MI 2026 Prelim P1 Ans

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 1 [4] Let $x, $y and $z be the price of each jar of strawberry jam, each bottle of honey and each can of sardines respectively. 5 4 6 59.80 (1)x y z    Supermarket A 5(0.75 ) 3 6 8 41.05 3.75 3 6 49.05 (2) x y z x y z         Supermarket B 5(0.6 ) (2 1) 6( 0.2) 44.70 3 3 6 1.2 44.70 3 3 6 45.90 (3) x y z x y z x y z              Using GC, 4.20, 5.50, 2.80x y z   The usual retail price of each jar of strawberry jam, each bottle of honey and each can of sardines is $4.20, $5.50 and $2.80 respectively.

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 2(i) [2]      1 First ThirdSecond Last termterm termterm when 1 1 1 1 1 ...1 1(2) 2(3) 3(4) ( 1) 1 1 1 1 ...2 6 12 ( 1) n r r n r r n n n n                     For 4n ,   4 1 1 1 1 1 1 4LHS 1 2 6 12 4(5) 5r r r      1 4RHS 1 (verified)4 1 5   2(ii) [2]       4 5 1 1 1 1 1 1 1 1 1 41 1 5 1 1 5 1 N N r r r r r r r r r N N                       2(iii) [2] As N  , 1 01N  and  5 1 11 5Nr r r  . Hence, it is convergent.  5 1 1 1 5r r r  

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 3(i) [4] 3 2 2 2 3i 6 8 ( ( i))( ) ( i)( ) z z z i z az bz c z az bz c             Method 1: Comparing coefficients Comparing constant term, 8c  Comparing coefficient of 3z , 1a  Comparing coefficient of z , 6 i i 6 8 2 2 2ii c b b b         Method 2: Long Division 3 2 2 3i 6 8 i z z z iz az b z            2 3 2 3 2 2 2 2i 8 i 3i 6 8i i 2i 6 8i 2i 2 8 8i 8 8i 0 z z z z z z z z z z z z z z                ----------------------------------------------------------------- 3 2 23i 6 8 ( i)( 2i 8)z z z i z z z        To find other roots, Let 2 2i 8 0z z  

2026 PU3 H2 MATHEMATICS (PRELIMS P1)      22i (2i) 4 1 8 2 1 2i 4 32 2 2i 36 2 2i 36 2 2i 6 2 2i, 4i z i i                 The other roots are 2i and 4iz z    . 3(ii) [2] 3 2 3i 6 8 0z z z i    Let 1z w 3 2 3 2 1 1 1 3i 6 8 0 1 3i 6 8i 0 iw w w w w w                        Hence, replace 1z w in the previous answers: 1 i, 2i, 4i 1 1 1, ,2 4 i ii, , 2 4 w w i i i w        

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 4(a) [2] 1 2e cosy xy x x    Differentiate implicitly with respect to x, 1 d de sin 2d dy y y x y x xx x     Differentiate implicitly with respect to x,   2 2 2 1 1 2 2 2 2 2 1 1 2 2 d d d d de e cos 2d d d d d d d d de e 2 cos 2 d d d d shown y y y y y y y y y x xx x x x x y y y y x xx x x x                        4(b) [4] When 0x , 1e 1 1 ln1 1y y y       . Substituting into 1 d de sin 2d dy y y x y x xx x     , d 1 0d d 1d y x y x    Substituting into 2 2 2 1 1 2 2 d d d de e 2 cos 2 d d d dy y y y y y x xx x x x          ,      2 2 2 2 2 2 2 d1 1 2 1 1 2 d d1 2 3d d 2d y x y x y x             2 2 11 2 ... 2 1 ... y x x y x x         

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 5(a)(i) [1] Description: The sequence has constant terms with a value of 3. 5(a) (ii) [1] The sequence has positive terms and the ( 1n )th term increases and tends to infinity. Hence, it diverges. 5(b) [4]     1 1 2 1 3 2 4 3 3 2 1 2 2 : 1, 2 2 , 2 1 1 2 2 2 2 , 2 3 2 3 2 3 2 3 2 3 2 3 2 2 (3) 2(3) 3 ... 2 3 2(3) 2 (3) .. 1 2 2 2(3) 3 2 2(3) 3 2 (3) ... 2 (3) 2(3) 3 2 2 3 2 ... . 2 2 3 ) 2 2( 3 (3 n n n n n n a n G r n n P k u k u u u u u u u u k u k k k k k k                                                            1 term 1 1 s 1 1 1 1 1 1 1 2 1 2 ) 2 3 2 3 1 2 2 3 3 2 2 3 3 n n n n n n n n k k k k                                 5(c) [1] For 2n ,   12 3 3nnu k    . Since   12 3n k  is even and 3 is odd,  12 3 3nnu k    is odd.

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Alternative: 12 3n nu u   . Since 12 nu  is even and 3 is odd, nu is odd.

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 6(i) [5] When 12x , 2 4 1 3 2 3y        . Point of intersection is 1 ,2 3    . By making x the subject of the equation:  1 : cosC x y 2 3: 4 yC x  . Area of R       3 0 3 3 0 0 /33/2 /3 0 0 3/2 3/2 2 3cos d 4 cos d d 3sin 4 3 / 2 3 3 2 2 4 3 3 3 uni 3 ts2 9 4 yy y y y y y yy                                  6(ii) [4] Volume of revolution

2026 PU3 H2 MATHEMATICS (PRELIMS P1)           2 23 0 23 3 0 0 3 3 0 0 from double-angle formula /32 3 0 0 /3 2 0 2 3cos d 4 3cos d d 4 cos 2 1 3 d d2 4 3cos 2 1 d2 4 2 sin 2 3 2 2 4 18 3 3 2 4 3 4 18 yy y yy y y y y y y yy y y y                                                                  2 2 2 3 3 8 6 24 3 units8 8               6(iii) [2] Volume of revolution      21 1 22 12 10 2 3 4 d cos d3 1.18 units to 2 d.p. x x x x          

2026 PU3 H2 MATHEMATICS (PRELIMS P1) Qn Solution 7(a) [2]       2 2 22 2 1 2 1 cos2sin cossin 2 d d 1 cos 1 1 1 co cos s 1 x x x x x x Cx x x C                   7(b) [3] 1 2 1 2d 1 1 d 2 2 d 2d "d 2 d 2 d " u x x u xx x x xu x u u x u                  2 2 21 2 2 21 2 1 1 1 2 2 1 1 1 1 1 21 2 2 d1 2 2 d 2 1 d 1 tan 2 tan 2 2 tan 2 d 2 d u ux x uu u u u u u c x c x x u u c                                 Alternative way to integrate     22 1 1 1 2 2 d d1 2 1 2 2 tan 2 2 tan 2 2 tan 2 2 2 u uu u u c x c x c              2 d 2 d Sub , , u x u x x u u    Factorise to make the coefficient of 2u to be 1 to use MF27 result.