MI 2026 Prelim P2 Ans

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 1 [4] *( 2i) 3i 2z z   Let iz a b  * * ( i 2i) 3i( i) 2 ( i( 2)) 3i( i) 2 i( 2) 3 i 3 2 a b a b a b a b a b a b              Comparing real and imaginary parts,   3 2 (1) 2 3 (2) Sub (1) into (2): 2 3( 3 2) 2 9 6 8 8 1 1 a b b a b b b b b b a                       1z i  

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 2(i) [3] Correct order: replace with replace with replace with 1 2 f ( ) f ( ) f ( 1) f ( 1) f (2 1) f (2 1) f (2 1) 1 2 y x x y xxx x x x y y y y x x yy y x                      2(ii) [3] Original graph: x y x y x y

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 3(i) [1]     1 2 1 2 1 1 2 2 1 2 1 4 3 4 3 34 1 4 34 1 4 1 312 4 y x x x x x                             Hence 2k  . 3(ii) [3] 11 22 2 3 2 3 1 1 3 1 1 12 4 24 3 1 3 1 2 211 2 2! 2 1 3 5 2 2 3 4 3 3 4 4 3 2 ...3! 1 3 27 1351 ...2 8 128 1024 4 x x x x x x x x x                                                                                   2 31 3 27 135 ...2 16 256 2048x x x       The x" in this formula from MF27 must be replaced by the appropriate expression in the given question.

2026 PU3 H2 MATHEMATICS (PRELIMS P2) 3(iii) [2] Expansion is valid for 3 14 3 14 4 3 x x x     Hence 4 43 3 x   3(iv) [1] 0.8 0.8 2 3 0 0 0.8 2 3 4 0 1 1 3 27 135 d 1 d 2 8 128 10244 3 1 3 9 135 2 16 128 4096 0.4848 (4 d.p.) x x x x x x x x x x                   3(v) [2]   0.8 0 1 d 0.4901 to 4d.p. 4 3 x x   The approximation in (iv) is not very good because 0.8 is not close enough to 0, OR Not enough terms in the binomial expansion was used. If you have troubles solving inequalities with modulus, can simply sketch a graph. y = |x| y = 43 4 3 43 For a series expansion to provide a good approximation, we need either to use sufficient terms or use a value of x that is very close to 0.

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 4(a) [3]    2 3   v w v w 0 Method 1: Simplify, then apply result for cross product                               3 2 3 2 3 6 2 3 6 2 6 7                                      v v v w w v w w 0 v v v w w v w w 0 0 v w v w 0 0 0 v w v w 0 0 v w 0 v w 0 v // w OR v = 0 OR w = 0 Since v ≠ 0 and w ≠ 0, v and w are parallel. Method 2: Apply result for cross product, then simplify    2 3   v w v w 0 v ‒ 2w // 3v + w OR v ‒ 2w = 0 OR 3v + w = 0 v ‒ 2w = k(3v + w) for some k ≠ 0 OR v = 2w OR v = 13 w (1 ‒ 3k)v = (2 + k)w, k ≠ 0 OR v = 2w OR v = 13 w (1 ‒ 3k)v = (2 + k)w OR v = 13 w Since v ≠ 0 and w ≠ 0, k ≠ 2 and k ≠ 13 . v = 21 3 kk w, k ≠ 2 , k ≠ 13 OR v = 13 w Hence, we have v = λw for some λ ≠ 0, i.e. v and w are parallel. 4(b)(i) [3] 2a and 3b , 3 a b Let θ be the angle between a and b. Method 1: Find θ, then a b   o cos 3 2 3 cos 1cos 120 2            a b a b     o sin 32 3 sin120 2 3 3 3 2           a b a b

