JPJC 2026 Prelim P1 Solutions

12 6 r h r Jurong Pioneer Junior College H2 Mathematics JC2-2026 Preliminary Exam Paper 1 Solution Q1 From graph, 0 1 k  Q2 By similar triangles, 6 1 12 2 r h    1 2r h 2 31 1 3 2 12 hV h h        2d 1 d 4 V hh   Now, d d d d d d V V h t h t    21 d2 4 4 d h t   d 1 d 2 h t    – 0.159 cm / min (3 s.f.) Rate at which the h is decreasing = 1 2 cm / min Q3 (i) Area of R =     4 2 0   5 25 dxx x   5.59 (ii) Volume of solid = 3 2 5 25 d y y    + Volume of cylinder  Volume of cone =         3 2 22 5 125 d + 4 2 4 4 3y y       = 88.0

2 Q4 2 2 2 Surface area, 1500 2 2 1500 2 750 2 r rh r rh r r            2 2 3 750Volume, 750 rV r r r r           At stationary point, 2d 750 3 0d V rr    23 750r  250 250 or (NA since 0)r r r     2 2 d 6 0 (since 0)d V r rr    . Hence, V is maximum. When 250r  , 3750V r r   2(750 ) 250 250750 250500 102500 r r                     ( a = 2500) Q5 (i)(a) a b c         is perpendicular to the plane. (i)(b) d is the perpendicular distance from the origin to the plane. (ii)(a) ( ) ( ) ( ) ( ) ( ) ( )              u v u v 0 u u v v v u u v 0 Since    u u v v 0 and ( )   v u u v , we have 2( ) u v 0 . Since  u v 0 , u and v are parallel. (ii)(b) ( ) ( ) 2 2 sin 90 2       u v u v u v u v

3 Q6(a) Since 2 3i  is a root, 3 2( 2 3i) ( 2 3i) 7( 2 3i) 0a b          . 60 12i ( 5 12i) 0a b       Comparing real parts, 5 60a b   ----------(1) Comparing imaginary parts, 12 12a  1a  Sub into (1): 65b Method 1 Using GC, the other two roots are 2 3i  and 5. Method 2 (When use of calculator not allowed) Since 2 3i  is a root, and the coefficients in the cubic equation are all real, 2 3i  is another root. The last root is a real number c 3 2 2 7 65 [ ( 2 3i)][ ( 2 3i)][ ] ( 4 13)( ) z z z z z z c z z z c                Comparing constant term, last root = c = 5 (b) (c) Area of triangle PQR 1 (6)(2 5) 212   units 2 5 O Im Re 5 O Im Re 5 2 3 3

4 Q7 (a)(i) 120 17 14n nv v    1 17 14 20 n n vv   From GC, 15 4.495v  (3 decimal places) (ii) As n , nv L 1and nv L  20 17 14 3 14 14 (exact)3 L L L L     (b)(i) 2 2 2 21 2 3 ...... 4 n    = 2 2 2 21 2 3 ...... (2 ) n    = 2 2 1 n r r   =  1 2 (2 1)(4 1)6 n n n   = 1 (2 1)(4 1)3 n n n  (b)(ii) 2 2 2( 1) ( 2) ........ 4n n n     = 2 2 1 n r n r    = 2 2 2 1 1 n n r r r r     = 1 1(2 1)(4 1) ( 1)(2 1)3 6n n n n n n     =  1 (2 1) 2(4 1) 16 n n n n      =  1 2 1 (7 1)6 n n n  2 2 2 21 3 5 ........ (2 1) n     = 2 2 2 2 2 2 1 2 4 6 ...... (2 ) n r r n        using (i) = 2 2 2 2 2 2 2 1 2 1 2 3 ...... n r r n        = 2 1 1 (2 1)(4 1) 43 n r n n n r     

5 = 1 1(2 1)(4 1) 4 ( 1)(2 1)3 6n n n n n n          1 2(2 1)(4 1) ( 1)(2 1)3 3n n n n n n       1 (2 1) (4 1) 2( 1)3 1 (2 1)(2 1)3 n n n n n n n         Q8 (a)(i) 2 1 cos 4cos 2 d d 2 xx x x   1 sin 42 8 x x C   (a)(ii)   2 2 2 2 2 1 12 e 1 2 e d d 2e 2 e 2 e 21 12 12 x x x x x x x x x C                 2 e 2 x C   (a)(iii)   1 2 2 4 d = tan = d d 2 d 1 2 vu x x x u x x vx x      1 2tan dx x x    2 2 1 2 4 2tan d 2 2 1 x x x x x x          2 3 1 2 4 1 4tan d 2 4 1 x x x x x        2 1 2 4 1tan ln 12 4 x x x C    (b) 24cosx  d 8sin cosd x   

6 When x = 2, 24cos 2  1cos 2 4     When x = 3, 24cos 3  3cos 2 6       3 2 6 2 2 4 1 d 4 1 8sin cos d 4cos 4 4cos x x x                4 2 2 6 1 8sin cos d 4 cos 4 4cos          4 2 6 sin d cos 1 cos         4 6 1 dcos    4 6 sec d     4 6 ln sec tan        ln sec tan ln sec tan4 4 6 6        = 2 1ln 2 1 ln 3 3    =  ln 2 1 ln 3  = 2 1ln 3  (exact) Q9 (i)

7 2 3 2 1 9 493 4 3 2 2 2 18x t t y t t t       d 6 4 2(3 2)d x t tt     2d 9 9 2 (3 1)(3 2)d y t t t tt       d (3 1)(3 2) 1 2 (3 1),d 2(3 2) 2 3 y t t t tx t      (ii) At the stationary point, d 0d y x  1 (3 1) 02 1 3 t t    When 1 3t , 2 3 2 1 1 1 13 43 3 2 2 1 9 1 1 493 2 33 2 3 3 18 x y                                  Coordinates of Q 1 ,32     Equation of tangent is y = 3 (iii) When y = 3, 3 2 9 493 2 32 18t t t    Using GC , 5 1 or 6 3t t  (NA) (Reject since it is pt Q) When t = 5 6 , 2 5 5 1 33 46 6 2 4x              coordinates of R is 3, 34     .

8 d 1 5 3 3 1d 2 6 4 y x          Equation of normal at R: 4 33 3 4y x       4 23y x  (iv) Coordinates of N 0, 2 . Area of QRN  1 3 1 ( ) 3 22 4 2          1 1 12 4      1 8 unit2 Q10 (a) Acute angle between 1L and 2L 1 1 2 0 1 1 3cos 1 2 0 1 1 3                                      

9 1 5cos 2 14 19.1    (b) Suppose 1L and 2L intersect. Then 4 1 4 2 1 0 3 1 0 1 3 c                                          . 2 0 (1) 2 (2) 3 (3) c              Sub (2) into (1): 4 Sub into (3): 2c Since 1L and 2L do not intersect, 2c . (c) Let B be the point (4, 1, 0) on 1L . Perpendicular distance from A to 1L 1 0 1 1 0 1 BA                  0 1 2 0 1 2 c                  2 2 2 c         24 4 2 c    21 82 c  (d) Perpendicular distance from A to  1 0 1 1 0 1 OA                    4 1 3 0 1 2 c                    4 2 c Since   2 41 82 2 cc   , 2 2 8 8 16c c c    . 1c 

10 Q11(i) (a) (b) (ii) 1a , 1 3b 1g( ) g(3 ) g 3 g(3 1) 3y x y x y x x                1 3c , 1 3d  x y O x y O