2017 A Level H2 FM 9649 P1 (Suggested Solutions)

1 2017 GCE A Level H2 FM 9649 Paper 1 Solutions [Solution] Let 9 1w 9 10w     33 10w       23 3 31 1 0w w w       3 6 31 1 0w w w     3 6 31 0 or 1 0w w w       6 roots of 63 10zz   are among the nine 9th roots of unity. Since 9 10w  has the 9 roots as shown on the Argand diagram and 3 10w  has roots  2 /3 ie  ,  4 /3 ie  and 1. Hence, 6 roots of 63 10zz   are  2 /9 ie  ,  4 /9 ie  ,  8 /9 ie  ,  10 /9 ie  ,  14 /9 ie  &  16 /9 ie  . [Solution] s a c b d    -------- (1) AI ab cd       is an eigenvalue of A  det A I 0      0a d bc      2 0a d ad bc          2 0a d ad s d s a        from (1)     22 0a d ad s a d s ad            22 0a d s a d s            0s s a d      [by long division] ors a d s      s is an eigenvalue of A and the second eigenvalue a d s   ab 1

2 [Solution]   2fe xxx (i)      1 2 2 2f 2 e e 2 1 e x x xx x x          2 2 2 2 2 2 2f 2 e 2e 2e 2 2 2 ex x x xx x x             3 3 2 2 2 2 2 2 2 3 2 2f 2 e 2 e 2 e 2 e 2 3 2 ex x x x xx x x            4 4 2 3 2 3 2 3 2 3 2 4 3 2f 2 e 2 e 2 e 2 e 2 e 2 4 2 ex x x x x xx x x       Conjecture:      12f 2 2 e n n n xx x n  , for all n  (ii) Let Let Pn be the statement:      12f 2 2 e n n n xx x n  , for all n  When n = 1, LHS =      1 2f 2 1 e xxx ; RHS     1 0 2 2= 2 2 e 2 1 e xxxx   = LHS Hence 1P is true. Assume Pk is true for some k  , i.e.,      12f 2 2 e k k k xx x k  Need to show that 1Pk is true, i.e.,      1 12f 2 1 2 e k k k xx x k     LHS =       1 dff d kk xx x     12d 2 2 ed k k xxkx     1 2 22 2 2 e 2 ek k x k xxk      1 2 22 2 e 2 ek k x k xxk     122 1 2 ek k xxk   = RHS Thus 1P is true P is also true.kk  Since 11 is true and is true is true,kkP P P  by mathematical induction, nP is true for all .n 

3 [Solution] T: 2 cos3 θ, 0 θ 2r     (i) 0r No solution. Max r =3, when cos3θ 1 3θ 0, 2 , 4 24θ 0, , 33  Min r =1, when cos3θ 1 3θ , 3 , 5   5θ , ,33  Lines of symmetry: x-axis ( θ 0,  ), The line 4θ , 33  The line 25θ , 33  (ii) Area enclosed by T   2 0 12 d θ2 r      2 0 2 cos3 θ dθ    2 0 4 4cos3 θ cos 3θ dθ       00 414θ sin 3θ 1 cos6θ dθ32       0 114 θ sin 6θ26       94 22    3 1

4 [Solution] Given that f ( )f ( ) 0ab  , a < b. (i) Case (1): 3 distinct real roots in (a, b). Case (2): No real root in (a, b). In both cases above, f ( )f ( ) 0ab  but f ( ) 0x  does not have exactly one real root. Alternatively: Case 1 Case 2 a f(a) f(b) b y = f(x) x y f(a) f(b) b a y = f(x)

