CJC 2026 Prelim P2 Ans

1 CATHOLIC JUNIOR COLLEGE H2 MATHEMATICS 2026 JC2 PRELIM EXAMINATION SOLUTION 1 Suggested Solution (a) Since ( 2 ) (5 4 ) 0     v w a b a b , we have 5 6 8 0     a a a b b b 2 2 5 6 8 0 3 6 0        a a b b a b 1 1 cos2 2    a b a b 120   2 Suggested Solution (a) 12 3 3 a d a d a d a    2( 12 ) ( 3 ) 3 (3 2 ) 0 a a d a d d d a      Since d is non-zero, 23d a Common ratio = 2 3a aa  (b) Given 22 ( 1) 1002 3n a n a a       Simplify to 2 2 300 0n n   Using GC (table) n 2 2 300n n  16 -12 17 23 18 60 Using GC (app) or quadratic formula, Considering 2 2 300 0n n   , n = 16.349 or -18.349 solving, we have 16.349 or 18.349n n  least n = 17. (c) From (a), we have r = 3 Since 1 11 1 1 1 3 r r r r r r r r r r r r k u T T T k k kku T k T T                 which is a constant.

2 This is another geometric series with common ratio 3k . (d) 1 r r u    = 1 3 (3 )1 3k k ka a k          , 3k 3 Suggested Solution (a) 2(3+ i)4 – 14(3 + i)3 + a(3 + i)2 + b(3 + i) + 10 = 0 2(28 + 96i) – 14(18 + 26i) + a(8 + 6i) + 3b + bi + 10 = 0 Comparing real and imaginary parts, 8a + 3b = 186 ------(1) and 6a + b = 172 ------(2) Solving simultaneously gives a = 33 and b = -26. (b) Since all the coefficients of 2z4 – 14z3 + 33z2 -26z + 10 = 0 are all real, 3 – i is a second root. (c) We write 2z4 – 14z3 + 33z2 -26z + 10 = 2( (3 i))( (3 i))(2 )z z z cz d      Simplifying, we have 2z4 – 14z3 + 33z2 -26z + 10 = 2 2( 6 10)(2 )z z z cz d    Comparing coefficients of like terms: d = 1 and c = -2 Solving 22 2 1 0z z   , we have 2 4 4(2)(1) 2 4 4 4z      1 1 i2 2z  . (d) The quadrilateral is a trapezium since one pair of opposite sides is parallel to each other.

3 4 Suggested Solution (a) The normal vector n is parallel to 8 0 4 0 6 2 1 2 0 1 1 1 AB AC                                                             = 1 4 8         Scalar product form: 1 0 1 4 2 4 16 8 1 8 r                                Cartesian Equation: 4 8 16x y z    . (b) Method 1 (using length of projection) Shortest Distance = 2 2 2 1 4 8 ( 1) 4 8 AT             = 1 1 0 4 1 8 1 8 8 9 9 k k                      = 8 9 9 k  (shown) Method 2: (using foot of perpendicular) Let F be the foot of perpendicular from T to 1 So vector equation of line passing through T and F is 1 1 2 4 8k                   r Since F lies on the line, 1 2 4 8 OF k            Since F lies on 1 , 1 1 2 4 4 16 8 8k                             1 1 4 2 4 8 8 16 k         9 8 81 k 

4 Distance FT   1 1 2 2 4 8 4 8 1 4 8 k k                                               9 8 1 16 6481 8 9 Shown9 k k     Method 3 (using derived formula for perpendicular distance of point from plane) Shortest Distance   2 2 2 1 1 2 4 16 8 ( 1) 4 8 1 1 8 8 169 1 8 9 Shown9 dOT n n OT n d n k k k                                        (c) Method 1 (using length of projection) By guess and check, P (0, -7, 0) lies on 2 . 2 1 28 9 9 4 1 4 PT k             8 9 11 2 9 3 k k  8 9 3 11 2k k   Removing modulus, 8 9 33 6 or 8 9 (33 6 )k k k k       3, 12k  

