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     = 0   2 2 2 1 1 1 12 2 1x m x y m y    = 0 (Shown) (d) Let the gradients of the two tangents be 1m and 2m respectively. For the two tangents to be perpendicular, 12mm = 1 2 1 2 1 2 y x   = 1 2 1 1y  = 22 x 22 11xy  = 3 As  11,P x y varies, the locus of P at which the two tangents are perpendicular is 22 3xy , which is a circle with centre O and radius 3 .

5 5 1 1 1 1 2 1 4 7 8 1 7 11 13 1 2 5 5      M and 2 2 0 1 1 5 1 3 3 3 1 1 1 13 1 6 6           M (a) Method 1 Use 1M to find column space of 1M 1 2 1 1 2 rref 1 1 1 2 1 0 0 1 4 7 8 0 1 0 1 1 7 11 13 0 0 1 1 2 5 5 0000          M A basis for 1R is 1 1 1 1 4 7,,1 7 11 1 2 5                                 . Method 2 Use T 1M to find row space of T 1M first rref 1 1 3 7 3 1 1 1 1 1 0 0 1 4 7 2 0 1 0 1 7 11 5 0 0 1 1 2 8 13 5 0 0 0 0 T          M A basis for 1R = a basis for the column space of 1M = basis for the row space of 1 TM = 3 0 0 0 3 0,,0 0 1 1 7 1                                 (b) Let 2Mx = 0. 2 0 1 1 5 1 3 3 3 1 1 1 13 1 6 6 x y z w                      = 0 0 0 0       By GC (simultaneous eqn solver), x y z w       = 11 22 11 22 10 01 ts                       

6 A basis for 2K is 11 11,20 02                     . Let 1 1 2 0       = 1 2 3 1 1 1 1 4 7 1 7 11 1 2 5 k k k                                    By GC (simultaneous eqn solver) (or any correct method, e.g., by inspection) 1 1 2 0       = 1 1 1 1 4 711 1 7 1122 1 2 5                                    Let 1 1 0 2       = 1 2 3 1 1 1 1 4 7 1 7 11 1 2 5 m m m                                    By GC (simultaneous eqn solver), 1 1 0 2       = 1 1 1 1 4 731 1 7 1122 1 2 5                                    Since each basis vector for 2K can be expressed as a linear combination of the basis vectors for 1R ,  all vectors in 2K are in 1R , hence 2K is a subspace of 1R . (Shown) (c) The set of vectors which belong to 1R but do not belong to 2K is denoted by W. Since both 1R and 2K are linear spaces, both spaces contain 0, which is in 1 2 2 2 1 ( )R K K K R   . Hence W0 .  W is not a vector space. (Shown) (d) 44 3 :T  is represented by 21MM . Method 1 Find rref MM21 21MM = cle rre l f ar y 0 7 14 14 0 1 2 2 0 18 36 36 0 0 0 0 0 10 20 20 0 0 0 0 0 45 90 90 0 0 0 0                              21rank MM = 1  3rank T +  3nullity T  4 (= dim of 4 , the domain of 3T ) [Or:    2 1 2 1rank nullity M M M M = 4 (= number of columns of 21MM )] 1+  3nullity T = 4

7  3nullity T = 3 Method 2 Solve M M x 021 21MM = 0 7 14 14 0 18 36 36 0 10 20 20 0 45 90 90              Let 21M M x = 0. By GC (simultaneous eqn solver), x y z w       = 1 0 0 0 2 2 0 1 0 0 0 1 t s w                                   the null space of 3T has 3 basis vectors,   3nullity T = 3.

8 6 C: 2,2x at y at , 0a is a constant, t Observe that   2,2P at at is on C.  ,0Fa . (a) PF =     2 22 2at a at = 2 4 2 2 2 2 224a t a t a a t   = 2 4 2 2 22a t a t a =   22at a = 2at a   2 0at a (b) Distance of P from the line xa = 2 ()at a = 2at a   2 0at a = PF  C is a parabola. (with F as its focus and xa as its directrix) [Recall definition: A parabola is the set of al l points in the plane that are equidistant from a fixed point (focus) and a fixed line (directrix of the parabola) that does not contain the focus.] Alternative method: find Cartesian equation of C C: x = 2at (1) y = 2at (2) 0a is a constant, t From (2), t = 2 y a Sub. into (1), x = 2 2 ya a   = 2 4 y a 2y = 4ax  C is a parabola. (with F as its focus and xa as its directrix) (c) dd 2 , 2dd xy at att d d d d d 2 1 d2 y t x t ya x at t   The equation of tangent at   2,2P at at is 2y at =   21 x att  1y x att

9 When y = 0, 1 x att  = 0 x = 2at  this tangent meets the x-axis at   2,0Q at . (Shown) (d) Method 1   2,0Q at ,  ,0Fa QF = 22()a