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 isin 5cos isin isi n= + + + + +          5 4 3 2 2 3 4 5cos 5icos sin 10cos sin 10icos sin 5cos sin isin= + − − + +          ( ) 5 3 2 4 4 2 3 5cos 10cos sin 5cos sin i 5cos sin 10cos sin sin= − + + − +          Comparing the imaginary parts: 4 2 3 5sin 5 5cos sin 10cos sin sin= − +      Comparing the imaginary parts: 5 3 2 4cos5 cos 10cos sin 5cos sin     = − + 5 3 2 4 5 4 2 3 5 5 53 42 53 42 1 cos 10cos sin 5cos sin sincot 5 15cos sin 10cos sin sin sin cot 10cot 5cot 5cot 10cot 1 10 5 where cot5 10 1 u u u uuu                  −+= −+ −+= −+ −+== −+ (iii) Let 22cotzu == 2 4 2 5 310 5 0 10 5 0 10 5 0z z u u u u u− + =  − + =  − + = Using (ii), cot 5 0 = Using (i), 2 2 2 3cot , cot , cot10 10 2z        =             z 22 3cot , cot since 010 10 z   =       

8 (iv) 2 10 5 0zz− + = 210 10 4(1)(5) 5 2 5 2z − = =  Since 2cot  is a decreasing function over 0, 2   , 22 3cot 5 2 5 and cot 5 2 510 10    = + = −       Thus ( ) 22 3 1 1 1tan tan + = 5 2 5 5 2 5 = 2 10 10 5 5 2 5 5 2 5    + = − + +    +−   

9 [Solution] (i) 1 ,1nnC pC k n+ = +  (ii) 1 100C = Method 1: ( ) 21 2 32 2 3 2 43 100 100 100 (1 ) 100 (1 ) 100 (1 ) C pC k p k C pC k p p k k p k p C pC k p p k p k p k p p = + = + = + = + + = + + = + = + + + = + + + : 1 2 2 1 1 1 100 (1 ... ) 1100 1 100 11 nn n n n n C p k p p p ppk p kk ppp −− − − − = + + + + +  −=+  − = − + −−

10 Method 2: Let 1 ()nnC p C+ − = − (1 ) 1 nn kpC k pC p p k p   + − = −  − =  = − 1n n C pC   + − =− Let nnvC =− . Then 1 n n v pv − = is a constant and the sequence  nv is a geometric progression with first term 11 100 1 vC k p = − = − − and common ratio p. The general term of the sequence is given by 1100 1 n nv k pp −=−  − . 1100 11 n n kkCp pp − = − + −− (iii) k =30 For 11 12 30 30200, 100 20011cp pp = − + = −− From the GC, the graph of 1130 30100 11yp pp = − + −− and y = 200 intersect at p = 0.868  87% Hence the company needs to retain approximately 87% of customers, week by week. (iv) If p = 0.75 For 11 12 200, 100 0.75 2000.25 0.25 kkc  − +  11 11[4 4(0.75 )] 200 100(0.75 ) 51.1 k k −  −  Least k = 52 (iv) If p = 0.75, k = 100 11100 100100 0.75 400 300(0.75 )0.25 0.25 nn nC −−= − + = − As , 0.75 0, 400n nnC→ →  → The number of customers will increase and approach 400 in the long run. Hence, based on the model, the company will succeed in maintaining a stabilised number of customers at 400, but will not experience any further growth. (To comment on the long term prospect, we need to mention about the growth of the company in the long run based on this model.)