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. Fr