such that the distance from P to the line 4 = 0 is always equal to twice the distance between P and the point (, 0). (a) Show that the equation of the locus of P is + 4y = 0. (b) Sketch the locus of P. (5 marks) 6. y y = sin y = cos O Figure π Figure shows the curves the shaded region. y = sin and y = cos. Find the area of (5 marks) 7. Solve the following inequalities : (a) > ; (b) y >. (5 marks) 00-CE- MTH 保留版權 ll Rights Reserved 00 Go on to the net page

4 8. Given π 0 < <. tan sin 4 Show that =. tan + sin + sin Hence, or otherwise, find the least value of tan sin tan + sin. (5 marks) 9. Let α, β be the roots of the equation z + z + = 0. Find α and β in polar form. Hence, or otherwise, find 6 6 α + β. (5 marks) 0. y (, 4) θ C(5, ) O Figure Figure shows a parallelogram OC. The position vectors of the points and C are i + 4j and 5 i + j respectively. (a) Find O and C. (b) Let θ be the acute angle between O and C. Find θ correct to the nearest degree. (6 marks) 00-CE- MTH 4 保留版權 ll Rights Reserved 00

5 . y y = g() y = f() O C Figure Let f ( ) = 6 and g ( ) = + 6. The graphs of y = f ( ) and y = g( ) intersect at points and (see Figure ). C is the verte of the graph of y = f ( ). (a) Find the coordinates of points, and C. (b) Write down the range of values of such that f ( ) g( ). Hence write down the value(s) of k such that the equation f ( ) = k has only one real root in this range. (7 marks). (a) Prove, by mathematical induction, that n n+ () + ( ) + 4( ) + L + ( n + ) ( ) = n( for all positive integers n. ) (b) Show that () + ( ) + ( ) + L + 98( 98 ) = 97( 99 ) +.(8 marks) 00-CE- MTH 5 4 保留版權 ll Rights Reserved 00 Go on to the net page

6 Section (48 marks) nswer any FOUR questions in this section. Each question carries marks.. E O π Figure 4 In Figure 4, O is a triangle. Point E is the foot of perpendicular from O to. Let O = a and O = b. It is given that O =, O = and π O =. (a) Find a b. (b) Find OE in terms of a and b. [Hint : Let E : E = t : ( t).] ( marks) (5 marks) (c) F is a variable point on OE. student says that F is always a constant. Eplain whether the student is correct or not. If you agree with the student, find the value of that constant. If you do not agree with the student, find two possible values of F. (5 marks) 00-CE- MTH 6 5 保留版權 ll Rights Reserved 00

7 4. P D 4 C Figure 5 Figure 5 shows an isosceles triangle C with = C and C = 4. D is the foot of perpendicular from to C and P is a point on D. Let PD = and r = P + P + PC, where 0 D. (a) Suppose that D =. dr (i) Show that =. d + 4 (ii) Find the range of values of for which () r is increasing, () r is decreasing. Hence, or otherwise, find the least value of r. (iii) Find the greatest value of r. (b) Suppose that D =. Find the least value of r. (9 marks) ( marks) 00-CE- MTH 7 6 保留版權 ll Rights Reserved 00 Go on to the net page

8 5. (a) E Figure 6 r C D F DEF is a triangle with perimeter p and area. circle C of radius r is inscribed in the triangle (see Figure 6). Show that = pr. (4 marks) (b) y R (, 5) C S (5, ) Q (, ) O Figure 7 In Figure 7, a circle C is inscribed in a right-angled triangle QRS. The coordinates of Q, R and S are (, ), (, 5) and (5, ) respectively. (i) Using (a), or otherwise, find the radius of C. (ii) Find the equation of C. (8 marks) 00-CE- MTH 8 7 保留版權 ll Rights Reserved 00

9 6. (a) + y = r y r h Figure 8 O In Figure 8, the shaded region is bounded by the circle + y = r, the y-ais and the line y = h, where 0 h < r. The shaded region is revolved about the y-ais. Show that the volume π of the solid generated is (r r h + h ). (4 marks) (b) y O + y = 96 Stopper Container C Figure 9 Figure 0 In Figure 9, and C are points on the y-ais, C is an arc of the circle + y = 96 and is a segment of the line y =. pot is formed by revolving C about the y-ais. (i) (ii) Find the capacity of the pot. Figure 0 shows a perfume bottle. The container is in the shape of the pot described above and the stopper is a solid sphere of radius 6. Find the capacity of the perfume bottle. (8 marks) 00-CE- MTH 9 8 保留版權 ll Rights Reserved 00 Go on to the net page

10 7. F D C Figure Figure shows a tetrahedron CD such that = 8, CD = 0, C = D = 5 and C = D = 40. F is the foot of perpendicular from C to D. (a) Find FC, giving your answer correct to the nearest degree. (8 marks) (b) student says that FC represents the angle between the planes CD and D. Eplain whether the student is correct or not. (4 marks) 00-CE- MTH 0 9 保留版權 ll Rights Reserved 00

11 8. (a) Let z = cosθ + i sinθ, where π < θ π. Show that z + = ( + cos θ ). Hence, or otherwise, find the greatest value of z. (5 marks) (b) w is a comple number such that w =. + + (i) Show that the greatest value of w 9 is 8. (ii) Eplain why the equation w 4 8 = 00 i ( w 9) has only two roots. (7 marks) END OF PPER 00-CE- MTH 0 保留版權 ll Rights Reserved 00

