JPJC_9649_2023_Prelim_P1_Solutions

H2 Further Mathematics 2023 JPJC JC2 Prelim Examination Paper 1 Solutions 1 (a) Ax x and Βx x .  A B x = Ax Bx = xx =   x   is an eigenvalue of AB with x as a corresponding eigenvector. (Shown) (b) Ax = x 3 1 0 1 4 6 6 1 5 11 10 1                = 1 1 1     = 4 (satisfying all 3 eqns) (c) Eigenvalue Eigenvector A B AB 1 1 1   4 2 6 1 2 3    1 3 4 1 1 2    2 1 3 Since AB is a 33 matrix with 3 distinct eigenvalues, so AB is diagonalisable. AB = 1PEP where P = 1 1 1 1 2 1 1 3 2    and E = 600 0 4 0 0 0 3      3 AB = 31 PE P = 1PDP , where D = 3E = 3 3 3 6 0 0 0 4 0 0 0 3     = 216 0 0 0 64 0 0 0 27   

2 2 (a) 1 1 0.6 1.2 14000 0.72 14000 nn nn pp pp       (b)  0.72 n np c d 14000 500001 0.72d   0.72 50000 n npc Given that 0 220000p  ,   0 220000 0.72 50000 170000 c c   Hence,  170000 0.72 50000 n np  .   60000 170000 0.72 50000 60000 n n p   Using GC, 8 9 10 62277 60000 58840 60000 56365 60000 p p p    Hence, should consider closing after 9 months of business.

3 3 (a) 5 16 16 3 iz  = 0 5z =  16 1 3 i  16 1 3 i = 16 1 3 32   arg 16 1 3 i = 1 3tan 1  ( 5z lies in Q3) = 2 3   5z = 2i 332e  = 2i2 332e k  z = 2 i 55 2 3 32 e , 0, 1, 2 k k       =  i 22 5152e k  = (6 2)i 152e k   = 14 8 2 4 2i i i i i15 15 15 15 32e , 2e , 2e , 2e , 2e        (b) For 2 z to be the least, look for the root that is closest to the complex number represented by  2, 0 . 2i 152ez  

4 4 Ellipse: 22 22xy (1)  11,P x y is external to the ellipse. (a) The equation of the line that has gradient m and passing through  11,P x y is  11y y m x x   . (b) Line: y =  11m x x y (2) For points of intersection, sub. (2) into (1),   22 112x m x x y   = 2     2 2 2 2 2 1 1 1 1 12 2 2x m x x x x m x x y y      = 2 2 2 2 2 2 2 2 1 1 1 1 1 12 4 2 4 4 2x m x m x x m x my x mx y y      = 2     2 2 2 2 2 1 1 1 1 1 11 2 4 2 2 4 2m x m y mx x y m x mx y       = 0 (Shown) (c) For the line to be tangent to the ellipse, discriminant = 0      2 2 2 2 2 1 1 1 1 1 14 4 1 2 2 2 4 2m y mx m y m x mx y      = 0      2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 116 2 8 1 2 2 1m y mx y m x m y m x mx y       = 0      2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 12 2 1 2 2 1m y mx y m x m y m x mx y       = 0 4 2 4 2 11 222 2 2 2 11 2233 1 1 1 1 1 1 1 1222 44 21 2 2y m x mmm x y m x y x m xmy myx my    = 0 2 2 2 2 1 1 1 1 2 1 2y x m x y m m  