SAJC 9758 2025 Prelim P1 Solutions

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 1 of 20 No Solution 1 32 2 22 32 22 2 4 3 2 2 f ( ) f (2) 8 4 2 10 (1) f ( 2) 8 4 2 2 (2) f '( ) 3 2 f '( 1) 3 2 0 (3) f ( ) d 16 ( ) d 16 1 1 1 164 3 2 16 4 16 (4)3 Usin x ax bx cx d a b c d a b c d x ax bx c a b c x x ax bx cx d x ax bx cx dx bd −− − = + + + = + + + = − − =− + − + =− − = + + − = − + = − =  + + + =  + + + = + = −  g GC to solve (1), (2), (3), (4) 3, 0, 9, 4 a b c d= = =− = 2 Let r and h be the radius and the height of the cylinder respectively. 2 2 External Surface Area 2 2 2 22 p r rh p r phr rr    =+ − = = − Volume 2 2 3 2 2 V r h prr r pr r     = =−  =− 2d 3d2 Vp rr =− For stationary values, 1 22d 30d 2 6 π V p p rrr  = − =  =  2 2 d 6d V rr =− When 1 2 6π pr = , 1 2 2 2 d 60d6 Vp r   =−    V is a maximum when 1 2 6π pr = cm.

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 2 of 20 3(i) 12 1 2 0 2 : 2 1 , : 5 2 , & 0 2 8 3 22 1 2 , where is a scalar 23 ll kk     −               = + = + −                −        −       −          rr 1l is not parallel to 2l . Equate the two lines: 1 2 0 2 2 1 5 2 0 2 8 3  −               + = + −              −        2 2 1 (1) 2 3 (2) 2 3 8 (3) No solution from GC OR 5Solution for equations (1) and (2) 2 , . 2 check that it     += − − =− − + = =− = 12 satisfies (3). 57LHS = 2(2) 3 22 RHS=8 Hence (3) is not satisfied. Since and are non-parallel and non-intersect ing, they are skew lines.ll − + =  (ii) Let F be foot of perpendicular from P to 1 . Equation of line through P and F: 1 0 1 5 1 5 , 01                = + = +           −−      r Substitute into equation of plane 1 : 10 5 1 2 1          + • =      −−    F P

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 3 of 20 1 375 2 2 22 3 2 OF    + =  =−  =     2 1 1 ' 2 7 5 2 3 0 3 Coordinates of ' are (1, 2,3). OP OF OP P            = − = − =                 Alternative method to find foot of perpendicular: Q(0,2,0) is a point on plane  00 11 11 1 1 1 1 11 1 0 0 0 13 3 1 1 122 0 1 1 1 101 37 1522 10 3 2 PF PQ OF          =•    ++    −−     −         −       = − • =               − − −             −     = + =       −−       2 1 1 ' 2 7 5 2 3 0 3 Coordinates of ' are (1, 2,3). OP OF OP P            = − = − =                 (iii) Let ( )1 ,2,0A that lies on 1l . 1 1 0 5 2 3 0 0 0 AP OP OA            = − = − =                uu u r uu