12 00 dditional Mathematics Section. 6. y = tan y π (a) > or < (b) y > or y < π π π π cos + i sin, cos ( ) + i sin ( ) 0. (a) 6i + 6j, 4i j (b) 7. (a) (, ), (6, 8), C(, 7) (b) 6 < k 8 or k = 7 00-CE- MTH 保留版權 ll Rights Reserved 00

13 Section Q. (a) a b = a b cos O π = () () cos E (b) (c) to + ( t) O OE = t + ( t) = Since ta + ( t)b OE, OE = 0 [ t a + ( t) b ] ( b a) = 0 t a b ta a + ( t) b b ( t) a b = 0 t 9t + 4( t) ( t) = 0 7t = 0 t = 7 6 OE = a + b 7 7 F = F cos F = E = a constant (since and E are constants) the student is correct. y Cosine Law, = O + O O O cos O = = = 7 π 7 + O () () cos π 00-CE- MTH 保留版權 ll Rights Reserved 00

14 From (b), E = 7 E = 7 7 = 7 F = E 7 = 7 ( ) = 7 00-CE- MTH 4 保留版權 ll Rights Reserved 00

15 + Q.4 (a) (i) P = PC = 4, P = ( ) r = P + P + PC = ( ) dr = ( ) = d P dr (ii) () 0 d r is increasing on. dr () 0 d r is decreasing on 0 r is the least at =. Least value of r = ( ) ( ) = + (iii) The greatest value of r occurs at the end-points. t = 0, r = ( 0) = 7 t =, r = ( ) = the greatest value of r is. 00-CE- MTH 5 4 保留版權 ll Rights Reserved 00 D 4 C

16 (b) r = ( ) dr = d + 4 From (a), r is decreasing on 0. r is the least at =. Least value = ( ) = 5 00-CE- MTH 6 5 保留版權 ll Rights Reserved 00

17 Q.5 (a) = rea of ODE + area of OEF + area of ODF (where O is centre of C ) = ( DE ) ( r) + ( EF) ( r) + ( DF) ( r) = ( DE + EF + FD) ( r) = pr D E O r C F (b) (i) QR = ( ) + ( 5) = y R (, 5) RS = ( 5) + (5 ) = 8 SQ = p = = ( 5) ( ) 50 = Q (, ) 50 rea of QRS = ( QR) ( RS) = ( = Using (a), = pr ) ( 8) O C S (5, ) = ( r = ) r 00-CE- MTH 7 6 保留版權 ll Rights Reserved 00

18 (ii) Equation of QR y 5 5 = ( ) y + = 0 Let (h, k) be the centre of C. h k + ( ) = h k + = () Equation of RS y 5 5 = 5 + y 7 = 0 h + k 7 ( ) = h + k 5 = () Solve () and (), h =, k =. the equation of C is ( ) + ( y ) =. 00-CE- MTH 8 7 保留版權 ll Rights Reserved 00

19 r h r h Q.6 (a) Volume = π d y = π ( r y ) d y r = π y y = π r r r h + h = π (r r h + h ) (b) (i) Using the result in (a), substitute r = 4, h = : Capacity of the pot π = π (4) [(4) (4) () + () ] = 645 π r h (ii) Let O be the centre of the stopper. O = 4, O = 6, O = O 6 = O O = 4 = 7 M D 4 O = O = 6 7 O M = Capacity of the perfume bottle = Capacity of pot found in (i) Volume of the stopper lying inside the pot π = 645π [(6) (6) () + () ] = 645π 45π = 600 π 00-CE- MTH 9 8 保留版權 ll Rights Reserved 00

20 Q.7 (a) F D Consider CD : Let E be the foot of perpendicular from to CD. DE 5 cos DE = = = D CF = CD sin DE F 5 5 = 0( ) 5 D E = 4 0 C Consider D : Consider CF : 5 4 F C D 8 cos D = F = C = 5 F = = (8) (5) CF 4 = 49 C 00-CE- MTH 0 9 保留版權 ll Rights Reserved 00

21 Consider F : 8 7 F = + F ( ) ( F) cos F 9 = (8) (7) ( ) 400 F F = Consider CF : C 4 F F + CF C cos FC = ( F) ( CF) = ( ) (4) = FC = 96 (correct to the nearest degree) (b) Consider F : F + F = = F F 90 Since CF D but F is not perpendicular to D, FC does not represent the angle between the two planes. The student is incorrect. 00-CE- MTH 0 保留版權 ll Rights Reserved 00

22 Q.8 (a) z = cos θ + i sin θ z + = (cos θ + ) + i sin θ z + = (cos θ + ) + sin θ = cos θ + cos θ + + sin θ = ( + cos θ ) Since π < θ π, π < θ π. cos θ the greatest value of z + = ( + ) = (b) (i) w = z w + 9 = (z) = 9 z From (a), z +. greatest value of w + 9 = 9() = 8 4 (ii) w 8 = 00i ( w 9) ( w + 9) ( w 9) 00 i( w 9) = 0 ( w 9) ( w i) = 0 w 9 = 0 () or w i = 0 () Consider () : w = ± which satisfies the condition w = Consider () : w + 9 = 00i From (i), w but 00i = 00. So equation () has no solutions the equation has only two roots. 00-CE- MTH 保留版權 ll Rights Reserved 00

HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 2000 MATHEMATICS PAPER 2

000-CE MATH PAPER HONG KONG EXAMINATIONS AUTHORITY HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 000 MATHEMATICS PAPER 5 am 45 pm (½ hours) Subject Code 80 Read carefully the instructions on the Answer