EJC 2023 JC1 Promos 9649 (Solutions)

1 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 [Turn over Q1 Solution (a) (b) Since r is minimum at 11 π6  , d1 311 πd tan 6 y x   At 11 π6  , 3 2x and 1 2y . Hence the equation is 13 33 222yx y x         r = 3 r = 1

2 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 Q2 Solution (a) dd dd yxyx yx x  22 22 dd d 2dd d yyxy xx xx  33 2 33 2 dd d 3dd d yyxy xx xx Conjecture:  1 1 dd d dd d nn n nn n yyxy x nx xx   , n   (b) Let nP be the proposition  1 1 dd d dd d nn n nn n yyxy x nx xx   where n   . When n = 1:  dRHS d d LHSd yxy x xyx   1P is true. Assume kP is true for some k   :  1 1 dd d dd d kk k kk k yyxy x kx xx   To prove 1kP is true:   1 1 1 1 1 1 1 1 dd d dd d dd d dd d ddd ddd dd (1 )dd kk kk kk kk kkk kkk kk kk xy xyxx x yyxkxx x yy yxk x xx yyxk x x                 Since 1P is true and kP is true  1kP is true, by Mathematical Induction, nP is true for all n   . (c) Let e xy  , d 1ed n n n xy x          1 1 1 dd d ee edd d 1e 1 e 1e x xx nn n nn nn x x n n x xx nxx x xn xn          

3 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 [Turn over Q3 Solution (a) 3 2 2 1 31 3 2 2 (2 5) 15 1 1 15 1 1 21 1 4 01 22 1 2 16 11 0 2 5 15 12 15 1 1 01 2 2 0 02 9542 15 1 1 01 2 2 0 0 (2 1)( 5) 2(2 1) RR RR Ra R R aa aa a aa a a aa a                             For system to be consistent, (2 1)( ) 0 5aa  OR (2 1)( 5) 0 2(2 1)aa a    1 ,52a OR 1 2a 5a (b) When 1 2a , 5 2 15 1 1 01 2 000 0    5 2 51 2 x yz yz    Let z  , 5 22y    2 5 7 2 2215 9 x         92 7 25 0 2 2 x y z                   ,   (c) 5( 1 ) 1 21 1 (1 ) 4 2 16( 1) 11 xy z xy a z xa y z        Substituting 1z into solution above, 92 7 25 1 2 2 x y z                   ,  

4 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 Q4 Solution (a) Since   33f(1)f(2) 1 3 1 2 6 1 3 0       , there is at least one root in the interval. Since 2f() 3 3xx  , which is positive over the interval (1, 2), there will be exactly one root in the interval. (b) (1)f(2) (2)f(1) (1)(1) (2)( 3) 7 1.75f(2) f(1) (1) ( 3) 4       Since 2f() 3 3xx  and f() 6x x  , are both positive over the interval (1, 2), the curve is upward sloping and concave upwards, hence the chord used for linear interpolation will lie above the actual curve, resulting in an underestimate. (c) (i) 33 1 22 31 2 1 or 33 33 nn n nn nn xx xxx xx      (c) (ii) For 1 1x  , f( 1 ) 0  and so g(1) is undefined, causing the iterative process to fail. From the GC, using 1 2x  , we have 1.879 (3 d.p.)  n nx 1 2 2 17/9 3 10555/5616 4 1.879385 5 1.879385 (d)    3 3 1 33 12 3 1 27 2 2313 1 nn n nn n nnnn uu u uuu uuuu      Using this formula, 45 1.879303, 1.879378uu . This suggests that the linear interpolation converges almost as fast as the Newton-Raphson method in this particular context.

5 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 [Turn over Q5 Solution (a) 3 :2 4 0 a Sb a b c c         Note that S0 , hence S is non-empty. Let , Suv such that 12 12 12 , aa bb cc        uv , and   Then 12 12 12 a b c a b c        uv 121 2 1 2 11 1 2 2 2 2( ) ( ) 4( ) (22) 0 44 ab c ab c b ac c b a           Suv As 3S   is closed under addition and scalar multiplication, S is a linear space. (b) Let , Suv such that 12 12 12 , aa bb cc        uv , and   12 12 12 a b c a b c        uv    12 12 12 12 1 2 12 1 2 1 2 1 2 12 12 2 2 11 2 111 2 2 211 2 21 TT 3( ) 6( ) 4( ) 2( ) () 33 642 642 TT a b c ac abc ab c ac ac abc abc ab a b c ac a ca b c bc ab c                                        u u v v Therefore T is a linear transformation.