5 Alternative method: Use GC to sketch modulus graphs to solve for k. Method 2: (using foot of perpendicular) Let E be the foot of perpendicular from T to 2 So vector equation of line passing through T and E is 1 2 2 1 2k                    r Since F lies on the line, 1 2 2 2 OF k            Since E lies on 2 , 1 2 2 2 1 7 2 2k                             2 1 2 2 2 2 7 k        2 11 9 k  Distance ET   1 1 2 2 2 2 2 2 2 1 2 k k                                             2 11 4 1 49 2 11 3 k k     Since distance of T from the two planes is the same, 8 9 11 2 9 3 k k  8 9 3 11 2k k   Removing modulus, 8 9 33 6 or 8 9 (33 6 )k k k k       3, 12k  

6 Alternative method: Use GC to sketch modulus graphs to solve for k. Method 3 (using derived formula for perpendicular distance of T from 2 ) Shortest Distance     22 2 1 2 2 1 7 2 (2) 1 2 1 2 2 2 73 1 11 23 dOT n n OT n d n k k k                                        Since distance of T from the two planes is the same, 8 9 11 2 9 3 k k  8 9 3 11 2k k   Removing modulus, 8 9 33 6 or 8 9 (33 6 )k k k k       3, 12k   Alternative method: Use GC to sketch modulus graphs to solve for k. (d) Using GC to solve the equations −x + 4y + 8z = 16 2x + y – 2z = −7, Equation of line of intersection: 44 16 9 9 25 14 9 9 0 1 r                                     where R Or 44 9 16 25 149 90r                        where R

7 Alternative Method: Direction vector of line = normal of 1 x normal of 2 = ൭ −1 4 8 ൱ × ൭ 2 1 −2 ൱ = ൭ −16 14 −9 ൱ To find point on the line of intersection, since solving two equations with 3 unknowns, one degree of freedom, so set z = 0 So, -x + 4y + 0 = 16 2x + y – 0 = -7 Solving, we get ݔ= − ସସ ଽ and y= ଶହ ଽ Hence, line of intersection is 44 9 16 25 149 90r                        where R 5 Suggested Solution (a)        E 1 0.8 0 0.15 9 0.05 0.35 Y       OR        E 0 0.8 1 0.15 10 0.05 1 0.35 Y                    2 2 22E 1 0.8 0 0.15 9 0.05 4.85 Y          2 Var 4.85 0.35 4.7275 Y     (b)       1 2 10E number of $1 coins E 10 ... 10 10 0.35 6.5 Y Y Y         (c)           10 10 P at least ten $1 coins P draws ten times P w ins at least one time 0.15 1 0.95 0.401       Alternative Method: Binomial Distribution Let U be the number of coins obtained in 10 games, excluding what Vik already have. ~ B(10,0.05)U ( 1) 1 ( 0) 0.401P U P U     P(draw all 10 games) = (0.15)10 Required probability = 0.401 + (0.15)10 = 0.40126 = 0.401.

8 6 Suggested Solution (a) Whether a ticket holder turns up for the journey is independent of other ticket holders. The probability that each ticket holder turns up for the journey is constant at 0.88. (b) It is likely that some ticket holders are traveling together, so whether a ticket holder turns up for the journey is not independent of other ticket holders. (c)  ~ B 200,0.88X  P 180 0.836X   (d)         P 183 P 183 P 180 1 P 180 0.0281227531 1 0.8362182376 0.172 X X X X         (e) Let largest number of tickets that the company can sell be n.  ~ B ,0.88X n  P 180 0.95X   Using G.C. n  P 180X  195 0.980 196 0.967 197 0.947 Hence largest number of tickets to sell is 196.

9 7 Suggested Solution (a) 6! 720 OR 6 6P , 6 6 6!C  , 6 5 5!C  , 6 3 3! 3!C   , 6 4 2 2 2 2 2C C    (b)  33! 2! 48  OR   33 2 2! 2! 48C    ,   33 2 2! 48P   OR     3 2 2 1 2! 2! 2! 48C C   (c) 6! 4 4! 62