2018 A Level H2 FM 9649 P1 (Qns & Solutions) (29 Sep 2024)

1 2018 GCE A Level H2 Further Maths 9649 Paper 1 Solutions [Solution] Let f ( ) tan 1x x x=− 2f '( ) sec tanx x x x=+ Applying Newton-Raphson method, taking 1 = 3.5, 1 2 tan 1 sec tan nn nn n n n     + −=− + 2  3.42875 3  3.42562 4  3.42562 Thus  = 3.42562 (correct to 5 d.p.) Key in recurrence relation in GC to generate the approximations

2 [Solution] (i) When P0 = 0.33, the deer population decreases to zero within 1 unit of time. (ii) A stable population cycle occurs for a period of (26.25 − 20) or 6.25 The units in which t is measured is 365 58.4 586.25= days (iii) During the stable cycle, 14 cos2Pt− or 124 cos2 6.25Pt − 

3 [Solution] (i) Let 2 1f ( ) 2x x= + ( ) ( ) ( )1 1 1 3 1 1 4 1 4 1f 0 4f 2f 1 4f f 2 4 2 43 2 2 2 6 2 9 3 17 6 310 459 I                + + + + = + + + +                             = (ii) 2 2 11 20 0 1 1 1 d tan tan 22 2 2 2 xIx x −− = = =  +  (iii) Using (i) and (ii) results, 1 99 310 341tan 2 2 70 459 357I− =   = (iv) Error  1 341 1tan 2 0.00134545 0.0001351357 7400 − − =  =

4 [Solution] (i) Coordinates of F is (a, 0) 2 d d 24 2 4 dd y y ay ax y a x x y=  =  = At P, d 2 1 d2 ya x ap p== Equation of tangent at P: ( ) 212y ap x ap p− = − 1y x app=+ When y = 0, x = −ap2 Thus the coordinates of Q is (−ap2, 0)

5 (ii) (a) Let the angle between the beam and the tangent be . Using the property of corresponding angle, PQF = 22 ( 1)QF ap a a p= + = + ( ) ( ) 2 2 2 4 2 2 4 2 222 ( ) (2 ) 2 1 4 2 1 1 ( 1) PF ap a ap a p p p a p p a p a p = − + = − + + = + + = + = + Since QF = PF, QPF PQF  = = Thus angle between the beam and the tangent is equal to QPF . (b) All satellite beams are reflected at the dish and are directed to the focus of the dish.

6 [Solution] e sin ,txt= e cos ,tyt= 0. 2t  ( ) ( ) 22 22dd e cos sin e cos sindd ttyx t t t ttt        + = + + −           ( ) ( ) 2 2 2 2 2 2e cos 2cos sin sin e cos 2cos sin sintt t t t t t t t t= + + + − + ( ) 22e 2 cos sin 2 ett tt= + = Required surface area ( ) 2222 00 dd2 d 2 e cos 2 e ddd ttyxy t t t tt      = + =        ( ) 2 2 0 2 2 e cos dt tt=    ( ) ( ) ( ) ( ) ( ) ( ) 22 22 2 2 000 2222 0 0 2 2 0 2 2 0 2 2 0 11e cos d e cos e sin d22 1 1 1 1 e sin e cos d2 2 2 2 1 1 1 e e cos d2 4 4 51 e cos d (e 2)44 1e cos d (e 2) 5 t t t tt t t t t t t t t t t t tt tt tt =+  =− + −  =− + − =− =−               ( )1 Area = 2 2 e 2 5 −  

7 [Solution] (i) cot 5 0 = 355 , , ,... 2 2 2 3, , since 10 10 2 2 2         =   =  −   (ii) cos5 isin 5+ ( ) 5 cos isin=+  by de Moivre’s Theorem ( ) ( ) ( ) ( ) ( ) 2 3 4 55 4 3 2cos 5cos isin 10cos isin 10cos