6 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 (c) 10 3 64 2 11 1    A 10 3 0 64 2 0 11 1 0 x y z                        Using GC, 10 3 rref ( ) 0 1 4 00 0   A Let z  , ,34xy   Basis for the null space of A = 3 4 1   (d) As nullity( ) 1A , 3rank( ) dim( ) nullity( ) 31 2   AA  (e) as 3 T6 4 2 aa c ba b c ca b c          , and 2( 3 ) (6 4 2 ) 4( ) (2 6 4) ( 4 4) (6 2 4) 0 ac abc a b c ab c           Range(T) S . As dim(Range(T)) rank( ) 2 A , and dim( ) 2S  , Then Range(T) S . Alternatively, Column Space of 10 Span 6 , Rang 1 e(T 4 ) 1          A As 2(1) 1(6) 4(1) 0 and 2(0) 1(4) 4(1) 0 ,

7 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 [Turn over Range(T) S . As dim(Range(T)) rank( ) 2 A , and dim( ) 2S  , Then Range(T) S . (f) 210 3 06 42 11 1 1                             12 T1 0 11   Subset of 3 required is 13 14 , 11            

8 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 Q6 Solution (a) 1240nnnuuu   Auxiliary equation: 2 41 0  42 0 2 25    General solution:    25 25 n n n uA B  , where A, B are constants When 0n , 0 1 ----- (1)ABu   When 1n ,  1 ----- (2)2525ABu     25 (1) – (2):  22 55 B   5 5 2 2 B    1 52 25 A B       22 22 55 55 0 25 5 , 2 nn nun              Alternatively:      (1) (2) 2 5 : 25 1 0 4 5 12 5 9 4 52 5 , 1 0 45 1 0 45 1 A AB            (b) As 21 5 , 25 n   as n . As 21 5 , 25 0 n  as n . Therefore nu converges if 0 2 25 5   25 .

9 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 [Turn over (c)(i) 1 1 1 1 1 1 1RHS 1 4(1 ) 11 241 2 21 4(4 ) 21 4 2 2 LHS n nn n n nn n n nn n n s uu u u uu u u uu u s                  (c)(ii) 1 11 4(1 ) n n s s    Let the limit of the sequence be l. Then as n , 2151 4(1 ) 4ll l   0l  , since 0 1u  , 1 0u  0 for 0nun  0 for 2nsn  5 2l (c)(iii) 8 8 7 8 2 2 uus u  Using GC, 87 5473, 1292uu 8 12238 10946s  8 5 2s  122385 5473

10 2023 JC1 H2 Further Mathematics Promotional Examination Paper 1 Q7 Solution (a) Consider the triangle PQR, with RPQ  , PRr , 2PQc and 2RQ r a . Using cosine rule,     22 2 2c o sRQ PQ PR PQ R R PQP         22 2 22 2 2 22 22 22 2 2 c o s 444 4 c o s 1c o s , ra c r c r ra r ac rc r ca cr aa ca cuv aa             (b) ce a (c)        2 2 22 22 222 2 2 2 22 2 242 2 2 22 2 22 22 4 2 22 2 2cos 1 cos cos from (a) Substituting cos , Substituting , ,s i n 2 22 ,rx c y ca cra r c r a caa ra r c x c a ca r a c x r a xc y a a c xc x ax ac x ac ay a ac x xc r y r cx a                             22 2 2 22 2cx a y aa c  (d)   2 22 2 2 232 3 3 32 1 3 xxy y      At intersection with ym xk ,      2 2 22 2 2 22 2 13 2 33 31 6 33 0 x mx k x m x mkx k mxm k x k          Since ym xk is a tangent, there is only 1 solution, i.e.     2 22 22 22 2 2 22 22 2 2 22 64 3 1 3 3 0 36 4 9 9