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 u r uu r Normal of plane 2 :

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 4 of 20 1 2 2 0 6 1 1 3 0 6 0 2 0 6 1 Equation of plane containing and : 1 1 1 . 0 0 . 2 1 1 1 0 Cartesian equation of is 1 n Pl r xz −−               =  = =−              −                   ==                += 4(a) (i) (ii) 2 2 2101 3f f 50 f2 2 2 2 4 a a a a a aa       = + = = − =               (b)(i) For hg to exist, ghRD g h R ( ,1] D (0,4) = − = Since ghRD , hg does not exist. (ii) Since h is a quadratic function with maximum point at x = 1, by symmetry, when h(x) = e 2 , x = 0 and x = 2. The other value of x is 2. Since h-1 exists, h is one to one and 1h h eR = D 0, 2 − =  (given) From the graph of h, restricted hD [2,4)= 1 1 e y x x=0 0 2a a 2a 3a a− 2a− 2( 2 , )aa− 0 f( )yx= (3 ,0)a

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 5 of 20 Alternative Method 1 to find x: Since h is a quadratic function, its roots are at 2− and 4. Let h( ) ( 2)( 4)x a x x= + − eh(0) 2= ee 82 16 aa=−  =− eh( ) ( 2)( 4)16x x x=− + − Solving ( ) eh 2x = , 0 or 2x= The other value of x is 2. Alternative Method 2 to find x: Since maximum point at x = 1, Let ( ) 2 h( ) 1x a x b= − + 90 e (2)2 e(1) (2) h(4) 0 (1) eh(0) 2 e8 2 e e 9e(2) 21 : 6 6 16 1 ab a a b a b =−  + =  + = −−− − −  =− = − = =− += ( ) 2e 9eh( ) 1 16 16xx=− − + Solving ( ) eh 2x = , 0 or 2x= The other value of x is 2. Alternative Method 3 to find x: Let 2h( )x ax bx c= + + eh(0) 2= , e 2c= h(4) 0= , e16 4 0 2ab+ + = ----- (1) h'(1) 0= , h'( ) 2x ax b=+ 20ab+= ----- (2) Solving (1) & (2), (by GC or manually) ee,16 8ab=− = Or 0.1698926143, 0.3397852286ab=− =

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 6 of 20 2e e eh( ) 16 8 2x x x =− + + Solving ( ) eh 2x = , 0 or 2x= The other value of x is 2. (iii) To find range of gh: hg h h gh hg D R R e[2, 4) 0, ( ,1 ln 2]2 ⎯⎯ → ⎯⎯ → ⎯⎯ → ⎯⎯ → − −  gh e R ,ln or ( ,1 ln 2] or ( ,0.307]2  = − − − −  5(a) 2 2 2 2 2 2 8 10 28 ( 8 ) ( 2 )( 10 ) 16 64 12 20 4 44 0 4 ( 11 ) 0 0 or 11 0 (reject since terms of AP, GP are distinc a d a d a d a d a d a d a d a ad d a ad d ad d d a d d a d ++ =++ + = + + + + = + + += += = + = 1t) 11da=− y x x=0 0 e 2ln()

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 7 of 20 (b)     1 143 1 (143) 1311 2 ( 1)2 286 13( 1)2 n Sa d nS a n d n n == =− =− = + − = − − From GC, largest 858nS = (c) 8 11 8 3 1 2 11 2 9 3 1Since = 1 ,the geometric series is conver gent.3 a d d d dr a d d d d r + − + −= = = =+ − + −  (d) Let nu be the nth term of the geometric progression. 12 ... (1 ) 11 1 1 3 11 3 nn n n n n uu SS b b r rr br r b ++  ++ =− −=−−− = − = − 11 31 23 1 1 1 since 12 3 2 3 n nn b b b −− =     =         Alternative:

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 8 of 20 1 2 3 12 11 ... 1 1 1 ...3 3 3 1 3 11 3 31 23 11 since 12 3 2 3 n n n n n n n n nn u u u b b b b b bb + + + ++ −− + + +      = + + +           = − =     =         6 (i) 1. Reflection of the graph in the x-axis followed by 2. Scaling of the resulting graph by a factor 1 3 parallel to the x-axis. (ii) From (i), g( ) f (3 )xx=−  a = −1, b = 3 , c = 0, d = 0. (iii) g ( ) 3f (3 )xx =− g (2) 3f (6) =− Since f (6) 4 = g (2) 3f (6) 3(4) 12 =− =− =−

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 9 of 20 7(a) (i) (a) (ii) Area of region bounded by both curves and y = 8 8 6 2 d33 1.13 units (3 s.f.) yy yy =−  − =  (0,0) (2,6) y = 3 8 6 0 1.6 2 2.6667 x y

2025 JC2 H2 Maths Prelim Paper 1 Solutions Page 10 of 20 Alternative Method: Area of region bounded by both curves and y = 8 2 2.6667 1.6 2 2 3(2.6667 1.6)(8) d 3 d 1 8.5336 2.7325 4.6669 1.13 units (3 s.f.) x x x xx= − − − − = − − =  7(b) (i) If we split the area of the region enclosed by the curve 2f ( ) 1 y x x= = + , the x-axis and the lines x = 0 and x = 1 into n rectangles, each of width 1 n as shown, the sum of the area of the n rectangles is given by ( ) ( ) 2 2 2 2 1 1 1 1 1f 0 f ... f 1 1 1f 0 f ... f 1 1 4 ( 1)1 1 1 ... 1 nA n n n n n n n n n n n n n n −   =  +  + +         −     = + + +   