5 (ii) Let 2f ( ) sec e xxx .  is a root of f ( ) 0x  ,  1.5, 2.5 . By linear interpolation,        1 1.5 f 2.5 2.5 f 1.5 f 2.5 f 1.5x   1.5 10.62445 2.5 195.36835 10.62445 195.36835   2.4484 2.45 (to 3sf) Since 1f ( ) f (2.4484) 9.8797 0x    and f (1.5) 195.36835 0 , the root  1.5, 2.45 . Since f is a continuous function in the interval [1.5, 2.5], linear interpolation is a suitable method to find the root. However, the first approximation lies in the interval  1.5, 2.45 which is only slightly shorter than the interval  1.5, 2.5 . Hence this method of approximating the root is slow, not an efficient method. (iii)  is a root of f ( ) 0x  ,  β 0.1, 0.9 . By Newton-Raphson method, 0 0.5x  , 2f '( ) 2sec tan e xx x x     1 ' 2 f f sec 2sec tan n n n nn n xn n n x nn xxx x xex x x e        1 ' f 0.5 0.350274860.5 0.5 1.0227 0.2300322f 0.5x       2 0.75526x  3 0.40306x 

6 As ,0 nnx  . nx converges to the other root 0 instead of . As shown in the diagram, since the initial approximation 0 0.5x  lies on the LHS of the minimum point. The tangent line has negative gradient and cuts the x-axis to the left of origin. All subsequent approximations will converge to the other root 0 instead of . 0.561 y 0 0.9 x

7 [Solution] Hyperbola: 22 22 1xy ab , a > 0, b > 0. (i) secθ,xa tanθyb Sun at focus (ae, 0). (a) Point on the orbital path at which it is closest to the sun is the point (a, 0). p ae a   ( 1)ae (b) Given that e = 1.04, p = 1.312,  1.312 0.04a 32.8a Since, 2 2 2 ()a b ae 2 87.788544b  9.37b ( 3s.f) (ii) Circle:     22 20 1.52x ae y        22 2sec tan 1.52a ae b    2 2 2 2 2 2 2 2sec 2 sec sec 1 1.52a a e a e b          2 2 2 2 2 2 2 2sec 2 sec 1.52a b a e a e b     2 2 2 2 2 2sec 2 sec 1.52a e a e a     2 2sec 1.52ae a  Since ( , )22  , sec 0  .  1.52sec cos 0.994 1.52 a ae ae a      0.110  x y a F(ae,0) 1.52 (asec, btan)

8 Arc length = 22 0.11 0 d dy2 dt dt x dt              0.11 22 2 0 2 32.8sec tan 9.37sec dt   =1.0599 2.11986 2.12

9 [Solution] (i)  1 33000 1 0.075 0.05 100000g      33000 0.875 100000    2 21 33000 0.875 33000 1 0.875 0.875 100000gg        23 32 33000 0.875 33000 1 0.875 0.875 0.875 100000gg       133000 0.875nngg  is the recurrence relation. Also,    2133000 1 0.875 0.875 0.875 0.875 100000nn ng        1 0.87533000 0.875 1000001 0.875 n n       264000 1 0.875 0.875 100000nn   7 199597.872 200000g   ( to 3 sf) (ii) 0 200000g    1 0 0 1 0.075 0.025 0.9g g t g t         2 2 1 0 00.9 0.9 0.9 0.9 1 0.9g g t g t t g t        So,  7 2 6 70 0.9 1 0.9 0.9 0.9 0g g t         7 71 0.9 0.9 2000001 0.9t    18335.98t

10 [Solution]   d 1 1 1d 3 4 P P P ht    (i) When h = 0,   d 1 1 1d 3 4 P PPt  11 dd(4 ) 12 PtPP    1 1 1 1 d4 4 12 P t cPP    11ln ln 44 12 P P t c    1ln '43 P tcP  /3e4 tP AP  where 'ecA /3 /3 /3 4 e 4 1 e e t tt AAP AA  As /3, e 0 tt   . 4P (millions). OR   d 1 1 1d 3 4 P PPt  Let   d 1 1 10d 3 4 equilibrium values : 0 or 4 P PPt PP       For 04 P , d 0d P t  , population of the fish increases and stabilizes at 4 millions. For P = 4, d 0d P t  , the population of the fish remains constant at 4 millions. For 4P , d 0d P t  , the population of the fish decreases and stabilizes at 4 millions.