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Method 2: Use the identity 2 2sin cos 1   cos cos      a ba b a b a b … (1) sin sin       a ba b a b a b … (2) 2 2sin cos 1   … (3) Put (1) and (2) into (3) 2 2 2 2 2 2 2 2 2 2 1 ( ) (2 )(3 ) 36 ( 3) 36 36 9 27 3 3.                                a b a b a b a b a b a b a b a b a b 4(b) (ii) [2] Method 1: Use vector addition     3 2 3 2 3 2 AC BC          c a c b c a b Method 2: Use Ratio Theorem 3 2 : 2 : 3 AC BC AC BC        By Ratio Theorem, 2 1 3 23OA      b c c a b 4(b) (iii) [3]  p a b O, C and P are collinear OC   // with as the common pointOP O   , for some 0OC OP         3 2 By comparison, 3 22 3               a b a b

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 5(a) [4] G = Area of rectangle ABDE + Area of Triangle BCD Step 1: Find Area of rectangle ABDE Method 1: Use isosceles triangle BCD Let M be the mid point of BD. Area of rectangle ABDE = 10(BD) = 10(2BM) = 20[BCsin( BCM)] = 20(8)sin 2 = 160 sin 2 Method 2: Use Sine Rule By sine rule, sin sin 2 8 2sin cos8sin 2 2 16sin 2cossin 22 2 BD BC BD                        Area of rectangle ABDE = 10(BD) = 10(16sin 2 ) = 160sin 2 Method 3: Use Cosine Rule By cosine rule,       2 2 2 2 2 2 2 2 2 2 cos 8 8 2 8 8 cos 128 128cos 64(2 2cos ) 8 2 2cos 8 2 2 1 2sin 8 4sin 16sin2 2 2 BD BC CD BC CD BD BD BD BD                             Area of rectangle ABDE = 10(BD) = 10(16sin 2 ) = 160 sin 2 ----------------------------------------------------------------------- Step 2: Find Area of triangle BCD Method 1: Use 1 sin2 ab C formula for area of triangle Area of Triangle BCD = 1 ( )( )sin( )2 BC CD BCD  = 1 (8)(8)sin( ) 32sin2  

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Method 2: Use 1 ( )( )2 base height formula for area of triangle Area of Triangle BCD = 1 ( )( )2 BD CM = 1 (16sin )(8cos )2 2 2   = 64 (2sin cos )2 2 2   = 32sin  ----------------------------------------------------------------------- G = Area of rectangle ABDE + Area of Triangle BCD = 160 sin 2 + 32sin  (from Step 1 and 2) (shown) 5(b) (i) [4] For stationary value of G, we set d 80cos 32 cos 0d 2G      ,      2 2 2 2 80cos 32 2cos 1 02 2 80cos 64cos 32 02 2 4cos 5cos 2 02 2 5 5 4 4 2cos 2 2 4 5 57cos 2 8                           Since 0    , we have 0 0 cos 12 2 2       Therefore 5 57cos 2 8   . 5(b) (ii) [2] 2 2 d 40 sin 32 sind 2G     For 0    , sin 0  and sin 02  . Hence, 2 2 d 40 sin 32 sin 0d 2G      By second derivative test, this stationary value of G is maximum.

2026 PU3 H2 MATHEMATICS (PRELIMS P2) Qn Solution 6(i) [3] Note: Maximum number of silver coins taken out before removing a gold coin is 2. Given: T denotes the total number of coins that the player removes from the box, the possible values of T is 1, 2 or 3.   first coin gold P 1 2xT x          first coin silver, 2nd gold 2 2P 2 2 1 2 1x xT x x x x                   first 2 coins silver, 2nd gold 2 1 2P 3 2 1 2 1 xT x x x x x             t 1 2 3  P T t 2 x x   22 1xx x    22 1x x  6(ii) [2]              2 2E 1 2 3 2 2 1 2 1 1 4 67 6 2 1 x xT x x x x x x x x x x                               Using GC (table / graph), 11x  since 1x  . 6(iii) [1] Using GC,     2 Var 0.40562 0.1645299 0.165 (3 s.f.)T   

2026 PU3 H2 MATHEMATICS (PRELIMS P2) 7(i) [2] 3 E, 2 L, X, C, N, T Number of different arrangements = 9! 302403! 2!  7(ii) [3] Step 1: Group ‘L’s together, arrange all units except 3Es Step 2: Choose 3 out of 6 poss